Rutherford scattering conclusions may be wrong - nucleus can be big
FrankH
the atom is small. It appears the main reason for believing this is
the result of the Rutherford scattering experiments which postulated
that the nucleus is a small point object containing all the positive
charge of the nucleus. The data appear to match the experimental
results quite well, so we assume Rutherford's postulate was correct,
eventhough, this is still a very indirect measure of the size of the
nucleus. It is still like firing bullets into a dark room to figure
out what is inside of the room. In this post, I will show that there
can be alternative explanations for the scattering results and that
the nucleus doesn't necessarily have to be a tiny speck within the
atom.
I have proposed a new model of the atom which postulates that atoms
are simply formed out of alternating sequences of electrons and
protons. The protons and electrons are arranged in a very particular
geometric sequence. You could think of this model as 2 sheets of alpha
particles (helium atoms) intersecting forming an X in the shape of an
octahedral. I call this the cubic atomic model. The details of this
theory can be found at:
http://ourworld.compuserve.com/homepages/frankhu/buildatm.htm
This theory has been discussed at some length in the newsgroup:
http://groups.google.com/groups?q=g:thl1577218448d&dq=&hl=en&lr=&ie=UTF-8&oe=UTF-8&selm=46484c9f.0401151015.438ea93a%40posting.google.com
A consequence of this theory is that the electrons are not orbiting
the nucleus. They are bound into the nucleus of the atom. Since there
are no orbiting electrons, the size of the atom should be dependent
only on the size of the nucleus, since there aren't any electrons to
make the atom larger than the nucleus. This would mean that the
nucleus would be much, much larger than is commonly thought. In fact,
the nucleus should be about the same size as the measured diameter of
the atom. This is in apparent disagreement with the famous Rutherford
scattering experiment which showed that the nucleus is a tiny
positively charged speck in the center of the atom. However, I have
done some rough calculations using the cubic model to show that the
same Rutherford scattering results could be reproduced by a much
larger atom nucleus which follows the cubic theory. The basic premise
is that the cubic atomic model forms atoms which have very thin edges.
You can imagine this by taking 2 pieces of paper and have them
intersect. If you were to shoot bullets through these sheets, the
bullets would only have to pass through the thin sheets and would be
undeflected. If you examine the photographs of larger atoms like
Krypton on my website, you can see how the atoms form an X octahedral
shape (like raw diamond crystal). The cubic theory postulates that the
alpha particles are able to pass through the thin arms of the atom
with virtually no deflection since the arms are not thicker than an
alpha particle. The only place where the alpha particles can reflect
with high angles is if it hits and tries to pass through a very thick
part of the atom. This would be like trying to hit the edges of the
intersecting sheets. The chance of this happening is fairly remote. I
have done calculations to determine how frequently you would expect to
see these reflective collisions and I have compared them with the
original Rutherford scattering experiment results and I can show that
the predicted percentage for the alpha particles at particular angles
for the cubic model roughly match the experimental results. The data
for the original Rutherford scattering data by GigerMardsden can be
found at:
http://dbhs.wvusd.k12.ca.us/Chem-History/GeigerMarsden-1913/GeigerMarsden-1913.html
I began my calculations by collecting the statistics on the size of
the gold atom according to the cubic model. It is 14 units high and
the arms are 10 units wide from side to side. I approximate this as a
sphere with a radius of 7. This has a spherical surface of 615. Since
the cubic atom is symmetrical, the unique orientations are only
contained in 1 quadrant of the sphere or 1/8 of the surface. This
corresponds to a 90 degree turn through each of the x,y,z axis. This
means the area of investigation is only 76. The size of the atomic
unit representing the area of the top of the atom's core is a 2 X 2
square with an area of 4. This means there are 76/4 = 19 unique
orientations can roughly fit into this quadrant with no overlap. There
are basically only 2 orientations which would result in high angle
reflections. These are the head-on (alpha tries to pass through core)
and edge-on (alpha tries to pass through arm edge). For a head on
orientation, I calculated a 4% chance of hitting the core directly,
32% chance the arms get hit and 64% chance of a complete miss. For the
edge on orientation, I calculated a 20% of hitting an arm and 80%
chance of a miss or pass through. I plugged these into the 19 possible
slots with 11 orientations being edge on, 1 orientation being head on
and the remaining 7 as being orientations where the alpha basically
passes through. The angles of deflection are based purely on a
classical elastic collision with the atom. Because the atom is
effectively a neutral matrix of joined helium atoms, the effect of the
columb forces deflecting the alpha are negligible.
The calculations show the percentage chance for:
A complete miss or pass through 86.3% Would expect angle <
5 degrees
An arm gets hit 13.1% Would expect any angle 0 -
180
A direct hit of the core .21% Would expect angle 90 -
180
This compares to the experimental data which shows:
Deflections less than 5 degrees 79.2%
Deflections 5 - 22 20.4%
Deflections greater than 22 .35%
The details of this calculation can be found on an excel spreadsheet:
http://ourworld.compuserve.com/homepages/frankhu/ruther.xls
The predictions from the cube model and the actual experimental
results are not exactly the same by any means, however, they are in
the same rough ballpark. The main point you should observe is that the
cubic model is able to predict a scattering pattern whereby the vast
majority of the alpha particles pass right through (86%), while a tiny
fraction (.21%) gets deflected through high angles. This scattering
pattern does not necessarily have to be created by the atom postulated
by Rutherford as a tiny compact nucleus containing all the positive
charge. You can basically get the same result from the cubic atomic
model. A better calculation using a computer model to consider random
orientation and random alpha may produce results more comparable (or
not) with the actual experimental data and would provide a more
detailed range of angles to expect when an arm is struck.
Unfortunately, I do not have the resources to commit to such a
calculation.
There would be other experiments to confirm or deny the cubic model
with Rutherford scattering. Perhaps some of these have already been
done. I would like to see the Rutherford scattering experiments
repeated but instead of using gold foil, use a form of crystallized
gold (octahedral crystal) where we are reasonably sure that the gold
atoms are all aligned in the same direction, and see how the high
angle scattering depends on the orientation of the crystal. Based on
the cubic model, I would predict that the crystal would present very
little scattering in most directions, since the alpha particles are
able to pass through the thin parts of the atom, but when the crystal
is oriented so it hits the very edge of the atom and tries to pass
through the core or arms, it will bounce back strongly. I would
predict that we would see lots of scattering at 90 degree crystal
orientations, which would not be explainable by the Rutherford
formula. The Rutherford formula would predict the same scattering
pattern/amounts no matter what the orientation.
Another possible experiment would be to use low speed alpha particles.
At some point, if the speed of the alpha particles were slow enough,
it wouldn't be able to penetrate the atoms thin arms and you would see
almost all of the alphas being deflected at high angles. Rutherford
would predict that all of the alphas would penetrate no matter how
slow the alphas were going since there isn't much for the aphas to run
into and an electron isn't likely to deflect an alpha very much.
If anybody knows the results of these experiments, please post them to
help confirm or deny the plausability of the cubic atomic model. In
conclusion, the results of the Rutherford scattering experiments do
not conclusively prove the notion of the nucleus being a tiny speck in
the atom with surrounding electron clouds. Thus far, arguments against
Rutherford have lacked an alternative solid model to base calculations
on. The cubic atomic theory provides this solid model which you can
run calculations on to show that it can return results similar to the
experimental results of the Rutherford scattering experiment.
Uncle Al
[snip]
> I have proposed a new model of the atom which postulates that atoms
> are simply formed out of alternating sequences of electrons and
> protons.
Ignorant idiot.
> The protons and electrons are arranged in a very particular
> geometric sequence.
vida supra.
If empirical reality says that you are an idiot, then you are an
empirical idiot. Not only are you an idiot as a physicist, you are an
ididot as a chemist. Hey stooopid, how do you reconcile atomic
spectroscopy?
--
Uncle Al
http://www.mazepath.com/uncleal/
(Toxic URL! Unsafe for children and most mammals)
"Quis custodiet ipsos custodes?" The Net!
FrankH
> FrankH wrote:
>
> [snip]
> > I have proposed a new model of the atom which postulates that atoms
> > are simply formed out of alternating sequences of electrons and
> > protons.
>
> Ignorant idiot.
>
> > The protons and electrons are arranged in a very particular
> > geometric sequence.
>
> vida supra.
>
> If empirical reality says that you are an idiot, then you are an
> empirical idiot. Not only are you an idiot as a physicist, you are an
> ididot as a chemist. Hey stooopid, how do you reconcile atomic
> spectroscopy?
If you look at my website, you can see that the cubic atomic model
presents empirical evidence to help explain:
The reactivity of an atom is related to symmetry
Why the electron shell energy levels are arranged in the way that they
are
Explains where neutrons are contained in the atom and why they don't
contribute to the chemical properties
Eliminates the problem of why should an electron orbit the nucleus
Eliminates the need for a strong nuclear force
Explains the newest pictures of atoms using STM
Explains the most common fission products of uranium
Explains how an alternating series of protons and electrons are stable
and why helium is so stable
It correctly predicts the relative difference for the first ionization
energies for hydrogen and helium
It correctly predicts the relative ionization energies for the
remaining electrons in an atom
Finally, it correctly accounts for the results of the Rutherford
scattering experiment
I would say that empirical reality is saying I'm not an idiot and it
has thus far explained many of the things I have thrown at it. There
will be more to come ... stay tuned.
As far as atomic spectroscopy is concerned, the spectrum produced by
an atom is not due to anything normally contained in an atom at ground
state. The spectrum is due to an electron which has escaped the normal
confines of the atom ground state and is technically not even part of
the atom anymore. It really doesn't tell us much about the internal
structure of the atom. Data like the ionization energies for each of
the electrons gives us a better idea of the internal structure since
it shows how tightly each electron is bound in the atom. But once an
electron has left its normal ground state, it is governed by an
entirely different set of rules. For the cubic atomic model, these
excited electrons would be governed by the rules of spherical
harmonics. A crude decription would be an electron attached by a
spring to the atom and it vibrates and has harmonics and resonant
frequencies. Quantum mechanics appears to use these spherical
harmonics in the spectroscopy equations, so I presume that they are
correct in predicting the expected spherical harmonics and spectral
lines. However, the interpretation of the equations as showing the
actual behavior of the electron being contained in a probability cloud
is misleading. The electron is free to move about randomly around the
atom, but the resonant frequencies and emmited photons correspond to
the wave functions. That is my guess about what happens when an atom
is excited beyond ground level. Technically, the cubic atomic model
says nothing about what happens to an electron once it leaves the
ground state. The cubic model details the bulk static properties of an
atom in the ground state.
Bjoern Feuerbacher
Oh, you are back? And apparently you are still ignoring all the
counterarguments to your ideas in the old thread - and you are still
ignoring that there is a *plethora* of evidence for QM and QFT. I gave
some examples in the last thread - you completely ignored them.
> We all know that the size of the nucleus compared to the total size of
> the atom is small.
Yes.
> It appears the main reason for believing this is
> the result of the Rutherford scattering experiments which postulated
> that the nucleus is a small point object containing all the positive
> charge of the nucleus.
No. That's not the "main" reason, that was only the first piece of
evidence. Nowadays we have *far* more experimental data on scattering of
nuclei and individual protons and neutrons (not Rutherford scattering -
the experiments have long ago gone far beyond that simple experiment of
Rutherford). Every scattering argument shows that these things *are*
small.
> The data appear to match the experimental
> results quite well, so we assume Rutherford's postulate was correct,
> eventhough, this is still a very indirect measure of the size of the
> nucleus.
Yes. Care to propose a more direct type of measurement of the size of
the nucleus?
> It is still like firing bullets into a dark room to figure
> out what is inside of the room.
That's a nice analogy, but you should be aware that analogies can be
misleading often.
> In this post, I will show that there
> can be alternative explanations for the scattering results and that
> the nucleus doesn't necessarily have to be a tiny speck within the
> atom.
>
> I have proposed a new model of the atom which postulates that atoms
> are simply formed out of alternating sequences of electrons and
> protons. The protons and electrons are arranged in a very particular
> geometric sequence. You could think of this model as 2 sheets of alpha
> particles (helium atoms) intersecting forming an X in the shape of an
> octahedral. I call this the cubic atomic model. The details of this
> theory can be found at:
>
> http://ourworld.compuserve.com/homepages/frankhu/buildatm.htm
Do you explain there also where the repulsive force comes from which
keeps the electrons from falling into the protons? And why he have never
found experimental evidence for this repulsive force?
> This theory has been discussed at some length in the newsgroup:
> http://groups.google.com/groups?q=g:thl1577218448d&dq=&hl=en&lr=&ie=UTF-8&oe=UTF-8&selm=46484c9f.0401151015.438ea93a%40posting.google.com
I recommend every reader of this thread here to go back to that thread
and see how many arguments FrankH has ignored there.
> A consequence of this theory is that the electrons are not orbiting
> the nucleus.
Well, the standard theory doesn't say, too, that electrons orbit the
nucleus, so this is not big news.
> They are bound into the nucleus of the atom.
In order to be bound, you have to have a combination of an attractive
and a repulsive force (if you don't want to invoke QM, and you don't
want the electrons to orbit). What is this repulsive force, and why
haven't we ever found any evidence for it?
> Since there
> are no orbiting electrons, the size of the atom should be dependent
> only on the size of the nucleus, since there aren't any electrons to
> make the atom larger than the nucleus.
Non sequitur. QM says, too, that the electrons aren't orbiting, but
nevertheless, the size of the atom is *not* determined by the size of
the nucleus - but by the wavelenghts of the stationary states.
> This would mean that the
> nucleus would be much, much larger than is commonly thought. In fact,
> the nucleus should be about the same size as the measured diameter of
> the atom. This is in apparent disagreement with the famous Rutherford
> scattering experiment which showed that the nucleus is a tiny
> positively charged speck in the center of the atom.
Nice that you see that.
> However, I have
> done some rough calculations using the cubic model to show that the
> same Rutherford scattering results could be reproduced by a much
> larger atom nucleus which follows the cubic theory. The basic premise
> is that the cubic atomic model forms atoms which have very thin edges.
> You can imagine this by taking 2 pieces of paper and have them
> intersect. If you were to shoot bullets through these sheets, the
> bullets would only have to pass through the thin sheets and would be
> undeflected. If you examine the photographs of larger atoms like
> Krypton on my website, you can see how the atoms form an X octahedral
> shape (like raw diamond crystal). The cubic theory postulates that the
> alpha particles are able to pass through the thin arms of the atom
> with virtually no deflection since the arms are not thicker than an
> alpha particle. The only place where the alpha particles can reflect
> with high angles is if it hits and tries to pass through a very thick
> part of the atom. This would be like trying to hit the edges of the
> intersecting sheets. The chance of this happening is fairly remote. I
> have done calculations to determine how frequently you would expect to
> see these reflective collisions and I have compared them with the
> original Rutherford scattering experiment results and I can show that
> the predicted percentage for the alpha particles at particular angles
> for the cubic model roughly match the experimental results.
"roughly". Interesting.
BTW, shouldn't the result of the scattering depend on the projectiles,
too, according to your model? Nevertheless, experiments show that one
gets the same scattering cross section for alpha particles and electrons
(the only deviations are at small angles, where the non-zero size of the
nucleus becomes important).
Additionally, does your model reproduce the observed dependence of the
scattering cross section on the charge of the nucleus, Z? Does it
explain why the scattering cross section does *not* depend on the number
of neutrons, N, in the nucleus? Rutherford's formula can do all of this
without any problems.
> The data
> for the original Rutherford scattering data by GigerMardsden can be
> found at:
>
> http://dbhs.wvusd.k12.ca.us/Chem-History/GeigerMarsden-1913/GeigerMarsden-1913.html
Thanks for the reference (it's always nice to have the opportunity to
read an original paper on a big discovery! :-) ) - but why don't you use
more modern sources, which have far more accurate data?
> I began my calculations by collecting the statistics on the size of
> the gold atom according to the cubic model. It is 14 units high and
> the arms are 10 units wide from side to side. I approximate this as a
> sphere with a radius of 7.
Why is this approximation justified? What happened to the "sharp edges"
you argued about above?
And what are your "units" here? (I know that this is irrelevant, because
you only want to calculate percentages, but nevertheless I would like to
know this here).
> This has a spherical surface of 615. Since
> the cubic atom is symmetrical, the unique orientations are only
> contained in 1 quadrant of the sphere or 1/8 of the surface. This
> corresponds to a 90 degree turn through each of the x,y,z axis. This
> means the area of investigation is only 76.
Agreed so far, although you use rather strange wording.
> The size of the atomic
> unit representing the area of the top of the atom's core is a 2 X 2
> square with an area of 4. This means there are 76/4 = 19 unique
> orientations can roughly fit into this quadrant with no overlap.
Right - but why is the "with no overlap" relevant?
And again, very confusing wording! I had to read this several times
until I understood what you try to say.
> There
> are basically only 2 orientations which would result in high angle
> reflections. These are the head-on (alpha tries to pass through core)
> and edge-on (alpha tries to pass through arm edge).
Why would these result in high angle deflection? What's there in the
core or in the arm edge which deflects the alpha particle? What force is
acting there?
> For a head on
> orientation, I calculated a 4% chance of hitting the core directly,
> 32% chance the arms get hit and 64% chance of a complete miss. For the
> edge on orientation, I calculated a 20% of hitting an arm and 80%
> chance of a miss or pass through. I plugged these into the 19 possible
> slots with 11 orientations being edge on, 1 orientation being head on
> and the remaining 7 as being orientations where the alpha basically
> passes through. The angles of deflection are based purely on a
> classical elastic collision with the atom.
You *do* know that classically, objects have a hard surface and
therefore bounce off each other when they collide because of
electrostatic forces, don't you?
> Because the atom is
> effectively a neutral matrix of joined helium atoms, the effect of the
> columb forces deflecting the alpha are negligible.
Are you sure about that? Even when the alpha particle hits "head on"?
> The calculations show the percentage chance for:
>
> A complete miss or pass through 86.3% Would expect angle <
> 5 degrees
> An arm gets hit 13.1% Would expect any angle 0 -
> 180
> A direct hit of the core .21% Would expect angle 90 -
> 180
>
> This compares to the experimental data which shows:
>
> Deflections less than 5 degrees 79.2%
> Deflections 5 - 22 20.4%
> Deflections greater than 22 .35%
Well, this *is* a "rough" agreement. ***VERY*** rough. It doesn't even
agree with the *original* results of Geiger and Marsden very well. You
have no clue how accurate scattering experiments nowadays are, have you?
IIRC, I recommended some books to you, among them one by Povh on nuclear
physics, which lists a *very* small amount of experimental results from
the plethora of experiments done in nuclear and particle physics (hint:
the book is nevertheless quite thick). Have you looked into that book in
the meantime?
Please go to the more recent experiments, which have shown again and
again (hey, I even did this experiment *myself* at one time!!!) that the
scattering cross section depends on the angle like 1/sin^4(theta/2). As
soon as you are able to reproduce this formula (or at least some
numerical results which agree with it on a level of, say, 1%), I will
admit that your model has some merit. Good luck.
(oh, BTW, shouldn't your model give a dependence of the scattering cross
section on phi?)
> The details of this calculation can be found on an excel spreadsheet:
>
> http://ourworld.compuserve.com/homepages/frankhu/ruther.xls
I don't know if I should love or cry here.....
You haven't got the *slightest* clue of how many experiments were done
in nuclear and particle physics in the past decades, how accurate they
are, how much work is spent in analysing the results. Sorry if this
insults you, put pretending that an *Excel spreadsheet* somehow is able
to reproduce these results based on your model is just pathetic, plain
and simple.
> The predictions from the cube model and the actual experimental
> results are not exactly the same by any means,
Nice that you see that.
> however, they are in the same rough ballpark.
BFD. I think *every* model of the atom and the nucleus, no matter how
contrived, will manage to come up with some or even *lots* of results
which roughly agree with the experimental results. However, so far
*none* of this plethora of "alternative models" has been able to
1) explain all the things which standard QM explains
2) explain even *one* thing which QM explains with a comparable accuracy
as QM.
You should try reading up on the "successes" of other "alternative
models". For example, you could look at Porat's model (right here in
sci.physics - the guy who either can't spell properly or doesn't bother
to do it), at the ideas of Black Light Power, or at Common Sense
science. All three have models which are far more developed than yours,
which strongly disagree with your, which include many strange ideas -
and nevertheless all three are able to reproduce a fair amount of
experimental evidence, sometimes with quite amazing accuracy. However,
all of them suffer the same faults - the two outlined above.
You have still a long way to go. Good luck again.
> The main point you should observe is that the
> cubic model is able to predict a scattering pattern whereby the vast
> majority of the alpha particles pass right through (86%), while a tiny
> fraction (.21%) gets deflected through high angles.
BFD. You are *still* doing a mainly qualitative analysis, instead of the
*quantitative* one done by Rutherford 90 years ago. He derived a
*formula* which described the results with amazing accuracy. Your
results so far are mainly handwaving: "See, I can reproduce the
qualitative behaviour in the right way!!!".
> This scattering
> pattern does not necessarily have to be created by the atom postulated
> by Rutherford as a tiny compact nucleus containing all the positive
> charge.
As long as you stay on this qualitative level, you are right. But as
soon as you become quantitative, i.e. take Rutherford's *formula* and
compare it to the experimental results, it's clear whose model is
better.
> You can basically get the same result from the cubic atomic
> model.
The crucial word here is "basically".
> A better calculation using a computer model to consider random
> orientation and random alpha may produce results more comparable (or
> not) with the actual experimental data and would provide a more
> detailed range of angles to expect when an arm is struck.
> Unfortunately, I do not have the resources to commit to such a
> calculation.
Why do you insist on finding another explanation for the results,
although the existing explanation already works rather well?
Hint: new theories in science are only invented when there is
experimental evidence that the old theories are wrong. Not simply
because someone doesn't like or doesn't understand the old theories.
> There would be other experiments to confirm or deny the cubic model
> with Rutherford scattering. Perhaps some of these have already been
> done.
Go read Povh. There is lots of stuff about scattering in it.
> I would like to see the Rutherford scattering experiments
> repeated but instead of using gold foil, use a form of crystallized
> gold (octahedral crystal) where we are reasonably sure that the gold
> atoms are all aligned in the same direction, and see how the high
> angle scattering depends on the orientation of the crystal.
The experiment has been done with all kinds of stuff, AFAIK - not only
with gold foils, but also with other elements, and as well with isolated
nuclei and even isolated protons and neutrons (well, obviously the
latter experiments shouldn't be called Rutherford scattering anymore).
Additionally, instead of alpha particles, also electrons have been used,
and protons, and positrons, and neutrinos. In *all* of these experiments
(I estimate there have been several millions of those - hey, AFAIK there
were already several millions of created Z bosons registered at LEP,
which was *also* a scattering experiment!), the predictions of the
standard theories were proven - often with an amazing accuracy.
So why on earth do (the one who knows nothing about even one of all of
these experiments) think that there is a problem with the standard
theories, and we need a new one?
> Based on
> the cubic model, I would predict that the crystal would present very
> little scattering in most directions, since the alpha particles are
> able to pass through the thin parts of the atom, but when the crystal
> is oriented so it hits the very edge of the atom and tries to pass
> through the core or arms, it will bounce back strongly. I would
> predict that we would see lots of scattering at 90 degree crystal
> orientations, which would not be explainable by the Rutherford
> formula. The Rutherford formula would predict the same scattering
> pattern/amounts no matter what the orientation.
Well, for starters you could try to predict the observed behaviour of
the scattering cross section on Z. There should be a heap of
experimental data on this to compare with. Good luck again.
> Another possible experiment would be to use low speed alpha particles.
> At some point, if the speed of the alpha particles were slow enough,
> it wouldn't be able to penetrate the atoms thin arms and you would see
> almost all of the alphas being deflected at high angles. Rutherford
> would predict that all of the alphas would penetrate no matter how
> slow the alphas were going since there isn't much for the aphas to run
> into and an electron isn't likely to deflect an alpha very much.
If the alpha particles are slow enough, the electrons *will* be able to
deflect them.
And, BTW, Rutherford scattering *has* be done at a *lot* of different
energies (speeds) now. So far, every experiment agreed with his formula.
> If anybody knows the results of these experiments, please post them to
> help confirm or deny the plausability of the cubic atomic model.
Why don't you do what I recommended to you - try getting a book on
experimental nuclear and particle physics and read up for yourself what
experimental evidence exists?
> In
> conclusion, the results of the Rutherford scattering experiments do
> not conclusively prove the notion of the nucleus being a tiny speck in
> the atom with surrounding electron clouds.
They do, if one uses the quantitative formula instead of your mostly
qualitative hand waving.
> Thus far, arguments against
> Rutherford have lacked an alternative solid model to base calculations
> on.
Wrong. Lots of people have tried to build alternative models (for
examples, see above) and claim to be able to explain stuff like
Rutherford scattering (well, to be fair, Porat avoids this topic - but
the other two models I mentioned adress this, IIRC). You are by far not
the first one.
> The cubic atomic theory provides this solid model which you can
> run calculations on to show that it can return results similar to the
> experimental results of the Rutherford scattering experiment.
Quit the handwaving. Rederive Rutherford's formula from your model.
Or, alternatively, explain the spectrum of the hydrogen atom.
Quantitatively.
Bye,
Bjoern
Bjoern Feuerbacher
>
> Uncle Al <Uncl...@hate.spam.net> wrote in message news:<403BC690...@hate.spam.net>...
> > FrankH wrote:
> >
> > [snip]
> > > I have proposed a new model of the atom which postulates that atoms
> > > are simply formed out of alternating sequences of electrons and
> > > protons.
> >
> > Ignorant idiot.
> >
> > > The protons and electrons are arranged in a very particular
> > > geometric sequence.
> >
> > vida supra.
> >
> > If empirical reality says that you are an idiot, then you are an
> > empirical idiot. Not only are you an idiot as a physicist, you are an
> > ididot as a chemist. Hey stooopid, how do you reconcile atomic
> > spectroscopy?
>
> If you look at my website, you can see that the cubic atomic model
> presents empirical evidence to help explain:
>
> The reactivity of an atom is related to symmetry
Standard QM can explain the reactivity of an atom, too.
Quantitatively, BTW.
> Why the electron shell energy levels are arranged in the way that they
> are
Standard QM explain this, too.
Quantitatively, BTW.
And another BTW: it also can do this for the magical numbers of nuclei.
How do you explain those?
> Explains where neutrons are contained in the atom
Evidence that this explanation is right?
And how do you reconcile this with the wave-like behaviour of quanta,
including neutrons? How can a wave be located at a certain position?
> and why they don't contribute to the chemical properties
Standard QM explains this, too: chemistry is based on electromagnetic
interactions, and neutrons are electrically neutral.
> Eliminates the problem of why should an electron orbit the nucleus
This hasn't been a problem since 1926.
> Eliminates the need for a strong nuclear force
Explain the binding energies of nuclei. Quantitatively, please, like the
standard theories of nuclei do.
> Explains the newest pictures of atoms using STM
Didn't you notice that the author of that paper claims that his results
are a verification of QM?
(oh, and BTW, I *still* think you have got a problem with your eyes -
these pictures *are* fuzzy, no matter how often you claim to see sharp
edges there)
> Explains the most common fission products of uranium
Standard nuclear theory can do this, too. Quantitatively. And IIRC, your
model gives only a rough estimate which disagrees with the actual
results. Apparently a rough agreement is enough for you to claim that
your model "explains" something.
> Explains how an alternating series of protons and electrons are stable
What do you mean here?
> and why helium is so stable
Standard theories explain this, too. Quantitatively, again.
> It correctly predicts the relative difference for the first ionization
> energies for hydrogen and helium
Can you point me to the exact point on your web page where you
demonstrate this?
And, BTW, can it explain the spectra of hydrogen and helium, too?
> It correctly predicts the relative ionization energies for the
> remaining electrons in an atom
Same question.
> Finally, it correctly accounts for the results of the Rutherford
> scattering experiment
This is close to a lie. In the original post, you yourself admitted that
you get only a rough agreement, and that further work is needed. Your
model does account only for the *qualitative* results of the Rutherford
experiment - it is still a *very* long way from a *quantitative*
explanation. In contrast, Rutherford had right from the start a formula
which agreed with the experimental results with amazing accuracy.
> I would say that empirical reality is saying I'm not an idiot and it
> has thus far explained many of the things I have thrown at it. There
> will be more to come ... stay tuned.
Judging from your performance wrt the fission products of uranium and
Rutherford's experiment, you will count every rough agreement as a
success for your model, and you will ignore that standard theories
explain all of this, too - but with a *far* better accuracy than your
attempts. I wouldn't say that you are an idiot - but you obviously are
1) rather fast in dismissing the standard explanations, although you
know close to nothing about the experimental evidence for them
2) rather fast in accepting rough agreement as a confirmation of your
ideas.
I would term this a "double standard"...
> As far as atomic spectroscopy is concerned, the spectrum produced by
> an atom is not due to anything normally contained in an atom at ground
> state.
Well, it is due to the electrons. Last I looked, the electrons *are*
contained in an atom at ground state...
> The spectrum is due to an electron which has escaped the normal
> confines of the atom ground state and is technically not even part of
> the atom anymore.
As long as it bound to it, it *is* part of the atom. You are making no
sense.
Do you claim, too, that the earth is not a part of the solar system
because it is not bound to the surface of the sun?
> It really doesn't tell us much about the internal structure of the atom.
Spectra don't tell us much about the internal structure of the atom???
This has to be one of the most ignorant, nonsensical sentences I've
heard in a while (well, ignoring Porat's statements here - he is simply
not comparable to anyone else).
> Data like the ionization energies for each of
> the electrons gives us a better idea of the internal structure since
> it shows how tightly each electron is bound in the atom.
Hint: standard QM predicts these ionization energies with good accuracy.
> But once an
> electron has left its normal ground state, it is governed by an
> entirely different set of rules.
Why should it be?
> For the cubic atomic model, these
> excited electrons would be governed by the rules of spherical
> harmonics.
What rules are those?
And why should they apply?
And why shouldn't they apply in the ground state?
> A crude decription would be an electron attached by a
> spring to the atom and it vibrates and has harmonics and resonant
> frequencies.
Ouch! No more comment.
> Quantum mechanics appears to use these spherical
> harmonics in the spectroscopy equations,
I don't know what you exactly mean by "spectroscopy equations", but yes,
standard QM uses spherical harmonics for describing bound states of
electrons in atoms. There are (rough estimate) several thousands of
these standard QM calculations which have been verified experimentally.
How do you explain this, if standard QM is wrong?
> so I presume that they are
> correct in predicting the expected spherical harmonics and spectral
> lines.
But why should standard QM be only right for the excited states, but not
for the ground state?????
> However, the interpretation of the equations as showing the
> actual behavior of the electron being contained in a probability cloud
> is misleading.
Thanks for showing again that you don't know what you are talking about.
Nobody says that the electron is contained in a probability cloud. And,
BTW, you were corrected on this before, so repeating this nonsense is
again close to a lie.
IIRC, I recommended Styer's book "The strange world of Quantum
Mechanics" to you. Have you read it in the meantime?
> The electron is free to move about randomly around the atom,
*Very* roughly right. (well, it isn't entirely free - there is the
electrostatic attraction to the nucleus, after all!)
> but the resonant frequencies and emmited photons correspond to
> the wave functions.
"correspond" in what way? Your sentence makes no sense.
> That is my guess about what happens when an atom
> is excited beyond ground level.
Sorry, I didn't understand what you tried to say.
> Technically, the cubic atomic model
> says nothing about what happens to an electron once it leaves the
> ground state. The cubic model details the bulk static properties of an
> atom in the ground state.
Again, why should the description of the ground state be handled by
another model than the description of the excited states?
And, BTW, you *do* know that the QM description of the excited states
(which you seem to accept!!!) is based on a *pointlike* (i.e. very
small) nucleus? So why are these predictions correct, although,
according to you, the nucleus is *not* very small?
Bye,
Bjoern
Franz Heymann
"FrankH" <frank...@yahoo.com> wrote in message
news:46484c9f.04022...@posting.google.com...
> Uncle Al <Uncl...@hate.spam.net> wrote in message
news:<403BC690...@hate.spam.net>...
> > FrankH wrote:
> >
> > [snip]
> > > I have proposed a new model of the atom which postulates that atoms
> > > are simply formed out of alternating sequences of electrons and
> > > protons.
> >
> > Ignorant idiot.
> >
> > > The protons and electrons are arranged in a very particular
> > > geometric sequence.
> >
> > vida supra.
> >
> > If empirical reality says that you are an idiot, then you are an
> > empirical idiot. Not only are you an idiot as a physicist, you are an
> > ididot as a chemist. Hey stooopid, how do you reconcile atomic
> > spectroscopy?
>
> If you look at my website, you can see that the cubic atomic model
> presents empirical evidence to help explain:
Your cubic atomic model is in fact a model of a cubic lump of horse dung.
You have trawled this rubbish before and you were shown by many physicists
to be an ignoramus. Why do you try it yet again? Are you a masochist?
Franz
Franz Heymann
> the atom is small. It appears the main reason for believing this is
> the result of the Rutherford scattering experiments which postulated
> that the nucleus is a small point object containing all the positive
> charge of the nucleus. The data appear to match the experimental
> results quite well,
The data don't only appear to match the expected differential cross sections
well, they actually *do* match the expected values so well that one no
longer says that the nucleus is a small point object, which is all that
could be said in Rutheford's day. One now knows that the proton charge
distribution, for example, has a RMS radius of 0.80 fm. I forget the error
bar size, but the measurements are actually good enough to provide a figure
for the skin depth of the proton. (Crudely, the radial thickness over which
the charge density dies down from its central value to zero)
> so we assume Rutherford's postulate was correct,
> eventhough, this is still a very indirect measure of the size of the
> nucleus. It is still like firing bullets into a dark room to figure
> out what is inside of the room.
No. Nowadays it is more akin to obtaining a diffraction pattern of a
spherically symmetrical object by using an incident beam with a wavelength
small compared to the proton radius. The process is in fact related to
using X-ray diffraction to determine crystal structures.
> In this post, I will show that there
> can be alternative explanations for the scattering results and that
> the nucleus doesn't necessarily have to be a tiny speck within the
> atom.
That proposed explanation will be wrong.
>
> I have proposed a new model of the atom which postulates that atoms
> are simply formed out of alternating sequences of electrons and
> protons.
Indeed I am right. Before making public proposals about new atomic models,
you should make quite certain that you are fully aware of the state of play
in the field. You obviously are not in that position.
Your earlier remarks indicated that you are probably not au fait with
elastic scattering experiments.
The later remarks, which I am snipping, serve only to prove that my earlier
impressions were correct.
I will leave it to other more hardy souls to wade throught the mire to show
you how wrong you are.
[snip]
Franz
Mark Fergerson
> I have proposed a new model of the atom
Kindly address NMR and EPR. After that, we'll get to the
hard stuff.
Mark L. Fergerson
Starblade Darksquall
You've been reading too much Neo-Tech.
(...Starblade Riven Darksquall...)
FrankH
[snip
> >
> > If you look at my website, you can see that the cubic atomic model
> > presents empirical evidence to help explain:
> >
> > The reactivity of an atom is related to symmetry
>
> Standard QM can explain the reactivity of an atom, too.
>
> Quantitatively, BTW.
>
>
> > Why the electron shell energy levels are arranged in the way that they
> > are
>
> Standard QM explain this, too.
>
> Quantitatively, BTW.
>
> And another BTW: it also can do this for the magical numbers of nuclei.
> How do you explain those?
>
associated with atoms which are perfectly symmetrical in the cubic
model.
>
> > Explains where neutrons are contained in the atom
>
> Evidence that this explanation is right?
>
I am working on showing that the maximum number of neutrons or the
most common isotopes is related to the shape of the atoms core.
> And how do you reconcile this with the wave-like behaviour of quanta,
> including neutrons? How can a wave be located at a certain position?
>
Individual particles like electrons and neutrons appear to behave like
waves. I'm not sure I agree with the experiments showing wave behavior
for particles. When inside an atom, they act like particles not waves.
>
> > and why they don't contribute to the chemical properties
>
> Standard QM explains this, too: chemistry is based on electromagnetic
> interactions, and neutrons are electrically neutral.
>
>
>
> > Eliminates the problem of why should an electron orbit the nucleus
>
> This hasn't been a problem since 1926.
>
>
> > Eliminates the need for a strong nuclear force
>
> Explain the binding energies of nuclei. Quantitatively, please, like the
> standard theories of nuclei do.
I have added calculations of hydrogen, helium, and lithium to my
website.
>
>
> > Explains the newest pictures of atoms using STM
>
> Didn't you notice that the author of that paper claims that his results
> are a verification of QM?
>
> (oh, and BTW, I *still* think you have got a problem with your eyes -
> these pictures *are* fuzzy, no matter how often you claim to see sharp
> edges there)
Yes, I noticed. I did ask the original experimenters about the edges.
They acknowleged that they were unusual and probably due to some sharp
dropoff in the orbital configuration, but they didn't seem to have a
great explanation. Their results may be explainable in terms of the
cubic theory - but I'm still looking into that.
>
>
> > Explains the most common fission products of uranium
>
> Standard nuclear theory can do this, too. Quantitatively. And IIRC, your
> model gives only a rough estimate which disagrees with the actual
> results. Apparently a rough agreement is enough for you to claim that
> your model "explains" something.
Can you point me to a reference of this? The most I've see is the
waterdrop analogy which shows splitting of the nucleus like a
waterdrop which didn't seem like much of an explanation.
>
>
> > Explains how an alternating series of protons and electrons are stable
>
> What do you mean here?
See the other post about the stability calculation
>
>
> > and why helium is so stable
>
> Standard theories explain this, too. Quantitatively, again.
>
>
> > It correctly predicts the relative difference for the first ionization
> > energies for hydrogen and helium
>
> Can you point me to the exact point on your web page where you
> demonstrate this?
I recently added this, you probably didn't see it when you wrote this.
Check again.
>
> And, BTW, can it explain the spectra of hydrogen and helium, too?
>
Where can I find how you calculate the spectra for helium and lithium.
I've only seen hydrogen. I did read in a physics book that the
calculations cannot be done exactly for multi-electron atoms. Is that
true?
>
> > It correctly predicts the relative ionization energies for the
> > remaining electrons in an atom
>
> Same question.
>
>
> > Finally, it correctly accounts for the results of the Rutherford
> > scattering experiment
>
> This is close to a lie. In the original post, you yourself admitted that
> you get only a rough agreement, and that further work is needed. Your
> model does account only for the *qualitative* results of the Rutherford
> experiment - it is still a *very* long way from a *quantitative*
> explanation. In contrast, Rutherford had right from the start a formula
> which agreed with the experimental results with amazing accuracy.
OK, I got a little carried away ...
>
>
> > I would say that empirical reality is saying I'm not an idiot and it
> > has thus far explained many of the things I have thrown at it. There
> > will be more to come ... stay tuned.
>
> Judging from your performance wrt the fission products of uranium and
> Rutherford's experiment, you will count every rough agreement as a
> success for your model, and you will ignore that standard theories
> explain all of this, too - but with a *far* better accuracy than your
> attempts. I wouldn't say that you are an idiot - but you obviously are
>
> 1) rather fast in dismissing the standard explanations, although you
> know close to nothing about the experimental evidence for them
> 2) rather fast in accepting rough agreement as a confirmation of your
> ideas.
>
> I would term this a "double standard"...
>
Well, you've had 100 years to work on the standard theory, I've had
about 1 month, so yes, I have some catching up to do, but I'm not
doing so bad after a month.
>
>
> > As far as atomic spectroscopy is concerned, the spectrum produced by
> > an atom is not due to anything normally contained in an atom at ground
> > state.
>
> Well, it is due to the electrons. Last I looked, the electrons *are*
> contained in an atom at ground state...
>
>
> > The spectrum is due to an electron which has escaped the normal
> > confines of the atom ground state and is technically not even part of
> > the atom anymore.
>
> As long as it bound to it, it *is* part of the atom. You are making no
> sense.
>
> Do you claim, too, that the earth is not a part of the solar system
> because it is not bound to the surface of the sun?
>
More like if you took a rock and threw it off the earth, the rock is
no longer part of the earth.
>
> > It really doesn't tell us much about the internal structure of the atom.
>
> Spectra don't tell us much about the internal structure of the atom???
> This has to be one of the most ignorant, nonsensical sentences I've
> heard in a while (well, ignoring Porat's statements here - he is simply
> not comparable to anyone else).
>
Can you tell me what is the relation between the electron shell
configurations shown on www.webelements.com and the spectra generated
by excited electons? I'm missing something here.
>
> > Data like the ionization energies for each of
> > the electrons gives us a better idea of the internal structure since
> > it shows how tightly each electron is bound in the atom.
>
> Hint: standard QM predicts these ionization energies with good accuracy.
>
References?
>
> > But once an
> > electron has left its normal ground state, it is governed by an
> > entirely different set of rules.
>
> Why should it be?
>
>
> > For the cubic atomic model, these
> > excited electrons would be governed by the rules of spherical
> > harmonics.
>
> What rules are those?
>
> And why should they apply?
>
> And why shouldn't they apply in the ground state?
>
>
> > A crude decription would be an electron attached by a
> > spring to the atom and it vibrates and has harmonics and resonant
> > frequencies.
>
> Ouch! No more comment.
>
>
> > Quantum mechanics appears to use these spherical
> > harmonics in the spectroscopy equations,
>
> I don't know what you exactly mean by "spectroscopy equations", but yes,
> standard QM uses spherical harmonics for describing bound states of
> electrons in atoms. There are (rough estimate) several thousands of
> these standard QM calculations which have been verified experimentally.
> How do you explain this, if standard QM is wrong?
I'm saying QM may be right in this case.
>
>
> > so I presume that they are
> > correct in predicting the expected spherical harmonics and spectral
> > lines.
>
> But why should standard QM be only right for the excited states, but not
> for the ground state?????
For large atoms, the electrons are not transitioning states in the
ground state. Nor are they emmitting spectra.
For hydrogen, there is no difference in models, it is a single proton
with a charge. You have to use larger atoms to get a difference.
>
>
> Bye,
> Bjoern
Bjoern Feuerbacher
>
> "FrankH" <frank...@yahoo.com> wrote in message
> news:46484c9f.04022...@posting.google.com...
[snip
> One now knows that the proton charge
> distribution, for example, has a RMS radius of 0.80 fm. I forget the error
> bar size,
0.870 +- 0.008 fm, according to the PDG.
<http://pdg.lbl.gov/2003/s016.pdf>
> but the measurements are actually good enough to provide a figure
> for the skin depth of the proton. (Crudely, the radial thickness over which
> the charge density dies down from its central value to zero)
Looks strange to me. I know that nuclei have a "skin" (the roughly have
a charge distribution which looks like a Fermi function, so it makes
sense to talk about a "skin"), but for the proton, IIRC, the charge
distribution is exponentially decreasing. What's supposed to be the
"skin" in that case?
[snip]
> Before making public proposals about new atomic models,
> you should make quite certain that you are fully aware of the state of play
> in the field. You obviously are not in that position.
Nothing new...
I really wonder why people who envision *new* models don't go first to
the nearest library and try to find out what *is* known about the *old,
established* models. They always seem to think that only a very tiny bit
is known, and if they can reproduce this stuff within an error of, say,
10%, this somehow proves that their ideas are valid.
> Your earlier remarks indicated that you are probably not au fait with
> elastic scattering experiments.
And that despite my remarks in the earlier thread where he can read up
on the state of the art. Typical.
[snip]
Bye,
Bjoern
Bjoern Feuerbacher
is this so?
FrankH wrote:
>
> Bjoern Feuerbacher <bfeu...@ix.urz.uni-heidelberg.de> wrote in message
> news:<403C85FB...@ix.urz.uni-heidelberg.de>...
[snip]
> > > Why the electron shell energy levels are arranged in the way that they
> > > are
> >
> > Standard QM explain this, too.
> >
> > Quantitatively, BTW.
> >
> > And another BTW: it also can do this for the magical numbers of nuclei.
> > How do you explain those?
>
> I am working on this, so far I have noticed the magic numbers are
> associated with atoms which are perfectly symmetrical in the cubic
> model.
Didn't you say earlier that perfectly symmetrical atoms in your model
correspond to *chemical* stability? Now you say they correspond to
*nuclear* stability? Did I misunderstand you somewhere?
And, BTW, another question that you ignored so far: how do you explain
that the energies involved in nuclear reactions are so much greater than
the ones involved in chemical reactions?
> > > Explains where neutrons are contained in the atom
> >
> > Evidence that this explanation is right?
> >
> I am working on showing that the maximum number of neutrons or the
> most common isotopes is related to the shape of the atoms core.
This doesn't adress my question.
> > And how do you reconcile this with the wave-like behaviour of quanta,
> > including neutrons? How can a wave be located at a certain position?
> >
> Individual particles like electrons and neutrons appear to behave like
> waves. I'm not sure I agree with the experiments showing wave behavior
> for particles.
Well, how do you explain then the results of the double-slit experiment?
> When inside an atom, they act like particles not waves.
How do you know? Why should they act different there?
Standard theories explain everything with some basic laws which apply
everythere. You, on the other hand, seem to need to invent new laws for
every new phenomenon.
[snip]
> > > Eliminates the need for a strong nuclear force
> >
> > Explain the binding energies of nuclei. Quantitatively, please, like the
> > standard theories of nuclei do.
>
> I have added calculations of hydrogen, helium, and lithium to my
> website.
Err, I talked about binding energies of *nuclei* here. Not ionization
energies. Don't you understand the difference?
> > > Explains the newest pictures of atoms using STM
> >
> > Didn't you notice that the author of that paper claims that his results
> > are a verification of QM?
> >
> > (oh, and BTW, I *still* think you have got a problem with your eyes -
> > these pictures *are* fuzzy, no matter how often you claim to see sharp
> > edges there)
>
> Yes, I noticed. I did ask the original experimenters about the edges.
> They acknowleged that they were unusual and probably due to some sharp
> dropoff in the orbital configuration, but they didn't seem to have a
> great explanation. Their results may be explainable in terms of the
> cubic theory - but I'm still looking into that.
I still don't understand where you see "edges" in these pictures.
> > > Explains the most common fission products of uranium
> >
> > Standard nuclear theory can do this, too. Quantitatively. And IIRC, your
> > model gives only a rough estimate which disagrees with the actual
> > results. Apparently a rough agreement is enough for you to claim that
> > your model "explains" something.
>
> Can you point me to a reference of this?
hmmmm ... difficult. I remember this reading somewhere (IIRC, at several
different sources), but unfortunately I can't remember where in the
moment.
Perhaps it's in the book by Povh I already recommended several times to
you.
Oh, something about this can be found here:
<http://hyperphysics.phy-astr.gsu.edu/hbase/hframe.html>
They don't give the argument explicitly, but you can see that it follows
from the binding energy curve.
> The most I've see is the
> waterdrop analogy which shows splitting of the nucleus like a
> waterdrop which didn't seem like much of an explanation.
Do you know the formula by Weizsaecker for the nuclear binding energy?
It is based on this "waterdrop" analogy, but uses far more sophisticated
arguments and can be used quite well for quantitative predictions.
Again I *really* would recommend Povh's book to you!!!
> > > Explains how an alternating series of protons and electrons are stable
> >
> > What do you mean here?
>
> See the other post about the stability calculation
Oh, yes, I see. However, in that post, you didn't explain *why* this is
stable. In particular, you didn't explain what kind of force there is
which counteracts the electrostatic attraction between the two (hint:
simply claiming that "protons and electrons have hard surfaces, they
can't penetrate each other" makes no sense).
[snip]
> > > It correctly predicts the relative difference for the first ionization
> > > energies for hydrogen and helium
> >
> > Can you point me to the exact point on your web page where you
> > demonstrate this?
>
> I recently added this, you probably didn't see it when you wrote this.
> Check again.
You posted several URLs, IIRC. On which page exactly can I find this?
> > And, BTW, can it explain the spectra of hydrogen and helium, too?
> >
> Where can I find how you calculate the spectra for helium and lithium.
Any basic book about atomic physics. I've studied physics in Germany and
hence can only give you only references to German books (we used English
books only later in the study), but I hope you will perhaps find some
translations. Two good books I now are by:
1) Mayer-Kuckuk
2) Haken and Wolf
> I've only seen hydrogen.
I've asked you this before - IIRC, you never answered: What books have
you read so far?
> I did read in a physics book that the
> calculations cannot be done exactly for multi-electron atoms. Is that
> true?
Yes, this is true. The calculations can only be done with numerical
approximations. However, these approximations are very good (they often
agree with the experimental results better than 0,1% deviation), and
there are (rough estimate) several tens of thousands of them. All agree
with the experiments, AFAIK. Hey, whole journals (where hundreds of
papers are published each year) are devoted only to this subject!
[snip]
> > > Finally, it correctly accounts for the results of the Rutherford
> > > scattering experiment
> >
> > This is close to a lie. In the original post, you yourself admitted that
> > you get only a rough agreement, and that further work is needed. Your
> > model does account only for the *qualitative* results of the Rutherford
> > experiment - it is still a *very* long way from a *quantitative*
> > explanation. In contrast, Rutherford had right from the start a formula
> > which agreed with the experimental results with amazing accuracy.
>
> OK, I got a little carried away ...
Thanks for the admission. I hope you will keep being so honest.
> > > I would say that empirical reality is saying I'm not an idiot and it
> > > has thus far explained many of the things I have thrown at it. There
> > > will be more to come ... stay tuned.
> >
> > Judging from your performance wrt the fission products of uranium and
> > Rutherford's experiment, you will count every rough agreement as a
> > success for your model, and you will ignore that standard theories
> > explain all of this, too - but with a *far* better accuracy than your
> > attempts. I wouldn't say that you are an idiot - but you obviously are
> >
> > 1) rather fast in dismissing the standard explanations, although you
> > know close to nothing about the experimental evidence for them
> > 2) rather fast in accepting rough agreement as a confirmation of your
> > ideas.
> >
> > I would term this a "double standard"...
> >
> Well, you've had 100 years to work on the standard theory, I've had
> about 1 month, so yes, I have some catching up to do, but I'm not
> doing so bad after a month.
Err, I pointed out above that Rutherford, for example, got the right
result on the very first try. Well, I don't know about the exact history
- maybe there were two years between the experiment and the theoretical
explanation. But the crucial point here is that Rutherford didn't
publish an explanation *before* he had found a formula which matched the
results with amazing accuracy - in contrast to you, who claims that your
model can explain these results, although so far you have got only some
*very* crude estimates which disagree significantly with the
experimental results.
Do you understand what I mean? I say it again, so hopefully you will get
it: IT IS UNWISE TO SAY THAT YOU HAVE GOT AN ALTERNATIVE MODEL UNLESS IT
CAN REPRODUCE AT LEAST THE BASIC RESULTS (and Rutherford's scattering
cross section, as well as the hydrogen spectrum, *are* the most basic
ones!) WITH AN ACCURACY AT LEAST COMPARABLE TO THE EXISTING MODELS.
I hope you got it?
Oh, BTW, it is even *more* unwise to say that you have got an
alternative model if you know only about 0,1% (and no, that is not an
understatement) of the available evidence for the existing models.
Again: go read some basic textbooks. Povh and the two above on atomic
physics would be a good start. Additionally, I would like to *again*
recommend Styer's book on QM to you.
Please don't misunderstand me - this is not an attempt to get rid of you
(hte discussions with you were far nicer than with the heap of crackpots
in this group...). It is only an attempt to try to explain to you that
one shouldn't make pronouncements about a topic one knows close to
nothing about.
[snip]
> > > The spectrum is due to an electron which has escaped the normal
> > > confines of the atom ground state and is technically not even part of
> > > the atom anymore.
> >
> > As long as it bound to it, it *is* part of the atom. You are making no
> > sense.
> >
> > Do you claim, too, that the earth is not a part of the solar system
> > because it is not bound to the surface of the sun?
> >
> More like if you took a rock and threw it off the earth, the rock is
> no longer part of the earth.
False analogy. The "atom" does not correspond to the earth - it
corresponds to the whole earth-rock system in this case. The earth does
correspond to the *nucleus*, not to the atom.
> > > It really doesn't tell us much about the internal structure of the atom.
> >
> > Spectra don't tell us much about the internal structure of the atom???
> > This has to be one of the most ignorant, nonsensical sentences I've
> > heard in a while (well, ignoring Porat's statements here - he is simply
> > not comparable to anyone else).
> >
> Can you tell me what is the relation between the electron shell
> configurations shown on www.webelements.com and the spectra generated
> by excited electons? I'm missing something here.
I don't know what exactly you mean by "electron shell configurations"
here.
Where exactly did you find this on the page mentioned?
However, the spectra generated by excited electrons correspond directly
to the energy levels in the atom. And these energy levels correspond to
wavefunctions - i.e. to the structure of the atom.
> > > Data like the ionization energies for each of
> > > the electrons gives us a better idea of the internal structure since
> > > it shows how tightly each electron is bound in the atom.
> >
> > Hint: standard QM predicts these ionization energies with good accuracy.
> >
> References?
Any textbook on atomic physics, I would say.
I *know* that the ionization energy for helium can be found in the two
books mentioned above (they give a numerical approximation which already
is quite good, and explain how one can improve this approximation with
more work). I'm not sure if they deal with heavier elements, too - but I
*do* know that more advanced books and, most important, research papers
deal with this and explain how one can do this. One book I have got here
right by my side is:
A.Szabo, N.S.Ostlund, Modern Quantum Chemistry, Dover
Publications,Mineola (1996).
But I wouldn't recommend it to you - it is a book for advanced readers.
What you seem to need is more a basic text book on Quantum Mechanics.
Please try the books I already mentioned (Mayer-Kuckuk, Haken&Wolf,
Styer); you could also try to read Schwabl. And the Feynman lectures on
QM (volume 3 of his lecture notes) are also a good read.
> > > But once an
> > > electron has left its normal ground state, it is governed by an
> > > entirely different set of rules.
> >
> > Why should it be?
> >
> >
> > > For the cubic atomic model, these
> > > excited electrons would be governed by the rules of spherical
> > > harmonics.
> >
> > What rules are those?
> >
> > And why should they apply?
> >
> > And why shouldn't they apply in the ground state?
Hello? (for all four questions)
[snip]
> > > Quantum mechanics appears to use these spherical
> > > harmonics in the spectroscopy equations,
> >
> > I don't know what you exactly mean by "spectroscopy equations", but yes,
> > standard QM uses spherical harmonics for describing bound states of
> > electrons in atoms. There are (rough estimate) several thousands of
> > these standard QM calculations which have been verified experimentally.
> > How do you explain this, if standard QM is wrong?
>
> I'm saying QM may be right in this case.
So it is right for the excited states, but suddenly wrong when it comes
to the ground state? Why on earth should this be the case???
And why is it able to predict transition frequencies between excited
states and the ground state, if its description of the ground state
isn't right?
And what about Moseley's formula, which also deals with the ground
state?
> > > so I presume that they are
> > > correct in predicting the expected spherical harmonics and spectral
> > > lines.
> >
> > But why should standard QM be only right for the excited states, but not
> > for the ground state?????
>
> For large atoms, the electrons are not transitioning states in the
> ground state.
I don't understand what this is supposed to mean, sorry.
> Nor are they emmitting spectra.
Huh? Large atoms don't emit spectra??? Or what on earth did you want to
say here???
> > > However, the interpretation of the equations as showing the
> > > actual behavior of the electron being contained in a probability cloud
> > > is misleading.
> >
> > Thanks for showing again that you don't know what you are talking about.
> > Nobody says that the electron is contained in a probability cloud. And,
> > BTW, you were corrected on this before, so repeating this nonsense is
> > again close to a lie.
> >
> > IIRC, I recommended Styer's book "The strange world of Quantum
> > Mechanics" to you. Have you read it in the meantime?
Hello?
> > > The electron is free to move about randomly around the atom,
> >
> > *Very* roughly right. (well, it isn't entirely free - there is the
> > electrostatic attraction to the nucleus, after all!)
> >
> >
> > > but the resonant frequencies and emmited photons correspond to
> > > the wave functions.
> >
> > "correspond" in what way? Your sentence makes no sense.
Hello?
> > > That is my guess about what happens when an atom
> > > is excited beyond ground level.
> >
> > Sorry, I didn't understand what you tried to say.
Hello?
> > > Technically, the cubic atomic model
> > > says nothing about what happens to an electron once it leaves the
> > > ground state. The cubic model details the bulk static properties of an
> > > atom in the ground state.
> >
> > Again, why should the description of the ground state be handled by
> > another model than the description of the excited states?
> >
> > And, BTW, you *do* know that the QM description of the excited states
> > (which you seem to accept!!!) is based on a *pointlike* (i.e. very
> > small) nucleus? So why are these predictions correct, although,
> > according to you, the nucleus is *not* very small?
>
> For hydrogen, there is no difference in models, it is a single proton
> with a charge.
In your model, the electron is "sitting" *on* the proton. In standard
QM, it isn't. How on earth can you say "there is no difference in
models"???
> You have to use larger atoms to get a difference.
Well, QM predicts the behaviour (spectra, binding abilities and so on)
*also* with amazing accuracy. So my question still stands.
Bye,
Bjoern
FrankH
>
> No. That's not the "main" reason, that was only the first piece of
> evidence. Nowadays we have *far* more experimental data on scattering of
> nuclei and individual protons and neutrons (not Rutherford scattering -
> the experiments have long ago gone far beyond that simple experiment of
> Rutherford). Every scattering argument shows that these things *are*
> small.
References?
> > I have proposed a new model of the atom which postulates that atoms
> > are simply formed out of alternating sequences of electrons and
> > protons. The protons and electrons are arranged in a very particular
> > geometric sequence. You could think of this model as 2 sheets of alpha
> > particles (helium atoms) intersecting forming an X in the shape of an
> > octahedral. I call this the cubic atomic model. The details of this
> > theory can be found at:
> >
> > http://ourworld.compuserve.com/homepages/frankhu/buildatm.htm
>
> Do you explain there also where the repulsive force comes from which
> keeps the electrons from falling into the protons? And why he have never
> found experimental evidence for this repulsive force?
I admit that my model postulates the existence of this as yet unproven
repulsive force that keeps a proton and electron apart. Thomas Lockyer
has a QVP model (he sent you his DVD) which shows that the magnetic
forces repel the electrostatic at close distances, which is a possible
and sensible explanation. See:
http://members.aol.com/tnlockyer/elposnull.gif. The evidence for this
repulsive force may already be out there, I just haven't found it.
Although I still like the intuitive idea that even at the subatomic
level, a particle occupies a fixed amount of space (like a little
marble) and must maintain a fixed distance apart no matter the
attractive forces between the two particles. I don't believe that the
reason why we have hard surfaces is because of the repulsion of the
outer electron shells. I an atom were really 99.999% empty space, then
you'd think that you could easily compress it 99.99%. But atoms are
not that compressable in a solid. How does QM explain that?
By comparison, is it more believeable that there is a strong nuclear
force which causes 2 protons in the nucleus to experience 1000 pounds
of binding force? See:http://www.geocities.com/quantum_00_2000/4forces/strongforce.html
The description of the gluons responsible for this force also has
properties which are simply postulated. See
http://www.factmonster.com/ce6/sci/A0821053.html and the evidence is
only indirect.
I think it is far more believeable that a proton and an electron
maintain a fixed position from each other.
[snip]
I
> > have done calculations to determine how frequently you would expect to
> > see these reflective collisions and I have compared them with the
> > original Rutherford scattering experiment results and I can show that
> > the predicted percentage for the alpha particles at particular angles
> > for the cubic model roughly match the experimental results.
>
> "roughly". Interesting.
So you would agree that it "roughly" matches and the the calculations
don't contain some fatal flaw? I did have to redo the calculations
serveral times to get it accurate. I would be surprised if there
weren't more mistakes.
>
>
> BTW, shouldn't the result of the scattering depend on the projectiles,
> too, according to your model? Nevertheless, experiments show that one
> gets the same scattering cross section for alpha particles and electrons
> (the only deviations are at small angles, where the non-zero size of the
> nucleus becomes important).
>
No it shouldn't. The cubic model is a neutral matrix, so it doesn't
matter what the charge of the object is that you throw at it.
Electrons, protons, alphas, whatever. All that matters is that the
scattering cross section is going to be the same and you should get
the same percentage chance of scattering no matter what you hit it
with.
>
> Additionally, does your model reproduce the observed dependence of the
> scattering cross section on the charge of the nucleus, Z? Does it
> explain why the scattering cross section does *not* depend on the number
> of neutrons, N, in the nucleus? Rutherford's formula can do all of this
> without any problems.
I can work on this calculation, but I can see that the atomic radius
of silver is approximately the same as gold. However in the cubic
atomic model, the silver atom is smaller and the cross section will be
smaller. The net percentage of scattering at angles will be the same,
but since there is less target area, you should see a reduction in the
overall number of reflected alphas for the same experimental setup.
So, yes, the Z dependence can probably be modeled, I do need to do the
math.
>
>
> > The data
> > for the original Rutherford scattering data by GigerMardsden can be
> > found at:
> >
> > http://dbhs.wvusd.k12.ca.us/Chem-History/GeigerMarsden-1913/GeigerMarsden-1913.html
>
> Thanks for the reference (it's always nice to have the opportunity to
> read an original paper on a big discovery! :-) ) - but why don't you use
> more modern sources, which have far more accurate data?
>
I scoured the internet trying to find a more recent modern source and
came up empty handed except for one paper from MIT published in 2002
using the most up to date equimpent science has to offer. Surely, this
must absolutely confirm the accuracy of the Rutherford experiment.
See:
http://web.mit.edu/tkfocht/www/rutherfordpaper.pdf
But guess what? Their experiments found a significant deviation at
angles below 10 degrees and at angles above 30 degrees, they couldn't
get an accurate data due to the infrequency of seeing alphas at above
this angle. Now if scientists in 2002 can't get accurate results over
30 degrees, how could GeigerMarsden get them in 1911? I smell a rat!
Is this the remarkable accuracy predicted by the Rutherford formula?
Curiosly, the authors go on to say that their experiments confirmed
the Rutherford formula despite the results and explained away the low
angle disagreements by saying electron sheilding and multiple
scattering events were the cause without justifying this statement.
Look at the actual graph. If I showed you experimental results like
this and tried to say it followed the Rutherford formula, you would
laugh. It looks a lot more like a linear dropoff. I can draw a neat
straight line through just about all of their data points.
>
> > I began my calculations by collecting the statistics on the size of
> > the gold atom according to the cubic model. It is 14 units high and
> > the arms are 10 units wide from side to side. I approximate this as a
> > sphere with a radius of 7.
>
> Why is this approximation justified? What happened to the "sharp edges"
> you argued about above?
This is only to calculate the possible orientations of the atom within
a spherical confinement.
>
> And what are your "units" here? (I know that this is irrelevant, because
> you only want to calculate percentages, but nevertheless I would like to
> know this here).
Units are in the width of the average distance between a proton and an
electron. I have previously guessed this dimension at 187pm based on
STM pictures of silicon.
>
>
> > This has a spherical surface of 615. Since
> > the cubic atom is symmetrical, the unique orientations are only
> > contained in 1 quadrant of the sphere or 1/8 of the surface. This
> > corresponds to a 90 degree turn through each of the x,y,z axis. This
> > means the area of investigation is only 76.
>
> Agreed so far, although you use rather strange wording.
>
>
> > The size of the atomic
> > unit representing the area of the top of the atom's core is a 2 X 2
> > square with an area of 4. This means there are 76/4 = 19 unique
> > orientations can roughly fit into this quadrant with no overlap.
>
> Right - but why is the "with no overlap" relevant?
>
> And again, very confusing wording! I had to read this several times
> until I understood what you try to say.
Thanks for hanging in there and trying to understand it. I had to
invent my own way of describing the possible orientations. I finally
got a ball and put stickers on it to represent the possible
orientations in a single quadrant. If you take a ball and place 5
stickers from the pole to the equator and then put 4 more along the
equator, and then 4 more heading up to the poles and then spread 7
more stickers in the space between, you get an idea of the
orientations and why I don't have overlap. I can't possiblely
calculate every possible orientation, so I only did the 19
orientations that can fit into a quadrant using a 2x2 target.
>
>
> > There
> > are basically only 2 orientations which would result in high angle
> > reflections. These are the head-on (alpha tries to pass through core)
> > and edge-on (alpha tries to pass through arm edge).
>
> Why would these result in high angle deflection? What's there in the
> core or in the arm edge which deflects the alpha particle? What force is
> acting there?
Simple elastic collision with a helium atom. Two alpha particles would
certainly bounce of each other and the arms of the cubic model are
basically built out of helium atoms which are nothing more than alpha
particles with their electrons. The edges present a thick layer (at
least 5 units deep) of helium atoms which cannot easily be pushed
aside. The cubic theory postulates that at angles of less than 90
degrees, the helium atoms can be pushed aside since they have to
penetrate 1 helium atom.
>
>
> > For a head on
> > orientation, I calculated a 4% chance of hitting the core directly,
> > 32% chance the arms get hit and 64% chance of a complete miss. For the
> > edge on orientation, I calculated a 20% of hitting an arm and 80%
> > chance of a miss or pass through. I plugged these into the 19 possible
> > slots with 11 orientations being edge on, 1 orientation being head on
> > and the remaining 7 as being orientations where the alpha basically
> > passes through. The angles of deflection are based purely on a
> > classical elastic collision with the atom.
>
> You *do* know that classically, objects have a hard surface and
> therefore bounce off each other when they collide because of
> electrostatic forces, don't you?
I have seen this theory, but disagree with it as I have stated before.
Surfaces are hard because they are space filling as shown in the cubic
model.
>
>
> > Because the atom is
> > effectively a neutral matrix of joined helium atoms, the effect of the
> > columb forces deflecting the alpha are negligible.
>
> Are you sure about that? Even when the alpha particle hits "head on"?
Doesn't matter where it hits, the particle always sees a closely
alternating series of protons and electrons in a balanced neutral
matrix.
>
>
> > The calculations show the percentage chance for:
> >
> > A complete miss or pass through 86.3% Would expect angle <
> > 5 degrees
> > An arm gets hit 13.1% Would expect any angle 0 -
> > 180
> > A direct hit of the core .21% Would expect angle 90 -
> > 180
> >
> > This compares to the experimental data which shows:
> >
> > Deflections less than 5 degrees 79.2%
> > Deflections 5 - 22 20.4%
> > Deflections greater than 22 .35%
>
> Well, this *is* a "rough" agreement. ***VERY*** rough. It doesn't even
> agree with the *original* results of Geiger and Marsden very well. You
> have no clue how accurate scattering experiments nowadays are, have you?
Can you find me some more references on the internet besides the MIT
one which disagreed with Rutherford?
>
> IIRC, I recommended some books to you, among them one by Povh on nuclear
> physics, which lists a *very* small amount of experimental results from
> the plethora of experiments done in nuclear and particle physics (hint:
> the book is nevertheless quite thick). Have you looked into that book in
> the meantime?
>
I looked for the author Povh in the seattle public library and came up
empty - did you spell this right?
> Please go to the more recent experiments, which have shown again and
> again (hey, I even did this experiment *myself* at one time!!!) that the
> scattering cross section depends on the angle like 1/sin^4(theta/2). As
> soon as you are able to reproduce this formula (or at least some
> numerical results which agree with it on a level of, say, 1%), I will
> admit that your model has some merit. Good luck.
Gee, what do you think of the MIT experiment then???
I noticed the rutherford formula could also be approximated by
1/theta^3 (roughly). For a fancy formula, its results are close to
being just exponential.
>
> (oh, BTW, shouldn't your model give a dependence of the scattering cross
> section on phi?)
What is phi?
>
>
>
> > The details of this calculation can be found on an excel spreadsheet:
> >
> > http://ourworld.compuserve.com/homepages/frankhu/ruther.xls
>
> I don't know if I should love or cry here.....
>
> You haven't got the *slightest* clue of how many experiments were done
> in nuclear and particle physics in the past decades, how accurate they
> are, how much work is spent in analysing the results. Sorry if this
> insults you, put pretending that an *Excel spreadsheet* somehow is able
> to reproduce these results based on your model is just pathetic, plain
> and simple.
If my calulations are wrong, then lets here it. Lots of physics is
done with math as many have pointed out and is perfectly valid
starting place.
>
>
> > The predictions from the cube model and the actual experimental
> > results are not exactly the same by any means,
>
> Nice that you see that.
>
>
> > however, they are in the same rough ballpark.
>
> BFD. I think *every* model of the atom and the nucleus, no matter how
> contrived, will manage to come up with some or even *lots* of results
> which roughly agree with the experimental results. However, so far
> *none* of this plethora of "alternative models" has been able to
> 1) explain all the things which standard QM explains
> 2) explain even *one* thing which QM explains with a comparable accuracy
> as QM.
>
> You should try reading up on the "successes" of other "alternative
> models". For example, you could look at Porat's model (right here in
> sci.physics - the guy who either can't spell properly or doesn't bother
> to do it), at the ideas of Black Light Power, or at Common Sense
> science. All three have models which are far more developed than yours,
> which strongly disagree with your, which include many strange ideas -
> and nevertheless all three are able to reproduce a fair amount of
> experimental evidence, sometimes with quite amazing accuracy. However,
> all of them suffer the same faults - the two outlined above.
>
> You have still a long way to go. Good luck again.
>
Thanks for wishing me luck, I'll need it. Supporting a new theory is a
David vs. Goliath task, but hey, David won didn't he? I asked Porat
for his paper, he never replied and his web site doesn't exist anymore
- Hey Porat, you out there????
Thanks for the other references, I will have to take a look. Maybe we
should consider these other theories as well if they predict with the
same accuracy. But the key would be if you could present experments
which contractict QM while supporting another theory like the ones I
have described.
>
> > The main point you should observe is that the
> > cubic model is able to predict a scattering pattern whereby the vast
> > majority of the alpha particles pass right through (86%), while a tiny
> > fraction (.21%) gets deflected through high angles.
>
> BFD. You are *still* doing a mainly qualitative analysis, instead of the
> *quantitative* one done by Rutherford 90 years ago. He derived a
> *formula* which described the results with amazing accuracy. Your
> results so far are mainly handwaving: "See, I can reproduce the
> qualitative behaviour in the right way!!!".
>
>
> > This scattering
> > pattern does not necessarily have to be created by the atom postulated
> > by Rutherford as a tiny compact nucleus containing all the positive
> > charge.
>
> As long as you stay on this qualitative level, you are right. But as
> soon as you become quantitative, i.e. take Rutherford's *formula* and
> compare it to the experimental results, it's clear whose model is
> better.
>
Not even Rutherford's formula can give exact numbers. You always have
to see if you can match the curve predicted by the rutherford formula
eventhough it appears to calculate exact numbers. If I stick my model
in a computer, I can get a finer breakout of the angles - my model
results cannot be calculated as the result of a simplistic formula.
Would that be quantitative enough for you?
>
> > You can basically get the same result from the cubic atomic
> > model.
>
> The crucial word here is "basically".
>
>
> > A better calculation using a computer model to consider random
> > orientation and random alpha may produce results more comparable (or
> > not) with the actual experimental data and would provide a more
> > detailed range of angles to expect when an arm is struck.
> > Unfortunately, I do not have the resources to commit to such a
> > calculation.
>
> Why do you insist on finding another explanation for the results,
> although the existing explanation already works rather well?
>
> Hint: new theories in science are only invented when there is
> experimental evidence that the old theories are wrong. Not simply
> because someone doesn't like or doesn't understand the old theories.
New theories happen all the time, the only time new theories get
accepted is if you perform an experiment in which only one of the
theories can survive. A theory is accepted if it can explain
everything the old one did, plus lots more. A theory can also be
accepted if it explains everything but is simpler.
>
Lots of the books on quantum mechanics have titles like "The odd
quantum" or the "Mysterious world of quantum mechanics", but loosely
paraphrased, these titles all say, "The wacky unbelieveable world of
quantum mechanics". Because the theory is counter intuitive and
difficult to understand or believe. We believe in all the weird
quantum mechnical stuff primarily because it matches experimental
results. That's fine, but it doesn't make quantum mechanics any less
weird. The reason why I want to find a new theory is to remove the
weirdness and replace it with a model in which all of our classical
mechanics still apply and all of the ideas are intuitive and easily
understood - like the cubic model. I would like to see quantum
mechanic books replaced by one which was simply titled, "The elegant
structure of the atom".
>
> > There would be other experiments to confirm or deny the cubic model
> > with Rutherford scattering. Perhaps some of these have already been
> > done.
>
> Go read Povh. There is lots of stuff about scattering in it.
>
>
>
>
> > I would like to see the Rutherford scattering experiments
> > repeated but instead of using gold foil, use a form of crystallized
> > gold (octahedral crystal) where we are reasonably sure that the gold
> > atoms are all aligned in the same direction, and see how the high
> > angle scattering depends on the orientation of the crystal.
>
> The experiment has been done with all kinds of stuff, AFAIK - not only
> with gold foils, but also with other elements, and as well with isolated
> nuclei and even isolated protons and neutrons (well, obviously the
> latter experiments shouldn't be called Rutherford scattering anymore).
> Additionally, instead of alpha particles, also electrons have been used,
> and protons, and positrons, and neutrinos. In *all* of these experiments
> (I estimate there have been several millions of those - hey, AFAIK there
> were already several millions of created Z bosons registered at LEP,
> which was *also* a scattering experiment!), the predictions of the
> standard theories were proven - often with an amazing accuracy.
You sure about that, or are there lots more papers like the MIT one
which declares it confirms Rutherford, but really contradicts it or
simply ignore the mismatched data as experimental error?
>
> So why on earth do (the one who knows nothing about even one of all of
> these experiments) think that there is a problem with the standard
> theories, and we need a new one?
>
See the above comments on the "odd quantum".
>
>
> > Based on
> > the cubic model, I would predict that the crystal would present very
> > little scattering in most directions, since the alpha particles are
> > able to pass through the thin parts of the atom, but when the crystal
> > is oriented so it hits the very edge of the atom and tries to pass
> > through the core or arms, it will bounce back strongly. I would
> > predict that we would see lots of scattering at 90 degree crystal
> > orientations, which would not be explainable by the Rutherford
> > formula. The Rutherford formula would predict the same scattering
> > pattern/amounts no matter what the orientation.
>
> Well, for starters you could try to predict the observed behaviour of
> the scattering cross section on Z. There should be a heap of
> experimental data on this to compare with. Good luck again.
>
Yup, I have to plug this in as I have described above.
>
> > Another possible experiment would be to use low speed alpha particles.
> > At some point, if the speed of the alpha particles were slow enough,
> > it wouldn't be able to penetrate the atoms thin arms and you would see
> > almost all of the alphas being deflected at high angles. Rutherford
> > would predict that all of the alphas would penetrate no matter how
> > slow the alphas were going since there isn't much for the aphas to run
> > into and an electron isn't likely to deflect an alpha very much.
>
> If the alpha particles are slow enough, the electrons *will* be able to
> deflect them.
They'd have to be pretty slow since a proton/electron collison is like
a speck of sand hitting a bowling ball. The bowling ball's trajectory
isn't likely to be affected. I suppose you would have to calcualte
beforehand at what this point would occur and then test for it
experimentally to see if there are higher speeds than this which still
cause full reflectivity of the alphas.
>
> And, BTW, Rutherford scattering *has* be done at a *lot* of different
> energies (speeds) now. So far, every experiment agreed with his formula.
As near as I can tell, it has always been with faster particles. Any
refrences for slower particles?
>
>
> > If anybody knows the results of these experiments, please post them to
> > help confirm or deny the plausability of the cubic atomic model.
>
> Why don't you do what I recommended to you - try getting a book on
> experimental nuclear and particle physics and read up for yourself what
> experimental evidence exists?
>
>
> > In
> > conclusion, the results of the Rutherford scattering experiments do
> > not conclusively prove the notion of the nucleus being a tiny speck in
> > the atom with surrounding electron clouds.
>
> They do, if one uses the quantitative formula instead of your mostly
> qualitative hand waving.
>
>
> > Thus far, arguments against
> > Rutherford have lacked an alternative solid model to base calculations
> > on.
>
> Wrong. Lots of people have tried to build alternative models (for
> examples, see above) and claim to be able to explain stuff like
> Rutherford scattering (well, to be fair, Porat avoids this topic - but
> the other two models I mentioned adress this, IIRC). You are by far not
> the first one.
Sorry, didn't have time to read the entire internet. I was referring
to a post by Jim Carr where he basically proposed the same thing, but
didn't have a concrete model.
>
>
> > The cubic atomic theory provides this solid model which you can
> > run calculations on to show that it can return results similar to the
> > experimental results of the Rutherford scattering experiment.
>
> Quit the handwaving. Rederive Rutherford's formula from your model.
>
> Or, alternatively, explain the spectrum of the hydrogen atom.
> Quantitatively.
That would be my next stop, to show that the spectrum is compatible
with the cubic model and that the electron shell structure can be
derived from the model.
-Thanks for all of the thoughful replies. I do appreciate the time you
took to understand this fairly lengthy and complex post. I do
apologize if I seem arrogant, I really am trying to figure out if the
cubic model will work and would be willing to accept if this theory
doesn't work if shown the evidence. I barely have time to work on this
stuff since I have a full time job, a long commute and little kids to
take care of at home. My hope is to show enough promise in the cubic
theory, that others will take an interest and perform or find the
results of some of the experiments I have proposed. If the theory
survives the skeptics on the internet, and if I can write it up in a
sensible scientific manner, I can publish in a peer reviewed journal
and let real science take its course. Either that, or you and Franz
shoot it down in a flaming wreck and that will be the end of that. But
so far, I haven't seen a fatal shot.
By the way, is there anybody else out there reading these posts on the
cubic model and agree with it? Or is it just Franz and Bjoern out
there. I could use a little support.
-thanks all
Franklin
Franz Heymann
news:<403C85FB...@ix.urz.uni-heidelberg.de>...
> [snip
> > >
> > > If you look at my website, you can see that the cubic atomic model
> > > presents empirical evidence to help explain:
> > >
> > > The reactivity of an atom is related to symmetry
> >
> > Standard QM can explain the reactivity of an atom, too.
> >
> > Quantitatively, BTW.
> >
> >
> > > Why the electron shell energy levels are arranged in the way that they
> > > are
> >
> > Standard QM explain this, too.
> >
> > Quantitatively, BTW.
> >
> > And another BTW: it also can do this for the magical numbers of nuclei.
> > How do you explain those?
> >
> I am working on this, so far I have noticed the magic numbers are
> associated with atoms which are perfectly symmetrical in the cubic
> model.
While you're at it, you might look into the explanation of both the singly
and the doubly magic nucleii
> > > Explains where neutrons are contained in the atom
> >
> > Evidence that this explanation is right?
> >
> I am working on showing that the maximum number of neutrons or the
> most common isotopes is related to the shape of the atoms core.
>
> > And how do you reconcile this with the wave-like behaviour of quanta,
> > including neutrons? How can a wave be located at a certain position?
> >
> Individual particles like electrons and neutrons appear to behave like
> waves. I'm not sure I agree with the experiments showing wave behavior
> for particles.
It is easily possible to cease being "not sure" on this matter by actually
reading the papers in which particle diffraction experiments are described
and analysed.
> When inside an atom, they act like particles not waves.
I take that as an admission that you have never read anything about the
behaviour of electrons in a crystal.
> > > and why they don't contribute to the chemical properties
> >
> > Standard QM explains this, too: chemistry is based on electromagnetic
> > interactions, and neutrons are electrically neutral.
> >
> >
> >
> > > Eliminates the problem of why should an electron orbit the nucleus
> >
> > This hasn't been a problem since 1926.
> >
> >
> > > Eliminates the need for a strong nuclear force
> >
> > Explain the binding energies of nuclei. Quantitatively, please, like the
> > standard theories of nuclei do.
> I have added calculations of hydrogen, helium, and lithium to my
> website.
You have hardly started. The atomic physicists have completed the QM
calculations for all the ground states as well as the important excited
states of all the atoms of interest in studying stellar structure. That
means every atom in the periodic table from H to Fe. The results for
spectral frequencies and intensities vs temperature are in excellent
agreement with observations, and are in daily use all over the world by folk
interested in the structure of stars.
I regret to inform you that if you were as fast as those physicists in doing
the calculations, you will need the services of a supercomputer and somewhat
more than 100 years of spare time on your part.
I am saying this just to illustrate to you just one tiny aspect of the sheer
size and magnificence of the edifice you are attacking with your puny
weapons.
>
> >
> >
> > > Explains the newest pictures of atoms using STM
> >
> > Didn't you notice that the author of that paper claims that his results
> > are a verification of QM?
> >
> > (oh, and BTW, I *still* think you have got a problem with your eyes -
> > these pictures *are* fuzzy, no matter how often you claim to see sharp
> > edges there)
> Yes, I noticed. I did ask the original experimenters about the edges.
> They acknowleged that they were unusual and probably due to some sharp
> dropoff in the orbital configuration, but they didn't seem to have a
> great explanation. Their results may be explainable in terms of the
> cubic theory - but I'm still looking into that.
>
> >
> >
> > > Explains the most common fission products of uranium
> >
> > Standard nuclear theory can do this, too. Quantitatively. And IIRC, your
> > model gives only a rough estimate which disagrees with the actual
> > results. Apparently a rough agreement is enough for you to claim that
> > your model "explains" something.
> Can you point me to a reference of this? The most I've see is the
> waterdrop analogy which shows splitting of the nucleus like a
> waterdrop which didn't seem like much of an explanation.
It seems that you have not ever seen the waterdrop analysis performed. You
appear not to be aware that it gives quantitative results for many aspects
of the behaviour of very large nucleii.
> > > Explains how an alternating series of protons and electrons are stable
> >
> > What do you mean here?
> See the other post about the stability calculation
>
> >
> >
> > > and why helium is so stable
> >
> > Standard theories explain this, too. Quantitatively, again.
> >
> >
> > > It correctly predicts the relative difference for the first ionization
> > > energies for hydrogen and helium
> >
> > Can you point me to the exact point on your web page where you
> > demonstrate this?
> I recently added this, you probably didn't see it when you wrote this.
> Check again.
>
> >
> > And, BTW, can it explain the spectra of hydrogen and helium, too?
> >
> Where can I find how you calculate the spectra for helium and lithium.
> I've only seen hydrogen. I did read in a physics book that the
> calculations cannot be done exactly for multi-electron atoms. Is that
> true?
Utterly untrue. The international "Atomic Wavefunctions Group" was led by
a physicist in my Department. It involved a few dozen physicists from
various countries including France, Germany, the UK and the USA. They set
themselves the task of calculating all the important wavefunctions of all
the atoms of stellar interest. They succeeded magnificently, but it took
more than a decade to get all the results. Those have now been in use for
the best part of twenty years.
Yes, actually you are, because all of your work up to now has been shown to
be built on an erroneous foundation, in this and earlier threads.
> > > As far as atomic spectroscopy is concerned, the spectrum produced by
> > > an atom is not due to anything normally contained in an atom at ground
> > > state.
> > Well, it is due to the electrons. Last I looked, the electrons *are*
> > contained in an atom at ground state...
> > > The spectrum is due to an electron which has escaped the normal
> > > confines of the atom ground state and is technically not even part of
> > > the atom anymore.
> >
> > As long as it bound to it, it *is* part of the atom. You are making no
> > sense.
> > Do you claim, too, that the earth is not a part of the solar system
> > because it is not bound to the surface of the sun?
> More like if you took a rock and threw it off the earth, the rock is
> no longer part of the earth.
>
> >
> > > It really doesn't tell us much about the internal structure of the
atom.
> > Spectra don't tell us much about the internal structure of the atom???
> > This has to be one of the most ignorant, nonsensical sentences I've
> > heard in a while (well, ignoring Porat's statements here - he is simply
> > not comparable to anyone else).
> Can you tell me what is the relation between the electron shell
> configurations shown on www.webelements.com and the spectra generated
> by excited electons? I'm missing something here.
Yes. You are right. You are missing almost everything here. Has the
following not occurred to you?
The electron shell configurations shown there are all purely the
configurations for the ground state. You need to get an electron into an
excited state before it can radiate. Such a state is not part of the ground
state configuration.
> > > Data like the ionization energies for each of
> > > the electrons gives us a better idea of the internal structure since
> > > it shows how tightly each electron is bound in the atom.
> >
> > Hint: standard QM predicts these ionization energies with good accuracy.
> >
> References?
Any of the hundreds of papers produced by the Atomic Wavefunctions Group.
M.J. Seaton, or D. Hummer, or W.Eisner are good names to look out for.
These are only the first 3 that came to mind out of a few dozen. Their
wavefunctions are all good to better than 1:10,000
> > > But once an
> > > electron has left its normal ground state, it is governed by an
> > > entirely different set of rules.
> >
> > Why should it be?
> >
> >
> > > For the cubic atomic model, these
> > > excited electrons would be governed by the rules of spherical
> > > harmonics.
> >
> > What rules are those?
Did you see that question?
> > And why should they apply?
Did you see that question?
> > And why shouldn't they apply in the ground state?
Did you see that question?
> > > A crude decription would be an electron attached by a
> > > spring to the atom and it vibrates and has harmonics and resonant
> > > frequencies.
> >
> > Ouch! No more comment.
> >
> >
> > > Quantum mechanics appears to use these spherical
> > > harmonics in the spectroscopy equations,
> >
> > I don't know what you exactly mean by "spectroscopy equations", but yes,
> > standard QM uses spherical harmonics for describing bound states of
> > electrons in atoms. There are (rough estimate) several thousands of
> > these standard QM calculations which have been verified experimentally.
> > How do you explain this, if standard QM is wrong?
>
> I'm saying QM may be right in this case.
As indeed it is in all tests to which it has been subjected.
> > > so I presume that they are
> > > correct in predicting the expected spherical harmonics and spectral
> > > lines.
> >
> > But why should standard QM be only right for the excited states, but not
> > for the ground state?????
>
> For large atoms, the electrons are not transitioning states in the
> ground state. Nor are they emmitting spectra.
Could you help by telling us what "transitioning states in the ground
state" means?
For your information, QM has produced the right answers for everything
connected with the ground states as well as the most important excited
states of all the elements for which calculations have been made.
Franz
Bjoern Feuerbacher
[snip lots]
> The international "Atomic Wavefunctions Group" was led by
> a physicist in my Department. It involved a few dozen physicists from
> various countries including France, Germany, the UK and the USA. They set
> themselves the task of calculating all the important wavefunctions of all
> the atoms of stellar interest. They succeeded magnificently, but it took
> more than a decade to get all the results. Those have now been in use for
> the best part of twenty years.
Franz, can you give me a link to this group? (they have published their
results on the web somewhere, I hope?) A Google search with the term
"Atomic Wavefunctions Group" unfortunately yielded no hit.
[snip]
> > > > Data like the ionization energies for each of
> > > > the electrons gives us a better idea of the internal structure since
> > > > it shows how tightly each electron is bound in the atom.
> > >
> > > Hint: standard QM predicts these ionization energies with good accuracy.
> > >
> > References?
>
> Any of the hundreds of papers produced by the Atomic Wavefunctions Group.
> M.J. Seaton, or D. Hummer, or W.Eisner are good names to look out for.
> These are only the first 3 that came to mind out of a few dozen. Their
> wavefunctions are all good to better than 1:10,000
Well, a Google search for "Seaton Hummer Eisner" brought 4 hits, but
none of them was helpful... :-(
Were their results only published in papers? Aren't they available
somewhere free on the Web?
[snip rest]
Bye,
Bjoern
Bjoern Feuerbacher
>
> Bjoern Feuerbacher <bfeu...@ix.urz.uni-heidelberg.de> wrote in message news:<403C816D...@ix.urz.uni-heidelberg.de>...
> >
> > No. That's not the "main" reason, that was only the first piece of
> > evidence. Nowadays we have *far* more experimental data on scattering of
> > nuclei and individual protons and neutrons (not Rutherford scattering -
> > the experiments have long ago gone far beyond that simple experiment of
> > Rutherford). Every scattering argument shows that these things *are*
> > small.
>
> References?
Your questions for references somehow begin to make me suspicious. I
*repeatedly* mentioned the book by Povh, for example, which contains
*lots* of information about scattering (including references) - why
don't you go and read it?
Are you willing to learn something, or aren't you?
> > > I have proposed a new model of the atom which postulates that atoms
> > > are simply formed out of alternating sequences of electrons and
> > > protons. The protons and electrons are arranged in a very particular
> > > geometric sequence. You could think of this model as 2 sheets of alpha
> > > particles (helium atoms) intersecting forming an X in the shape of an
> > > octahedral. I call this the cubic atomic model. The details of this
> > > theory can be found at:
> > >
> > > http://ourworld.compuserve.com/homepages/frankhu/buildatm.htm
> >
> > Do you explain there also where the repulsive force comes from which
> > keeps the electrons from falling into the protons? And why he have never
> > found experimental evidence for this repulsive force?
>
> I admit that my model postulates the existence of this as yet unproven
> repulsive force that keeps a proton and electron apart.
Well, nice that you admit this.
Now please answer the question: why haven't we (for example, in
scattering experiments) *ever* seen experimental evidence for this
repulsive force?
You *do* know that scattering experiments today can "look" far inside
the proton, don't you? That their spatial resolution is around 10^(-18)
m?
(see e. g. Povh)
> Thomas Lockyer has a QVP model (he sent you his DVD)
Yes. I haven't looked at it so far - what he wrote in this newsgroup was
nonsensical enough. (for example, IIRC, he totally misunderstood the
quantization of angular momentum - even after repeated explanations)
> which shows that the magnetic
> forces repel the electrostatic at close distances, which is a possible
> and sensible explanation. See:
> http://members.aol.com/tnlockyer/elposnull.gif.
Magnetic forces can only occur to the magnetic moments of the particles.
Magnetic moments are closely associated with angular momentum. Since
Thomas Lockyer doesn't understand angular momentum, I don't think his
model has any value.
Additionally, standard physics does *also* take the magnetic forces due
to the magnetic moments of the particles into account when analysing
scattering experiments (see, for example, "Mott scattering cross
section" and "Rosenbluth formula"). Their theoretical predictions are
totally different from the nonsense which Thomas Lockyer proposes - and,
hint: they agree with the experimental results. For the 10th time: see
e.g. Povh.
> The evidence for this
> repulsive force may already be out there, I just haven't found it.
Well, why don't you stop posting here and do some *reading* first? Start
by looking into basic books on atomic, molecular and nuclear physics;
include a basic book about QM; and when you have finished those, you can
go to the nearest university library and try looking into the tens of
thousands of papers published on this (with lots of theoretical
predictions, based on the standard theories, which have been verified by
experiments with often astounding accuracy).
> Although I still like the intuitive idea that even at the subatomic
> level, a particle occupies a fixed amount of space (like a little
> marble)
This contradicts basic QM. How would you explain the results of the
double slit experiments, based on this idea?
> and must maintain a fixed distance apart no matter the
> attractive forces between the two particles.
They can only maintain their distance in the presence of an attractive
force if there is an equal repulsive force. You *are* familiar with
Newton's second law, aren't you?
> I don't believe that the
> reason why we have hard surfaces is because of the repulsion of the
> outer electron shells.
Then where does the necessary repulsive force comes from, in your
opinion?
Gravity is a force which is accelerating you downwards. Nevertheless,
you don't move downwards - you keep standing firmly on the ground.
Hence, according to Newton's laws, there must be an *other* force, equal
to the gravitational force, which accelerates you upwards.
Exactly the same type of argument can be made if you push your hand
against the wall or other hard objects.
The only way to avoid the conclusion that there is a repulsive force is
to claim that Newton's law don't apply in these situation. Do you
*really* want to claim this?
> I an atom were really 99.999% empty space, then
> you'd think that you could easily compress it 99.99%.
Well, based on macroscopic physics, one could think this. But, hint:
electrons don't behave like macroscopic particles. You keep ignoring the
ugly fact that they have wave-like properties.
> But atoms are not that compressable in a solid. How does QM explain that?
By pointing out that (essentially because of the wave-like properties of
electrons), there is a lower limit to the size of an atom (essentially
there are limits on the possible wave lengths of an electron in a
Coulomb potential).
> By comparison, is it more believeable that there is a strong nuclear
> force which causes 2 protons in the nucleus to experience 1000 pounds
> of binding force?
What problem do you have with this?
> See:http://www.geocities.com/quantum_00_2000/4forces/strongforce.html
> The description of the gluons responsible for this force also has
> properties which are simply postulated.
Again you show that you have not the slightest clue of the experimental
evidence. This stuff isn't simply postulated - *lots* of experiments
were done on this, with very good results. For the 11th time: read Povh.
> See
> http://www.factmonster.com/ce6/sci/A0821053.html and the evidence is
> only indirect.
Do you have a problem with indirect evidence?
Hint: the evidence for the existence of electrons is only indirect, too.
> I think it is far more believeable that a proton and an electron
> maintain a fixed position from each other.
In science, it's not important what is more "believable" (for the simple
reason that different people find different things "believable"!!!!!).
It's important what the experimental *evidence* says.
> > > I
> > > have done calculations to determine how frequently you would expect to
> > > see these reflective collisions and I have compared them with the
> > > original Rutherford scattering experiment results and I can show that
> > > the predicted percentage for the alpha particles at particular angles
> > > for the cubic model roughly match the experimental results.
> >
> > "roughly". Interesting.
>
> So you would agree that it "roughly" matches and the the calculations
> don't contain some fatal flaw?
I haven't redone your calculations, but mostly they looked sensible. SO
WHAT??????????? I have repeatedly pointed out that you are by far not
the only person with an "alternative model", and that other people have
done *far* more than you wrt agreement between their models and the
data.
> I did have to redo the calculations
> serveral times to get it accurate. I would be surprised if there
> weren't more mistakes.
I don't see any reason to do your work for you.
> > BTW, shouldn't the result of the scattering depend on the projectiles,
> > too, according to your model? Nevertheless, experiments show that one
> > gets the same scattering cross section for alpha particles and electrons
> > (the only deviations are at small angles, where the non-zero size of the
> > nucleus becomes important).
> >
> No it shouldn't. The cubic model is a neutral matrix, so it doesn't
> matter what the charge of the object is that you throw at it.
Err, I was referring to the *size* of the projectile, not its charge.
Or don't you agree that electrons and alpha particles have different
sizes?
Nevertheless, according to Rutherford's formula (you know, the one which
has been tested ad nauseaum?), the charge of the projectile *does*
matter (more on this below). In your model, the result is *independent*
of the charge of the projectile. Seems like a problem for your model,
don't you think?
> Electrons, protons, alphas, whatever. All that matters is that the
> scattering cross section is going to be the same and you should get
> the same percentage chance of scattering no matter what you hit it
> with.
Why isn't the size of the projectile important, in your opinion?
> > Additionally, does your model reproduce the observed dependence of the
> > scattering cross section on the charge of the nucleus, Z? Does it
> > explain why the scattering cross section does *not* depend on the number
> > of neutrons, N, in the nucleus? Rutherford's formula can do all of this
> > without any problems.
>
> I can work on this calculation, but I can see that the atomic radius
> of silver is approximately the same as gold. However in the cubic
> atomic model, the silver atom is smaller and the cross section will be
> smaller. The net percentage of scattering at angles will be the same,
> but since there is less target area, you should see a reduction in the
> overall number of reflected alphas for the same experimental setup.
> So, yes, the Z dependence can probably be modeled, I do need to do the
> math.
Please do. According to Rutherford's formula, the scattering cross
section is proportional to the charge of the projectile squared (that's
where the big problem for your model is, which according to you above
says that the charge of the projectile doesn't matter), and proportional
to the charge of the target squared.
I'm interested in how you will manage to get this behaviour from your
model. Especially I will be interested to see how you get from your
model that the scattering cross section for an alpha particle is four
times greater than the one for an electron (provided both have the same
energy).
Additionally, I would like to see you to reproduce the behaviour of
Rutherford's scattering cross section wrt energy (it is proportional to
1/E^2).
Good luck.
Oh, BTW, what possible experiment could disprove your model?
> > > The data
> > > for the original Rutherford scattering data by GigerMardsden can be
> > > found at:
> > >
> > > http://dbhs.wvusd.k12.ca.us/Chem-History/GeigerMarsden-1913/GeigerMarsden-1913.html
> >
> > Thanks for the reference (it's always nice to have the opportunity to
> > read an original paper on a big discovery! :-) ) - but why don't you use
> > more modern sources, which have far more accurate data?
> >
> I scoured the internet trying to find a more recent modern source and
> came up empty handed except for one paper from MIT published in 2002
> using the most up to date equimpent science has to offer.
The paper doesn't say this. In contrast, it says that Rutherford's
scattering set up was used. Apparently only for the data processing they
used modern equipment - but even there, not "high-end" stuff.
> Surely, this
> must absolutely confirm the accuracy of the Rutherford experiment.
> See:
>
> http://web.mit.edu/tkfocht/www/rutherfordpaper.pdf
This paper doesn't look like a scientific article to me, but rather as
if a student reported the results of his lab work. I have written
similar stuff during the lab work required in my study, too.
Going to
<http://web.mit.edu/tkfocht/www/>,
my suspicion is confirmed. The link to this paper is found under the
heading "Junior Lab Papers and Presentations".
So instead of going to read some of the things I directed you to (like
Povh) you did a web search and found a report on a lab work experiment.
And you seem to think this is can be used as if it is a valid scientific
paper. Say, are you *really* unwilling to learn *anything*?
> But guess what? Their experiments found a significant deviation at
> angles below 10 degrees and at angles above 30 degrees, they couldn't
> get an accurate data due to the infrequency of seeing alphas at above
> this angle.
You seem to have looked in this paper only for problems, but apparently
ignored all the counterarguments and good results. I give some quotes:
"... Rutherford's nuclear model fit the experimental data, even
accounting for extremely large angle scattering. [1]"
The reference here is to a book: MIT Department of Physics, "Rutherford
scattering" [2002]. May I suggest to you to try getting this book?
"At high angles, the discrepancies between measurement and prediction
are largely due to error in the measurement."
"... we found that the Rutherford cross-section fails at low angles due
to electron shielding and multiple scattering events."
So they gave sensible explanations for the deviations. What problem do
you have with this?
And, BTW, deviations are large angles are *expected* - because of the
finite size of the nucleus, it's not really point-like. Hey, I used this
myself in an experiment to measure the size of the gold nucleus!!!
> Now if scientists in 2002 can't get accurate results over
> 30 degrees, how could GeigerMarsden get them in 1911?
I don't know. Look into their paper and try to find out. Perhaps they
simply used longer measurement times.
> I smell a rat!
Oh, interesting. You apparently accuse scientists of fraud.
> Is this the remarkable accuracy predicted by the Rutherford formula?
Yes. Hint: this experiment has been done countless times - and the
agreement with the formula is *always* great (in the range of angles
where it makes sense - because of the reasons outlined above, it doesn't
work well for very small and very large angles).
Additionally, the "improved" formulas, which take things like magnetic
moments of target and projectile into account (see above), have *also*
been tested countless times with amazing accuracy.
> Curiosly, the authors
Only one author.
> go on to say that their experiments confirmed
> the Rutherford formula despite the results
There is nothing curious about that. They explained quite clearly why
they get agreement only for this range of angles. Apparently you don't
understand how experimental confirmations of theoretical predictions
work. You seem to think that as soon there is the slightest discrepancy,
the model has to be thrown out - apparently you don't know that there is
something called "experimental error", whose effect has to be
considered. In contrast, you don't apply the same level of scrutiny to
your own ideas. I smell a double standard. Again.
> and explained away the low
> angle disagreements by saying electron sheilding and multiple
> scattering events were the cause without justifying this statement.
Why on earth do you think this statement needs "justifying"????? It is a
perfectly sensible statement. What don't you understand about it?
> Look at the actual graph.
I presume you mean figure 5. What about it? It shows nice agreement in
the range of
> If I showed you experimental results like
> this and tried to say it followed the Rutherford formula, you would
> laugh.
If you additionally gave the explanations why it doesn't fit at the
other angles, and pointed out that this comes from lab work, not from a
real, high-end scientific experiment, nobody would laugh. I have
produced *lots* of plots like this one during my own lab work - some of
them even worse, even in areas of science where no one would ever think
that something is wrong with the theory (for example, oscillations of a
pendulum).
> It looks a lot more like a linear dropoff. I can draw a neat
> straight line through just about all of their data points.
So what??? There is no theory which predicts such a straight line, so
the fact that you could draw such a line proves precisely nothing.
[snip]
> > And what are your "units" here? (I know that this is irrelevant, because
> > you only want to calculate percentages, but nevertheless I would like to
> > know this here).
>
> Units are in the width of the average distance between a proton and an
> electron. I have previously guessed this dimension at 187pm based on
> STM pictures of silicon.
Is this the distance between their centers of their surfaces? If it's
the distance between their centers, what radii do you assume for them?
[snip]
> > > The size of the atomic
> > > unit representing the area of the top of the atom's core is a 2 X 2
> > > square with an area of 4. This means there are 76/4 = 19 unique
> > > orientations can roughly fit into this quadrant with no overlap.
> >
> > Right - but why is the "with no overlap" relevant?
> >
> > And again, very confusing wording! I had to read this several times
> > until I understood what you try to say.
>
> Thanks for hanging in there and trying to understand it. I had to
> invent my own way of describing the possible orientations. I finally
> got a ball and put stickers on it to represent the possible
> orientations in a single quadrant. If you take a ball and place 5
> stickers from the pole to the equator and then put 4 more along the
> equator, and then 4 more heading up to the poles and then spread 7
> more stickers in the space between, you get an idea of the
> orientations
Yes.
> and why I don't have overlap.
No.
> I can't possiblely
> calculate every possible orientation, so I only did the 19
> orientations that can fit into a quadrant using a 2x2 target.
You said you used an Excel spread sheet, IIRC. What stops you from
calculating far more orientations using this?
> > > There
> > > are basically only 2 orientations which would result in high angle
> > > reflections. These are the head-on (alpha tries to pass through core)
> > > and edge-on (alpha tries to pass through arm edge).
> >
> > Why would these result in high angle deflection? What's there in the
> > core or in the arm edge which deflects the alpha particle? What force is
> > acting there?
>
> Simple elastic collision with a helium atom.
A deflection is an acceleration - surely you agree with that. According
to Newton, accelerations are caused by forces. So again - what *force*
is acting there?
> Two alpha particles would
> certainly bounce of each other
Why? (beside the Coulomb repulsion)
> and the arms of the cubic model are
> basically built out of helium atoms which are nothing more than alpha
> particles with their electrons. The edges present a thick layer (at
> least 5 units deep) of helium atoms which cannot easily be pushed
> aside.
Why? (again, beside the Coulomb repulsion)
> The cubic theory postulates that at angles of less than 90
> degrees, the helium atoms can be pushed aside since they have to
> penetrate 1 helium atom.
How does this work?
[snip]
> > You *do* know that classically, objects have a hard surface and
> > therefore bounce off each other when they collide because of
> > electrostatic forces, don't you?
>
> I have seen this theory, but disagree with it as I have stated before.
> Surfaces are hard because they are space filling as shown in the cubic
> model.
See above. Your ideas contradict Newton's laws.
> > > Because the atom is
> > > effectively a neutral matrix of joined helium atoms, the effect of the
> > > columb forces deflecting the alpha are negligible.
> >
> > Are you sure about that? Even when the alpha particle hits "head on"?
>
> Doesn't matter where it hits, the particle always sees a closely
> alternating series of protons and electrons in a balanced neutral
> matrix.
A "closely alternating series" may have a total charge of zero, but
obviously all the higher multipole moments are *not* zero. What about
their effects?
> > > The calculations show the percentage chance for:
> > >
> > > A complete miss or pass through 86.3% Would expect angle <
> > > 5 degrees
> > > An arm gets hit 13.1% Would expect any angle 0 -
> > > 180
> > > A direct hit of the core .21% Would expect angle 90 -
> > > 180
> > >
> > > This compares to the experimental data which shows:
> > >
> > > Deflections less than 5 degrees 79.2%
> > > Deflections 5 - 22 20.4%
> > > Deflections greater than 22 .35%
> >
> > Well, this *is* a "rough" agreement. ***VERY*** rough. It doesn't even
> > agree with the *original* results of Geiger and Marsden very well. You
> > have no clue how accurate scattering experiments nowadays are, have you?
>
> Can you find me some more references on the internet
Why do you insist on the internet? Lots of things you find there are not
very reliable (as you yourself proved by confusing the results of lab
work with a serious scientific experiment). Why don't you got reading
some books instead?
> besides the MIT one which disagreed with Rutherford?
That's a *strong* misrepresentation. It agreed with the theoretical
prediction in the range of angles where one expects it to agree.
> > IIRC, I recommended some books to you, among them one by Povh on nuclear
> > physics, which lists a *very* small amount of experimental results from
> > the plethora of experiments done in nuclear and particle physics (hint:
> > the book is nevertheless quite thick). Have you looked into that book in
> > the meantime?
> >
> I looked for the author Povh in the seattle public library and came up
> empty - did you spell this right?
Yes. For example, see here:
<http://www.amazon.com/exec/obidos/tg/detail/-/3540661158/qid=1077886225//ref=sr_8_xs_ap_i3_xgl14/102-1302641-0151367?v=glance&s=books&n=507846>
I don't think that you will find such specific science book in public
libraries - try going to a university library!
> > Please go to the more recent experiments, which have shown again and
> > again (hey, I even did this experiment *myself* at one time!!!) that the
> > scattering cross section depends on the angle like 1/sin^4(theta/2). As
> > soon as you are able to reproduce this formula (or at least some
> > numerical results which agree with it on a level of, say, 1%), I will
> > admit that your model has some merit. Good luck.
>
> Gee, what do you think of the MIT experiment then???
See above: very nice agreement within the range of angles where on
expects this.
> I noticed the rutherford formula could also be approximated by
> 1/theta^3 (roughly).
Huh??? Why on earth do you think so?
> For a fancy formula, its results are close to being just exponential.
Sorry, I don't understand what this is supposed to mean.
> > (oh, BTW, shouldn't your model give a dependence of the scattering cross
> > section on phi?)
>
> What is phi?
**********gaping mouth**********
**********astonishment**********
**********finally: BIG sigh************
And *you* want to tell *us* something about scattering?????
Try reading up on "spherical coordinates"!!!!!
This is sooooooooooo basic, it's simply unbelievable that someone who
doesn't know this has the *arrogance* to think that his model is better
than the established ones!!!
> > > The details of this calculation can be found on an excel spreadsheet:
> > >
> > > http://ourworld.compuserve.com/homepages/frankhu/ruther.xls
> >
> > I don't know if I should love or cry here.....
> >
> > You haven't got the *slightest* clue of how many experiments were done
> > in nuclear and particle physics in the past decades, how accurate they
> > are, how much work is spent in analysing the results. Sorry if this
> > insults you, put pretending that an *Excel spreadsheet* somehow is able
> > to reproduce these results based on your model is just pathetic, plain
> > and simple.
>
> If my calulations are wrong, then lets here it. Lots of physics is
> done with math as many have pointed out and is perfectly valid
> starting place.
*sigh*
You didn't understand at all what I said in the paragraph above. Try
reading it again.
> > > The predictions from the cube model and the actual experimental
> > > results are not exactly the same by any means,
> >
> > Nice that you see that.
> >
> >
> > > however, they are in the same rough ballpark.
> >
> > BFD. I think *every* model of the atom and the nucleus, no matter how
> > contrived, will manage to come up with some or even *lots* of results
> > which roughly agree with the experimental results. However, so far
> > *none* of this plethora of "alternative models" has been able to
> > 1) explain all the things which standard QM explains
> > 2) explain even *one* thing which QM explains with a comparable accuracy
> > as QM.
> >
> > You should try reading up on the "successes" of other "alternative
> > models". For example, you could look at Porat's model (right here in
> > sci.physics - the guy who either can't spell properly or doesn't bother
> > to do it), at the ideas of Black Light Power, or at Common Sense
> > science. All three have models which are far more developed than yours,
> > which strongly disagree with your, which include many strange ideas -
> > and nevertheless all three are able to reproduce a fair amount of
> > experimental evidence, sometimes with quite amazing accuracy. However,
> > all of them suffer the same faults - the two outlined above.
> >
> > You have still a long way to go. Good luck again.
> >
> Thanks for wishing me luck, I'll need it. Supporting a new theory is a
> David vs. Goliath task, but hey, David won didn't he?
Einstein won, too. Hint: maybe because his ideas could explain things
which weren't explainable before?
I haven't seen *your* model doing anything like this so far...
> I asked Porat
> for his paper, he never replied and his web site doesn't exist anymore
> - Hey Porat, you out there????
Strange. Usually he jumps at every opportunity to send someone his book.
Did you send him an e-mail?
> Thanks for the other references, I will have to take a look. Maybe we
> should consider these other theories as well if they predict with the
> same accuracy.
"Black light power" has an absolutely amazing collection of results
which agree with the experimental results. However, they are still
utterly wrong, and their model is still utter nonsense.
> But the key would be if you could present experments
> which contractict QM while supporting another theory like the ones I
> have described.
Sorry, I don't know any experiments that contradict QM. Hey, that's
precisely the reason why I don't see any reason to give your model any
big thoughts!
[snip]
> > > This scattering
> > > pattern does not necessarily have to be created by the atom postulated
> > > by Rutherford as a tiny compact nucleus containing all the positive
> > > charge.
> >
> > As long as you stay on this qualitative level, you are right. But as
> > soon as you become quantitative, i.e. take Rutherford's *formula* and
> > compare it to the experimental results, it's clear whose model is
> > better.
> >
> Not even Rutherford's formula can give exact numbers. You always have
> to see if you can match the curve predicted by the rutherford formula
> eventhough it appears to calculate exact numbers.
Huh? Sorry, I have no clue what this is supposed to mean.
> If I stick my model
> in a computer, I can get a finer breakout of the angles - my model
> results cannot be calculated as the result of a simplistic formula.
> Would that be quantitative enough for you?
Please try.
[snip]
> > Why do you insist on finding another explanation for the results,
> > although the existing explanation already works rather well?
> >
> > Hint: new theories in science are only invented when there is
> > experimental evidence that the old theories are wrong. Not simply
> > because someone doesn't like or doesn't understand the old theories.
>
> New theories happen all the time,
New so-called theories by crackpots happen all the time. New theories by
scientists happen only when there are problems with the older ones (and
there *are* problems in General Relativity as well as in Quantum Field
Theory - not experimental discrepancies, but theoretical problems). Oh,
drop the word "experimental" before "evidence" in my paragraph above,
please.
> the only time new theories get
> accepted is if you perform an experiment in which only one of the
> theories can survive.
Right.
Hence you should try to get a testable prediction out of your model
which disagrees with the predictions of standard QM. Can you do this?
> A theory is accepted if it can explain
> everything the old one did, plus lots more.
Right.
> A theory can also be
> accepted if it explains everything but is simpler.
I don't know any example for this.
> Lots of the books on quantum mechanics have titles like "The odd
> quantum" or the "Mysterious world of quantum mechanics", but loosely
> paraphrased, these titles all say, "The wacky unbelieveable world of
> quantum mechanics".
Yes. Humans base their world view on the macroscopic world - so it's no
big wonder that if we look at the microscopic world, we see lots of
things which seem counter-intuitive and weird to us.
> Because the theory is counter intuitive and
> difficult to understand or believe.
No quibble about that. However, that's *not* an argument for "it is
probably wrong".
> We believe in all the weird
> quantum mechnical stuff primarily because it matches experimental
> results.
Yes - that's science. (although I would change the "believe" into
"accept" or something like that - QM is not faith)
> That's fine, but it doesn't make quantum mechanics any less
> weird.
No quibble about that.
> The reason why I want to find a new theory is to remove the
> weirdness
That's not a scientific reason.
> and replace it with a model in which all of our classical
> mechanics still apply and all of the ideas are intuitive and easily
> understood - like the cubic model.
Good luck.
You could start by explaining the results of the double slit experiment
(with light and with electrons) and the photo effect. These were the
primary results which led to QM. Additionally, you could look at the
Stern-Gerlach experiment.
As already mentioned: Styer's book on QM and Feynman's lectures, volume
3, are *really* recommend stuff to read for you.
> I would like to see quantum
> mechanic books replaced by one which was simply titled, "The elegant
> structure of the atom".
Well, *I* think that QM *is* elegant (do you think that "weird" and
"elegant" contradict each other?). Do you know Dirac's formulation of
QM, with all of the bra-ket stuff?
[snip]
> > > I would like to see the Rutherford scattering experiments
> > > repeated but instead of using gold foil, use a form of crystallized
> > > gold (octahedral crystal) where we are reasonably sure that the gold
> > > atoms are all aligned in the same direction, and see how the high
> > > angle scattering depends on the orientation of the crystal.
> >
> > The experiment has been done with all kinds of stuff, AFAIK - not only
> > with gold foils, but also with other elements, and as well with isolated
> > nuclei and even isolated protons and neutrons (well, obviously the
> > latter experiments shouldn't be called Rutherford scattering anymore).
> > Additionally, instead of alpha particles, also electrons have been used,
> > and protons, and positrons, and neutrinos. In *all* of these experiments
> > (I estimate there have been several millions of those - hey, AFAIK there
> > were already several millions of created Z bosons registered at LEP,
> > which was *also* a scattering experiment!), the predictions of the
> > standard theories were proven - often with an amazing accuracy.
>
> You sure about that,
Yes.
> or are there lots more papers like the MIT one
> which declares it confirms Rutherford,
Well, it did.
> but really contradicts it
Absolute utter nonsense. It agreed with the theoretical prediction in
the range of angles where one expects agreement.
And again, this was the result of the *lab work* of a student - it was
not a real scientific experiment!
> or simply ignore the mismatched data as experimental error?
I ignore nothing. *You* ignore the perfectly sensible explanations for
the deviations.
> > So why on earth do (the one who knows nothing about even one of all of
> > these experiments) think that there is a problem with the standard
> > theories, and we need a new one?
> >
> See the above comments on the "odd quantum".
"I don't understand and don't like QM" is not an argument for "we need a
new theory".
[snip]
> > > Another possible experiment would be to use low speed alpha particles.
> > > At some point, if the speed of the alpha particles were slow enough,
> > > it wouldn't be able to penetrate the atoms thin arms and you would see
> > > almost all of the alphas being deflected at high angles. Rutherford
> > > would predict that all of the alphas would penetrate no matter how
> > > slow the alphas were going since there isn't much for the aphas to run
> > > into and an electron isn't likely to deflect an alpha very much.
> >
> > If the alpha particles are slow enough, the electrons *will* be able to
> > deflect them.
>
> They'd have to be pretty slow since a proton/electron collison is like
> a speck of sand hitting a bowling ball.
Not if you look at the electrostatic forces.
> The bowling ball's trajectory
> isn't likely to be affected.
False analogy.
> I suppose you would have to calcualte
> beforehand at what this point would occur and then test for it
> experimentally to see if there are higher speeds than this which still
> cause full reflectivity of the alphas.
Good luck.
> > And, BTW, Rutherford scattering *has* be done at a *lot* of different
> > energies (speeds) now. So far, every experiment agreed with his formula.
>
> As near as I can tell, it has always been with faster particles. Any
> refrences for slower particles?
Depends on what exactly you mean by "faster" and "slower".
[snip]
> > > The cubic atomic theory provides this solid model which you can
> > > run calculations on to show that it can return results similar to the
> > > experimental results of the Rutherford scattering experiment.
> >
> > Quit the handwaving. Rederive Rutherford's formula from your model.
> >
> > Or, alternatively, explain the spectrum of the hydrogen atom.
> > Quantitatively.
>
> That would be my next stop, to show that the spectrum is compatible
> with the cubic model and that the electron shell structure can be
> derived from the model.
Good luck.
I expect to see more handwaving and claims that rough agreements are
evidence for your model, and more ignoring the vast literature out there
which shows that the predictions of standard physics agree with the
experimental results *far* better.
> -Thanks for all of the thoughful replies.
You are welcome.
It's nice to discuss with someone who is at least a *bit* willing to try
to learn something new - instead of the usual crackpot behaviour
(usually they simply say "this stuff is nonsense, I don't want to bother
with it" and refuse to learn anything about the actual evidence).
> I do appreciate the time you
> took to understand this fairly lengthy and complex post. I do
> apologize if I seem arrogant,
Yes. For someone who is only aware of about 0.1% of the available
evidence (probably the figure is far lower), you seem to be fairly
arrogant in proposing an alternative model.
> I really am trying to figure out if the
> cubic model will work and would be willing to accept if this theory
> doesn't work if shown the evidence.
Well, what would disprove your model?
> I barely have time to work on this
> stuff since I have a full time job, a long commute and little kids to
> take care of at home. My hope is to show enough promise in the cubic
> theory, that others will take an interest and perform or find the
> results of some of the experiments I have proposed.
I think you will have no problem to get interest in people who know
little about physics. However, I am willing to bet 100$ dollar that you
won't find any physicist (who is knowledgable about the evidence) who
agrees with you, sorry.
> If the theory survives the skeptics on the internet,
That's a strange argument. Who judges if it survives? You, right?
> and if I can write it up in a
> sensible scientific manner, I can publish in a peer reviewed journal
> and let real science take its course.
Sorry, but I don't think you will ever be able to "polish up" your stuff
so far that any peer reviewed paper will accept it. You seem to have no
clue of the standards which are applied there. (and no, I don't talk
about "censorship of alternative ideas" or something like that)
> Either that, or you and Franz
> shoot it down in a flaming wreck and that will be the end of that. But
> so far, I haven't seen a fatal shot.
Well, in our opinion, we have provided *several* fatal shots. See what I
meant above about your argument "if the theory survives the skeptics on
the internet"?
> By the way, is there anybody else out there reading these posts on the
> cubic model and agree with it? Or is it just Franz and Bjoern out
> there. I could use a little support.
Well, *perhaps* you will get agreement from Porat - but I don't think
you will want to have his agreement after you have seen some of his more
weird claims. And perhaps Thomas Lockyer could also take your side - I
haven't seen him here for a while...
Bye,
Bjoern
Franz Heymann
news:<403C816D...@ix.urz.uni-heidelberg.de>...
> >
> > No. That's not the "main" reason, that was only the first piece of
> > evidence. Nowadays we have *far* more experimental data on scattering of
> > nuclei and individual protons and neutrons (not Rutherford scattering -
> > the experiments have long ago gone far beyond that simple experiment of
> > Rutherford). Every scattering argument shows that these things *are*
> > small.
>
> References?
How about this one for one of the early review papers:
Weber, Proc 1067 Int Sym on Electron and Photon interactions at High
Energies, Stanford.
That contains a discussion of the size determinations of both the proton and
the neutron.
If you do not already know the answer to that, then a short soundbite is not
going to give it to you. Open a book on QM and read it from cover to cover,
and you will have the answer (and a lot more besides).
> By comparison, is it more believeable that there is a strong nuclear
> force which causes 2 protons in the nucleus to experience 1000 pounds
> of binding force?
See:http://www.geocities.com/quantum_00_2000/4forces/strongforce.html
> The description of the gluons responsible for this force also has
> properties which are simply postulated. See
> http://www.factmonster.com/ce6/sci/A0821053.html and the evidence is
> only indirect.
>
> I think it is far more believeable that a proton and an electron
> maintain a fixed position from each other.
You are welcome to harbour such a thought. But you have not one shred of
evidence for it, and you are never going to get any such evidence. You seem
to be suffering from a time slippage, causing you to be wandering around in
the corridors of physics as they were 90 years ago.
> [snip]
> I
> > > have done calculations to determine how frequently you would expect to
> > > see these reflective collisions and I have compared them with the
> > > original Rutherford scattering experiment results and I can show that
> > > the predicted percentage for the alpha particles at particular angles
> > > for the cubic model roughly match the experimental results.
> >
> > "roughly". Interesting.
> So you would agree that it "roughly" matches and the the calculations
> don't contain some fatal flaw? I did have to redo the calculations
> serveral times to get it accurate. I would be surprised if there
> weren't more mistakes.
>
> >
> >
> > BTW, shouldn't the result of the scattering depend on the projectiles,
> > too, according to your model? Nevertheless, experiments show that one
> > gets the same scattering cross section for alpha particles and electrons
> > (the only deviations are at small angles, where the non-zero size of the
> > nucleus becomes important).
> >
> No it shouldn't. The cubic model is a neutral matrix, so it doesn't
> matter what the charge of the object is that you throw at it.
By now you should surely have learnt that a system of particles subject to
electrostatic forces only cannot be at rest in a stable configuration? Your
cubic model is therefore dead on arrival.
> Electrons, protons, alphas, whatever. All that matters is that the
> scattering cross section is going to be the same and you should get
> the same percentage chance of scattering no matter what you hit it
> with.
No. That is actually not in agreement with observation. Not by a long
chalk. There is a Z^2 factor involved. And there is also a factor
dependent on the energy of the projectile.
> > Additionally, does your model reproduce the observed dependence of the
> > scattering cross section on the charge of the nucleus, Z? Does it
> > explain why the scattering cross section does *not* depend on the number
> > of neutrons, N, in the nucleus? Rutherford's formula can do all of this
> > without any problems.
> I can work on this calculation,
That is surely only by way of speaking. I will bet you a penny to a pound
that you can *not* work on that calculation and come up with a prediction
which will match existing data.
> but I can see that the atomic radius
> of silver is approximately the same as gold. However in the cubic
> atomic model, the silver atom is smaller and the cross section will be
> smaller. The net percentage of scattering at angles will be the same,
That last statement is garbage.
> but since there is less target area,
I guess you do not yet know that the total Rutherford scattering cross
section is infinite, and is in reality only limited by the shielding arising
from the electrons of the atom.
> you should see a reduction in the
> overall number of reflected alphas for the same experimental setup.
And that is quite precisely wrong.
> So, yes, the Z dependence can probably be modeled, I do need to do the
> math.
Your time would be better occupied by learning some modern physics instead.
I presume you did not realise that that URL points to a student's write-up
of an undergraduate experiment. I am quite surprised that he got results as
good as that in an undergrad lab.
Please do not omit to look up the reference I gave you above, which shows
the state of the art nearly forty years earlier. Note the good agreement
between theory and experiment everywhere except in those limiting cases
where the differential cross sections wese so small that the statistics were
somewhat wanting.
You will realise, I hope, that that undergrad experiment was done with alpha
particles with an energy of 4 MeV or so At such low energies, multiple
scattering does indeed affect the results, and electron screening is a
serious matter.
The actual professional experiments have been done with much more
sophisticated equipment, using electrons and protons as projectiles, at
energies anywhere between 5 MeV and 300 GeV in the CM. At the higher
energies, very, very much cleaner results are obtained.
[snip]
Franz
FrankH
> > associated with atoms which are perfectly symmetrical in the cubic
> > model.
>
> While you're at it, you might look into the explanation of both the singly
> and the doubly magic nucleii
Yes, I will look into the doubly magic nuclei. Would you be impressed
if I found a reason for the magic numbers? Does QM explain the magic
numbers?
[snip]
> > > standard theories of nuclei do.
> > I have added calculations of hydrogen, helium, and lithium to my
> > website.
>
> You have hardly started. The atomic physicists have completed the QM
> calculations for all the ground states as well as the important excited
> states of all the atoms of interest in studying stellar structure. That
> means every atom in the periodic table from H to Fe. The results for
> spectral frequencies and intensities vs temperature are in excellent
> agreement with observations, and are in daily use all over the world by folk
> interested in the structure of stars.
> I regret to inform you that if you were as fast as those physicists in doing
> the calculations, you will need the services of a supercomputer and somewhat
> more than 100 years of spare time on your part.
> I am saying this just to illustrate to you just one tiny aspect of the sheer
> size and magnificence of the edifice you are attacking with your puny
> weapons.
How simple and elegant can quantum mecahnics be if it takes an army of
physicists and supercomputers to do fundamental calculations? Out of
curiosity, why would stellar astronomy need the calculated values,
when the observed spectra values are easy to get by experiment?
I really don't understand this concept of a ground state QM
calculation. Do you just take the equations for hydrogen and assume
they work for carbon since it shouldn't exeed the shell levels
calculated for hydrogen?
[snip]
>
> Yes, actually you are, because all of your work up to now has been shown to
> be built on an erroneous foundation, in this and earlier threads.
Actually, I don't see where you showed this - lack of evidence due to
lack of research, yes, but erroneous foundation no. I am diligently
working on the areas pointed out to me. Please summarize these
erroneous foundation points.
> Yes. You are right. You are missing almost everything here. Has the
> following not occurred to you?
> The electron shell configurations shown there are all purely the
> configurations for the ground state. You need to get an electron into an
> excited state before it can radiate. Such a state is not part of the ground
> state configuration.
Yes, it has, that is why it doesn't make any sense to me. How can they
predict ground states based on excited states. (This is also the same
problem I have with QM ground state predictions). I thought this was
based on spectra analysis or is the ground state determined in some
other manner.
[snip]
> > > > For the cubic atomic model, these
> > > > excited electrons would be governed by the rules of spherical
> > > > harmonics.
> > >
> > > What rules are those?
>
> Did you see that question?
Yes, I said "spherical harmonics"
>
> > > And why should they apply?
>
> Did you see that question?
Yes, it should apply because you are doing an analysis of a particle
in a spherical field which is "spherical harmonics".
>
> > > And why shouldn't they apply in the ground state?
>
> Did you see that question?
Yes, because in my model, if an atom is in the ground state, all
particles are contained within the atom and there are no particles
that could be in involved in spherical harmonics.
> >
> > I'm saying QM may be right in this case.
>
> As indeed it is in all tests to which it has been subjected.
Are you sure you aren't being a bit boastful? From some of the
references that Bejorn gave me, it looks like there are plenty of
things that QM fails at.
See:http://www.blacklightpower.com/bookdownload.shtml
>
> > > > so I presume that they are
> > > > correct in predicting the expected spherical harmonics and spectral
> > > > lines.
> > >
> > > But why should standard QM be only right for the excited states, but not
> > > for the ground state?????
> >
> > For large atoms, the electrons are not transitioning states in the
> > ground state. Nor are they emmitting spectra.
>
> Could you help by telling us what "transitioning states in the ground
> state" means?
As I mentioned above, electrons are not in motion in the ground state.
They are bound and not in an excited state where they can transition
between shells and emmit light.
>
> For your information, QM has produced the right answers for everything
> connected with the ground states as well as the most important excited
> states of all the elements for which calculations have been made.
>
> > > > However, the interpretation of the equations as showing the
> > > > actual behavior of the electron being contained in a probability cloud
> > > > is misleading.
> > >
> > > Thanks for showing again that you don't know what you are talking about.
> > > Nobody says that the electron is contained in a probability cloud. And,
> > > BTW, you were corrected on this before, so repeating this nonsense is
> > > again close to a lie.
Did you and Bjorn ever figure out what an electron is doing around the
nucleus? Radial movement or not? Everywhere at once or not? Seems to
me that even you two can't agree with what is going on. I continue to
suspect that the real answer is "don't know" and "don't care" or at
the very least "it calculates corectly, therefore it must be". This is
a big hole in QM.
> > >
> > > IIRC, I recommended Styer's book "The strange world of Quantum
> > > Mechanics" to you. Have you read it in the meantime?
> > >
Hmmm, can't seem to find it at my local library .... Got a pile of
other books though.
-Franklin
Franz Heymann
> Franz Heymann wrote:
> >
> > "FrankH" <frank...@yahoo.com> wrote in message
> > news:46484c9f.04022...@posting.google.com...
>
> [snip
>
>
> > One now knows that the proton charge
> > distribution, for example, has a RMS radius of 0.80 fm. I forget the
error
> > bar size,
>
> 0.870 +- 0.008 fm, according to the PDG.
> <http://pdg.lbl.gov/2003/s016.pdf>
>
>
> > but the measurements are actually good enough to provide a figure
> > for the skin depth of the proton. (Crudely, the radial thickness over
which
> > the charge density dies down from its central value to zero)
>
> Looks strange to me. I know that nuclei have a "skin" (the roughly have
> a charge distribution which looks like a Fermi function, so it makes
> sense to talk about a "skin"), but for the proton, IIRC, the charge
> distribution is exponentially decreasing. What's supposed to be the
> "skin" in that case?
I have done a Google, and cannot find this skin depth of which I spoke. I
think I may have confused the skin depth of a nucleus with that of the
proton itself, as you suggested.
I will go on trying to get a handle on it.
I have to say that at first sight I am unhappy with your exponentiial drop
with radius, because, after all, there are 3 quarks swarming around in
there.
Is there a QCD prediction?
[snip]
Franz
Franz Heymann
> Franz Heymann wrote:
>
> [snip lots]
>
>
> > The international "Atomic Wavefunctions Group" was led by
> > a physicist in my Department. It involved a few dozen physicists from
> > various countries including France, Germany, the UK and the USA. They
set
> > themselves the task of calculating all the important wavefunctions of
all
> > the atoms of stellar interest. They succeeded magnificently, but it
took
> > more than a decade to get all the results. Those have now been in use
for
> > the best part of twenty years.
>
> Franz, can you give me a link to this group? (they have published their
> results on the web somewhere, I hope?)
I doubt it, since the work was essentially completed about twenty years ago.
A Google search with the term
> "Atomic Wavefunctions Group" unfortunately yielded no hit.
The three leading lights were M.J. Seaton, now retired, then UCLondon, P.
Burke, originally UCLondon, later QU Belfast, probably also now retired and
D. Hummer, probably also now retired, then Boulder, Colorado.
I don't know how widely their results are available. What I do know is that
they are available on-line to all the UK participants in the Starlink
network. I am certai that similar organisations on the Continent and in the
USA will have similar forms of access in their countries. There will have
been many dozens of papers in the normal press under the names of any or all
of them, with and without collaborators, on the theoretical considerations
behind the computations. If you want to try and locate some of the work by
looking in Physics Abstracts, the really active period was roughly 1965 to
1980
>
>
> [snip]
>
>
> > > > > Data like the ionization energies for each of
> > > > > the electrons gives us a better idea of the internal structure
since
> > > > > it shows how tightly each electron is bound in the atom.
> > > >
> > > > Hint: standard QM predicts these ionization energies with good
accuracy.
> > > >
> > > References?
> >
> > Any of the hundreds of papers produced by the Atomic Wavefunctions
Group.
> > M.J. Seaton, or D. Hummer, or W.Eisner are good names to look out for.
> > These are only the first 3 that came to mind out of a few dozen. Their
> > wavefunctions are all good to better than 1:10,000
>
> Well, a Google search for "Seaton Hummer Eisner" brought 4 hits, but
> none of them was helpful... :-(
>
> Were their results only published in papers? Aren't they available
> somewhere free on the Web?
As I explained, their work essentially predates the Web.
I'll contact Seaton to try and get some info from him.
Franz
Big Bird
> I'm not sure I agree with the experiments showing wave behavior
> for particles.
Whenever you disagree with expriment ...
... you're wrong.
I'm sorry, but I don't think there's a more polite way to state this.
Franz Heymann
You realise, of course, that solving for the wavefunctions of, say, Na, just
as an example, is equivalent, in classical terms, to solving the 11 body
problem. I suppose you are aware that even the 3 body problem in classical
mechanics is a pretty hard nut to crack.
> Out of
> curiosity, why would stellar astronomy need the calculated values,
> when the observed spectra values are easy to get by experiment?
No. The relative intensities of single spectral lines versus temperature
are in fact not measurable experimentally in the temperature ranges needed
for astronomical work.
> I really don't understand this concept of a ground state QM
> calculation. Do you just take the equations for hydrogen and assume
> they work for carbon since it shouldn't exeed the shell levels
> calculated for hydrogen?
No. That would be crass.
They calculate the complete wavefunctions for each required state of each
atom right from first principles
> [snip]
> > Yes, actually you are, because all of your work up to now has been shown
to
> > be built on an erroneous foundation, in this and earlier threads.
>
> Actually, I don't see where you showed this - lack of evidence due to
> lack of research, yes, but erroneous foundation no. I am diligently
> working on the areas pointed out to me. Please summarize these
> erroneous foundation points.
Only two are needed for your edifice to come tumbling down:
(1) Look back at what a number of us have told you umpteen times about the
stability of a system of charges under the influence of electrostatic forces
only.
(2) Classical mechanics are not applicable to the motion of particles in the
quantum domain.
> > Yes. You are right. You are missing almost everything here. Has the
> > following not occurred to you?
> > The electron shell configurations shown there are all purely the
> > configurations for the ground state. You need to get an electron into
an
> > excited state before it can radiate. Such a state is not part of the
ground
> > state configuration.
> Yes, it has,
Nonsense. An excited state is not a ground state. Full stop.
> that is why it doesn't make any sense to me.
The reason why what we have to say to you doesn't make sense to you is that
you have *no* knowledge of quantum mechanics and of atomic structure.
> How can they
> predict ground states based on excited states.
They don't.
> (This is also the same
> problem I have with QM ground state predictions). I thought this was
> based on spectra analysis or is the ground state determined in some
> other manner.
You have the cart before the horse. You calculate the details of the
relevant states, and use that to predict the spectra.
>
> [snip]
> > > > > For the cubic atomic model, these
> > > > > excited electrons would be governed by the rules of spherical
> > > > > harmonics.
> > > >
> > > > What rules are those?
> >
> > Did you see that question?
> Yes, I said "spherical harmonics"
> >
> > > > And why should they apply?
> >
> > Did you see that question?
> Yes, it should apply because you are doing an analysis of a particle
> in a spherical field which is "spherical harmonics".
> >
> > > > And why shouldn't they apply in the ground state?
> >
> > Did you see that question?
> Yes, because in my model, if an atom is in the ground state, all
> particles are contained within the atom and there are no particles
> that could be in involved in spherical harmonics.
Your model is crap.
> > > I'm saying QM may be right in this case.
> >
> > As indeed it is in all tests to which it has been subjected.
> Are you sure you aren't being a bit boastful? From some of the
> references that Bejorn gave me, it looks like there are plenty of
> things that QM fails at.
> See:http://www.blacklightpower.com/bookdownload.shtml
>
> >
> > > > > so I presume that they are
> > > > > correct in predicting the expected spherical harmonics and
spectral
> > > > > lines.
> > > >
> > > > But why should standard QM be only right for the excited states, but
not
> > > > for the ground state?????
> > >
> > > For large atoms, the electrons are not transitioning states in the
> > > ground state. Nor are they emmitting spectra.
> >
> > Could you help by telling us what "transitioning states in the ground
> > state" means?
> As I mentioned above, electrons are not in motion in the ground state.
> They are bound and not in an excited state where they can transition
> between shells and emmit light.
If they are not in m otion in the ground state, then they are in unstable
equilibrium in your model.
> > For your information, QM has produced the right answers for everything
> > connected with the ground states as well as the most important excited
> > states of all the elements for which calculations have been made.
> >
> > > > > However, the interpretation of the equations as showing the
> > > > > actual behavior of the electron being contained in a probability
cloud
> > > > > is misleading.
> > > >
> > > > Thanks for showing again that you don't know what you are talking
about.
> > > > Nobody says that the electron is contained in a probability cloud.
And,
> > > > BTW, you were corrected on this before, so repeating this nonsense
is
> > > > again close to a lie.
>
> Did you and Bjorn ever figure out what an electron is doing around the
> nucleus? Radial movement or not? Everywhere at once or not? Seems to
> me that even you two can't agree with what is going on.
We are in full agreement about the status of the electron in te ground state
of Hydrogen. If you failed to see that in our correspondence, your powers of
cognition are even worse than I thought.
> I continue to
> suspect that the real answer is "don't know" and "don't care" or at
> the very least "it calculates corectly, therefore it must be".
You are wrong in all those suspicions.
> This is
> a big hole in QM.
There are no holes of any description in QM.
Franz
Bjoern Feuerbacher
Thanks for all the valuable information! I'll see what I can find...
Bye,
Bjoern
Bjoern Feuerbacher
I don't know if there are any actual calculations on this (I haven't
seen one in the books on QFT I read), but I think an exponential dropoff
looks sensible. The three valence quarks are supposed to be all in an
s-state, so one should expect an exponential decrease of the charge
density.
Bye,
Bjoern
Bjoern Feuerbacher
>
> "Franz Heymann" <notfranz...@btopenworld.com> wrote in message news:<c1lqve$ndj$2...@titan.btinternet.com>...
> > "FrankH" <frank...@yahoo.com> wrote in message
> > > >
> > > I am working on this, so far I have noticed the magic numbers are
> > > associated with atoms which are perfectly symmetrical in the cubic
> > > model.
>
> >
> > While you're at it, you might look into the explanation of both the singly
> > and the doubly magic nucleii
>
> Yes, I will look into the doubly magic nuclei. Would you be impressed
> if I found a reason for the magic numbers? Does QM explain the magic
> numbers?
As I already said several times: YES.
Don't you listen?
[snip]
> How simple and elegant can quantum mecahnics be if it takes an army of
> physicists and supercomputers to do fundamental calculations?
Don't confuse a theory with mathematical elegancy with a theory where
one can easily do "fundamental calculations".
And, BTW, there *are* some fundamental calculations which can be done
quite easily - the spectrum of the H atom and the Rutherford scattering
cross section.
> Out of
> curiosity, why would stellar astronomy need the calculated values,
> when the observed spectra values are easy to get by experiment?
I don't know exactly, but one of the reasons which comes to mind at once
is that there are environments in space (high vacuum in some places,
high density and temperature) which can't be easily simulated in
experiments in the laboratory.
> I really don't understand this concept of a ground state QM
> calculation.
What don't you understand about it?
> Do you just take the equations for hydrogen and assume
> they work for carbon since it shouldn't exeed the shell levels
> calculated for hydrogen?
They take the Schroedinger equation with terms for all the
electron-nucleus and electron-electron interactions and solve it
numerically.
It has close to nothing to do with the shells of hydrogen. The only
thing it has in common is that the same type of Schroedinger equation
(but with a lot more terms!) is used.
> [snip]
>
> >
> > Yes, actually you are, because all of your work up to now has been shown
> > to be built on an erroneous foundation, in this and earlier threads.
>
> Actually, I don't see where you showed this - lack of evidence due to
> lack of research, yes, but erroneous foundation no. I am diligently
> working on the areas pointed out to me. Please summarize these
> erroneous foundation points.
For example, you keep assuming that one can treat elementary particles
as small objects with hard surfaces. This contradicts not only
scattering experiments which show that these "particles" have wave-like
behaviours, but also begs the question which force provides the hard
surfaces.
> > Yes. You are right. You are missing almost everything here. Has the
> > following not occurred to you?
> > The electron shell configurations shown there are all purely the
> > configurations for the ground state. You need to get an electron into an
> > excited state before it can radiate. Such a state is not part of the
> > ground state configuration.
>
> Yes, it has, that is why it doesn't make any sense to me.
Huh?
> How can they
> predict ground states based on excited states.
Huh? No one does do anything like this. What on earth are you talking
about?
> (This is also the same
> problem I have with QM ground state predictions).
Sorry, I don't understand the problem.
> I thought this was
> based on spectra analysis or is the ground state determined in some
> other manner.
Do you mean experiments or theory here?
> [snip]
> > > > > For the cubic atomic model, these
> > > > > excited electrons would be governed by the rules of spherical
> > > > > harmonics.
> > > >
> > > > What rules are those?
> >
> > Did you see that question?
>
> Yes, I said "spherical harmonics"
And *I* asked essentially what you mean by "governed by the rules of
spherical harmonics". Care to answer that question?
> > > > And why should they apply?
> >
> > Did you see that question?
>
> Yes, it should apply because you are doing an analysis of a particle
> in a spherical field which is "spherical harmonics".
Very strange wording. Are you sure you know what you are talking about?
> > > > And why shouldn't they apply in the ground state?
> >
> > Did you see that question?
>
> Yes, because in my model, if an atom is in the ground state, all
> particles are contained within the atom
In excited states, all particles are contained within the atom, too - so
what's the difference?
> and there are no particles
> that could be in involved in spherical harmonics.
What on earth is "involved in spherical harmonics" supposed to mean?
>
> > > I'm saying QM may be right in this case.
> >
> > As indeed it is in all tests to which it has been subjected.
>
> Are you sure you aren't being a bit boastful?
No.
> From some of the
> references that Bejorn gave me,
I gave you those in order to illustrate that there are other people like
you out there who also have developed alternative models.
> it looks like there are plenty of
> things that QM fails at.
> See:http://www.blacklightpower.com/bookdownload.shtml
So you rely on a crackpot instead of what scientists say?
And I've looked at the book - but I don't remember that it tells us
where QM failed (experimentally). It is only another bunch of whining
that QM looks so nonsensical.
> > > > > so I presume that they are
> > > > > correct in predicting the expected spherical harmonics and spectral
> > > > > lines.
> > > >
> > > > But why should standard QM be only right for the excited states, but
> > > > not for the ground state?????
> > >
> > > For large atoms, the electrons are not transitioning states in the
> > > ground state. Nor are they emmitting spectra.
> >
> > Could you help by telling us what "transitioning states in the ground
> > state" means?
>
> As I mentioned above, electrons are not in motion in the ground state.
> They are bound and not in an excited state where they can transition
> between shells and emmit light.
That's still an unsupported claim, and this helps nothing for explaining
why QM shouldn't apply for the ground state.
[snip]
> > > > Thanks for showing again that you don't know what you are talking
> > > > about.
> > > > Nobody says that the electron is contained in a probability cloud.
> > > > And,
> > > > BTW, you were corrected on this before, so repeating this nonsense is
> > > > again close to a lie.
>
> Did you and Bjorn ever figure out what an electron is doing around the
> nucleus? Radial movement or not?
I think we agreed on "no radial movement", or, better "state of movement
isn't clearly defined".
I recommended Styer's book to you on picturing an electron. When will
you finally look at it?
> Everywhere at once or not?
In a sense, it *is* everywhere at once. Similar to the double-slit
experiment.
> Seems to
> me that even you two can't agree with what is going on.
We have agreed in the meantime.
> I continue to
> suspect that the real answer is "don't know" and "don't care" or at
> the very least "it calculates corectly, therefore it must be". This is
> a big hole in QM.
You think that "we can't understand it with common sense, but we can
calculate the results of experiments with great accuracy" is a "big
hole" in a physical theory? Why?
> > > > IIRC, I recommended Styer's book "The strange world of Quantum
> > > > Mechanics" to you. Have you read it in the meantime?
> > > >
> Hmmm, can't seem to find it at my local library .... Got a pile of
> other books though.
You could order it at Amazon.
Bye,
Bjoern
FrankH
[snip]
> > Although I still like the intuitive idea that even at the subatomic
> > level, a particle occupies a fixed amount of space (like a little
> > marble)
>
> This contradicts basic QM. How would you explain the results of the
> double slit experiments, based on this idea?
>
distribution for particles coming from a double slit setup. Everyone
seems to think it should act like two spray cans and the resulting
distribution should be smooth. I think it would be like two slits
spitting out bb pellets and the pellets collide and head off at
particular angles. I wanted to see if this analysis shows any tendency
for it to show diffraction like properties. This would show that
particles can have wave like properties while remaining particles the
entire time. But I haven't seen any such analysis, although such a
thing would explain the double slit experiment.
[snip]
> The only way to avoid the conclusion that there is a repulsive force is
> to claim that Newton's law don't apply in these situation. Do you
> *really* want to claim this?
>
The concept of a hard surface seems clear enough to me. The repulsive
force is an inherent property of a particle that cannot be compressed.
Since it cannot be compressed, it must therefore be able to generate
an opposing force to prevent compression.
[snip]
[snip]
>
> Please do. According to Rutherford's formula, the scattering cross
> section is proportional to the charge of the projectile squared (that's
> where the big problem for your model is, which according to you above
> says that the charge of the projectile doesn't matter), and proportional
> to the charge of the target squared.
>
> I'm interested in how you will manage to get this behaviour from your
> model. Especially I will be interested to see how you get from your
> model that the scattering cross section for an alpha particle is four
> times greater than the one for an electron (provided both have the same
> energy).
According to the cubic model, an alpha is a cube which presents a
surface area of 2 x 2 = 4 units. An electron, presents a surface area
of 1 unit. So an alpha has a 4 times better chance of hitting
anything. So size does matter and this would also help confirm that
the size of an electron is about the same size as the proton/electrons
in the alpha.
I was suprised to see that the scattering pattern for electrons and
alphas are apparently identical. According to the hyperphysics
formulas, the electron would create a negative impact parameter:
http://hyperphysics.phy-astr.gsu.edu/hbase/nuclear/rutsca2.html#c5
That doesn't seem to make a lot of sense and the electron would
definitely take a different path than an alpha. But the term is
squared, so the negative value apparently doesn't make any difference.
What does the scattering data for neutrons tell us? If the nucleus is
small we should probably see most alphas go through with very little
scattering at other angles except for the rare direct hit of the
nucleus. Is this the case? The cubic model would likely predict no
difference is relative scattering distribution.
>
> Additionally, I would like to see you to reproduce the behaviour of
> Rutherford's scattering cross section wrt energy (it is proportional to
> 1/E^2).
OK, you got me there. On first glance the cubic model should not show
any dependence on E since the scattering cross section remains
constant for any E at any particular orientation. I will have to
investigate more since this would be a big problem. Does Povh show
experimental data showing the dependence on E? I may have to take your
advice and get it from Amazon since I have no access to otherwise. The
only other things I can think of here is that at close range, lower
energy alphas would be deflected more strongly due to near misses as
it is in the Rutherford model, but I would think the effect would be
relatively small since the alternating proton/electron sequence would
tend to null out the electrostatic field.
>
> Oh, BTW, what possible experiment could disprove your model?
>
The experiments I had mentioned and other experimental results showing
incorrect dependencies as you have pointed out with 1/e^2. The theory
is very testable.
> This paper doesn't look like a scientific article to me, but rather as
> if a student reported the results of his lab work. I have written
> similar stuff during the lab work required in my study, too.
>
> Going to
> <http://web.mit.edu/tkfocht/www/>,
> my suspicion is confirmed. The link to this paper is found under the
> heading "Junior Lab Papers and Presentations".
>
Nice bit of detective work - I did miss that.
> So instead of going to read some of the things I directed you to (like
> Povh) you did a web search and found a report on a lab work experiment.
> And you seem to think this is can be used as if it is a valid scientific
> paper. Say, are you *really* unwilling to learn *anything*?
>
I'm willing to learn, but have limited resources and time.
[snip]
>
> > Is this the remarkable accuracy predicted by the Rutherford formula?
>
> Yes. Hint: this experiment has been done countless times - and the
> agreement with the formula is *always* great (in the range of angles
> where it makes sense - because of the reasons outlined above, it doesn't
> work well for very small and very large angles).
>
> Additionally, the "improved" formulas, which take things like magnetic
> moments of target and projectile into account (see above), have *also*
> been tested countless times with amazing accuracy.
>
Seems to me that matching on only 11% of the range (10-30 degrees)of
the Rutherford formula is not a great deal of accuracy. The MIT paper
may have been a simple lab result, but is it an incorrect or odd
result, or is it typical of the high-end experimental results? If I go
look at Povh, will I see these same results and conclusions?
> > > (oh, BTW, shouldn't your model give a dependence of the scattering cross
> > > section on phi?)
> >
> > What is phi?
>
> **********gaping mouth**********
>
> **********astonishment**********
>
> **********finally: BIG sigh************
>
> And *you* want to tell *us* something about scattering?????
>
> Try reading up on "spherical coordinates"!!!!!
>
> This is sooooooooooo basic, it's simply unbelievable that someone who
> doesn't know this has the *arrogance* to think that his model is better
> than the established ones!!!
I don't know that it is better - I just came up with a new model and I
want to see how it compares with experimental data and current theory.
All I can say is that my arrogance is caused more by ignorance than by
actual arrogance:)
>
Looking at spherical coordinates, phi is the lattitude. There is a
dependency of the cross section on the atom orientation (all 3
orientations rho, phi and theta). I only calculated 2 cross sections
for the angles producing the highest high angle scattering events. I'd
need a computer program to calculate the rest of the cross sections.
However, these are not expected to contribute much to the result since
the alphas go through the atom at these orientations.
-Franklin
Bjoern Feuerbacher
I notice that you ignored (and snipped) my question why you don't go and
read some books - e.g. Povh.
FrankH wrote:
>
> Bjoern Feuerbacher <bfeu...@ix.urz.uni-heidelberg.de> wrote in message news:<403F4343...@ix.urz.uni-heidelberg.de>...
> [snip]
> > > Although I still like the intuitive idea that even at the subatomic
> > > level, a particle occupies a fixed amount of space (like a little
> > > marble)
> >
> > This contradicts basic QM. How would you explain the results of the
> > double slit experiments, based on this idea?
> >
> I've been looking for an analysis of the expected particle
> distribution for particles coming from a double slit setup.
The Feynman lectures, volume 3, discuss the double slit experiment in
detail. Try reading it.
Doing the necessary computations is a fairly easy exercise - IIRC, I did
them first in the second or third semester of my study.
> Everyone
> seems to think it should act like two spray cans and the resulting
> distribution should be smooth.
Huh? What do you mean by "spray can", and what do you mean by "smooth"
here?
> I think it would be like two slits
> spitting out bb pellets and the pellets collide and head off at
> particular angles.
Err, one gets the *same* interference pattern if the "pellets"
(electrons or photons) are sent one by one - the next is only sent when
the previous one has been detected. So your "pellets collide"
explanation holds no water. Beside, I wonder how you would get the
observed interference pattern from these collisions - minima and maxima
at well-defined positions.
> I wanted to see if this analysis shows any tendency
> for it to show diffraction like properties.
I don't think this would be possible. For starters, you would have to
get the right dependencies on the distance between the two slits and the
momenta of the "pellets".
> This would show that
> particles can have wave like properties while remaining particles the
> entire time. But I haven't seen any such analysis, although such a
> thing would explain the double slit experiment.
Well, since the experiment works also with single particles, without any
possibility of collisions (see above), such an analysis seems to be
rather unnecessary, don't you think?
> [snip]
>
> > The only way to avoid the conclusion that there is a repulsive force is
> > to claim that Newton's law don't apply in these situation. Do you
> > *really* want to claim this?
> >
> The concept of a hard surface seems clear enough to me.
It's a macroscopic concept, based on your everyday life, on common
sense. Microscopically, it doesn't make any sense.
> The repulsive
> force is an inherent property of a particle that cannot be compressed.
What's the reason why it can't be compressed?
What's the source of this force?
Why do we get agreement with observations if we calculate, e.g., binding
lenghts in crystals based on the assumption that there is an electric
repulsion between the atoms? (source: any textbook on solid state
physics)
> Since it cannot be compressed, it must therefore be able to generate
> an opposing force to prevent compression.
Where does this force come from?
What rules does it obey?
> > Please do. According to Rutherford's formula, the scattering cross
> > section is proportional to the charge of the projectile squared (that's
> > where the big problem for your model is, which according to you above
> > says that the charge of the projectile doesn't matter), and proportional
> > to the charge of the target squared.
> >
> > I'm interested in how you will manage to get this behaviour from your
> > model. Especially I will be interested to see how you get from your
> > model that the scattering cross section for an alpha particle is four
> > times greater than the one for an electron (provided both have the same
> > energy).
>
> According to the cubic model, an alpha is a cube which presents a
> surface area of 2 x 2 = 4 units.
Where do you get this cube from? An alpha particle contains only four
particles - for a cube with a side length of two units, you need eight
particls.
> An electron, presents a surface area of 1 unit.
The electron has a diameter half of the diameter of a proton? Say, how
often do you plan to contradict experimental observations? (here e.g.
electron-positron scattering)
> So an alpha has a 4 times better chance of hitting
> anything. So size does matter and this would also help confirm that
> the size of an electron is about the same size as the proton/electrons
> in the alpha.
The observation that electrons are scattering only one forth as much as
alpha particles is already explained by Rutherford's formula. Why do you
feel the need to invent a new explanation?
> I was suprised to see that the scattering pattern for electrons and
> alphas are apparently identical.
What do you mean by "pattern"? The general shape of the curve?
And what's so surprising about this? The scattering cross section
depends only on the charge *squared*.
> According to the hyperphysics
> formulas, the electron would create a negative impact parameter:
> http://hyperphysics.phy-astr.gsu.edu/hbase/nuclear/rutsca2.html#c5
Unfortunately I don't get a connection to that web site in the moment,
so I can't look up what they are doing there.
> That doesn't seem to make a lot of sense and the electron would
> definitely take a different path than an alpha.
It would bend into the other direction - it is attracted rather than
repelled. But both particles "travel" on hyperbolas.
> But the term is
> squared, so the negative value apparently doesn't make any difference.
Right.
> What does the scattering data for neutrons tell us?
AFAIK, it is totally different from the scattering data for electrons,
protons and alpha particles (neutrons have no electric charge, therefore
no Rutherford scattering can take place). I don't have a reference handy
for this, sorry - but as already mentioned lots of times, you could try
looking into the book by Povh.
What I *do* know is that the mean free path length of neutrons in matter
(i.e. their penetration depth) is quite long, compared to the one of
electrons, protons and alpha particles. How does your model explain
this?
> If the nucleus is
> small we should probably see most alphas go through with very little
> scattering at other angles except for the rare direct hit of the
> nucleus.
I think you meant "neutrons" here, not "alphas", right?
> Is this the case?
I don't know exactly - if I have time, I will try reading up on this.
A quick Google search for "neutron scattering" yielded these web sites:
<http://www.neutron.anl.gov/>
<http://neutrons.ornl.gov/>
Perhaps you can find some valuable information there.
> The cubic model would likely predict no
> difference is relative scattering distribution.
Yes, I would have guessed this, too, from what I know about your model.
> > Additionally, I would like to see you to reproduce the behaviour of
> > Rutherford's scattering cross section wrt energy (it is proportional to
> > 1/E^2).
>
> OK, you got me there.
*grin*
> On first glance the cubic model should not show
> any dependence on E since the scattering cross section remains
> constant for any E at any particular orientation. I will have to
> investigate more since this would be a big problem.
Please do.
> Does Povh show experimental data showing the dependence on E?
IIRC, he does. I haven't looked at the book quite a time...
> I may have to take your
> advice and get it from Amazon since I have no access to otherwise.
Please do. I hope you will understand it - it requires some knowledge of
QM...
> The only other things I can think of here is that at close range, lower
> energy alphas would be deflected more strongly due to near misses as
> it is in the Rutherford model, but I would think the effect would be
> relatively small since the alternating proton/electron sequence would
> tend to null out the electrostatic field.
You haven't explained so far how the deflection really works in your
model. You said that the particles have hard surfaces, and that this
provides a force, but you haven't specified anything about this force so
far. For example, how big is it? Infinitely big? Or can the particles be
compressed a bit, and the more they are compressed, the bigger the force
gets - like a rubber ball?
> > Oh, BTW, what possible experiment could disprove your model?
> >
> The experiments I had mentioned and other experimental results showing
> incorrect dependencies as you have pointed out with 1/e^2. The theory
> is very testable.
I don't think that your theory isn't testable. But I fear that for every
experimental result which would disprove your model, you will come up
with a new fancy explanation why in reality it doesn't disprove it -
inventing some new mechanisms when needed, and so on.
> > This paper doesn't look like a scientific article to me, but rather as
> > if a student reported the results of his lab work. I have written
> > similar stuff during the lab work required in my study, too.
> >
> > Going to
> > <http://web.mit.edu/tkfocht/www/>,
> > my suspicion is confirmed. The link to this paper is found under the
> > heading "Junior Lab Papers and Presentations".
> >
> Nice bit of detective work - I did miss that.
Well, I have some knowledge how a scientific paper looks like (I have
written some myself), and this simply didn't "look right". But the main
reason for my suspicion was that no journal would bother today to
publish a paper about something which is so well-established. This would
be somehow akin to a journal publishing an experiment which demonstrates
that antennas give off electromagnetic radiation!
> > So instead of going to read some of the things I directed you to (like
> > Povh) you did a web search and found a report on a lab work experiment.
> > And you seem to think this is can be used as if it is a valid scientific
> > paper. Say, are you *really* unwilling to learn *anything*?
> >
> I'm willing to learn, but have limited resources and time.
Well, I have pointed you repeatedly to possible sources.
BTW, I've asked you several times what you have read so far on QM and
atomic and nuclear physics. I don't remember you responding ever...
> > > Is this the remarkable accuracy predicted by the Rutherford formula?
> >
> > Yes. Hint: this experiment has been done countless times - and the
> > agreement with the formula is *always* great (in the range of angles
> > where it makes sense - because of the reasons outlined above, it doesn't
> > work well for very small and very large angles).
> >
> > Additionally, the "improved" formulas, which take things like magnetic
> > moments of target and projectile into account (see above), have *also*
> > been tested countless times with amazing accuracy.
> >
>
> Seems to me that matching on only 11% of the range (10-30 degrees)of
> the Rutherford formula is not a great deal of accuracy.
*sigh* The paper as well as I have explained in great detail *why* it
doesn't match there. Also I and Franz pointed out that if one uses
better experimental equipment, one *does* get better agreement. Your
behaviour begins to look like plain denial to me.
> The MIT paper
> may have been a simple lab result, but is it an incorrect or odd
> result, or is it typical of the high-end experimental results?
It is absolutely atypical!
> If I go
> look at Povh, will I see these same results and conclusions?
You will see some selected results and a lot of references to papers
where you can find more.
> > > > (oh, BTW, shouldn't your model give a dependence of the scattering
> > > > cross section on phi?)
> > >
> > > What is phi?
> >
> > **********gaping mouth**********
> >
> > **********astonishment**********
> >
> > **********finally: BIG sigh************
> >
> > And *you* want to tell *us* something about scattering?????
> >
> > Try reading up on "spherical coordinates"!!!!!
> >
> > This is sooooooooooo basic, it's simply unbelievable that someone who
> > doesn't know this has the *arrogance* to think that his model is better
> > than the established ones!!!
>
> I don't know that it is better - I just came up with a new model and I
> want to see how it compares with experimental data and current theory.
> All I can say is that my arrogance is caused more by ignorance than by
> actual arrogance:)
Why didn't you come here at sci.physics (or went into a decent library)
and asked first where you can read up on the available evidence, instead
of concluding at once that there is insufficient evidence?
> Looking at spherical coordinates, phi is the lattitude.
Yes. (well, beside the small spelling mistake ;-) )
> There is a
> dependency of the cross section on the atom orientation (all 3
> orientations rho, phi and theta).
The Rutherford cross section depends only on theta. Your model seems to
predict an additional dependence on phi, if I understand it correctly.
> I only calculated 2 cross sections
> for the angles producing the highest high angle scattering events. I'd
> need a computer program to calculate the rest of the cross sections.
Didn't you say you used an Excel spreadsheet? I don't think it should be
very hard to change it accordingly.
> However, these are not expected to contribute much to the result since
> the alphas go through the atom at these orientations.
Hence your cross section *does* depend on phi? Or do I misunderstand
something here?
Bye,
Bjoern
Franz Heymann
> Bjoern Feuerbacher <bfeu...@ix.urz.uni-heidelberg.de> wrote in message
news:<403F4343...@ix.urz.uni-heidelberg.de>...
> [snip]
> > > Although I still like the intuitive idea that even at the subatomic
> > > level, a particle occupies a fixed amount of space (like a little
> > > marble)
> >
> > This contradicts basic QM. How would you explain the results of the
> > double slit experiments, based on this idea?
> >
> I've been looking for an analysis of the expected particle
> distribution for particles coming from a double slit setup.
You appear not to know how or where to look.
Actually, why don't you work it out for yourself?
> Everyone
> seems to think it should act like two spray cans and the resulting
> distribution should be smooth.
Foe "everyone", read "nobody, except possibly FrankH"
> I think it would be like two slits
> spitting out bb pellets and the pellets collide and head off at
> particular angles
You think wrongly.
> I wanted to see if this analysis shows any tendency
> for it to show diffraction like properties.
It does not do so.
This would show that
> particles can have wave like properties while remaining particles the
> entire time. But I haven't seen any such analysis, although such a
> thing would explain the double slit experiment.
You are talking nonsense.
>
> [snip]
>
> > The only way to avoid the conclusion that there is a repulsive force is
> > to claim that Newton's law don't apply in these situation. Do you
> > *really* want to claim this?
> >
> The concept of a hard surface seems clear enough to me. The repulsive
> force is an inherent property of a particle that cannot be compressed.
> Since it cannot be compressed, it must therefore be able to generate
> an opposing force to prevent compression.
>
> [snip]
> [snip]
> >
> > Please do. According to Rutherford's formula, the scattering cross
> > section is proportional to the charge of the projectile squared (that's
> > where the big problem for your model is, which according to you above
> > says that the charge of the projectile doesn't matter), and proportional
> > to the charge of the target squared.
> >
> > I'm interested in how you will manage to get this behaviour from your
> > model. Especially I will be interested to see how you get from your
> > model that the scattering cross section for an alpha particle is four
> > times greater than the one for an electron (provided both have the same
> > energy).
>
> According to the cubic model, an alpha is a cube which presents a
> surface area of 2 x 2 = 4 units.
And if you are looking a bit skew on, what does the projected area become?
> An electron, presents a surface area
> of 1 unit. So an alpha has a 4 times better chance of hitting
> anything. So size does matter and this would also help confirm that
> the size of an electron is about the same size as the proton/electrons
> in the alpha.
>
> I was suprised to see that the scattering pattern for electrons and
> alphas are apparently identical. According to the hyperphysics
> formulas, the electron would create a negative impact parameter:
You are once again talking nonsense.
Do you know what an impact parameter is?
Do you know that an impact parameter is essentially a classical concept
which is occasionally used by modern physicists to lend a little colour to
their statements?
> http://hyperphysics.phy-astr.gsu.edu/hbase/nuclear/rutsca2.html#c5
> That doesn't seem to make a lot of sense and the electron would
> definitely take a different path than an alpha. But the term is
> squared, so the negative value apparently doesn't make any difference.
Neither the alpha particle nor the electron follow "trajectories" during a
scattering process. The fact that a classical and a QM mechanical
calculation for the specific case of an elastic scatter from a point target
under the action of an inverse square law is coincidental.
>
> What does the scattering data for neutrons tell us?
It would give information entirely on processes which depepnd on the nature
of the nuclear forces. The alpha particle experiments to which you referred
give information essentially about electromagnetic forces.
>If the nucleus is
> small we should probably see most alphas go through with very little
> scattering at other angles except for the rare direct hit of the
> nucleus. Is this the case? The cubic model would likely predict no
> difference is relative scattering distribution.
You will fail miserably to come to grips with reality if you insist on these
qualitative handwavings.
Your knowledge of scatteing cross sections is entirely absent. For
instance, did you know that the Rutherford total scattering cross section is
infinite, despite the point like nature of the nucleus as far as the
projectile beams represented by alpha particles is concerned?
Concepts like geometrical crossectional areas are of very nearly vanishing
interest in studying electromagnetic interactions.
>
> >
> > Additionally, I would like to see you to reproduce the behaviour of
> > Rutherford's scattering cross section wrt energy (it is proportional to
> > 1/E^2).
> OK, you got me there. On first glance the cubic model should not show
> any dependence on E since the scattering cross section remains
> constant for any E at any particular orientation. I will have to
> investigate more since this would be a big problem. Does Povh show
> experimental data showing the dependence on E? I may have to take your
> advice and get it from Amazon since I have no access to otherwise. The
> only other things I can think of here is that at close range, lower
> energy alphas would be deflected more strongly due to near misses as
> it is in the Rutherford model, but I would think the effect would be
> relatively small since the alternating proton/electron sequence would
> tend to null out the electrostatic field.
>
> >
> > Oh, BTW, what possible experiment could disprove your model?
> >
> The experiments I had mentioned and other experimental results showing
> incorrect dependencies as you have pointed out with 1/e^2. The theory
> is very testable.
The world is truly awash with data on electromagnetic elastic interactions
(refined versions of Rutherford scattering experiments). Experiments have
been done with setups involving projectile beams with energies as low as a
few hundred eV to CM energies as high as 200,000,000,000 eV. Without
exception, all of these have been interpreted satisfactorily by relying
purely on the QM descriptions of elastic interactions.
> > This paper doesn't look like a scientific article to me, but rather as
> > if a student reported the results of his lab work. I have written
> > similar stuff during the lab work required in my study, too.
> >
> > Going to
> > <http://web.mit.edu/tkfocht/www/>,
> > my suspicion is confirmed. The link to this paper is found under the
> > heading "Junior Lab Papers and Presentations".
> >
> Nice bit of detective work - I did miss that.
>
> > So instead of going to read some of the things I directed you to (like
> > Povh) you did a web search and found a report on a lab work experiment.
> > And you seem to think this is can be used as if it is a valid scientific
> > paper. Say, are you *really* unwilling to learn *anything*?
> >
> I'm willing to learn, but have limited resources and time.
In that case, you would be well advised not to spend any of your valuable
time on reinventing the wheel.
>
> [snip]
> >
> > > Is this the remarkable accuracy predicted by the Rutherford formula?
> >
> > Yes. Hint: this experiment has been done countless times - and the
> > agreement with the formula is *always* great (in the range of angles
> > where it makes sense - because of the reasons outlined above, it doesn't
> > work well for very small and very large angles).
> >
> > Additionally, the "improved" formulas, which take things like magnetic
> > moments of target and projectile into account (see above), have *also*
> > been tested countless times with amazing accuracy.
> >
>
> Seems to me that matching on only 11% of the range (10-30 degrees)of
> the Rutherford formula is not a great deal of accuracy. The MIT paper
> may have been a simple lab result, but is it an incorrect or odd
> result, or is it typical of the high-end experimental results? If I go
> look at Povh, will I see these same results and conclusions?
There are litle hundreds of papers i the professional journals in which
elastic scattering experimnts with accuracies better than 0.1% are
described.
FrankH, you simply cannot criticise physics form the sidelines.
> > > > (oh, BTW, shouldn't your model give a dependence of the scattering
cross
> > > > section on phi?)
> > >
> > > What is phi?
> >
> > **********gaping mouth**********
> >
> > **********astonishment**********
> >
> > **********finally: BIG sigh************
> >
> > And *you* want to tell *us* something about scattering?????
> >
> > Try reading up on "spherical coordinates"!!!!!
> >
> > This is sooooooooooo basic, it's simply unbelievable that someone who
> > doesn't know this has the *arrogance* to think that his model is better
> > than the established ones!!!
> I don't know that it is better - I just came up with a new model and I
> want to see how it compares with experimental data and current theory.
> All I can say is that my arrogance is caused more by ignorance than by
> actual arrogance:)
> >
> Looking at spherical coordinates, phi is the lattitude. There is a
> dependency of the cross section on the atom orientation (all 3
> orientations rho, phi and theta). I only calculated 2 cross sections
> for the angles producing the highest high angle scattering events. I'd
> need a computer program to calculate the rest of the cross sections.
> However, these are not expected to contribute much to the result since
> the alphas go through the atom at these orientations.
Oh dear
Franz
Bjoern Feuerbacher
>
> I notice that you ignored (and snipped) my question why you don't go and
> read some books - e.g. Povh.
>
> FrankH wrote:
> >
[snip lots]
> > According to the hyperphysics
> > formulas, the electron would create a negative impact parameter:
> > http://hyperphysics.phy-astr.gsu.edu/hbase/nuclear/rutsca2.html#c5
>
> Unfortunately I don't get a connection to that web site in the moment,
> so I can't look up what they are doing there.
Finally I got the connection. Looking at that page, it seems to me that
they derived the formula for b only for the case of a repelling force
(look at the picture), not for an attractive force. Hence this formula
doesn't apply to the electron.
Looking at the derivation of the formula at
<http://hyperphysics.phy-astr.gsu.edu/hbase/nuclear/impar.html#c1>,
I think the problem may be the limits on the angle phi_i and phi_f - for
an attractive force, the direction where phi = 0 is different than the
one given in the figure on this page (the symmetry plane has a different
orientation), and therefore one gets other values for the limits of the
angle. I can try to figure this out in more detail, if you want.
[snip rest]
> Bye,
> Bjoern
Bjoern Feuerbacher
I tried to read your web page yesterday, but encountered a big problem:
my web browsers (Netscape and Mozilla) displayed only the text until the
sentence "The white cubes are not part of the model, they just hold up
other parts.", but nothing afterwards. In contrast, trying later with
Internet Explorer, the whole page was shown. There is something strange
about your web page - could you try to fix this problem, please?
Anyway, now that I've looked at all you wrote, I've found sooo many
problems with your ideas that I really don't know where to start.
Perhaps at first only one rather simple question - could you please
address this when you post the next time to sci.physics?
Well, the question is: what on earth is the difference between a
Hydrogen atom and a neutron in your model??? If I understand it
correctly, both are simply a "proton cube" and an "electron cube" which
are somehow bound to each other. There seems to be no difference at all
in your model between the two particles. But what we find in reality is
that these particles are *totally* different:
1) They have different masses.
2) They have different magnetic moments.
3) They have different sizes (well, the sizes are determined by
scattering, so you will probably argue against this).
4) A hydrogen atom is stable, a neutron decays with a half life of about
9 minutes.
5) They have totally different penetration depths (mean free path
lengths): neutrons can pass easily through matter, hydrogen atoms can't.
6) Hydrogen atoms can be excited and even ionized, they show spectra -
neutrons don't do this.
7) The spins don't add up - electrons, protons and neutrons all have
spin 1/2. But according to QM, you can't build a composite of two
particles with spin 1/2 which again has spin 1/2. (well, you don't
accept QM, so this argument probably won't bother you)
8) When a neutron decays, not only a proton and and electron emerge, but
also an electron-antineutrino.
I suspect that you will now claim that the difference between a neutron
and a hydrogen atom is that the neutron has additionally also an
antineutrino bound to it somehow. But please consider:
1) I don't see how the addition of an antineutrino would solve all of
the eight problems outlined above - this would solve only 8, 7 and
perhaps 4.
2) This hypothesis will only get you into more trouble. Hint: other
particles emit also neutrinos when they decay. For example, beside
beta-minus decay (the one of the neutron), there is also beta-plus
decay, where a nucleus with Z protons and N neutrons decays into one
with Z-1 protons, N+1 neutrons, a positron and an electron-neutrino...
Bye,
Bjoern
FrankH
> Hi again, Frank!
>
> I tried to read your web page yesterday, but encountered a big problem:
> my web browsers (Netscape and Mozilla) displayed only the text until the
> sentence "The white cubes are not part of the model, they just hold up
> other parts.", but nothing afterwards. In contrast, trying later with
> Internet Explorer, the whole page was shown. There is something strange
> about your web page - could you try to fix this problem, please?
>
Microsoft Word where it was originally generated.
cubic model, but that they would have to have radically different
properties. The difference must be in the way they are bound. Hydrogen
is bound by normal Columnb force. I make no claim to how a neutron
binds together an electron and proton. As you point out the bond must
produce a mass defect and be of a completely different type of bond.
My model would hypothesize that the bond is such that it keeps the
proton and neutron components of the atom together, while the electron
is loosely bound with normal electrostatic force. My pictures show the
neutron as being the same size as a hydrogen atom, but this isn't
necessarily the case. If they are about the same size, then it helps
to explain how the atom sticks together, since you can see a neat
alternation of proton/electron.
The cubic model would explain Tritium as a hydrogen with 3 neutrons
and this might look like N
HN
N
This is looking from above with H=hydrogen and N=neutron. I suppose
the model would predict that a maxium of 4 neutrons could be attached,
although we never see it (perhaps too unstable). The nature of the
neutron bond is such that it doesn't give up its electron and
therefore contributes little to chemical reactions.
So definitely, neutrons are different from hydrogen atoms. Same
components but different kind of binding involved.
>
> Bye,
> Bjoern
Franz Heymann
What is the nature of this new force?
Does it obey an inverse square law?
On which property of a particle does it act, in the sense that an electric
field acts on a charge?
Are there both attractive and a repulsive modes for this force?
Franz
Bjoern Feuerbacher
>
> Bjoern Feuerbacher <bfeu...@ix.urz.uni-heidelberg.de> wrote in message news:<4045CDD6...@ix.urz.uni-heidelberg.de>...
> > Hi again, Frank!
> >
> > I tried to read your web page yesterday, but encountered a big problem:
> > my web browsers (Netscape and Mozilla) displayed only the text until the
> > sentence "The white cubes are not part of the model, they just hold up
> > other parts.", but nothing afterwards. In contrast, trying later with
> > Internet Explorer, the whole page was shown. There is something strange
> > about your web page - could you try to fix this problem, please?
> >
> I'll have to check on that, it has a lot of weird tags added by
> Microsoft Word where it was originally generated.
Let this be a lesson to you - never trust Microsoft! ;-)
Good. Didn't you see this problem ever before?
> The difference must be in the way they are bound.
How could this account for all the 8 problems outlined above?
> Hydrogen
> is bound by normal Columb force. I make no claim to how a neutron
> binds together an electron and proton.
But you simply assert that such a binding force has to exist,
apparently?
> As you point out the bond must
> produce a mass defect
So you accept E = mc^2 for the mass defect due to the binding energy?
Nice.
> and be of a completely different type of bond.
> My model would hypothesize that the bond is such that it keeps the
> proton and neutron components of the atom together, while the electron
> is loosely bound with normal electrostatic force.
So your model *does* need an additional force after all? You boasted on
your web site that your model makes the strong nuclear force unnecessary
- apparently this was false, wasn't it?
> My pictures show the
> neutron as being the same size as a hydrogen atom, but this isn't
> necessarily the case.
Why not?
Oh, BTW, where is your evidence that the electron has the same size as
the proton? How do you explain Bhabha scattering?
> If they are about the same size, then it helps
> to explain how the atom sticks together, since you can see a neat
> alternation of proton/electron.
"neat" doesn't imply "right" necessarily.
> The cubic model would explain Tritium as a hydrogen with 3 neutrons
> and this might look like N
> HN
> N
What's the additional, third N up there on the right?
> This is looking from above with H=hydrogen and N=neutron. I suppose
> the model would predict that a maxium of 4 neutrons could be attached,
> although we never see it (perhaps too unstable).
Where do you get the number 4 from?
> The nature of the
> neutron bond is such that it doesn't give up its electron and
> therefore contributes little to chemical reactions.
>
> So definitely, neutrons are different from hydrogen atoms. Same
> components but different kind of binding involved.
I haven't seen you explain how this idea solves any of the 8 problems
mentioned above...
Bye,
Bjoern
FrankH
objections to the cubic atomic model. I have included these questions
in the newsgroup since others will probably have the same questions
and this newsgroup serves as my "science notebook" to keep track of my
theories and discoveries.
> Dear Frank,
>
> I've now further looked at your web site and would like to point out a
> few
> more problems I've found with your model. If you don't have time to
> address my comments in the moment, no problem - I can wait. ;-) The
> only
> thing I *really* would like you to address is my previous question
> about
> the difference between a H atom and a neutron.
The cubic model postulates (meaning unjustified and unexplained) a
different bonding configuration for the neutron. By definition, it has
all of the characteristics that you mention. This is a weak answer, I
know, but it is required to get the cubic theory off the ground and
the theory will fail if no answer is ever provided, but if I run
across something that looks good, I'll let you know. So far I've only
seen Tom Lockneyers QVP model and it might work.
>
> So, let's start:
>
> 1) I have a problem with your rule (2):
> "The smallest unit that can be added to build an atom is a combination
> of
> 2 electrons and 2 protons. One electron/proton pair can be considered
> to
> be a neutron. The extra proton is bound to the neutron while the extra
> electron can be ionized from the atom."
>
> This rule would imply that every existing atom resp. nuclid had an
> equal
> number of protons and neutrons. From observation we know that this
> obviously isn't true - for every elements, there are usually lots of
> different isotopes, with different numbers of neutrons. How do you
> account
> for those? The easiest example here would be Deuterium and Tritium -
> according to the list of elements you provide in photos lower down,
> these
> both don't exist in your model...
>
> Even if you change your model to include all of the isotopes, you still
> have to explain why all of the isotopes of an element have the same
> chemical properties. Especially your claim about a relation between
> symmetry and reactiveness seems to get into trouble by this...
I have updated the web site to explain isotopes as neutron elements
appearing between the arms of the atoms. The reason why they do not
change the chemical properites are 1) The neutron is bound such that
the electron portion of the neutron cannot escape and provide a basis
of chemical reactions. 2) The neturons are placed between the arms
such that if an atom were to react with another atom, it would likely
react with the tips of the atoms and ignore anything else. This is why
only the outer electron shells (which correspond to the outer most
tips of the atom in the cubic model) appear to be involved in
chemical reactions. I need to do much more research in this area of
chemical reaction to see if the cubic model helps to explain
chemistry.
>
>
> 2) Your model also seems to imply that the masses of all nuclides
> should
> be equal to the mass of a proton times the number of protons in the
> nuclid
> plus the mass of a neutron times the number of neutrons in it. This
> also
> contradicts observations - have you ever heard of "nuclear binding
> energies" and "mass deficit"? The "waterdrop model" of standard nuclear
> physics can explain the observed masses of the nuclids quantitatively -
> can your model do this, too?
This is another postulated area of the model where the neutron has
mass defect and it explains this to the extent that if you assume
this, the total mass works out.
>
> 3) Your rules (3) and (4) seem to contradict each other - something
> which arms stretching outwards obviously occupies more space than a
> more
> compact object.
Yes, it does. With no arms, it would form a sphere, so rule 4 is a
further restriction to prevent this. These work together to form the
proper shape.
>
>
> 4) I don't understand what your rule (5) is supposed to mean - what
> does
> "all available space ... on the arms" mean?
>
>
> Additionally, all five rules seem to be made up out of nowhere, whereas
> the rules of standard QM were made up based on experimental
> observations...
>
All avalible space means take the size of the core N, then the next
levels can contain N-1, N-2, N-3 until nothing more fits.
Yes, these rule were made out of nowhere, based only upon the
geometric stacking pattern of the bricks I was playing with and a
little knowlege of the electron shell configurations. This is
definitely a theory in search of experimental data.
>
>
> 4) Your sentence "Adding a minimum atomic unit of 2 protons and
> electrons,
> we get a cube shape for Helium." is obviously wrong - you added more
> than
> such a minimum atomic unit to the H atom in order to get the Helium
> atom. This is in line with my first comment above.
>
Yup, there is always an exception to the rule. I have updated by
website to include this exception.
>
>
> 5) It's not clear to me why Beryllium looks like a stack with height 4,
> instead of a stack with height 3 with another unit attached to its
> side. Does this follow somehow from your rule (3)? I don't see how,
> sorry.
I have updated my web site to indicate that the ionization energy
pattern and symettry would indicate that it is a stack. Otherwise, it
could go another way.
>
>
> 6) Oxygen looks *very* symmetrical to me in your model - doesn't this
> contradict your claim that symmetrical atoms are usually very stable
> and
> chemically inactive?
Yup, I mention this in my web site. Symmetry is not the whole picture
with regard to chemical inactivity.
>
>
> 7) Why do we get Neon by adding a unit to exactly the side opposite of
> the side where the last unit was added to get Fluorine? Does this
> perhaps
> follow from your rule (3) somehow? Also, Neon doesn't look very
> symmetric
> to me, contrary to your claim that symmetry is usually associated with
> inertness.
Getting the model to work past Neon was actually one of the greatest
hurtles since it is so unsymmetric. But I had to push past it to get
the rest of the atoms to work. The core must always take an even
number of units before arms can be built up again. This was done to
keep electron pairs shown in the electron configuration chart (always
added in pairs before you go to the next energy level). I suppose this
should be another rule. Another reason why the core must have an even
number is because the atomic units are a 2 x 2 square and in order to
fit symettrically on the sides, the core must be even.
>
>
> 8) In order to get Sodium, you apparently either split the added unit
> and
> added one half below and one half above an already existing unit, or
> you
> moved the already existing unit downwards or upwards before adding the
> new
> one. Why did you do this? How does this follow from your rules?
>
This is to remain as symmetric as possible. Adding it anywhere else
would be more lobsided. It needs to remain as flat as possible, so it
pushes aside
the arm atom. This is somewhat arbitrary and may not necessarily be
the case.
>
> 9) Still sodium: why wasn't the next unit simply added on one of the
> four
> shorter arms? Your rule (5) says apparently that a new unit has to be
> added to the arms, if this is possible, but you nevertheless added the
> new
> unit here directly to the vertical core. Why don't you follow your own
> rules? Or do I misunderstand something about them? Similar questions as
> 8
> and 9 here apply to Mg, Al and Si.
I think you may misunderstand here, since a unit is being added to the
arms for Si, Mg, Al and Si. The vertical core is the same for all of
these atoms (six high).
>
> 10) "The outer sides of Silicon are tall enough to allow another unit
> to
> be attached to the sides..." What do you mean by "tall enough"?
>
This has to do with the N, N-1, N-2 progression, where N is a 2 x 2 x
2 helium nuclei in the core. N=3 for neon, so it can take N-2=2 and
N-1=1. Tall enough in this case means the arms of a vertical length of
4 and so can take another atomic unit of 2 and maintain the stacking
shape of the arms.
>
> 11) "All of the 8 electrons surrounding the core would presumably have
> the
> same energy level..." This contradicts observation - see the ionizazion
> enthalpies of argon you yourself quote lower down.
>
I should say relative difference between electrons. The difference
should be about the same for electrons with similar geometric
configurations. I have updated my web site to make this clearer.
>
> 12) You yourself point out that Krypton is not fully symmetric - this
> is
> yet another counterexample to your rule "symmetric implies not very
> reactive".
Yup, this is a very, very, very rough observation and obviously, not a
rule.
>
>
> 13) "Expanding this by adding 2 more units to the core, we get Hafnium
> at
> 72 electrons. Interestingly, this is not the next nobel gas, so there
> is a
> break in the geometric sequence. Some other rule must apply past
> Xenon."
> Hint: standard QM explains this quite easily.
What is this something? I'd be curious in seeing how QM solves large
atom problems.
>
>
> 14) wrt Radon "What is interesting to note about this atom is that it
> is
> entirely constructed of helium nuclei which is a cube composed of 4
> protons and 4 neutrons."
>
> Yes, that's somehow interesting. But so what??? Doesn't it bother you
> in
> the least that this again is a counterexample to your claim that
> symmetry
> is associated with inertness? More on this below in 16.
Yup, lots of counterexamples, not a hard/fast rule. Need more research
here. The alternate constuction of Radon was an effort to keep the
rule.
>
>
> 15) "Otherwise, I was unable to stack more than 8 cubes before my model
> started to fall over. You could say my model lost stability if it
> wasn't
> grouped into larger units."
>
> Err, what on earth has stability wrt toppling other to do with chemical
> stability???
>
Nothing really, I just though that it was curious that problems that I
was having in the macroscopic world (stacking sugar cubes) could be
solved by a solution in the microscopic world (bonding He atoms).
>
> 16) "There seems to be a loose relationship between at atom being made
> up
> of helium atoms and being a noble gas. The rule is loose because
> Krypton
> and Xenon contain significant non-helium components."
>
> There are only six noble gases, and your rule holds only for four of
> them. I wouldn't call this even a "rule" - I would call this
> "coincidence". Especially because you have no physical reason to get
> from
> "is made up from helium atoms" to "is a noble gas".
I would say more than coincidence, but certainly not a rule, lots more
research on chemical reactivity to do here.
>
>
> "There do appear to be some inconsistencies in the relationship between
> the atom structure and whether it is a noble element. This is an area
> of
> further study."
>
> So you even recognize yourself that there is a problem...
>
Yup, can't explain everything yet. The theory is still young.
>
>
> 17) "You will notice a striking pattern of repeating energy levels.
> Each
> of the 7 primary electron shells is like an inverted pyramid. And one
> noble element is found in each shell. Why should it be arranged like
> this?"
>
> See the standard QM solutions for bound states in a Coulomb potential.
>
Will look for that when I get Povh, haven't seen an explation in other
books so far. My primary sources are the hyperphysics web site, I read
a college physics text book, a book called "From atoms to quarks" and
"The odd quantum".
>
>
> 18) "You can imagine that if you shake a cubic atom by adding energy to
> it, some electrons will be looser than others. The electrons on the
> outside would take less energy to ionize than electrons contained in
> the
> vertical core."
>
> Well, in your model, I can't see how one could *ever* get an electron
> from
> the inside to become ionized without breaking the whole core! Could you
> please explain how this is supposed to work? How do you get the
> electrons
> from inside to the outside without destroying the whole structure?
The electron is always loosely bound and its appearance in the atom is
optional which means that there may be a special bond between the
neutron and proton as well (yes, additional bonding types) which keeps
the atom together when it is ionized. I explain that an electron in
the core can be ionzied since its path to free space is very short. I
would have to squeeze past the arms, by going between them, but isn't
blocked on all sides.
>
>
> 19) "This simple mechanical model shows why we see a triangular pattern
> in
> the electron shells."
>
> Sorry, I don't understand at all where you get a triangular pattern
> from
> in your model.
Ok, lots more research to justify this statement. The only thing I can
say now is that the bottom 2 electrons of the triangle represent the 2
electrons in the vertical core. The next level of 6 electrons
represent the 6 faces of a cube.
>
>
> 20) "The number of electrons in the next higher energy level are in
> groups
> of four (because of the four sides of a cube),..."
>
> Huh? A cube has six faces, as you yourself said just a sentence
> before. Additionally, there is no grouping of four in the ionization
> energies of the electrons - only a grouping of 10.
OK, bad terminology. How about saying the 4 vertical sides of a cube?
I'm not sure we even see groupings of 10. In the ionziation chart of
argon, it appears to have a group of 6, 2, 8, 2 (as far as relative
ioniztion energies).
>
> 21) "It is also interesting to note that the atom builds up narrow arms
> instead of filling in the spaces between the arms."
>
> Again: huh? Your model seems to show the exact opposite to me - there
> are
> no narrow arms, but the arms are more like pyramids, i.e. the space
> between the arms *is* filled up.
I think the terminology is poor, the pictures do a better job of
explaining it. I think you have the concept correct where the arms
fill up like pyramids. The space I am referring to is the space
between the tips of the arms.
>
>
> 22) "However, we know that the larger atoms can take additional
> neutrons
> to form isotopes."
>
> Oh, so you *do* know this? However, what you apparently don't know is
> that
> *all* atoms can do this (ever heard of Deuterium, Tritium and 3He?),
> not
> only the larger ones, and that there are also isotopes with *less*
> neutrons than the stable ones have.
Yes, all atoms can do this including hydrogen. Placement is less
certain with hydrogen but possible.
>
>
> 23) "I would think that the neutrons fill in the spaces between the
> arms."
>
> This is already rule number six. Again without justification.
>
Yup, I have added rule 6 to my website. The only justification is that
maintains the other 5 rules and gives a place for the neutrons to
hide.
>
> 24) "Since they are effectively sheilded by the extended arms,..."
>
> Shielded against what? And how does this shielding work, specifically?
>
As an atom approaches another atom, it is likely to bump into the
extended arms of the atom rather than reach into the spaces between.
Also, the tips of the atoms create a bonding area like docking port
that allows them to combine. The alternating series of
electron/protons means there is always a way that the atoms can be
oriented to create an attractive bond. There is no such site between
the arms.
>
> 25) "...they do not have an effect on the chemical bonding properties
> of
> the atoms."
>
> What *has* an effect on these bonding properties, then?
>
Much more research on chemistry needed here, but the looseness of the
free electron and the ability of atoms to "dock" with each other as
described above.
>
> "This is an area of further study to see if there is a specific
> relationship between the size of the arms and the number of neutrons an
> atom can take."
>
> Good luck. Please pay special attention to:
> * there are also isotopes with less neutrons
> * please explain why from a certain number of excess neutrons or
> protons, the atom becomes unstable and undergoes beta decay
> * standard nuclear physics can explain all of this, and it can give the
> masses of all these nuclids
Will keep these in mind
>
>
> 26) "Eliminates the problem of why should an electron orbit the
> nucleus"
>
> Since no one claims that it does, I see no need to comment on that
> section.
>
I can still only conclude that QM doesn't explain what the electron is
doing. Lots of different opinions on this still. How an electron can
be both a wave and a particle also appears to be controversial. I
think these 2 areas still remain a fundamental area of fuzzyness for
QM.
>
>
> 27) "Eliminates the need for a strong nuclear force.
>
> Since all of the protons and neutrons are not packed into a dense speck
> in
> the middle of the atom, there is no need to presume that there must be
> some extremely strong force holding them in there."
>
> Thanks for showing that you have no clue about the reasons why the
> nuclear
> force was invented (this "packed together" was only one of several
> reasons).
What other reasons are there, yes I have no clue.
>
>
> 28) "Explains the newest pictures of atoms using STM"
>
> We have already discussed this. I don't remember you explaining why the
> authors wrote that the results agree with the predictions of QM - how
> could this be possible, if QM is wrong?
>
Yes, the authors believe their results confirm QM. I think alternate
explainations of their results could exist. The pictures are an
experimental result, results are always interpreted.
>
> 29) "Explains the most common fission products of uranium"
>
> IIRC, we already discussed that, too - and your so-called explanation
> was
> found to be insufficient, in contrast to standard nuclear physics.
>
I don't think it was insufficient, considering the randome statistical
nature of fission, I doubt that any answer matching with precision is
possible. I'd say the results were close enough to warrarnt further
investigation. It isn't proof.
>
> 30) "Explains how an alternating series of protons and electrons are
> stable and why helium is so stable."
>
> We have already discussed that, too - and you are still missing a
> repulsive force.
This is another postulate (yes, unjustified and unexplained) that is
required to get the theory off the ground, but it isn't so hard to
believe if you think macroscopically. QM is fond of saying that the
world of the microscopic is totally different from the macroscopic,
but one of the motivators of this theory is to prove this wrong.
>
>
> 31) "It correctly predicts the relative difference for the first
> ionization energies for hydrogen and helium"
>
> What on earth has the net force on an element to do with the ionization
> energy? You *do* understand the difference between "force" and
> "energy",
> don't you?
>
> In the newsgroup you said that your model can predict both of these
> ionization energies - but here you present only a comparison of the
> forces. Did you lie?
>
> And what about the deviation of 9%? Do you think this is insignificant
> or
> what?
I have updated my web site to futher clarify this. This is only
relative difference and I have assumed a relationship between how
tightly an element is bound in a system and the ionization energy. The
9% difference is significant and certanly not like the .0000000000001
accuracy claimed elsewhere. But once again, close enough to warrant
further investigation. There are so many unknowns about the bonding
type and strength, that it is suprising that it is this close.
>
> You completely ignore that standard QM can predict both of these
> ionization energies with very good accuracy. How is this possible, if
> standard QM is wrong for the ground state?
>
As you have stated yourself, just because a theory can predict a great
number of things with great accuracy, that doesn't prove it is
correct. Otherwise you would belive Tom's QVP theory which predicts
stuff with incredible accuracy.
>
> 32) "It correctly predicts the relative ionization energies for the
> remaining electrons in an atom"
>
> Problems with this were already discussed above.
I have added additional clarification in the web site.
>
>
> 33) "It correctly accounts for the results of the Rutherford scattering
> experiment"
>
> This was already discussed - in the meantime you admitted yourself that
> this claim is overstated. Please correct it.
I have reduced the claim on the website.
>
>
> 35) In the newsgroup, you also claimed that your model can explain
> the magical numbers for nuclei. I don't find this on your page. Did you
> perhaps not even know what these magical numbers are, and thought we
> talked about ionization energies and shells or something like that?
> Please read up in Povh on what the magical numbers are.
>
I only have a rough idea for the magic numbers. I will read up on Povh
when I get it. (still waiting on Borders books).
>
> 36) I once made an argument that the differences between chemical and
> nuclear energies are a factor of 1000 or even 1 000 000, and asked you
> how
> your model explains this. I don't remember you ever answering this...
>
I would explain that chemical reactions largely involve just bonds and
attractions created by the Columb forces. Nuclear forces require
rearranging bonds between the neutrons and protons which could have
energies many times higher. Of course, I admit that I have no clue as
to the binding mechanisims, so I cannot explain or characterize it.
But such a difference would explain the different energies between
chemical and nuclear reactions.
>
>
> Have fun with all of the questions! ;-)
>
>
> Bye,
> Bjoern
>
> Hi again, Frank!
>
> Just this morning I discovered a new problem with
> your model - again
> something with the neutron (well, this applies to
> the hydrogen atom, too,
> but there I don't know what the experimental results
> are).
>
> The problem is: your neutron consists of an electron
> and a proton with a
> distance of about 10^(-10) m between their centers
> (is this right, or is
> the distance actually smaller? I base this on your
> claim that the neutron
> looks like a H-atom, and this is the diameter of a
> H-atom).
>
> The electron has a charge of -1 e (elementary
> charge), the proton +1 e.
> Such a configuration has an electric dipole moment
> of 10^(-10) e * m
> = 10^(-8) e * cm.
>
> However, the experimental upper limit for the
> electric dipole
> moment of the neutron is 0.63 * 10^(-25) e * cm
> (source: Particle Data
> Group, <http://pdg.lbl.gov/2003/s017.pdf>).
>
> How do you explain this discrepancy? This electric
> dipole moment could in
> your model only result if the distance between the
> centers of electron and
> proton were around 10^(-25) cm = 10^(-27) m. Do you
> *really* want to claim
> this? (if yes, you will have some problems to
> explain why an H atom is
> so much bigger, around 10^/-10) m) Or will you claim
> that the charges of
> electron or proton change in your model? Or that
> there is something wrong
> with the experimental determinations of the electric
> dipole moment of the
> neutron?
>
>
>
> Bye,
> Bjoern
>
>
I make no claim about how a neutron is built or shaped. It could be
that a neutron is the same size as a proton and the fundamental atomic
unit looks more like a triangle of proton,neutron,electron. Maybe the
proton/neutron form a pair which leaves the electron loose to be
ionized. The only thing the cubic model requires is that there be an
atomic unit that can be stacked as I have shown in my model. That is
easier to explain if everything is cubic and the same size, but not
required. Until I can come up with the portion of the theory that
explains atomic particle construction, like the QVP model, I cannot
say how this works.
-thanks for all the thoughful questions
I have updated my web site to claim that my theory is probably wrong,
but I'll work hard to prove it right!
Bjoern Feuerbacher
>
> Thanks to Bjoern Feuerbacher for providing the following interesting
> objections to the cubic atomic model. I have included these questions
> in the newsgroup since others will probably have the same questions
> and this newsgroup serves as my "science notebook" to keep track of my
> theories and discoveries.
>
> > Dear Frank,
> >
> > I've now further looked at your web site and would like to point out a
> > few
> > more problems I've found with your model. If you don't have time to
> > address my comments in the moment, no problem - I can wait. ;-) The
> > only
> > thing I *really* would like you to address is my previous question
> > about
> > the difference between a H atom and a neutron.
>
> The cubic model postulates (meaning unjustified and unexplained) a
> different bonding configuration for the neutron. By definition, it has
> all of the characteristics that you mention. This is a weak answer, I
> know,
This is *very* weak - especially in the light of the fact that the
standard model *can* explain all of these difference between the H atom
and the neutron!!!
For example, how would a different bonding configuration resolve the
problems with the magnetic and electric dipole moments of the neutron I
pointed out to you?
> but it is required to get the cubic theory off the ground and
> the theory will fail if no answer is ever provided, but if I run
> across something that looks good, I'll let you know.
Well, don't you think that a model with no sound fundament is a bit
problematic?
> So far I've only
> seen Tom Lockneyers QVP model and it might work.
Tom's model is a non-starter. He doesn't understand quantization of
angular momentum, for starters.
> > So, let's start:
> >
> > 1) I have a problem with your rule (2):
> > "The smallest unit that can be added to build an atom is a combination
> > of
> > 2 electrons and 2 protons. One electron/proton pair can be considered
> > to
> > be a neutron. The extra proton is bound to the neutron while the extra
> > electron can be ionized from the atom."
> >
> > This rule would imply that every existing atom resp. nuclid had an
> > equal
> > number of protons and neutrons. From observation we know that this
> > obviously isn't true - for every elements, there are usually lots of
> > different isotopes, with different numbers of neutrons. How do you
> > account
> > for those? The easiest example here would be Deuterium and Tritium -
> > according to the list of elements you provide in photos lower down,
> > these both don't exist in your model...
> >
> > Even if you change your model to include all of the isotopes, you still
> > have to explain why all of the isotopes of an element have the same
> > chemical properties. Especially your claim about a relation between
> > symmetry and reactiveness seems to get into trouble by this...
>
> I have updated the web site to explain isotopes as neutron elements
> appearing between the arms of the atoms.
Err, there are also isotopes which have *less* neutrons than the stable
isotopes. What about them?
> The reason why they do not
> change the chemical properites are 1) The neutron is bound such that
> the electron portion of the neutron cannot escape and provide a basis
> of chemical reactions.
In other words: yet another unfounded postulate, right?
> 2) The neturons are placed between the arms
> such that if an atom were to react with another atom, it would likely
> react with the tips of the atoms and ignore anything else.
> This is why only the outer electron shells (which correspond to the outer
> most tips of the atom in the cubic model) appear to be involved in
> chemical reactions.
Speaking about reactions - what number of possible bindings would your
model suggest for every element?
> I need to do much more research in this area of
> chemical reaction to see if the cubic model helps to explain
> chemistry.
Good luck.
Things you could research on are e.g. numbers of possible bonds, bond
lengths and bond angles. And energy released in the reactions.
> > 2) Your model also seems to imply that the masses of all nuclides
> > should
> > be equal to the mass of a proton times the number of protons in the
> > nuclid
> > plus the mass of a neutron times the number of neutrons in it. This
> > also
> > contradicts observations - have you ever heard of "nuclear binding
> > energies" and "mass deficit"? The "waterdrop model" of standard nuclear
> > physics can explain the observed masses of the nuclids quantitatively -
> > can your model do this, too?
>
> This is another postulated area of the model where the neutron has
> mass defect and it explains this to the extent that if you assume
> this, the total mass works out.
Say, how many unfounded postulates do you need to rescue your model?
Additionally, I'm not sure what "the neutron has mass defect" is
supposed to mean...
[snip]
> > 4) I don't understand what your rule (5) is supposed to mean - what
> > does "all available space ... on the arms" mean?
> >
> >
> > Additionally, all five rules seem to be made up out of nowhere, whereas
> > the rules of standard QM were made up based on experimental
> > observations...
> >
> All avalible space means take the size of the core N,
Do you mean the side length of the "cube" in the core? Or the number of
"alpha cubes" in it? Or what?
> then the next
> levels can contain N-1, N-2, N-3 until nothing more fits.
N-1, N-2, ... what?
> Yes, these rule were made out of nowhere, based only upon the
> geometric stacking pattern of the bricks I was playing with and a
> little knowlege of the electron shell configurations. This is
> definitely a theory in search of experimental data.
Well, I still don't see why we need a new model (especially one which
isn't based on observational evidence, but on guesswork) - QM works
quite well...
> > 4) Your sentence "Adding a minimum atomic unit of 2 protons and
> > electrons,
> > we get a cube shape for Helium." is obviously wrong - you added more
> > than
> > such a minimum atomic unit to the H atom in order to get the Helium
> > atom. This is in line with my first comment above.
> >
>
> Yup, there is always an exception to the rule.
Shouldn't happen if the model is really valid, IMO.
> I have updated by website to include this exception.
Thanks.
> > 5) It's not clear to me why Beryllium looks like a stack with height 4,
> > instead of a stack with height 3 with another unit attached to its
> > side. Does this follow somehow from your rule (3)? I don't see how,
> > sorry.
>
> I have updated my web site to indicate that the ionization energy
> pattern and symettry would indicate that it is a stack. Otherwise, it
> could go another way.
In other words, you had to add *again* another rule, so that your model
won't contradict observations. Doesn't this trouble you in the least?
> > 6) Oxygen looks *very* symmetrical to me in your model - doesn't this
> > contradict your claim that symmetrical atoms are usually very stable
> > and chemically inactive?
>
> Yup, I mention this in my web site. Symmetry is not the whole picture
> with regard to chemical inactivity.
To me, it looks as if symmetry had close to nothing to do with chemical
inactivity.
> > 7) Why do we get Neon by adding a unit to exactly the side opposite of
> > the side where the last unit was added to get Fluorine? Does this
> > perhaps
> > follow from your rule (3) somehow? Also, Neon doesn't look very
> > symmetric
> > to me, contrary to your claim that symmetry is usually associated with
> > inertness.
>
> Getting the model to work past Neon was actually one of the greatest
> hurtles since it is so unsymmetric.
Well, I asked you above how you did arrive *at* Neon, not how you went
beyond it.
> But I had to push past it to get
> the rest of the atoms to work. The core must always take an even
> number of units before arms can be built up again. This was done to
> keep electron pairs shown in the electron configuration chart (always
> added in pairs before you go to the next energy level). I suppose this
> should be another rule.
How many rules will you need? 10? 20?
> Another reason why the core must have an even
> number is because the atomic units are a 2 x 2 square and in order to
> fit symettrically on the sides, the core must be even.
Why do they have to fit symmetrically on the sides?
> > 8) In order to get Sodium, you apparently either split the added unit
> > and
> > added one half below and one half above an already existing unit, or
> > you
> > moved the already existing unit downwards or upwards before adding the
> > new
> > one. Why did you do this? How does this follow from your rules?
> >
> This is to remain as symmetric as possible.
Why is this necessary? None of your rules say this.
Also, different people may have different ideas here what looks more
symmetric.
> Adding it anywhere else
> would be more lobsided. It needs to remain as flat as possible, so it
> pushes aside the arm atom.
Why?
> This is somewhat arbitrary and may not necessarily be
> the case.
In other words, you need yet another rule?
> > 9) Still sodium: why wasn't the next unit simply added on one of the
> > four
> > shorter arms? Your rule (5) says apparently that a new unit has to be
> > added to the arms, if this is possible, but you nevertheless added the
> > new
> > unit here directly to the vertical core. Why don't you follow your own
> > rules? Or do I misunderstand something about them? Similar questions as
> > 8 and 9 here apply to Mg, Al and Si.
>
> I think you may misunderstand here, since a unit is being added to the
> arms for Si, Mg, Al and Si. The vertical core is the same for all of
> these atoms (six high).
No, you misunderstood what I meant. You added the new square directly
*at* the core (you didn't make the core higher, but glued the new stuff
on the side of it). My question was: why didn't you add the new square
on an arm directly, pointing further outwards? (so that one of the arms
would then have a length of two) Do you understand the question now?
> > 10) "The outer sides of Silicon are tall enough to allow another unit
> > to be attached to the sides..." What do you mean by "tall enough"?
> >
> This has to do with the N, N-1, N-2 progression, where N is a 2 x 2 x
> 2 helium nuclei in the core. N=3 for neon, so it can take N-2=2 and
> N-1=1. Tall enough in this case means the arms of a vertical length of
> 4 and so can take another atomic unit of 2 and maintain the stacking
> shape of the arms.
Why do they arms need to have a "stacking shape"? This also wasn't
mentioned in your rules.
> > 11) "All of the 8 electrons surrounding the core would presumably have
> > the
> > same energy level..." This contradicts observation - see the ionizazion
> > enthalpies of argon you yourself quote lower down.
> >
> I should say relative difference between electrons.
Huh? What is this supposed to mean?
> The difference
> should be about the same for electrons with similar geometric
> configurations. I have updated my web site to make this clearer.
Thanks.
> > 12) You yourself point out that Krypton is not fully symmetric - this
> > is
> > yet another counterexample to your rule "symmetric implies not very
> > reactive".
>
> Yup, this is a very, very, very rough observation and obviously, not a
> rule.
Well, what is the support for this observation? I remember seeing lots
of examples on your web site contradicting this observation, but close
to none agreeing with it.
> > 13) "Expanding this by adding 2 more units to the core, we get Hafnium
> > at
> > 72 electrons. Interestingly, this is not the next nobel gas, so there
> > is a
> > break in the geometric sequence. Some other rule must apply past
> > Xenon."
> >
> > Hint: standard QM explains this quite easily.
>
> What is this something? I'd be curious in seeing how QM solves large
> atom problems.
Solving the Schroedinger equation for the H atom, you get "shells", i.e.
groups of orbitals with the same energy. The sizes of these shells
increases: the first has 2 orbitals, the second 8, the third 18, the
fourth 32... (in general: 2*n^2; for more on this, especially
sub-shells, see lower down). For atoms with more electrons, one
obviously can't use the solutions of the hydrogen atom directly, but a
closer examination reveals that although there are far more electrons,
which all interact with each other, the shells are still roughly valid
(with some exceptions - try reading up on "Hund's rules", for example).
Now the only remaining piece is that if all occupied shells in an
electron are "closed", i.e. fully occupied, the atom is a noble gas
(because closed shells are very stable, and therefore chemical reactions
won't happen easily). Using these rules (which can be derived from QM,
as roughly explained here), it turns out that Hafnium is a noble gas.
This is the "rough rules" approach - works in most cases, but there may
be exceptions (e.g. for really heavy atoms like gold, one needs to take
relativistic effects into account). But OTOH, one can also solve the
Schroedinger equation for atoms directly (by numerical approximations),
and one usually gets the same results - only with much better accuracy.
[snip]
> > 15) "Otherwise, I was unable to stack more than 8 cubes before my model
> > started to fall over. You could say my model lost stability if it
> > wasn't grouped into larger units."
> >
> > Err, what on earth has stability wrt toppling other to do with chemical
> > stability???
> >
>
> Nothing really, I just though that it was curious that problems that I
> was having in the macroscopic world (stacking sugar cubes) could be
> solved by a solution in the microscopic world (bonding He atoms).
Read up on "coincidence".
> > 16) "There seems to be a loose relationship between at atom being made
> > up
> > of helium atoms and being a noble gas. The rule is loose because
> > Krypton
> > and Xenon contain significant non-helium components."
> >
> > There are only six noble gases, and your rule holds only for four of
> > them. I wouldn't call this even a "rule" - I would call this
> > "coincidence". Especially because you have no physical reason to get
> > from "is made up from helium atoms" to "is a noble gas".
>
> I would say more than coincidence,
Why? Did you calculate the probability?
> but certainly not a rule, lots more
> research on chemical reactivity to do here.
Well, seems like if you have a lot to do...
> > "There do appear to be some inconsistencies in the relationship between
> > the atom structure and whether it is a noble element. This is an area
> > of further study."
> >
> > So you even recognize yourself that there is a problem...
> >
>
> Yup, can't explain everything yet. The theory is still young.
I would prefer if you would call it a "model". "theory" is used in
science in general only for well-tested stuff, not for things like your
model which even according to yourself has still lots of unresolved
questions.
> > 17) "You will notice a striking pattern of repeating energy levels.
> > Each
> > of the 7 primary electron shells is like an inverted pyramid. And one
> > noble element is found in each shell. Why should it be arranged like
> > this?"
> >
> > See the standard QM solutions for bound states in a Coulomb potential.
> >
>
> Will look for that when I get Povh,
Povh is about nuclear and particle physics. Stuff like bound states in a
Coulomb potential can be found in books about atomic and molecular
physics.
(although Povh discusses the bound states in a Coulomb potential briefly
when he talks about quarkonium)
> haven't seen an explation in other
> books so far.
This follows from the 2*n^2-rule I mentioned above - which is a
straightforward result of the solution of the Schroedinger equation in a
Coulomb potential.
> My primary sources are the hyperphysics web site,
Quite well written, but has *far* too few information and references to
be really helpful. Can only serve as an introduction to the main ideas.
> I read a college physics text book,
College? Hmmm - I don't know if there is much stuff in there which could
help you...
> a book called "From atoms to quarks" and "The odd quantum".
Could you give the authors, please?
> > 18) "You can imagine that if you shake a cubic atom by adding energy to
> > it, some electrons will be looser than others. The electrons on the
> > outside would take less energy to ionize than electrons contained in
> > the vertical core."
> >
> > Well, in your model, I can't see how one could *ever* get an electron
> > from
> > the inside to become ionized without breaking the whole core! Could you
> > please explain how this is supposed to work? How do you get the
> > electrons
> > from inside to the outside without destroying the whole structure?
>
> The electron is always loosely bound and its appearance in the atom is
> optional which means that there may be a special bond between the
> neutron and proton as well (yes, additional bonding types)
Next unsupported rule/postulate.
> which keeps
> the atom together when it is ionized. I explain that an electron in
> the core can be ionzied since its path to free space is very short. I
> would have to squeeze past the arms, by going between them, but isn't
> blocked on all sides.
There are enough gaps so that it can go out of the core? Sorry, I have
problems visualizing this.
> > 19) "This simple mechanical model shows why we see a triangular pattern
> > in the electron shells."
> >
> > Sorry, I don't understand at all where you get a triangular pattern
> > from in your model.
>
> Ok, lots more research to justify this statement. The only thing I can
> say now is that the bottom 2 electrons of the triangle represent the 2
> electrons in the vertical core. The next level of 6 electrons
> represent the 6 faces of a cube.
If you can't explain it, then you should drop the sentence above.
> > 20) "The number of electrons in the next higher energy level are in
> > groups of four (because of the four sides of a cube),..."
> >
> > Huh? A cube has six faces, as you yourself said just a sentence
> > before. Additionally, there is no grouping of four in the ionization
> > energies of the electrons - only a grouping of 10.
>
> OK, bad terminology. How about saying the 4 vertical sides of a cube?
Oh, much better. I already suspected this. But why is this relevant for
the energy levels?
> I'm not sure we even see groupings of 10.
Again, standard QM can explain this - the crucial term here is
"sub-shells". I omitted this detail above when talking about shells -
they are divided into sub-shells. The shell n contains sub-shells of
sizes 2, 6, 10, 14, ... 4n-2. Summing this up, you get the total size 2
n^2 mentioned above. The sub-shells are called s-orbitals, p-orbitals,
d-orbitals, f-orbitals (and more - but those are in general not
relevant). And again, this follows straightforwardly from solving the
Schroedinger equation in a Coulomb potential.
> In the ionziation chart of
> argon, it appears to have a group of 6, 2, 8, 2 (as far as relative
> ioniztion energies).
Well, looking at where Argon is located in the periodic table, it should
show groups of 6, 2, 6, 2, 2 (roughly - for giving more detail, I would
have to do an actual calculation), with larger jumps after both (6,2).
Looking at the ionization energies you yourself provide, the
*differences* between the energies are (in eV):
1145.2
1265.2
1840
1467
1543
3214 *
1847
26918 **
5426
5816
7651
6545
6720
9554 *
6104
309028 **
29461
I've marked smaller jumps with * and large jumps with **. What do we
see? A small jump after 6, a large one after two more, a smaller one
after six more, and large one again after two more. Nice agreement with
the predictions of standard QM, don't you think? Now, how does your
model explain this grouping?
> > 21) "It is also interesting to note that the atom builds up narrow arms
> > instead of filling in the spaces between the arms."
> >
> > Again: huh? Your model seems to show the exact opposite to me - there
> > are no narrow arms, but the arms are more like pyramids, i.e. the space
> > between the arms *is* filled up.
>
> I think the terminology is poor, the pictures do a better job of
> explaining it. I think you have the concept correct where the arms
> fill up like pyramids. The space I am referring to is the space
> between the tips of the arms.
Could you try to change the terminology, please?
> > 22) "However, we know that the larger atoms can take additional
> > neutrons to form isotopes."
> >
> > Oh, so you *do* know this? However, what you apparently don't know is
> > that
> > *all* atoms can do this (ever heard of Deuterium, Tritium and 3He?),
> > not
> > only the larger ones, and that there are also isotopes with *less*
> > neutrons than the stable ones have.
>
> Yes, all atoms can do this including hydrogen. Placement is less
> certain with hydrogen but possible.
How do you explain that the more neutrons there are, the less stable an
isotope is? And how do you explain that the *less* neutrons there are,
also the less stable an isotope is? Please note that I'm *not* referring
to "stable" in the sense of "doesn't break apart" - I'm talking about
*beta* decay! (beta minus and beta plus).
(hint: standard nuclear physics explains this - by using the "fermi gas
model" for the nucleus, it turns out that it is most stable when it has
about the same number of neutrons as protons)
[snip]
> > 24) "Since they are effectively sheilded by the extended arms,..."
> >
> > Shielded against what? And how does this shielding work, specifically?
> >
>
> As an atom approaches another atom, it is likely to bump into the
> extended arms of the atom rather than reach into the spaces between.
Why? Simple geometry tells me that the space between the arms is
approximately equally large than the arms themselves.
> Also, the tips of the atoms create a bonding area like docking port
> that allows them to combine.
How?
And doesn't this imply that the number of bonds an atom can make
corresponds somehow to the number of its longest arms, or something like
that?
> The alternating series of
> electron/protons means there is always a way that the atoms can be
> oriented to create an attractive bond. There is no such site between
> the arms.
So the chemical bonds are simply electrostatic forces in your model, or
what?
[snip]
> > 26) "Eliminates the problem of why should an electron orbit the
> > nucleus"
> >
> > Since no one claims that it does, I see no need to comment on that
> > section.
> >
>
> I can still only conclude that QM doesn't explain what the electron is
> doing.
It doesn't explain it in terms of common-day language - but it can
describe it mathematically. Lots of people have a problem with the
simple ugly fact that we can describe parts of nature only with maths,
that our "common sense" doesn't apply everywhere. But that's a simple
fact - sorry for you.
If you want to get a "picture" of how QM "describes" an electron, try
reading "The strange world of Quantum Mechanics" by Styer.
> Lots of different opinions on this still.
I don't think so. E.g. Franz and me have agreed on this.
> How an electron can
> be both a wave and a particle also appears to be controversial.
I don't know about any controversy about this among physicists. Only
laymen who haven't ever studied this argue against this.
And, BTW, it's better to say that it is neither wave nor particle, but a
"quantum" - a thing which can show both properties of a particle and of
a wave, and which is described properly by QM.
> I think these 2 areas still remain a fundamental area of fuzzyness for
> QM.
Says the layman.
> > 27) "Eliminates the need for a strong nuclear force.
> >
> > Since all of the protons and neutrons are not packed into a dense speck
> > in the middle of the atom, there is no need to presume that there must be
> > some extremely strong force holding them in there."
> >
> > Thanks for showing that you have no clue about the reasons why the
> > nuclear
> > force was invented (this "packed together" was only one of several
> > reasons).
>
> What other reasons are there, yes I have no clue.
See Povh - he has quite a bit on this.
What immediately springs to mind is the fact that the volume of the
nucleus was observed to be directly proportional to the number of
nucleons in it. This implies the existence of a short-range attractive
force. The same thing follows from a study of the nuclear binding
energies.
> > 28) "Explains the newest pictures of atoms using STM"
> >
> > We have already discussed this. I don't remember you explaining why the
> > authors wrote that the results agree with the predictions of QM - how
> > could this be possible, if QM is wrong?
> >
>
> Yes, the authors believe their results confirm QM. I think alternate
> explainations of their results could exist. The pictures are an
> experimental result, results are always interpreted.
But how do you explain that the picture *predicted* by QM (5d, IIRC)
agrees so closely with the observed picture (5c, IIRC)? How could this
be possible, if QM is wrong? Coincidence?
[snip]
> > 30) "Explains how an alternating series of protons and electrons are
> > stable and why helium is so stable."
> >
> > We have already discussed that, too - and you are still missing a
> > repulsive force.
>
> This is another postulate (yes, unjustified and unexplained) that is
> required to get the theory off the ground,
Now we have at least 10 postulates/rules, probably more.
> but it isn't so hard to
> believe if you think macroscopically.
Why should I? We are discussing microscopic effects.
Do you doubt that the atoms in your hand electrically repulse the atoms
in the table when your hand rests on the table? If yes, why do you doubt
this? If no, why can't you accept that this nicely explains why the
table seems "hard" and impenetrable to your hand?
> QM is fond of saying that the
> world of the microscopic is totally different from the macroscopic,
Well, that's what the experiments (see below) said. Also try reading
Styer.
> but one of the motivators of this theory is to prove this wrong.
You haven't achieved this so far.
The motivations for the invention of QM were mainly:
1) atomic spectra
2) wave-like behaviour of particles, e.g. electron-double-slit
experiment
3) particle-like behaviour of waves, e.g. photo- and Compton-effect.
You have explained so far none of these phenomena. Rutherford scattering
also played its part here - again something which you have only *very*
roughly addressed.
An additional piece of evidence which came a bit later, only after QM
was already invented (but not yet brought into its final, elegant form,
and not yet fully interpreted) were the experiments by Stern and Gerlach
which showed the quantization of magnetic moments. You haven't commented
anything on these experiments so far. See Styer for more on this.
> > 31) "It correctly predicts the relative difference for the first
> > ionization energies for hydrogen and helium"
> >
> > What on earth has the net force on an element to do with the ionization
> > energy? You *do* understand the difference between "force" and
> > "energy", don't you?
> >
> > In the newsgroup you said that your model can predict both of these
> > ionization energies - but here you present only a comparison of the
> > forces. Did you lie?
> >
> > And what about the deviation of 9%? Do you think this is insignificant
> > or what?
>
> I have updated my web site to futher clarify this. This is only
> relative difference and I have assumed a relationship between how
> tightly an element is bound in a system and the ionization energy.
You are not only assuming a "relationship" - you are assuming a
*PROPORTIONALITY*. What's the basis for this assumption?
In reality, this is simply not true. The Coulomb attraction between an
electron and a helium nucleus in a helium anion (He^+) is twice as large
as that between an electron and a hydrogen nucleus in a hydrogen anion
(H^+, i.e. simply a proton), but the ionization energy is not twice as
large, but *four times* as large.
Quote: "(A lot of assumptions, not necessarily true.)"
Don't you think that a model which needs to make a lot of unfounded
assumptions has some problems?
> The
> 9% difference is significant and certanly not like the .0000000000001
> accuracy claimed elsewhere.
Nice that you admit this.
> But once again, close enough to warrant
> further investigation.
Well, Schroedinger got it right to a far better accuracy on the first
try. That should tell you something...
> There are so many unknowns about the bonding
> type and strength, that it is suprising that it is this close.
Well, then why is standard QM able to predict it with such great
accuracy?
> > You completely ignore that standard QM can predict both of these
> > ionization energies with very good accuracy. How is this possible, if
> > standard QM is wrong for the ground state?
> >
>
> As you have stated yourself, just because a theory can predict a great
> number of things with great accuracy, that doesn't prove it is
> correct.
What did I say, specifically? I don't remember this - can you give a
quote?
> Otherwise you would belive Tom's QVP theory which predicts
> stuff with incredible accuracy.
It does? That's new to me. Last time I looked, his predictions about
angular momenta were nonsense.
> > 32) "It correctly predicts the relative ionization energies for the
> > remaining electrons in an atom"
> >
> > Problems with this were already discussed above.
>
> I have added additional clarification in the web site.
And I have outlined above that standard QM explains the observations far
better (jumps after 6 and 2, not after 8 or 4).
> > 33) "It correctly accounts for the results of the Rutherford scattering
> > experiment"
> >
> > This was already discussed - in the meantime you admitted yourself that
> > this claim is overstated. Please correct it.
>
> I have reduced the claim on the website.
Thanks.
[snip]
> > 36) I once made an argument that the differences between chemical and
> > nuclear energies are a factor of 1000 or even 1 000 000, and asked you
> > how
> > your model explains this. I don't remember you ever answering this...
> >
>
> I would explain that chemical reactions largely involve just bonds and
> attractions created by the Columb forces. Nuclear forces require
> rearranging bonds between the neutrons and protons which could have
> energies many times higher. Of course, I admit that I have no clue as
> to the binding mechanisims, so I cannot explain or characterize it.
> But such a difference would explain the different energies between
> chemical and nuclear reactions.
So you *do* need an additional type of force between nucleons? Doesn't
this contradict your claim that your model eliminates the need for the
strong force?
[snip]
Quote from rule 2:
"One electron/proton pair can be considered to be a neutron."
This isn't claiming how a neutron is built or shaped??? And in all your
photos, you keep using the neutron as if it is built exactly like a H
atom.
> It could be
> that a neutron is the same size as a proton
This does contradict your photos.
> and the fundamental atomic
> unit looks more like a triangle of proton,neutron,electron.
Oh, a new model? Could you please decice on this rather basic question
before your proceed?
Models should be tested first on the easy cases before one goes on to
use them on more complicated cases. The easiest cases here are the H
atom and the neutron. So I would really recommend to you to try to get
an adequate description of those first before you use them as building
blocks to go on!
> Maybe the
> proton/neutron form a pair which leaves the electron loose to be
> ionized.
What has all of this to do with the simple fact that your model could
only produce the observed upper limit on the electric dipole moment if
the proton and the electron in the pair had a distance of only 10^(-27)
m?
> The only thing the cubic model requires is that there be an
> atomic unit that can be stacked as I have shown in my model. That is
> easier to explain if everything is cubic and the same size, but not
> required. Until I can come up with the portion of the theory that
> explains atomic particle construction, like the QVP model, I cannot
> say how this works.
And you don't see a problem in the fact that you can't explain the
difference between your two basic building blocks, the H atom and the
neutron?
> -thanks for all the thoughful questions
>
> I have updated my web site to claim that my theory is probably wrong,
> but I'll work hard to prove it right!
Thanks for all the additions and clarifications.
I hope that you will first take time to study what standard physics has
to say about all this, and especially *why* it says this (i.e. which
experiments lead to the theories we have nowadays), before you proceed
to invent even more rules and postulates to "save" your model.
Bye,
Bjoern
FrankH
> > Seems to me that matching on only 11% of the range (10-30 degrees)of
> > the Rutherford formula is not a great deal of accuracy.
>
> *sigh* The paper as well as I have explained in great detail *why* it
> doesn't match there. Also I and Franz pointed out that if one uses
> better experimental equipment, one *does* get better agreement. Your
> behaviour begins to look like plain denial to me.
>
>
> > The MIT paper
> > may have been a simple lab result, but is it an incorrect or odd
> > result, or is it typical of the high-end experimental results?
>
> It is absolutely atypical!
>
>
> > If I go
> > look at Povh, will I see these same results and conclusions?
>
> You will see some selected results and a lot of references to papers
> where you can find more.
>
OK, I finally got a hold of "Particles and Nuclei, 4th Edition" by
Povh, Rith, Scholz, Zetsche.
Looking through the book, I do not see any experimental results from
the standard Rutherford experiment. Section 5.2 on the Rutherford
cross section doesn't contain any experimental data. There are other
types of scattering and some experimental results, but nothing
directly addressing the dependence on E and Z for Rutherford
scattering.
I think I will have to remain skeptical of the claims of remarkable
accuracy for the Rutherford formula. The MIT result was a grad student
result, but the equipment is as modern as it needs to be for accurate
results. Certainly it was much better than in Rutherford's day. Aside
from the detector technology, there isn't much more high end you could
make this relatively simple experiment. If the conclusions in the
paper are true about the high end low end of the range, then I would
suspect that the data should be comparable no matter how accurately
you did the experiment, since the conditions of multiple scattering
would be present in any experimental setup. So a match on 11% of the
range is probably an accurate result. I would say that the scatter
data roughly matches the Rutherford formula, but not to the extent
that you would have to conclude that the theory behind the formula
must be correct.
In investigating the dependence on E, I ran across a cute little web
applet on the web which models protons scattering off a group of
protons/electrons:
http://www.explorelearning.com/index.cfm?method=cResource.dspView&ResourceID=16
I was experimenting with scattering using a group of protons verus
using a alternating series of protons/electrons (like in the cubic
model). By changing the proton speed (E), I could certainly see a
dependence on E on the scattering angle. The proton/electron model
scatters in a much more unpredicatable manner than for a group of
protons, but the relationship of more energy means less scattering
seemed apparent.
>
> > There is a
> > dependency of the cross section on the atom orientation (all 3
> > orientations rho, phi and theta).
>
> The Rutherford cross section depends only on theta. Your model seems to
> predict an additional dependence on phi, if I understand it correctly.
I think we are mixing terminology. Theta is angle at which you measure
scattering, phi doesn't make any sense in this experimental setup. For
the cubic model, there is a dependence on the atom's orientation for
single scattering events (which includes phi), but it would only show
a dependence on theta in an experimental setup.
If you used a gold crystal where all the atoms are lined up the same
way, you could conceiveably control phi and show a dependence on phi.
This is why I suggested doing an experiment with a gold crystal to
determine whether the Rutherford or the cubic model is correct.
Rutherford scattering shouldn't matter no matter the orientation, but
the cubic atomic model should show a strong correlation to
orientation.
>
>
> > I only calculated 2 cross sections
> > for the angles producing the highest high angle scattering events. I'd
> > need a computer program to calculate the rest of the cross sections.
>
> Didn't you say you used an Excel spreadsheet? I don't think it should be
> very hard to change it accordingly.
I would need to do a Monte Carlo type simulation with a program like
the applet I found. Plus, you are right that I don't know enough about
the nature of the repulsive force involved in the scattering to make
an accurate model.
-Franklni
Bjoern Feuerbacher
>
> Bjoern Feuerbacher <bfeu...@ix.urz.uni-heidelberg.de> wrote in message news:<404484FE...@ix.urz.uni-heidelberg.de>...
> > > Seems to me that matching on only 11% of the range (10-30 degrees)of
> > > the Rutherford formula is not a great deal of accuracy.
> >
> > *sigh* The paper as well as I have explained in great detail *why* it
> > doesn't match there. Also I and Franz pointed out that if one uses
> > better experimental equipment, one *does* get better agreement. Your
> > behaviour begins to look like plain denial to me.
> >
> >
> > > The MIT paper
> > > may have been a simple lab result, but is it an incorrect or odd
> > > result, or is it typical of the high-end experimental results?
> >
> > It is absolutely atypical!
> >
> >
> > > If I go
> > > look at Povh, will I see these same results and conclusions?
> >
> > You will see some selected results and a lot of references to papers
> > where you can find more.
> >
>
> OK, I finally got a hold of "Particles and Nuclei, 4th Edition" by
> Povh, Rith, Scholz, Zetsche.
Good! :-)
> Looking through the book, I do not see any experimental results from
> the standard Rutherford experiment.
Sorry.
> Section 5.2 on the Rutherford
> cross section doesn't contain any experimental data.
Aren't there references in that section?
> There are other
> types of scattering and some experimental results, but nothing
> directly addressing the dependence on E and Z for Rutherford
> scattering.
Well, the other types of scattering are *based* on Rutherford scattering
- so if the other types of scattering agree with experiment, this is
essentially a demonstration of the validity of Rutherford's formula,
too.
> I think I will have to remain skeptical of the claims of remarkable
> accuracy for the Rutherford formula.
Or you will have to search a bit harder. After all, these experiments
were first done 90 years ago - no wonder that it's hard to find a
reference today for them!
Oh, BTW, even the original paper showed a nice agreement between theory
and experimental data - a far better agreement than you have achieved so
far...
> The MIT result was a grad student
> result, but the equipment is as modern as it needs to be for accurate
> results.
No. The only really modern stuff there was the equipment used to analyze
the data. The data taking itself wasn't done with really modern
instruments.
> Certainly it was much better than in Rutherford's day.
The data processing, yes. The data taking, no.
And, BTW, I have already explained the discrepancies in that experiment.
Hey, it was even explained in the paper itself!!! You simply dismissed
the explanations as "handwaving" - you didn't even try to understand
them!
> Aside
> from the detector technology, there isn't much more high end you could
> make this relatively simple experiment. If the conclusions in the
> paper are true about the high end low end of the range, then I would
> suspect that the data should be comparable no matter how accurately
> you did the experiment, since the conditions of multiple scattering
> would be present in any experimental setup.
No. Multiple scattering occurs only
1) at low energies and
2) thick targets.
Both parameters can be changed.
> So a match on 11% of the
> range is probably an accurate result.
Well, I did the experiment myself during some lab work, and I got better
results. In contrast to the experiment presented in that paper, I
included also the beam profile in the data analysis, which improved the
agreement between theory and data.
> I would say that the scatter
> data roughly matches the Rutherford formula, but not to the extent
> that you would have to conclude that the theory behind the formula
> must be correct.
What about the plot you yourself presented some weeks ago, which showed
that the energy dependence of the cross section agrees with the
theoretical prediction? There was a much better accuracy in that plot!!!
> In investigating the dependence on E, I ran across a cute little web
> applet on the web which models protons scattering off a group of
> protons/electrons:
>
> http://www.explorelearning.com/index.cfm?method=cResource.dspView&ResourceID=16
>
> I was experimenting with scattering using a group of protons verus
> using a alternating series of protons/electrons (like in the cubic
> model). By changing the proton speed (E), I could certainly see a
> dependence on E on the scattering angle. The proton/electron model
> scatters in a much more unpredicatable manner than for a group of
> protons, but the relationship of more energy means less scattering
> seemed apparent.
In other words, instead of doing a *quantitative* analysis, you are
again only handwaving.
> > > There is a
> > > dependency of the cross section on the atom orientation (all 3
> > > orientations rho, phi and theta).
> >
> > The Rutherford cross section depends only on theta. Your model seems to
> > predict an additional dependence on phi, if I understand it correctly.
>
> I think we are mixing terminology. Theta is angle at which you measure
> scattering, phi doesn't make any sense in this experimental setup.
Wrong. It makes a lot of sense. The experimental setup has a specific
axis (the direction of the incident beam), and phi is the angle of
rotation around that axis.
> For
> the cubic model, there is a dependence on the atom's orientation for
> single scattering events (which includes phi), but it would only show
> a dependence on theta in an experimental setup.
Why?????
> If you used a gold crystal where all the atoms are lined up the same
> way, you could conceiveably control phi and show a dependence on phi.
Well, in the experiment, a gold foil is used. A gold foil *is* a gold
crystal.
> This is why I suggested doing an experiment with a gold crystal to
> determine whether the Rutherford or the cubic model is correct.
Err, the experiment *was* done with a gold crystal. It was done with a
gold foil.
> Rutherford scattering shouldn't matter no matter the orientation, but
> the cubic atomic model should show a strong correlation to
> orientation.
In other words: Rutherford's observations proved you wrong.
[snip]
Bye,
Bjoern
alistair
found evidence that the electronic charge of quarks is made of lots of
partial charges averaging a distance of 10 ^ -13 metres from the
centre of the quark.
However three quarks together in a proton would have partial charges
only
10 ^ -17 metres from one another. I think I'm right but I don't expect
them to publish it.Even though my model predicts the coulomb constant
for electric force k = 9.9 x 10 ^ 10
Bjoern Feuerbacher
>
> I have submitted an article to physical review D saying that I have
> found evidence that the electronic charge of quarks is made of lots of
> partial charges averaging a distance of 10 ^ -13 metres from the
> centre of the quark.
Since quarks are known to have a diameter of less than 10^(-18) metres,
this proposal doesn't look very promising. Hey, even nucleons have only
a diameter of around 10^(-15) metres - how on earth could a quark be 100
times as large as a nucleon???
> However three quarks together in a proton would have partial charges
> only 10 ^ -17 metres from one another.
What about three quarks together in a neutron? In a Lambda? A quark and
an antiquark in a pion? And so on...
> I think I'm right but I don't expect them to publish it.
What's your evidence?
> Even though my model predicts the coulomb constant
> for electric force k = 9.9 x 10 ^ 10
Your physical units are missing here.
Bye,
Bjoern
FrankH
it doesn't need to be justified.
>
> > There are other
> > types of scattering and some experimental results, but nothing
> > directly addressing the dependence on E and Z for Rutherford
> > scattering.
>
> Well, the other types of scattering are *based* on Rutherford scattering
> - so if the other types of scattering agree with experiment, this is
> essentially a demonstration of the validity of Rutherford's formula,
> too.
>
>
> > I think I will have to remain skeptical of the claims of remarkable
> > accuracy for the Rutherford formula.
>
> Or you will have to search a bit harder. After all, these experiments
> were first done 90 years ago - no wonder that it's hard to find a
> reference today for them!
>
> Oh, BTW, even the original paper showed a nice agreement between theory
> and experimental data - a far better agreement than you have achieved so
> far...
>
>
> > The MIT result was a grad student
> > result, but the equipment is as modern as it needs to be for accurate
> > results.
>
> No. The only really modern stuff there was the equipment used to analyze
> the data. The data taking itself wasn't done with really modern
> instruments.
>
>
> > Certainly it was much better than in Rutherford's day.
>
> The data processing, yes. The data taking, no.
I would say that having a solid state electronic detector is much
better than staring through a microscope waiting for flashes on a
screen. This must have been a mind-numbing experiment in Rutherford's
day. I would go as far as to say that it was so difficult to collect
data, that it is nearly impossible to get good results with the
equipment used in Rutherford's day. Only by using modern electronic
detectors/counters which do not have a fallible human component, can
good results be obtained easiliy.
Yes, that particular plot was much more convincing, which is why I am
continuing to investigate the dependence on E for the cubic model.
>
> > In investigating the dependence on E, I ran across a cute little web
> > applet on the web which models protons scattering off a group of
> > protons/electrons:
> >
> > http://www.explorelearning.com/index.cfm?method=cResource.dspView&ResourceID=16
> >
> > I was experimenting with scattering using a group of protons verus
> > using a alternating series of protons/electrons (like in the cubic
> > model). By changing the proton speed (E), I could certainly see a
> > dependence on E on the scattering angle. The proton/electron model
> > scatters in a much more unpredicatable manner than for a group of
> > protons, but the relationship of more energy means less scattering
> > seemed apparent.
>
> In other words, instead of doing a *quantitative* analysis, you are
> again only handwaving.
>
Yes, I'm waving my hands - sorry that's all I can do with the time
I've got! But there is merit to qualitative analysis as well.
>
>
> > > > There is a
> > > > dependency of the cross section on the atom orientation (all 3
> > > > orientations rho, phi and theta).
> > >
> > > The Rutherford cross section depends only on theta. Your model seems to
> > > predict an additional dependence on phi, if I understand it correctly.
> >
> > I think we are mixing terminology. Theta is angle at which you measure
> > scattering, phi doesn't make any sense in this experimental setup.
>
> Wrong. It makes a lot of sense. The experimental setup has a specific
> axis (the direction of the incident beam), and phi is the angle of
> rotation around that axis.
>
>
> > For
> > the cubic model, there is a dependence on the atom's orientation for
> > single scattering events (which includes phi), but it would only show
> > a dependence on theta in an experimental setup.
>
> Why?????
I think I now understand what you mean. Theta is only measured in the
horizontal plane (typically). The detector is placed in a horizontal
plane to the foil. It isn't placed out of that plane, above or below
the foil (phi). The cubic model would still only show a theta
dependence due to the random arrangement of atoms in the gold foil
(assuming it is random). The amount of scattering coming off at any
theta angle would be a matter of statistical chance.
>
>
> > If you used a gold crystal where all the atoms are lined up the same
> > way, you could conceiveably control phi and show a dependence on phi.
>
> Well, in the experiment, a gold foil is used. A gold foil *is* a gold
> crystal.
>
I suppose that would depend on how the foil is made. I thought most
gold foil was made by pounding into ever thinner sheets. This would
certainly lead to a random arrangement. Gold is one of the few
elements that this can be done with. If you made the foil through some
sort of gas deposition process, then this would be crystalline. I've
not seen any reference to the method of manufacture for the gold foil
in what I've read so far.
>
> > This is why I suggested doing an experiment with a gold crystal to
> > determine whether the Rutherford or the cubic model is correct.
>
> Err, the experiment *was* done with a gold crystal. It was done with a
> gold foil.
>
>
> > Rutherford scattering shouldn't matter no matter the orientation, but
> > the cubic atomic model should show a strong correlation to
> > orientation.
>
> In other words: Rutherford's observations proved you wrong.
>
Thus far, I've only seen experimental setups where the detector was on
a horizontal plane with the foil. I haven't seen any results of
experiments where they move the detector around a 360 degree circle at
a fixed theta angle to see if there are any differences in the phi
angle. Ideally, you'd want to place detectors all around in a sphere
surrounding the target so you can record scattering at all possible
angles at once. Do you know of any experiments where that was done
with a crystalline foil? I would think that the cubic model would
predict that you would see wide variances in scattering as you move
about the 360 degree circle, while the rutherford model would predict
no differences using a gold crystal foil. So if you know of such an
experiment, that would prove the cubic model wrong.
>
>
> [snip]
>
> Bye,
> Bjoern
Franz Heymann
Do, if it should please you.
You obviously have not read any of the experiments in which
elsctrons have been scattered off protons and many heavier
nucleii, with results with accuracies in the region of 1%
over essentially the whole angular range.
You have also obviously not read any of the experiments in
which such scattering experiments have been done at
sufficiently high energies to enable *deviations* from the
simple Rutherford formula to be measured,, and hence to
determine both nuclear radii and radiative corrections to
elastic scattering cross sections.
> The MIT result was a grad student
> result,
It was not. It was an undergrad student result.
> but the equipment is as modern as it needs to be for
accurate
> results. Certainly it was much better than in Rutherford's
day.
It may haver had bells and whistles, but boy, did the folk
in Rutherford's time know how to wring the best out of their
equipment!!
> Aside
> from the detector technology, there isn't much more high
end you could
> make this relatively simple experiment. If the conclusions
in the
> paper are true about the high end low end of the range,
It was not a paper. It was an undergrad lab report.
Undergrads are not renowned for their experience at
performing good experiments. I have encountered many who
found it difficult to obtain a good result for measuring the
density of a lump of aluminium or the refractive index of a
piece of glass..
The rest of your letter was mainly babble, so I snipped it.
[snip]
Franz
Franz Heymann
"alistair" <alis...@goforit64.fsnet.co.uk> wrote in message
news:861c1b21.04032...@posting.google.com...
> I have submitted an article to physical review D saying
that I have
> found evidence that the electronic charge of quarks is
made of lots of
> partial charges averaging a distance of 10 ^ -13 metres
from the
> centre of the quark.
How do three quarks with radii of that order fit into a
proton whose radius is about 100 times smaller?
> However three quarks together in a proton would have
partial charges
> only
> 10 ^ -17 metres from one another.
Just above you said 10^-13 metres. Those quarks must have
been inside protons, because no free quarks have ever been
observed.
> I think I'm right
No, you are categorically not right.
> but I don't expect
> them to publish it.
Yes. This time you are right, unless they publish it as an
April fool joke.
> Even though my model predicts the
coulomb constant
> for electric force k = 9.9 x 10 ^ 10
What might the coulomb constant for electric force be?
What was the experiment you did which yielded these results?
Who produced the theory with which you compared the results?
Franz
FrediFizzx
| alistair wrote:
| >
| > I have submitted an article to physical review D saying that I have
| > found evidence that the electronic charge of quarks is made of lots of
| > partial charges averaging a distance of 10 ^ -13 metres from the
| > centre of the quark.
|
| Since quarks are known to have a diameter of less than 10^(-18) metres,
| this proposal doesn't look very promising. Hey, even nucleons have only
| a diameter of around 10^(-15) metres - how on earth could a quark be 100
| times as large as a nucleon???
How big is the effective field of a quark at "normal" energy? It seems to
me that for the trio of quarks in a proton, it would have to be on the order
of the size of a hydrogen atom. Is the effective field part of what we call
quantum objects? I would think that it would be.
FrediFizzx
Bjoern Feuerbacher
>
> Bjoern Feuerbacher <bfeu...@ix.urz.uni-heidelberg.de> wrote in message news:<40616B8E...@ix.urz.uni-heidelberg.de>...
> > FrankH wrote:
[snip]
> > > Section 5.2 on the Rutherford
> > > cross section doesn't contain any experimental data.
> >
> > Aren't there references in that section?
> >
> Abosolutely none. Seems that this is such a foregone conclusion, that
> it doesn't need to be justified.
Seems to me that the stuff is simply so old (and therefore
well-established) that no one bothers anymore to give any references to
experimental data for it.
Have you seen a reference for the measurement of the hydrogen spectrum
lately?
> > > There are other
> > > types of scattering and some experimental results, but nothing
> > > directly addressing the dependence on E and Z for Rutherford
> > > scattering.
> >
> > Well, the other types of scattering are *based* on Rutherford scattering
> > - so if the other types of scattering agree with experiment, this is
> > essentially a demonstration of the validity of Rutherford's formula,
> > too.
Did you get this?
> > > I think I will have to remain skeptical of the claims of remarkable
> > > accuracy for the Rutherford formula.
> >
> > Or you will have to search a bit harder. After all, these experiments
> > were first done 90 years ago - no wonder that it's hard to find a
> > reference today for them!
And this?
> > Oh, BTW, even the original paper showed a nice agreement between theory
> > and experimental data - a far better agreement than you have achieved so
> > far...
And this?
> > > The MIT result was a grad student
> > > result, but the equipment is as modern as it needs to be for accurate
> > > results.
> >
> > No. The only really modern stuff there was the equipment used to analyze
> > the data. The data taking itself wasn't done with really modern
> > instruments.
> >
> >
> > > Certainly it was much better than in Rutherford's day.
> >
> > The data processing, yes. The data taking, no.
>
> I would say that having a solid state electronic detector is much
> better than staring through a microscope waiting for flashes on a
> screen.
This ensures that fewer errors are made, right. So what? For example, in
this (undergraduate!) lab work experiment, apparently no attention was
paid to the beam profile, which is a crucial factor for the outcome.
AFAIK, Geiger and Marsden *did* pay attention to this. High-end
equipment isn't all that's necessary to get a nice outcome of an
experiment - you also need a lot of experience, you need to pay
attention to seemingly small details, and so on. All stuff which can't
be expected from an undergraduate.
> This must have been a mind-numbing experiment in Rutherford's
> day.
Yes.
> I would go as far as to say that it was so difficult to collect
> data, that it is nearly impossible to get good results with the
> equipment used in Rutherford's day.
How do you explain then the nice results in the paper by Geiger and
Marsden?
> Only by using modern electronic
> detectors/counters which do not have a fallible human component, can
> good results be obtained easiliy.
Well, look into all the results reported in Povh (on other scattering
experiments). There you will see the good results you want to have. E.g.
Bhabha scattering should be discussed there, IIRC.
> > And, BTW, I have already explained the discrepancies in that experiment.
> > Hey, it was even explained in the paper itself!!! You simply dismissed
> > the explanations as "handwaving" - you didn't even try to understand
> > them!
Did you get this?
> > > Aside
> > > from the detector technology, there isn't much more high end you could
> > > make this relatively simple experiment. If the conclusions in the
> > > paper are true about the high end low end of the range, then I would
> > > suspect that the data should be comparable no matter how accurately
> > > you did the experiment, since the conditions of multiple scattering
> > > would be present in any experimental setup.
> >
> > No. Multiple scattering occurs only
> > 1) at low energies and
> > 2) thick targets.
> >
> > Both parameters can be changed.
And this?
> > > So a match on 11% of the
> > > range is probably an accurate result.
> >
> > Well, I did the experiment myself during some lab work, and I got better
> > results. In contrast to the experiment presented in that paper, I
> > included also the beam profile in the data analysis, which improved the
> > agreement between theory and data.
And this?
> > > I would say that the scatter
> > > data roughly matches the Rutherford formula, but not to the extent
> > > that you would have to conclude that the theory behind the formula
> > > must be correct.
> >
> > What about the plot you yourself presented some weeks ago, which showed
> > that the energy dependence of the cross section agrees with the
> > theoretical prediction? There was a much better accuracy in that plot!!!
> >
> Yes, that particular plot was much more convincing, which is why I am
> continuing to investigate the dependence on E for the cubic model.
Well, plots with such an accurary are the norm, not the exception in
scattering experiments.
Remember the scanned plots I sent you?
> > > In investigating the dependence on E, I ran across a cute little web
> > > applet on the web which models protons scattering off a group of
> > > protons/electrons:
> > >
> > > http://www.explorelearning.com/index.cfm?method=cResource.dspView&ResourceID=16
> > >
> > > I was experimenting with scattering using a group of protons verus
> > > using a alternating series of protons/electrons (like in the cubic
> > > model). By changing the proton speed (E), I could certainly see a
> > > dependence on E on the scattering angle. The proton/electron model
> > > scatters in a much more unpredicatable manner than for a group of
> > > protons, but the relationship of more energy means less scattering
> > > seemed apparent.
> >
> > In other words, instead of doing a *quantitative* analysis, you are
> > again only handwaving.
> >
> Yes, I'm waving my hands - sorry that's all I can do with the time
> I've got! But there is merit to qualitative analysis as well.
Only little merit. As long as one is doing only qualitative analyses,
one can "explain" a lot of data with a lot of quite different models.
Only when one starts to become quantitative, one can decide which models
is the right one.
> > > > > There is a
> > > > > dependency of the cross section on the atom orientation (all 3
> > > > > orientations rho, phi and theta).
> > > >
> > > > The Rutherford cross section depends only on theta. Your model seems to
> > > > predict an additional dependence on phi, if I understand it correctly.
> > >
> > > I think we are mixing terminology. Theta is angle at which you measure
> > > scattering, phi doesn't make any sense in this experimental setup.
> >
> > Wrong. It makes a lot of sense. The experimental setup has a specific
> > axis (the direction of the incident beam), and phi is the angle of
> > rotation around that axis.
Did you get this?
> > > For
> > > the cubic model, there is a dependence on the atom's orientation for
> > > single scattering events (which includes phi), but it would only show
> > > a dependence on theta in an experimental setup.
> >
> > Why?????
>
> I think I now understand what you mean. Theta is only measured in the
> horizontal plane (typically).
Maybe typical for Rutherford scattering (although I doubt this), but
absolutely atypical in other scattering experiments (e.g. X-ray
scattering in order to determine crystal structure, or think of the 4
pi-detectors in modern experimental particle physics!)
> The detector is placed in a horizontal
> plane to the foil. It isn't placed out of that plane, above or below
> the foil (phi).
Err, no. Phi doesn't measure "above or below the foil".
Think about a cone starting from the target and ending somewhere on a
plane "behind" the target (seen from the projectile) or a sphere around
the target. Obviously the intersection of the cone with the plane or the
sphere is a circular line. The opening angle of the cone is theta, and
phi tells you where on the circular line you are.
> The cubic model would still only show a theta
> dependence due to the random arrangement of atoms in the gold foil
> (assuming it is random). The amount of scattering coming off at any
> theta angle would be a matter of statistical chance.
Why should it be random? Gold is a metal. Metals have lattices with an
ordered structure.
> > > If you used a gold crystal where all the atoms are lined up the same
> > > way, you could conceiveably control phi and show a dependence on phi.
> >
> > Well, in the experiment, a gold foil is used. A gold foil *is* a gold
> > crystal.
> >
> I suppose that would depend on how the foil is made.
I suppose that you have no idea of metals.
> I thought most
> gold foil was made by pounding into ever thinner sheets. This would
> certainly lead to a random arrangement.
Why on earth do you think so??? Even a very thin foil is still a metal.
Metals have ordered lattices. Simple chemistry.
> Gold is one of the few
> elements that this can be done with. If you made the foil through some
> sort of gas deposition process, then this would be crystalline.
Gold is *always* "crystalline" (it has an ordered lattice), no matter
how you produce it.
> I've
> not seen any reference to the method of manufacture for the gold foil
> in what I've read so far.
Because it's irrelevant.
[snip]
> > > Rutherford scattering shouldn't matter no matter the orientation, but
> > > the cubic atomic model should show a strong correlation to
> > > orientation.
> >
> > In other words: Rutherford's observations proved you wrong.
> >
> Thus far, I've only seen experimental setups where the detector was on
> a horizontal plane with the foil. I haven't seen any results of
> experiments where they move the detector around a 360 degree circle at
> a fixed theta angle to see if there are any differences in the phi
> angle.
See above - in modern experiments, often 4pi-detectors are used
(although I don't know if they were used for tests of Rutherford
scattering ever).
> Ideally, you'd want to place detectors all around in a sphere
> surrounding the target so you can record scattering at all possible
> angles at once.
In other words, 4pi-detectors.
> Do you know of any experiments where that was done
> with a crystalline foil?
Sorry, no. I'm working in theoretical physics and don't have much
knowledge about experiments. But I would assume that this was indeed
done. Almost every scattering experiments one can think of (and a lot
one didn't ever think of) have already been done, according to my
experience.
> I would think that the cubic model would
> predict that you would see wide variances in scattering as you move
> about the 360 degree circle, while the rutherford model would predict
> no differences using a gold crystal foil.
Yes.
> So if you know of such an
> experiment, that would prove the cubic model wrong.
I think your model is already proven wrong by the vast difference
between a neutron and a H atom. Your handwaving about "different
binding" explains nothing.
Bye,
Bjoern
Bjoern Feuerbacher
What do you mean by "effective field" and "normal energy"?
Bye,
Bjoern
Bjoern Feuerbacher
>
> Bjoern Feuerbacher <bfeu...@ix.urz.uni-heidelberg.de> wrote in message news:<40616B8E...@ix.urz.uni-heidelberg.de>...
> > FrankH wrote:
> > >
[snip]
> > > Section 5.2 on the Rutherford
> > > cross section doesn't contain any experimental data.
> >
> > Aren't there references in that section?
> >
> Abosolutely none. Seems that this is such a foregone conclusion, that
> it doesn't need to be justified.
Well, a Google search yielded this PDF-File:
<http://bcs.whfreeman.com/tiplermodernphysics4e/content/cat_020/RutherfordPrediction.pdf>
Look at Fig. 4-11 on page 3 of that file. There you see a plot which
shows really nice agreement between the data of Geiger and Marsden and
Rutherford's formula, much better than the plot you found in that
undergraduate lab work report
Also look at Fig. 4-12. Again, good agreement with the predictions of
Rutherford's formula.
I'm waiting for you to reproduce such well-obeyed theoretical
predictions from your model...
[snip]
> The cubic model would still only show a theta
> dependence due to the random arrangement of atoms in the gold foil
> (assuming it is random). The amount of scattering coming off at any
> theta angle would be a matter of statistical chance.
It just occured to me that you destroyed your own model here with this
"random arrangement" claim:
The foils used have a thickness of at least several micrometers - I
think you agree with that. Atoms have a diameter of around 10^(-10) to
10^(-9) meters - I think you agree with that, too? Hence such a foil
consists of at least one thousand, probably more than tens of thousands
of layers of atoms. Agreed so far?
If yes: if the atoms are arranged randomly in the foil, this would mean
that there are no free straight "tunnels" through the foil (because of
the large number of layers, on every straight line through the foil,
there lies at least one, probably many, atoms). Alpha particles which
fly at the foil would hence sooner or later hit the "body" of an atom
(instead of flying through the free space between the arms) - and in
your model, every hit means that it should be deflected. Hence we can
conclude that virtually none of the alpha particles should be able to
pass straight through the foil. Obviously this is totally contrary to
what is observed!
How do you explain that?
You may want to read up on "mean free path lenght", "total absorption
cross section" and stuff like that. Please pay special attention to the
fact that this quantities are energy dependant, too. And please consider
that standard physics has no problem at all explaining this
(*quantitatively*) - essentially this follows all from Rutherford
scattering.
[snip]
Bye,
Bjoern
Franz Heymann
Why should this be so? And if you propose to tell us that
speed is gained, please count in the time to develop and
construct the modern equipment and compare it with the time
taken to set up a classical Rutherford experiment.
> This must have been a mind-numbing experiment in
Rutherford's
> day.
No. Physicists at the Cavendish loved that kind of
experiment.
And just look at the payoff: Goodbye to the plum pudding
model of the atom and hello to the modern concept of atomic
structure.
> I would go as far as to say that it was so difficult to
collect
> data, that it is nearly impossible to get good results
with the
> equipment used in Rutherford's day.
You are wrong. They got enough data to prove the point
anout the existence of a very small atomic nucleus. Better
data would have been overkill. The Rutherford scattering
experiments hailed the birth of modern atomic physics.
> Only by using modern electronic
> detectors/counters which do not have a fallible human
component, can
> good results be obtained easiliy.
Please show in detail where thr original Rutherford
scattering experiments gave bad results due to human
fallibility. And don't just waffle about it. Give chapter
and verse.
Hint: Look at the publications of Geiger and Marsden, who,
if my memory serves me correctly, actually performed the
original experiment.
[snip much bull]
Franz
FrediFizzx
As opposed to the part of the field that runs off to infinity. IOW, the
"near" field that can capture an electron and make an atom. And energy
levels say at room temperature and low speeds. Well, even considering the
entire field that runs off to infinity. Is this part of the quantum object?
FrediFizzx
FrankH
> FrankH wrote:
> >
> > Bjoern Feuerbacher <bfeu...@ix.urz.uni-heidelberg.de> wrote in message news:<40616B8E...@ix.urz.uni-heidelberg.de>...
> > > FrankH wrote:
> > > >
>
> [snip]
>
>
> > > > Section 5.2 on the Rutherford
> > > > cross section doesn't contain any experimental data.
> > >
> > > Aren't there references in that section?
> > >
> > Abosolutely none. Seems that this is such a foregone conclusion, that
> > it doesn't need to be justified.
>
> Well, a Google search yielded this PDF-File:
> <http://bcs.whfreeman.com/tiplermodernphysics4e/content/cat_020/RutherfordPrediction.pdf>
>
> Look at Fig. 4-11 on page 3 of that file. There you see a plot which
> shows really nice agreement between the data of Geiger and Marsden and
> Rutherford's formula, much better than the plot you found in that
> undergraduate lab work report
>
> Also look at Fig. 4-12. Again, good agreement with the predictions of
> Rutherford's formula.
>
This paper IS using the 1911 Geigner and Marsden data and isn't the
result of a separate experiment. As I had said before, it would be
nearly impossible to get good results in 1911, and I would have to say
that they may have fudged the data or selectively picked from their
data to get results that match the theory. In fact, one should always
be suspicious of scientific results that precisely match theory since
you would think experimental error would tend to put the data points
around the theoretical line, not on them as is shown in most of the
Geigner and Marsden data.
I believe I have read that the thickness amounts to several hundred
atoms. Yes, the alpha would end up hitting nearly every one of these
atoms, but just like if you fired a bullet into a room full of packing
peanunts, the bullet isn't going to be deflected significantly,
eventhough it hits nearly every peanut along the path. In my model, a
deflection occurs only in the rare instances where the atom is lined
up just right so it has to pass through the very thick part of the
atom. As I calculated, the would happen rarely. It does, however, slow
down due to the atoms absorbing energy as it passes by. I believe this
helps to explain the stopping power of Gold since an alpha cannot
penetrate very far into a block of Gold. If the gold is thick enough
(I don't think it takes much), nearly all the alphas will get blocked.
I don't think this is very well explained by the Rutherford model
since 99.999% of gold is apparently empty space (according to the
standard model), you'd think an alpha could easily pass through a
block of gold with no problem, but it doesn't. I haven't see an
analysis of the standard model predicting the stopping power of a
particular substance. Somehow, the electrons would have to slow the
alpha down. But considering, we can't even physically model what the
electron is doing around the atom, - it is a quantum wave neither wave
nor particle - or whatever, I don't see how you could do the
calculation.
Timothy Golden
> We all know that the size of the nucleus compared to the total size of
> the atom is small. It appears the main reason for believing this is
> the result of the Rutherford scattering experiments which postulated
> that the nucleus is a small point object containing all the positive
> charge of the nucleus. The data appear to match the experimental
> results quite well, so we assume Rutherford's postulate was correct,
> eventhough, this is still a very indirect measure of the size of the
> nucleus. It is still like firing bullets into a dark room to figure
> out what is inside of the room. In this post, I will show that there
> can be alternative explanations for the scattering results and that
> the nucleus doesn't necessarily have to be a tiny speck within the
> atom.
>
I also have trouble with the Rutherford conclusion and have heard
people say that there is much more evidence but haven't heard them say
anything convincing.
The simplest experiment that I can think of is to measure the
rotational moment of inertia of an atom or a small molecule. For the
accepted model this should be smaller than for the alternative. For
example, suppose a water molecule was spun by an oscillatory electric
field. This would no doubt be difficult to measure but seems far more
substantial than the path of a subatomic particle interacting with an
unknown structure. There may be some way to infer the measurement from
thermodynamics but it is beyond me.
If the accepted model is correct then the water molecule can spin
about its center of mass very easily. The difference in rotational
moment is drastic for your alternative.
> I have proposed a new model of the atom which postulates that atoms
> are simply formed out of alternating sequences of electrons and
> protons. The protons and electrons are arranged in a very particular
> geometric sequence. You could think of this model as 2 sheets of alpha
> particles (helium atoms) intersecting forming an X in the shape of an
> octahedral. I call this the cubic atomic model. The details of this
> theory can be found at:
>
> http://ourworld.compuserve.com/homepages/frankhu/buildatm.htm
>
> This theory has been discussed at some length in the newsgroup:
> http://groups.google.com/groups?q=g:thl1577218448d&dq=&hl=en&lr=&ie=UTF-8&oe=UTF-8&selm=46484c9f.0401151015.438ea93a%40posting.google.com
>
> A consequence of this theory is that the electrons are not orbiting
> the nucleus. They are bound into the nucleus of the atom. Since there
> are no orbiting electrons, the size of the atom should be dependent
> only on the size of the nucleus, since there aren't any electrons to
> make the atom larger than the nucleus. This would mean that the
> nucleus would be much, much larger than is commonly thought. In fact,
> the nucleus should be about the same size as the measured diameter of
> the atom. This is in apparent disagreement with the famous Rutherford
> scattering experiment which showed that the nucleus is a tiny
> positively charged speck in the center of the atom. However, I have
> done some rough calculations using the cubic model to show that the
> same Rutherford scattering results could be reproduced by a much
> larger atom nucleus which follows the cubic theory. The basic premise
> is that the cubic atomic model forms atoms which have very thin edges.
> You can imagine this by taking 2 pieces of paper and have them
> intersect. If you were to shoot bullets through these sheets, the
> bullets would only have to pass through the thin sheets and would be
> undeflected. If you examine the photographs of larger atoms like
> Krypton on my website, you can see how the atoms form an X octahedral
> shape (like raw diamond crystal). The cubic theory postulates that the
> alpha particles are able to pass through the thin arms of the atom
> with virtually no deflection since the arms are not thicker than an
> alpha particle. The only place where the alpha particles can reflect
> with high angles is if it hits and tries to pass through a very thick
> part of the atom. This would be like trying to hit the edges of the
> intersecting sheets. The chance of this happening is fairly remote. I
> have done calculations to determine how frequently you would expect to
> see these reflective collisions and I have compared them with the
> original Rutherford scattering experiment results and I can show that
> the predicted percentage for the alpha particles at particular angles
> for the cubic model roughly match the experimental results. The data
> for the original Rutherford scattering data by GigerMardsden can be
> found at:
>
> http://dbhs.wvusd.k12.ca.us/Chem-History/GeigerMarsden-1913/GeigerMarsden-1913.html
>
> I began my calculations by collecting the statistics on the size of
> the gold atom according to the cubic model. It is 14 units high and
> the arms are 10 units wide from side to side. I approximate this as a
> sphere with a radius of 7. This has a spherical surface of 615. Since
> the cubic atom is symmetrical, the unique orientations are only
> contained in 1 quadrant of the sphere or 1/8 of the surface. This
> corresponds to a 90 degree turn through each of the x,y,z axis. This
> means the area of investigation is only 76. The size of the atomic
> unit representing the area of the top of the atom's core is a 2 X 2
> square with an area of 4. This means there are 76/4 = 19 unique
> orientations can roughly fit into this quadrant with no overlap. There
> are basically only 2 orientations which would result in high angle
> reflections. These are the head-on (alpha tries to pass through core)
> and edge-on (alpha tries to pass through arm edge). For a head on
> orientation, I calculated a 4% chance of hitting the core directly,
> 32% chance the arms get hit and 64% chance of a complete miss. For the
> edge on orientation, I calculated a 20% of hitting an arm and 80%
> chance of a miss or pass through. I plugged these into the 19 possible
> slots with 11 orientations being edge on, 1 orientation being head on
> and the remaining 7 as being orientations where the alpha basically
> passes through. The angles of deflection are based purely on a
> classical elastic collision with the atom. Because the atom is
> effectively a neutral matrix of joined helium atoms, the effect of the
> columb forces deflecting the alpha are negligible.
>
> The calculations show the percentage chance for:
>
> A complete miss or pass through 86.3% Would expect angle <
> 5 degrees
> An arm gets hit 13.1% Would expect any angle 0 -
> 180
> A direct hit of the core .21% Would expect angle 90 -
> 180
>
> This compares to the experimental data which shows:
>
> Deflections less than 5 degrees 79.2%
> Deflections 5 - 22 20.4%
> Deflections greater than 22 .35%
>
>
> The details of this calculation can be found on an excel spreadsheet:
>
> http://ourworld.compuserve.com/homepages/frankhu/ruther.xls
>
> The predictions from the cube model and the actual experimental
> results are not exactly the same by any means, however, they are in
> the same rough ballpark. The main point you should observe is that the
> cubic model is able to predict a scattering pattern whereby the vast
> majority of the alpha particles pass right through (86%), while a tiny
> fraction (.21%) gets deflected through high angles. This scattering
> pattern does not necessarily have to be created by the atom postulated
> by Rutherford as a tiny compact nucleus containing all the positive
> charge. You can basically get the same result from the cubic atomic
> model. A better calculation using a computer model to consider random
> orientation and random alpha may produce results more comparable (or
> not) with the actual experimental data and would provide a more
> detailed range of angles to expect when an arm is struck.
> Unfortunately, I do not have the resources to commit to such a
> calculation.
>
> There would be other experiments to confirm or deny the cubic model
> with Rutherford scattering. Perhaps some of these have already been
> done. I would like to see the Rutherford scattering experiments
> repeated but instead of using gold foil, use a form of crystallized
> gold (octahedral crystal) where we are reasonably sure that the gold
> atoms are all aligned in the same direction, and see how the high
> angle scattering depends on the orientation of the crystal. Based on
> the cubic model, I would predict that the crystal would present very
> little scattering in most directions, since the alpha particles are
> able to pass through the thin parts of the atom, but when the crystal
> is oriented so it hits the very edge of the atom and tries to pass
> through the core or arms, it will bounce back strongly. I would
> predict that we would see lots of scattering at 90 degree crystal
> orientations, which would not be explainable by the Rutherford
> formula. The Rutherford formula would predict the same scattering
> pattern/amounts no matter what the orientation.
>
> Another possible experiment would be to use low speed alpha particles.
> At some point, if the speed of the alpha particles were slow enough,
> it wouldn't be able to penetrate the atoms thin arms and you would see
> almost all of the alphas being deflected at high angles. Rutherford
> would predict that all of the alphas would penetrate no matter how
> slow the alphas were going since there isn't much for the aphas to run
> into and an electron isn't likely to deflect an alpha very much.
>
> If anybody knows the results of these experiments, please post them to
> help confirm or deny the plausability of the cubic atomic model. In
> conclusion, the results of the Rutherford scattering experiments do
> not conclusively prove the notion of the nucleus being a tiny speck in
> the atom with surrounding electron clouds. Thus far, arguments against
> Rutherford have lacked an alternative solid model to base calculations
> on. The cubic atomic theory provides this solid model which you can
> run calculations on to show that it can return results similar to the
> experimental results of the Rutherford scattering experiment.
Franz Heymann
[snip]
> This paper IS using the 1911 Geigner and Marsden data and
isn't the
> result of a separate experiment. As I had said before, it
would be
> nearly impossible to get good results in 1911, and I would
have to say
> that they may have fudged the data or selectively picked
from their
> data to get results that match the theory. In fact, one
should always
> be suspicious of scientific results that precisely match
theory since
> you would think experimental error would tend to put the
data points
> around the theoretical line, not on them as is shown in
most of the
> Geigner and Marsden data.
Before presuming to call Geiger and Marsden fraudsters, it
would behoove you to do a statistical analysis of their data
to see if you have a leg to stand on.
Do you have the temerity to think that in 90 years folk
with much more knowledge of the statistical analysis of data
than you would not have spotted any skullduggery?
You are quite despicable. You have run out of steam, so you
start to accuse others of fudging their data, without even
the slightest shred of evidence.
Go and take a running jump at yourself.
[snip]
Franz
Bjoern Feuerbacher
Huh??? What field are you talking about? The electrostatic field of the
quark? The color field? Or something completely different?
And how do you plan to partition the field into one part which runs off
to infinity and one part which doesn't?
> IOW, the
> "near" field that can capture an electron and make an atom.
Electrons are captured by the electrostatic field of the proton - which
is composed of quarks, true. Btu why do you call that electrostatic
field the "near" field??? Its force depends like 1/r^2 on distance, i.e.
it runs off to infinity.
> And energy
> levels say at room temperature and low speeds.
Sorry, I don't understand what this is supposed to mean.
> Well, even considering the
> entire field that runs off to infinity. Is this part of the quantum object?
Since I still not know what field you are talking about (the
electrostatic field?), I can't answer this question, sorry.
Bye,
Bjoern
Franz Heymann
"Timothy Golden" <tttp...@yahoo.com> wrote in message
news:5c7f8ade.04032...@posting.google.com...
[snipo]
> I also have trouble with the Rutherford conclusion and
have heard
> people say that there is much more evidence but haven't
heard them say
> anything convincing.
That is entirely your problem. Why don't you read the
professional journals in which elastic scattering
experiments are described and analysed? There are thousands
of such papers. I do mean *thousands*, involving the
scattering of electrons off a vast number of target
particles ranging from other electrons, positrons, protons,
deuterons and a large selection of nucleii all the way up to
gold.
Don't come back and say "but electrons are not alpha
particles", because electrons obey the *same* Rutherford
scattering law as alpha particles.
By the 1950's, the Rutherford scattering phenomenon had been
established so thoroughly that it became possible to
interpret the deviations from Rutherford scattering at high
energies in terms of the finite-sized charge distributions
of the target particles.
>
> The simplest experiment that I can think of is to measure
the
> rotational moment of inertia of an atom or a small
molecule. For the
> accepted model this should be smaller than for the
alternative. For
> example, suppose a water molecule was spun by an
oscillatory electric
> field. This would no doubt be difficult to measure but
seems far more
> substantial than the path of a subatomic particle
interacting with an
> unknown structure. There may be some way to infer the
measurement from
> thermodynamics but it is beyond me.
> If the accepted model is correct then the water molecule
can spin
> about its center of mass very easily. The difference in
rotational
> moment is drastic for your alternative.
That little lot has sweet fanny adams to do with determining
the sizes of atomic nucleii.
Nuclear magnetic resonance, which is what you appear to be
waffling about, is a well understood phenomenon. The radius
of the nucleus does not appear in the expression for the
resonance frequency.
[snip]
Franz
Bjoern Feuerbacher
>
> Bjoern Feuerbacher <bfeu...@ix.urz.uni-heidelberg.de> wrote in message news:<4062C793...@ix.urz.uni-heidelberg.de>...
> > FrankH wrote:
> > >
> > > Bjoern Feuerbacher <bfeu...@ix.urz.uni-heidelberg.de> wrote in message news:<40616B8E...@ix.urz.uni-heidelberg.de>...
> > > > FrankH wrote:
> > > > >
> >
> > [snip]
> >
> >
> > > > > Section 5.2 on the Rutherford
> > > > > cross section doesn't contain any experimental data.
> > > >
> > > > Aren't there references in that section?
> > > >
> > > Abosolutely none. Seems that this is such a foregone conclusion, that
> > > it doesn't need to be justified.
> >
> > Well, a Google search yielded this PDF-File:
> > <http://bcs.whfreeman.com/tiplermodernphysics4e/content/cat_020/RutherfordPrediction.pdf>
> >
> > Look at Fig. 4-11 on page 3 of that file. There you see a plot which
> > shows really nice agreement between the data of Geiger and Marsden and
> > Rutherford's formula, much better than the plot you found in that
> > undergraduate lab work report
> >
> > Also look at Fig. 4-12. Again, good agreement with the predictions of
> > Rutherford's formula.
> >
>
> This paper IS using the 1911 Geigner and Marsden data and isn't the
> result of a separate experiment.
Err, they did do *several* experiments back then. The data shown there
comes from a *later* experiment, done to *test* Rutherford's formula,
not the original one, which led to the development of that formula.
> As I had said before, it would be
> nearly impossible to get good results in 1911,
Why?
And how do you explain the good agreement between theoretical prediction
and experimental observations then?
And didn't it ever occur to you that if it had *really* been nearly
impossible to get such good results back then, all the other scientists
would have *known* this and therefore the results by Geiger and Marsden
wouldn't have gotten accepted ever? Essentially you accuse not only
Geiger and Marsden of fraud - you also accuse all the other scientists
back then who accepted their results as genuine of incompetence!!!
> and I would have to say
> that they may have fudged the data or selectively picked from their
> data to get results that match the theory.
Congratulations! You have wandered off from "someone how is simply
ignorant about physics and has therefore proposed a model which doesn't
fit the observations, but is willing to learn and willing to be
disproved by the exerimental data" into pure crackpottery. That's sooooo
typical for crackpots: if the experimental data disproves your model,
don't admit that you are wrong. Simply claim that the experimental data
is wrong, that the experimentalists were stupid, incompetent or
dishonest and commited fraud.
Would you please withdraw this completely unsupport assault on Geiger
and Marsden's integrity? Or at least provide a bit of evidence to
support it, instead of only "I don't like their results, hence they have
to be wrong" and "they surely weren't able to make accurate measurements
back then"?
> In fact, one should always
> be suspicious of scientific results that precisely match theory
Err, the data there doesn't "precisely" match theory. There still are
small deviations. So what are you talking about?
I see that you are using a "head - I win; tails - you lose" approach. If
the data doesn't agree with the theory well, you claim that this is
evidence that the theory is wrong. If the data does agree with the
theory well, you claim that such good agreement isn't possible, and
hence there was probably fraud involved.
> since
> you would think experimental error would tend to put the data points
> around the theoretical line,
> not on them as is shown in most of the Geigner and Marsden data.
Wrong. The data do lie scattered around the curve. The deviations are
small, but they are there.
> > I'm waiting for you to reproduce such well-obeyed theoretical
> > predictions from your model...
Hello?
> > > The cubic model would still only show a theta
> > > dependence due to the random arrangement of atoms in the gold foil
> > > (assuming it is random). The amount of scattering coming off at any
> > > theta angle would be a matter of statistical chance.
> >
> > It just occured to me that you destroyed your own model here with this
> > "random arrangement" claim:
> >
> > The foils used have a thickness of at least several micrometers - I
> > think you agree with that. Atoms have a diameter of around 10^(-10) to
> > 10^(-9) meters - I think you agree with that, too? Hence such a foil
> > consists of at least one thousand, probably more than tens of thousands
> > of layers of atoms. Agreed so far?
> >
> > If yes: if the atoms are arranged randomly in the foil, this would mean
> > that there are no free straight "tunnels" through the foil (because of
> > the large number of layers, on every straight line through the foil,
> > there lies at least one, probably many, atoms). Alpha particles which
> > fly at the foil would hence sooner or later hit the "body" of an atom
> > (instead of flying through the free space between the arms) - and in
> > your model, every hit means that it should be deflected. Hence we can
> > conclude that virtually none of the alpha particles should be able to
> > pass straight through the foil. Obviously this is totally contrary to
> > what is observed!
> >
> > How do you explain that?
>
> I believe I have read that the thickness amounts to several hundred
> atoms.
Apparently either there were far thinner foils used, or you read wrong.
For foils with a diameter of several micrometers, it should be several
thousand layers, not only several hundreds.
> Yes, the alpha would end up hitting nearly every one of these
> atoms, but just like if you fired a bullet into a room full of packing
> peanunts, the bullet isn't going to be deflected significantly,
> eventhough it hits nearly every peanut along the path.
Err, a bullet would go through the peanuts. Are the alpha particles able
to go through the atoms, too, in your model? If not, that's a totally
false analogy.
> In my model, a
> deflection occurs only in the rare instances where the atom is lined
> up just right so it has to pass through the very thick part of the
> atom.
If you have thousands of layers, this isn't a rare instance anymore.
This is almost certainly to occur for *every* alpha particle.
> As I calculated, the would happen rarely.
Your calculation applied to one atom, not to thousands of layers!
> It does, however, slow
> down due to the atoms absorbing energy as it passes by. I believe this
> helps to explain the stopping power of Gold since an alpha cannot
> penetrate very far into a block of Gold. If the gold is thick enough
> (I don't think it takes much), nearly all the alphas will get blocked.
Can you reproduce the dependence of the penetration depth on the energy
with your model? Standard physics can.
> I don't think this is very well explained by the Rutherford model
> since 99.999% of gold is apparently empty space (according to the
> standard model), you'd think an alpha could easily pass through a
> block of gold with no problem, but it doesn't.
Err, ever heard of electrostatic forces? *They* do most of the stopping.
You may try to read up on "Bethe-Bloch formula". Again a formula based
on standard physics which agrees very nicely with the experimental
observations.
You could look here, for example (that's the first page I found in a
Google search, and it looks really nice - complete with experimental
data):
<http://besch2.physik.uni-siegen.de/~depac/DePAC/DePAC_tutorial_database/grupen_istanbul/node9.html>
> I haven't see an
> analysis of the standard model predicting the stopping power of a
> particular substance.
Then you haven't looked very hard. The Bethe-Bloch formula is rather
basic knowledge in particle physics.
> Somehow, the electrons would have to slow the
> alpha down.
The electrostatic forces, to be more precise.
> But considering, we can't even physically model what the
> electron is doing around the atom,
We can. Why do you think QM isn't a physical description?
>- it is a quantum wave neither wave
> nor particle -
Yes. So what? What's "unphysical" about that?
> or whatever, I don't see how you could do the calculation.
See the page above. Essentially you can do a calculation like this
classically - you don't need to know what the electrons are doing in the
atom, you only need to know how electrostatic forces and energy and
momentum conservation work.
> > You may want to read up on "mean free path lenght", "total absorption
> > cross section" and stuff like that. Please pay special attention to the
> > fact that this quantities are energy dependant, too. And please consider
> > that standard physics has no problem at all explaining this
> > (*quantitatively*) - essentially this follows all from Rutherford
> > scattering.
Did you get this?
Bye,
Bjoern
alistair
The mass can have a diameter of 10^ -18 metres.I think that charge is
a kind of mass and both are particles.
Bjoern Feuerbacher
>
> frank...@yahoo.com (FrankH) wrote in message news:<46484c9f.04022...@posting.google.com>...
> > We all know that the size of the nucleus compared to the total size of
> > the atom is small. It appears the main reason for believing this is
> > the result of the Rutherford scattering experiments which postulated
> > that the nucleus is a small point object containing all the positive
> > charge of the nucleus. The data appear to match the experimental
> > results quite well, so we assume Rutherford's postulate was correct,
> > eventhough, this is still a very indirect measure of the size of the
> > nucleus. It is still like firing bullets into a dark room to figure
> > out what is inside of the room. In this post, I will show that there
> > can be alternative explanations for the scattering results and that
> > the nucleus doesn't necessarily have to be a tiny speck within the
> > atom.
> >
>
> I also have trouble with the Rutherford conclusion
Why?
> and have heard
> people say that there is much more evidence but haven't heard them say
> anything convincing.
What's unconvincing about the data taken by Geiger and Marsden?
> The simplest experiment that I can think of is to measure the
> rotational moment of inertia of an atom or a small molecule.
Rotational moments of inertia *were* measured (by looking at the
rotational spectra of the molecules). The results agreed with the
predictions of QM for the binding lengths, AFAIK.
> For the
> accepted model this should be smaller than for the alternative.
Do you mean Franklin's alternative?
> For
> example, suppose a water molecule was spun by an oscillatory electric
> field.
How should that work, specifically?
> This would no doubt be difficult to measure
How do you plan to measure this, specifically?
> but seems far more
> substantial than the path of a subatomic particle interacting with an
> unknown structure.
Well, Rutherford scattering has been done with different projectiles and
different targets. The results always agreed with the predictions. So
where is your problem?
And why is measuring the rotational moment of inertia more "substantial"
than this?
> There may be some way to infer the measurement from
> thermodynamics but it is beyond me.
Huh?
> If the accepted model is correct then the water molecule can spin
> about its center of mass very easily.
You mean, its rotational moment of inertia should be small, or what?
> The difference in rotational
> moment is drastic for your alternative.
Probably yes.
[snip]
Bye,
Bjoern
Bjoern Feuerbacher
>
> I think that the charge and mass are in two separate locations.
How should that work?
> The mass can have a diameter of 10^ -18 metres.
Mass is a property of particles. Physical properties have no diameter.
You are making no sense.
> I think that charge is a kind of mass
In what sense?
>and both are particles.
Mass and charge are *properties* of particles. You make no sense.
Bye,
Bjoern
Franz Heymann
I think you are trying to say that charge and mass may
possibly not have the same spatial distribution.
That might well be right. In the case of the electron, for
instance, its charge appears to be concentrated at a point,
and yet its mass may be distribured in space, because it is
caused by fields distributed around the centre where the
charge is located.
Franz
Timothy Golden
> Timothy Golden wrote:
> >
> > frank...@yahoo.com (FrankH) wrote in message news:<46484c9f.04022...@posting.google.com>...
> > > We all know that the size of the nucleus compared to the total size of
> > > the atom is small. It appears the main reason for believing this is
> > > the result of the Rutherford scattering experiments which postulated
> > > that the nucleus is a small point object containing all the positive
> > > charge of the nucleus. The data appear to match the experimental
> > > results quite well, so we assume Rutherford's postulate was correct,
> > > eventhough, this is still a very indirect measure of the size of the
> > > nucleus. It is still like firing bullets into a dark room to figure
> > > out what is inside of the room. In this post, I will show that there
> > > can be alternative explanations for the scattering results and that
> > > the nucleus doesn't necessarily have to be a tiny speck within the
> > > atom.
> > >
> >
> > I also have trouble with the Rutherford conclusion
>
> Why?
>
>
> > and have heard
> > people say that there is much more evidence but haven't heard them say
> > anything convincing.
>
> What's unconvincing about the data taken by Geiger and Marsden?
>
It's not so much the data as it is the conclusion.
>
> > The simplest experiment that I can think of is to measure the
> > rotational moment of inertia of an atom or a small molecule.
>
> Rotational moments of inertia *were* measured (by looking at the
> rotational spectra of the molecules). The results agreed with the
> predictions of QM for the binding lengths, AFAIK.
>
I am not aware of these measurements. I'll see what I can find.
>
> > For the
> > accepted model this should be smaller than for the alternative.
>
> Do you mean Franklin's alternative?
Yes
>
>
> > For
> > example, suppose a water molecule was spun by an oscillatory electric
> > field.
>
> How should that work, specifically?
The water molecule is net neutral but has two dipoles on it. Placing a
cold one in an electric field should orient the molecule without
translating it very much. changing the field the molecule should track
that change. Enough molecules in an E field could produce a measurable
impedance. This impedance would be due to the rotation of the
molecules. This could just be two plates with a reversing field.
> > This would no doubt be difficult to measure
> How do you plan to measure this, specifically?
I don't really know if that impedance is measurable. It probably
isn't.
>
>
> > but seems far more
> > substantial than the path of a subatomic particle interacting with an
> > unknown structure.
>
> Well, Rutherford scattering has been done with different projectiles and
> different targets. The results always agreed with the predictions. So
> where is your problem?
It's the conclusion of the tiny concentrated nucleus.
>
> And why is measuring the rotational moment of inertia more "substantial"
> than this?
If the rotational moment can be measured then it can either agree or
disagree with the atomic model. The two extremes are a shell model
with larger rotational inertia and the accepted (Rutherford) model
with smaller rotational inertia (none if the nucleus doesn't respond
to changes in the molecular orientation)
>
>
> > There may be some way to infer the measurement from
> > thermodynamics but it is beyond me.
>
> Huh?
Part of kinetic energy is rotational and part is translational.
There are certain properties of matter such as freezing point that
should relate to these kinetic energies. A crystal of ice has some
kinetic energy but it is just vibrating translationally and
rotationally. The seperability of the rotational and translational
properties is what I would hunt for.
>
>
> > If the accepted model is correct then the water molecule can spin
> > about its center of mass very easily.
>
> You mean, its rotational moment of inertia should be small, or what?
Yes, relative to the alternative.
> > The difference in rotational
> > moment is drastic for your alternative.
>
> Probably yes.
Thank You.
>
>
> [snip]
>
>
> Bye,
> Bjoern
-Tim
Franz Heymann
wrote in message
news:<406409B4...@ix.urz.uni-heidelberg.de>...
[snip]
> > What's unconvincing about the data taken by Geiger and
Marsden?
> It's not so much the data as it is the conclusion.
The data is in good agreement with what was predicted by the
Rutherford scattering theory.
What is left to be unconvinced about?
I thought the experiment was supposed to be simpler than a
plain old elastic scattering experiment.
So what does your experiment tell you about the charge
distribution of an atomic nucleus?
> > > This would no doubt be difficult to measure
> > How do you plan to measure this, specifically?
>
> I don't really know if that impedance is measurable. It
probably
> isn't.
Yes, indeed, you are right. The main problem is that
because of collissions with neighbouring molecules, the
effective "Q" of your oscillating molecules would be so
feeble as to wipe out the resonance for practical purposes.
> > > but seems far more
> > > substantial than the path of a subatomic particle
interacting with an
> > > unknown structure.
> >
> > Well, Rutherford scattering has been done with different
projectiles and
> > different targets. The results always agreed with the
predictions. So
> > where is your problem?
>
> It's the conclusion of the tiny concentrated nucleus.
This is the second time you said that. I am not going to
reiterate why you are quite unjustified about doubting the
conclusions.
>
> >
> > And why is measuring the rotational moment of inertia
more "substantial"
> > than this?
>
> If the rotational moment can be measured then it can
either agree or
> disagree with the atomic model.
Remember we are talking about the charge distribution of the
nucleus. Your experiment has absolutely nothing to do with
that.
> The two extremes are a shell model
> with larger rotational inertia and the accepted
(Rutherford) model
> with smaller rotational inertia (none if the nucleus
doesn't respond
> to changes in the molecular orientation)
The plain unadorned Rutheford differential cross section
refers specifically to a point nucleus. Any experiment
which gives results in agreement with the Rutherford cross
section effectively provides the answer that the target
nucleus has its charge located at a point, limited by the
effective resolution of that experiment.
What you folk who stand and shout from the side lines appear
to be blissfully unaware of, is that 55 years ago, the
theory of elastic scattering by a Coulomb field was
developed further, to the extent of providing quantitative
predictions of how the differential cross section would
differ from the simple Rutherford formula, particularly at
larger scattering angles, if the charge was not all located
at a point. These effects become more and more marked as
the energy of the projectile is increased.
The current position is that the RMS radius of the charge
distribution is now known very accurately for a large number
of nucleii, from the proton itself, to as far up as Gold.
In the latter case, the scattering experiments are so
refined that one can even begin to see the radial variations
in charge density associated with the shell model.
> > > There may be some way to infer the measurement from
> > > thermodynamics but it is beyond me.
> >
> > Huh?
>
> Part of kinetic energy is rotational and part is
translational.
> There are certain properties of matter such as freezing
point that
> should relate to these kinetic energies. A crystal of ice
has some
> kinetic energy but it is just vibrating translationally
and
> rotationally. The seperability of the rotational and
translational
> properties is what I would hunt for.
The length scales characteristic of such things as freezing,
or molecular rotations are typically molecular sizes, i.e.
somewhere in the ball park of
10^-10 metres. Nuclear sizes, of which we are speaking, are
typically in the ball park of 10^-15 metres, which is
100,000 times smaller. Draw your own conclusions.
[snip]
Franz
FrediFizzx
Well, of course it has to be the coulomb field further away since the color
field is very short ranged.
| And how do you plan to partition the field into one part which runs off
| to infinity and one part which doesn't?
It is energy dependent. I would think it goes like this; shoot positrons at
a proton at lower energies, then they are deflected by the coulomb field.
When you get to a certain level energy-wise, the electrons are no longer
deflected by the coulomb field of course taking into consideration their
trajectory compared to where the proton is. So it seems that there is a
balance point of energy to "size" so it seems you should be able to specify
and near field size with respect to energy level that we could call a near
field aspect.
| > IOW, the
| > "near" field that can capture an electron and make an atom.
|
| Electrons are captured by the electrostatic field of the proton - which
| is composed of quarks, true. Btu why do you call that electrostatic
| field the "near" field??? Its force depends like 1/r^2 on distance, i.e.
| it runs off to infinity.
Well, actually I was calling it "effective" field instead of near field
which is maybe a better description. IOW, effective field size is atomic
scale size.
| > And energy
| > levels say at room temperature and low speeds.
|
| Sorry, I don't understand what this is supposed to mean.
Well you must know what room temperature means so I guess you are having
trouble with low speeds? See above for energy level.
| > Well, even considering the
| > entire field that runs off to infinity. Is this part of the quantum
object?
|
| Since I still not know what field you are talking about (the
| electrostatic field?), I can't answer this question, sorry.
Whoops; never mind about this part. I lost my mind for a minute. ;-)
FrediFizzx
Bjoern Feuerbacher
>
> Bjoern Feuerbacher <bfeu...@ix.urz.uni-heidelberg.de> wrote in message news:<406409B4...@ix.urz.uni-heidelberg.de>...
> > Timothy Golden wrote:
> > >
> > > frank...@yahoo.com (FrankH) wrote in message news:<46484c9f.04022...@posting.google.com>...
> > > > We all know that the size of the nucleus compared to the total size of
> > > > the atom is small. It appears the main reason for believing this is
> > > > the result of the Rutherford scattering experiments which postulated
> > > > that the nucleus is a small point object containing all the positive
> > > > charge of the nucleus. The data appear to match the experimental
> > > > results quite well, so we assume Rutherford's postulate was correct,
> > > > eventhough, this is still a very indirect measure of the size of the
> > > > nucleus. It is still like firing bullets into a dark room to figure
> > > > out what is inside of the room. In this post, I will show that there
> > > > can be alternative explanations for the scattering results and that
> > > > the nucleus doesn't necessarily have to be a tiny speck within the
> > > > atom.
> > > >
> > >
> > > I also have trouble with the Rutherford conclusion
> >
> > Why?
> >
> >
> > > and have heard
> > > people say that there is much more evidence but haven't heard them say
> > > anything convincing.
> >
> > What's unconvincing about the data taken by Geiger and Marsden?
> >
>
> It's not so much the data as it is the conclusion.
If you can present another atom model which describes the data as well
as Rutherford's model, feel free to give it here.
> > > The simplest experiment that I can think of is to measure the
> > > rotational moment of inertia of an atom or a small molecule.
> >
> > Rotational moments of inertia *were* measured (by looking at the
> > rotational spectra of the molecules). The results agreed with the
> > predictions of QM for the binding lengths, AFAIK.
> >
>
> I am not aware of these measurements. I'll see what I can find.
IIRC, I read it in the book by Haken and Wolf on atomic and molecular
physics, and also in the book by Mayer-Kuckuk. Unfortunately, both are
German, and I don't know if English translations are available...
> > > For the
> > > accepted model this should be smaller than for the alternative.
> >
> > Do you mean Franklin's alternative?
>
> Yes
It's not immediately clear to me why in his model the inertial moment of
rotation should be greater, sorry.
> > > For
> > > example, suppose a water molecule was spun by an oscillatory electric
> > > field.
> >
> > How should that work, specifically?
>
> The water molecule is net neutral but has two dipoles on it. Placing a
> cold one in an electric field should orient the molecule without
> translating it very much. changing the field the molecule should track
> that change. Enough molecules in an E field could produce a measurable
> impedance. This impedance would be due to the rotation of the
> molecules. This could just be two plates with a reversing field.
Yes, that would probably work... Although one would have to do a closer
analysis of this, paying attention to the quantization of angular
momentum.
But it's still not entirely clear to me how you plan to measure the
inertial moment with this experiment. By measuring how fast the
molecules follow the changes of the E field, or what???
[snip]
> > > but seems far more
> > > substantial than the path of a subatomic particle interacting with an
> > > unknown structure.
> >
> > Well, Rutherford scattering has been done with different projectiles and
> > different targets. The results always agreed with the predictions. So
> > where is your problem?
>
> It's the conclusion of the tiny concentrated nucleus.
As said above: if you have a different model which reproduces the data
with the same accuracy, feel free to provide it.
> > And why is measuring the rotational moment of inertia more "substantial"
> > than this?
>
> If the rotational moment can be measured then it can either agree or
> disagree with the atomic model. The two extremes are a shell model
> with larger rotational inertia and the accepted (Rutherford) model
> with smaller rotational inertia (none if the nucleus doesn't respond
> to changes in the molecular orientation)
What do you mean by "respond" here?
> > > There may be some way to infer the measurement from
> > > thermodynamics but it is beyond me.
> >
> > Huh?
>
> Part of kinetic energy is rotational and part is translational.
Agreed.
> There are certain properties of matter such as freezing point that
> should relate to these kinetic energies.
You are losing me again. What do you mean by "relate" here?
> A crystal of ice has some
> kinetic energy but it is just vibrating translationally and
> rotationally. The seperability of the rotational and translational
> properties is what I would hunt for.
How?
[snip]
Bye,
Bjoern
Bjoern Feuerbacher
You talked about a "part" of the field. What was that supposed to mean?
> | And how do you plan to partition the field into one part which runs off
> | to infinity and one part which doesn't?
>
> It is energy dependent.
What is energy dependent?
> I would think it goes like this; shoot positrons at
> a proton at lower energies, then they are deflected by the coulomb field.
> When you get to a certain level energy-wise, the electrons are no longer
> deflected by the coulomb field
Why do you think so???
> of course taking into consideration their
> trajectory compared to where the proton is.
Sorry, I can't follow you here.
> So it seems that there is a
> balance point of energy to "size" so it seems you should be able to specify
> and near field size with respect to energy level that we could call a near
> field aspect.
I have no clue what you are trying to say here, sorry.
> | > IOW, the
> | > "near" field that can capture an electron and make an atom.
> |
> | Electrons are captured by the electrostatic field of the proton - which
> | is composed of quarks, true. Btu why do you call that electrostatic
> | field the "near" field??? Its force depends like 1/r^2 on distance, i.e.
> | it runs off to infinity.
>
> Well, actually I was calling it "effective" field instead of near field
> which is maybe a better description. IOW, effective field size is atomic
> scale size.
You still haven't explained what you *mean* by "effective field". And
why should the size of that field equal to the atomic scale?
> | > And energy
> | > levels say at room temperature and low speeds.
> |
> | Sorry, I don't understand what this is supposed to mean.
>
> Well you must know what room temperature means so I guess you are having
> trouble with low speeds? See above for energy level.
Energy levels of *what*? And what do you mean by "energy levels at room
temperature"? Do you mean "energy corresponding to room temperature, by
E = k_B T", or what???
[snip]
Bye,
Bjoern
FrediFizzx
Never mind. I think I have figured out what I am doing wrong here. But the
question still remains in my thinking that the "size" of a quantum object is
dependent on the momentum transferred in the interaction. In the case of
shooting positrons at a proton, the more momentum the positron has, the
closer it can get to the proton. So the question would be what is really
the "size" of a proton with respect to energy level of the interaction? A
proton is said to have a certain charge radius, but is this charge radius
dependent on a certain energy level of the interaction that measured it?
FrediFizzx
Franz Heymann
"FrediFizzx" <fredi...@hotmail.com> wrote in message
news:c42m5d$2e1hm8$1...@ID-185976.news.uni-berlin.de...
[snip]
> Well, of course it has to be the coulomb field further
away since the color
> field is very short ranged.
Oh? Asymptotic freedom and all that as the separation gets
smaller?
>
> | And how do you plan to partition the field into one part
which runs off
> | to infinity and one part which doesn't?
>
> It is energy dependent. I would think it goes like this;
shoot positrons at
> a proton at lower energies, then they are deflected by the
coulomb field.
> When you get to a certain level energy-wise, the electrons
are no longer
> deflected by the coulomb field of course taking into
consideration their
> trajectory compared to where the proton is.
Electrons are, for practical purposes, deflected *only* by
the Coulomb interaction.
[snip]
Franz
FrediFizzx
news:c46b7d$kp6$9...@sparta.btinternet.com...
Yes, I had this somewhat messed up. Sorry. The point I was trying to make
was that when shooting positrons at a proton, it takes more and more
momentum for the positron to get closer and closer to the proton. IOW, take
a positron with a certain momentum and after shooting a bunch of positrons
with the same momentum at the proton, you will get a certain radius that
depends on the momentum of the positron due to the coulomb replusion. So
how does one get around this momentum dependancy when measuring the charge
radius of the proton?
FrediFizzx
Steve Harris sbharris@ROMAN9.netcom.com
> Yes, I had this somewhat messed up. Sorry. The point I was trying to make
> was that when shooting positrons at a proton, it takes more and more
> momentum for the positron to get closer and closer to the proton. IOW, take
> a positron with a certain momentum and after shooting a bunch of positrons
> with the same momentum at the proton, you will get a certain radius that
> depends on the momentum of the positron due to the coulomb replusion. So
> how does one get around this momentum dependancy when measuring the charge
> radius of the proton?
COMMENT:
Well, you simply note that inside a certain radius for the proton,
Coulombic attraction/repulsion no longer holds up in calculating
momentum transfer. It looks like the proton charge is less than it
"should be" for the radius you're at, because actually the incoming
lepton has penetrated part of the proton's "charge-jelly," and doesn't
feel it.
Think of penetrating into the Earth-- gravity is maximal at the
surface, but it actually decreases as you go below the surface. If you
had an object like a mini-black hole which you were measuring the mass
of the earth with, by shooting it in on ever closer and closer
hyperbolic orbits, you'd get your maximal acceleration on
surface-grazing passes, but the effective mass of the primary you
measured by orbital parameters and radius from the center, would
DECREASE as you looked at passes which penetrated the Earth and went
on through. By a careful analysis of this, you could even generate a
density map of the Earth as a function of radius. That's what the
form/structure factors do for deep elastic scattering of leptons off
baryons.
Robert Hofstadter got the Nobel prize for this.
SBH
Bjoern Feuerbacher
>
[snip lots]
> Never mind. I think I have figured out what I am doing wrong here. But the
> question still remains in my thinking that the "size" of a quantum object is
> dependent on the momentum transferred in the interaction.
Well, the cross sections of interactions usually depend on the
transferred energy and momentum. If you want to interpret the cross
sections geometrically, then yes, the size of the quantum objects depend
on the energy and momentum transferred.
> In the case of
> shooting positrons at a proton, the more momentum the positron has, the
> closer it can get to the proton.
Right. It can even get inside it. But I don't see what this has to do
with what you said above...
> So the question would be what is really
> the "size" of a proton with respect to energy level of the interaction?
Sorry, I don't understand the question...
> A proton is said to have a certain charge radius, but is this charge radius
> dependent on a certain energy level of the interaction that measured it?
No, AFAIK.
Bye,
Bjoern
Franz Heymann
> FrediFizzx wrote:
> >
>
> [snip lots]
>
>
> > Never mind. I think I have figured out what I am doing
wrong here. But the
> > question still remains in my thinking that the "size" of
a quantum object is
> > dependent on the momentum transferred in the
interaction.
>
> Well, the cross sections of interactions usually depend on
the
> transferred energy and momentum. If you want to interpret
the cross
> sections geometrically, then yes, the size of the quantum
objects depend
> on the energy and momentum transferred.
That would say the electron is infinite in extent.
> > In the case of
> > shooting positrons at a proton, the more momentum the
positron has, the
> > closer it can get to the proton.
>
> Right. It can even get inside it. But I don't see what
this has to do
> with what you said above...
>
>
> > So the question would be what is really
> > the "size" of a proton with respect to energy level of
the interaction?
>
> Sorry, I don't understand the question...
>
>
> > A proton is said to have a certain charge radius, but is
this charge radius
> > dependent on a certain energy level of the interaction
that measured it?
>
> No, AFAIK.
Franz
Franz Heymann
[snip]
> Never mind. I think I have figured out what I am doing
wrong here. But the
> question still remains in my thinking that the "size" of a
quantum object is
> dependent on the momentum transferred in the interaction.
In the case of
> shooting positrons at a proton, the more momentum the
positron has, the
> closer it can get to the proton. So the question would be
what is really
> the "size" of a proton with respect to energy level of the
interaction? A
> proton is said to have a certain charge radius, but is
this charge radius
> dependent on a certain energy level of the interaction
that measured it?
I think you are misunderstanding the situation about
measuring the proton size. (That is, the radial
distribution of its charge density.)
In (correct) qualitative language, the proton radius is
determined by studying the diffraction of electrons by
protons. If the proton is a point particle, the angular
distribution of elastically scattered electrons (strictly
the differential crossection) turns out to be the Rutherford
differential cross section.
If the proton has a non-zero radius for its charge
distribution, the angular distribution changes.
Specifically, leads to less protons scattered at large
angles.
This deviation from Rutherford scattering is known
theoretically on a quantitative footing, and it enables the
proton's charge radius to be determined. This radius has
been measured at many different electron energies, and is in
fact
0.870 +- 0.008 fm. The "level of interaction", whatever
that may mean, is irrelevant.
Franz
Bjoern Feuerbacher
>
> "Bjoern Feuerbacher" <bfeu...@ix.urz.uni-heidelberg.de>
> wrote in message
> news:4067FD39...@ix.urz.uni-heidelberg.de...
> > FrediFizzx wrote:
> > >
> >
> > [snip lots]
> >
> >
> > > Never mind. I think I have figured out what I am doing
> > > wrong here. But the
> > > question still remains in my thinking that the "size" of
> > > a quantum object is
> > > dependent on the momentum transferred in the
> > > interaction.
> >
> > Well, the cross sections of interactions usually depend on
> > the
> > transferred energy and momentum. If you want to interpret
> > the cross
> > sections geometrically, then yes, the size of the quantum
> > objects depend
> > on the energy and momentum transferred.
>
> That would say the electron is infinite in extent.
Well, yes. I said *if* he wants to interpret the cross section
geometrically - I didn't recommend him to do this. ;-)
Bye,
Bjoern
FrankH
>
> I also have trouble with the Rutherford conclusion and have heard
> people say that there is much more evidence but haven't heard them say
> anything convincing.
Yes, the conclusion is based on the most indirect of evidence. While
the Rutherford formula is quite complex, it largely just predicts an
exponential like curve for the results. Lots of natural phenomenon
drop off exponentially, so drawing the tiny nucleus conclusion off of
this experimental result is pushing it.
>
> The simplest experiment that I can think of is to measure the
> rotational moment of inertia of an atom or a small molecule. For the
> accepted model this should be smaller than for the alternative. For
> example, suppose a water molecule was spun by an oscillatory electric
> field. This would no doubt be difficult to measure but seems far more
> substantial than the path of a subatomic particle interacting with an
> unknown structure. There may be some way to infer the measurement from
> thermodynamics but it is beyond me.
>
> If the accepted model is correct then the water molecule can spin
> about its center of mass very easily. The difference in rotational
> moment is drastic for your alternative.
>
Yes, I would think that if the atom is big, that it would spin
differently than if the nucleus is small, just like a skater that
brings their arms closer, spins faster. However, for a molecule like
water, it may not make any difference because the spin may depend more
on the bond length than the atom nucleus size. This is like an oxygen
skater holding out 2 hydrogen atoms on outstretched arms.
I have been looking at NMR since it speaks of spinning atoms. NMR
presumes that all atoms are already spinning and that they will align
in a magnetic field. I don't know if I believe that atoms are always
spinning. The cubic model would have a hard time holding atoms
together in molecules if they are constantly spinning. I would suspect
what they are actually measuring is somekind of harmonic motion of the
atom. But the cubic model does neatly explain how atoms align in a
magnetic field. If hydrogen is nothing more than a proton and
electron, then it simply aligns like a bar magnet with proton/electron
attracted to the poles. I supposed the standard model explains this by
saying the electron cloud is distorted so the electron spends more
time on one side than the other, but this picture gets less clear as
you look at larger atoms. The cubic model would predict that any
lobsidedness would lead to a stronger magnetic moment. Just how the
electron orbitals and nucleus of the standard model achieve this is
not clear to me.
-Franklin
Franz Heymann
> tttp...@yahoo.com (Timothy Golden) wrote in message
news:<5c7f8ade.04032...@posting.google.com>...
> > frank...@yahoo.com (FrankH) wrote in message
> [snip]
> >
> > I also have trouble with the Rutherford conclusion and
have heard
> > people say that there is much more evidence but haven't
heard them say
> > anything convincing.
>
> Yes, the conclusion is based on the most indirect of
evidence. While
> the Rutherford formula is quite complex, it largely just
predicts an
> exponential like curve for the results.
That is undiluted crap. No exponential functions occur
anywhere in the Rurherford scattering diferential cross
section.
Tell us in all honesty: Have you ever seen an expression
for the Rutherford diferential cross section?
> Lots of natural phenomenon
> drop off exponentially, so drawing the tiny nucleus
conclusion off of
> this experimental result is pushing it.
But there is no exponential of any description involved in
the Rutherford scattering formula.
You really should have familiarised yourself with this fact
before committing yourself to nonsense so firmly.
[snip]
The rest of the post was not worth repearing here, so I
snipped it.
Franz
Bjoern Feuerbacher
>
> tttp...@yahoo.com (Timothy Golden) wrote in message news:<5c7f8ade.04032...@posting.google.com>...
> > frank...@yahoo.com (FrankH) wrote in message
> [snip]
> >
> > I also have trouble with the Rutherford conclusion and have heard
> > people say that there is much more evidence but haven't heard them say
> > anything convincing.
>
> Yes, the conclusion is based on the most indirect of evidence.
BFD.
The existence of electrons is also based on indirect evidence.
The existence of alpha particles is also based on indirect evidence.
> While
> the Rutherford formula is quite complex,
Huh????? It's a totally simple formula. What on earth is complex about
it???
> it largely just predicts an
> exponential like curve for the results.
Huh??? What on earth is "exponential like" about this curve??? That it
goes to zero or what? That applies to a *lot* of curves - including
1/x^n!!!
And it doesn't only predict this curve - it doesn't only predict the
dependence on the angle. It also predicts the dependence on the energy
and on the charges of the projectile as well as the target. All of these
predictions have been checked, with great accuracy. And you *know* this
- you have seen graphs showing this. So stop lying!
> Lots of natural phenomenon
> drop off exponentially,
But Rutherford's formula has *nothing* to do with an exponential drop
off!!!!!!! Stop lying!
> so drawing the tiny nucleus conclusion off of
> this experimental result is pushing it.
For the 20th time: if you can reproduce the results with the same
accuracy as Rutherford's formula can, feel free to do this.
As long as you can't do this (and, hint: no one was able to do this in
the last 90 years), your argument is just baseless - and plainly
nonsensical.
> > The simplest experiment that I can think of is to measure the
> > rotational moment of inertia of an atom or a small molecule. For the
> > accepted model this should be smaller than for the alternative. For
> > example, suppose a water molecule was spun by an oscillatory electric
> > field. This would no doubt be difficult to measure but seems far more
> > substantial than the path of a subatomic particle interacting with an
> > unknown structure. There may be some way to infer the measurement from
> > thermodynamics but it is beyond me.
> >
> > If the accepted model is correct then the water molecule can spin
> > about its center of mass very easily. The difference in rotational
> > moment is drastic for your alternative.
> >
> Yes, I would think that if the atom is big, that it would spin
> differently than if the nucleus is small, just like a skater that
> brings their arms closer, spins faster.
Moments of inertia have been measured. Try looking into any book on
Molecular Physics (e.g. Haken&Wolf:
<http://www.amazon.com/exec/obidos/tg/detail/-/3540407928/qid=1080722542/sr=8-1/ref=sr_8_xs_ap_i1_xgl14/002-2442715-7955208?v=glance&s=books&n=507846>)
They agree with the theoretical predictions. A pity for your model,
isn't it?
> However, for a molecule like
> water, it may not make any difference because the spin may depend more
> on the bond length than the atom nucleus size. This is like an oxygen
> skater holding out 2 hydrogen atoms on outstretched arms.
How do you explain molecular bonds?
What size has the nucleus in your model? What sizes have a proton, a
neutron, an electron?
How do you explain the results of Bhabha scattering? And the results of
Hofstadter's experiments?
> I have been looking at NMR since it speaks of spinning atoms.
Don't confuse "spin" with an actually rotating movement.
> NMR
> presumes that all atoms are already spinning and that they will align
> in a magnetic field. I don't know if I believe that atoms are always
> spinning.
Well, if you have got an alternative explanation for their magnetic
moments (try reading up on Stern and Gerlach - see e.g. the book
mentioned above), feel free to provide it.
> The cubic model would have a hard time holding atoms
> together in molecules if they are constantly spinning. I would suspect
> what they are actually measuring is somekind of harmonic motion of the
> atom.
The measurements are consistent with "spinning" nuclei. If you have got
an alternative, *quantitative* description, feel free to provide it.
> But the cubic model does neatly explain how atoms align in a
> magnetic field. If hydrogen is nothing more than a proton and
> electron, then it simply aligns like a bar magnet with proton/electron
> attracted to the poles.
Huh???????? Are you sure you understand the difference between electric
and magnetic forces???????
And how do you explain the quantization of magnetic moment? Again, try
reading up on Stern and Gerlach (see the book mentioned above, or see
here:
<http://plato.stanford.edu/entries/physics-experiment/app5.html>; or try
reading the book by Styer that I have recommended to you at least 10
times now).
> I supposed the standard model explains this by
> saying the electron cloud is distorted so the electron spends more
> time on one side than the other,
Total absolute nonsense. The distribution of the electrons has nothing
to do with the alignment of atoms in a magnetic field.
> but this picture gets less clear as
> you look at larger atoms. The cubic model would predict that any
> lobsidedness would lead to a stronger magnetic moment.
Why? Charge distribution has nothing to do with magnetic moments. It
only determines the *ELECTRIC* dipole moment.
Perhaps you should try to learn some electrodynamics first before trying
to build a new atom model...
> Just how the
> electron orbitals and nucleus of the standard model achieve this is
> not clear to me.
This is explained in detail in essentially all books on atomic and
molecular physics. Try reading them.
At first, your ignorance was an excuse for your behaviour. But you have
been pointed countless times to literature where you can read up. So
far, you almost entirely refused to do that. You have gone from simple
ignorance to *willful* ignorance.
Bye,
Bjoern
Bjoern Feuerbacher
>
[snip]
> I have been looking at NMR since it speaks of spinning atoms.
Further comment: no, NMR does not talk about spinning atoms, it talks
about spinning *nuclei*. The magnetic moments associated with atoms and
nuclei are totally different.
> I would suspect
> what they are actually measuring is somekind of harmonic motion of the
> atom.
Harmonic motions of molecules produces infrared radiation - completely
other frequencies and a completely other spectrum than the one observed
in NMR. Harmonic motions of nuclei produces gamma radiation - completely
other frequencies and a completely other spectrum than the one observed
in NMR.
[snip]
Bye,
Bjoern
FrankH
--- Bjoern Feuerbacher <B.Feue...@thphys.uni-heidelberg.de> wrote:
> Hi again, Frank!
>
> Recently we discussed the results of scattering of
> gold foils, and the
> question appeared there what influence the thickness
> of the foil might
> have. I realized that it is difficult to discuss
> such questions within
> your model - since apparently you didn't tell
> anywhere how matter is built
> from the atoms!
>
> Do you think that in a crystal, your "cubes" or
> "stars" directly touch
> each other (the ends of the arms touch each other)?
> Or are there spaces
> between the atoms, i.e. the atoms hand around there
> on their places,
> without any direct connections betwenn them? (this
> is closely connected
> with another question you didn't address: what is
> the size of your atoms?)
>
> If the second: what holds them their in their
> places?
Since materials have different densities, I would imagine that atoms
do not immediately touch each other. If they did, then you would think
that all materials would have roughly the same density. I would think
that there would still be columb attraction for the various components
of the atoms depending on the orientation. If you look at how the
protons/electrons are arranged, you can always flip an atom around so
that it matches up proton/electron if they were touching. Obviously, I
need to come up with a better picture for how this works and possibly
relate it density and measured atomic radius. The model should predict
density.
>
> If the first: doesn't this mean that a big part of a
> crystal shoud be
> impenetrable? There are "tunnels" between the atoms
> - but the opening area
> of these tunnels is on the same order as the size of
> the atoms
> itself. Hence if one shoots particles at a crystal,
> your model seems to
> predict that a substantial amount should be absorbed
> or bounced back (I
> would say on the order of 10% to 40%!), and that the
> fraction which
> doesn't go through should be independent of the
> energy of the
> projecticles, right?
The key is that the alpha can pass through the "solid" part of the
atom, so even if all atoms are touching and there is no empty space at
all, the alpha passes through just like a bullet through packing
peanuts. I would say that at a certain low enough energy, you would
see a massive absorbtion of alphas, which is why I suggested that this
is a test to separate out the large vs. small nucleus models. But as
with anything, the faster the alpha, the more will get through, so
there is a dependency upon energy level. Remember, that the alpha only
has to pass through a layer which is only as thick as the alpha
itself, so it can easily pass through. It might even be a billiard
ball effect that causes the incoming alpha to replace the alpha in the
atom and the alpha that exits is not the same one that entered.
>
> You will probably know where this will lead: all of
> this doesn't agree
> with the experiments!!! In reality, most particles
> will go right through -
> and their range becomes larger when their energies
> increase (based on
> this, already Lenard, ten years *before* Rutherford,
> proposed that
> most of the atoms is essentially "empty space"). How
> do you explain that?
The reason why I went through the whole calculation of how an alpha
would interact with the cubic model was to show that the model
predicts that most alpha will sail through while reflecting a tiny
amount at high angles. My rough calculations show that this is
possible if you presume that an alpha will not be significantly
deflected if it doesn't hit a "thick" part of the atom edge on.
>
>
> Bye,
> Bjoern
>
>
To answer some of the other interesting questions you posed.
I have gone back and looked at the ionization data and you are indeed
correct that what turns out to by symettric can vary. In fact, I think
I have got the stacking wrong based on the ionization data. The
ionization data would more favor Neon being constructed entirely out
of alpha particles forming an X. I will have to completely review the
ionization data and see where I went wrong in the stacking. I am
thinking there is preference to forming atoms out of alpha particles.
To answer questions about gold foil - when metal is cooled from a
molten state, it will crystalize as you say, however, it forms grains
which are still randomly oriented and usually are small. When you take
this and make a foil film out of it by pounding it, this further
randomizes the gold atoms from the crystal form. To accurately test my
cubic model, I would need a gold foil where all of the atoms are
aligned in exactly the same way throughout the entire foil, so you
could change the direction of the foil and observe any differences.
The cubic model should show differences, the Rutherford model should
show no differences due to orientation.
Concerning molecular binding, the cubic model would place a definite
limit on the number of bonds that are possible - I would say 6 if all
the verticies of the atom are allowed to form bonds, but I don't see
atoms forming any more than 4 bonds like Carbon, so maybe the
top/bottom core vertices aren't allowed to form molecules. Bonds are
strictly formed through electrostatic attraction of the vertice.
I have commented that the Rutherford formula is actually not
quantitative. The experiment results always match a curve which is in
proportion to the rutherford formula. I haven't seen somebody actually
calculate the actual number of reflected angles that should be seen
and then plot that against the actual number observed. I seemed to
recall reading that the Rutherford formula only predicts the shape of
the experimental results and that the formula was off when it came to
predicting the actual quantity, so nobody puts a direct plot of the
predicted Rutherford result against actual experimental results. I
seemed to recall that it said it was off by a factor of 4 (I may have
been smoking something since I can't seem to find the reference
anymore). If you think of the cubic model in terms of Rutherford
scattering, you would basically have an alpha scattering off of
another alpha since that is what atoms are made up of and the rest of
the dependencies on the alpha energy would follow. So, I would say
that the cubic model matches the rutherford formula as long as you
assume N=2.
-thanks
Bjoern Feuerbacher
> More great questions from Bjoern
Oh, finally you are back! Apparently you don't have much time for answering?
Good argument - *if* all atoms have roughly the same size. Which isn't
true. (Unfortunately, I don't remember at the moment how atom sizes are
determined experimentally - didn't Povh have a section on this?)
> I would think
> that there would still be columb attraction for the various components
> of the atoms depending on the orientation.
Yes. But how could this Coulomb attraction lead to a stable structure,
without the atoms "touching each other"?
> If you look at how the
> protons/electrons are arranged, you can always flip an atom around so
> that it matches up proton/electron if they were touching. Obviously, I
> need to come up with a better picture for how this works and possibly
> relate it density and measured atomic radius. The model should predict
> density.
Good luck.
>>If the first: doesn't this mean that a big part of a
>>crystal shoud be
>>impenetrable? There are "tunnels" between the atoms
>>- but the opening area
>>of these tunnels is on the same order as the size of
>>the atoms
>>itself. Hence if one shoots particles at a crystal,
>>your model seems to
>>predict that a substantial amount should be absorbed
>>or bounced back (I
>>would say on the order of 10% to 40%!), and that the
>>fraction which
>>doesn't go through should be independent of the
>>energy of the
>>projecticles, right?
>
>
> The key is that the alpha can pass through the "solid" part of the
> atom, so even if all atoms are touching and there is no empty space at
> all, the alpha passes through just like a bullet through packing
> peanuts.
Well, how is this possible, when all the particles in your model have
got "hard surfaces"?
Wouldn't this destroy the atoms, at least?
> I would say that at a certain low enough energy, you would
> see a massive absorbtion of alphas, which is why I suggested that this
> is a test to separate out the large vs. small nucleus models.
What energy, specifically? You are *very* vague here.
> But as
> with anything, the faster the alpha, the more will get through,
Why?
Essentially you propose that passing through the nucleus is possible for
the alpha particles, but it is "slowed down" when doing this, right?
What mechanism causes this "slowing down"? What forces are acting there?
And can this idea explain the Bethe-Bloch formula?
> so there is a dependency upon energy level.
energy *level*?????
> Remember, that the alpha only
> has to pass through a layer which is only as thick as the alpha
> itself, so it can easily pass through.
On the arms, yes. But how do you explain that *most* alpha particles get
through and aren't deflected or absorbed at all? The amount of absorbed
alpha particles is *far* smaller than the area of the "thick" part of
your atom compared to the area of the "thin" parts!
> It might even be a billiard
> ball effect that causes the incoming alpha to replace the alpha in the
> atom and the alpha that exits is not the same one that entered.
Nice idea. Unfortunately, there is no evidence that this can happen (various
nuclear reactions were studied, some of them including shooting alpha
particles at nuclei - but AFAIK, none of them showed another alpha particle
emerging from the nucleus).
>>You will probably know where this will lead: all of
>>this doesn't agree
>>with the experiments!!! In reality, most particles
>>will go right through -
>>and their range becomes larger when their energies
>>increase (based on
>>this, already Lenard, ten years *before* Rutherford,
>>proposed that
>>most of the atoms is essentially "empty space"). How
>>do you explain that?
>
>
> The reason why I went through the whole calculation of how an alpha
> would interact with the cubic model was to show that the model
> predicts that most alpha will sail through while reflecting a tiny
> amount at high angles.
I'm sorry, but how does this reflecting work, if the alpha particles are
able to pass through the nucleus? OTOH, how does the passing through
work if the particles all have hard surfaces? Do you want to have it
both ways?
> My rough calculations show that this is
> possible if you presume that an alpha will not be significantly
> deflected if it doesn't hit a "thick" part of the atom edge on.
I don't see how this rough calculation is consistent with the simple
fact that *most* alpha particles go right through. *By far* most.
The simple argument that they are only absorbed by the thick part
doesn't seem to be able to explain this quantitatively. Additionally,
I think you will have troubles with explaining the energy dependency
of the mean free path length of the alpha particles.
> To answer some of the other interesting questions you posed.
>
> I have gone back and looked at the ionization data and you are indeed
> correct that what turns out to by symettric can vary.
Sorry, I don't know what exactly you are talking about here. The "shell
structure" of the ionization energies perhaps?
BTW, did I ever mention that there is also a shell structure for the
*nuclear* energies, and that the standard physics can describe that, too?
> In fact, I think
> I have got the stacking wrong based on the ionization data. The
> ionization data would more favor Neon being constructed entirely out
> of alpha particles forming an X. I will have to completely review the
> ionization data and see where I went wrong in the stacking. I am
> thinking there is preference to forming atoms out of alpha particles.
Please tell me (by e-mail - it's easy to miss post in the newsgroup!)
when you have updated your website accordingly, and have got pictures
of the new structures.
> To answer questions about gold foil - when metal is cooled from a
> molten state, it will crystalize as you say, however, it forms grains
> which are still randomly oriented and usually are small.
Right - but I don't know if that is relevant in a foil of that small
thickness.
> When you take
> this and make a foil film out of it by pounding it, this further
> randomizes the gold atoms from the crystal form.
Why?
> To accurately test my
> cubic model, I would need a gold foil where all of the atoms are
> aligned in exactly the same way throughout the entire foil, so you
> could change the direction of the foil and observe any differences.
> The cubic model should show differences, the Rutherford model should
> show no differences due to orientation.
Probably right.
> Concerning molecular binding, the cubic model would place a definite
> limit on the number of bonds that are possible - I would say 6 if all
> the verticies of the atom are allowed to form bonds, but I don't see
> atoms forming any more than 4 bonds like Carbon,
If you count double bounds, then the S in SO_3 has six bonds.
BTW, how do you explain the different bond lengths in ethan, ethen and
ethin?
> so maybe the
> top/bottom core vertices aren't allowed to form molecules. Bonds are
> strictly formed through electrostatic attraction of the vertice.
If they attract each other, why don't they move towards each other until
they touch?
> I have commented that the Rutherford formula is actually not
> quantitative. The experiment results always match a curve which is in
> proportion to the rutherford formula.
I don't know what this is supposed to mean, sorry. That one could
multiply the data with a constant, and still Rutherford's formula would
match, or what??? If yes, that's simply totally wrong. Why on earth do
you think so?
> I haven't seen somebody actually
> calculate the actual number of reflected angles that should be seen
> and then plot that against the actual number observed.
I don't know what this is supposed to mean, sorry. "actual number of
reflected angles"???
> I seemed to
> recall reading that the Rutherford formula only predicts the shape of
> the experimental results and that the formula was off when it came to
> predicting the actual quantity,
Well, that's wrong, plain and simple.
> so nobody puts a direct plot of the
> predicted Rutherford result against actual experimental results.
Wrong, plain and simple.
> I seemed to recall that it said it was off by a factor of 4 (I may have
> been smoking something since I can't seem to find the reference
> anymore).
Either you recall wrongly, or you misunderstood something.
> If you think of the cubic model in terms of Rutherford
> scattering, you would basically have an alpha scattering off of
> another alpha since that is what atoms are made up of and the rest of
> the dependencies on the alpha energy would follow.
Huh? How does any dependency on energy follow from that?
> So, I would say
> that the cubic model matches the rutherford formula as long as you
> assume N=2.
I still haven't seen you reproducing the energy dependence in
Rutherford's formula.
Neither the energy dependence of Bethe-Bloch's formula, BTW.
Bye,
Bjoern
FrankH
> FrankH wrote:
> > More great questions from Bjoern
>
> Oh, finally you are back! Apparently you don't have much time for answering?
>
- you may not hear much from me for a while.
> > I would think
> > that there would still be columb attraction for the various components
> > of the atoms depending on the orientation.
>
> Yes. But how could this Coulomb attraction lead to a stable structure,
> without the atoms "touching each other"?
>
The atoms may in fact be touching one another. Density may be related
to how efficiently they can be stacked together and still attract. I
was curious how QM explains the attraction of idential atoms (like in
a piece of iron). There is no covalent or ionic bonding of any type,
yet the attraction is extremely strong.
> > The key is that the alpha can pass through the "solid" part of the
> > atom, so even if all atoms are touching and there is no empty space at
> > all, the alpha passes through just like a bullet through packing
> > peanuts.
>
> Well, how is this possible, when all the particles in your model have
> got "hard surfaces"?
>
> Wouldn't this destroy the atoms, at least?
I am imagining that the hard surfaces get pushed aside. As the alpha
impacts the atom, it causes an alpha sized portion of the atom to open
up like a door by the impact. The impact slows the alpha down without
significantly changing direction (this is postulated, I don't see how
this would actually happen in practice since you'd think there'd be
some deflection). After the alpha passes, the door shuts again to
reform the atom. In this process, an electron can get ionized (which
is why alpha radiation is ionizing), so the atom is partially
destroyed, but quickly puts itself back together. Theoretically, if
you shot enough alphas into the atom, you could probably get it to
transmute into another element.
>
>
> > I would say that at a certain low enough energy, you would
> > see a massive absorbtion of alphas, which is why I suggested that this
> > is a test to separate out the large vs. small nucleus models.
>
> What energy, specifically? You are *very* vague here.
You could calculate the energy level at which rutherford would predict
almost total relfection of the alphas and then compare to see if it
meets experimental data. Any anomoly would favor the cubic model.
>
>
> > But as
> > with anything, the faster the alpha, the more will get through,
>
> Why?
The cubic model would essentially use the rutherford formula except
that instead of gold atoms, you use helium (alpha) atoms. All of the
properties of the rutherford formula will then be followed by the
cubic model including dependence on E.
>
> On the arms, yes. But how do you explain that *most* alpha particles get
> through and aren't deflected or absorbed at all? The amount of absorbed
> alpha particles is *far* smaller than the area of the "thick" part of
> your atom compared to the area of the "thin" parts!
>
Considering the loss of energy, I would say that most of the alphas
are hitting something. Alphas can barely pass through a thin gold foil
before being totally blocked. If there were little or no absorbtion,
you'd think alphas would be able to pass through a thick block of
gold, but they cannot.
>
>
>
>
> > It might even be a billiard
> > ball effect that causes the incoming alpha to replace the alpha in the
> > atom and the alpha that exits is not the same one that entered.
>
> Nice idea. Unfortunately, there is no evidence that this can happen (various
> nuclear reactions were studied, some of them including shooting alpha
> particles at nuclei - but AFAIK, none of them showed another alpha particle
> emerging from the nucleus).
>
How on earth could they tell one alpha from one another?
>
> >>You will probably know where this will lead: all of
> >>this doesn't agree
> >>with the experiments!!! In reality, most particles
> >>will go right through -
> >>and their range becomes larger when their energies
> >>increase (based on
> >>this, already Lenard, ten years *before* Rutherford,
> >>proposed that
> >>most of the atoms is essentially "empty space"). How
> >>do you explain that?
> >
> >
> > The reason why I went through the whole calculation of how an alpha
> > would interact with the cubic model was to show that the model
> > predicts that most alpha will sail through while reflecting a tiny
> > amount at high angles.
>
> I'm sorry, but how does this reflecting work, if the alpha particles are
> able to pass through the nucleus? OTOH, how does the passing through
> work if the particles all have hard surfaces? Do you want to have it
> both ways?
>
Reflections happen when the portion of the atom it is hitting simply
cannot swing away (like a door). This happens if it there is no where
for it to swing open as in the thick part of the atoms.
>
> > My rough calculations show that this is
> > possible if you presume that an alpha will not be significantly
> > deflected if it doesn't hit a "thick" part of the atom edge on.
>
> I don't see how this rough calculation is consistent with the simple
> fact that *most* alpha particles go right through. *By far* most.
As I mentioned, the alphas are not deflected in angle, but they
definitely loose a lot of energy which is consistent with them hitting
something. You did mention the other formula for electrons slowing the
alpha - and that is possible, but so is my argument that they are
actually hitting something.
> > Concerning molecular binding, the cubic model would place a definite
> > limit on the number of bonds that are possible - I would say 6 if all
> > the verticies of the atom are allowed to form bonds, but I don't see
> > atoms forming any more than 4 bonds like Carbon,
>
> If you count double bounds, then the S in SO_3 has six bonds.
>
> BTW, how do you explain the different bond lengths in ethan, ethen and
> ethin?
There are certainly different bond lengths, I would imagine that
double bonds form when 2 vertice of the atom are able to bind to
another atom. Since there is more bonding force created by double the
electrostatic attraction, the bond is shorter.
>
>
> > so maybe the
> > top/bottom core vertices aren't allowed to form molecules. Bonds are
> > strictly formed through electrostatic attraction of the vertice.
>
> If they attract each other, why don't they move towards each other until
> they touch?
They just might. Or perhaps they vibrate like closely spaced gas
molecules from thermal energy. The relative orientation of local
electrostatic forces hold the structure stable. To a certain extent,
solids may behave like gasses in that the same number of atoms occupy
the same amount of space at the same pressure. So 1000 Cu atoms in
solid may occupy roughly the same space as 1000 Au atoms, but since Cu
is lighter, it is less dense than Au.
>
>
> > I have commented that the Rutherford formula is actually not
> > quantitative. The experiment results always match a curve which is in
> > proportion to the rutherford formula.
>
> I don't know what this is supposed to mean, sorry. That one could
> multiply the data with a constant, and still Rutherford's formula would
> match, or what??? If yes, that's simply totally wrong. Why on earth do
> you think so?
>
The original Mardsen Geiger paper shows that the results are
proportional to the 1/sin^4(theta) function.
http://dbhs.wvusd.k12.ca.us/webdocs/Chem-History/GeigerMarsden-1913/GeigerMarsden-1913.html
I didn't think this was very interesting considering that if you
create a plot of 1/theta^3, you get something that matches the data
pretty well. Try it with the data I have in:
http://ourworld.compuserve.com/homepages/frankhu/ruther.xls
They do not attempt to match it up numerically with the rutherford
formula that predicts actual particle count.
http://hyperphysics.phy-astr.gsu.edu/hbase/rutsca.html#c2
The MIT paper mentions trying to do that, but they also normalized the
data so that the curves match up rather than providing a direct
comparison.
-Franklin
Bjoern Feuerbacher
> ganesh
FrankH wrote:
> Bjoern Feuerbacher <feue...@thphys.uni-heidelberg.de> wrote in message news:<c5mbvp$o2m$1...@news.urz.uni-heidelberg.de>...
>
>>FrankH wrote:
>>
>>>More great questions from Bjoern
>>
>>Oh, finally you are back! Apparently you don't have much time for answering?
>>
>
> Yes, I am getting busy on many other projects and don't have much time
> - you may not hear much from me for a while.
Sorry to hear that. In contrast to all the crackpots in this group, it's
rather nice to discuss with you - although I still think that your model
has not much merit and contradicts the evidence.
>>>I would think
>>>that there would still be columb attraction for the various components
>>>of the atoms depending on the orientation.
>>
>>Yes. But how could this Coulomb attraction lead to a stable structure,
>>without the atoms "touching each other"?
>>
>
> The atoms may in fact be touching one another.
Well, that contradicts what you said before:
"Since materials have different densities, I would imagine that atoms
do not immediately touch each other."
Could you please make up your mind?
> Density may be related
> to how efficiently they can be stacked together and still attract.
Huh? I don't understand this, sorry.
> I
> was curious how QM explains the attraction of idential atoms (like in
> a piece of iron). There is no covalent or ionic bonding of any type,
> yet the attraction is extremely strong.
The outermost electrons (valence electrons) aren't bound very strongly
in a metal atom. If many such atoms are grouped together in a metal
lattice, it is energetically much more favourable for them to be "shared"
between a lot (in general even all) atoms than staying at only one. One
could even say that every block of metal is one single huge molecule.
Oh, BTW: this model for the behaviour of electrons in metals leads, as
usual, to quantitative predictions, and as usual, these predictions
agree nicely with the experimental observations. See any book on solid
state physics. Especially interesting should be the sections on the
specific heat capacity of metals.
>>>The key is that the alpha can pass through the "solid" part of the
>>>atom, so even if all atoms are touching and there is no empty space at
>>>all, the alpha passes through just like a bullet through packing
>>>peanuts.
>>
>>Well, how is this possible, when all the particles in your model have
>>got "hard surfaces"?
>>
>>Wouldn't this destroy the atoms, at least?
>
>
> I am imagining that the hard surfaces get pushed aside. As the alpha
> impacts the atom, it causes an alpha sized portion of the atom to open
> up like a door by the impact.
This sounds more like magic than like physics! ("open sesame"...)
How on earth should an impact manage to do that???
> The impact slows the alpha down without
> significantly changing direction (this is postulated, I don't see how
> this would actually happen in practice since you'd think there'd be
> some deflection).
Yes.
> After the alpha passes, the door shuts again to
> reform the atom.
How on earth is this supposed to happen?
> In this process, an electron can get ionized
How, specifically?
> (which
> is why alpha radiation is ionizing), so the atom is partially
> destroyed, but quickly puts itself back together.
How?
> Theoretically, if
> you shot enough alphas into the atom, you could probably get it to
> transmute into another element.
Well, right, this indeed works. There is a wealth of literature on this.
Unfortunately for your model, if one assumes that alpha particles are
slowed down in matter simply by the Coulomb force, and calculates based
on that how much energy an alpha particle loses, depending on its energy
and the density of the matter, one gets results which, as usual, nicely
agree with observations (Bethe-Bloch formula). So here is yet again
another experimental result which is consistent with the standard atomic
theory (additional to Rutherford scattering, spectra and so on); I
wonder how long it will take you to reproduce *this* result?
Frank, don't you think that the fact that literally *thousands* of
experiments over the last about 80 years have *always* confirmed the
standard atomic theory says something about its validity? Do you think
all of these agreements between theory and experiments were merely
coincidence, or perhaps based on fraud?
>>>I would say that at a certain low enough energy, you would
>>>see a massive absorbtion of alphas, which is why I suggested that this
>>>is a test to separate out the large vs. small nucleus models.
>>
>>What energy, specifically? You are *very* vague here.
>
>
> You could calculate the energy level at which rutherford would predict
> almost total relfection of the alphas
There is no such energy "level". You can see this yourself by looking at
Rutherford's formula! The angular distribution stays always the same -
no matter what energy the alphas have.
> and then compare to see if it
> meets experimental data. Any anomoly would favor the cubic model.
In other words, you think it's the job of scientists to disprove your model,
not your job to provide experimental evidence for its validity?
>>>But as
>>>with anything, the faster the alpha, the more will get through,
>>
>>Why?
>
> The cubic model would essentially use the rutherford formula except
> that instead of gold atoms, you use helium (alpha) atoms.
Huh? Why??? This makes no sense at all. Rutherford's formula was derived
*specifically* for a tiny nucleus with *lots* of "void" around it. But
in your model, the "alpha particles" which constitute your nucleus are
"surrounded" by a *lot* of *other* "alpha particles"!!! That's a
*totally* different situation, and you can't simply claim that
Rutherford's formula applies there, too!
> All of the
> properties of the rutherford formula will then be followed by the
> cubic model including dependence on E.
I very much doubt that. E.g. how do you plan to reproduce the dependence
on the number of protons in the nucleus, using this approach?
(Rutherford's scattering cross section is proportional to Z^2)
>>Sorry, but that makes no sense at all. Are you sur
>>On the arms, yes. But how do you explain that *most* alpha particles get
>>through and aren't deflected or absorbed at all? The amount of absorbed
>>alpha particles is *far* smaller than the area of the "thick" part of
>>your atom compared to the area of the "thin" parts!
>>
>
> Considering the loss of energy, I would say that most of the alphas
> are hitting something.
Why are most able to pass through then, without any deflection?
> Alphas can barely pass through a thin gold foil
> before being totally blocked. If there were little or no absorbtion,
> you'd think alphas would be able to pass through a thick block of
> gold, but they cannot.
Depends on their energy.
And you can make exactly the same arguments about protons and electrons,
BTW. Which, according to your model, have the same size - but they
behave *differently* wrt to their range in matter. Hence any geometrical
interpretation of their range is doomed right from the start.
And we haven't even started talking about neutrons, which, according to
your model, have *double* the size of an electron or proton - but
nevertheless are able to pass through matter *far* more easily than a
proton or a neutron (so far, you have said that the difference between a
hydrogen atom and a neutron is the type of binding between the proton
and the neutron - but you haven't said *anything* about how this should
be able to influence its range in matter!).
Oh, and then are this ugly little fact that the cross section in
electron-electron and electron-positron scattering depends on energy -
and in the second case shows some resonance peaks.
I *really* would like to know how you explain all of this. In standard
physics, all of this can be explained easily - but I don't see how any
model which describes scattering essentially based only on geometry
would *ever* be able to do that!
>>>It might even be a billiard
>>>ball effect that causes the incoming alpha to replace the alpha in the
>>>atom and the alpha that exits is not the same one that entered.
>>
>>Nice idea. Unfortunately, there is no evidence that this can happen (various
>>nuclear reactions were studied, some of them including shooting alpha
>>particles at nuclei - but AFAIK, none of them showed another alpha particle
>>emerging from the nucleus).
>>
>
> How on earth could they tell one alpha from one another?
Not directly. But you could e.g. study the angular distribution of the
emerging
alphas. Obviously, if alphas are captured and others emitted, one should get
a totally different angular distribution than if the alphas are only
deflected!
[snip]
>>>The reason why I went through the whole calculation of how an alpha
>>>would interact with the cubic model was to show that the model
>>>predicts that most alpha will sail through while reflecting a tiny
>>>amount at high angles.
>>
>>I'm sorry, but how does this reflecting work, if the alpha particles are
>>able to pass through the nucleus? OTOH, how does the passing through
>>work if the particles all have hard surfaces? Do you want to have it
>>both ways?
>>
>
> Reflections happen when the portion of the atom it is hitting simply
> cannot swing away (like a door). This happens if it there is no where
> for it to swing open as in the thick part of the atoms.
Err, sorry for you, but if you hit a door with a ball, the ball will be
reflected, too - although the door can and does swing away. The ball will
not pass through the door, although it can swing away.
So again: how does th passing through work if the particles all have
hard surfaces? Your suggestion above that the impact simply "opens a
door" through which the particle passes is, as said above, magic, not
physics.
>>>My rough calculations show that this is
>>>possible if you presume that an alpha will not be significantly
>>>deflected if it doesn't hit a "thick" part of the atom edge on.
>>
>>I don't see how this rough calculation is consistent with the simple
>>fact that *most* alpha particles go right through. *By far* most.
>
>
> As I mentioned, the alphas are not deflected in angle,
Why should they? Simply saying "I postulate that" will convince no one.
You are violating things like conservation of momentum here, if you
didn't notice.
> but they
> definitely loose a lot of energy which is consistent with them hitting
> something.
I don't see how this addresses my argument that *most* (energetic)
charged particles simply go right through matter. *By far* most.
> You did mention the other formula for electrons slowing the
> alpha - and that is possible, but so is my argument that they are
> actually hitting something.
Without getting deflected. Rrrrigggght. Try to tell this to another one.
Oh, BTW, I've got a nice vineyard to sell in Antartica. Interested?
>>>Concerning molecular binding, the cubic model would place a definite
>>>limit on the number of bonds that are possible - I would say 6 if all
>>>the verticies of the atom are allowed to form bonds, but I don't see
>>>atoms forming any more than 4 bonds like Carbon,
>>
>>If you count double bounds, then the S in SO_3 has six bonds.
>>
>>BTW, how do you explain the different bond lengths in ethan, ethen and
>>ethin?
>
> There are certainly different bond lengths, I would imagine that
> double bonds form when 2 vertice of the atom are able to bind to
> another atom.
Could you try to draw a picture of this, please? I'm not sure if I
understand what you mean.
> Since there is more bonding force created by double the
> electrostatic attraction, the bond is shorter.
I don't see how this follows, sorry.
>>>so maybe the
>>>top/bottom core vertices aren't allowed to form molecules. Bonds are
>>>strictly formed through electrostatic attraction of the vertice.
>>
>>If they attract each other, why don't they move towards each other until
>>they touch?
>
>
> They just might.
Do they, or don't they? Make up your mind.
> Or perhaps they vibrate like closely spaced gas
> molecules from thermal energy.
Well, yes, molecules often vibrate - but why should that happen in your
model? For such an oscillation to occur, you need a force which
counteracts deviations from a "stable" point. What force would that be
in you model?
> The relative orientation of local
> electrostatic forces hold the structure stable.
Please elaborate. I don't see how that is supposed to work.
> To a certain extent,
> solids may behave like gasses in that the same number of atoms occupy
> the same amount of space at the same pressure.
This idea is *totally* contradicted by a lot of data. Start with "density".
BTW, Y.Porat proposed exactly the same idea in his book...
> So 1000 Cu atoms in
> solid may occupy roughly the same space as 1000 Au atoms, but since Cu
> is lighter, it is less dense than Au.
Unfortunately for you, the different densities of metals are in no way
correlated to the different weights of the atoms themselves.
>>>I have commented that the Rutherford formula is actually not
>>>quantitative. The experiment results always match a curve which is in
>>>proportion to the rutherford formula.
>>
>>I don't know what this is supposed to mean, sorry. That one could
>>multiply the data with a constant, and still Rutherford's formula would
>>match, or what??? If yes, that's simply totally wrong. Why on earth do
>>you think so?
>>
>
> The original Mardsen Geiger paper shows that the results are
> proportional to the 1/sin^4(theta) function.
>
> http://dbhs.wvusd.k12.ca.us/webdocs/Chem-History/GeigerMarsden-1913/GeigerMarsden-1913.html
Well, that may have been right in that original paper - but already in
the following paper, they did a *quantitative* analysis, not showing
only the proportionality, but the actual agreement between curve and
data. See here:
<http://bcs.whfreeman.com/tiplermodernphysics4e/content/cat_020/RutherfordPrediction.pdf>
And that'S the standard way it is done in essentially all experiments
- hey, even I, in my undergraduate lab work, did it that way!
> I didn't think this was very interesting considering that if you
> create a plot of 1/theta^3,
Did you mean 1/sin(theta)^3?
> you get something that matches the data
> pretty well. Try it with the data I have in:
>
> http://ourworld.compuserve.com/homepages/frankhu/ruther.xls
Have you tried doing a fit?
> They do not attempt to match it up numerically with the rutherford
> formula that predicts actual particle count.
Well, in essentially all following experiments (including my own lab
work), they *did* do this.
> http://hyperphysics.phy-astr.gsu.edu/hbase/rutsca.html#c2
I don't see how what is written on that page supports you.
> The MIT paper mentions trying to do that, but they also normalized the
> data so that the curves match up rather than providing a direct
> comparison.
BFD. We already established that that MIT paper came from an undergrad lab
work, and I've already mentioned several times that the results from my
own lab work were better. So, could you stop harping on that MIT paper,
please?
Bye,
Bjoern