Why should we believe electrons move about a central nucleus?
FrankH
central nucleus containing neutrons/protons and electrons surround
this nucleus. Quantum mechanics show the electrons in probability
clouds around the nucleus. As near as I can tell, the reason for this
was due to electron scattering experiments done nearly 100 years ago,
using instruments that would be considered incredibly crude by todays
standards. These experiments showed that the nucleus had to be compact
and the electrons extremely small which made an atom mostly empty
space. But did they really??? If I took a gun and fired bullets at a
stack of cardboard boxes filled with packing peanuts, I might also
conclude that the boxes were mostly empty space due to limited
backscattering - when in fact, the boxes and their contents are quite
large. So the logic used to determine the size of the nucleus escapes
me. I have seen new pictures generated with the latest STM technology
at http://arxiv.org/PS_cache/cond-mat/pdf/0305/0305103.pdf What is
remarkable is that we can now actually image the structure of
individual atoms. The picture shows the surface of a group of Silicon
atoms and what I see doesn't appear to be mostly space or a fuzzy
cloud. Instead, it looks a lot like lego building bricks to me with
very distinct edges. My question is, has physics done any recent
experiments to verify the well accepted values for the size of the
atom/nucleus and electron?
I am wondering this because I was thinking, if I were a bunch of
protons and electrons, how would I assemble into atoms? Naturally, the
answer would be to start sticking together like building blocks with
proton/electrons alternating like a NaCL crystal. This is totally
contrary to the standard atomic nucleus model. I took this further and
started building out this structure so that you have atoms which are
built out of a regular geometric sequence of ever increasing layers of
electrons/protons. Curiously, the model does match some of the
observed electronic states for the atoms I have built out. However,
this model would mean that the central nucleus wouldn't exist. All of
the proton/electrons would be spread out in roughly a Octrahetral
shape in a completely neutrally charged matrix. I have detailed this
model and some pictures from this at:
http://ourworld.compuserve.com/homepages/frankhu/buildatm.htm
This web site also has the STM images I was talking about before. I
have not seen anyone propose such a simple theory before, so I was
interested in seeing what people think of it since this is a new way
of thinking about the atom with no central nucleus, but uses regular
geometric progression to describe the atomic structure.
Sam Wormley
Quantum Numbers
http://scienceworld.wolfram.com/physics/QuantumNumbers.html
Hydrogen Atom
http://scienceworld.wolfram.com/physics/HydrogenAtom.html
Uncle Al
>
> The central theme of most models of the atom is that there is a
> central nucleus containing neutrons/protons and electrons surround
> this nucleus. Quantum mechanics show the electrons in probability
> clouds around the nucleus. As near as I can tell, the reason for this
> was due to electron scattering experiments done nearly 100 years ago,
You are an uneducated dunce. Hey stooopid, don't you think that in
the 21st century there is an overwhelmingly preponderant understanding
of quantum mechanics that allows calculations to 14 significant
figures to agree with equally precise observations? Look up the "Lamb
shift."
If you cannot manage Google, have a grade school kid help you.
[snip simplistic crap]
--
Uncle Al
http://www.mazepath.com/uncleal/
(Toxic URL! Unsafe for children and most mammals)
"Quis custodiet ipsos custodes?" The Net!
Franz Heymann
"FrankH" <frank...@yahoo.com> wrote in message
news:46484c9f.04011...@posting.google.com...
[snip]
You appear not to know any physics worth talking about.
Franz
Pyriform
Be careful now - you are pitting yourself against a profound thinker. I
quote from his website:
"What causes gravity?
Gravity is caused by a slight imbalance of positive/negative charges in
so called neutrally charged matter. The negative/positive charges in a
neutral atom do not exactly cancel each other out. There is a tiny
residual positive charge. These tiny positive charges added together
over the volume of the earth produce a very large positively charged
field at the surface of the earth. The diverging electrostatic field
created by the earth causes dipoles in neutrally charged matter to be
attracted to the source of the field"
So simple to see once a genius has lit the path.
--
Pyriform
galathaea
In case the others haven't scared you off from learning, you should really
read up on scattering theory. In particular, going from simple Rutherford
models to more complex atomic scattering models would really help you clear
up some of this confusion. It is quite a pretty field, and the problems of
inverse scattering calculations and objectives like determining charge
structure and field dynamics from scattering distributions in space and time
will give one a broad understanding of modern thought in this area.
--
-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-
galathaea: prankster, fablist, magician, liar
tj Frazir
pressure expands ,,a low forms around mass.
Thats why motion is a gain in mass .
It takes up more space per time unit of expansion so less eergy will
expand.
ELECTRON is in orbit.
Line all dipoles up so the orbits of atoms overlap and it will conduct.
If the orbits dont line up the electron wave or partical can not do a
figer 8 as it shares the orbit or be pushed into the next orbit.
WHY we know the electron orbits is the propties of conductivity.
FrankH
> Careful Frank, these might blowout your mind.
>
> Quantum Numbers
> http://scienceworld.wolfram.com/physics/QuantumNumbers.html
>
justify the periodicy that we observe in the elctron shells? My simple
cubic model justifies them as the primary quantum number as
corresponding the the electrons in the core of the atoms and the
shells result from the geometric sequence extending out.
> Hydrogen Atom
> http://scienceworld.wolfram.com/physics/HydrogenAtom.html
>
Gee, all this math to explain a Hydrogen atom? What was wrong with
saying that a hydrogen atom is composed of a proton and electron?
> Atom
> http://scienceworld.wolfram.com/physics/Atom.html
Yup, this is the rather limited history of the atomic model. As I
mentioned, my simplistic model is not part of this history. I would
have though this would have been the first to be proposed and then
thrown out. Certainly, it would be more believeable than the plum
pudding model. So what happened?
FrankH
> FrankH wrote:
> >
> > The central theme of most models of the atom is that there is a
> > central nucleus containing neutrons/protons and electrons surround
> > this nucleus. Quantum mechanics show the electrons in probability
> > clouds around the nucleus. As near as I can tell, the reason for this
> > was due to electron scattering experiments done nearly 100 years ago,
>
> You are an uneducated dunce. Hey stooopid, don't you think that in
> the 21st century there is an overwhelmingly preponderant understanding
> of quantum mechanics that allows calculations to 14 significant
> figures to agree with equally precise observations? Look up the "Lamb
> shift."
>
> If you cannot manage Google, have a grade school kid help you.
>
> [snip simplistic crap]
Well, thank you for the nice compliment - I am an MIT graduate, so I
am more educated than most - but I admit I no next to nothing when it
comes to particle physics which is why I need the assistance of
geniuses like yourself to help me ponder the nature of the universe.
I looked up the Lamb shift and found it quite interesting. I have
heard of this thing that got calculated to 14 digits of precision to
justify QM, but didn't know what it was. Thank you for clearing that
up. Although from what I read, it sounds a bit suspicious considering
science has a hard time measuring anything experimentally to 14 digits
of precision. If I wanted to measure the diameter of a penny, I
couldn't do that to 14 digits of precision, so how can something that
has to do with atoms we can't even see be measured with such
precision. But in any case, supposing that this result is correct, are
there any other such examples of QM coming out with an experimentally
verified prediction - I didn't see any in the web sites I visited. The
thing that they were measuring was also quite obscure - a slight
variation of a superfine spectrum line. How about something big and
obvious like the observed shell structure of atoms and why the shells
have the number of electrons and energy states that they do? Why do
they have such a geometric sequence? I have a book on quantum
mechanics and I don't find a chart of the electron shell sequence in
it. Does quantum mechanics explain why atoms combine into things like
H2O? Why not H4O instead? Surely there must be some calculation that
shows that the imbalance in the probability orbitals of the atoms
cause a lower energy state to exist or something like that. Although I
suspect the answer to all of these questions is No. But then again,
I'm an uneducated Newbie, and need the help from you folks who know
better.
FrankH
Ah yes, a famous quote from my own website. You laugh but some day
....
Actually there is no contradition between my lego block theory of
atoms an the electrostatic theory of gravity. All that is needed is a
slight imbalance of positive/negative charges on average. I have read
that it would take only 1 electron to be missing in 10^18 atoms to
create an electrostatic force equivalent to gravity. It isn't hard to
belive that out of a billion billion atoms that one of them might be
missing an electron, in fact I would be suprised if there weren't any
missing electrons considering how easily they can be ionized. The
Earth is also constantly bombarded by positively charged ions from the
solar wind, you'd think that alone would be enough to put a positive
charge on the earth.
FrankH
Damn straight! and proud of it. I don't know anything about your
physics filled to the brim with counter-intuitive ideas, miles of
nonsensical math, multi-dimension and generally defying all logic. You
all complain about not being able to find the one big elegant theory
everything, because you have to admit current theory can't do it, but
yet you go around like you already know how everything in the world
works down to the tiniest detail, but you don't. Excuse the flame, but
this forum exists to assist people understand and challange scientific
ideas, not to state the obvious that I am unfamiliar with the world of
conventional physics. I am presenting you with what appears to be an
entirely new concept of the atom never before seen. Shouldn't that be
the least bit interesting to you? How often do you run across unique
theories of atom formation?
FrankH
> In case the others haven't scared you off from learning, you should really
> read up on scattering theory. In particular, going from simple Rutherford
> models to more complex atomic scattering models would really help you clear
> up some of this confusion. It is quite a pretty field, and the problems of
> inverse scattering calculations and objectives like determining charge
> structure and field dynamics from scattering distributions in space and time
> will give one a broad understanding of modern thought in this area.
Thank you for your thoughful post. Can you recommend a web site that
might explain the more advanced scattering experiments? The books I've
been able to find devote about 1 sentence to the Rutherford experiment
with no further justification. Although fundamentally, I have a hard
time believing you can determine anything by scattering considering
that we don't know how "hard" the things that the electrons are
bouncing off. They could be billiard balls or puffs of gas. I'd have
to see more justification. For me, the new STM pictures of Silicon
atoms as little bricks is most convincing. If Rutherford had an STM, I
don't think he would be so quick to conclude that an atom is 99.999%
empty space. It looks 100% filled with sharply defined edges to me.
The picture on the web site also shows where an atom is missing and
you can peer down and see the clearly defined sides of the other
surrounding atoms - how can you explain this with an atom which is
nearly all empty space with the electrons wizzing about randomly? It
looks solid and space filling to me.
Franz Heymann
> Sam Wormley <swor...@mchsi.com> wrote in message
news:<4006DB47...@mchsi.com>...
> > Careful Frank, these might blowout your mind.
> >
> > Quantum Numbers
> > http://scienceworld.wolfram.com/physics/QuantumNumbers.html
> >
> Yes, very interesting these quantum numbers - Can Quantum Mechanics
> justify the periodicy that we observe in the elctron shells?
Quantum mechanics is capable of answering any question you wish about the
electromagnetic behaviour of atoms, and has already answered all the ones
which are currently regarded as being interesting.
My simple
> cubic model justifies them as the primary quantum number as
> corresponding the the electrons in the core of the atoms and the
> shells result from the geometric sequence extending out.
Your simple cubic model is a heap of rubbish, unless it can by any chance
tell us about the Lamb-Retherford frequency in the Hydrogen atom.
You have more confidence in your dung heap than is warranted by its value as
garden compost.
Franz
Franz Heymann
"FrankH" <frank...@yahoo.com> wrote in message
news:46484c9f.04011...@posting.google.com...
> "Franz Heymann" <notfranz...@btopenworld.com> wrote in message
news:<bu76kh$p31$1...@hercules.btinternet.com>...
> > "FrankH" <frank...@yahoo.com> wrote in message
> > news:46484c9f.04011...@posting.google.com...
> >
> > [snip]
> >
> > You appear not to know any physics worth talking about.
> >
> > Franz
>
> Damn straight! and proud of it.
That's it, then.
[snip]
Franz
Franz Heymann
If the depth of your ignorance is such that you think that the force between
a charged sphere and a polarisable dielectric will obey the inverse square
law, you should not be putting your nonsense on the net.
Franz
Franz Heymann
Note for Galathea:
Perhaps this last effort from FrankH should convince you that his mental
abilities are not up to what is required to study quantum mechanics..
Franz
Bjoern Feuerbacher
outdated
view (Bohr's model), which was replaced approx. 80 years ago.
FrankH wrote:
>
> The central theme of most models of the atom is that there is a
> central nucleus containing neutrons/protons and electrons surround
> this nucleus. Quantum mechanics show the electrons in probability
> clouds around the nucleus.
Oh, you *do* know this? Then why did you choose such a nonsensical title
for your thread?
> As near as I can tell, the reason for this
> was due to electron scattering experiments done nearly 100 years ago,
That's *one* of the reasons, by far not the only one.
> using instruments that would be considered incredibly crude by todays
> standards.
They were perfectly well suited for the task which was tried to achieve.
The results are reliable.
Further, this experiment (you *do* talk about Rutherford scattering,
don't you?) has been repeated countless times in the meantime, with far
better instrumental equipment. Hey, I even did it myself a few years
ago!
> These experiments showed that the nucleus had to be compact
Right.
> and the electrons extremely small
Wrong. The experiment didn't show anything about the size of electrons.
> which made an atom mostly empty
> space. But did they really???
Yes - provided that Coulomb's law is right.
> If I took a gun and fired bullets at a
> stack of cardboard boxes filled with packing peanuts, I might also
> conclude that the boxes were mostly empty space due to limited
> backscattering - when in fact, the boxes and their contents are quite
> large.
False analogy. Rutherford scattering is due to electrostatic forces.
> So the logic used to determine the size of the nucleus escapes
> me.
Well, then try to learn the physics behind it.
> I have seen new pictures generated with the latest STM technology
> at http://arxiv.org/PS_cache/cond-mat/pdf/0305/0305103.pdf
I heard a talk about these results, and IIRC, they were disputed.
> What is
> remarkable is that we can now actually image the structure of
> individual atoms.
Possibly, yes. I wouldn't be too sure about this. Does anybody know if
these results have been confirmed by other researchers in the meantime?
> The picture shows the surface of a group of Silicon
> atoms and what I see doesn't appear to be mostly space or a fuzzy
> cloud.
Huh? To me, the "clouds" look *very* fuzzy! Try looking at Fig. 5, for
example! Even in Fig. 6 d, which is a *very* clear picture, the "clouds"
*still* look fuzzy!
> Instead, it looks a lot like lego building bricks to me with
> very distinct edges.
Are you sure you are looking at the same pictures as me???
> My question is, has physics done any recent
> experiments to verify the well accepted values for the size of the
> atom/nucleus and electron?
Yes. Such experiments are done all the time. Sizes of the atoms are
measured by determining the distances between layers and the overall
geometry of crystal lattices, for examples. Sizes of nuclei are a
by-product in lots of experiments done on nuclear structure. The size of
electrons is measured (and again is a by-product) in experiments done
with scattering of electrons and positrons (LEP, for example).
> I am wondering this because I was thinking, if I were a bunch of
> protons and electrons, how would I assemble into atoms?
By attracting each other by the electrostatic force and finding a stable
configuration of "my" probability distribution.
> Naturally, the
> answer would be to start sticking together like building blocks with
> proton/electrons alternating like a NaCL crystal.
That's "natural" in the macroscopic world, but unfortunately, such
common sense doesn't work for such small particles.
> This is totally contrary to the standard atomic nucleus model.
Duh.
> I took this further and
> started building out this structure so that you have atoms which are
> built out of a regular geometric sequence of ever increasing layers of
> electrons/protons.
Can this model predict the Rutherford scattering cross section?
Can it predict the valences of atoms, i.e. their possibilities to bind
other atoms?
Can it predict ionization energies?
The standard model can do all of this nicely - all of these things were
incorporated in it right from the start. So, you have got a *lot* of
things to do if you want to replace it! Good luck.
> Curiously, the model does match some of the
> observed electronic states for the atoms I have built out.
In what way does it "match" them?
> However,
> this model would mean that the central nucleus wouldn't exist.
Well, then please explain the results of the Rutherford experiment.
And the fact that the standard model, with the central nucleus, does
correctly predict a *lot* of features of atoms (start with the spectra).
> All of
> the proton/electrons would be spread out in roughly a Octrahetral
> shape in a completely neutrally charged matrix.
Completely contrary to experimental results, sorry.
> I have detailed this
> model and some pictures from this at:
>
> http://ourworld.compuserve.com/homepages/frankhu/buildatm.htm
Try answering the questions above, then we'll see.
> This web site also has the STM images I was talking about before.
Thanks, I already looked at them in the original paper.
> I have not seen anyone propose such a simple theory before,
Have you looked at the model of Y. Porat? ;-)
> so I was
> interested in seeing what people think of it since this is a new way
> of thinking about the atom with no central nucleus, but uses regular
> geometric progression to describe the atomic structure.
I think it contradicts experimental results. If you think otherwise,
derive
the Rutherford scattering cross section and the atomic spectra from
your model.
Bye,
Bjoern
Bjoern Feuerbacher
>
> "Pyriform" <nob...@nowhere.com> wrote in message news:<40073268$0$2439$cc9e...@news.dial.pipex.com>...
> > Sam Wormley wrote:
> > > Careful Frank, these might blowout your mind.
> > >
> > > Quantum Numbers
> > > http://scienceworld.wolfram.com/physics/QuantumNumbers.html
> > >
> > > Hydrogen Atom
> > > http://scienceworld.wolfram.com/physics/HydrogenAtom.html
> > >
> > > Atom
> > > http://scienceworld.wolfram.com/physics/Atom.html
> >
> > Be careful now - you are pitting yourself against a profound thinker. I
> > quote from his website:
> >
> > "What causes gravity?
> > Gravity is caused by a slight imbalance of positive/negative charges in
> > so called neutrally charged matter. The negative/positive charges in a
> > neutral atom do not exactly cancel each other out. There is a tiny
> > residual positive charge. These tiny positive charges added together
> > over the volume of the earth produce a very large positively charged
> > field at the surface of the earth. The diverging electrostatic field
> > created by the earth causes dipoles in neutrally charged matter to be
> > attracted to the source of the field"
> >
> > So simple to see once a genius has lit the path.
>
> Ah yes, a famous quote from my own website. You laugh but some day
> ....
Well, what you neglect to consider is (among other things):
1) The charge of the electron and the proton have been shown to be equal
with *very* great accuracy - around 10^21! (see, for example, here:
<http://pdg.lbl.gov/2002/bxxxn.pdf>)
2) If the earth would have an overall positive charge (BTW, the
formulation "positively charge field" doesn't make sense), we would have
noticed this long ago - for example, by observing the trajectories of
charged cosmic rays.
3) The attraction force between a positive charge and dipoles depends
with 1/r^3 on distance, whereas the force of gravity depends on distance
with 1/r^2.
> Actually there is no contradition between my lego block theory of
> atoms an the electrostatic theory of gravity.
Well, unfortunately, there *are* a lot of contradictions with
observations.
See above.
> All that is needed is a
> slight imbalance of positive/negative charges on average.
See above for counterarguments.
> I have read
> that it would take only 1 electron to be missing in 10^18 atoms to
> create an electrostatic force equivalent to gravity.
Well, unfortunately for you, the charges of electrons and protons have
been shown to be equal to an accuracy of 10^21, so this charge of "1
electron missing per 10^18 atoms" can't be provided by your proposal.
> It isn't hard to
> belive that out of a billion billion atoms that one of them might be
> missing an electron,
It is neither hard to believe that out of a billion billion atoms, one
of them might have an electron too much, don't you think?
> in fact I would be suprised if there weren't any
> missing electrons considering how easily they can be ionized.
And *I* would be surprised if there weren't any additional electrons
considering how easily they can be attached.
> The
> Earth is also constantly bombarded by positively charged ions from the
> solar wind,
Oh, so you *know* that cosmic rays are charged? Why don't you see that
the observation of the trajectories of these rays instantly disproves
your "the earth is positively charged" idea?
> you'd think that alone would be enough to put a positive
> charge on the earth.
Hint: there are negatively charged cosmic rays, too...
Bye,
Bjoern
Bjoern Feuerbacher
>
> Sam Wormley <swor...@mchsi.com> wrote in message news:<4006DB47...@mchsi.com>...
> > Careful Frank, these might blowout your mind.
> >
> > Quantum Numbers
> > http://scienceworld.wolfram.com/physics/QuantumNumbers.html
> >
> Yes, very interesting these quantum numbers - Can Quantum Mechanics
> justify the periodicy that we observe in the elctron shells?
Depends on what you mean by "justify". QM *predicts* this periodicity.
It follows automatically when you solve the Schroedinger equation for
a Coulomb potential. BTW, the periodicity observed in *nuclear* shells
is *also* predicted, by the very same methods (only difference: anotjer
potential is used).
> My simple
> cubic model justifies them as the primary quantum number as
> corresponding the the electrons in the core of the atoms and the
> shells result from the geometric sequence extending out.
How does your model explain the spectra of the atoms, and the results
of the Rutherford scattering experiments?
> > Hydrogen Atom
> > http://scienceworld.wolfram.com/physics/HydrogenAtom.html
> >
> Gee, all this math to explain a Hydrogen atom?
Well, can you predict the spectrum of a Hydrogen atom with less math?
If yes, please show us!
> What was wrong with
> saying that a hydrogen atom is composed of a proton and electron?
This doesn't give much information about it's behaviour, for starters.
> > Atom
> > http://scienceworld.wolfram.com/physics/Atom.html
>
> Yup, this is the rather limited history of the atomic model. As I
> mentioned, my simplistic model is not part of this history.
Nice. Hint: lots of other models (by laymen like you) are also not part
of this history. Try thinking about why this is the case... (hint: it
has to do something with "predictive power" and "consistency with
experimental results")
> I would
> have though this would have been the first to be proposed and then
> thrown out. Certainly, it would be more believeable than the plum
> pudding model. So what happened?
This model of "Lego blocks" of yours reminds me a bit of Demokrit's view
of atoms - granted, he didn't think that atoms consist of Lego blocks,
but
he *did* think that there are some "elementary particles", looking
essentially like "blocks" of different forms, and these elementary
building blocks he called "atoms".
So, essentially, your model *was* first proposed and then thrown out.
Bye,
Bjoern
Pyriform
> Well, thank you for the nice compliment - I am an MIT graduate so
> I am more educated than most - but I admit I no next to nothing when
> it comes to particle physics
Engineering degree, right?
--
Pyriform
Bjoern Feuerbacher
>
> Uncle Al <Uncl...@hate.spam.net> wrote in message news:<4006E079...@hate.spam.net>...
> > FrankH wrote:
> > >
> > > The central theme of most models of the atom is that there is a
> > > central nucleus containing neutrons/protons and electrons surround
> > > this nucleus. Quantum mechanics show the electrons in probability
> > > clouds around the nucleus. As near as I can tell, the reason for this
> > > was due to electron scattering experiments done nearly 100 years ago,
> >
> > You are an uneducated dunce. Hey stooopid, don't you think that in
> > the 21st century there is an overwhelmingly preponderant understanding
> > of quantum mechanics that allows calculations to 14 significant
> > figures to agree with equally precise observations? Look up the "Lamb
> > shift."
> >
> > If you cannot manage Google, have a grade school kid help you.
> >
> > [snip simplistic crap]
>
> Well, thank you for the nice compliment - I am an MIT graduate, so I
> am more educated than most
What degrees do you have? It looks strange to me that a graduate of the
MIT should come up with such a strange model...
> - but I admit I no next to nothing when it
> comes to particle physics which is why I need the assistance of
> geniuses like yourself to help me ponder the nature of the universe.
Why didn't you try to learn something about atomic, nuclear and particle
physics first? About the available experimental evidence? About the
reasons
which lead to the current models?
Instead of trying to learn, you invented your own model - and now claim
that it's better than the standard models, although you don't even
*know*
much about them!
> I looked up the Lamb shift and found it quite interesting. I have
> heard of this thing that got calculated to 14 digits of precision to
> justify QM, but didn't know what it was.
It's new to me that the Lamb shift was calculated to so many digits.
Wasn't
this the anomalous magnetic moment of the electron instead?
> Thank you for clearing that
> up. Although from what I read, it sounds a bit suspicious considering
> science has a hard time measuring anything experimentally to 14 digits
> of precision. If I wanted to measure the diameter of a penny, I
> couldn't do that to 14 digits of precision, so how can something that
> has to do with atoms we can't even see be measured with such
> precision.
What can be measured with *very* great precision are frequencies (AFAIK,
it has to do something with "beats") - so if you want to measure
something with great precision, you only have to find a way to "convert"
it into a frequency. This can be done easily when measuring the magnetic
moment of the electron or the Lamb shift.
> But in any case, supposing that this result is correct, are
> there any other such examples of QM coming out with an experimentally
> verified prediction - I didn't see any in the web sites I visited.
*sigh* Try opening *any* textbook on QM or particle physics. You will
find *lots* of experimentally verified predictions.
Even better, go to the next university library and open an arbitrary
journal about experimental particle physics. I think most of the
articles you find in there (and there are tens of thousands of them!) do
contain an experimental
verification of predictions of QM resp. the Standard Model of Particle
Physics.
Some things to read about:
* anomalous magnetic moment of the electron and of the muon
* Casimir effect
* atomic spectra in general
* Moseley's formula specially
* resonances
* *countless* scattering experiments - among others, electron-positron
scattering
* neutrino oscillations (contradict the Standard Model (but this can be
easily solved), but are a *very* nice verification of principles of QM!)
* lasers
* the dependence of the heat capacity of metals on the temperature
These are only *very* few things which came to mind almost immediately.
There is *lots* of stuff more out there! Open your eyes!
> The thing that they were measuring was also quite obscure - a slight
> variation of a superfine spectrum line.
Yes, in a sense, that's obscure. So what??? Why should this be a *bad*
thing??? Isn't it amazing that we are able to predict accurately even
such
obscure things!?!
> How about something big and
> obvious like the observed shell structure of atoms and why the shells
> have the number of electrons and energy states that they do?
That *is* predicted by Schroedinger's equations. Complete with all of
the energies associated with these shells. Oh, add this to the list
above.
BTW, the observed shell structure of nuclei (I bet that you even haven't
*heard* so far that there is a shell model for nuclei, right?) is *also*
predicted by standard QM - again, in complete agreement with
experimental results. Oh, add this to the list above.
> Why do they have such a geometric sequence?
What's "geometric" about their sequence?
> I have a book on quantum
> mechanics and I don't find a chart of the electron shell sequence in
> it.
What book is this?
And what, precisely, do you want to see in such a chart?
> Does quantum mechanics explain why atoms combine into things like
> H2O?
Yes.
> Why not H4O instead?
Energetically unstable, plain and simple. You can't have a geometrically
stable configuration of an oxygen nucleus, four protons and 20
electrons.
> Surely there must be some calculation that
> shows that the imbalance in the probability orbitals of the atoms
> cause a lower energy state to exist or something like that.
Yes, you are close to it.
> Although I
> suspect the answer to all of these questions is No.
Well, you suspect wrong.
> But then again,
> I'm an uneducated Newbie, and need the help from you folks who know
> better.
OTOH, you are a graduate from MIT, on the other hand, you call yourself
"uneducated". Seems to be a bit contradictory...
Bye,
Bjoern
Bjoern Feuerbacher
>
> "Franz Heymann" <notfranz...@btopenworld.com> wrote in message news:<bu76kh$p31$1...@hercules.btinternet.com>...
> > "FrankH" <frank...@yahoo.com> wrote in message
> > news:46484c9f.04011...@posting.google.com...
> >
> > [snip]
> >
> > You appear not to know any physics worth talking about.
> >
> > Franz
>
> Damn straight! and proud of it.
In other words, you are unwilling to learn what our current models
say, why we think they are right, and what the experimental evidence
for them is?
> I don't know anything about your
> physics filled to the brim with counter-intuitive ideas, miles of
> nonsensical math, multi-dimension and generally defying all logic.
In other words "I don't understand it / like it, therefore it's wrong
and I don't bother learning anything about it."
> You all complain about not being able to find the one big elegant theory
> everything, because you have to admit current theory can't do it, but
> yet you go around like you already know how everything in the world
> works down to the tiniest detail, but you don't.
Why you think anyone here is "going around like [he] know[s] already how
everything in the world works"? I haven't seen anyone going around like
that in this thread. I, myself, am well aware that I know little about
the
world. OTOH, even knowing so little, it's clear that your model doesn't
work. You seem to think that anyone who critizes your model is doing
this simply because he is arrogant and close-minded...
> Excuse the flame, but
> this forum exists to assist people understand and challange scientific
> ideas, not to state the obvious that I am unfamiliar with the world of
> conventional physics.
The reaction of many people here is quite rude, granted, but that it
partly your fault. We have seen *far* to many people coming into this
newsgroup proudly shouting that they've found a new, simple model for
how xyz works, and that all standard theories are wrong. *Every* time,
these people knew little about what the standard theories even say (you
are a bit better here - you at least know that something like
"probability clouds" are supposed to exist!), but this lack of knowledge
never stops such people.
> I am presenting you with what appears to be an
> entirely new concept of the atom never before seen.
Sorry, we have seen stuff like this *far* too many times already...
> Shouldn't that be
> the least bit interesting to you?
It will get interesting as soon as you are able to predict the most
basic
things like atomic spectra and Rutherford's scattering cross section
from
your model.
> How often do you run across unique theories of atom formation?
*Lots* of times in this newsgroup... So far, none of it has been worth
looking at it. None of it even *tried* to predict the most basic
observable things like atomic spectra, Rutherford's scattering cross
section, and so on.
Bye,
Bjoern
Bjoern Feuerbacher
>
> >
> > In case the others haven't scared you off from learning, you should really
> > read up on scattering theory. In particular, going from simple Rutherford
> > models to more complex atomic scattering models would really help you clear
> > up some of this confusion. It is quite a pretty field, and the problems of
> > inverse scattering calculations and objectives like determining charge
> > structure and field dynamics from scattering distributions in space and time
> > will give one a broad understanding of modern thought in this area.
>
> Thank you for your thoughful post. Can you recommend a web site that
> might explain the more advanced scattering experiments?
I don't know about a web site, but I know a book which, in my opinion,
discusses this quite nicely.
<http://www.amazon.com/exec/obidos/tg/detail/-/0387594396/qid=1074253613//ref=sr_8_xs_ap_i0_xgl14/102-7930935-5172948?v=glance&s=books&n=507846>
If the link doesn't work: the author is B.Povh, the title is
"Particles and Nuclei: An Introduction to the Physical Concepts". I know
only the German version of this book, but I think the English version
should be nice, too... ;-)
> The books I've
> been able to find devote about 1 sentence to the Rutherford experiment
> with no further justification.
That's a pity. :-(
> Although fundamentally, I have a hard
> time believing you can determine anything by scattering considering
> that we don't know how "hard" the things that the electrons are
> bouncing off.
Scattering of electrons has little to do with "hardness". It works by
the
*Coulomb* force.
> They could be billiard balls or puffs of gas.
Scattering of electrons has little to nothing to do with scattering of
marbles of billiard balls. The mechanisms are completely different.
> I'd have to see more justification.
Well, *assuming* that there is indeed a "compact" nucleus with a
positive
charge equal to the number of electrons around it, and calculating what
number of particles should be scattered into which direction, and
comparing this prediction with experiment, it turns out that there is
*very* good agreement. (read up on "Rutherford scattering cross
section") I still have to see someone reproduce this formula with
another model...
> For me, the new STM pictures of Silicon
> atoms as little bricks is most convincing.
Derive Rutherford's scattering cross section formula from it.
> If Rutherford had an STM, I
> don't think he would be so quick to conclude that an atom is 99.999%
> empty space. It looks 100% filled
Err, did it ever occur to you that the brightness of the clouds in these
pictures is purely conventional? The density of the electrons in atoms
*is*
very, very, very small - these pictures don't contradict this in any
way.
> with sharply defined edges to me.
Are you sure we are looking at the same pictures????? I don't see sharp
edges there, but *fuzzy* edges.
> The picture on the web site also shows where an atom is missing and
> you can peer down and see the clearly defined sides of the other
> surrounding atoms - how can you explain this with an atom which is
> nearly all empty space with the electrons wizzing about randomly? It
> looks solid and space filling to me.
Look at the Figure 5. To me, the atoms look *very* fuzzy to me. Even
more, the lines drawn in there show you how the electron density
decreases from the center outwards. This is exactly what the theory
predicts! And, BTW, no one says that electrons are "wizzing [sic] around
randomly" there.
Bye,
Bjoern
OC
> The central theme of most models of the atom is that there is a
> central nucleus containing neutrons/protons and electrons surround
> this nucleus. Quantum mechanics show the electrons in probability
> clouds around the nucleus. As near as I can tell, the reason for this
> was due to electron scattering experiments done nearly 100 years ago,
> using instruments that would be considered incredibly crude by todays
> standards. These experiments showed that the nucleus had to be compact
> and the electrons extremely small which made an atom mostly empty
> space. But did they really??? If I took a gun and fired bullets at a
> stack of cardboard boxes filled with packing peanuts, I might also
> conclude that the boxes were mostly empty space due to limited
> backscattering - when in fact, the boxes and their contents are quite
> large. So the logic used to determine the size of the nucleus escapes
> me. I have seen new pictures generated with the latest STM technology
> at http://arxiv.org/PS_cache/cond-mat/pdf/0305/0305103.pdf What is
> remarkable is that we can now actually image the structure of
> individual atoms. The picture shows the surface of a group of Silicon
> atoms and what I see doesn't appear to be mostly space or a fuzzy
> cloud. Instead, it looks a lot like lego building bricks to me with
> very distinct edges.
Could it be that they are showing the surface of a Si crystal?
> My question is, has physics done any recent
> experiments to verify the well accepted values for the size of the
> atom/nucleus and electron?
Do you think that modern physics would work if the atomic model was
completly wrong? You cannot build a house on bad foundations.
> <snip>
OC
Sean Chapin
Please spare us of your pretentious garbage where you think you are a
cut above the rest because you went to MIT or any other school that
claims to only graduate the "best". I've been to enough conferences
and met enough graduates of institutions like yours, both in academia
and in industry, to know the only way any of you are a cut above the
rest is that you are better BS artists. The truly gifted ones are few
and far between. Aside from that, you certainly aren't any more
"educated" than anyone else, and arrogant comments like yours bear
that out quite clearly. Get over yourself. You and your degree ain't
that special.
If you actually want to discuss science, with some logic behind it,
and without flashing around where your degree comes from, then you'll
get a lot more ears here.
Gregory L. Hansen
You realize, don't you, that an STM is not like poking an atom with an ice
pick and drawing a dot where it goes "tink tink". The signal is an
electrical current that varies continuously as the distance between tip
and atom is changed. Where you draw the contour is arbitrary. If the
Leggos you're looking at are on the right on page 8, that looks to me like
artifacts caused by a finite resolution of the instrument. Download any
porno picture from the internet and zoom in enough, and you'll see the
same effect.
Scattering theory is rather complex. Most quantum mechanics textbooks
will get you far enough to distinguish "hard" from "soft" potentials. The
billiard ball model, i.e.
V(r) = V0, r<a
= 0, r>a
is a standard example of a particle with a clearly defined edge. Other
models, like V(r)=1/r or V(r)=exp(kr)/r give different scattering results.
You can theoretically model the scattering from a 1/r potential versus a
potential like
V(r) = V0, r<a
= 1/r, r>a
which would compare the scattering from a point-like particle versus a
particle with a finite charge radius. Internal structure, like the quarks
that make up a proton, will change your scattering results in complicated
ways. To really explore that theoretically is in the realm of quantum
field theory.
--
"In any case, don't stress too much--cortisol inhibits muscular
hypertrophy. " -- Eric Dodd
Franz Heymann
"Bjoern Feuerbacher" <bfeu...@ix.urz.uni-heidelberg.de> wrote in message
news:4007C797...@ix.urz.uni-heidelberg.de...
[snip]
> 3) The attraction force between a positive charge and dipoles depends
> with 1/r^3 on distance, whereas the force of gravity depends on distance
> with 1/r^2.
It is actually a lot worse than that. His dipole is not a permanent dipole,
but one whose moment is induced by the local field. thus P itself varies as
1/r^2, giving him a 1/r^5 interaction. Such a radial dependence is
incapable of supporting a stable orbit.
[snip]
Franz
Franz Heymann
> FrankH wrote:
> >
> > >
> > > In case the others haven't scared you off from learning, you should
really
> > > read up on scattering theory. In particular, going from simple
Rutherford
> > > models to more complex atomic scattering models would really help you
clear
> > > up some of this confusion. It is quite a pretty field, and the
problems of
> > > inverse scattering calculations and objectives like determining charge
> > > structure and field dynamics from scattering distributions in space
and time
> > > will give one a broad understanding of modern thought in this area.
> >
> > Thank you for your thoughful post. Can you recommend a web site that
> > might explain the more advanced scattering experiments?
>
> I don't know about a web site, but I know a book which, in my opinion,
> discusses this quite nicely.
>
<http://www.amazon.com/exec/obidos/tg/detail/-/0387594396/qid=1074253613//re
f=sr_8_xs_ap_i0_xgl14/102-7930935-5172948?v=glance&s=books&n=507846>
>
> If the link doesn't work: the author is B.Povh, the title is
> "Particles and Nuclei: An Introduction to the Physical Concepts". I know
> only the German version of this book, but I think the English version
> should be nice, too... ;-)
I don't think that book will be very suitable for FrankH. It is, after all,
only an introductory book, and he specifically asked for a book which would
"explain the more advanced scattering experiments". Perhaps he should
acquire Greiner's two excellent books on quantum mechanics. I have English
versions of them.
{:-))
[snip]
Franz
Franz Heymann
news:<4006E079...@hate.spam.net>...
> > FrankH wrote:
> > >
> > > The central theme of most models of the atom is that there is a
> > > central nucleus containing neutrons/protons and electrons surround
> > > this nucleus. Quantum mechanics show the electrons in probability
> > > clouds around the nucleus. As near as I can tell, the reason for this
> > > was due to electron scattering experiments done nearly 100 years ago,
> >
> > You are an uneducated dunce. Hey stooopid, don't you think that in
> > the 21st century there is an overwhelmingly preponderant understanding
> > of quantum mechanics that allows calculations to 14 significant
> > figures to agree with equally precise observations? Look up the "Lamb
> > shift."
> >
> > If you cannot manage Google, have a grade school kid help you.
> >
> > [snip simplistic crap]
>
> Well, thank you for the nice compliment - I am an MIT graduate, so I
> am more educated than most - but I admit I no next to nothing when it
> comes to particle physics which is why I need the assistance of
> geniuses like yourself to help me ponder the nature of the universe.
>
> I looked up the Lamb shift and found it quite interesting. I have
> heard of this thing that got calculated to 14 digits of precision to
> justify QM, but didn't know what it was.
It was the Lamb shift. I thought you said you looked it up?
> Thank you for clearing that
> up. Although from what I read, it sounds a bit suspicious considering
> science has a hard time measuring anything experimentally to 14 digits
> of precision.
The Lamb shift has been measured and agrees exquisitely with the
predictions. I thought you said you looked it up? If you did, you would
have kinown how it was done.
> If I wanted to measure the diameter of a penny, I
> couldn't do that to 14 digits of precision, so how can something that
> has to do with atoms we can't even see be measured with such
> precision.
Physicists are clever. Now go and read about the Lamb shift and come back
when you understand something about it. Hint: The Lamb shift is a very
small frequency difference between two extremely large frequencies.
> But in any case, supposing that this result is correct,
There is no reason to do any supposing. The experiment has been performed
by more than one group of physicists, and by now it so well understood that
it is probably an undergraduate lab experiment in some physics department
somewhere.
> are
> there any other such examples of QM coming out with an experimentally
> verified prediction - I didn't see any in the web sites I visited. The
> thing that they were measuring was also quite obscure - a slight
> variation of a superfine spectrum line.
Quite.
> How about something big and
> obvious like the observed shell structure of atoms and why the shells
> have the number of electrons and energy states that they do?
Been there. Done that. There was an international collaboration who
speciality was the calculation of the atomic wavefunctions of all the atoms
up to Iron. The calculations took about twenty physicists about a decade,
using the most powerful computers available at all stages of the work. They
now know the electronic configurations of all the electrons in all those
atoms for all the most important excited states, and have predicted the
frequencies and the relative intensities of the major spectral lines as
functions of temperature for all those atoms. Their work is in daily use by
astrophysicists who are concerned with studying the structures of stars.
Quantum mechanics has provided answers to *all* the problems to which it has
been put, and it has never turned out an incorrect answer.
> Why do
> they have such a geometric sequence?
Why don't you learn some physics?
> I have a book on quantum
> mechanics and I don't find a chart of the electron shell sequence in
> it. Does quantum mechanics explain why atoms combine into things like
> H2O?
Molecular structure is an extremely difficult study. Even at the
macroscopic level, the three-body problem does not have an analytical
solution. Having said that, there are groups of physicists who spend their
time doing the tedious computations necessary to solve for the wavefunctions
of molecules. They can in practice handle only two and three atom molecules
because of the extreme computational difficulties.
> Why not H4O instead? Surely there must be some calculation that
> shows that the imbalance in the probability orbitals of the atoms
> cause a lower energy state to exist or something like that.
See what I said above.
> Although I
> suspect the answer to all of these questions is No.
Your suspicions are quite unfounded and exist only because of your total
lack of knowledge.
> But then again,
> I'm an uneducated Newbie, and need the help from you folks who know
> better.
We cannot actually help you much, both because of the time which would be
involved and the difficulties associated with writing any serious
mathematics in ASCII.
If you are really interested, you would stop wasting your time trying to
reinvent the wheel, and acquire some really elementary modern physics books
to start your studies.
Franz
Franz Heymann
[snip]
I am an MIT graduate, so I am more educated than most
On the contrary, your writings make you appear to be both singularly
ignorant and singularly arrogant.
[snip]
Franz
Franz Heymann
Remember that the Lamb shift is a very small difference in frequency between
two very large frequencies. What Uncle Al meant was that the limit on the
computational accuracy of the *difference* is in the region of 10^-14 of the
frequency of one of the lines.
> Wasn't
> this the anomalous magnetic moment of the electron instead?
No. That has been computed and compared with experiment to something like 1
part in 10^8 or threreabouts.
FrankH
> > The picture shows the surface of a group of Silicon
> > atoms and what I see doesn't appear to be mostly space or a fuzzy
> > cloud.
>
> Huh? To me, the "clouds" look *very* fuzzy! Try looking at Fig. 5, for
> example! Even in Fig. 6 d, which is a *very* clear picture, the "clouds"
> *still* look fuzzy!
>
>
> > Instead, it looks a lot like lego building bricks to me with
> > very distinct edges.
>
> Are you sure you are looking at the same pictures as me???
is the high resolution picture of 6a. What I want you to specifically
look at is the spaces between the atoms. There is a distinct dropoff
between each atom such that there is an actual physical separation
between the atoms. Each atom looks suspiciously like a Duplo brick
which is basically a cube with a round nub at the top. A green arrow
in the picture shows an atomic defect where an atom is missing and you
can see the edges go all the way down. How can QM explain such sharp
dropoffs and edges between atoms? The picture you have pointed out as
6d, is not an STM image, it is the predicted shape of the atom based
on QM. So of course, this is a fuzzy cloud. You are supposed to look
at the real image in 6c and compare it with the theoretical shape in
6d and then conclude they are the same. Personally, I don't think it
is a good fit. You can almost seem to make out some substructure in 6c
like the front of the atom is made out of 2 lobes.
Thanks for providing such detailed posts, it is giving me a lot of
areas to look at - what fun!
FrankH
>
> Could it be that they are showing the surface of a Si crystal?
>
difference?
>
> Do you think that modern physics would work if the atomic model was
> completly wrong? You cannot build a house on bad foundations.
>
This reminds me of the situation before we figured out the planets
revolve around the sun: From: http://c2.com/cgi/wiki?AddingEpicycles
If you have a bad design (such as trying to work out the motion of
planets on paper while constrained by dogma to pretend that the sun
moves more than the earth), and if you then find you keep having to
add more bad design to add features to that design, then you are
"AddingEpicycles".
As Renaissance astronomers got better at recording the exact locations
of planets (relative to the surface of the earth at given times), they
kept trying to plot these locations back to presumed coordinates (and
offsets) on traditionally decreed "celestial spheres" that carried the
planets. But the more precise they got, the more complicated their
offsets were as they added epicycles. These eventually became
celestial spheres bearing epicycles bearing epicycles bearing
epicycles bearing planets. Obscured within the numbers, the planets
were - of course - really going around the Sun.
The cure, of course, is to fix the root bad design. If you get it
right, the change will ripple thru all the subsidiary designs and
simplify them, possibly by making them go poof.
The fact that lots of scientists believe a theory for a long time,
does not prove it is correct. Yes, the atomic model could be totally
wrong, yet still make predictions, especially if you try to build the
model to specifically fit the observations instead of starting with a
model and seeing if it matches observations.
Tnlockyer
>Date: 1/15/2004 10:32 PM Pacific Standard Time
Wrote in:
>Message-id: <46484c9f.04011...@posting.google.com>
>
>"Franz Heymann" <notfranz...@btopenworld.com> wrote in message
>news:<bu76kh$p31$1...@hercules.btinternet.com>...
>> news:46484c9f.04011...@posting.google.com...
>>
>> [snip]
>>
>>
>> Franz
>
>physics filled to the brim with counter-intuitive ideas, miles of
>all complain about not being able to find the one big elegant theory
>everything, because you have to admit current theory can't do it, but
>yet you go around like you already know how everything in the world
>this forum exists to assist people understand and challange scientific
>ideas, not to state the obvious that I am unfamiliar with the world of
>conventional physics.
Frank, well said: The sad fact is that physics does not have the answers after
100 years of theory and experiment. If you were to, as they suggest, go back
and see what has been proposed, you could spend (waste) your whole life, as
they have, and be no closer to the truth.
> I am presenting you with what appears to be an
>entirely new concept of the atom never before seen. Shouldn't that be
>the least bit interesting to you? How often do you run across unique
>theories of atom formation?
Frank, you make the mistake of all the previous theorists in trying to reverse
engineer or to invent models to try and explain the experimental evidence.
What you want is a model that goes straight forward and that puts itself
together, more or less automatically and precisely, and is supported by prior
experimental results.
The atom model must explain, as a minimum, known atomic spin states, binding
energy, beta decay sequences, and nuclear magnetic moments.
The standard model can do none of those things, because the SM does not have
the correct models for the nucleon structures. If the SM nucleon models were
correct, SM would give us the value for the (n-1H) neutrino energy so important
in beta decay calculations.
Don't be misled into thinking the Schroedinger equation does anything it was
not put up to by the act of inserted boundary conditions to beg the results.
Heisenberg could not understand why the proton and neutron were apparently
equally effected by the "nuclear strong force" so he invented "isotopic spin
space". So anytime you read isotopic spin is those physics textbooks, it is
pure BS.
The strong force is not "special". The strong force proves to be purely
electromagnetic in nature. No one previously realized that the near field
nucleon magnetic moments are superior to the nuclear electric moments.
Being an engineer gives you a distinct advantage because you insist on
explaining the physical world in particle terms. Unfortunately, particle
physics seems satisfied with just abstract jargon.
Hey, jargon pays their rent, and may get them elected for the Nobel for some
theory that perpetuates the business. (i.e. if the theory invents some
particle that requires building the next round of larger accelerators, all the
better.).
Case in point, the electroweak theory that was elected for and given the Nobel
before anyone had attempted to prove it.
This guaranteed an accelerator would be built and a Nobel awarded to the
experimenters, on any far fetched claims to have "found" the particle.
The particle physics business has spawned many educated charlatans and
mountebanks, that will try to defend the SM and, as you have seen , some of
them frequent these newsgroups.
Regards: Tom:
Thomas Lockyer (77 and retired)
"When you can measure what you are speaking about, and express it in numbers,
you know something about it. Lord Kelvin (1824-1907)
Edward Green
...
> Well, thank you for the nice compliment - I am an MIT graduate, so I
> am more educated than most - but I admit I no next to nothing when it
> comes to particle physics which is why I need the assistance of
> geniuses like yourself to help me ponder the nature of the universe.
>
> I looked up the Lamb shift and found it quite interesting. I have
> heard of this thing that got calculated to 14 digits of precision to
> justify QM, but didn't know what it was. <...> The
> thing that they were measuring was also quite obscure - a slight
> variation of a superfine spectrum line. How about something big and
> obvious like the observed shell structure of atoms and why the shells
> have the number of electrons and energy states that they do?
Are you a recent MIT graduate or a centenarian? If you graduated from
MIT when QM was first being developed then your ignorance of the most
elementary facts of quantum chemstry, which I think are at least
alluded to today in HS chemistry, might be understandable, if you are
a recent graudate, then
it is inexplicable.
Y.Porat
Bjoern
please tell them that there is
'The Porat model of the Atom and the nucleus'............
in which among the others
a heavier than Florine nucleus is a sort of
a hectagonal pipe ? (8 sides)(if you whant it more simplified and tangible
it is a 'rectangular pipe' (with 8 free edges 4 at the front
and 4 at the back pole ) etc etc etc
and tell them that this Model goes all along the periodic table
not just the light elements etc etc etc.....
TIA
-----------------
all the best
Y.Porat
>
Franz Heymann
"Edward Green" <null...@aol.com> wrote in message
news:2a0cceff.04011...@posting.google.com...
Not really. The examination procedure is not 100% foolproof.
Franz
OC
> >
> > Could it be that they are showing the surface of a Si crystal?
> >
> Yes, this is absolutely the surface of an Si crystal, does that make a
> difference?
From your post, you seem to infer the electronic structure of a single
Si atom using STM images of a Si crystal, where there are a lot of Si
atoms bound together. I do not think that this is a correct way to
proceed.
From the experiments there is no indication that there is a "bad
design" at the base of the atomic model.
Physicists have no reason to think that the atomic model is
fundamentally wrong, that is why they trust it.
By the way, why should your model be better?
OC
rsm...@york.ac.uk
> Sam Wormley <swor...@mchsi.com> wrote in message news:<4006DB47...@mchsi.com>...
> > Careful Frank, these might blowout your mind.
> >
> > Quantum Numbers
> > http://scienceworld.wolfram.com/physics/QuantumNumbers.html
> >
> Yes, very interesting these quantum numbers - Can Quantum Mechanics
> justify the periodicy that we observe in the elctron shells?
The periodicity of the electron shells is one of the most famous
successes of the quantum-mechanical atomic model. It predicts a
periodic arrangement of the shells exactly like the one we observe.
> > Hydrogen Atom
> > http://scienceworld.wolfram.com/physics/HydrogenAtom.html
> >
> Gee, all this math to explain a Hydrogen atom? What was wrong with
> saying that a hydrogen atom is composed of a proton and electron?
That's exactly what this theory does say. The mathematics is required
to obtain meaningful predictions of the atom's behaviour, predictions
which are in very close agreement with experiment.
If you're serious about going anywhere in physics or any related
field, I'd suggest you get used to maths. You'll see it a *lot*.
OC
> >
> > Could it be that they are showing the surface of a Si crystal?
> >
> Yes, this is absolutely the surface of an Si crystal, does that make a
> difference?
From your post, it seems that you infer the electronic structure of a
single atom from the STM images. But these are images of many Si atoms
bound together in the crystal. Why should this way to proceed correct?
So, do you think that physicists have made the atomic model more and
more complicated over time by "adding epicycles"? That the atomic
model is fundamentally flawed, but physicists do not want to give it
up because it is established by dogma?
Isn't it possible instead that all the experiments done in the last
hundred years do not give any indication that the atomic model is
basically wrong?
On what is your conclusion based?
OC
Edward Green
> > Gee, all this math to explain a Hydrogen atom? What was wrong with
> > saying that a hydrogen atom is composed of a proton and electron?
>
> That's exactly what this theory does say. The mathematics is required
> to obtain meaningful predictions of the atom's behaviour, predictions
> which are in very close agreement with experiment.
>
> If you're serious about going anywhere in physics or any related
> field, I'd suggest you get used to maths. You'll see it a *lot*.
He claims to be a graduate of MIT. Perhaps that's "Massachusetts
Interstate Trucking" school? (Feynman graduated from the other one).
Bjoern Feuerbacher
>
> "Bjoern Feuerbacher" <bfeu...@ix.urz.uni-heidelberg.de> wrote in message
> news:4007C797...@ix.urz.uni-heidelberg.de...
>
> [snip]
>
> > 3) The attraction force between a positive charge and dipoles depends
> > with 1/r^3 on distance, whereas the force of gravity depends on distance
> > with 1/r^2.
>
> It is actually a lot worse than that. His dipole is not a permanent dipole,
> but one whose moment is induced by the local field. thus P itself varies as
> 1/r^2, giving him a 1/r^5 interaction.
Yes, I know - I merely tried to be generous... ;-)
> Such a radial dependence is
> incapable of supporting a stable orbit.
Even worse: it's contradicted by observations. ;-)
Bye,
Bjoern
Bjoern Feuerbacher
>
> "Bjoern Feuerbacher" <bfeu...@ix.urz.uni-heidelberg.de> wrote in message
> news:4007CCD2...@ix.urz.uni-heidelberg.de...
> > FrankH wrote:
> > >
> > > Uncle Al <Uncl...@hate.spam.net> wrote in message
> news:<4006E079...@hate.spam.net>...
[snip]
> > > I looked up the Lamb shift and found it quite interesting. I have
> > > heard of this thing that got calculated to 14 digits of precision to
> > > justify QM, but didn't know what it was.
> >
> > It's new to me that the Lamb shift was calculated to so many digits.
>
> Remember that the Lamb shift is a very small difference in frequency between
> two very large frequencies. What Uncle Al meant was that the limit on the
> computational accuracy of the *difference* is in the region of 10^-14 of the
> frequency of one of the lines.
The "two very large frequencies" should be the frequencies corresponding
to the 2s and 2p - orbitals of hydrogen. The corresponding frequency is
around 10^15 Hz. Hence the Lamb shift has been calculated to a precision
of around 10 Hz? Again, that's new to me. I remember only that it has
been calculated to a precision of around 1000 or even 10 000 Hz. (I read
about it in chapter 14 of
Weinberg's book on QFT, but admittedly that was about a year ago...)
> > Wasn't
> > this the anomalous magnetic moment of the electron instead?
>
> No. That has been computed and compared with experiment to something like 1
> part in 10^8 or threreabouts.
Well, in the table of the Particle Data Group, 11 digits are given.
[snip rest]
Bye,
Bjoern
Bjoern Feuerbacher
>
> Bjoern Feuerbacher <bfeu...@ix.urz.uni-heidelberg.de> wrote in message news:<4007C797...@ix.urz.uni-heidelberg.de>...
> > FrankH wrote:
[snip]
> Bjoern
> please tell them that there is
> 'The Porat model of the Atom and the nucleus'............
I already mentioned this to FrankH in another post.
> in which among the others
> a heavier than Florine nucleus is a sort of
> a hectagonal pipe ? (8 sides)(if you whant it more simplified and tangible
> it is a 'rectangular pipe' (with 8 free edges 4 at the front
> and 4 at the back pole ) etc etc etc
I'm still waiting for an explanation *why* nuclei should have this
structure...
> and tell them that this Model goes all along the periodic table
> not just the light elements etc etc etc.....
Your model essentially "predicts" the masses of nuclei. Hint: formulas
in standard theories (Weizsaecker's formula, for example) can do this,
too. Also for *all* elements. Not just the light ones.
Bye,
Bjoern
Bjoern Feuerbacher
FrankH wrote:
> Bjoern Feuerbacher <bfeu...@ix.urz.uni-heidelberg.de> wrote in message news:<4007C535...@ix.urz.uni-heidelberg.de>...
> > > The picture shows the surface of a group of Silicon
> > > atoms and what I see doesn't appear to be mostly space or a fuzzy
> > > cloud.
> >
> > Huh? To me, the "clouds" look *very* fuzzy! Try looking at Fig. 5, for
> > example! Even in Fig. 6 d, which is a *very* clear picture, the "clouds"
> > *still* look fuzzy!
> >
> >
> > > Instead, it looks a lot like lego building bricks to me with
> > > very distinct edges.
> >
> > Are you sure you are looking at the same pictures as me???
>
> Just to make sure we're on the same page, the picture I am looking at
> is the high resolution picture of 6a.
And on that picture, the "clouds" don't look fuzzy to you? Say, have you
got some problems with you eyes?
> What I want you to specifically
> look at is the spaces between the atoms. There is a distinct dropoff
> between each atom such that there is an actual physical separation
> between the atoms.
Again: are you *sure* we are looking at the same picture? Even if I
magnify the picture, I am unable to find any spot on it which is
*completely* black. A little red or yellow is essentially *everywhere*.
I don't see any clear physical separations, I see "clouds" which get
thinner and thinner as one moves away from the centers of the atoms, but
don't disappear completely anywhere.
And even if they *would* completely disappear on this picture - BFD!
There is only a limited amount of colors available for this picture; a
black spot doesn't mean that the electron density at that place is zero,
only that it is below a certain level!
> Each atom looks suspiciously like a Duplo brick
> which is basically a cube with a round nub at the top.
Please go to the nearest oculist. I don't think the picture looks
anything like that!
> A green arrow
> in the picture shows an atomic defect where an atom is missing and you
> can see the edges go all the way down.
I can see the green arrow, but I can't see "edges" there. All I'm seeing
there is the color getting ***gradually*** blacker and blacker, but
never getting *totally* black - there is always a little red left (a
very dark red, admittedly, but nevertheless red).
> How can QM explain such sharp
> dropoffs and edges between atoms?
Where on earth do you see a *SHARP* dropoff in this picture??? The
dropoff at the green arrow is gradual and smooth!
> The picture you have pointed out as
> 6d, is not an STM image, it is the predicted shape of the atom based
> on QM.
Yes, sorry, I noticed this in the meantime.
> So of course, this is a fuzzy cloud. You are supposed to look
> at the real image in 6c and compare it with the theoretical shape in
> 6d and then conclude they are the same.
No. You are to conclude they are sufficiently similar that the remaining
discrepancies could be due to instrumental limitations.
But, BTW, do you agree that picture 6c is *more* fuzzy than picture 6d?
(if no, then there is really a serious problem with your eyes; if yes,
then you agree that the *prediction* of Quantum Mechanics is a picture
which is *less* fuzzy than the actual observation - contrary to your
previous assertions...)
> Personally, I don't think it is a good fit.
Personally, I think they look quite similar, but I'm not entirely
convinced - and as mentioned elsewhere, IIRC, these results were
disputed by some other scientists. So one shouldn't rely too deeply on
them.
> You can almost seem to make out some substructure in 6c
> like the front of the atom is made out of 2 lobes.
Sorry, I can't see anything like that in the picture.
> Thanks for providing such detailed posts, it is giving me a lot of
> areas to look at - what fun!
Well, I hope you will *learn* something!
Bye,
Bjoern
Bjoern Feuerbacher
>
> o.ch...@rhul.ac.uk (OC) wrote in message news:<aea4c614.04011...@posting.google.com>...
[snip]
> > Do you think that modern physics would work if the atomic model was
> > completly wrong? You cannot build a house on bad foundations.
> >
> This reminds me of the situation before we figured out the planets
> revolve around the sun: From: http://c2.com/cgi/wiki?AddingEpicycles
Completely false analogy. 1) There were *far* less scientists back then
(one could argue that there even there *no* scientists back then). 2) As
far as I know, no other (working!) scientific theory was ever built on
the geocentric
model of Ptolemy (sp?).
> If you have a bad design (such as trying to work out the motion of
> planets on paper while constrained by dogma to pretend that the sun
> moves more than the earth), and if you then find you keep having to
> add more bad design to add features to that design, then you are
> "AddingEpicycles".
>
> As Renaissance astronomers got better at recording the exact locations
> of planets (relative to the surface of the earth at given times), they
> kept trying to plot these locations back to presumed coordinates (and
> offsets) on traditionally decreed "celestial spheres" that carried the
> planets. But the more precise they got, the more complicated their
> offsets were as they added epicycles. These eventually became
> celestial spheres bearing epicycles bearing epicycles bearing
> epicycles bearing planets. Obscured within the numbers, the planets
> were - of course - really going around the Sun.
Again, false analogy. The epicycles were added back then for only one
reason: to make the observations consistent with the description. It
wasn't
possible to make any *new* predictions based on the supposed epicycles.
Essentially, they were untestable. The situation today is completely
different: every new hypothesis in particle physics is thoroughly tested
before it is accepted, even if it makes a lot of sense at face value and
solves a lot of long-standing problems.
> The cure, of course, is to fix the root bad design.
Fine. Now all that remains to show for you is that Quantum Mechanics is
a "root bad design". Good luck!
> If you get it
> right, the change will ripple thru all the subsidiary designs and
> simplify them, possibly by making them go poof.
Again, good luck! So far, you haven't even reproduced the most *basic*
predictions of QM, like atomic spectra and Rutherford's scattering cross
section.
> The fact that lots of scientists believe a theory for a long time,
> does not prove it is correct.
Completely right.
But, OTOH, the fact that someone who knows practically nothing about QM
and particle physics thinks that it has to be wrong because he doesn't
like all the abstract mathematics, does not prove it isn't correct.
> Yes, the atomic model could be totally
> wrong, yet still make predictions,
Yes, this is a possibility. But considering the *thousands* of tested
predictions, this isn't too likely...
> especially if you try to build the
> model to specifically fit the observations instead of starting with a
> model and seeing if it matches observations.
In general, science uses *both* of these approaches. One should use a
mixture of them.
Bye,
Bjoern
FrankH
[snip]
> He claims to be a graduate of MIT. Perhaps that's "Massachusetts
> Interstate Trucking" school? (Feynman graduated from the other one).
Yes, that would be Mens Institude of Typewriting - Well actually, I
graduated in 1986 with a batchelors in computer science from the
Massachusetts Institute of Technology. I have the brass rat to prove
it. Quantum physics was not a required course for CS and not everybody
at MIT is a physics nut. That explains why I don't know much about
quantum mechanics - but I'm reading more every day! But I don't want
this discussion to be about me, tell me why my atomic model is wrong
or old or whatever - lets talk science.
FrankH
[snip]
> Frank, you make the mistake of all the previous theorists in trying to reverse
> engineer or to invent models to try and explain the experimental evidence.
>
> What you want is a model that goes straight forward and that puts itself
> together, more or less automatically and precisely, and is supported by prior
> experimental results.
Actually, as has been noted by the other posters, I am quite ignorant
of the experimental evidence and built my model starting with a
thought experiment of what would be the most simplest way for
electrons/protons to assemble. Please take another look at my website.
My model's advantage is that it does go together straight forward with
a predicatable geometric sequence which you can do almost blindly, and
it would put itself together automatically. So far, I have seen that
this model can explain the observed electron shell structure with the
main quantum number representing the electrons in the vertical core
and you can see why we get the number of electrons in the outer shells
due to the geometric progression of the outer electrons fitting into
the vertical core. I am working on understanding the spectrum and
scattering experiments and I think it may be possible to explain the
results in terms of my atomic model. Spin appears to be due to the
fact that the atom assembles a mirror image of itself about the axes,
and the reversed electron/proton may explain the difference states. I
would like to learn more about decay and nuclear fission, since my
model with its four arms would tend to favor fission which would knock
off arms and you might think that there would be more products which
represent a quarter/half/3-quarter of the atom - or maybe not. My
models do show a pretty pattern of electron/proton pairs (looks like
an X viewed from the top). This may give it predictable magnetic
properties. I have also found that the most stable elements are built
out of mostly helium nuclei. I couldn't figure out why the noble gases
were not totally symmetric until I realized that the atom favors being
built out of helium nuclei. You can directly see this in my model.
> The strong force is not "special". The strong force proves to be purely
> electromagnetic in nature. No one previously realized that the near field
> nucleon magnetic moments are superior to the nuclear electric moments.
>
The strong force is required to keep a compact nucleus which contains
all of the protons/neutrons together. My model may eliminate the need
since the protons/electrons/neutrons are evenly spread across the
entire atom in a neutral matrix. I have some theories on the
scattering experiments which would allow an atomic model like mine.
FrankH
>
> Well, what you neglect to consider is (among other things):
> 1) The charge of the electron and the proton have been shown to be equal
> with *very* great accuracy - around 10^21! (see, for example, here:
> <http://pdg.lbl.gov/2002/bxxxn.pdf>)
electron/proton charges appear equal, but I have theorized that the
electron can get far enough away from the proton to leave it
unsheilded part of the time. Of course, this is a wild theory, but it
could also just be due to an imbalance of charges. Have we measured
what the average electric charge is on the average object you find on
the Earth?
> 2) If the earth would have an overall positive charge (BTW, the
> formulation "positively charge field" doesn't make sense), we would have
> noticed this long ago - for example, by observing the trajectories of
> charged cosmic rays.
Actually, we have noticed this. The Earth has an electric field
measuring about 120v/m at sea level. I believe the total charge if
concentrated at a point in the center the Earth has been estimated at
10^6 C. We also have the Earth's magnetic field to contended with
which appears to play the major factor in solar wind trajectories.
> 3) The attraction force between a positive charge and dipoles depends
> with 1/r^3 on distance, whereas the force of gravity depends on distance
> with 1/r^2.
>
Ah yes, this does seem to be an interesting objection, although the
formula for this sort of force is basically dialectric constant *
volume * E^2. I think that would make the force drop off with 1/r^4
which is definitely not gravity. Back to the drawing board for this
one. But I did make some interesting calculations on what the force
would be on a 1 kg cube of carbon based on a 120v/M field (see my
other gravity posts).
F = (1.6 * .000681 * 120^2)/2 = 7.8 Newtons
This is saying that about 80% of the force on a 1kg sample of carbon
is due to the electrostatic force acting on dipoles. Now this seem
screwy because electrostatic fields appear to be easily sheilded by a
Faraday cage, but I don't think my block of carbon would suddenly
loose 80% of it weight if I put it in a metal cage. Yet, the
electrostatic force is there and it should have some effect. So how
can we account for the fact that the electrostatic field of the earth
seemingly doesn't have any effect on the weight of things, when it
should?
I have also been doing some simple experiments with a charged balloon.
I have notice rather strange behavior that if I bring a strongly
charged balloon near a piece of styrofoam, it initially attracts, but
ocassionally a charge gets transferred and they repel, but if you
force them together again, they attract. The seemly repel at large
distances, yet attract at close. Very strange behavior.
I also tried to see if the electrostatic force on an object depended
on what it was made of. I got an equal weight of iron, aluminum and
paper. I hung them on a pendulum and measured how far I could deflect
each with my balloon. All appeared to deflect the same (within my
experimental error which could have been quite large). This does give
me some confirmation that an electrostatic gravity would at least
treat all masses the same and not be dependent on the dilectric
constant (which would be the other bust with the theory that uses
dipole attraction). So I would be inclined to believe that if gravity
is caused by the electrostatic force, that we are dealing with normal
charge interactions. I made a calculation that 1 atom out of 182
million would have an extra electron to account for the gravity on a
1kg sample of Carbon 12. As you mention, you'd think there'd be just
as good a chance that there'd be one missing, so there would have to
be some bias in the system favoring one charge state over another. I
don't know what that might be, but that doesn't mean it doesn't exist.
There might be something in the strange behavior of the styrofoam and
balloon. Although there would still be the problem with sheilding -
Faraday cages have no effect on gravity. But as I said, if sheilding
works the way it should, why don't we see the affect in a 120v/m
field?
[snip]
FrankH
> of the Rutherford scattering experiments?
>
OK, I'm going to take a wild-ass stab at these questions. I've only
gotten the barest of understanding from the hyper-physics site
http://hyperphysics.phy-astr.gsu.edu/hbase/nuclear/rutsca2.html.
Basically, the experiments assume right off the bat that the charge is
concentrated in a point charge which may not be true. The other
problem is the assumption that only hitting a heavy nuclei would
result in a scattering of greater than 90 degrees. The gold atoms are
supported by a sheet of glass which supports the gold atoms which help
it deflect any bombardment. If you hit a tennis ball glued to a wall
with a much heavier billiard ball, you would certaily get a
backscatter of greater than 90% even though the tennis ball isn't very
dense at all. The experiments appear to assume the gold atom is in
free space. I haven't done any math and can't say much about the
Rutherford formula matching observations. My model could potential
explain the results because the particles apparently pass right
through the glass made up of silicon and oxygen atoms without any
scattering. I would think that at least some of them would be hit.
Since none of them are, I presume that the particles can pass right
through them without interaction. The average cross-section of an atom
in my model doesn't have a cross section much larger than a silicon or
oxygen atom. The arms are only the thickness of an electron/proton
pair. I would expect most particles to fly straight through without
interaction, just like they do for the glass. The only area where the
atom is dense, is if you hit it straight on going through the vertical
core of the atom. There you would be hitting a stack 10 deep. I would
think the chance of hitting it in the manner to be very remote. This
matches the experimental result that very little gets backscattered.
It might also have to be backed up by the glass atoms, so it doesn't
have to rely on Columb force alone to repel the collision. I have no
idea what the math would say about my model, but that is my initial
guess on scattering.
The spectra of atoms can be largely explained by the Bohr model. My
model has shells like Bohr which you can predict spectra and I am
guessing it should work the same way. The spectra of hydrogen is
special because we're not talking about atoms here. We just have a
free electron bouncing back and forth. An extension to my model is
that space is quantizized, meaning a particle can't occupy just any
space. I read of an experiment trying to measure the effect of a
neutron dropping in a gravity field and they found that it fell in
discrete steps - not a continuous drop. I theorize an electron also
follows these rules and must therefore take measured steps away from
the proton. Space could be thought of as being a matrix of cubes where
a particle can only be in one cube at a time. The spectral lines are
generated by the electron moving between these cubes. These cube
locations would also correspond to the electron configurations in
fully populated atoms. Each cube represents one quantum number away
from the proton. Generally, the distance from one quantum shell to the
next is very much the same, although you could see that either an
electron could drop straight down perpendicular to the proton which
would produce the shortest path, or it might take a 45 degree angle
path to an adjacent cube which would make its path slightly larger.
This may explain some of the fine spectra lines and Lamb shift due to
the slightly different paths an electron could take through the cubic
structure to get to another quantum shell. One would expect the direct
path to be much brighter than the indirect path, producing a very
bright and a very faint line (is that what we see?) I couldn't figure
out if the distance between the electron shells was the same or
increasing exponentially. My model would favor them being the same.
But that's my guess on spectra based on the little bit of research
I've done so far. Please feel free to rip it apart. Thanks for bearing
with me. This is loads of fun!
rsm...@york.ac.uk
> "Franz Heymann" <notfranz...@btopenworld.com> wrote in message news:<bu76kh$p31$1...@hercules.btinternet.com>...
> > "FrankH" <frank...@yahoo.com> wrote in message
> > news:46484c9f.04011...@posting.google.com...
> >
> > [snip]
> >
> > You appear not to know any physics worth talking about.
> >
> > Franz
>
> Damn straight! and proud of it. I don't know anything about your
> physics filled to the brim with counter-intuitive ideas, miles of
> nonsensical math, multi-dimension and generally defying all logic.
That's strange, because I am unaware of any physics filled to the brim
with miles of nonsensical maths.[1] However, I do know of the physics
filled to the brim with miles of sensible and logical mathematical
deductions from a simple set of postulates, producing predictions in
amazing agreement with experiment. You should try studying it
sometime. You may even enjoy it.
Robert
[1] Some of the fringe theories that get posted to sci.physics
notwithstanding.
Bjoern Feuerbacher
>
> Bjoern Feuerbacher <bfeu...@ix.urz.uni-heidelberg.de> wrote in message
> [snip]
> >
> > Well, what you neglect to consider is (among other things):
> > 1) The charge of the electron and the proton have been shown to be equal
> > with *very* great accuracy - around 10^21! (see, for example, here:
> > <http://pdg.lbl.gov/2002/bxxxn.pdf>)
> This is an article on Baryons - did I miss something?
Perhaps that the proton *is* a baryon? Hey, it's the very first entry
in this table of particle properties!
> Sure electron/proton charges appear equal, but I have theorized that the
> electron can get far enough away from the proton to leave it
> unsheilded part of the time.
What do you mean by "unshielded", specifically?
> Of course, this is a wild theory, but it
> could also just be due to an imbalance of charges. Have we measured
> what the average electric charge is on the average object you find on
> the Earth?
Well, I have pointed out that the *total* electric charge of the earth
is close to zero (see cosmic rays). Why do you want to know the charge
of the "average" object on earth?
> > 2) If the earth would have an overall positive charge (BTW, the
> > formulation "positively charge field" doesn't make sense), we would have
> > noticed this long ago - for example, by observing the trajectories of
> > charged cosmic rays.
>
> Actually, we have noticed this. The Earth has an electric field
> measuring about 120v/m at sea level. I believe the total charge if
> concentrated at a point in the center the Earth has been estimated at
> 10^6 C.
References, please.
> We also have the Earth's magnetic field to contended with
> which appears to play the major factor in solar wind trajectories.
Yes, it play a major factor - but nevertheless, effects of an additional
electric field of the magnitude you propose should be observable, I
would think.
> > 3) The attraction force between a positive charge and dipoles depends
> > with 1/r^3 on distance, whereas the force of gravity depends on distance
> > with 1/r^2.
> >
> Ah yes, this does seem to be an interesting objection,
And as Frank Heymann pointed out, I'm even generous to you here, because
I don't take into account that your dipoles are formed by polarization.
> although the
> formula for this sort of force is basically dialectric constant *
> volume * E^2.
Huh??? This formula gives the *energy* in the field. It has *nothing* to
do with the force between a dipole and a point charge!
> I think that would make the force drop off with 1/r^4
> which is definitely not gravity.
Sorry, but where did you get this from?
> Back to the drawing board for this one.
Nice to see that you can admit that you have been wrong!
> But I did make some interesting calculations on what the force
> would be on a 1 kg cube of carbon based on a 120v/M field (see my
> other gravity posts).
>
> F = (1.6 * .000681 * 120^2)/2 = 7.8 Newtons
You seem to be using the wrong formula here - again the one for energy
instead of the one for force. Why don't you write the units behind your
numbers? If you would do this, your error would become obvious.
> This is saying that about 80% of the force on a 1kg sample of carbon
> is due to the electrostatic force acting on dipoles. Now this seem
> screwy because electrostatic fields appear to be easily sheilded by a
> Faraday cage, but I don't think my block of carbon would suddenly
> loose 80% of it weight if I put it in a metal cage.
Well, the explanation is easy: you used the wrong formula!
> Yet, the
> electrostatic force is there and it should have some effect.
Yes, but much smaller than the one you calculated above.
> So how
> can we account for the fact that the electrostatic field of the earth
> seemingly doesn't have any effect on the weight of things, when it
> should?
Wrong formula.
> I have also been doing some simple experiments with a charged balloon.
> I have notice rather strange behavior that if I bring a strongly
> charged balloon near a piece of styrofoam, it initially attracts, but
> ocassionally a charge gets transferred and they repel, but if you
> force them together again, they attract. The seemly repel at large
> distances, yet attract at close. Very strange behavior.
I don't know what's going on exactly there, but my guess would be that
this has something to do with polarization.
> I also tried to see if the electrostatic force on an object depended
> on what it was made of. I got an equal weight of iron, aluminum and
> paper.
Did they have the same shape, too? If not, the moment of inertia could
have an influence on your results!
> I hung them on a pendulum and measured how far I could deflect
> each with my balloon. All appeared to deflect the same (within my
> experimental error which could have been quite large).
Nice.
> This does give
> me some confirmation that an electrostatic gravity would at least
> treat all masses the same and not be dependent on the dilectric
> constant (which would be the other bust with the theory that uses
> dipole attraction). So I would be inclined to believe that if gravity
> is caused by the electrostatic force, that we are dealing with normal
> charge interactions.
In other words: all objects would have to be charged, and their charge
would have to be proportional to their masses. Hint: this isn't
observed.
> I made a calculation that 1 atom out of 182
> million would have an extra electron to account for the gravity on a
> 1kg sample of Carbon 12. As you mention, you'd think there'd be just
> as good a chance that there'd be one missing, so there would have to
> be some bias in the system favoring one charge state over another.
Right. Where is the evidence for such a bias, and why should it exist?
> I don't know what that might be, but that doesn't mean it doesn't exist.
Yes, absence of evidence is not evidence of absence. But why on earth
should we try to find a new theory of gravity, although the existing one
is very well tested and shows no problems so far? Just because you don't
like it, or what?
> There might be something in the strange behavior of the styrofoam and
> balloon.
See above.
> Although there would still be the problem with sheilding -
> Faraday cages have no effect on gravity. But as I said, if sheilding
> works the way it should, why don't we see the affect in a 120v/m
> field?
See above, too.
Bye,
Bjoern
Bjoern Feuerbacher
>
> null...@aol.com (Edward Green) wrote in message
> [snip]
> > He claims to be a graduate of MIT. Perhaps that's "Massachusetts
> > Interstate Trucking" school? (Feynman graduated from the other one).
>
> Yes, that would be Mens Institude of Typewriting - Well actually, I
> graduated in 1986 with a batchelors in computer science from the
> Massachusetts Institute of Technology. I have the brass rat to prove
> it. Quantum physics was not a required course for CS
Oh, that explains a bit why you know so little about physics!
But if you are interested in QM, why didn't you take the course anyway?
> and not everybody
> at MIT is a physics nut. That explains why I don't know much about
> quantum mechanics - but I'm reading more every day! But I don't want
> this discussion to be about me, tell me why my atomic model is wrong
> or old or whatever - lets talk science.
Well, as already mentioned: you seem to picture electrons as little
marbles or "lego blocks". This contradicts observations.
I would recommend the following book to you:
D. Styer, "The strange world of Quantum Mechanics"
Bye,
Bjoern
Bjoern Feuerbacher
>
> Bjoern Feuerbacher <bfeu...@ix.urz.uni-heidelberg.de> wrote in message
> > How does your model explain the spectra of the atoms, and the results
> > of the Rutherford scattering experiments?
> >
> OK, I'm going to take a wild-ass stab at these questions. I've only
> gotten the barest of understanding from the hyper-physics site
> http://hyperphysics.phy-astr.gsu.edu/hbase/nuclear/rutsca2.html.
Then you should stop right here and get a better understanding before
proceeding any further. The page you mention here gives only a *very*
crude,
popular science account of what was *really* observed back then, and
what lead Rutherford to his proposal!
Try going to the link "Rutherford formula":
<http://hyperphysics.phy-astr.gsu.edu/hbase/rutsca.html#c2>
This formula is a prediction from Rutherford's model, and it has been
shown
to be valid with very great accuracy. Hence one of your first tasks
should be to produce this formula from your model. After all, it was the
main reason to propose that the nucleus is compact and heavy!
> Basically, the experiments assume right off the bat that the charge is
> concentrated in a point charge which may not be true.
Err, this is called a "model". Do you have a problem with model
building? This model gives the right results, so why do you think there
is any problem with it? OTOH, *your* model hasn't been able to reproduce
*any* of these results so far... (see above - spectra and Rutherford's
scattering cross section, the two main reasons why the model was
invented!)
> The other
> problem is the assumption that only hitting a heavy nuclei would
> result in a scattering of greater than 90 degrees.
This isn't assumed. The scattering happens by the Coulomb force, not by
"hitting a heavy nuclei [sic]".
Even the page you cite above says nothing about "hitting" a nucleus - it
only says "they are scattering off something more massive than
themselves".
> The gold atoms are
> supported by a sheet of glass which supports the gold atoms which help
> it deflect any bombardment.
Are you sure about this sheet of glass? This is totally new to me - and
IIRC, such a sheet wasn't used when I did the experiment myself a few
years ago.
> If you hit a tennis ball glued to a wall
> with a much heavier billiard ball, you would certaily get a
> backscatter of greater than 90% even though the tennis ball isn't very
> dense at all.
Well, maybe, but obviously you wouldn't get the dependence of the amount
of scattered electrons / alpha particles on angle described by
Rutherford's formula.
> The experiments appear to assume the gold atom is in free space.
I think they use a kind of Born-Oppenheimer approximation - the
scattering
of the electron by the nucleus is much faster than any reaction of the
lattice
in which the atom is embedded.
> I haven't done any math and can't say much about the
> Rutherford formula matching observations.
If your model can't predict the basic observations which led Rutherford
to his model (and I talk about *quantitative* predictions here, not hand
waving), then it is of little value.
> My model could potential
> explain the results because the particles apparently pass right
> through the glass made up of silicon and oxygen atoms without any
> scattering.
I don't think so. Silicon and oxygen atoms scatter electrons / alpha
particles, too - but the scattering is much weaker (notice that the
scattering
cross section is proportional to Z^2!). Additionally, I still doubt that
there really *was* even glass anywhere.
> I would think that at least some of them would be hit.
Hit what? Neither atoms or nuclei are hard marbles. They interact with
electrons / alpha particles only by the electromagnetic force, not by
physical contact.
> Since none of them are, I presume that the particles can pass right
> through them without interaction.
You presume wrongly.
> The average cross-section of an atom
> in my model doesn't have a cross section much larger than a silicon or
> oxygen atom.
Pardon?
> The arms are only the thickness of an electron/proton pair.
> I would expect most particles to fly straight through without
> interaction, just like they do for the glass. The only area where the
> atom is dense, is if you hit it straight on going through the vertical
> core of the atom. There you would be hitting a stack 10 deep. I would
> think the chance of hitting it in the manner to be very remote.
Can you quantify this?
> This
> matches the experimental result that very little gets backscattered.
That's hand waving! Present a *quantitative* analysis, then you have a
point!
This shouldn't be sooo hard to do - don't you simply have to look how
many percent of your atom are "dense" enough?
> It might also have to be backed up by the glass atoms, so it doesn't
> have to rely on Columb force alone to repel the collision.
On what should it rely instead? How does your atom interact with the
electrons / the alpha particles which are used for scattering?
> I have no
> idea what the math would say about my model, but that is my initial
> guess on scattering.
In other words: you can't provide the answer.
> The spectra of atoms can be largely explained by the Bohr model.
Right, largely - but not exclusively.
> My model has shells like Bohr which you can predict spectra and I am
> guessing it should work the same way.
Could you please get a bit more quantitative than "I am guessing"?
> The spectra of hydrogen is
> special because we're not talking about atoms here. We just have a
> free electron bouncing back and forth.
Huh??? Hydrogen atoms are electrically neutral - how could this be if
they contained only a free electron?
> An extension to my model is
> that space is quantizized, meaning a particle can't occupy just any
> space.
This is getting stranger and stranger...
> I read of an experiment trying to measure the effect of a
> neutron dropping in a gravity field and they found that it fell in
> discrete steps - not a continuous drop.
I know of this experiment, and IIRC, the results were something
different. Where did you get this from?
> I theorize an electron also
> follows these rules and must therefore take measured steps away from
> the proton.
Does this mean that their distance can only have discrete values?
If yes, why isn't this quantization of space noticed elsewhere? For
example, why doesn't it have any effect on nuclei?
> Space could be thought of as being a matrix of cubes where
> a particle can only be in one cube at a time.
Nice idea. No evidence.
> The spectral lines are
> generated by the electron moving between these cubes.
How?
> These cube
> locations would also correspond to the electron configurations in
> fully populated atoms.
How?
> Each cube represents one quantum number away from the proton.
Sorry, I don't understand this.
> Generally, the distance from one quantum shell to the
> next is very much the same, although you could see that either an
> electron could drop straight down perpendicular to the proton which
> would produce the shortest path, or it might take a 45 degree angle
> path to an adjacent cube which would make its path slightly larger.
You are still picturing electrons as little marbles which move around.
Such a picture is contradicted by QM - for example, the double slit
experiment.
> This may explain some of the fine spectra lines and Lamb shift due to
> the slightly different paths an electron could take through the cubic
> structure to get to another quantum shell.
"may".
> One would expect the direct
> path to be much brighter than the indirect path, producing a very
> bright and a very faint line
Huh? Why???
> (is that what we see?)
No (AFAIK).
> I couldn't figure
> out if the distance between the electron shells was the same or
> increasing exponentially.
How did you try to figure this out?
> My model would favor them being the same.
Why?
> But that's my guess on spectra based on the little bit of research
> I've done so far.
"guess" is the crucial word here.
> Please feel free to rip it apart. Thanks for bearing
> with me. This is loads of fun!
It would be a lot more fun if you would bother to learn a bit about
the things you are talking about.
Bye,
Bjoern
Bjoern Feuerbacher
>
> tnlo...@aol.com (Tnlockyer) wrote in message >
> [snip]
> > Frank, you make the mistake of all the previous theorists in trying to
> > reverse
> > engineer or to invent models to try and explain the experimental evidence.
> >
> > What you want is a model that goes straight forward and that puts itself
> > together, more or less automatically and precisely, and is supported by
> > prior experimental results.
>
> Actually, as has been noted by the other posters, I am quite ignorant
> of the experimental evidence and built my model starting with a
> thought experiment of what would be the most simplest way for
> electrons/protons to assemble.
That's a strange way to build a model. Why didn't you learn about the
experimental evidence and *then* tried to build a model based on that?
> Please take another look at my website.
> My model's advantage is that it does go together straight forward with
> a predicatable geometric sequence which you can do almost blindly, and
> it would put itself together automatically.
Nice. So what? Solving the Schroedinger equation for the H atom is also
fairly straightforward.
> So far, I have seen that
> this model can explain the observed electron shell structure with the
> main quantum number representing the electrons in the vertical core
I still haven't seen you making any *quantitative* predictions - like
energies.
> and you can see why we get the number of electrons in the outer shells
> due to the geometric progression of the outer electrons fitting into
> the vertical core.
Well, we get this from the Schroedinger equation, too.
> I am working on understanding the spectrum and
> scattering experiments
Good luck.
> and I think it may be possible to explain the
> results in terms of my atomic model.
Please notice that hand waving is not enough. What we would like to see
is *quantitative* data.
> Spin appears to be due to the
> fact that the atom assembles a mirror image of itself about the axes,
> and the reversed electron/proton may explain the difference states.
How does this explain the results of the Stern-Gerlach experiment?
> I would like to learn more about decay and nuclear fission,
Try the book by Povh about nuclear physics - it has a lot about
scattering, too.
> since my
> model with its four arms would tend to favor fission which would knock
> off arms
A model of atoms/nuclei with "arms"? You really should talk to Porat!
> and you might think that there would be more products which
> represent a quarter/half/3-quarter of the atom - or maybe not. My
> models do show a pretty pattern of electron/proton pairs (looks like
> an X viewed from the top). This may give it predictable magnetic
> properties.
"may"
> I have also found that the most stable elements are built
> out of mostly helium nuclei.
Nuclear physics explains this. See the book by Povh mentioned above -
the crucial term is "pairing energy".
> I couldn't figure out why the noble gases
> were not totally symmetric until I realized that the atom favors being
> built out of helium nuclei. You can directly see this in my model.
Nice. I'm *still* waiting for the quantitative predictions...
> > The strong force is not "special". The strong force proves to be purely
> > electromagnetic in nature. No one previously realized that the near field
> > nucleon magnetic moments are superior to the nuclear electric moments.
> >
> The strong force is required to keep a compact nucleus which contains
> all of the protons/neutrons together. My model may eliminate the need
> since the protons/electrons/neutrons are evenly spread across the
> entire atom in a neutral matrix.
Have you calculated if this system is stable? I would think that the
Coulomb forces would destroy it.
> I have some theories on the
> scattering experiments which would allow an atomic model like mine.
Don't give us even more theories - give us quantitative predictions from
your model.
Bye,
Bjoern
Franz Heymann
> Bjoern Feuerbacher <bfeu...@ix.urz.uni-heidelberg.de> wrote in message
> [snip]
> >
> > Well, what you neglect to consider is (among other things):
> > 1) The charge of the electron and the proton have been shown to be equal
> > with *very* great accuracy - around 10^21! (see, for example, here:
> > <http://pdg.lbl.gov/2002/bxxxn.pdf>)
> This is an article on Baryons - did I miss something?
Yes. This line in that URL :-
|qp + qe| < 10^-21
> Sure
> electron/proton charges appear equal, but I have theorized that the
> electron can get far enough away from the proton to leave it
> unsheilded part of the time.
What is that supposed to mean?
> Of course, this is a wild theory, but it
> could also just be due to an imbalance of charges.
The electron and proton charges are equal to some phenomenal accuracy. It
is as certain as one could be in scientific matters that they are equal.
> Have we measured
> what the average electric charge is on the average object you find on
> the Earth?
Yes. Frequently. It depends strongly on the recent history of the object.
>
> > 2) If the earth would have an overall positive charge (BTW, the
> > formulation "positively charge field" doesn't make sense), we would have
> > noticed this long ago - for example, by observing the trajectories of
> > charged cosmic rays.
> Actually, we have noticed this. The Earth has an electric field
> measuring about 120v/m at sea level. I believe the total charge if
> concentrated at a point in the center the Earth has been estimated at
> 10^6 C. We also have the Earth's magnetic field to contended with
> which appears to play the major factor in solar wind trajectories.
>
> > 3) The attraction force between a positive charge and dipoles depends
> > with 1/r^3 on distance, whereas the force of gravity depends on distance
> > with 1/r^2.
> >
> Ah yes, this does seem to be an interesting objection, although the
> formula for this sort of force is basically dialectric constant *
> volume * E^2. I think that would make the force drop off with 1/r^4
> which is definitely not gravity.
Still wrong. 1/r^5 actually
> Back to the drawing board for this
> one. But I did make some interesting calculations on what the force
> would be on a 1 kg cube of carbon based on a 120v/M field (see my
> other gravity posts).
>
> F = (1.6 * .000681 * 120^2)/2 = 7.8 Newtons
>
> This is saying that about 80% of the force on a 1kg sample of carbon
> is due to the electrostatic force acting on dipoles.
Total rubbish.
> Now this seem
> screwy because electrostatic fields appear to be easily sheilded by a
> Faraday cage, but I don't think my block of carbon would suddenly
> loose 80% of it weight if I put it in a metal cage. Yet, the
> electrostatic force is there and it should have some effect. So how
> can we account for the fact that the electrostatic field of the earth
> seemingly doesn't have any effect on the weight of things, when it
> should?
Balderdash. The effect is utterly, utterly negligible
The rest is undiluted bulldust so I snipped it.
[snip]
Franz
Franz Heymann
How does your model explain the spectra of the atoms, and the results
> > of the Rutherford scattering experiments?
> >
> OK, I'm going to take a wild-ass stab at these questions. I've only
> gotten the barest of understanding from the hyper-physics site
> http://hyperphysics.phy-astr.gsu.edu/hbase/nuclear/rutsca2.html.
> Basically, the experiments assume right off the bat that the charge is
> concentrated in a point charge which may not be true. The other
> problem is the assumption that only hitting a heavy nuclei would
> result in a scattering of greater than 90 degrees. The gold atoms are
> supported by a sheet of glass
No.
> which supports the gold atoms which help
> it deflect any bombardment. If you hit a tennis ball glued to a wall
> with a much heavier billiard ball, you would certaily get a
> backscatter of greater than 90% even though the tennis ball isn't very
> dense at all. The experiments appear to assume the gold atom is in
> free space.
It is. The nearest atom is another gold atom. It is about ************
away. That is a huge distance compared to the dimensions of the projectile,
which has a radius in the region of **************
> I haven't done any math
Then you have failed to answer the question. Why do you waste your readers'
time?
> and can't say much about the
> Rutherford formula matching observations.
That much is obvious.
> My model could potential
> explain the results
No, it can not. Your say so is not physics.
because the particles apparently pass right
> through the glass made up of silicon and oxygen atoms without any
> scattering.
What glass?
There was no substrate involved in the scattering experiment my
undergraduates did.
> I would think that at least some of them would be hit.
> Since none of them are, I presume that the particles can pass right
> through them without interaction. The average cross-section of an atom
> in my model doesn't have a cross section much larger than a silicon or
> oxygen atom. The arms are only the thickness of an electron/proton
> pair. I would expect most particles to fly straight through without
> interaction, just like they do for the glass. The only area where the
> atom is dense, is if you hit it straight on going through the vertical
> core of the atom. There you would be hitting a stack 10 deep. I would
> think the chance of hitting it in the manner to be very remote. This
> matches the experimental result that very little gets backscattered.
> It might also have to be backed up by the glass atoms, so it doesn't
> have to rely on Columb force alone to repel the collision. I have no
> idea what the math would say about my model, but that is my initial
> guess on scattering.
>
> The spectra of atoms can be largely explained by the Bohr model.
Not at all. It copes only with the simplest aspects of the Hydrogen
spectrum and is quite useless for more complicated atoms. The Bohr model is
as dead as a duck for all but the most elementary introduction ro atomic
physics.
> My
> model has shells like Bohr which you can predict spectra and I am
> guessing it should work the same way.
Please give a quantitative explanation of what you mean. Or are you just
waffling?
> The spectra of hydrogen is
> special because we're not talking about atoms here. We just have a
> free electron bouncing back and forth.
A free electron cannot just bounce back and forth. It can onlt move at a
constant velocity.
> An extension to my model is
> that space is quantizized,
Please show quantitatively how space is quantised in your model. Or are you
just waffling?
There is no physics as yet which requires space to be quantised, so which
phenomena require this in your model?
> meaning a particle can't occupy just any
> space. I read of an experiment trying to measure the effect of a
> neutron dropping in a gravity field and they found that it fell in
> discrete steps - not a continuous drop.
Reference please.
> I theorize
You don't know what "theorise" means. It does not mean talking through a
hole in the side of your neck.
> an electron also
> follows these rules
You have not given any rules yet.
> and must therefore take measured steps away from
> the proton. Space could be thought of as being a matrix of cubes where
> a particle can only be in one cube at a time.
What are the dimensions of this cube, and why?
> The spectral lines are
> generated by the electron moving between these cubes.
You are drivelling.
> These cube
> locations would also correspond to the electron configurations in
> fully populated atoms. Each cube represents one quantum number away
> from the proton. Generally, the distance from one quantum shell
I thought you were talking about cubes. What are these shells?
> to the
> next is very much the same, although you could see that either an
> electron could drop straight down perpendicular to the proton which
> would produce the shortest path, or it might take a 45 degree angle
> path to an adjacent cube which would make its path slightly larger.
> This may explain some of the fine spectra lines and Lamb shift due to
> the slightly different paths an electron could take through the cubic
> structure to get to another quantum shell. One would expect the direct
> path to be much brighter than the indirect path, producing a very
> bright and a very faint line (is that what we see?) I couldn't figure
> out if the distance between the electron shells was the same or
> increasing exponentially. My model would favor them being the same.
> But that's my guess on spectra based on the little bit of research
> I've done so far. Please feel free to rip it apart. Thanks for bearing
> with me. This is loads of fun!
You are a bog ignorant fool. Every utterance of yours smells of a midden
heap Why don't you take up politics instead?
Franz
Franz Heymann
> tnlo...@aol.com (Tnlockyer) wrote in message >
> [snip]
> > Frank, you make the mistake of all the previous theorists in trying to
reverse
> > engineer or to invent models to try and explain the experimental
evidence.
> >
> > What you want is a model that goes straight forward and that puts itself
> > together, more or less automatically and precisely, and is supported by
prior
> > experimental results.
>
> Actually, as has been noted by the other posters, I am quite ignorant
> of the experimental evidence and built my model starting with a
> thought experiment of what would be the most simplest way for
> electrons/protons to assemble.
That was, as has been pointed out by a number of contributors, a stupid
starting point. You would be better served if you were to start by learning
physics, so that you can have some idea of what its strengths and weaknesses
are. At present you are only making a laughing stock of yourself.
> Please take another look at my website.
Your website is a heap of garbage.
[snip]
Franz
FrankH
> FrankH wrote:
> >
> > tnlo...@aol.com (Tnlockyer) wrote in message >
> > [snip]
> > > Frank, you make the mistake of all the previous theorists in trying to
> > > reverse
> > > engineer or to invent models to try and explain the experimental evidence.
> > >
> > > What you want is a model that goes straight forward and that puts itself
> > > together, more or less automatically and precisely, and is supported by
> > > prior experimental results.
> >
> > Actually, as has been noted by the other posters, I am quite ignorant
> > of the experimental evidence and built my model starting with a
> > thought experiment of what would be the most simplest way for
> > electrons/protons to assemble.
>
> That's a strange way to build a model. Why didn't you learn about the
> experimental evidence and *then* tried to build a model based on that?
>
theorists in reverse engineering a model. I just wanted to come up
with a physical model of an atom. So I used what I had avaliable and
found a geometric sequence - now I do need to see if it matches
experimental results.
>
> > Please take another look at my website.
> > My model's advantage is that it does go together straight forward with
> > a predicatable geometric sequence which you can do almost blindly, and
> > it would put itself together automatically.
>
> Nice. So what? Solving the Schroedinger equation for the H atom is also
> fairly straightforward.
Uhhh, not straightforward to me. How can you say a full page of
equations is straightforward. I think most people look at that and
scratch their heads. The diagrams of the electron orbits is even more
bizzare. Just how electrons manage to confine themselves to the
strange donut shapes required in QM is unintuitive. The higher level
shells can't even be reasonably drawn. On the other hand, the
geometric model I have can be grasped without any math by anybody and
you can build out all the elements with ease. It has a simplicity that
the Schroedinger equations do not and science tends to favor simpler
and more elegant solutions.
>
>
> > So far, I have seen that
> > this model can explain the observed electron shell structure with the
> > main quantum number representing the electrons in the vertical core
>
> I still haven't seen you making any *quantitative* predictions - like
> energies.
I am going to work on that - but I don't know exactly where to begin.
I was hoping I could get someone who knows this type of math
interested enough in my model to run the quantitative predictions and
pull some numbers out. Can you point me in the right direction on how
to calculate energies based on my geometric model?
>
>
> > and you can see why we get the number of electrons in the outer shells
> > due to the geometric progression of the outer electrons fitting into
> > the vertical core.
>
> Well, we get this from the Schroedinger equation, too.
>
The equations don't graphically show why only a limited number of
electrons can fit into each shell. You can clearly see the limitations
on my geometric model since you can only fit on so many electrons
before you have to expand the central core. It provides a very
intuituive and visual explanation for the shell structure.
>
> > I am working on understanding the spectrum and
> > scattering experiments
>
> Good luck.
>
>
> > and I think it may be possible to explain the
> > results in terms of my atomic model.
>
> Please notice that hand waving is not enough. What we would like to see
> is *quantitative* data.
>
>
> > Spin appears to be due to the
> > fact that the atom assembles a mirror image of itself about the axes,
> > and the reversed electron/proton may explain the difference states.
>
> How does this explain the results of the Stern-Gerlach experiment?
>
It doesn't. I was thinking about explaining the fine spectral lines
between electrons with same energy state but with a different spin.
>
> > I would like to learn more about decay and nuclear fission,
>
> Try the book by Povh about nuclear physics - it has a lot about
> scattering, too.
>
>
> > since my
> > model with its four arms would tend to favor fission which would knock
> > off arms
>
> A model of atoms/nuclei with "arms"? You really should talk to Porat!
>
>
> > and you might think that there would be more products which
> > represent a quarter/half/3-quarter of the atom - or maybe not. My
> > models do show a pretty pattern of electron/proton pairs (looks like
> > an X viewed from the top). This may give it predictable magnetic
> > properties.
>
> "may"
I just looked up what are the primary products of U238 fission and the
largest components are generally about 1/2 and 1/4 of the original
atom size. Looks like it knocks off an the vertical core plus the
electrons in the arms.
See http://www.chem.uidaho.edu/~honors/decay.html
>
>
> > I have also found that the most stable elements are built
> > out of mostly helium nuclei.
>
> Nuclear physics explains this. See the book by Povh mentioned above -
> the crucial term is "pairing energy".
>
>
> > I couldn't figure out why the noble gases
> > were not totally symmetric until I realized that the atom favors being
> > built out of helium nuclei. You can directly see this in my model.
>
> Nice. I'm *still* waiting for the quantitative predictions...
Isn't this a type of quantitative prediction? The model shows the role
of paring energy in the production of noble gases. The prediction
being that if you build out a symettrical atom built mostly out of
helium nuclei, you get a nobel gas. Predicting the properties of
elements based solely on geometric symmettry is a significant
prediction.
>
>
> > > The strong force is not "special". The strong force proves to be purely
> > > electromagnetic in nature. No one previously realized that the near field
> > > nucleon magnetic moments are superior to the nuclear electric moments.
> > >
> > The strong force is required to keep a compact nucleus which contains
> > all of the protons/neutrons together. My model may eliminate the need
> > since the protons/electrons/neutrons are evenly spread across the
> > entire atom in a neutral matrix.
>
> Have you calculated if this system is stable? I would think that the
> Coulomb forces would destroy it.
How would I calculate the force so that it would satisfy you. I don't
know if I believe the radius given for electrons and protons, so I'm
not certain what to plug in for distance. Why would you think the
Colulob forces would destroy it? Every electron is touching a proton
and vice-versa. I would think it would be highly stable. As a
comparison, I have some round magnets and they attract 1/r^2 with
opposite poles on opposite sides, so they roughly model the
interaction of proton/electrons. When I stack them together like I
have them in my model, they stick together quite strongly.
FrankH
> FrankH wrote:
> >
> > Bjoern Feuerbacher <bfeu...@ix.urz.uni-heidelberg.de> wrote in message
> > [snip]
> > >
> > > Well, what you neglect to consider is (among other things):
> > > 1) The charge of the electron and the proton have been shown to be equal
> > > with *very* great accuracy - around 10^21! (see, for example, here:
> > > <http://pdg.lbl.gov/2002/bxxxn.pdf>)
> > This is an article on Baryons - did I miss something?
>
> Perhaps that the proton *is* a baryon? Hey, it's the very first entry
> in this table of particle properties!
>
>
> > Sure electron/proton charges appear equal, but I have theorized that the
> > electron can get far enough away from the proton to leave it
> > unsheilded part of the time.
>
> What do you mean by "unshielded", specifically?
>
totally neutral is if they occupied exactly the same place in space.
Since they do not, there must always be some imbalance where an object
in the vicinity will see either a more positive field or more negative
field. Since electrons are free to move about and protons are not,
this gives a bias towards seeing the positive field more often than
the negative - leading to an imbalance when measured some distance
away from the electron/proton pair. This means that an electron/proton
could have exactly the same charge, but yet still produce an
observable positive charge from moment to moment.
>
> > Of course, this is a wild theory, but it
> > could also just be due to an imbalance of charges. Have we measured
> > what the average electric charge is on the average object you find on
> > the Earth?
>
> Well, I have pointed out that the *total* electric charge of the earth
> is close to zero (see cosmic rays). Why do you want to know the charge
> of the "average" object on earth?
>
>
> > > 2) If the earth would have an overall positive charge (BTW, the
> > > formulation "positively charge field" doesn't make sense), we would have
> > > noticed this long ago - for example, by observing the trajectories of
> > > charged cosmic rays.
> >
> > Actually, we have noticed this. The Earth has an electric field
> > measuring about 120v/m at sea level. I believe the total charge if
> > concentrated at a point in the center the Earth has been estimated at
> > 10^6 C.
>
> References, please.
See:
http://hypertextbook.com/facts/1998/TreshaEdwards.shtml
There is some disagreement about the value, but most are in the
ballpark of 120v/m. Also see the blazelabs reference I have below.
>
>
> > We also have the Earth's magnetic field to contended with
> > which appears to play the major factor in solar wind trajectories.
>
> Yes, it play a major factor - but nevertheless, effects of an additional
> electric field of the magnitude you propose should be observable, I
> would think.
>
>
> > > 3) The attraction force between a positive charge and dipoles depends
> > > with 1/r^3 on distance, whereas the force of gravity depends on distance
> > > with 1/r^2.
> > >
> > Ah yes, this does seem to be an interesting objection,
>
> And as Frank Heymann pointed out, I'm even generous to you here, because
> I don't take into account that your dipoles are formed by polarization.
>
>
> > although the
> > formula for this sort of force is basically dialectric constant *
> > volume * E^2.
>
> Huh??? This formula gives the *energy* in the field. It has *nothing* to
> do with the force between a dipole and a point charge!
>
>
> > I think that would make the force drop off with 1/r^4
> > which is definitely not gravity.
>
> Sorry, but where did you get this from?
This is the dielectrophoretic force
See this page:
http://www.blazelabs.com/f-efield.asp
This provided the basis for the exploration of electrostatic gravity.
I'm just trying to answer the question that was asked.
[snip]
>
> Yes, absence of evidence is not evidence of absence. But why on earth
> should we try to find a new theory of gravity, although the existing one
> is very well tested and shows no problems so far? Just because you don't
> like it, or what?
Wait, I didn't know we had an adequate theory of gravity since we
can't seem to unite it with the other forces. The appeal of an
electrostatic gravity is that it would provide that unification
instantly by electrostatic force = gravity force. We do want to unify
the forces, don't we? I don't know if I can get there, but it's worth
a shot!
[snip]
Edward Green
Good answer to my provocative post.
I feel MIT has done you a technical disservice, though. The material
I allude to doesn't live in a specialized course in "quantum physics",
but in freshman chemistry. It's one of those things which might first
be introduced in high school chemistry, then successively developed
and refined in various passes -- certainly by freshman chem one should
be aware that description of atomic and molecular structure is a major
triumph of non-relativisitic QM. If they allow even CS majors to
graduate without this knowledge ... well, they do call it a school of
"technology". :-/ Demand a partial tuition refund -- one expects such
specialization from a trade school, not MIT.
I first say the detailed mathematical solution to the hydrogenic atom
in "physical chemistry", which is "hard" third year chemstry with a
strong flavor of physics. As such, the h-atom solution to
Schroedinger's equation was a hard thing to wow undergraduates, and
separate the boys from the men from the pre-meds, but in retrospect
was an easy thing, since it relied purely on methodical application of
19th century diff-eq style mathematics: a tour-de-force, but
straightforward; like Mr. Woodworking showing you how to make a
reverse double miter rabbit warren joint on TV.
Anyway, being a Good Hard Thing, and unfortunately, the hardest thing
in this line we can solve exactly (which shows the feeling of
limitless power one might have in first touring its force might be
naive), the hydrogenic atom serves as a basis for all sorts of
approximate wave functions for poly-electronic atoms and molecules,
the general adequacy of which as a whole, in explaining atomic and
molecular physics, is a sustained major triumph of quantum mechanics,
as already mentioned.
Franz Heymann
We look forward to the results of your work. Do remember that quantum
electrodynamics has produced results for the frequencies and relative
intensities of *all* the important spectral lines of *all* the elements up
to Fe. There is excellent agreement between these and the observed values.
>
> >
> > > Please take another look at my website.
> > > My model's advantage is that it does go together straight forward with
> > > a predicatable geometric sequence which you can do almost blindly, and
> > > it would put itself together automatically.
> >
> > Nice. So what? Solving the Schroedinger equation for the H atom is also
> > fairly straightforward.
>
> Uhhh, not straightforward to me.
In that case you should shut up.
> How can you say a full page of
> equations is straightforward. I think most people look at that and
> scratch their heads.
The rest of the world is not as dumb as you.
> The diagrams of the electron orbits is even more
> bizzare. Just how electrons manage to confine themselves to the
> strange donut shapes required in QM is unintuitive.
Of course it is counterintuitive. Intuition is invariably wrong when it is
used to extrapolate into regions where intuition has no right to be used.
> The higher level
> shells can't even be reasonably drawn. On the other hand, the
> geometric model I have can be grasped without any math by anybody and
> you can build out all the elements with ease.
And as a bonus, it hhas not yielded one solitary numerical result worth
spitting on.
Come on, you gas bag, calculate the frequency of the Lyman alpha line in the
Hydrogen spectrum to six figure accuracy.
> It has a simplicity that
> the Schroedinger equations do not and science tends to favor simpler
> and more elegant solutions.
Your version is neither elegant, nor simple, nor correct.
> > > So far, I have seen that
> > > this model can explain the observed electron shell structure with the
> > > main quantum number representing the electrons in the vertical core
> >
> > I still haven't seen you making any *quantitative* predictions - like
> > energies.
> I am going to work on that - but I don't know exactly where to begin.
How about shutting up then until you have learnt how to begin. You have
waffled far more than enough by now.
> I was hoping I could get someone who knows this type of math
> interested enough in my model to run the quantitative predictions and
> pull some numbers out. Can you point me in the right direction on how
> to calculate energies based on my geometric model?
There is no right way of achieving your object. Your approach is dead on
arrival.
>
> >
> >
> > > and you can see why we get the number of electrons in the outer shells
> > > due to the geometric progression of the outer electrons fitting into
> > > the vertical core.
> >
> > Well, we get this from the Schroedinger equation, too.
> >
> The equations don't graphically show why only a limited number of
> electrons can fit into each shell.
You don't need much by way of equations for that. The exclusion principle
and a very elementary knowledge of quantised angular momentum is enough.
> You can clearly see the limitations
> on my geometric model
Yes. Very clearly. More clearly than you think.
> since you can only fit on so many electrons
> before you have to expand the central core. It provides a very
> intuituive and visual explanation for the shell structure.
Balls
Bollocks. Go and look at the periodic table and a table of atomic numbers.
> Predicting the properties of
> elements based solely on geometric symmettry is a significant
> prediction.
You have not predicted anything.
> > > > The strong force is not "special". The strong force proves to be
purely
> > > > electromagnetic in nature. No one previously realized that the near
field
> > > > nucleon magnetic moments are superior to the nuclear electric
moments.
> > > >
> > > The strong force is required to keep a compact nucleus which contains
> > > all of the protons/neutrons together. My model may eliminate the need
> > > since the protons/electrons/neutrons are evenly spread across the
> > > entire atom in a neutral matrix.
> >
> > Have you calculated if this system is stable? I would think that the
> > Coulomb forces would destroy it.
> How would I calculate the force so that it would satisfy you.
That is entirely your problem.
> I don't
> know if I believe the radius given for electrons and protons, so I'm
> not certain what to plug in for distance.
Your beliefs on this matter are about as relevant as your views on what
manure to put on a vegetable plot. Go and look at the evidence..
> Why would you think the
> Colulob forces would destroy it? Every electron is touching a proton
> and vice-versa. I would think it would be highly stable. As a
> comparison, I have some round magnets and they attract 1/r^2 with
> opposite poles on opposite sides, so they roughly model the
> interaction of proton/electrons. When I stack them together like I
> have them in my model, they stick together quite strongly.
Classical models can never explain the structure of atoms. It has been
tried and it failed.
Franz
Bjoern Feuerbacher
>
> "FrankH" <frank...@yahoo.com> wrote in message
> news:46484c9f.04012...@posting.google.com...
[snip]
> > The gold atoms are
> > supported by a sheet of glass
>
> No.
Oh, good. I was beginning to doubt my memory! ;-)
[snip]
> > meaning a particle can't occupy just any
> > space. I read of an experiment trying to measure the effect of a
> > neutron dropping in a gravity field and they found that it fell in
> > discrete steps - not a continuous drop.
>
> Reference please.
I think he means this:
<http://www.physi.uni-heidelberg.de/~abele/nature.pdf>
What he totally ignores is that this effect has *nothing* to do with
"quantization of space", but is yet another confirmation of the
predictions of standard QM. Nevertheless, he said just a few days ago
that he knows of no experimental verifications of QM, IIRC...
[snip rest]
Bye,
Bjoern
Bjoern Feuerbacher
>
> Bjoern Feuerbacher <bfeu...@ix.urz.uni-heidelberg.de> wrote in message news:<400E6FF2...@ix.urz.uni-heidelberg.de>...
> > FrankH wrote:
> > >
> > > Bjoern Feuerbacher <bfeu...@ix.urz.uni-heidelberg.de> wrote in message
> > > [snip]
> > > >
> > > > Well, what you neglect to consider is (among other things):
> > > > 1) The charge of the electron and the proton have been shown to be equal
> > > > with *very* great accuracy - around 10^21! (see, for example, here:
> > > > <http://pdg.lbl.gov/2002/bxxxn.pdf>)
> > > This is an article on Baryons - did I miss something?
> >
> > Perhaps that the proton *is* a baryon? Hey, it's the very first entry
> > in this table of particle properties!
Did you get this?
> > > Sure electron/proton charges appear equal, but I have theorized that the
> > > electron can get far enough away from the proton to leave it
> > > unsheilded part of the time.
> >
> > What do you mean by "unshielded", specifically?
> >
> My theory is that the only way that an electron and proton could be
> totally neutral is if they occupied exactly the same place in space.
You mean "...the only way that a *system* consisting of an electron and
a proton could be totally neutral...", right?
Well, according to the standard solutions to the Schroedinger equation,
you are partly right - such a system is only *completely* neutral if one
looks at it from an infinite distance. If the distance to the proton is
finite,
one should see a slight positive charge. I think this is one of the
causes
of the van-der-Waals force.
> Since they do not, there must always be some imbalance where an object
> in the vicinity will see either a more positive field or more negative
> field.
Always a positive *charge* ("positive field" doesn't make sense),
according to the Schroedinger equation. But this should be absolutely
negligible for pretty much all effects - the positive charge is *very*
small.
> Since electrons are free to move about and protons are not,
> this gives a bias towards seeing the positive field more often than
> the negative - leading to an imbalance when measured some distance
> away from the electron/proton pair.
Right.
> This means that an electron/proton
> could have exactly the same charge, but yet still produce an
> observable positive charge from moment to moment.
Right again. So what? Where does this lead to?
Didn't you want to explain gravitational attraction, based on
electrostatic forces? If yes, how does it help you that atoms should
show a slight positive charge? This would lead to a *repelling* force,
not an attracting force!
[snip]
> > > > 2) If the earth would have an overall positive charge (BTW, the
> > > > formulation "positively charge field" doesn't make sense), we would
> > > > have
> > > > noticed this long ago - for example, by observing the trajectories of
> > > > charged cosmic rays.
> > >
> > > Actually, we have noticed this. The Earth has an electric field
> > > measuring about 120v/m at sea level. I believe the total charge if
> > > concentrated at a point in the center the Earth has been estimated at
> > > 10^6 C.
> >
> > References, please.
> See:
> http://hypertextbook.com/facts/1998/TreshaEdwards.shtml
Thanks! The page looks quite reliable, so I think I can trust them.
> There is some disagreement about the value, but most are in the
> ballpark of 120v/m. Also see the blazelabs reference I have below.
I also note that the say that the field points downward - i.e., positive
particles are attracted towards the earth. In other words, the charge
of the earth's surface is negative! This seems to contradict several
of your assertions...
Do you know of any explanation for this charge? Unfortunately, this
topic is
fairly new to me...
[snip]
> > > although the
> > > formula for this sort of force is basically dialectric constant *
> > > volume * E^2.
> >
> > Huh??? This formula gives the *energy* in the field. It has *nothing* to
> > do with the force between a dipole and a point charge!
Hello?
> > > I think that would make the force drop off with 1/r^4
> > > which is definitely not gravity.
> >
> > Sorry, but where did you get this from?
> This is the dielectrophoretic force
>
> See this page:
> http://www.blazelabs.com/f-efield.asp
Who wrote this? At first sight, it looks suspicously like a crackpot
web site...
However, the formula they give there for the "dielectrophoretic force"
is *not* the one you gave above! What *you* wrote was
F = 1/2 * dielectric constant * Volume * E^2.
What *they* write is
F = 1/2 * polarisability * Volume * grad E^2!
Confusing the dielectric constant with polarisability is an
understandable
error (although a strange one) - but leaving out the gradient makes the
formula totally nonsensical!
> This provided the basis for the exploration of electrostatic gravity.
> I'm just trying to answer the question that was asked.
Which question?
> > Yes, absence of evidence is not evidence of absence. But why on earth
> > should we try to find a new theory of gravity, although the existing one
> > is very well tested and shows no problems so far? Just because you don't
> > like it, or what?
>
> Wait, I didn't know we had an adequate theory of gravity
General Relativity?
> since we
> can't seem to unite it with the other forces.
So what??? The strong force hasn't be shown to be united with the other
forces, too! (although there are some hints that this may be the case)
What
strange sort of criterion *is* this???
> The appeal of an
> electrostatic gravity is that it would provide that unification
> instantly by electrostatic force = gravity force.
Yes, that would be nice - but unfortunately, the evidence speaks
*strongly* against this.
Just consider the ugly fact that electrostatic forces can be attracting
and repelling, but gravity is always attracting. How on earth do you
plan to explain this discrepancy? (hint: Quantum Field Theory *can*
explain this)
> We do want to unify the forces, don't we?
Yes. But your approach doesn't look very promising, sorry.
> I don't know if I can get there, but it's worth a shot!
Einstein tried to unify electrodynamics and gravity, too, with far
better ideas than your, for several decades. He failed.
And, hint: modern theories like string theory *do* unite gravity with
the other forces, in a sense. There are a lot of problems still with
string theory (among others that there is no experimental evidence for
it so far and that it isn't clear if it will be possible in the future
to find such evidence), but nevertheless, it looks promising. And in
contrast to your model, string theory is strongly based on known, well
tested physical theories, and doesn't contradict any known experimental
evidence.
Bye,
Bjoern
Bjoern Feuerbacher
>
> Bjoern Feuerbacher <bfeu...@ix.urz.uni-heidelberg.de> wrote in message news:<400EAC16...@ix.urz.uni-heidelberg.de>...
> > FrankH wrote:
> > >
[snip]
> > > Actually, as has been noted by the other posters, I am quite ignorant
> > > of the experimental evidence and built my model starting with a
> > > thought experiment of what would be the most simplest way for
> > > electrons/protons to assemble.
> >
> > That's a strange way to build a model. Why didn't you learn about the
> > experimental evidence and *then* tried to build a model based on that?
> >
> As the other poster noted, that has been the mistake of previous
> theorists in reverse engineering a model.
This doesn't make much sense. One can't build a model without knowing
the results of some experiments, without knowing how the world reacts
if one "asks" it certain things. Calling this "reverse engineering"
is nonsense.
> I just wanted to come up
> with a physical model of an atom. So I used what I had avaliable and
> found a geometric sequence - now I do need to see if it matches
> experimental results.
Well, why don't you open a few books on atomic and nuclear physics and
see for yourself? IIRC, you have got quite a few recommendations
already.
We have far too many crackpots here with "great new models" who first
ask
us to find errors in their models, and then, when we point out the
errors,
refuse to admit that they might be wrong - they either completely ignore
the evidence, misunderstand it, or make up a few new ad hoc hypotheses
in
order to take them in account. All the while completely ignoring that
Rutherford, Bohr and Schroedinger had *reasons* to come up with their
models - all these people seem to think that Rutherford et al. dreamed
up their models
out of nowhere, without any justification or logical reasoning.
> > > Please take another look at my website.
> > > My model's advantage is that it does go together straight forward with
> > > a predicatable geometric sequence which you can do almost blindly, and
> > > it would put itself together automatically.
> >
> > Nice. So what? Solving the Schroedinger equation for the H atom is also
> > fairly straightforward.
>
> Uhhh, not straightforward to me. How can you say a full page of
> equations is straightforward. I think most people look at that and
> scratch their heads.
Please don't confuse "easy" with "straightforward".
> The diagrams of the electron orbits is even more
> bizzare.
What diagrams? What orbits????? There are no orbits of electrons in
standard
QM.
> Just how electrons manage to confine themselves to the
> strange donut shapes required in QM is unintuitive.
Well, picture them as standing waves - I think then it becomes much
easier
to understand.
And, BTW: do you equate "orbits" with these "donut shapes" here??? If
yes, then you are really thoroughly confused about what QM says!
> The higher level shells can't even be reasonably drawn.
So what??? There are clear mathematical formulas which describe them.
Essentially their shapes are described by the spherical harmonics -
and these functions are well-known in physics and mathematics and used
in a *lot* of places - they weren't simply "made up" just for the H
atom!
They are not only important in atomic physics, but essentially
everywhere
where one wants to look at a (often only approximately) spherically
symmetric problem - for example, they are used in electrostatics as well
as in electrodynamics.
> On the other hand, the
> geometric model I have can be grasped without any math by anybody and
> you can build out all the elements with ease.
Nice. So what? That's not evidence that it's right!
> It has a simplicity that
> the Schroedinger equations do not and science tends to favor simpler
> and more elegant solutions.
"simpler solution" in science doesn't necessarily mean "solution which
is based on playing around with lego blocks". And you totally ignore
that
having evidence is *far* more important in science than having a simple
solution.
> > > So far, I have seen that
> > > this model can explain the observed electron shell structure with the
> > > main quantum number representing the electrons in the vertical core
> >
> > I still haven't seen you making any *quantitative* predictions - like
> > energies.
>
> I am going to work on that
Nice. Please come back when you have achieved this! Until then, this
discussion is quite futile.
> - but I don't know exactly where to begin.
Well, why don't you try to compute Rutherford's scattering cross
section? Or the spectrum of the hydrogen atom? Both are given by fairly
simple formulas.
> I was hoping I could get someone who knows this type of math
What type of math do you need, specifically?
> interested enough in my model to run the quantitative predictions and
> pull some numbers out.
In other words: you invent some strange ideas and then want someone
other to do the hard work for you.
> Can you point me in the right direction on how
> to calculate energies based on my geometric model?
So far, you haven't told us *anything* *quantitatively* about your
model, so it's hard for me to tell you how to do something with it...
After all, it's *your* model, isn't it?
> > > and you can see why we get the number of electrons in the outer shells
> > > due to the geometric progression of the outer electrons fitting into
> > > the vertical core.
> >
> > Well, we get this from the Schroedinger equation, too.
> >
> The equations don't graphically show why only a limited number of
> electrons can fit into each shell.
No equation can show anything graphically - this makes no sense. Hence
this
objections seems rather strange.
> You can clearly see the limitations
> on my geometric model since you can only fit on so many electrons
> before you have to expand the central core. It provides a very
> intuituive and visual explanation for the shell structure.
Unfortunately for you, "common sense" explanations have often shown
to be wrong in science.
Have you ever looked at the double slit experiment with electrons?
Feynman describes it fairly nice in his "Lectures on Physics", Volume
III. Also there is a discussion of this (or something equivalent) in the
book "The strange world of Quantum Mechanics" by D. Styer. I think this
experiment makes it fairly hard to think of an electron as a small "lego
block"...
[snip]
> > > Spin appears to be due to the
> > > fact that the atom assembles a mirror image of itself about the axes,
> > > and the reversed electron/proton may explain the difference states.
> >
> > How does this explain the results of the Stern-Gerlach experiment?
> >
> It doesn't.
Then you don't have an explanation for spin! The Stern-Gerlach
experiment
was the *reason* to develop the *concept* of spin - before, no one had
the slightest idea that spin exists! Hence if you want to explain spin,
you
*have* to explain the results of this experiment!
> I was thinking about explaining the fine spectral lines
> between electrons with same energy state but with a different spin.
This sentence doesn't make sense. Are you sure you know what you are
talking about? (I think I know what you *mean*, but the way you express
it contains several grave errors!)
[snip]
> > > since my
> > > model with its four arms would tend to favor fission which would knock
> > > off arms
> >
> > A model of atoms/nuclei with "arms"? You really should talk to Porat!
Did you notice this?
> > > and you might think that there would be more products which
> > > represent a quarter/half/3-quarter of the atom - or maybe not. My
> > > models do show a pretty pattern of electron/proton pairs (looks like
> > > an X viewed from the top). This may give it predictable magnetic
> > > properties.
> >
> > "may"
>
> I just looked up what are the primary products of U238 fission and the
> largest components are generally about 1/2 and 1/4 of the original
> atom size. Looks like it knocks off an the vertical core plus the
> electrons in the arms.
> See http://www.chem.uidaho.edu/~honors/decay.html
This looks like retrospective fitting to me - or could you have
predicted
this beforehand from your model?
BTW, talking about fission: can your model explain why iron is the most
stable element?
> > > I have also found that the most stable elements are built
> > > out of mostly helium nuclei.
> >
> > Nuclear physics explains this. See the book by Povh mentioned above -
> > the crucial term is "pairing energy".
Did you get this?
> > > I couldn't figure out why the noble gases
> > > were not totally symmetric until I realized that the atom favors being
> > > built out of helium nuclei. You can directly see this in my model.
> >
> > Nice. I'm *still* waiting for the quantitative predictions...
>
> Isn't this a type of quantitative prediction?
Huh? Why on earth do you think so?
> The model shows the role
> of paring energy in the production of noble gases.
Sorry, I don't understand this sentence. In what way does your model
show this?????
> The prediction
> being that if you build out a symettrical atom built mostly out of
> helium nuclei, you get a nobel gas.
How does your model arrive at this prediction?
> Predicting the properties of
> elements based solely on geometric symmettry is a significant
> prediction.
Well, how does your model predict the properties of "symmetrical atoms
built mostly out of helium nuclei"?
And, BTW, your model doesn't have a nucleus, apparently - the protons
are distributed over all of the atom, right? If yes, then how do you
explain that one can experimentally detect not only excitations of the
atom, but also excitations of the nucleus? With energies which are about
1 million times greater? This is in complete agreement with the facts
that in the standard model, this is about the size difference between
atom and nucleus, and excitation energies should increase when the size
is smaller, according to Heisenberg's uncertainty.
[snip]
> > > The strong force is required to keep a compact nucleus which contains
> > > all of the protons/neutrons together. My model may eliminate the need
> > > since the protons/electrons/neutrons are evenly spread across the
> > > entire atom in a neutral matrix.
> >
> > Have you calculated if this system is stable? I would think that the
> > Coulomb forces would destroy it.
>
> How would I calculate the force so that it would satisfy you.
Well, you know the charges, so where is the problem with calculating the
force??? Aren't you able to do some fairly basic calculation with
Coulomb's law? Granted, this becomes complicated because in your model,
the particles themselves apparently aren't spherical symmetric - but as
a first approximation, you could simply pretend they are.
> I don't
> know if I believe the radius given for electrons and protons,
Well, the radii given surely contradict your model of "lego blocks",
don't they?
> so I'm not certain what to plug in for distance.
What have the radii of the particles to do with their distances to each
other?
> Why would you think the
> Colulob forces would destroy it?
Because pretty much all configurations of positive and negative charges
form an instable system. I can't think of any stable system offhand -
the stabilization can only be provided if you take quantum effects into
account.
> Every electron is touching a proton and vice-versa.
So what? Why should this make the system stable?
And what do you mean by "touching"? Do electrons and protons have
something like a "hard surface" in your model?
> I would think it would be highly stable.
Hand waving - or "waffling", as Franz said.
> As a
> comparison, I have some round magnets and they attract 1/r^2 with
> opposite poles on opposite sides,
It's new to me that magnets attract each other with 1/r^2 - this can't
be true. Magnets are dipoles, and dipoles don't attract each other with
such a law.
> so they roughly model the
> interaction of proton/electrons. When I stack them together like I
> have them in my model, they stick together quite strongly.
Hint: magnets have a hard surface.
Or, microscopically seen: the magnetic force which is pulling the two
together is counteracted by the electrostatic force of the atoms in the
magnets, which repel each other. And even more important for this
repulsion is the Pauli principle and other quantum effects - if these
wouldn't exists, the magnets wouldn't stick together, they would merge
with another.
[snip]
Bye,
Bjoern
FrankH
coming from a proton/electron system? If there were positive charge at
all, no matter how tiny, it would add up over the entire mass of the
earth and could produce a massive force. If you believe in
Van-der-waals forces, maybe the addition of these tiny forces makes up
gravity.
>
> > Since electrons are free to move about and protons are not,
> > this gives a bias towards seeing the positive field more often than
> > the negative - leading to an imbalance when measured some distance
> > away from the electron/proton pair.
>
> Right.
>
>
> > This means that an electron/proton
> > could have exactly the same charge, but yet still produce an
> > observable positive charge from moment to moment.
>
> Right again. So what? Where does this lead to?
>
> Didn't you want to explain gravitational attraction, based on
> electrostatic forces? If yes, how does it help you that atoms should
> show a slight positive charge? This would lead to a *repelling* force,
> not an attracting force!
You'd think that if everything were positively charged, everything
would repel each other. However, one way of solving this problem is
that the dielectrophoretic force is greater than the positive charge
repulsion. If all the neutral atoms in your body are attracted to
Earth by the dielectrophoretic force and only a few positively charged
atoms are repelling, the neutral atoms win. Basically, any
sufficiently high charge source attracts all neutral matter. You can
see this if you charge up a comb and place it near a non-charged
material. It doesn't matter what - tiny bits of paper, metal, rocks,
dirt - they are neutrally charged, yet are strongly attracted to the
charged comb. Another solution would be that everything on the Earth
has a bias to be negatively charged. Then we would get normal
electrostatic attraction.
>
>
>
> [snip]
>
>
> > > > > 2) If the earth would have an overall positive charge (BTW, the
> > > > > formulation "positively charge field" doesn't make sense), we would
> > > > > have
> > > > > noticed this long ago - for example, by observing the trajectories of
> > > > > charged cosmic rays.
> > > >
> > > > Actually, we have noticed this. The Earth has an electric field
> > > > measuring about 120v/m at sea level. I believe the total charge if
> > > > concentrated at a point in the center the Earth has been estimated at
> > > > 10^6 C.
> > >
> > > References, please.
> > See:
> > http://hypertextbook.com/facts/1998/TreshaEdwards.shtml
>
> Thanks! The page looks quite reliable, so I think I can trust them.
>
>
> > There is some disagreement about the value, but most are in the
> > ballpark of 120v/m. Also see the blazelabs reference I have below.
>
> I also note that the say that the field points downward - i.e., positive
> particles are attracted towards the earth. In other words, the charge
> of the earth's surface is negative! This seems to contradict several
> of your assertions...
Yes, I have noted that the polarity appears reversed. I'm checking
this, but I think it may be a problem with the point of reference
(where do you start zero volts - in space or on the ground). We know
that clouds have a charge separation where the bottom is negatively
charged and the top is positive. This would be consisitent with a
positively charged ground.
>
> Do you know of any explanation for this charge? Unfortunately, this
> topic is
> fairly new to me...
I have already provided two possible answers - either there is a
residual positive charge coming out of an electron/proton system, or
the Earth gets charged by being in the solar wind. But I'd say it was
still anyones guess.
>
>
> [snip]
>
>
> > > > although the
> > > > formula for this sort of force is basically dialectric constant *
> > > > volume * E^2.
> > >
> > > Huh??? This formula gives the *energy* in the field. It has *nothing* to
> > > do with the force between a dipole and a point charge!
>
> Hello?
>
>
>
> > > > I think that would make the force drop off with 1/r^4
> > > > which is definitely not gravity.
> > >
> > > Sorry, but where did you get this from?
> > This is the dielectrophoretic force
> >
> > See this page:
> > http://www.blazelabs.com/f-efield.asp
>
> Who wrote this? At first sight, it looks suspicously like a crackpot
> web site...
Here is a less crackpot site.
http://www.ibmm-microtech.co.uk/pages/science/dep.htm
>
> However, the formula they give there for the "dielectrophoretic force"
> is *not* the one you gave above! What *you* wrote was
> F = 1/2 * dielectric constant * Volume * E^2.
> What *they* write is
> F = 1/2 * polarisability * Volume * grad E^2!
>
> Confusing the dielectric constant with polarisability is an
> understandable
> error (although a strange one) - but leaving out the gradient makes the
> formula totally nonsensical!
>
Yes, I oversimplified. I made the original calculation in the
newsgroup article:
I was confused about how the del operator worked. I found the page:
http://www-math.mit.edu/~djk/18_022/chapter07/section01.html
It looked like to me that for the electrostatic gravity case, there
are no X or Y components and only an E in the Z direction, so grad E^2
would be the same as E^2. I also had a question about the 120v/M
figure being the value of the gradient such that grad E = 120v/m. It
would seem to be as it measures the potential difference between 1
meter of space. The guy I was debating this with in the newsgroup
appeared to agree with my calculation, although I think I still must
be missing something since I don't see how divergence of the field
gets into the picture, since a 120v/m could also represent a uniform
electric field. Can you help me out? I think the other
dielectrophoretic web site has a more scientific derivation and the
form of the equation looks different. Looks like it divides by 2E
instead of just 2. Maybe it drops off by 1/r^2???? The force would
have to be proportional to E which drops off at 1/r^2.
I would like to make a proper calculation for the force on a 1kg cube
of carbon in a 120v/m field on the Earth. Thanks for your help.
[snip]
FrankH
> >
> > I just looked up what are the primary products of U238 fission and the
> > largest components are generally about 1/2 and 1/4 of the original
> > atom size. Looks like it knocks off an the vertical core plus the
> > electrons in the arms.
> > See http://www.chem.uidaho.edu/~honors/decay.html
>
> This looks like retrospective fitting to me - or could you have
> predicted
> this beforehand from your model?
It is not restrospective fitting, the reason why I wanted to find the
fission product data is because my model's geometric shape would make
it so that it would be most vulnerable to splitting along the vertical
core. I made the prediction that we should see 1/4, 1/2, 3/4
fragments. Doing a further analysis on uranium, my model would predict
that the core would contain 14 atomic units (a square of
electron,proton,neutron) in the core and the arms would contain 19-20
units in each of the arms for a total atomic weight of 92. I would
predict that the fission products should contain the core plus parts
of the arms. So you would expect to see a 1/4 fraction at 14+19 = 33,
1/2 fraction at 14 + 38=52, 3/4 fraction at 14 + 57 = 71.
The experiment results show the most common fission products being Br,
Kr and Rb at atomic weights 35, 36, 37 and I, Xe, CS at atomic weights
53, 54 and 55. This approximately corresponds to the predicted 1/4 and
1/2 fractions. The predictions are in the bottom of the range, but I
think it is reasonable to think that you could get a few extra atomic
units attached when you split an atom. What do you think of that?
>
> BTW, talking about fission: can your model explain why iron is the most
> stable element?
>
Looking at my model for iron (which is a partially poplated krypton
atom), I can see that the core is built out of helium nuclei (divisble
by 2) and is made up of 8 atomic units. With the exception of the very
top/bottom of the arms are also made out of helium nuclei. According
to my model, this atom is symmettric and contains the maximum number
of helium nuclei. So I am not suprised that this is a very stable
element. The next element meeting this criteria is Krypton. Are you
saying iron is more stable than krypton? Where do I get a chart of
these stabilities? It would be interesting to run a comparison against
many elements in my model to see if any trends show up.
>
> And, BTW, your model doesn't have a nucleus, apparently - the protons
> are distributed over all of the atom, right? If yes, then how do you
> explain that one can experimentally detect not only excitations of the
> atom, but also excitations of the nucleus? With energies which are about
> 1 million times greater? This is in complete agreement with the facts
> that in the standard model, this is about the size difference between
> atom and nucleus, and excitation energies should increase when the size
> is smaller, according to Heisenberg's uncertainty.
Do you have any references? I haven't heard of this and would like to
investigate.
[snip]
Bjoern Feuerbacher
>
> Bjoern Feuerbacher <bfeu...@ix.urz.uni-heidelberg.de> wrote in message news:<400FBF16...@ix.urz.uni-heidelberg.de>...
> > FrankH wrote:
> > >
> > > Bjoern Feuerbacher <bfeu...@ix.urz.uni-heidelberg.de> wrote in message news:<400E6FF2...@ix.urz.uni-heidelberg.de>...
> > > > FrankH wrote:
> > > > >
> > > > > Bjoern Feuerbacher <bfeu...@ix.urz.uni-heidelberg.de> wrote in message
> > > > > [snip]
> > > > > >
> > > > > > Well, what you neglect to consider is (among other things):
> > > > > > 1) The charge of the electron and the proton have been shown to be
> > > > > > equal
> > > > > > with *very* great accuracy - around 10^21! (see, for example,
> > > > > > here: <http://pdg.lbl.gov/2002/bxxxn.pdf>)
> > > > >
> > > > > This is an article on Baryons - did I miss something?
> > > >
> > > > Perhaps that the proton *is* a baryon? Hey, it's the very first entry
> > > > in this table of particle properties!
> >
> > Did you get this?
Hello?????
> > > > > Sure electron/proton charges appear equal, but I have theorized that
> > > > > the
> > > > > electron can get far enough away from the proton to leave it
> > > > > unsheilded part of the time.
> > > >
> > > > What do you mean by "unshielded", specifically?
> > > >
> > > My theory is that the only way that an electron and proton could be
> > > totally neutral is if they occupied exactly the same place in space.
> >
> > You mean "...the only way that a *system* consisting of an electron and
> > a proton could be totally neutral...", right?
Hello? It would be nice if you wouldn't ignore most of what I write!
> > Well, according to the standard solutions to the Schroedinger equation,
> > you are partly right - such a system is only *completely* neutral if one
> > looks at it from an infinite distance. If the distance to the proton is
> > finite, one should see a slight positive charge. I think this is one of
> > the causes of the van-der-Waals force.
Hello???
> > > Since they do not, there must always be some imbalance where an object
> > > in the vicinity will see either a more positive field or more negative
> > > field.
> >
> > Always a positive *charge* ("positive field" doesn't make sense),
> > according to the Schroedinger equation. But this should be absolutely
> > negligible for pretty much all effects - the positive charge is *very*
> > small.
> >
> So you would agree that we should see some small positive charge
> coming from a proton/electron system?
A *very* small charge, and it would depend on the distance.
Exponentially!
> If there were positive charge at
> all, no matter how tiny, it would add up over the entire mass of the
> earth and could produce a massive force.
A *repelling* force. Not like gravity...
> If you believe in Van-der-waals forces,
I don't "believe" in them - I know from experimental evidence that they
exist.
> maybe the addition of these tiny forces makes up gravity.
Hint: van-der-Waals forces have a completely different dependence on
distance compared to gravity.
> > > Since electrons are free to move about and protons are not,
> > > this gives a bias towards seeing the positive field more often than
> > > the negative - leading to an imbalance when measured some distance
> > > away from the electron/proton pair.
> >
> > Right.
> >
> >
> > > This means that an electron/proton
> > > could have exactly the same charge, but yet still produce an
> > > observable positive charge from moment to moment.
> >
> > Right again. So what? Where does this lead to?
> >
> > Didn't you want to explain gravitational attraction, based on
> > electrostatic forces? If yes, how does it help you that atoms should
> > show a slight positive charge? This would lead to a *repelling* force,
> > not an attracting force!
>
> You'd think that if everything were positively charged, everything
> would repel each other.
Well, maybe it does - the forces are *sooooo* small that one shouldn't
be able to detect them, I think. There are a lot of other, *much* bigger
forces counteracting them!
> However, one way of solving this problem is
> that the dielectrophoretic force is greater than the positive charge
> repulsion. If all the neutral atoms in your body are attracted to
> Earth by the dielectrophoretic force and only a few positively charged
> atoms are repelling, the neutral atoms win.
Err, didn't we agree that this force has the wrong dependence on
distance,
so it can't explain gravity?
> Basically, any
> sufficiently high charge source attracts all neutral matter. You can
> see this if you charge up a comb and place it near a non-charged
> material. It doesn't matter what - tiny bits of paper, metal, rocks,
> dirt - they are neutrally charged, yet are strongly attracted to the
> charged comb.
I still don't see how this could be relevant to gravity!
> Another solution would be that everything on the Earth
> has a bias to be negatively charged. Then we would get normal
> electrostatic attraction.
Well, let's say that Earth is negatively charged and the sun positively.
Then you could explain the attraction between them with electrostatic
forces. However, what about the other planets? Let's look at Venus, for
example: if it is negatively charged, then it is attracted by the sun
and repelled by the earth. If it is positively charge, then it is
repelled by the sun and attracted by the earth. But what we observe in
reality is that Venus is attracted by *both* the sun *and* the earth!
(reference: any text book on orbital mechanics) How do you explain this?
[snip]
> > > > > Actually, we have noticed this. The Earth has an electric field
> > > > > measuring about 120v/m at sea level. I believe the total charge if
> > > > > concentrated at a point in the center the Earth has been estimated
> > > > > at 10^6 C.
> > > >
> > > > References, please.
> > > See:
> > > http://hypertextbook.com/facts/1998/TreshaEdwards.shtml
> >
> > Thanks! The page looks quite reliable, so I think I can trust them.
> >
> >
> > > There is some disagreement about the value, but most are in the
> > > ballpark of 120v/m. Also see the blazelabs reference I have below.
> >
> > I also note that the say that the field points downward - i.e., positive
> > particles are attracted towards the earth. In other words, the charge
> > of the earth's surface is negative! This seems to contradict several
> > of your assertions...
>
> Yes, I have noted that the polarity appears reversed. I'm checking
> this, but I think it may be a problem with the point of reference
> (where do you start zero volts - in space or on the ground).
No, this makes no sense at all! The choice of point of reference (the
"zero") for voltage doesn't have any influence on the direction of an
electrostatic field!
> We know
> that clouds have a charge separation where the bottom is negatively
> charged and the top is positive. This would be consisitent with a
> positively charged ground.
Do we know this? IIRC, the opposite polarization of clouds occurs, too.
> > Do you know of any explanation for this charge? Unfortunately, this
> > topic is fairly new to me...
>
> I have already provided two possible answers - either there is a
> residual positive charge coming out of an electron/proton system, or
> the Earth gets charged by being in the solar wind. But I'd say it was
> still anyones guess.
My question was more in the line of "Do you know if *standard physics*
has any explanation for this?", *not* of "Can you find an explanation
for this yourself?".
> > > > > although the
> > > > > formula for this sort of force is basically dialectric constant *
> > > > > volume * E^2.
> > > >
> > > > Huh??? This formula gives the *energy* in the field. It has *nothing*
> > > > to do with the force between a dipole and a point charge!
> >
> > Hello?
Hello?????
> > > > > I think that would make the force drop off with 1/r^4
> > > > > which is definitely not gravity.
> > > >
> > > > Sorry, but where did you get this from?
> > >
> > > This is the dielectrophoretic force
> > >
> > > See this page:
> > > http://www.blazelabs.com/f-efield.asp
> >
> > Who wrote this? At first sight, it looks suspicously like a crackpot
> > web site...
>
> Here is a less crackpot site.
> http://www.ibmm-microtech.co.uk/pages/science/dep.htm
Thanks. I notice that they have got a quite different formula there...
> > However, the formula they give there for the "dielectrophoretic force"
> > is *not* the one you gave above! What *you* wrote was
> > F = 1/2 * dielectric constant * Volume * E^2.
> > What *they* write is
> > F = 1/2 * polarisability * Volume * grad E^2!
> >
> > Confusing the dielectric constant with polarisability is an
> > understandable
> > error (although a strange one) - but leaving out the gradient makes the
> > formula totally nonsensical!
> >
> Yes, I oversimplified. I made the original calculation in the
> newsgroup article:
>
> http://groups.google.com/groups?q=g:thl1731381782d&dq=&hl=en&lr=&ie=UTF-8&oe=UTF-8&selm=46484c9f.0308270959.32d838e9%40posting.google.com
Apparently you ignored the gradient there, too.
> I was confused about how the del operator worked.
Didn't they teach you about it at the MIT?
> I found the page:
>
> http://www-math.mit.edu/~djk/18_022/chapter07/section01.html
>
> It looked like to me that for the electrostatic gravity case, there
> are no X or Y components and only an E in the Z direction, so grad E^2
> would be the same as E^2.
Huh?????? How on earth do you arrive at this conclusion?????
If E is in z direction (better: in r direction) and depends only on
z (or: on r), then grad E^2 = E * dE/dz !!!!!
> I also had a question about the 120v/M
> figure being the value of the gradient such that grad E = 120v/m.
Huh??? Sorry, this again makes no sense at all! E = 120 V/m, not
grad E = 120 V/m!!!
> It would seem to be as it measures the potential difference between 1
> meter of space.
Yes, the electric field measures the potential difference with respect
to changes in position.
> The guy I was debating this with in the newsgroup
> appeared to agree with my calculation,
Then he probably didn't look very hard at it!
> although I think I still must
> be missing something since I don't see how divergence of the field
> gets into the picture,
Huh? Where on earth do you see a "divergence" here??? You need to
calculate the *gradient* of E^2!
> since a 120v/m could also represent a uniform
> electric field.
Well, this number alone isn't enough to calculate the gradient - you
need
to have some numbers which tell you how the electric field depends on
height above ground!
> Can you help me out?
Sorry, no. As already mentioned, the topic "electric field of the earth"
is fairly new to me.
> I think the other
> dielectrophoretic web site has a more scientific derivation and the
> form of the equation looks different. Looks like it divides by 2E
> instead of just 2.
What you missed, apparently, is that it also multiplies by m
(apparerently the dipole moment) instead of by the polarisability. Seems
to be a completely different formula to me.
> Maybe it drops off by 1/r^2????
I see no reason at all why it should.
> The force would have to be proportional to E
Why should it?
> which drops off at 1/r^2.
Why should it?
> I would like to make a proper calculation for the force on a 1kg cube
> of carbon in a 120v/m field on the Earth. Thanks for your help.
As already pointed out: if you want to calculate the dielectrophoretic
force, these numbers are not sufficient. You need the dependence of E on
height.
*If* you are right and E is proportional to 1/r^2, then E*dE/dr is
proportional to 1/r^5. Doesn't look very promising to me...
> [snip]
Bye,
Bjoern
Bjoern Feuerbacher
>
> Bjoern Feuerbacher <bfeu...@ix.urz.uni-heidelberg.de> wrote in message
> [snip]
> > >
> > > I just looked up what are the primary products of U238 fission and the
> > > largest components are generally about 1/2 and 1/4 of the original
> > > atom size. Looks like it knocks off an the vertical core plus the
> > > electrons in the arms.
> > > See http://www.chem.uidaho.edu/~honors/decay.html
> >
> > This looks like retrospective fitting to me - or could you have
> > predicted this beforehand from your model?
>
> It is not restrospective fitting, the reason why I wanted to find the
> fission product data is because my model's geometric shape would make
> it so that it would be most vulnerable to splitting along the vertical
> core. I made the prediction that we should see 1/4, 1/2, 3/4
> fragments.
Oh, then that's really a nice prediction from your model! Does it work
for other fissions and fusions, too? And where are the 3/4 fragments in
this case?
And, BTW, where does the page you cite above say that "the
largest components are generally about 1/2 and 1/4 of the original
atom size"? I don't find anything like that in that web page. In
contrast, I find the following:
"In reality, the exact clevage into two Pd nuclei is a lower probability
that the fissioning into two unequal particles. It is found that the
most
common immediate fission products are the set Br, Kr, Rb and I, Xe, and
Cs with everything in between also formed to varying degrees."
How does your model explain this?
> Doing a further analysis on uranium, my model would predict
> that the core would contain 14 atomic units (a square of
> electron,proton,neutron) in the core and the arms would contain 19-20
> units in each of the arms for a total atomic weight of 92. I would
> predict that the fission products should contain the core plus parts
> of the arms.
Why?
> So you would expect to see a 1/4 fraction at 14+19 = 33,
> 1/2 fraction at 14 + 38=52, 3/4 fraction at 14 + 57 = 71.
>
> The experiment results show the most common fission products being Br,
> Kr and Rb at atomic weights 35, 36, 37
That's not 33.
> and I, Xe, CS at atomic weights
> 53, 54 and 55.
That's not 52.
And how do you explain the missing 71?
> This approximately corresponds to the predicted 1/4 and
> 1/2 fractions.
Yes. *APPROXIMATELY*. How do you explain the deviations? Finding models
that *approximately* reproduce experimental results isn't very hard (you
*really* should try reading Porat's books! He has got *lots* of examples
for this!). A theory is only reliable if it either makes *precise*
predictions or at least can explain why the observed results deviate (a
bit) from its predictions.
> The predictions are in the bottom of the range, but I
> think it is reasonable to think that you could get a few extra atomic
> units attached when you split an atom.
Why should this be reasonable?
> What do you think of that?
I think that the *approximate* coincidence between your model and some
data is simply that - coincidence. I have seen *lots* of strange models
in my life, and pretty much *all* of them were able to predict some data
with relative good accuracy. Do you think that all of these conflicting
models are right at once? (again, you *really* should try reading
Porat's book!)
> > BTW, talking about fission: can your model explain why iron is the most
> > stable element?
> >
> Looking at my model for iron (which is a partially poplated krypton
> atom), I can see that the core is built out of helium nuclei (divisble
> by 2) and is made up of 8 atomic units. With the exception of the very
> top/bottom of the arms are also made out of helium nuclei. According
> to my model, this atom is symmettric and contains the maximum number
> of helium nuclei.
Didn't you say that such symmetric atoms represent noble gases?
> So I am not suprised that this is a very stable
> element.
The question was not about "very stable". It was about "the most
stable".
And how do you explain that the elements *below* iron don't undergo
fission ever, that they can only give energy by *fusion*?
> The next element meeting this criteria is Krypton. Are you
> saying iron is more stable than krypton?
Yes!!! Again I wonder how little they teach you at MIT...
> Where do I get a chart of these stabilities?
A quick Google search for "nuclear binding energies" gives:
<http://hyperphysics.phy-astr.gsu.edu/hbase/nucene/nucbin.html>
Look at the figure in "Fission and fusion can yield energy".
> It would be interesting to run a comparison against
> many elements in my model to see if any trends show up.
Well, good luck.
Hint: nuclear physics can explain the shape of this curve.
> > And, BTW, your model doesn't have a nucleus, apparently - the protons
> > are distributed over all of the atom, right? If yes, then how do you
> > explain that one can experimentally detect not only excitations of the
> > atom, but also excitations of the nucleus? With energies which are about
> > 1 million times greater? This is in complete agreement with the facts
> > that in the standard model, this is about the size difference between
> > atom and nucleus, and excitation energies should increase when the size
> > is smaller, according to Heisenberg's uncertainty.
>
> Do you have any references? I haven't heard of this and would like to
> investigate.
*sigh* They really don't teach much at MIT, apparently...
Have you ever heard of gamma radiation? That's the (in the public) best
known example of what I said above. You *do* know that the energy of
gamma radiation is about 1 million times greater than normal light
(which comes, according to standard theory, from the electron
excitations), don't you?
You can also look here for another example:
<http://hyperphysics.phy-astr.gsu.edu/hbase/hframe.html>
Bye,
Bjoern
Franz Heymann
> [snip]
> > He claims to be a graduate of MIT. Perhaps that's "Massachusetts
> > Interstate Trucking" school? (Feynman graduated from the other one).
>
> Yes, that would be Mens Institude of Typewriting - Well actually, I
> graduated in 1986 with a batchelors in computer science from the
> Massachusetts Institute of Technology.
What makes you think that entitles you to talk rubbish about physics?
[snip]
> lets talk science.
Then go and learn some physics and come back after four years if you want to
be argumentative rather than being in a learning mode.
Franz
Franz Heymann
[snip]
> Well actually, I
> graduated in 1986 with a batchelors in computer science from the
> Massachusetts Institute of Technology.
They could not have taught you much if they did not even teach you how to
spell your qualification correctly. How did you manage to slip through the
net and graduate?
[snip]
Franz
jmfb...@aol.com
Using the net requires individual interaction; he only worked
ala batch jobs. That implies that his I/O medium was cards.
/BAH
Subtract a hundred and four for e-mail.
FrankH
[snip]
>
>
> > > > The strong force is required to keep a compact nucleus which contains
> > > > all of the protons/neutrons together. My model may eliminate the need
> > > > since the protons/electrons/neutrons are evenly spread across the
> > > > entire atom in a neutral matrix.
> > >
> > > Have you calculated if this system is stable? I would think that the
> > > Coulomb forces would destroy it.
> >
> > How would I calculate the force so that it would satisfy you.
>
> Well, you know the charges, so where is the problem with calculating the
> force??? Aren't you able to do some fairly basic calculation with
> Coulomb's law? Granted, this becomes complicated because in your model,
> the particles themselves apparently aren't spherical symmetric - but as
> a first approximation, you could simply pretend they are.
>
>
protons and electrons are stable.
To calculate if the cubic model of atom construction is stable, we
will start with a square and build our way up.
Assuming we start with a square with vertices:
A - b
| |
c - D
Where A & D are protons and c & b are electrons.
We calculate stability based on the net forces exerted on each
component. By convention consider attractive forces as being negative
and repulsive forces as being positive. A stable configuration should
be one where all net forces in the X/Y axes are negative and equal.
Since this structure is completely symmettric, we can calculate the
force on one component and it will be the same for the rest of the
components. No matter which component you look at, there will always
be 2 adjacent components which are attracting and diagonal component
repelling.
We assume that the proton and electron are separated by some minimum
distance. The measured width of a Silicon atom using STM photographs
is ~750pm. According to the cubic atomic model, this corresponds to a
width of 4 protons/electrons, so the size of 1 proton/electron
according to the cubic model is ~187 pm or 187 X 10^-12 meters.
Force between adjacent pairs A-B and A-C using Coulumb's Law:
9 x 10^9 * 1.602 X 10^-19 * -1.602 X 10^-19/ (187 X 10^-12)^2 = -6.605
X 10^-9 Newtons
Force between diagonal pair A-D.
Distance from A-D by Pythagoras theorem = 264 X 10^-12 meters
9 x 10^9 * 1.602 X 10^-19 * 1.602 X 10^-19/ (264 X 10^-12)^2 = 3.314 X
10^-9 Newtons
Force in the X & Y directions by Pythagoras theorem = 2.352 X 10^-9
Newtons
X Y
A-B -6.605 X 10^-9 0
A-D 2.352 X 10^-9 2.352 X 10^-9
A-C 0 -6.605 X 10^-9
Net -3.291 X 10^-9 -3.291 X 10^-9
Force is negative and equal in the X & Y axes for all components and
therefore is stable.
Next we consider a cube where e & g are electrons and H & F are
protons
A -- b Y
| \ |\ |
| e | F |
c -- D | |
\ \ | |_____ X
H -- g \
\
Z
Components in capital letters are protons, lower case are electron.
This is also symmettric, so we only need to consider one component.
Distance between A-g = sqrt( (264 X 10^-12)^2 + (187 X 10^-12)^2) =
323 X 10^-12 meters
Force =9 x 10^9 * 1.602 X 10^-19 * -1.602 X 10^-19/ (323 X 10^-12)^2 =
-2.213 X 10^-9 Newtons
Force = sqrt (X^2 + Y^2 + Z^2) where X, Y & Z = -1.277 x 10^-9 Newtons
X Y Z
A-b -6.605 X 10^-9 0 0
A-D 2.352 X 10^-9 2.352 X 10^-9 0
A-c 0 -6.605 X 10^-9 0
A-e 0 0 -6.605 X 10^-9
A-F 2.352 X 10^-9 0 2.352 X 10^-9
A-g -1.277 x 10^-9 -1.277 x 10^-9 -1.277 x 10^-9
A-H 0 2.352 X 10^-9 2.352 X 10^-9
Net -3.178 X 10^-9 -3.178 X 10^-9 -3.178 X 10^-9
Force is negative and equal in the X, Y & Z axes for all components
and therefore is stable. This represents the Helium atom.
A -- b -- I
| \ |\ |\
| e | F | j
c -- D --l |
\ \ \|
H -- g -- K
Components A, c, H, e, I, j, K, l all see the same force
Components b, D, F, g all see the same force
Calculate forces on A like components
X Y Z
A-B -6.605 X 10^-9 0 0
A-D 2.352 X 10^-9 2.352 X 10^-9 0
A-C 0 -6.605 X 10^-9 0
A-E 0 0 -6.605 X 10^-9
A-F 2.352 X 10^-9 0 2.352 X 10^-9
A-G -1.277 x 10^-9 -1.277 x 10^-9 -1.277 x 10^-9
A-H 0 2.352 X 10^-9 2.352 X 10^-9
A-I 1.650 X 10^-9 0 0
A-l -1.181 X 10^-9 -.5909 X 10^-9 0
A-K .8988 X 10^-9 .4277 X 10^-9 .4277 X 10^-9
A-j -1.181 X 10^-9 0 -.5909 X 10^-9
Net -2.991 X 10^9 -4.938 X 10^-9 -4.938 X 10^-9
Calculate forces in middle b like components
All X axes forces cancel due to symmettry
Components which are the same as the cube above.
b-I 0 0 0
b-l 0 2.352 X 10^-9 0
b-D 0 -6.605 X 10^-9 0
b-F 0 0 -6.605 X 10^-9
b-j 0 0 2.352 X 10^-9
b-K 0 -1.277 x 10^-9 -1.277 x 10^-9
b-g 0 2.352 X 10^-9 2.352 X 10^-9
Additional components
b-A 0 0 0
b-c 0 2.352 X 10^-9 0
b-e 0 0 2.352 X 10^-9
b-H 0 -1.277 x 10^-9 -1.277 x 10^-9
Net 0 -2.103 X 10^-9 -2.103 X 10^-9
This appears to be negative and equal in the Y & Z axes and therefore
appears to be stable. Adding additional atomic units (2 protons, 2
electrons) would be similar to the above calculation and I think it
would keep on resulting in attractive and equal forces (although I'm
not sure).
What do you think of that? Did I do the math right???
-thanks
Franklin
Franz Heymann
You have not a clue as to the meaning of "stability" in a system of
particles. You have not touches on the question of looking at whether the
system will return to its original configuration when a perturbation is
applied and removed, or whether the application of a perturbation will
simply cause the whole stgructure to fly apart. I can assure you right now
that your system is unstable if you insist on analysing it in terms of
classical mechanics. You are obviously not aware of a completely general
theorem which states that *no* static arrangement whatsoever of charges is
in stable equilibrium
I snip that infantile calculation of yours without further ado.
[snip]
Franz
FrankH
useful to know before I went through the exercise of applying Columbs
Law which is what I assumed you meant by the statement:
> > > Well, you know the charges, so where is the problem with calculating the
> > > force??? Aren't you able to do some fairly basic calculation with
> > > Coulomb's law?
Didn't I ask you what you what would satisfy you for a stability
calculation?
Although, I would have to say that since the force driving the
particles in a cube is about half as strong as the force between two
oppositely charged particles, that this would be able to resist
peturbations since it is not weakly attractive. Do you have any
references on how to do proper peturbation calculations?
Although, I must comment that even if I did such calculations, that
may not be the end of the story since it takes some energy to get
electrons/protons to join into atoms and the same energy to break them
apart again, so the components in an atom are probably bound even
tighter than what one might expect from normal Columb forces.
Naturally, we don't see atoms of helium spontaneously form from just
free floating electrons/protons although my simple calcuations show
that they could form a cubic matrix and not immediately fly apart due
to the spacing and Columnb forces.
Franz Heymann
No, not at the moment. But I will give you a hint: It may be deduced from
the nature of Laplace's equation, namely that every point in an
electrostatic field is a saddle point in the potential distribution.
> That would have been
> useful to know before I went through the exercise of applying Columbs
> Law which is what I assumed you meant by the statement:
It is taught as a matter of course, as an interesting throw-away, in any
self-respecting undergraduate course in electrostatics.
> > > > Well, you know the charges, so where is the problem with calculating
the
> > > > force??? Aren't you able to do some fairly basic calculation with
> > > > Coulomb's law?
>
> Didn't I ask you what you what would satisfy you for a stability
> calculation?
>
> Although, I would have to say that since the force driving the
> particles in a cube is about half as strong as the force between two
> oppositely charged particles, that this would be able to resist
> peturbations since it is not weakly attractive. Do you have any
> references on how to do proper peturbation calculations?
You must learn to distinguish between unstable end stable equilibrium. In
the case of a stable equilibrium, a small perturbation of the position of
any element leads a force in such directions as to restore the status quo.
The converse holds for unstable equilibrium.
>
> Although, I must comment that even if I did such calculations, that
> may not be the end of the story since it takes some energy to get
> electrons/protons to join into atoms and the same energy to break them
> apart again, so the components in an atom are probably bound even
> tighter than what one might expect from normal Columb forces.
Only Coulomb forces are involved in the interactions under consideration.
What else did you have in mind?
> Naturally, we don't see atoms of helium spontaneously form from just
> free floating electrons/protons although my simple calcuations show
> that they could form a cubic matrix and not immediately fly apart due
> to the spacing and Columnb forces.
For the second time,you should abandon your calculations. Your system of
particles is in unstable equilibrium.
[snip]
Franz
Franz Heymann
Just in case a reader picks a fight about my argument, and asks why a NaCl
crystal is stable, please remember that we are here talking about FrankH's
*classical* calculation.
Franz
FrankH
[snip]
> > For the second time,you should abandon your calculations. Your system of
> > particles is in unstable equilibrium.
>
> Just in case a reader picks a fight about my argument, and asks why a NaCl
> crystal is stable, please remember that we are here talking about FrankH's
> *classical* calculation.
>
calculations that Rutherford used for his scattering experiments. And
what about the NaCl stability? That would seem to be a stable system
of ionic charges. If there is no stable arrangement of charges, then
what is hydrogen? How about H2 which appears to be nothing more than a
group of 2 protons and 2 electrons? That appears to be stable. You
might say that this isn't a static arrangement, since the normal
picture of this is the 2 electrons whizzing about in a covalent bond,
however, the cubic model would predict that H2 is a static square
stucture. Rather than sharing electrons, the electrons are in a static
position attracted to the nearby proton. That certainly adds a new
spin on possible chemical bonding mechanisms.
Bjoern Feuerbacher
>
> Bjoern Feuerbacher <bfeu...@ix.urz.uni-heidelberg.de> wrote in message news:<400FC82B...@ix.urz.uni-heidelberg.de>...
> [snip]
> >
> >
> > > > > The strong force is required to keep a compact nucleus which contains
> > > > > all of the protons/neutrons together. My model may eliminate the need
> > > > > since the protons/electrons/neutrons are evenly spread across the
> > > > > entire atom in a neutral matrix.
> > > >
> > > > Have you calculated if this system is stable? I would think that the
> > > > Coulomb forces would destroy it.
> > >
> > > How would I calculate the force so that it would satisfy you.
> >
> > Well, you know the charges, so where is the problem with calculating the
> > force??? Aren't you able to do some fairly basic calculation with
> > Coulomb's law? Granted, this becomes complicated because in your model,
> > the particles themselves apparently aren't spherical symmetric - but as
> > a first approximation, you could simply pretend they are.
> >
> >
> OK, I'm going to give a shot at calculating if a cubic matrix of
> protons and electrons are stable.
>
> To calculate if the cubic model of atom construction is stable, we
> will start with a square and build our way up.
Well, even a square shouldn't be stable, so you will have some
problems...
> Assuming we start with a square with vertices:
>
> A - b
> | |
> c - D
>
> Where A & D are protons and c & b are electrons.
>
> We calculate stability based on the net forces exerted on each
> component.
Yes, this makes sense (I can think of no other criterium of "stability"
here).
> By convention consider attractive forces as being negative
> and repulsive forces as being positive.
That makes sense, too.
> A stable configuration should
> be one where all net forces in the X/Y axes are negative and equal.
Huh??? Wrong. A stable configuration would be one where all net forces
are *zero*! If all forces are negative, this would mean that all
particles are attracted to each other, hence the square would shrink -
it wouldn't remain stable!
> Since this structure is completely symmettric, we can calculate the
> force on one component and it will be the same for the rest of the
> components.
Same magnitude, different direction.
> No matter which component you look at, there will always
> be 2 adjacent components which are attracting and diagonal component
> repelling.
Right.
> We assume that the proton and electron are separated by some minimum
> distance. The measured width of a Silicon atom using STM photographs
> is ~750pm.
You could simply have used a parameter here for the distance - the
result
doesn't depend on the actual magnitude of the distances, only on the
geometry.
> According to the cubic atomic model, this corresponds to a
> width of 4 protons/electrons, so the size of 1 proton/electron
> according to the cubic model is ~187 pm or 187 X 10^-12 meters.
>
> Force between adjacent pairs A-B and A-C using Coulumb's Law:
> 9 x 10^9 * 1.602 X 10^-19 * -1.602 X 10^-19/ (187 X 10^-12)^2 = -6.605
> X 10^-9 Newtons
>
> Force between diagonal pair A-D.
>
> Distance from A-D by Pythagoras theorem = 264 X 10^-12 meters
> 9 x 10^9 * 1.602 X 10^-19 * 1.602 X 10^-19/ (264 X 10^-12)^2 = 3.314 X
> 10^-9 Newtons
>
> Force in the X & Y directions by Pythagoras theorem = 2.352 X 10^-9
> Newtons
>
> X Y
> A-B -6.605 X 10^-9 0
> A-D 2.352 X 10^-9 2.352 X 10^-9
> A-C 0 -6.605 X 10^-9
>
> Net -3.291 X 10^-9 -3.291 X 10^-9
Conclusion: unstable. As I said.
> Force is negative and equal in the X & Y axes for all components and
> therefore is stable.
Nonsensical conclusion. Your square will shrink, it won't remain stable.
[snip cube example - same basic errors]
> What do you think of that? Did I do the math right???
Math right, apparently, physical conclusions complete nonsense. Why on
earth do you think that if the net negative forces are negativ, this
means that the square is stable?!?!?
Bye,
Bjoern
Bjoern Feuerbacher
[snip]
> You are obviously not aware of a completely general
> theorem which states that *no* static arrangement whatsoever of charges is
> in stable equilibrium
Interesting! I don't remember hearing about this theorem during my
education (although it is intuitively obvious); can you tell me where to
read up on this?
[snip]
Bye,
Bjoern
Bjoern Feuerbacher
>
> "Franz Heymann" <notfranz...@btopenworld.com> wrote in message news:<bvfuic$g8h$9...@sparta.btinternet.com>...
[snip]
> Although, I would have to say that since the force driving the
> particles in a cube is about half as strong as the force between two
> oppositely charged particles, that this would be able to resist
> peturbations since it is not weakly attractive.
Sorry, I don't understand what this is supposed to mean.
What do you mean by "weakly attractive"?
And why do you think this helps resisting perturbations?
> Do you have any
> references on how to do proper peturbation calculations?
First learn how to spell the word correctly. ;-)
Well, the usual way to do perturbative calculations is to start with a
static configuration, then change one of the parameters and look if the
system will return to the stable configuration or not. Unfortunately for
you, you don't have a static configuration here to start with...
> Although, I must comment that even if I did such calculations, that
> may not be the end of the story since it takes some energy to get
> electrons/protons to join into atoms
Huh? Pardon??? Energy is *released* when electrons and protons join to
from an atom, not *needed*!
> and the same energy to break them apart again,
Don't you notice that you contradict either yourself or conservation of
energy here? You say that you have to put in energy to "join" electrons
and protons, and you have to put in the same energy again to break them
apart!!! Do you *really* think that this makes sense?
> so the components in an atom are probably bound even
> tighter than what one might expect from normal Columb forces.
How on earth did you arrive at this strange conclusion?????
> Naturally, we don't see atoms of helium spontaneously form from just
> free floating electrons/protons
Well, obviously not, because for a helium atom, you also need two
neutrons!!!
> although my simple calcuations show
> that they could form a cubic matrix and not immediately fly apart due
> to the spacing and Columnb forces.
Wrong. Your calculation shows that your cube is *unstable*, although you
didn't understand this.
[snip]
Bye,
Bjoern
Bjoern Feuerbacher
>
> "Franz Heymann" <notfranz...@btopenworld.com> wrote in message news:<bvjror$q7$1...@hercules.btinternet.com>...
> [snip]
> > > For the second time,you should abandon your calculations. Your system of
> > > particles is in unstable equilibrium.
> >
> > Just in case a reader picks a fight about my argument, and asks why a NaCl
> > crystal is stable, please remember that we are here talking about FrankH's
> > *classical* calculation.
> >
> What's wrong with classical calculations?
That they give wrong results here perhaps???
> That's the same type of
> calculations that Rutherford used for his scattering experiments.
Yes. Rutherford scattering is (fortunately!) one of the very few cases
where a classical calculation gives essentially the same results as a
quantum mechanics calculation.
> And
> what about the NaCl stability? That would seem to be a stable system
> of ionic charges.
Earth to Franklin: Franz Heymann just told you that according to
classical mechanics and electrodynamics, such a configuration is *not*
stable.
> If there is no stable arrangement of charges, then
> what is hydrogen?
Hint: if you treat it using classical mechanics and electrodynamics,
hydrogen is *not* stable. Only quantum mechanics explains its stability.
> How about H2 which appears to be nothing more than a
> group of 2 protons and 2 electrons? That appears to be stable.
Same argument.
> You might say that this isn't a static arrangement, since the normal
> picture of this is the 2 electrons whizzing about in a covalent bond,
Ouch! No, this is *not* the normal picture of it - this is a popular
science picture which, unfortunately, is still taught in some
undergraduate classes.
> however, the cubic model would predict that H2 is a static square
> stucture.
Wrong. Your calculations about the cubic model clearly show it to be
*unstable*.
> Rather than sharing electrons, the electrons are in a static
> position attracted to the nearby proton.
If they are attracted, this can't be a static configuration. Hint:
attraction is a force. Forces imply an acceleration. A system in which a
particle is accelerated isn't static.
> That certainly adds a new
> spin on possible chemical bonding mechanisms.
That certainly adds a new spin on your understanding of basic classical
mechanics - saying that a system in which an attracting force is acting
is "static" is *really* a nice way to show your ignorance.
Bye,
Bjoern
Franz Heymann
> "Franz Heymann" <notfranz...@btopenworld.com> wrote in message
news:<bvjror$q7$1...@hercules.btinternet.com>...
> [snip]
> > > For the second time,you should abandon your calculations. Your system
of
> > > particles is in unstable equilibrium.
> >
> > Just in case a reader picks a fight about my argument, and asks why a
NaCl
> > crystal is stable, please remember that we are here talking about
FrankH's
> > *classical* calculation.
> >
> What's wrong with classical calculations?
They don't usually give the right answers in the quiantum domain
> That's the same type of
> calculations that Rutherford used for his scattering experiments.
Correct. Rutherford was dead lucky. It happens that by pure fluke, the
angular distribution of the scatteing of a charged particle by a point
charge at low energies is correctly given by the classical expression. That
sort of luck does not occur often.
> And
> what about the NaCl stability? That would seem to be a stable system
> of ionic charges.
According to classical electroststics, it should not be a stable structure.
Thatwas my point. Its stability can only be discussed in QM terms.
> If there is no stable arrangement of charges, then
> what is hydrogen?
Your "classical" hydrogen is not an asembly of static charges.
And your classical hydrogen is not stable in any case, because classically
it would radiate em waves at such a rate that the eledctron would fall into
the proton within about 10 nanoseconds or thereabouts.
> How about H2 which appears to be nothing more than a
> group of 2 protons and 2 electrons? That appears to be stable.
It is. So what? What is your point?
> You
> might say that this isn't a static arrangement, since the normal
> picture of this is the 2 electrons whizzing about in a covalent bond,
I am uninterested in any of your examples, since, in your classical terms
they *all* have accelerated electrons which radiate continuously, thereby
losing energy continuously.
> however, the cubic model would predict that H2 is a static square
> stucture.
QM certainly does. Classical mechanics does not.
> Rather than sharing electrons, the electrons are in a static
> position attracted to the nearby proton. That certainly adds a new
> spin on possible chemical bonding mechanisms.
Crap.
You bore me stiff and I have spent enough time proving to you that you mnow
next to nothing.
Goodbye.
Franz
Franz Heymann
It is actually trivial: Each charge is in the electric field produced at
that point by all the other charges. When in equilibrium, that particle
must be at a point where locally the field is zero. So far, so good. Our
friend could possibly get this far with his calculation. But now comes the
rub: The Laplace equation says effectively that each point in an
electrostatic field is a saddle point in the potential distribution. And for
a stable equilibrium, the particle has to sit at a potential *minimum*, not
at a saddle point, since in the latter case there is at least one direction
in which a perturbation of position would result i the particle escaping.
I.e. there is no configuration of charges which can be in static equilibrium
under the influence of electrostatic charges alone.
Franz
FrankH
>
> > Force is negative and equal in the X & Y axes for all components and
> > therefore is stable.
>
> Nonsensical conclusion. Your square will shrink, it won't remain stable.
>
certainly shrink. But it can't keep on shrinking forever and get
infintely small, which is why the cubic model must postulate that
there is a certain minimum distance that a proton and electron can
have. This appears to be the part that is causing the confusion. At
some point, the internal components of the protons/electrons cannot
get any closer. I guessed that this distance is about 187pm. At this
point, the proton/electron effectively have a "hard" spherical shell
with a radius of 93pm. It is this hard shell which provides the
opposite force which causes the net force to be zero. This is just
like the Earth providing an opposing "up" force to the "down" force of
gravity. Without this hard shell, Franz is correct that a charged
particle could not possibly have a minimal energy, but with a hard
shell, this becomes possible. So is this structure stable assuming the
components do have this hard shell?
There also seems to be some confusion over the model I am using since
you mentioned that an electron must be moving and emitting radiation
in my model. As you noted any classical Bohr model would have the
electron falling into the proton in no time at all. My cubic model is
actually all about what happens after the electron finishes falling
directly onto the proton. It goes "clunk" and electron has fallen onto
the proton and is basically motionless. This is my model that I am
studying. As long as there is some hard limit on the distance between
the proton/electron, this should be a stable system.
I was wondering how QM explains this situation where a low speed
electron meets up with a low speed proton and somehow the electron
transforms from a particle into this object described by wavefunctions
whizzing randomly around the proton. Does QM explain why they just
don't go "clunk" as one would intuitively think they would?
I also never understood that if QM still recognizes the electron as a
particle moving about the proton, how does this solve the radiation
problem? I would think the electrons would still undergo accelerations
and such accelerations should still give off energy no matter how
randomly it is orbiting. I have read that the atom emits only if the
electron goes between various levels, but that doensn't excuse why an
electron moving at any level shouldn't still be emitting energy. Only
non-moving electrons should not emmit anything.
ThomasL283
>Date: 2/4/2004 1:14 PM Pacific Standard Time
Wrote in:
>Message-id: <46484c9f.0402...@posting.google.com>
>
>Bjoern Feuerbacher <bfeu...@ix.urz.uni-heidelberg.de> wrote in message
>news:<401F9331...@ix.urz.uni-heidelberg.de>...
You said:
>> > Force is negative and equal in the X & Y axes for all components and
>> > therefore is stable.
Bjoern said:
>> Nonsensical conclusion. Your square will shrink, it won't remain stable.
You said:
>Yes, if only electrostatic forces are involved here, it would
>certainly shrink. But it can't keep on shrinking forever and get
>infintely small, which is why the cubic model must postulate that
>snip<
Frank, binding is not a simple one electrostatic force, it is a confluence
between the electric and magnetic moment forces.
When you have had an opportunity to play the DVD, I just sent you, review part
5.
There has to be a photon released to create a mass defect, as a result of the
electron binding to the proton.
The electric and magnetic confluence provides the EM to constitute the photon ,
at the null of forces.
This null keeps the electron bound to the proton. The bond is elastic until
broken, and this causes the harmonic oscillations seen in the spectroscopy.
Bjoern also has a copy of the DVD. I hope he has taken the opportunity to
seriously consider the working models.
Regards: Tom:
Tom Lockyer (77 and retired)
"If you want to do the impossible, don't hire an expert because he knows
it can't be done." Henry Ford.
Franz Heymann
news:<401F9331...@ix.urz.uni-heidelberg.de>...
> >
> > > Force is negative and equal in the X & Y axes for all components and
> > > therefore is stable.
> >
> > Nonsensical conclusion. Your square will shrink, it won't remain stable.
> >
> Yes, if only electrostatic forces are involved here, it would
> certainly shrink. But it can't keep on shrinking forever and get
> infintely small, which is why the cubic model must postulate that
> there is a certain minimum distance that a proton and electron can
> have. This appears to be the part that is causing the confusion. At
> some point, the internal components of the protons/electrons cannot
> get any closer. I guessed that this distance is about 187pm.
Would it help you to know that the proton "hard core" radius is about 1 fm,
or 187,000 times smaller than you thought, and the electron size is at
least another four orders of magnitude smaller?
> At this
> point, the proton/electron effectively have a "hard" spherical shell
> with a radius of 93pm. It is this hard shell which provides the
> opposite force which causes the net force to be zero. This is just
> like the Earth providing an opposing "up" force to the "down" force of
> gravity. Without this hard shell, Franz is correct that a charged
> particle could not possibly have a minimal energy, but with a hard
> shell, this becomes possible. So is this structure stable assuming the
> components do have this hard shell?
>
> There also seems to be some confusion over the model I am using since
> you mentioned that an electron must be moving and emitting radiation
> in my model. As you noted any classical Bohr model would have the
> electron falling into the proton in no time at all. My cubic model is
> actually all about what happens after the electron finishes falling
> directly onto the proton. It goes "clunk" and electron has fallen onto
> the proton and is basically motionless.
Thus giving you a neutron and a neutrino. This is a well studied process
which is known to happen occasionally in the case of atoms somewhat heavier
than hydrogen. It is known as k-capture.
Bang goes your model.
> This is my model that I am
> studying. As long as there is some hard limit on the distance between
> the proton/electron, this should be a stable system.
I have indicated to you that whatever you do, no static system of electric
charges can be in stable equilibrium under the action of electroststic
forces alone.
I sincerely recommend that you should drop this nonsense and learn some
physics instead.
>
> I was wondering how QM explains this situation where a low speed
> electron meets up with a low speed proton and somehow the electron
> transforms from a particle into this object described by wavefunctions
> whizzing randomly around the proton. Does QM explain why they just
> don't go "clunk" as one would intuitively think they would?
Yes. In exquisite detail. All the calculations have been done in complete
detail for all the atoms from hydrogen up to Iron. The results are in daily
use by astronomers who study stellar structure via their spectra.
>
> I also never understood that if QM still recognizes the electron as a
> particle moving about the proton, how does this solve the radiation
> problem?
The long and short of it is that according to QM, the electron does not do
that. The fact that you ask this question is an absolutely clear example of
the ststement that you should stop posting here and learn some elementary
quantum mechanics.
> I would think the electrons would still undergo accelerations
> and such accelerations should still give off energy no matter how
> randomly it is orbiting.
No, and no in the case of electrons in the ground state.
> I have read that the atom emits only if the
> electron goes between various levels,
That is correct.
> but that doensn't excuse why an
> electron moving at any level shouldn't still be emitting energy.
That is precisely what happens if the electron is in any level other than
the ground state.
> Only
> non-moving electrons should not emmit anything.
No. And that statement is not even true classically. In classical
mechanics, it is the acceleration which matters.
Franz
FrankH
> "FrankH" <frank...@yahoo.com> wrote in message
> news:46484c9f.0402...@posting.google.com...
> > Bjoern Feuerbacher <bfeu...@ix.urz.uni-heidelberg.de> wrote in message
> news:<401F9331...@ix.urz.uni-heidelberg.de>...
> > >
> > > > Force is negative and equal in the X & Y axes for all components and
> > > > therefore is stable.
> > >
> > > Nonsensical conclusion. Your square will shrink, it won't remain stable.
> > >
> > Yes, if only electrostatic forces are involved here, it would
> > certainly shrink. But it can't keep on shrinking forever and get
> > infintely small, which is why the cubic model must postulate that
> > there is a certain minimum distance that a proton and electron can
> > have. This appears to be the part that is causing the confusion. At
> > some point, the internal components of the protons/electrons cannot
> > get any closer. I guessed that this distance is about 187pm.
>
> Would it help you to know that the proton "hard core" radius is about 1 fm,
> or 187,000 times smaller than you thought, and the electron size is at
> least another four orders of magnitude smaller?
>
observations have been indirect and require a number of assumptions to
come up with a hard number. Some of these assumptions may not be true.
> > At this
> > point, the proton/electron effectively have a "hard" spherical shell
> > with a radius of 93pm. It is this hard shell which provides the
> > opposite force which causes the net force to be zero. This is just
> > like the Earth providing an opposing "up" force to the "down" force of
> > gravity. Without this hard shell, Franz is correct that a charged
> > particle could not possibly have a minimal energy, but with a hard
> > shell, this becomes possible. So is this structure stable assuming the
> > components do have this hard shell?
> >
> > There also seems to be some confusion over the model I am using since
> > you mentioned that an electron must be moving and emitting radiation
> > in my model. As you noted any classical Bohr model would have the
> > electron falling into the proton in no time at all. My cubic model is
> > actually all about what happens after the electron finishes falling
> > directly onto the proton. It goes "clunk" and electron has fallen onto
> > the proton and is basically motionless.
>
> Thus giving you a neutron and a neutrino. This is a well studied process
> which is known to happen occasionally in the case of atoms somewhat heavier
> than hydrogen. It is known as k-capture.
> Bang goes your model.
As you say, this only happens occasionally and wouldn't pose a problem
to my model as long as it doesn't occur most of the time.
>
> > This is my model that I am
> > studying. As long as there is some hard limit on the distance between
> > the proton/electron, this should be a stable system.
>
> I have indicated to you that whatever you do, no static system of electric
> charges can be in stable equilibrium under the action of electroststic
> forces alone.
> I sincerely recommend that you should drop this nonsense and learn some
> physics instead.
You can keep on saying that, but it is hardly a convincing argument. I
keep on saying that there is MORE involved here than just
electrostatic forces alone. There is a MINIMUM distance and I would
think based on the calculations I made that it would be obvious that
this system is extremely stable. There was also another similar
theorem stating that it was impossible to create a stable magnetic
suspension system - that was shown false by the invention of the
Levitron and stable magnetic suspension of magenets over
superconducting materials. I wouldn't put so much faith in theorems.
>
> >
> > I was wondering how QM explains this situation where a low speed
> > electron meets up with a low speed proton and somehow the electron
> > transforms from a particle into this object described by wavefunctions
> > whizzing randomly around the proton. Does QM explain why they just
> > don't go "clunk" as one would intuitively think they would?
>
> Yes. In exquisite detail. All the calculations have been done in complete
> detail for all the atoms from hydrogen up to Iron. The results are in daily
> use by astronomers who study stellar structure via their spectra.
What are you talking about? You can draw up equations to generate the
spectra of an excited atom, but that doesn't say anything about why an
electron just doesn't get captured by the proton. It somehow avoids
this capture and that is the question I am asking.
> >
> > I also never understood that if QM still recognizes the electron as a
> > particle moving about the proton, how does this solve the radiation
> > problem?
>
> The long and short of it is that according to QM, the electron does not do
> that. The fact that you ask this question is an absolutely clear example of
> the ststement that you should stop posting here and learn some elementary
> quantum mechanics.
>
What kind of answer is that? "The long and short of it" is hardly an
explanation. By now, I have been reading a couple of quantum mechanics
books and none of them has come close to addressing this question. It
is glossed over or just completely ignored - however - this is a
central point if quantum mechanical models of the atom are not to
share the same fate as the Bohr orbital models. Do quantum mechanics
have an answer?
> > I would think the electrons would still undergo accelerations
> > and such accelerations should still give off energy no matter how
> > randomly it is orbiting.
>
> No, and no in the case of electrons in the ground state.
OK, explain how an electron in a quantum mechanical shell can move
around in its shell without changing direction and thereby avoid any
acceleration. Doesn't sound possible to me.
>
> > I have read that the atom emits only if the
> > electron goes between various levels,
>
> That is correct.
>
> > but that doensn't excuse why an
> > electron moving at any level shouldn't still be emitting energy.
>
> That is precisely what happens if the electron is in any level other than
> the ground state.
Explain why even in ground state where the electrons are still moving
and changing directions, how they can avoid emitting energy.
>
> > Only
> > non-moving electrons should not emmit anything.
>
> No. And that statement is not even true classically. In classical
> mechanics, it is the acceleration which matters.
>
Yes, that is true and as far as I know, a change of direction requires
a change in acceleration.
> Franz
FrankH
Here is another interesting tidbit about this calculation. It appears
to correctly predict the relative first ionization energies for
Hydrogen and Helium. If we calculate the net force on an element of a
hydrogen atom (just a proton/electron pair at 187pm, we get 3.314 X
10^-9. If we do a similar calculation using the x,y,z forces for an
element of the cube helium, we come up with 5.504 X 10^-9. This
compares with the first ionization energy of hydrogen at 1312 kJ mol
and 2372 kJ mol for helium (from www.webelements.com). The ratio for
the predicted difference in force is (5.504/3.314) = 1.66. The actual
ratio is (2372/1312) =1.80 which agrees to within 9%.
This doesn't extend to lithium which would have a force of either
7.696 x 10-9 for an outside component or 2.974 x 10-9 for an inside
component. The first ionization energy for lithium is only 520 kJ mol.
The ratios are (3.313/2.974) = .83 and (1312/520)=2.52 which do not
agree at all. Since this is a completely asymmetric atom, there may be
other factors which make an accurate calculation difficult. But I
still think the calculations for the helium and hydrogen are
significant since I am unaware of any other theory that would explain
why the ionization energy is so much higher for helium than for
hydrogen.
Another interesting note about ionization energies is that if you look
at the ionization energies for an atom like argon:
http://www.webelements.com/webelements/elements/text/Ar/ionz.html
You will notice that there are natural breaks in the energy levels.
The first six have about the same energy difference. This corresponds
to the six faces of a cube which make up the outer layer of argon.
Then there is a grouping of 2 for the 7th and 8th ionization. This
potentially represents the ionization of the next lower layer of the
core of the cubic atomic model. The next 8 ionizations represent the 8
electrons directly surrounding the core. The last 2 represent the very
center core electrons. See the picture of argon:
http://ourworld.compuserve.com/homepages/frankhu/Argon-18.jpg
This is a cutaway, so you can view the atoms core:
http://ourworld.compuserve.com/homepages/frankhu/AUT_1188.jpg
You can see that the cubic atomic model gives a reasonable explanation
for the differences in ionization energies and when they occur.
Franz Heymann
> "Franz Heymann" <notfranz...@btopenworld.com> wrote in message
news:<bvuce0$ad6$1...@titan.btinternet.com>...
> > "FrankH" <frank...@yahoo.com> wrote in message
> > news:46484c9f.0402...@posting.google.com...
> > > Bjoern Feuerbacher <bfeu...@ix.urz.uni-heidelberg.de> wrote in
message
> > news:<401F9331...@ix.urz.uni-heidelberg.de>...
> > > >
> > > > > Force is negative and equal in the X & Y axes for all components
and
> > > > > therefore is stable.
> > > >
> > > > Nonsensical conclusion. Your square will shrink, it won't remain
stable.
> > > >
> > > Yes, if only electrostatic forces are involved here, it would
> > > certainly shrink. But it can't keep on shrinking forever and get
> > > infintely small, which is why the cubic model must postulate that
> > > there is a certain minimum distance that a proton and electron can
> > > have. This appears to be the part that is causing the confusion. At
> > > some point, the internal components of the protons/electrons cannot
> > > get any closer. I guessed that this distance is about 187pm.
> >
> > Would it help you to know that the proton "hard core" radius is about 1
fm,
> > or 187,000 times smaller than you thought, and the electron size is at
> > least another four orders of magnitude smaller?
> >
> Have you ever seen a picture of a proton with a ruler next to it?
No. Intelligent people are, however, endowed with deductive powers.
> All
> observations have been indirect and require a number of assumptions to
> come up with a hard number. Some of these assumptions may not be true.
The analysis involved in arriving at a proton charge distribution is based
on the same concepts as the analysis which allows one to determine the
charge distributions in a crystal from X-ray diffraction data.
The conclusion that the charge radius of the proton is 1.27 fm (if my memory
serves me correctly) is very firmly based. Actually, even the skin depth,
over which the charge density dies out to zero, has been measured.
>
> > > At this
> > > point, the proton/electron effectively have a "hard" spherical shell
> > > with a radius of 93pm. It is this hard shell which provides the
> > > opposite force which causes the net force to be zero. This is just
> > > like the Earth providing an opposing "up" force to the "down" force of
> > > gravity. Without this hard shell, Franz is correct that a charged
> > > particle could not possibly have a minimal energy, but with a hard
> > > shell, this becomes possible. So is this structure stable assuming the
> > > components do have this hard shell?
> > >
> > > There also seems to be some confusion over the model I am using since
> > > you mentioned that an electron must be moving and emitting radiation
> > > in my model. As you noted any classical Bohr model would have the
> > > electron falling into the proton in no time at all. My cubic model is
> > > actually all about what happens after the electron finishes falling
> > > directly onto the proton. It goes "clunk" and electron has fallen onto
> > > the proton and is basically motionless.
> >
> > Thus giving you a neutron and a neutrino. This is a well studied
process
> > which is known to happen occasionally in the case of atoms somewhat
heavier
> > than hydrogen. It is known as k-capture.
> > Bang goes your model.
>
> As you say, this only happens occasionally and wouldn't pose a problem
> to my model as long as it doesn't occur most of the time.
In your model it would happen *every time* in the case of the Hydrogen atom,
and the atom would have a mean free life of around a few nanoseconds.
> > > This is my model that I am
> > > studying. As long as there is some hard limit on the distance between
> > > the proton/electron, this should be a stable system.
> >
> > I have indicated to you that whatever you do, no static system of
electric
> > charges can be in stable equilibrium under the action of electroststic
> > forces alone.
> > I sincerely recommend that you should drop this nonsense and learn some
> > physics instead.
>
> You can keep on saying that, but it is hardly a convincing argument.
The argument is totally correct. The fact that you do not comprehend it is
your private problem
I
> keep on saying that there is MORE involved here than just
> electrostatic forces alone.
Weak interactioin?
Strong interaction?
Gravitational interaction?
Your system is a static arrangement of charges and will therefore not be a
stable structure. It really is as simplre as that.
There is a MINIMUM distance and I would
> think based on the calculations I made that it would be obvious that
> this system is extremely stable.
No. On the contrary, it is obvious that it would be unstable.
Do you understand what a saddle point in a potential distribution means?
> There was also another similar
> theorem stating that it was impossible to create a stable magnetic
> suspension system
That theorem is true and it *really* is not possible to hold a system of
permanent magnets in stable equilibrium under the action of magnetostatic
forces alone.
> - that was shown false by the invention of the
> Levitron and stable magnetic suspension of magenets over
> superconducting materials.
For your information, the current argument is about *electrostatic* fields.
However, the theorem for the magnetostatic case was distinctly *not* shown
to be false by the experiments to which you refer. I invite you to read the
theorem properly and then to look in detail at the experiments in order to
convince yourself that such levitation schemes did not violate the theorem.
> I wouldn't put so much faith in theorems.
A theorem is a theorem, your lack of faith notwithstanding.
In the case of the electroststic field, the theorem is based on one premise
only, namely that the electroststic field obeys an inverse square law. All
follows from that alone, by correctly executed steps of deductive argument.
> > > I was wondering how QM explains this situation where a low speed
> > > electron meets up with a low speed proton and somehow the electron
> > > transforms from a particle into this object described by wavefunctions
> > > whizzing randomly around the proton. Does QM explain why they just
> > > don't go "clunk" as one would intuitively think they would?
> >
> > Yes. In exquisite detail. All the calculations have been done in
complete
> > detail for all the atoms from hydrogen up to Iron. The results are in
daily
> > use by astronomers who study stellar structure via their spectra.
>
> What are you talking about? You can draw up equations to generate the
> spectra of an excited atom, but that doesn't say anything about why an
> electron just doesn't get captured by the proton. It somehow avoids
> this capture and that is the question I am asking.
You are misunderstanding the situation utterly. The same theory which
determines the spectra also determines that the ground states of atoms are
stable.
>
> > >
> > > I also never understood that if QM still recognizes the electron as a
> > > particle moving about the proton, how does this solve the radiation
> > > problem?
> >
> > The long and short of it is that according to QM, the electron does not
do
> > that. The fact that you ask this question is an absolutely clear
example of
> > the ststement that you should stop posting here and learn some
elementary
> > quantum mechanics.
> >
> What kind of answer is that? "The long and short of it" is hardly an
> explanation.
It was not an explanation. It was a statement of fact.
If you want it in detail, read Schiff "Quantum Mechanics". That will occupy
you at least six months.
> By now, I have been reading a couple of quantum mechanics
> books and none of them has come close to addressing this question.
What question? You have raised more than one.
> It
> is glossed over or just completely ignored - however - this is a
> central point if quantum mechanical models of the atom are not to
> share the same fate as the Bohr orbital models. Do quantum mechanics
> have an answer?
Yes, indeed it does "have the answer".
In fact it has only correct answers. It has successfully answered every
single question which has ever been asked of it completely correctly.
> > > I would think the electrons would still undergo accelerations
> > > and such accelerations should still give off energy no matter how
> > > randomly it is orbiting.
> >
> > No, and no in the case of electrons in the ground state.
> OK, explain how an electron in a quantum mechanical shell can move
> around in its shell without changing direction and thereby avoid any
> acceleration. Doesn't sound possible to me.
>
> >
> > > I have read that the atom emits only if the
> > > electron goes between various levels,
> >
> > That is correct.
> >
> > > but that doensn't excuse why an
> > > electron moving at any level shouldn't still be emitting energy.
> >
> > That is precisely what happens if the electron is in any level other
than
> > the ground state.
>
> Explain why even in ground state where the electrons are still moving
> and changing directions, how they can avoid emitting energy.
Who said they are moving and changing direction in the ground state?
>
>
> >
> > > Only
> > > non-moving electrons should not emmit anything.
> >
> > No. And that statement is not even true classically. In classical
> > mechanics, it is the acceleration which matters.
> >
> Yes, that is true and as far as I know, a change of direction requires
> a change in acceleration.
That is incorrect. A constant acceleration migh also produce a continuing
change in direction.
However, it is irrelevant in the case of an atom. You really have to learn
that you cannot think of the electron in an atom as executing a motion on
some trajectory.
Franz
Big Bird
It is a property of holomorphic functions in general to have no local
maxima. See e.g. Arfken "Mathematical Methods for Physicists",
exercises 6.2.1 and especially 6.2.3. Any decent math book should have
something about this under "analytic functions" or "holomorphic
functions" or there abouts. Maybe under "Laplace's equation". A quick
google for these terms only gives me this:
www.physics.umd.edu/courses/Phys604/
Fall03/Notes/AnalyticFunctions.pdf
Which only says (on page 21) that it is "easy to show" this.
All you have to note then is that in a volume that contains no
charges, the potential obeys Laplace's equation. Which means that it
cannot have a local minimum. (the same is true for masses and even
static configurations of magnets, emaning that no static
configurations of these can be in a stable equilibrium as this would
require for each part of the system to be in a local potential
minimum).
FrankH
[Snip]
> > > I have indicated to you that whatever you do, no static system of
> electric
> > > charges can be in stable equilibrium under the action of electroststic
> > > forces alone.
> > > I sincerely recommend that you should drop this nonsense and learn some
> > > physics instead.
> >
> > You can keep on saying that, but it is hardly a convincing argument.
>
> The argument is totally correct. The fact that you do not comprehend it is
> your private problem
>
> I
> > keep on saying that there is MORE involved here than just
> > electrostatic forces alone.
>
> Weak interactioin?
> Strong interaction?
> Gravitational interaction?
>
> Your system is a static arrangement of charges and will therefore not be a
> stable structure. It really is as simplre as that.
>
> There is a MINIMUM distance and I would
> > think based on the calculations I made that it would be obvious that
> > this system is extremely stable.
>
> No. On the contrary, it is obvious that it would be unstable.
> Do you understand what a saddle point in a potential distribution means?
Let me try to make this as clear as possible - I keep on saying that
my model postulates that electrons and protons have a "hard" surface.
Whatever makes up electrons/protons (call them quarks or whatever)
take up a finite amount of space and at a certain point, they cannot
get any closer despite the electrostatic force they are generating. If
you still have trouble with this concept that take 8 ping pong balls
on the space shuttle. Charge 4 of them up with a positive
electrostatic charge, and the other 4 with a negative charge. Now if
you let them float free and interact with each other, I would predict
that they would form a stable cube. This "hardness" is a postulate,
and may not neccessarly be true, but it is well within reason to think
it is this way since this is how macroscopic objects work - so it
follows intuition and doesn't defy logic. Do you get it? HARD
SURFACE!!!! The other force involved is not any kind of field force
(e.g. weak, strong, etc.) It is the physical reality that solid
objects cannot merge together no matter the degree of force applied to
it.
[snip]
Is this a statement of fact, or is this just postulated? At least Bohr
postulated that the laws of electrostatics simply didn't apply at the
level of atoms. I also saw a website which was explaining the quantum
mechanical system of electron shells and at least it was honest enough
to say that "we don't know what's going on with the electron and we
don't care, so we ignore the problem". The explanations I have seen
seem to indicate that the uncertainty principle makes is impossible to
know where the electron is located and where it might be in the next
moment, thereby making it impossible to calculate whether it is
undergoing any accelerations. However, not being able to do the
calulations doesn't mean that the electron doesn't undergo
accelerations.
>
> > By now, I have been reading a couple of quantum mechanics
> > books and none of them has come close to addressing this question.
>
> What question? You have raised more than one.
How can an electron remain in its orbital cloud without radiation
losses?
>
> > It
> > is glossed over or just completely ignored - however - this is a
> > central point if quantum mechanical models of the atom are not to
> > share the same fate as the Bohr orbital models. Do quantum mechanics
> > have an answer?
>
> Yes, indeed it does "have the answer".
> In fact it has only correct answers. It has successfully answered every
> single question which has ever been asked of it completely correctly.
Out of curiosity, does it explain why the most common products of
Uranium fissions are a bit bigger than about 1/4 or 1/2 the mass of
the original atom? Does it provide a precise calculation for the
ionization energy for hydrogen and helium?
>
> > > > I would think the electrons would still undergo accelerations
> > > > and such accelerations should still give off energy no matter how
> > > > randomly it is orbiting.
> > >
> > > No, and no in the case of electrons in the ground state.
> > OK, explain how an electron in a quantum mechanical shell can move
> > around in its shell without changing direction and thereby avoid any
> > acceleration. Doesn't sound possible to me.
> >
> > >
> > > > I have read that the atom emits only if the
> > > > electron goes between various levels,
> > >
> > > That is correct.
> > >
> > > > but that doensn't excuse why an
> > > > electron moving at any level shouldn't still be emitting energy.
> > >
> > > That is precisely what happens if the electron is in any level other
> than
> > > the ground state.
> >
> > Explain why even in ground state where the electrons are still moving
> > and changing directions, how they can avoid emitting energy.
>
> Who said they are moving and changing direction in the ground state?
> >
[snip]
Based on your responses, you are saying that an electron isn't moving
in its probability cloud. This isn't the picture I got from reading
about it, but it certainly is possible that the electron is remaining
mostly static, but may wander from point to point like it was stuck in
a bottle. Is this the correct quantum mechanical picture?
But if this is correct, then how does the electron resist the pull of
the nucleus if it doesn't have any momentum to keep it away from the
nucleus? By any calculation, the static electron should fall straight
into the nucleus. This is a much worse situation that the Bohr atom
which at least used momentum and centripital force to keep the
electron out of the nucleus.
If this is incorrect, then the electron does have some momentum and
since it is confined to being located in a probability cloud, it must
have to change direction and lose energy. The only way out would be if
the electron had a Star Trek transporter and could magically move
itself from place to place without having to change directions. This
seems highly unlikely.
Of course, I'm going to guess that you're going to say that it is just
a point of fact that the electron remains in its probability cloud
without losing energy. This point of fact sounds just like what was
proposed for epicycles whereby, there had to be an imaginary point
about which the planets looped around. It was a point of fact that
these points had to exist to explain the rest of the phenomenon. It
seems to me that the quantum mechanical house of cards also has its
foundations in a similar point of fact.
Since it seems that an electron cannot logically maintain its position
about the nucleus, the only logical solution is that the electrons are
actually part of the nucleus. If we assume that the electrons are part
of the nucleus in a Thompson like (plum pudding) atom, then the
problems of keeping the electrons "in the air" go away. The main
problem with the Thompson model is that it didn't specify a distinct
structure for the atom nucleus. A random arrangement cannot explain
Rutherford scattering. However, a model which shows specific
arrangements could explain the backscatter as the result of alpha
particles bouncing off specific structures within the atom. The
effective cross section for such specific structures may turn out to
be as small as what we calculate for a point nucleus.
OC
As far as I know, there is nothing in the experimental resultats that
can suggest such a "hard surface" for protons and electrons. Your
assumption is therefore not justified. It does not matter if it seems
reasonable: if the experiments show that an assumption is incorrect,
you have to abandon it. Nature is not necessarily intuitive.
And extrapolating to microscopic objects what we know about
macroscopic objects has been often misleading.
Actually, the main assumption was that the angular momentum of the
electron is quantized (which is not in contradiction with Coulomb's
law).
> I also saw a website which was explaining the quantum
> mechanical system of electron shells and at least it was honest enough
> to say that "we don't know what's going on with the electron and we
> don't care, so we ignore the problem". The explanations I have seen
> seem to indicate that the uncertainty principle makes is impossible to
> know where the electron is located and where it might be in the next
> moment, thereby making it impossible to calculate whether it is
> undergoing any accelerations. However, not being able to do the
> calulations doesn't mean that the electron doesn't undergo
> accelerations.
>
> >
> > > By now, I have been reading a couple of quantum mechanics
> > > books and none of them has come close to addressing this question.
> >
> > What question? You have raised more than one.
> How can an electron remain in its orbital cloud without radiation
> losses?
Microscopic objects do not behave necessarily as classical objects.
> >
> > > It
> > > is glossed over or just completely ignored - however - this is a
> > > central point if quantum mechanical models of the atom are not to
> > > share the same fate as the Bohr orbital models. Do quantum mechanics
> > > have an answer?
> >
> > Yes, indeed it does "have the answer".
> > In fact it has only correct answers. It has successfully answered every
> > single question which has ever been asked of it completely correctly.
>
> Out of curiosity, does it explain why the most common products of
> Uranium fissions are a bit bigger than about 1/4 or 1/2 the mass of
> the original atom? Does it provide a precise calculation for the
> ionization energy for hydrogen and helium?
As far as I know, the calculation have been made for most atoms, not
only fo hydrogen and helium. And they seem to agree very well with the
experimental result.
OC
Bjoern Feuerbacher
FrankH wrote:
>
> Bjoern Feuerbacher <bfeu...@ix.urz.uni-heidelberg.de> wrote in message news:<401F9331...@ix.urz.uni-heidelberg.de>...
> >
> > > Force is negative and equal in the X & Y axes for all components and
> > > therefore is stable.
> >
> > Nonsensical conclusion. Your square will shrink, it won't remain stable.
> >
> Yes, if only electrostatic forces are involved here, it would
> certainly shrink.
Nice that we agree here.
> But it can't keep on shrinking forever and get
> infintely small,
Why not?
> which is why the cubic model must postulate that
> there is a certain minimum distance that a proton and electron can
> have.
Present evidence that such a minimum distance exists, please.
I would say that already beta decay contradicts this!
> This appears to be the part that is causing the confusion. At
> some point, the internal components of the protons/electrons cannot
> get any closer.
Please present evidence for this assertion.
And please present evidence that electrons even *have* internal
components.
> I guessed that this distance is about 187pm.
Based on what?
> At this
> point, the proton/electron effectively have a "hard" spherical shell
> with a radius of 93pm.
Where does this "hard" shell come from? Which repulsive force provides
this hardnees?
> It is this hard shell which provides the
> opposite force
This makes no sense - you have it exactly backwards! A surface on its
own cannot provide a force, it's exactly the other way round: because
there is a repulsive force, objects have hard surfaces!
Oh, BTW, please present evidence for this hard surface and this force.
And explain why no evidence for this repulsive force has ever been seen
in electron-proton scattering experiments.
> which causes the net force to be zero. This is just
> like the Earth providing an opposing "up" force to the "down" force of
> gravity.
This "up" force results from the electrostatic forces of the atoms in
it, and partly from the Pauli principle. Where does your repulsive force
come from?
> Without this hard shell, Franz is correct that a charged
> particle could not possibly have a minimal energy, but with a hard
> shell, this becomes possible. So is this structure stable assuming the
> components do have this hard shell?
Why should we assume this? There is no evidence for, but lots of
evidence against this.
> There also seems to be some confusion over the model I am using since
> you mentioned that an electron must be moving and emitting radiation
> in my model.
I don't remember mentioning this.
> As you noted any classical Bohr model would have the
> electron falling into the proton in no time at all.
Right. (assuming that you meant the "no time at all" metaphorically)
> My cubic model is
> actually all about what happens after the electron finishes falling
> directly onto the proton. It goes "clunk" and electron has fallen onto
> the proton and is basically motionless.
This would only work if there were repulsive forces - and again, there
is no evidence for, but lots of evidence against this.
> This is my model that I am studying.
Please explain the results of all the scattering experiments which were
done on atoms and protons. And by "explain", I don't mean handwavy
arguments, I mean actual calculations which reproduce the observed
scattering cross sections.
> As long as there is some hard limit on the distance between
> the proton/electron, this should be a stable system.
*IF* there is such a hard limit...
> I was wondering how QM explains this situation where a low speed
> electron meets up with a low speed proton and somehow the electron
> transforms from a particle into this object described by wavefunctions
> whizzing randomly around the proton.
1) Electrons are *always* described by wave functions, not just when
they are bound to a proton
2) Saying that wave functions "whizz randomly around the proton" makes
no sense.
> Does QM explain why they just
> don't go "clunk" as one would intuitively think they would?
Heisenberg's uncertainty principle. If it went "clunk" to the proton,
its momentum and its position would be fixed simulataneously. This isn't
possible.
Or in other words: this is caused by the wave-like aspects of electrons.
You get a "standing wave".
> I also never understood that if QM still recognizes the electron as a
> particle moving about the proton,
It doesn't. Why on earth do you think so?
> how does this solve the radiation problem?
By realizing that the electron does *not* move around the proton.
> I would think the electrons would still undergo accelerations
> and such accelerations should still give off energy no matter how
> randomly it is orbiting.
Hint: according to QM, the electron has no trajectory.
> I have read that the atom emits only if the
> electron goes between various levels, but that doensn't excuse why an
> electron moving at any level shouldn't still be emitting energy.
It doesn't move. Plain and simple.
> Only non-moving electrons should not emmit anything.
Right.
Bye,
Bjoern
Robert J. Kolker
Bjoern Feuerbacher wrote:
>
>
> Hint: according to QM, the electron has no trajectory.
How does an electron get from the heated element of a T.V. tube to the
phosphorescent screen? First it is there and now it is here. Just
because no measurement will tell you excactly where it is in between,
does not mean it isn't somewhere.
Everyone who has ever died by electrocution is living testimony to the
fact that electrons move.
Bob Kolker
Bjoern Feuerbacher
>
> "Franz Heymann" <notfranz...@btopenworld.com> wrote in message news:<bvuce0$ad6$1...@titan.btinternet.com>...
> > "FrankH" <frank...@yahoo.com> wrote in message
> > news:46484c9f.0402...@posting.google.com...
> > > Bjoern Feuerbacher <bfeu...@ix.urz.uni-heidelberg.de> wrote in message
> > news:<401F9331...@ix.urz.uni-heidelberg.de>...
> > > >
> > > > > Force is negative and equal in the X & Y axes for all components and
> > > > > therefore is stable.
> > > >
> > > > Nonsensical conclusion. Your square will shrink, it won't remain stable.
> > > >
> > > Yes, if only electrostatic forces are involved here, it would
> > > certainly shrink. But it can't keep on shrinking forever and get
> > > infintely small, which is why the cubic model must postulate that
> > > there is a certain minimum distance that a proton and electron can
> > > have. This appears to be the part that is causing the confusion. At
> > > some point, the internal components of the protons/electrons cannot
> > > get any closer. I guessed that this distance is about 187pm.
> >
> > Would it help you to know that the proton "hard core" radius is about 1
> > fm,
> > or 187,000 times smaller than you thought, and the electron size is at
> > least another four orders of magnitude smaller?
> >
> Have you ever seen a picture of a proton with a ruler next to it?
No. Obviously this isn't possible.
However, this directly follows, without problems, from scattering
experiments - even Rutherford scattering provides this data, if it is
analyzed closely.
> All
> observations have been indirect and require a number of assumptions to
> come up with a hard number. Some of these assumptions may not be true.
For example?
The only assumption *I* know of is that there are electrostatic forces
between the projectile (alpha particle or electron) and the proton.
Magnetic forces are also taken into account, but they are only important
at higher energies.
> > > At this
> > > point, the proton/electron effectively have a "hard" spherical shell
> > > with a radius of 93pm. It is this hard shell which provides the
> > > opposite force which causes the net force to be zero. This is just
> > > like the Earth providing an opposing "up" force to the "down" force of
> > > gravity. Without this hard shell, Franz is correct that a charged
> > > particle could not possibly have a minimal energy, but with a hard
> > > shell, this becomes possible. So is this structure stable assuming the
> > > components do have this hard shell?
> > >
> > > There also seems to be some confusion over the model I am using since
> > > you mentioned that an electron must be moving and emitting radiation
> > > in my model. As you noted any classical Bohr model would have the
> > > electron falling into the proton in no time at all. My cubic model is
> > > actually all about what happens after the electron finishes falling
> > > directly onto the proton. It goes "clunk" and electron has fallen onto
> > > the proton and is basically motionless.
> >
> > Thus giving you a neutron and a neutrino. This is a well studied process
> > which is known to happen occasionally in the case of atoms somewhat
> > heavier than hydrogen. It is known as k-capture.
> > Bang goes your model.
>
> As you say, this only happens occasionally and wouldn't pose a problem
> to my model as long as it doesn't occur most of the time.
Can you explain in your model *when* this happens, and *why* this
happens only occasionally and not always?
> > > This is my model that I am
> > > studying. As long as there is some hard limit on the distance between
> > > the proton/electron, this should be a stable system.
> >
> > I have indicated to you that whatever you do, no static system of electric
> > charges can be in stable equilibrium under the action of electroststic
> > forces alone.
> > I sincerely recommend that you should drop this nonsense and learn some
> > physics instead.
>
> You can keep on saying that, but it is hardly a convincing argument.
It isn't a convincing argument to tell you that you should learn first
1) what the evidence is
2) what the theories really say, instead of the straw men you attach?
Interesting world view.
> I keep on saying that there is MORE involved here than just
> electrostatic forces alone.
You keep on *asserting* this. Where is your evidence? As already said
several times, you could start by deriving Rutherford's scattering cross
section formula from your model. Alternatively, you could try explaining
the spectrum of hydrogen. Both are in standard QM fairly straightforward
calculations, and the predictions agree perfectly with the observations.
Your model so far has given only some very vague predictions, and all of
these can be obtained from standard QM, too...
> There is a MINIMUM distance
Evidence?
> and I would
> think based on the calculations I made that it would be obvious that
> this system is extremely stable.
Where can we find these calculations?
> There was also another similar
> theorem stating that it was impossible to create a stable magnetic
> suspension system - that was shown false by the invention of the
> Levitron and stable magnetic suspension of magenets over
> superconducting materials. I wouldn't put so much faith in theorems.
Please give some references for
1) the existence of this theorem
2) that these inventions disproved this theorem.
> > > I was wondering how QM explains this situation where a low speed
> > > electron meets up with a low speed proton and somehow the electron
> > > transforms from a particle into this object described by wavefunctions
> > > whizzing randomly around the proton. Does QM explain why they just
> > > don't go "clunk" as one would intuitively think they would?
> >
> > Yes. In exquisite detail. All the calculations have been done in
> > complete
> > detail for all the atoms from hydrogen up to Iron. The results are in
> > daily use by astronomers who study stellar structure via their spectra.
>
> What are you talking about? You can draw up equations to generate the
> spectra of an excited atom, but that doesn't say anything about why an
> electron just doesn't get captured by the proton. It somehow avoids
> this capture and that is the question I am asking.
Heisenberg's uncertainty principle - or, in other words, the wave-like
properties of an electron.
Hint: according to QM, an electron isn't like a little marble which
moves around on trajectories.
> > > I also never understood that if QM still recognizes the electron as a
> > > particle moving about the proton, how does this solve the radiation
> > > problem?
> >
> > The long and short of it is that according to QM, the electron does not do
> > that. The fact that you ask this question is an absolutely clear example
> > of the ststement that you should stop posting here and learn some
> > elementary quantum mechanics.
> >
> What kind of answer is that?
The appropriate answer. If you don't know the answer to elementary
questions like the one above, you should learn some basic QM.
> "The long and short of it" is hardly an explanation.
This newsgroup isn't intended for teaching basic QM to people. Hint: for
that purpose, there do exist some things called "books".
> By now, I have been reading a couple of quantum mechanics
> books and none of them has come close to addressing this question.
Either you read bad books (which did you read?), or you have reading
comprehension problems.
You should try reading "The strange world of Quantum Mechanics" by Styer
- maybe that would help you.
> It is glossed over or just completely ignored - however - this is a
> central point if quantum mechanical models of the atom are not to
> share the same fate as the Bohr orbital models. Do quantum mechanics
> have an answer?
Yes. The point is quite simple: QM doesn't say that electrons move
around the nucleus. Hence obviously there is no radiation.
> > > I would think the electrons would still undergo accelerations
> > > and such accelerations should still give off energy no matter how
> > > randomly it is orbiting.
It isn't "orbiting randomly" - it isn't orbiting at all. Again:
electrons aren't like little marbles which move around on well-defined
trajectories.
> > No, and no in the case of electrons in the ground state.
>
> OK, explain how an electron in a quantum mechanical shell can move
> around in its shell without changing direction and thereby avoid any
> acceleration. Doesn't sound possible to me.
It doesn't move around it its shell. Where did you get this weird idea
from?
> > > I have read that the atom emits only if the
> > > electron goes between various levels,
> >
> > That is correct.
> >
> > > but that doensn't excuse why an
> > > electron moving at any level shouldn't still be emitting energy.
> >
> > That is precisely what happens if the electron is in any level other than
> > the ground state.
>
> Explain why even in ground state where the electrons are still moving
> and changing directions, how they can avoid emitting energy.
They aren't moving and changing directions.
[snip]
Bye,
Bjoern
Bjoern Feuerbacher
>
> "Franz Heymann" <notfranz...@btopenworld.com> wrote in message news:<c0aenk$b2t$6...@titan.btinternet.com>...
[snip]
> > > There is a MINIMUM distance and I would
> > > think based on the calculations I made that it would be obvious that
> > > this system is extremely stable.
> >
> > No. On the contrary, it is obvious that it would be unstable.
> > Do you understand what a saddle point in a potential distribution means?
>
> Let me try to make this as clear as possible - I keep on saying that
> my model postulates that electrons and protons have a "hard" surface.
You can postulate whatever you like - as long as you present no evidence
for this (which includes explaining why all the existing evidence from
scattering experiment doesn't contradict you), this has no value at all.
> Whatever makes up electrons/protons (call them quarks or whatever)
What makes you think that electrons are composed of something?
> take up a finite amount of space
Why do you think so?
> and at a certain point, they cannot
> get any closer despite the electrostatic force they are generating.
Present evidence for this assertion.
> If you still have trouble with this concept that take 8 ping pong balls
> on the space shuttle.
The concept is clear - what'
s missing is the evidence.
> Charge 4 of them up with a positive
> electrostatic charge, and the other 4 with a negative charge. Now if
> you let them float free and interact with each other, I would predict
> that they would form a stable cube. This "hardness" is a postulate,
> and may not neccessarly be true, but it is well within reason to think
> it is this way since this is how macroscopic objects work
*sigh* Do you have got *any* clue *why* macroscopic objects are hard?
Apparently not!
Hint: electrostatic repulsion.
> - so it
> follows intuition and doesn't defy logic.
It only follows intuition and doesn't defy logic if one doesn't
understand that the hardness of macroscopic objects is not some
mysterious intrinsic properties, but follows straightforwardly from
electrostatic repulsion (and partly the Pauli principle).
> Do you get it? HARD SURFACE!!!!
Do you get it? HARD SURFACE FOLLOWS FROM ELECTROSTATIC REPULSION!!!!
> The other force involved is not any kind of field force
> (e.g. weak, strong, etc.)
What is "field force" supposed to mean?
> It is the physical reality that solid
> objects cannot merge together no matter the degree of force applied to
> it.
It is physical reality that this property follows simply from
electrostatic repulsion (and partly the Pauli principle). It's not an
intrinsic property of objects to be "hard" - how should that work? Where
should the force necessary for this come from?
[snip]
> > It was not an explanation. It was a statement of fact.
> > If you want it in detail, read Schiff "Quantum Mechanics". That will
> > occupy you at least six months.
> >
> Is this a statement of fact, or is this just postulated?
What? That the ground state is stable? This follows from the simple fact
that it is the state with the lowest possible energy - as can be derived
from Schroedinger's equation.
An electron sitting motionless on the proton would strongly contradict
Heisenberg's uncertainty relation.
> At least Bohr
> postulated that the laws of electrostatics simply didn't apply at the
> level of atoms.
Fortunately QM has improved since then - Schroedinger's equation
explains it without such postulates.
> I also saw a website which was explaining the quantum
> mechanical system of electron shells and at least it was honest enough
> to say that "we don't know what's going on with the electron and we
> don't care, so we ignore the problem".
Reference, please.
> The explanations I have seen
> seem to indicate that the uncertainty principle makes is impossible to
> know where the electron is located and where it might be in the next
> moment, thereby making it impossible to calculate whether it is
> undergoing any accelerations.
That's only *VERY* vaguely right.
Much closer to the truth would be saying that electrons don't *have*
something like a well-defined trajectory!
Perhaps Feynman's view of QM would be better understandable to you? Try
looking into his lectures, volume 3. Especially interesting for you
should be the concept of "sum over paths".
> However, not being able to do the
> calulations doesn't mean that the electron doesn't undergo
> accelerations.
Well, the explanation isn't right, plain and simple. The electron simple
*has* no well-defined trajectory. Having such a trajectory would already
contradict the well-known double slit experiment.
> > > By now, I have been reading a couple of quantum mechanics
> > > books and none of them has come close to addressing this question.
> >
> > What question? You have raised more than one.
>
> How can an electron remain in its orbital cloud without radiation
> losses?
Why should it loose radiation? It doesn't move.
> > > It
> > > is glossed over or just completely ignored - however - this is a
> > > central point if quantum mechanical models of the atom are not to
> > > share the same fate as the Bohr orbital models. Do quantum mechanics
> > > have an answer?
> >
> > Yes, indeed it does "have the answer".
> > In fact it has only correct answers. It has successfully answered every
> > single question which has ever been asked of it completely correctly.
>
> Out of curiosity, does it explain why the most common products of
> Uranium fissions are a bit bigger than about 1/4 or 1/2 the mass of
> the original atom?
AFAIK, yes.
It should have something to do with the binding energies
(Bethe-Weizsaecker formula), but I would have to read up on this in
order to find out.
> Does it provide a precise calculation for the
> ionization energy for hydrogen and helium?
YES!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
Say, how ignorant *ARE* you??????????????????????????????
I can't believe your claim that you have read several books on QM
anymore. These are such *basic* calculations, they are given in detail
in *every* book on QM!!!!!!! How on earth did you manage to miss
them?!?!?!?!?!?!?!?
[snip]
> > > Explain why even in ground state where the electrons are still moving
> > > and changing directions, how they can avoid emitting energy.
> >
> > Who said they are moving and changing direction in the ground state?
> > >
> [snip]
> Based on your responses, you are saying that an electron isn't moving
> in its probability cloud.
Err, the probability cloud *is* the electron, in a sense.
> This isn't the picture I got from reading about it,
Then you have either read the wrong things (what have you read?), or you
have got severe reading comprehension problems.
> but it certainly is possible that the electron is remaining
> mostly static,
Not "mostly". Completely.
> but may wander from point to point like it was stuck in
> a bottle.
Huh?????
> Is this the correct quantum mechanical picture?
No!!!!!
The correct QM picture is that the ability to find an electron at a
precise location is described by its wave function (and this wave
function is static). In a sense, the electron is at every point
*simultaneously* - just as Schroedinger's cat is alive and dead
*simultaneously*, and just as the electron or the photon in a
double-slit experiment goes to *both* slits *simultaneously*.
Your *big* problem is that you picture the electron like a little marble
which flies around in space on a well-defined path. QM (and not only the
theory, the experiments as well - e. g. double slit) that this picture
is wrong.
For more on this, I recommend "The strange world of Quantum Mechanics"
by Daniel F. Styer to you.
> But if this is correct,
It isn't.
> then how does the electron resist the pull of
> the nucleus if it doesn't have any momentum to keep it away from the
> nucleus?
Again, your problem is that you picture the electron to be like a little
marble which has a well-defined position and momentum. This simply
doesn't work.
> By any calculation, the static electron should fall straight
> into the nucleus.
If it would behave like a classical particle (macroscopic object),
right. Hint: it doesn't.
> This is a much worse situation that the Bohr atom
> which at least used momentum and centripital force to keep the
> electron out of the nucleus.
There is nothing "worse" here if one gives up the picture of the
electron as a little marble.
> If this is incorrect, then the electron does have some momentum
The expectation value for its momentum is zero. The expectation value
for its momentum *squared* isn't. Try to understand that... (hint: one
*can* understand this if one gives up the picture of the electron as a
little marble)
> and
> since it is confined to being located in a probability cloud,
This makes no sense. You really don't understand what this "probability
cloud" means.
> it must
> have to change direction and lose energy.
Only if it would act like a little marble. It doesn't. You completely
ignore the wave-like properties of the electron.
> The only way out would be if
> the electron had a Star Trek transporter and could magically move
> itself from place to place without having to change directions. This
> seems highly unlikely.
You are caught in a classical picture of the electron. It's no wonder
that you keep arriving at nonsensical results.
> Of course, I'm going to guess that you're going to say that it is just
> a point of fact that the electron remains in its probability cloud
> without losing energy.
No, he won't say that - because "the electron remains in its probability
cloud" makes no sense. If you have understood why this statement makes
no sense, you probably have understood the solutions to your other
problems, too.
You have got a totally mixed picture in your mind, apparently - OT1H,
you know that the electron is described by a wave function and that
there is something like a "probability cloud", but OTOH, you picture the
electron like a small marble which sits somewhere in this cloud. This
makes no sense at all - you are mixing QM and classical concepts here.
You haven't understood what this "probability cloud" really *means*.
> This point of fact sounds just like what was
> proposed for epicycles whereby, there had to be an imaginary point
> about which the planets looped around. It was a point of fact that
> these points had to exist to explain the rest of the phenomenon.
What you miss is that in the model with epicycles, one had to use
different sets of epicycles for solving different types of problems -
this model wasn't intended to describe the real world, but merely as a
tool for calculations (unfortunately most people didn't understand this
and used the model as a picture for the real world...). In QM, the
situation is totally different.
> It seems to me that the quantum mechanical house of cards also has its
> foundations in a similar point of fact.
This only seems to be so to you because you don't understand QM yet.
Again, try reading the book by Styer - or maybe Feynman's lectures,
volume 3.
> Since it seems that an electron cannot logically maintain its position
> about the nucleus,
It *has* no fixed position. In a sense, it is at all positions at once.
Just like in the double slit experiment, where it goes through both
slits at once, in a sense. If you understand this crucial point, your
problems will go away.
> the only logical solution is that the electrons are
> actually part of the nucleus.
Contradicts Heisenberg's uncertainty relation and a plethora of
scattering experiments. And doesn't explain atomic spectra, as QM does.
> If we assume that the electrons are part
> of the nucleus in a Thompson like (plum pudding) atom, then the
> problems of keeping the electrons "in the air" go away.
Please explain the atomic spectra in this model.
And please explain the results of the double slit experiment, if
electrons really can have a fixed position and a fixed momentum at once.
> The main
> problem with the Thompson model is that it didn't specify a distinct
> structure for the atom nucleus.
No. The main problem with the Thompson model is that it is unable to
correctly predict atomic spectra and Rutherford's formula for
scattering, among a lot of other things.
> A random arrangement cannot explain
> Rutherford scattering.
Your model apparently can't do this, too. Or did I somehow miss your
calculation?
OTOH, the standard model can do this quite easily...
> However, a model which shows specific
> arrangements could explain the backscatter as the result of alpha
> particles bouncing off specific structures within the atom.
Please present your calculations.
And please explain how this backscattering works - what forces are
involved there?
> The
> effective cross section for such specific structures may turn out to
> be as small as what we calculate for a point nucleus.
No one claims that the nucleus is point-like. What on earth are you
talking about???
Bye,
Bjoern
Bjoern Feuerbacher
>
> Bjoern Feuerbacher wrote:
> >
> >
> > Hint: according to QM, the electron has no trajectory.
>
> How does an electron get from the heated element of a T.V. tube to the
> phosphorescent screen?
I think this is best answered using Feynman's "sum over paths"-view of
QM: it moves on all possible paths going from the cathode to the screen
at once.
Summing up these paths, it turns out that if you try to detect it, you
will most probably find it on the direct line between the cathode and
the screen.
> First it is there and now it is here.
Yes. However, this doesn't imply that it followed one fixed trajectory.
> Just
> because no measurement will tell you excactly where it is in between,
> does not mean it isn't somewhere.
Through which slit does the electron go in a double-slit experiment?
> Everyone who has ever died by electrocution is living testimony to the
> fact that electrons move.
They "move" only if you take the sum over all paths, i. e. a statistical
average.
Bye,
Bjoern
Franz Heymann
If the topic at this point is still a Hydrogen atom, I would beg to differ,
for the following reason:
Consider the ground state of the H atom.
The wave function is spherically symmetrical and is dependent only on radial
position.
Performing an [Integral Psistar *operaror* Psi*dv] calculation with various
choices of the operator to determine expectation values gives the following
results:
Use operator for tangential momentum gives answer 0 for expectation value.
But, that might just mean that it is equally likely to be found moving
leftwards as rightwards, so to check by
Using the operator for tangential momentum squared. The answer is still 0
So, the electron does not have a mean square tangential velocity either. If
it is moving at all, it must be in the radial direction.
Use the operator for radial momentum.
The result is 0 for the expectation value, so it has no average radial
momentum. Now do the crucial test.
Operate with the operator for p^2
Lo and behold, the result is NOT zero.
So the electron does have a non-zero mean square radial momentum
The zero radial momentum simply means that, crudely, the electron is equally
likely to move outwards as inwards.
But it is moving all right, with a finite mean square radial momentum, and
therefore also a non zero kinetic energy, since (non-relativistically), the
KE is just p^2/(2*m).
In short, when the exp(iwt) factor is brought into the wave function (since
the ground state is an energy eigenstate), the motion may be described as
a radial pulsation of the wave, in a manner leading to an electron moving
purely radially.
It is the fact that the electron is moving purely radially, with a
spherically symmetrical wavefunction, thus effectively executing an electric
*monopole* oscillation, which prevents it from radiating itself to death.
> > Only non-moving electrons should not emmit anything.
>
> Right.
No. I have just shown otherwise.
{:-))
Franz
Bjoern Feuerbacher
[snip most]
Hmmm... I agree totally with your calculation (IIRC, in another post, I
even pointed out myself to FrankH that <P^2> is not zero). But what you
do here looks to me like a classical interpretation of a quantum
mechanical calculation. Are you sure this makes sense?
> But it is moving all right, with a finite mean square radial momentum, and
> therefore also a non zero kinetic energy, since (non-relativistically), the
> KE is just p^2/(2*m).
> In short, when the exp(iwt) factor is brought into the wave function (since
> the ground state is an energy eigenstate), the motion may be described as
> a radial pulsation of the wave, in a manner leading to an electron moving
> purely radially.
I would say that it would make only sense to describe the motion in this
way if the calculation of the expectation value of the radius would give
the result that this expectation value oscillates (depends on time in a
periodic manner).
Would you say that a particle in the ground state of a harmonic
oscillator potential is moving, too? If yes, then what's the reason for
introducing the coherent states at all?
> It is the fact that the electron is moving purely radially, with a
> spherically symmetrical wavefunction, thus effectively executing an electric
> *monopole* oscillation, which prevents it from radiating itself to death.
Sounds like a nice interpretation of the ground state wave function -
but I am not sure if this really makes sense...
> > > Only non-moving electrons should not emmit anything.
> >
> > Right.
>
> No. I have just shown otherwise.
> {:-))
Sorry - I withdraw my comment. Only electrons which aren't *accerated*
emit anything - and even this applies only to classical movement.
Bye,
Bjoern