Electron Model Bohr Magneton:
ThomasL283
model, to the structures of composite particles, to the structures of atomic
nuclei.
What sets these VPP models apart from the current models is the VPP geometry
can be used to calculate the related fundamental physical constants.
Because of the VPP good results, and the current SM lack of calculation
ability, the VPP models, I have no doubt VPP will be the model for the twenty
first century. Will I live to see it? Probably not.
Here is an example of calculations involving the electron's Bohr magneton
magnetic moment.
UB = 9.27400899 E-24 Joule per Tesla
Magnetic moment is the turning torque in a magnetic field. If the electron was
in a 1 Tesla field, the torque would be 9.27400899 E-24 Joule.
As most of you probably know, torque and energy have the same units of Joule,
but torque is a vector quantity, and energy is a scalar quantity.
But how can we model Joule per Tesla? If we cancel out all of the units
possible in J/T we get an area (m^2) times a current (A).
Physically, the magnetic moment is realized by a current in a wire loop
enclosing an area. If we have a two turn coil, the area is effectively
doubled.
How can we get the current value? Well the VPP electron model two loop areas
are: 4.68471083662493 E-25 (m^2).
Giving a current of: 19.7963317601877 (A).
That's cool, but what is the electron's charge? Well the VPP electron model
shows that the charge is: (sqr 2 lambda / sqr 2 velocity of light) times
19.7963317601877 Ampere, or e1=1.602176462 E-19 (A s). See:
http://members.aol.com/tnlockyer/Bohramps.gif
Regards: Tom:
P.S. The electron size can be determined by Compton scattering photon off the
eelctron, giving an apparent electron size of 2.426310215 E-12 meter.
While it is true that the electron is traditionally considered a point charge,
and HEP cannot show a structure up to very high energies, all characteristics
can be shown from a finite electron size.
Anyway, if the electron is a point, and the positron is a point, and the muon
is a point, "What's the point." ;-)
Tom Lockyer (75 and retired) See "Vector Partcles and Nuclear Models"
0963154680 at http://www.amazon.com
"When you can measure what you are speaking about and express it in numbers,
you know something about it." Lord Kelvin (1824-1907)
Jacques Distler
<thoma...@aol.com> wrote:
>Here is an example of calculations involving the electron's Bohr magneton
>magnetic moment.
[Snip yet another rehash of Thomas's "x=x" proofs. Exercise for the
reader: find something in something in this proof which was not
debunked -- at length -- in previous posts in this newsgroup.]
>P.S. The electron size can be determined by Compton scattering photon off the
>eelctron, giving an apparent electron size of 2.426310215 E-12 meter.
No, Compton scattering DOES NOT imply that the "size" of an electron is
equal to its Compton wavelength.
Do you even know what the phenomenon of Compton scattering *is*, let
alone how the "Compton wavelength" appears in the analysis thereof?
> While it is true that the electron is traditionally considered a point charge,
>and HEP cannot show a structure up to very high energies, all characteristics
>can be shown from a finite electron size.
No. In your VPP "x=x" proofs, the size of the electron cancels out of
the equations. You could use any size you wanted, including one which
was within the experimental bounds (ie, a million times smaller than
you claim).
>Anyway, if the electron is a point, and the positron is a point, and the muon
>is a point, "What's the point." ;-)
Sorry that Mother Nature does not live up to your requirements.
I'm sure it's a bitter disappointment to you.
But repeating the same drivel in post after post, with different
subject headings each time will *not* make it any less incorrect.
JD
--
PGP public key: http://golem.ph.utexas.edu/~distler/distler.asc
FrediFizzx
news:20020420143351...@mb-ce.aol.com...
|
| P.S. The electron size can be determined by Compton scattering photon off
the
| eelctron, giving an apparent electron size of 2.426310215 E-12 meter.
"Apparent electron size" is the key phrase here. What do you exactly mean
by "apparent"? I don't see how you are making the jump from the Compton
scattering effect to the "apparent" size of the electron. When you use
Compton wavelength, you are then looking at the electron as a wave instead
of a particle as Mike pointed out. You can't just use the wave properties
of a particle to give it an actual physical size like you are doing. See
the following link for the Compton Effect.
http://scienceworld.wolfram.com/physics/ComptonEffect.html
| While it is true that the electron is traditionally considered a point
charge,
| and HEP cannot show a structure up to very high energies, all
characteristics
| can be shown from a finite electron size.
You are mixing up the wave properties of a particle with an actual size. So
your geometry is "possibly" representing something else besides actual
physical size. I think your mistake is that you are then taking your "wave
size" and applying properties that do depend on an actual physical size. At
least that is what it looks like to me.
FrediFizzx
Jacques Distler
<FrediFi...@HaHahotmail.com> wrote:
>"ThomasL283" <thoma...@aol.com> wrote in message
>news:20020420143351...@mb-ce.aol.com...
>|
>| P.S. The electron size can be determined by Compton scattering photon off
>the
>| eelctron, giving an apparent electron size of 2.426310215 E-12 meter.
>
>"Apparent electron size" is the key phrase here. What do you exactly mean
>by "apparent"? I don't see how you are making the jump from the Compton
>scattering effect to the "apparent" size of the electron. When you use
>Compton wavelength, you are then looking at the electron as a wave instead
>of a particle as Mike pointed out.
No. No such assumption is made about the electron.
It is the *photon* which is a wave of wavelength hc/E.
Since the photon transfers energy to the electron it scatters off of,
its wavelength shifts.
The amount of the shift is proportional to the "Compton Wavelength" of
the electron. The constant of proportionality depends on the scattering
angle; at 90^o, the shift in the photon wavelength is precisely the
Compton Wavelength of the electron.
Se the derivation at the link you provided:
>>You can't just use the wave properties
>of a particle to give it an actual physical size like you are doing. See
>the following link for the Compton Effect.
>
>http://scienceworld.wolfram.com/physics/ComptonEffect.html
FrediFizzx
news:200420021612523538%dis...@golem.ph.utexas.edu...
| In article <3cc1d5e4$0$30651$8252...@news.compuserve.de>, FrediFizzx
| <FrediFi...@HaHahotmail.com> wrote:
|
| >"ThomasL283" <thoma...@aol.com> wrote in message
| >news:20020420143351...@mb-ce.aol.com...
| >|
| >| P.S. The electron size can be determined by Compton scattering photon
off
| >the
| >
| >"Apparent electron size" is the key phrase here. What do you exactly
mean
| >by "apparent"? I don't see how you are making the jump from the Compton
| >scattering effect to the "apparent" size of the electron. When you use
| >Compton wavelength, you are then looking at the electron as a wave
instead
| >of a particle as Mike pointed out.
|
| No. No such assumption is made about the electron.
|
| It is the *photon* which is a wave of wavelength hc/E.
| Since the photon transfers energy to the electron it scatters off of,
| its wavelength shifts.
|
| The amount of the shift is proportional to the "Compton Wavelength" of
| the electron. The constant of proportionality depends on the scattering
| angle; at 90^o, the shift in the photon wavelength is precisely the
| Compton Wavelength of the electron.
|
| Se the derivation at the link you provided:
|
| >>You can't just use the wave properties
| >of a particle to give it an actual physical size like you are doing. See
| >the following link for the Compton Effect.
| >
| >http://scienceworld.wolfram.com/physics/ComptonEffect.html
Now I am a little confused here. Compton wavelength of the electron doesn't
imply a "wave" property of the electron? So the Compton wavelength is only
a property of a particle by which it will shift the wavelength of a
scattered photon and nothing more?
FrediFizzx
FrediFizzx
news:3cc1d5e4$0$30651$8252...@news.compuserve.de...
So I stand corrected here thanks to Jacques. It is not "wave" properties at
all. Compton wavelength is just a property of a particle by which it will
shift a scattered photon's wavelength. See the link above. That means
Thomas' geometry really is just a fantasy construction. Sorry dude.
FrediFizzx
Jacques Distler
<FrediFi...@HaHahotmail.com> wrote:
>So the Compton wavelength is only
>a property of a particle by which it will shift the wavelength of a
>scattered photon and nothing more?
Correct.
This combination, h/mc, does appear in other, related contexts.
E.g., consider a massive particle which mediates some force (eg, the
W-bosons of the weak interactions). Then the length scale over which
the force operates is also given by h/mc. So the weak interactions are
a very short-range force, whereas electromagnetism -- mediated by the
massless photon -- is a long (infinte) range force.
Jon Bell
FrediFizzx <FrediFi...@HaHahotmail.com> wrote:
>
>imply a "wave" property of the electron?
No it's not.
> So the Compton wavelength is only
>a property of a particle by which it will shift the wavelength of a
>scattered photon and nothing more?
Correct. The name "Compton wavelength" is rather unfortunate because it's
not actually the wavelength of anything. A better name would be something
like "Compton wavelength-shift factor."
--
Jon Bell <jtb...@presby.edu> Presbyterian College
Dept. of Physics and Computer Science Clinton, South Carolina USA
FrediFizzx
Well, after looking more into how Thomas was getting Compton wavelength, I
have gone and confused myself again. Compton wavelength showed up for him
when he set the magnitude of the magnetic and electric vector lengths equal
to each other thus was setting the forces equal and solving for the length.
Compton wavelength, of course, pops out of that equation, I guess, as the
vector lengths? Is this at all interesting? Or is he wrong in how he is
equating them? So I guess the question I am asking is: can the
rationalized Compton wavelength actually be related to the magnetic and
electric vector lengths? See the following link.
http://www.flashrock.com/upload/equalforces.pdf
If so, then it would give more significance to Compton wavelength other than
just being a particle property that shifts the scattered photon's
wavelength. Where am I going wrong here?
FrediFizzx
FrediFizzx
news:Guw39...@presby.edu...
| In article <3cc1e12b$0$30668$8252...@news.compuserve.de>,
| FrediFizzx <FrediFi...@HaHahotmail.com> wrote:
| >
| >Now I am a little confused here. Compton wavelength of the electron
doesn't
| >imply a "wave" property of the electron?
|
| No it's not.
|
| > So the Compton wavelength is only
| >a property of a particle by which it will shift the wavelength of a
| >scattered photon and nothing more?
|
| Correct. The name "Compton wavelength" is rather unfortunate because it's
| not actually the wavelength of anything. A better name would be something
| like "Compton wavelength-shift factor."
Well, if I set h*f equal to m_e*c^2, Compton wavelength pops right out of
that. So Compton wavelength is exactly equal to the wavelength of a photon
with energy equal to the electron. So maybe it should have been called
"photon equivalent energy wavelength". So, I guess, in a way it is the
wavelength of something. It is the wavelength of an energy equivalent
photon. I think this is actually a better description of how we are using
it than in Compton scattering.
FrediFizzx
FrediFizzx
| FrediFizzx <FrediFi...@HaHahotmail.com> wrote:
| >
| >Now I am a little confused here. Compton wavelength of the electron
doesn't
| >imply a "wave" property of the electron?
|
| No it's not.
Well, now I am really confused by your "no it's not" Check out below. It
is a reply about why the electron is not a black hole. Is this guy wrong?
"However, as a subatomic particle the electron is also a quantum-mechanical
object. Recall the wave-particle duality hypothesis of de Broglie. All
objects have a wave function which represents the probability of locating
that object at a particular point in space. During a collision this wave
function momentarily collapses and the particle is truly at one "point" in
space, but it immediately starts to spread out again after the instant of
collision. The typical spread of the wave-function of a point particle is
given by the Compton wavelength: [lambda = h/mc]
and this can be considered the true quantum-mechanical "size" of the object.
Notice that this size gets smaller as the mass gets larger. For you or I or
the sun this quantum-mechanical size is essentially zero (there's not much
uncertainty as to where the sun is!) but for an electron the size is 2.42
x10-12 m. Though still small, this is much, much larger than the
Schwarzschild radius. So quantum-mechanically most of the electron is
"outside" its event horizon. That's why it and other subatomic particles are
not black holes."
Answered by: Brent Nelson, M.A., Physics Ph.D. student, UC Berkeley
http://www.physlink.com/Education/AskExperts/ae191.cfm
FrediFizzx
FrediFizzx
Well, it looks like I might have to correct myself again. Seems like this
subject is more complex than I realized at first. Compton wavelength of the
electron is related to the "wave function" spread of the electron's quantum
mechanical properties and is considered its quantum mechanical size
according to:
http://www.physlink.com/Education/AskExperts/ae191.cfm
So it looks to me like Thomas is not wrong if he replaces "apparent" with
"quantum mechanical". And the electron's Compton wavelength is exactly
equal to the wavelength of an energy equivalent photon. So there is still
the problem of how to go from this quantum mechanical size to a physical
size. If not possible, then maybe some of the VPP geometry is actually
representing quantum mechanical properties of the electron somehow.
However, I think Thomas is taking something that is not an actual physical
size and doing operations on it like it was. Bizarre! Or is it? So if the
electron interacts with another particle, its wave function collapses and it
looks like a "point". What the heck is its "true" size when it is not
interacting? Is it still a point? Or is it the size from its Compton
wavelength? Always when we try to measure the electron's size, it will come
out as a point according to QM. But what is its size if we don't measure
it?
FrediFizzx
Jacques Distler
<FrediFi...@HaHahotmail.com> wrote:
>Well, now I am really confused by your "no it's not" Check out below. It
>is a reply about why the electron is not a black hole. Is this guy wrong?
>
>"However, as a subatomic particle the electron is also a quantum-mechanical
>object. Recall the wave-particle duality hypothesis of de Broglie. All
>objects have a wave function which represents the probability of locating
>that object at a particular point in space. During a collision this wave
>function momentarily collapses and the particle is truly at one "point" in
>space, but it immediately starts to spread out again after the instant of
>collision. The typical spread of the wave-function of a point particle is
>given by the Compton wavelength: [lambda = h/mc]
>and this can be considered the true quantum-mechanical "size" of the object.
>Notice that this size gets smaller as the mass gets larger. For you or I or
>the sun this quantum-mechanical size is essentially zero (there's not much
>uncertainty as to where the sun is!) but for an electron the size is 2.42
>x10-12 m. Though still small, this is much, much larger than the
>Schwarzschild radius. So quantum-mechanically most of the electron is
>"outside" its event horizon. That's why it and other subatomic particles are
>not black holes."
>
>Answered by: Brent Nelson, M.A., Physics Ph.D. student, UC Berkeley
>
>http://www.physlink.com/Education/AskExperts/ae191.cfm
Sorry, but that is a completely silly "explanation" (with the electron
collapsing to a point during collisions and then spreading out again).
QED is not a single-particle theory for a good reason. You cannot
localize an electron in a box smaller than a Compton wavelength, h/mc.
If you try to localize an electron in a box much smaller than h/mc, you
pump enough energy into your system to allow the creation of
electron-positron pairs. Instead of a single electron, you'll find that
your box contains, say, 3 electrons, 2 positrons, and some number of
photons.
This effect (particle creation if you hit the system with enough energy
<=> try to probe it at short enough distances) is cleanly separable
from the particle having a nonzero size.
*Your* Compton wavelength is much smaller than your Schwarzschild
radius. That DOES NOT mean that you are a black hole and it DOES NOT
mean that you are pointlike. Your *size* -- a quantity easily measured
by scattering photons off you -- is much large than either your Compton
wavelength or your Schwarzschild radius.
[If you don't think the laws of physics apply to you, then substitute
the phrase "a proton" for "you", and the same statements hold.]
Jacques Distler
<FrediFi...@HaHahotmail.com> wrote:
>Compton wavelength of the
>electron is related to the "wave function" spread of the electron's quantum
>mechanical properties and is considered its quantum mechanical size
>according to:
>
>http://www.physlink.com/Education/AskExperts/ae191.cfm
>
>So it looks to me like Thomas is not wrong if he replaces "apparent" with
>"quantum mechanical".
No. That "explanation" is seriously misguided. In no reasonable sense
does the Compton wavelength represent the "size" of a particle.
If it did, you would be a black hole.
Jon Bell
FrediFizzx <FrediFi...@HaHahotmail.com> wrote:
>Answered by: Brent Nelson, M.A., Physics Ph.D. student, UC Berkeley
>
>http://www.physlink.com/Education/AskExperts/ae191.cfm
[snip]
>"However, as a subatomic particle the electron is also a quantum-mechanical
>object. Recall the wave-particle duality hypothesis of de Broglie. All
>objects have a wave function which represents the probability of locating
>that object at a particular point in space. During a collision this wave
>function momentarily collapses and the particle is truly at one "point" in
>space, but it immediately starts to spread out again after the instant of
>collision. The typical spread of the wave-function of a point particle is
>given by the Compton wavelength: [lambda = h/mc]
I'm afraid I don't see how he arrived at the last statement. To me,
"spread of the wave-function" means delta-x in the Heisenberg
uncertainty principle: (delta-x)(delta-p) = hbar/2. But I can't see how
a delta-x of h/mc is particularly "typical", seeing as the value of
delta-x for a particular particle depends on delta-p for that particle.
Plugging h/mc for delta-x in the HUP and solving for delta-p (taking into
account hbar = h/(2*pi)) gives delta-p = mc/(4*pi). What's typical about
that?
I'm not saying he's *wrong* because he could be referring to something
else entirely when he talks about "typical spread of the wave-function."
But I don't know what that could be, and a quick Google search didn't turn
up anything else along these lines.
Note that "spread of the wave-function" does not necessarily mean
"wavelength of the wave-function". The usual wave-function that we
construct for a particle is something called a "wave packet", obtained by
adding an infinite number of waves with differing wavelengths. The range
of wavelengths (delta-lambda) is related to the range in possible momentum
values (delta-p). The result looks like a single wave with a more or less
uniform wavelength and an amplitude which is maximum at the center and
tapers off to zero on the sides, with an overall width of delta-x. So
typically you have some (possibly large) number of wavelengths contained
within the width of the wave packet.
FrediFizzx
| In article <3cc32306$0$30658$8252...@news.compuserve.de>, FrediFizzx
| <FrediFi...@HaHahotmail.com> wrote:
|
|
| >Compton wavelength of the
| >electron is related to the "wave function" spread of the electron's
quantum
| >mechanical properties and is considered its quantum mechanical size
| >according to:
| >
| >http://www.physlink.com/Education/AskExperts/ae191.cfm
| >
| >So it looks to me like Thomas is not wrong if he replaces "apparent" with
| >"quantum mechanical".
|
| No. That "explanation" is seriously misguided. In no reasonable sense
| does the Compton wavelength represent the "size" of a particle.
|
| If it did, you would be a black hole.
Aaakkk! Yes, it did seem a bit odd.
OK, so Thomas still has to explain the jump he is making to his electron
size from Compton scattering. I think it has do to actually with something
else and Thomas just thew that in there out of the blue. His explanation in
his book is better. Set the magnetic and electric forces equal and Compton
wavelength falls out when using the Bohr magneton for "magnetic charge". It
goes along with his premise that for the electron-positron pair to form, the
magnetic and electric vectors lengths of the creating photon have to equal
each other and its Poynting vector length. And they are related to Compton
wavelength. Or something like that. Is it actually possible to calculate
the physical vector lengths of the photon with energy equal to 2*m_e*c^2?
FrediFizzx
Jacques Distler
<FrediFi...@HaHahotmail.com> wrote:
>>| No. That "explanation" is seriously misguided. In no reasonable sense
>| does the Compton wavelength represent the "size" of a particle.
>|
>| If it did, you would be a black hole.
>
>Aaakkk! Yes, it did seem a bit odd.
It's more than odd, it's infuriatingly wrong.
1) Whenever you talk about "the electron wave function", you are
assuming that a single-particle quantum mechanical description is
valid.
2) In that context, if you say that the wave function is concentrated
in a region of size L, you mean that the particle is certain to be
found inside that region (*not* that the *size* of the particle is L).
The particle is said to be "localized" to within a distance L.
3) The "explanation" goes on to assert that the electron wave function
is "typically" localized in a region of size L=hc/m (except when the
electron undergoes "collisions", at which time it asserts the electron
wave function momentarily collapses to a point -- which is bizarre, and
not the way things work in QM).
But that is *exactly* the length scale at which assumption 1) *fails*.
As I explained, hc/m is the length scale at which the single-particle
description breaks down. You *need* to use quantum field theory to
describe physics at such length scales.
If the author of this "explanation" were correct, you could *never* use
single-particle quantum mechanics -- ever!
ThomasL283
Wrote:
In article <20020420143351...@mb-ce.aol.com>, ThomasL283
<thoma...@aol.com> wrote:
>Here is an example of calculations involving the electron's Bohr magneton
>magnetic moment.
[Snip yet another rehash of Thomas's "x=x" proofs. Exercise for the
reader: find something in something in this proof which was not
debunked -- at length -- in previous posts in this newsgroup.]
No one came up with numbers, just proof that the equations were dimensionally
correct. If x did not eqaul x the equations would be wrong, they are not. (They
even used the VPP geometric relationships to prove x=x without deriving them)
see the pdf's.
I'll call your bluff, and give you the three numbers VPP used to get the
fundamental charge from the Bohr.
(1) electron Compton
2.426310215 E-12 m
(2) velocity of light
2.99792458 E8 m/s
(3) Bohr magneton.
9.27400899 E-24 J/T
Can you get the fundamental charge NUMBERS from those three above given
numbers?
Play hell if you can.
The VPP model geometry did get 1.602176452 E-19 (A s) and there is no other
way.
Your failure to get 1.602176462 E-19 (A s) above, from the three given numbers,
and the VPP success,
is proof the VPP geometry gives a correct electron model.
Regards: Tom:
larry shultis
> > Jacques Distler dis...@golem.ph.utexas.edu
> Wrote:
>
> In article <20020420143351...@mb-ce.aol.com>, ThomasL283
> <thoma...@aol.com> wrote:
>
> >Here is an example of calculations involving the electron's Bohr magneton
> >magnetic moment.
>
> [Snip yet another rehash of Thomas's "x=x" proofs. Exercise for the
> reader: find something in something in this proof which was not
> debunked -- at length -- in previous posts in this newsgroup.]
>
> No one came up with numbers, just proof that the equations were dimensionally
> correct. If x did not eqaul x the equations would be wrong, they are not.
(They
You still have no clue about the difference between dimensions and units.
> even used the VPP geometric relationships to prove x=x without deriving them)
> see the pdf's.
>
> I'll call your bluff, and give you the three numbers VPP used to get the
> fundamental charge from the Bohr.
>
> (1) electron Compton
> 2.426310215 E-12 m
> (2) velocity of light
> 2.99792458 E8 m/s
> (3) Bohr magneton.
> 9.27400899 E-24 J/T
Sorry Thomas, but the numbers, except for the speed of light, are combinations
of elementary constants as you could see if you will write them out as
such combinations. You cannot get the Bohr magneton except by combining
the constants (the units which stand for the constants have specific values
which are combined to give the above number) because that is how it is defined.
Same for Compton wavelength for the electron, it is a quantity defined in terms
of elementary constants.
By the way, where do think those numebers come from? They are not measured.
They
are all defined and (1) and (3) are defined in terms of other constants.
Larry
Jim Panetta
Bullshit.
Electron Compton wavelength == lambda_e == h / (m_e c)
Velocity of light == c
Bohr Magneton == mu_B == e h / (4 pi me)
(NOTE: the symbol == means "is defined as")
Pretty obviously:
e = mu_B * 4 pi / ( lambda_e c )
e = e h 4 pi me c 1
------- * * ---- * -
4 pi me h c
e = 9.27400899 E-24 J/T * 4 * 3.141592653589793238 /
(2.426310215 E-12 m * 2.99792458 E8 m/s)
= 1.6021765 E-19 Coulombs
The electron compton wavelength is not a physical constant. It
is constructed from the physical constants h, m_e, and c.
The Bohr magneton is not a physical constant. It is constructed
from the physical constants e, h and m_e.
>Your failure to get 1.602176462 E-19 (A s) above, from the three
>given numbers, and the VPP success, is proof the VPP geometry gives a
>correct electron model.
Thomas, please take a basic physics course. Learn about dimensional
analysis. Learn about energy. Learn about significant figures.
And most of all, learn about the proper ways of testing theories.
--Jim
--
My opinions are mine...not SLAC's...not Penn's...not DOE's...mine.
(except by random, unforseeable coincidences)
pan...@slac.stanford.edu -- Save the whales! Free the mallocs!
FrediFizzx
news:aa4to4$p99$1...@usenet.Stanford.EDU...
Jim, we have been telling him this for the last five or so threads now. He
just doesn't get it. I even showed him that his fantasy model geometry was
derivable from these equations by multiplying one side them by x/x. Maybe
that was a mistake on my part because now he thinks that makes his model
true.
http://www.flashrock.com/upload/bohring.pdf
http://www.flashrock.com/upload/studycarefully.pdf
FrediFizzx
Jim Panetta
>"Jim Panetta" <pan...@moa.SLAC.Stanford.EDU> wrote in message
>news:aa4to4$p99$1...@usenet.Stanford.EDU...
>|
>| is constructed from the physical constants h, m_e, and c.
>|
>| The Bohr magneton is not a physical constant. It is constructed
>| from the physical constants e, h and m_e.
>
>Jim, we have been telling him this for the last five or so threads now. He
>just doesn't get it. I even showed him that his fantasy model geometry was
>derivable from these equations by multiplying one side them by x/x. Maybe
>that was a mistake on my part because now he thinks that makes his model
>true.
Freddi, I've been peripherally involved in these discussions for
*years*. So far, the best quote I've seen on Thomas's "model"
was from Dan Evens in 1996:
"Absolutely brimming over with wrongability."
In this discussion, the players come and go. As far as I remember,
Jim Carr has been around about the longest. I was in it in '97-'99,
then I left for a while. Thomas's stuff has been debunked ad nauseam.
For giggles, take a look on google groups for "Clockwork Neutrons" in
sci.physics.particle. It's one of the most entertaining "Proof by
Vigorous Hand Waving" examples I've ever seen.
Jacques Distler
<thoma...@aol.com> wrote:
>I'll call your bluff, and give you the three numbers VPP used to get the
>fundamental charge from the Bohr.
>
>(1) electron Compton
> 2.426310215 E-12 m
This is not a measured quantity. It is *defined* to be equal to h/mc
>(2) velocity of light
> 2.99792458 E8 m/s
This one is measured.
>(3) Bohr magneton.
> 9.27400899 E-24 J/T
Again, this is not a measured quantity. It is *defined* to be equal to
eh/(4pi mc)
>Can you get the fundamental charge NUMBERS from those three above given
>numbers?
Any 5th grader can extract "e" from the above quantities, using nothing
fancier than elementary algebra (in fact, he doesn't even need (2)).
If you cannot do 5th grade algebra without MathCAD, then there is not
much we can do to help you.
larry shultis
Thomas seems to believe that if you use algebra, cancelling names of
numerical values in an equation, rather than arithmetic on numerical values,
then
you are doing dimensional analysis. That is the only thing I can think of
considering his comment that we had only done dimensional analysis to
show that he had extracted e from the Bohr magneton.
Larry
ThomasL283
Wrote in:
>Message-id: <230420022153409050%dis...@golem.ph.utexas.edu>
>>I'll call your bluff, and give you the three numbers VPP used to get the
>>fundamental charge from the Bohr.
>>
>>(1) electron Compton
>> 2.426310215 E-12 m
>
>This is not a measured quantity. It is *defined* to be equal to h/mc
You cheated and brought in h and m, all I gave you was J/T, c and lambda
>> 2.99792458 E8 m/s
>
>This one is measured.
>
Yes, but now it has been set exact by international agreement (1975).
>
>>(3) Bohr magneton.
>> 9.27400899 E-24 J/T
>
>Again, this is not a measured quantity. It is *defined* to be equal to
>eh/(4pi mc)
>
You cheated and have to use (e) (h) and (mass) in the definition. I wanted you
to derive the value of (e) from only the three numbers, (c), (lambda) and (UB)
constants.
>>Can you get the fundamental charge NUMBERS from those three above given
>>numbers?
>
>Any 5th grader can extract "e" from the above quantities, using nothing
>fancier than elementary algebra (in fact, he doesn't even need (2)).
>
Nope:
My original requirement was the only thing known were those three constants,
i.e. the electron's Compton wavelength, the velocity of light, and the Bohr,
to get the charge (e).
You fellows brought in:
(4) plancks constant (h)
(5) the electron's mass (m)
(6) and the electron's charge (e) explicitly.
And item (6) was the fundamental charge number I asked for from ONLY the three
numbers I gave you.
That is cheating, pure and simple.
You and Panetta did not use ONLY the three given constants.
You guys still don't get it. The VPP geometry ties all those constants
together.
http://members.aol.com/tnlockyer/Bohramps.gif
Look at the variables list.
Then look at the calculations that get the fundamental charge. (e1)
Notice VPP geometry is doing all the work, with just THREE given constants.
(No h, No electron mass, No explicite charge (e) has to be inserted to get
results.)
Sure the constants are ALL related, as Larry keeps pointing out. But why?
Because VPP is MORE than a model, VPP geometry is the reality behind those
definitions you guys are so proud of.
Jacques Distler
<gold...@charter.net> wrote:
>> Any 5th grader can extract "e" from the above quantities, using nothing
>> fancier than elementary algebra (in fact, he doesn't even need (2)).
>>
>> If you cannot do 5th grade algebra without MathCAD, then there is not
>> much we can do to help you.
>
>Thomas seems to believe that if you use algebra, cancelling names of
>numerical values in an equation, rather than arithmetic on numerical values,
>then
>you are doing dimensional analysis. That is the only thing I can think of
>considering his comment that we had only done dimensional analysis to
>show that he had extracted e from the Bohr magneton.
Ah!
So Thomas not only cannot *do* dimensional analysis, he does not even
know what dimensional analysis *is*.
Thanks for the clarification.
Jacques Distler
<thoma...@aol.com> wrote:
Only if you have not progressed beyond 2nd grade math.
If you had mastered 5th grade math, you would know that
eh/(4pi mc)
------------ x 4pi = e
h/mc
without *ever* needing recourse to the values of "m", "c" or "h". (They
cancel out of the ratio, a concept which, evidently, you never mastered
in elementary school.)
I did cheat, though, in needing the value of "pi".
But then *you*, for reasons only a second grader would understand, need
to know the value of "c" above.
[The larger point (aside from your evident failure to master 5th grade
algebra), is that the numbers (for the Bohr Magneton and Compton
Wavelength) that you bandy about with totemic conviction are NOT
MEASURED QUANTITITIES.
The quoted values are CALCULATED from the known values of "e", "m", "c"
and "h". So extracting the value of one of those *inputs* is merely
getting out something that was already put *in* by hand.]
larry shultis
> >Jacques Distler dis...@golem.ph.utexas.edu
> Wrote in:
> >Message-id: <230420022153409050%dis...@golem.ph.utexas.edu>
> >
> In article <20020423190913...@mb-fh.aol.com>, ThomasL283
> <thoma...@aol.com> wrote:
>
> >>I'll call your bluff, and give you the three numbers VPP used to get the
> >>fundamental charge from the Bohr.
> >>
> >>(1) electron Compton
> >> 2.426310215 E-12 m
> >
> >This is not a measured quantity. It is *defined* to be equal to h/mc
>
> You cheated and brought in h and m, all I gave you was J/T, c and lambda
As Jim showed you e=(4 x pi x mu_e ) / (lambda_e x c)
Insert the numbers and do the arithmetic. Or maybe the (4 x pi) conversion
factor from the unit hbar to the unit h has your dander up?
It is also extremely irrational to demand that derived units not be converted
to the underlying fundamental units used in definining the units. After all
you are using a standard set of fundamental units and derived units when you
write J/T.
Larry
FrediFizzx
| Somewhere in the newsfeed, FrediFizzx said:
| >"Jim Panetta" <pan...@moa.SLAC.Stanford.EDU> wrote in message
| >news:aa4to4$p99$1...@usenet.Stanford.EDU...
| >|
| >| The electron compton wavelength is not a physical constant. It
| >| is constructed from the physical constants h, m_e, and c.
| >|
| >| The Bohr magneton is not a physical constant. It is constructed
| >| from the physical constants e, h and m_e.
| >
| >Jim, we have been telling him this for the last five or so threads now.
He
| >just doesn't get it. I even showed him that his fantasy model geometry
was
| >derivable from these equations by multiplying one side them by x/x.
Maybe
| >that was a mistake on my part because now he thinks that makes his model
| >true.
|
| Freddi, I've been peripherally involved in these discussions for
| *years*. So far, the best quote I've seen on Thomas's "model"
| was from Dan Evens in 1996:
|
| "Absolutely brimming over with wrongability."
Yes, I know you have been involved. I thought I would bring you up to speed
on the latest. Thomas has got himself "brain-locked" on some real *stupid*
stuff now though. Usually, he was a bit more imaginative in the past. That
is a great quote. I also like the "Borg Ship electrons" that Jacques
mentioned in a previous thread.
| In this discussion, the players come and go. As far as I remember,
| Jim Carr has been around about the longest. I was in it in '97-'99,
| then I left for a while. Thomas's stuff has been debunked ad nauseam.
| For giggles, take a look on google groups for "Clockwork Neutrons" in
| sci.physics.particle. It's one of the most entertaining "Proof by
| Vigorous Hand Waving" examples I've ever seen.
I have been in and out since about '98 and sometimes I would try to be on
his side (thru my own ignorance of the subject) but I have actually learned
a lot from the discussions so there is some value to be gained from it.
Looked up "Clockwork Neutrons"--what a blast from the past!
FrediFizzx
FrediFizzx
Thomas, you really should get off this dead end street. Please!
As soon as you say electron Compton wavelength you are saying h/m_e*c. As
soon as you say Bohr magneton, you are saying e*h/4*pi*m_e. It is that
simple. They are *exactly* equal by definition. We can freely substitute
them anytime we wish. The numerical values of those things don't mean jack.
Use the equations only and you will stay out of trouble. Hopefully.
I know you can come up with something better than this stupid stuff to
discuss. Well, maybe not.
Compton wavelength, in how it is used here, is simply the wavelength of an
energy equivalent photon. Nothing more nothing less. It is directly
derivable from h*f = m_e*c^2. All this expression applies to is the photon.
I know you would like your electron to be made of a photon, but there is
absolutely no evidence for that much less extreme difficulty in the
mechanics of the construction. There would have to be *new* physics which
should have been seen by now for your construction to exist. Now, maybe if
your model was a million times smaller, you might have been able to get away
with it. But it is not. So you can't.
FrediFizzx
sch...@gefen.cc.biu.ac.il
:>(2) velocity of light
:> 2.99792458 E8 m/s
:
: This one is measured.
No it is not -- the speed of light is *defined* as 299 792 458 m/s (cf.
last August's _Physics Today_ where they have the Big List o' Fundamental
Constants). The meter was redefined in terms of the speed of light because
it is possible to measure the speed of light and the second to greater
precision than it is to measure a length.
-----
Richard Schultz sch...@mail.biu.ac.il
Department of Chemistry, Bar-Ilan University, Ramat-Gan, Israel
Opinions expressed are mine alone, and not those of Bar-Ilan University
-----
"Logic is a wreath of pretty flowers which smell bad."
Jacques Distler
wrote:
>:>(2) velocity of light
>:> 2.99792458 E8 m/s
>:
>: This one is measured.
>
>No it is not -- the speed of light is *defined* as 299 792 458 m/s (cf.
>last August's _Physics Today_ where they have the Big List o' Fundamental
>Constants). The meter was redefined in terms of the speed of light because
>it is possible to measure the speed of light and the second to greater
>precision than it is to measure a length.
The speed of light is a measured quantity (unlike the electron compton
wavelength or the bohr magneton).
Whether it is used to define the meter or the other way 'round is not
the *point* here. The point is that it is an experimental input, unlike
the latter two quantities which are *calculated* outputs.
larry shultis
"FrediFizzx" <FrediFi...@HaHahotmail.com> wrote in message
As I understand it, the electric field is primary and the magnetic field is
due to motion relative to the electric field (virtual photons). Thomas seems to
make both the electric and the magnetic fields primaries along with reifying
the Poynting vector as an actual existing energy rather than a description of
energy (a relationship not some kind of stuff) transfer. I think you also said
something like "real physical vectors". "Vector" is a mathematical description
and should not be reified. Also Jacques tends to reify the mathematical
description
"wave function" and its metaphoric "collapse". It would be better for all to
stick to mathematical descriptions of "objective reality" as used by Einstein
and others rather than getting into some Platonic shadow-world in which
mathematics is reified.
Larry
larry shultis
"larry shultis" <gold...@charter.net> wrote in message
news:uccep58...@corp.supernews.com...
>
> "ThomasL283" <thoma...@aol.com> wrote in message
> news:20020424001352...@mb-cf.aol.com...
> > >Jacques Distler dis...@golem.ph.utexas.edu
> > Wrote in:
> > >Message-id: <230420022153409050%dis...@golem.ph.utexas.edu>
> > >
> > In article <20020423190913...@mb-fh.aol.com>, ThomasL283
> > <thoma...@aol.com> wrote:
> >
> > >>I'll call your bluff, and give you the three numbers VPP used to get the
> > >>fundamental charge from the Bohr.
> > >>
> > >>(1) electron Compton
> > >> 2.426310215 E-12 m
> > >
> > >This is not a measured quantity. It is *defined* to be equal to h/mc
> >
> > You cheated and brought in h and m, all I gave you was J/T, c and lambda
>
> As Jim showed you e=(4 x pi x mu_e ) / (lambda_e x c)
Sorry for the typo. That should read:
e=(4 x pi x mu_B ) / (lambda_e x c)
Larry
Michael Moroney
>>>I'll call your bluff, and give you the three numbers VPP used to get the
>>>fundamental charge from the Bohr.
What an exercise in uselessness. The Bohr is _defined_ as (e h)/(4 pi Me)
so you need e to get the Bohr in the first place. It is not a measured
value.
>>>(1) electron Compton
>>> 2.426310215 E-12 m
>>
>>This is not a measured quantity. It is *defined* to be equal to h/mc
>You cheated and brought in h and m, all I gave you was J/T, c and lambda
No, _you_ brought in h and m by bring in the Compton length. Which, like
the Bohr, is a defined, not a measured quantity.
>>>(3) Bohr magneton.
>>> 9.27400899 E-24 J/T
>>
>>Again, this is not a measured quantity. It is *defined* to be equal to
>>eh/(4pi mc)
>>
>You cheated and have to use (e) (h) and (mass) in the definition. I wanted you
>to derive the value of (e) from only the three numbers, (c), (lambda) and (UB)
>constants.
Again, it is you who bring in e h and Me, by bringing in the Bohr. It
is defined as above.
>My original requirement was the only thing known were those three constants,
>i.e. the electron's Compton wavelength, the velocity of light, and the Bohr,
>to get the charge (e).
>You fellows brought in:
>(4) plancks constant (h)
>(5) the electron's mass (m)
>(6) and the electron's charge (e) explicitly.
Again, it is YOU who bring in h me and e, by bringing in two constants
_defined_ in terms of them. You are only pretending that you are not.
Nobody is being fooled.
>That is cheating, pure and simple.
No, they're in the definitions.
> Notice VPP geometry is doing all the work, with just THREE given constants.
>(No h, No electron mass, No explicite charge (e) has to be inserted to get
>results.)
Because e is in the definition of the Bohr, extracting it from the Bohr
is nothing more than algebra of x=x "proofs".
-Mike
ThomasL283
>Date: 4/23/2002 9:31 PM Pacific Daylight Time
>Message-id: <230420022331422161%dis...@golem.ph.utexas.edu>
>
>
>In article <20020424001352...@mb-cf.aol.com>, ThomasL283
><thoma...@aol.com> wrote:
>
>>Nope:
>>
>>My original requirement was the only thing known were those three constants,
>>i.e. the electron's Compton wavelength, the velocity of light, and the
>Bohr,
>>to get the charge (e).
>>
>>You fellows brought in:
>>
>>(4) plancks constant (h)
>>
>>(5) the electron's mass (m)
>>
>>(6) and the electron's charge (e) explicitly.
>>
>>And item (6) was the fundamental charge number I asked for from ONLY the
>three
>>numbers I gave you.
>>
>If you had mastered 5th grade math, you would know that
>
>
> eh/(4pi mc)
> ------------ x 4pi = e
> h/mc
>
>without *ever* needing recourse to the values of "m", "c" or "h". (They
>cancel out of the ratio, a concept which, evidently, you never mastered
>in elementary school.)
That has been my point. But, once you ring the bell YOU can't "unring it" by
retrieving those units of (h) (m_e) or (c) from e= (A s) without VPP.
What I have been trying to show is the particle geometry makes those
constants play out in their ratios.
And the constants magnitudes are given by the dimensions of the model
geometry.
http://members.aol.com/tnlockyer/Bohramps.gif
If you have an alternate reason, let's hear it.
>
>[The larger point (aside from your evident failure to master 5th grade
>algebra), is that the numbers (for the Bohr Magneton and Compton
>Wavelength) that you bandy about with totemic conviction are NOT
>MEASURED QUANTITITIES.
>
Yes, that you are correct, the Bohr is a calculated value.
But that fact does not negate the VPP geometric ratios that allow the Bohr
magneton (m^2 A) to calculate (e) from the geometry.
>The quoted values are CALCULATED from the known values of "e", "m", "c"
>and "h". So extracting the value of one of those *inputs* is merely
>getting out something that was already put *in* by hand.]
Of course. I say as much in both of the books.
BTW. The Bohr is also given by:
UB= 1/2 (e c (lambda/2pi))
(lambda/2pi) happens to be the edge lengths of the VPP electron model.
But to get the constants back from the Bohr as (m^2 A) WITHOUT inserting the
constants back *in* by hand cannot be done by YOUR methods.
BTW, the VPP model can calculate the fine structure constant from the Bohr.
Are you saying the fine structure is put *in* the Bohr by hand ?
http://members.aol.com/tnlockyer/VPPfine.gif
Please try to make your points by reason not innuendo.
Jacques Distler
<thoma...@aol.com> wrote:
>>If you had mastered 5th grade math, you would know that
>>
>>
>> eh/(4pi mc)
>> ------------ x 4pi = e
>> h/mc
>>
>>without *ever* needing recourse to the values of "m", "c" or "h". (They
>>cancel out of the ratio, a concept which, evidently, you never mastered
>>in elementary school.)
>
>That has been my point. But, once you ring the bell YOU can't "unring it" by
>retrieving those units of (h) (m_e) or (c) from e= (A s) without VPP.
That has *not* been your point. You have been claiming that your model
allows you to *compute* e.
That is utter HOKUM. The ratio of the bohr magneton to the compton
wavelength is *by definition of those quantities* equal to e/(4pi).
You have brought nothing, zippo, nada to the table that was not there
already in the definitions of these quantities.
This is another stupid "x=x" proof, dressed up with some geometrical
mumbo jumbo and a frightening inability to keep straight what is input
and what is calculated output from your "model".
>>[The larger point (aside from your evident failure to master 5th grade
>>algebra), is that the numbers (for the Bohr Magneton and Compton
>>Wavelength) that you bandy about with totemic conviction are NOT
>>MEASURED QUANTITITIES.
>
>Yes, that you are correct, the Bohr is a calculated value.
>
>But that fact does not negate the VPP geometric ratios that allow the Bohr
>magneton (m^2 A) to calculate (e) from the geometry.
There *is* no geometry.
Your spinning cubes are a figment of your imagination.
All you have are stupid "x=x" proofs and an inability to do simple
dimensional analysis (the volt, I see, is still a unit of energy for
you).
>
>>The quoted values are CALCULATED from the known values of "e", "m", "c"
>>and "h". So extracting the value of one of those *inputs* is merely
>>getting out something that was already put *in* by hand.]
>
>Of course. I say as much in both of the books.
>
>BTW. The Bohr is also given by:
>
>UB= 1/2 (e c (lambda/2pi))
Can't even get that one straight, can you?
>(lambda/2pi) happens to be the edge lengths of the VPP electron model.
A mere million times larger than the experimental upper limit on the
size of the electron.
'nuff said.
Tnlockyer
>Date: 4/25/2002 11:33 AM Pacific Daylight Time
>Message-id: <250420021333277736%dis...@golem.ph.utexas.edu>
>In article <20020425140453...@mb-cf.aol.com>, ThomasL283
><thoma...@aol.com> wrote:
>>But that fact does not negate the VPP geometric ratios that allow the Bohr
>>magneton (m^2 A) to calculate (e) from the geometry.
>
>There *is* no geometry.
>
J D, you simply refuse to really read what my computer tells me.
I find that the VPP model volume and VPP current loop areas allow calcualting
the Bohr without your charge you think is has to be used.
http://members.aol.com/tnlockyer/Bohr.gif
You address your arguments to the math, point by point, or shut up.
-Tom:
Thomas Lockyer (75 and retired) See "Vector Particles and Nuclear Models"
0963154680 at http://www.amazon.com
"When you can measure what you are speaking about, and express it in numbers,
Michael Moroney
>>>But that fact does not negate the VPP geometric ratios that allow the Bohr
>>>magneton (m^2 A) to calculate (e) from the geometry.
>>
>>There *is* no geometry.
>>
>J D, you simply refuse to really read what my computer tells me.
>I find that the VPP model volume and VPP current loop areas allow calcualting
>the Bohr without your charge you think is has to be used.
There is no "calculating" the Bohr. It is *defined* to be (e h)/(4 pi Me).
Your "calculation" of the Bohr is just a very convoluted method of coming
up with that same equation. In other words, yet another "x=x" proof.
-Mike
larry shultis
What makes up the circulating current in your model? As I indicated elsewhere,
current and charge cannot be separated. A current is made up of charge, but
you seem to want to create charge from a current, so what is the current?
Larry
Jacques Distler
<tnlo...@aol.com> wrote:
>J D, you simply refuse to really read what my computer tells me.
>
>I find that the VPP model volume and VPP current loop areas allow calcualting
>the Bohr without your charge you think is has to be used.
>
>http://members.aol.com/tnlockyer/Bohr.gif
>
>You address your arguments to the math, point by point, or shut up.
Been there, done that. (As any 5th grader could easily follow.)
In all your voluminous ramblings, you have yet to come up with a single
correct "prediction" that does not reduce to a stupid "x=x" proof.
And yet you manage, without so much as batting an eyelash, to pass off
mountains of incorrect predictions (like an electron a million times
larger that the experimental upper limit on its size).
You have ceased to be entertaining, Thomas.
Have you simply reached the limit of your repertoire?
Jim Panetta
>i.e. the electron's Compton wavelength, the velocity of light, and the Bohr,
>to get the charge (e).
>
>You fellows brought in:
>
>(4) plancks constant (h)
>
>(5) the electron's mass (m)
>
>(6) and the electron's charge (e) explicitly.
Wrong:
e = mu_B * 4 pi / ( lambda_e c )
>You and Panetta did not use ONLY the three given constants.
One can always use any normal mathematical constant. You put them
in for no apparrent reason, why can't we?
>Sure the constants are ALL related, as Larry keeps pointing out. But why?
Because they're *defined* that way. The definitions are useful
to simplify certain equations, that's all. (For example, the
calculation of energy levels in spin-orbit interaction.)
BTW, the Bohr magneton is *not* the magnetic moment of the electron.
That quantity is measured to be about 0.1% larger. The accuracy on
this measurement is to four parts in 10^12. From your model, the
magnetic moment of the electron is exactly equal to the Bohr magneton.
This is *wrong*.
>Because VPP is MORE than a model, VPP geometry is the reality behind those
>definitions you guys are so proud of.
VPP is the product of handwaving and algebraic sleight-of-hand. It
has no more basis in reality than miracle weight loss cures and
supply side economics.