[General] Quantisation of classical electromagnetism
John Duffield
johnduffield at btconnect.com
Tue May 12 13:37:47 PDT 2015
Martin:
IMHO the electron in a spindle-sphere
<http://www.antiprism.com/album/860_tori/index.html> S orbital is only a
little larger than a free electron. It exists as a 511keV standing wave when
it’s a free electron. When it’s a standing wave in an atomic orbital
<http://en.wikipedia.org/wiki/Atomic_orbital#Electron_properties> it’s a
mere 13.6ev less energy. So its Compton wavelength has only increased a
little. IMHO you should remember the photon in the box for this
<http://www.tardyon.de/mirror/hooft/hooft.htm> . Try to imagine jiggling the
box around just so, such that the photon wavelength increases a little
because the box is effectively a little bit bigger. Or imagine the electron
around your waist, and you’re playing hula-hoop.
Regards
John D
From: General
[mailto:general-bounces+johnduffield=btconnect.com at lists.natureoflightandpar
ticles.org] On Behalf Of Mark, Martin van der
Sent: 12 May 2015 08:32
To: Nature of Light and Particles - General Discussion
Subject: Re: [General] Quantisation of classical electromagnetism
Hi Richard,
You raise an important question. I cannot answer it yet, as I will try to
explain.
The structure of an electron inside an atom or in the conduction band of a
metal or semi-conductor is different from that of the free electron. It
manages to behave as it were much larger than the free electron, it can be
measured that the charge is distributed over the whole atom or even the
whole conductor.
Does the electron really “disintegrate”? I do not believe so. But believing
is quite different from knowing. What would have to happen is that the
environment makes up for the binding forces inside the electron. I think
those forces are far too weak. Moreover, the speed of light would come in
the way to keep coherence over larger distances than the size of the free
electron (at the Compton wavelength scale)
A proper theory must solve this. For the moment I think that the electron
hardly changes its structure inside an atom or conductor and that only the
external fiels have just the de Broglie wavelength that fits their imposed
quantum state, as in the pilot wave picture.
Cheers, Martin
Dr. Martin B. van der Mark
Principal Scientist, Minimally Invasive Healthcare
Philips Research Europe - Eindhoven
High Tech Campus, Building 34 (WB2.025)
Prof. Holstlaan 4
5656 AE Eindhoven, The Netherlands
Tel: +31 40 2747548
From: General
[mailto:general-bounces+martin.van.der.mark=philips.com at lists.natureoflighta
ndparticles.org] On Behalf Of Richard Gauthier
Sent: dinsdag 12 mei 2015 8:23
To: Nature of Light and Particles - General Discussion
Subject: Re: [General] Quantisation of classical electromagnetism
Andrew and Martin,
I think it would be a good challenge for anyone with a single-looped or
double-looped photon model of an electron to model their electron in the 1s
atomic state of hydrogen (where n=1, l=0, ml=0 and ms = + or - 1/2 hbar)
where the electron has zero hbar atomic angular momentum even though it has
internal electron spin 1/2 hbar. I model the electron here as oscillating
back and forth linearly through the center of the atom as a charged photon
with a helical trajectory of variable pitch and radius, with a total energy
of E=mc^2 -13.6 eV and a maximum kinetic energy when the helically
circulating charged photon passes the nucleus, generating a variable de
Broglie wavelength along its trajectory (because its longitudinal momentum
is changing as it oscillates in the atom) and making one complete de Broglie
path per oscillation. The most probable position of the charged photon (the
electron) to be detected is at 1 Bohr radius ao (as predicted by QM for the
hydrogen atom) because the charged photon obeys the Schrodinger equation (in
the non-relativistic approximation). If the 1s electron (charged photon)
absorbs an uncharged photon of energy 13.6 eV, the hydrogen atom is ionized
with the charged photon now having energy E=mc^2 .
Richard
On May 11, 2015, at 1:30 PM, Mark, Martin van der
<martin.van.der.mark at philips.com <mailto:martin.van.der.mark at philips.com> >
wrote:
Dear Andrew, I have been away for a few days.
In your previous replies you tried to challenge me to be precise, and have
questioned the correctness of my statements. That is very good.
But unfortunately you have put me in a position where I, in turn, have to
correct you on all the things you say that are half baked or wrong. It is
difficult to remain very polite since most of your brown replies, need
“attention”. So brace, but remember that I only put in the effort because I
respect you and because I think that your opinion matters.
I hope no one is color blind, yet another color will be used: purple! (just
in case some haze troubles the mind
)
So: purple (Martin) responds to brown (Andrew) responds to green (Martin)
responds to red (Andrew)
Dr. Martin B. van der Mark
Principal Scientist, Minimally Invasive Healthcare
Philips Research Europe - Eindhoven
High Tech Campus, Building 34 (WB2.025)
Prof. Holstlaan 4
5656 AE Eindhoven, The Netherlands
Tel: +31 40 2747548
From: General [
<mailto:general-bounces+martin.van.der.mark=philips.com at lists.natureoflighta
ndparticles.org>
mailto:general-bounces+martin.van.der.mark=philips.com at lists.natureoflightan
dparticles.org]On Behalf Of Andrew Meulenberg
Sent: donderdag 7 mei 2015 7:40
To: Nature of Light and Particles - General Discussion
Subject: Re: [General] Quantisation of classical electromagnetism
Dear Martin,
It is great communicating with someone who has also thought about the issue.
My comments are sometimes too cryptic because I assume that you would have
come to the same conclusions. Let me try (in brown) to identify some of the
differences below.
On Thu, May 7, 2015 at 3:50 AM, Mark, Martin van der <
<mailto:martin.van.der.mark at philips.com> martin.van.der.mark at philips.com>
wrote:
Andrew, thanks, please see below, in green
Dr. Martin B. van der Mark
Principal Scientist, Minimally Invasive Healthcare
From: General [mailto: <mailto:general-bounces%2Bmartin.van.der.mark>
general-bounces+martin.van.der.mark=
<mailto:philips.com at lists.natureoflightandparticles.org>
philips.com at lists.natureoflightandparticles.org]On Behalf Of Andrew
Meulenberg
Sent: woensdag 6 mei 2015 19:46
To: Nature of Light and Particles - General Discussion
Subject: Re: [General] Quantisation of classical electromagnetism
Dear John W, Martin, et al.,
I don't think that a week together in San Diego would be enough to transfer
the information that we all need to share. And, I will probably miss even
that. I am already learning so much and have so much to contribute that I
feel frustrated that I have to divide my time.
Just this morning (my time), I changed my idea of the electron radius. After
seeing it expressed many times by various members of this group as 1/2 the
Compton radius (and considering that to be wrong), it finally hit me, when
reading it again, that, even within my own model, I had been wrong and this
smaller radius is probably correct.
some comments below:
On Wed, May 6, 2015 at 3:28 PM, Mark, Martin van der <
<mailto:martin.van.der.mark at philips.com> martin.van.der.mark at philips.com>
wrote:
Andrew, John, all
John W is quite right as well, just a small remark on the hydrogen atom.
By the virial theorem, for a 1/r potential, potential energy is minus two
times the kinetic energy and kinetic energy is equal to the binding energy
(13.6 eV in the ground state).
For the structure of the atom there are three conditions, one of
electromagnetic, and two of inertial nature.
1) The coulomb potential runs to minus infinity, that is very deep. It comes
from the charge of proton and electron.
2) Then the centrifugal force (depends on mass of proton and electron) must
balance the Coulomb force, this could have been in a continuum of orbits if
the electron and proton were just particles (without a wave nature) (see
gravitation and solar system for an exact analogy),
3) The mass of proton and electron set the scale of the de Broglie
wavelength (which, incidentally, is exactly the same for proton and electron
in the bound state), and hence the bound state has a finite size, 0.1 nm
diameter for the ground state. The particle’s waves must interfere
constructively within the boundary conditions: quantized energy levels
appear.
Cheers, Martin
I also have three basic conditions:
* The QM description of the mechanical resonance of a body confined in
a potential well. The reason for this resonance is not the interference with
the nucleus (which does not appear in the fundamental equations). There is a
simple physical and mathematical basis that is taught in 1st year calculus.
* The classical description of the orbiting electron creates an EM
field that is evolving into a photon as the electron decays to a deeper
level. The resonance between the electron and emitted-photon frequencies,
along with the virial theorem and conservation of energy and ang. mom.,
determine the allowed energy levels. The fact that these levels agree with
the mechanical levels gives a double resonance.
I do not really understand what you mean by the double resonance.
The levels identified with the classical description agree with those of
the mechanical (QM) system.
To be precise: the mechanical part equals the mechanical part. No
information in that at all, it is not a coincidence it is an incidence.
* the ground state is established by the requirement of a photon to
have an angular momentum of hbar.
Why is this fundamental? It depends on the system you are looking at.
Circular orbits are a confusing thing o look at too, if you want to look at
angular momentum. The real hydrogen atom has NO angular momentum (spin=0) in
the ground state, contrary to the Bohr model of it!!!!!
[Exactly, therefore it cannot radiate a photon except to a system w ang.
mom. = hbar. No levels below gnd state have that value.]
Wrong answer to the given question. The ground state has IN THE FIRST PLACE
nothing to do with the emitted energy part, but with what remains. What
remains must be a mode of the system, a quantum state they call this in
quantum mechanics. It means that a single wavelength, 3D resonance must be
maintained. For the hydrogen atom (or for that matter any spherical system)
must have simultaneous solutions for r, theta and phi, such that at least
one has a single wave in it. It appears to be the radial solution, a
breather (not the Bohr type phi solution with one wavelength on the
circumference).
To get from a higher state into this state the right combination of energy,
momentum and angular momentum must be emitted, hence the selection rules as
the are, and that is what your answer is about.
The groundstate of the atom is the groundstate because it is the lowest
energy state, with just the fundamental tone (one wavelength) fitting to the
boundary conditions.
There is no reason given that the wavelength cannot fit 2 cycles rather than
one. Is it more difficult than having n wavelengths in a single cycle? (ref
Lissajou figures) This is the basis of Mill's hydrino states. The limiting
factor is the photon.
See previous answer and understand that the reason is implicitly given there
already. A mode of a resonator or waveguide, a quantum state, these ar the
same sort of things. They have Q or quality factor that tells you how broad
or narrow the resonance is and how long it lives. For a very long live time
such as the ground state hydrogen atom, the Q is very very high, and the
level is very precisely defined. The wave must therefore fit very very very
very precisely on the mode. Not a factor of two difference, no, minus dozens
of orders of magnitude precise!
The one wavelength holds for the complete atom, electron and proton: the
electron is light and moving fast, the proton id heavy and slow, but both
have the same momentum! Hence they have the same de Broglie wavelength
The solution assuming an infinite mass nucleus (hence no deBroglie
wavelength & no resonance) still produces discrete levels. Therefore that
issue cannot be causal.
Bullshit, sorry, but here you should really know better since:
1) I have told you already you are referring to the BORN APPROXIMATION
2) It is in every basic textbook, second year physics
3) The so-called reduced mass gives a slightly different set of levels,
the correct levels. Remember the Rydbergconstant? And its slightly different
numbers?
Just do your homework, and go to Wikipedia, that will be good enough.
Here another remark regarding your earlier question, whether I was talking
physics or mathematics. (see next question below)
Here you see I get the physics right because I think physics. Then I get the
numbers right because I understand the approximations, the calculus and the
details. That can only be if you know how to the mathematics right.
Sorry I just hate mediocrity.
Looking at your conditions produced other thoughts.
1. The statement that "the coulomb potential runs to minus infinity" is
a mathematician, not a physicist talking.
True. The Coulomb potential is a mathematical concept that models reality
quite perfectly. Mathematics is the language of physics. Further, the
electron has an almost 1/r dependent potential still at TeV collision
energies, this is why people say it has point-like behavior.
And, it will be considered valid until experimental evidence show otherwise.
Then, 60 years of mathematical 'proofs' will immediately disappear. Do you
know of any nuclear physicist who would consider the nuclear Coulomb
potential to be a singularity? Even before the quark model became popular?
What is reality, a singular potential that contains all of the energy in the
universe, or a presently measured finite charge density of the proton and
neutron? Feynman jested that the whole universe consisted of a single
electron oscillating back and forth in time. If it was singular and
contained all of the energy in the universe, maybe Feynman's jest was
correct.
The Coulomb potential is BY DEFINITION a 1/r potential associated with a
POINT CHARGE.
It therefore runs, by definition, to minus infinity.
Real charges, like the proton or a charged party balloon, rubbed-up on a
cat, have a cut-off at their respective charge radii of 0.87 fm and 10 cm:
at smaller size the potential may be or is quite constant. There is always a
reason in the physics of things that will avoid infinities naturally.
I know you want to make that point, and I fully agree. But please do not
misuse the argument or definition of things.
Further your remark is irrelevant because the binding energy of the ground
state of hydrogen (or any atom) is alpha^2 times smaller than the electron’s
classical radius (which incidentally is close to the charge radius of the
proton)
Do you consider the proton charge radius to be its field-energy density or
the major extent of itsCoulomb potential (e.g., the electron Compton
radius)? We haven't defined charge yet have we? [Am I being the
mathematician now for insisting on a valid definition?]
It is the proton’s START of the Coulomb potential! Look it Up in a book on
high energy physics if you really want to know exactly what they mean by it.
The potential energy PE must come from the energies, as expressed by the
mass and charge, of proton and electron. Since the largest energy is the
mass, the PE is limited to a GeV. Therefore, the electrical potential cannot
exceed this value.
Yes it can exceed its RESTmass , and will be precisely gamma m0, see above,
not exceeding its relativistic mass, of course.
In some frame of reference, the relativistic mass is infinite. However, the
charge field changes in that frame also. In the rest frame, PE is finite and
1/r must be limited.
This, like relativity, makes a big difference in some fields of physics.
2. The source of the wave nature of the electron is never defined in
QM. Is it the 'hidden variable"?
No it is not, but almost, what the real structure is, well we have our
ideas,,,, The hidden variable has to do with phase coherence in the
measurement process. Will explain that over a glass of beer, it is worth a
good set of papers.
It can be defined classically, if spin is a real angular momentum, not just
a Q#, and relativity is more than just a mind game.
Quantum spin is angular momentum, but not that of a rigid body. For spin ½
you need something like a fluid that is circulating in 2 directions at the
same time, like a spinning, rotating , twisting torus. Think of a smoke ring
with a twist and rotating like a wheel
More beer required here too
I think of the surface of a ball of yarn! Only photons can pass thru each
other (or itself), thus the electron is more than a fluid. It is circulating
in all directions. Uniquely, it can have an infinity of angular momenta.
That is why it can have spin 1/2 in any direction you wish to chose. That
question puzzled since college days; but, I was too 'young' to properly
question the 'cant' being fed us. I don't think that the professors, bright
as they were, could have understood my question, much less answered it.
(What about a spin axis along the time direction?)
Who are you trying to convince here? Surely not me. Look at the 1997
Williamson van der Mark paper and find all you try to say. Incidentally the
word fluid refers to something you apparently do not know: It is well known
that electromagnetic fields behave like a so-called INVISCID FLUID.
Other fluids are useful too: more beer required ;-) I do agree that spin ½
is not easy, I know the problems and the salient details. It is one of the
key things in physics to understand!
3. I do not believe that looking at the system in center-of-mass
(momentum) coordinates introduces quantized levels in two dimensions. Can
only adding 1 or 2 more dimensions produce the fixed levels? Can you
describe how such levels might occur? If you define a bell as quantized,
then the levels can be quantized. However, they still can have a continuum
of values unless the structure is fixed. I have to admit that this is like
my condition 1 and both are weak w/o a better reason for discrete values.
The 'standing wave' concept is attractive, but misleading.
More below:
From: General [mailto: <mailto:general-bounces%2Bmartin.van.der.mark>
general-bounces+martin.van.der.mark=
<mailto:philips.com at lists.natureoflightandparticles.org>
philips.com at lists.natureoflightandparticles.org] On Behalf Of John
Williamson
Sent: woensdag 6 mei 2015 11:12
To: Nature of Light and Particles - General Discussion
Cc: Nick Bailey; Kyran Williamson; Michael Wright; Manohar .; Ariane Mandray
Subject: Re: [General] Quantisation of classical electromagnetism
Hihi,
A lot of questions there Andrew.
All quantised means is "countable".
QM is certainly putting a lot more weight to the word than that. Pointing
out resonances has a physical meaning that can be useful.
Yes there are exceptions. Mostly exceptions! The quantised electron charge
comes, for me, from an interaction rate. Hence the reason all charges in
contact have the same value.
I would say that this looks at effect, not cause or definition.
Other quantum numbers may just be an intrinsic sign- such as the lepton
number difference between the positron and the electron. Quantised states in
atoms and quantum wells are resonant states, indeed. In the FQHE these are
bound quasi-particle-flux-quantum states. These are more musical ratios,
than integer numbers. Quantised conductance, for example, is simply a
rate-per-single-electron. The popular press and Wikipedia tends to sweep all
the unknowns into one big unknown. That thing which cannot be known - the
great UNCERTAINTY! Assigning a quantum number to something is tantamount to
putting all your lack of understanding into a single number. Too much of
this kind of shit passes as understanding!
Agreed!
The ground state of the Hydrogen atom is that energy where potential=
kinetic, and the de Broglie wavelength of the electron equals the de Broglie
wavelength of the proton. A single wavelength with periodic boundary
conditions - for both! What a beautiful resonance! Simple, singing resonance
- with no dissipation. Physics tries indeed to mystify this, but it is
really a simple congruence. Engineers know better!
For a 1/r potential the virial thm states that KE = PE/2. You and Martin
agree about the relationship between proton and electron as being important.
Is this a claim of QM or something that you both simply agreed on? The basic
Schrodinger equation assumes an infinite proton mass.
This has nothing to do with the Schroedinger equation but with the Born
approximation, which is not necessary to make, the proton mass is finite,
and it can be taken into account by introducing the reduced mass: m_p x
m_e/(m_p + m_e)
Oh and KE = -PE/2, PE is negative!!!
Introducing the reduced mass (and a nuclear deBroglie wavelength for
'resonance') changes the values, but not the nature, of the discrete energy
levels. The nucleus travels~2000 orbits before it completes a single
deBroglie wavelength. How come the electron is only allowed a maximum of a
single cycle?
I have given you all the clues, now please do the work! The nucleus is slow
and heavy, the electron light and fast, oscillating about their common
centre of mass with EXACTLY the same momentum and hence the same de Broglie
wavelength.
There is no nuclear wavelength, yet the solution has discrete levels. You
are correct about a resonance between two wavelengths (frequencies). But I
think that they are between the electron and EM wave becoming a photon.
Indeed the Coulomb potential goes way down (as you argue so beautifully in
your paper). Shorter lengths, however, are less than one wavelength and
hence, though they could be resonant, actually at a higher energy, through
interference. The one wavelength state is the ground state. For this state
the Coulomb field, cancelled outside the Bohr radius corresponds exactly to
the 13.6 eV binding energy of the Hydrogen atom. All very simple and very
beautiful!
What prevents the 1/2 wavelength state from existing and being occupied? (Or
for 1/n, with a single wavelength being completed in n orbits.)
If you do that, it simply interferes away, then that energy has to go
somewhere, it cannot be destroyed, so it will be radiated. This is why atoms
radiate while an electron changes “orbit” : temporarily there is no fit, but
energy must be conserved.
Are you assuming that the electron is a wave and not localized? That its
wave function, distributed around the atom and extended to 2 orbits per
cycle, would cancel because of phase reversal? Then what about 3 (or any odd
integer) orbits?
The electron is a wave and a particle at the same time, right?!
More below:
Martin is, as usual, right in (pretty much) everything he says. Especially
in that it is very important!
Regards, John W.
_____
From: General [general-bounces+john.williamson=
<mailto:glasgow.ac.uk at lists.natureoflightandparticles.org>
glasgow.ac.uk at lists.natureoflightandparticles.org] on behalf of Mark, Martin
van der [ <mailto:martin.van.der.mark at philips.com>
martin.van.der.mark at philips.com]
Sent: Wednesday, May 06, 2015 8:48 AM
To: Nature of Light and Particles - General Discussion
Cc: Nick Bailey; Kyran Williamson; Michael Wright; Manohar .; Ariane Mandray
Subject: Re: [General] Quantisation of classical electromagnetism
Dear Andrew,
I have good answers to most of your questions, but have no time right now to
write them down,
we must come back to this, it is very important indeed.
In any case it comes down to the following:
* Quantization comes from any wave equation with imposed boundary
conditions. [if you can establish standing waves?] stationary waves may do
already
* Uncertainty is no more than what the Fourier limit tells you.
[agreed]
* Copenhagen interpretation is Copenhagen mystification: although it
is not very wrong at the simple level, it takes away any possibility for
improvement by dogma.[agreed]
* Wave/particle dualism is the consequence of special relativity,
see Louis de Broglie. [do you have a particular reference? I have not seen
this statement before.] Thesis of Louis de Broglie, more beer would also
help. Niels Bohr and his gang were successful enough to make people forget
about this or so, it is a mystery why it has not become common knowledge
among physicists.
I think that I had heard that before, but not really registered on it. I had
thought of relativity as applied to the deBroglie wavelength rather than
being fundamental to it. I'm finding that my youthful disinterest in the
history of physics is extracting a penalty now.
It is not your fault, it is mentioned rarely. But since I have, now you
know.
I don't have his Thesis handy; but, in his "Theory of Quanta," he does
provide support for your earlier statement about resonance between the
nucleus and electron.
"This is exactly BOHR’s formula that he deduced from the theorem mentioned
above
and which again can be regarded as a phase wave resonance condition for an
electron in
orbit about a proton."
If this is also considered resonance, rather than just strict mechanics,
then the energy levels have 3 resonances in coincidence.
Andrew
Dear Andrew I hope to have cleared up a few things, surely we should talk
about them some more,
Very best regards, Martin
_____
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