<div dir="ltr"><div>Dear Martin,<br><br></div>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.<br><div class="gmail_extra"><br><div class="gmail_quote">On Thu, May 7, 2015 at 3:50 AM, Mark, Martin van der <span dir="ltr"><<a href="mailto:martin.van.der.mark@philips.com" target="_blank">martin.van.der.mark@philips.com</a>></span> wrote:<br><blockquote class="gmail_quote" style="margin:0px 0px 0px 0.8ex;border-left:1px solid rgb(204,204,204);padding-left:1ex">
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<p class="MsoNormal"><span style="font-size:11pt;font-family:"Calibri","sans-serif";color:rgb(119,147,60)">Andrew, thanks, please see below, in green</span></p><span style="font-size:11pt;font-family:"Calibri","sans-serif";color:rgb(31,73,125)"></span><span class=""><p class="MsoNormal"><span style="font-size:11pt;font-family:"Calibri","sans-serif";color:rgb(31,73,125)"> </span></p>
<p class="MsoNormal"><span style="font-size:10pt;font-family:"Arial","sans-serif";color:navy" lang="DE">Dr. Martin B. van der Mark</span><span style="color:navy" lang="DE"></span></p>
<p class="MsoNormal"><span style="font-size:10pt;font-family:"Arial","sans-serif";color:navy">Principal Scientist, Minimally Invasive Healthcare</span><span style="font-size:11pt;font-family:"Calibri","sans-serif";color:navy"></span></p>
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</span><p class="MsoNormal"><font size="1"><b><span style="font-family:"Tahoma","sans-serif"">From:</span></b><span style="font-family:"Tahoma","sans-serif""> General [mailto:<a href="mailto:general-bounces%2Bmartin.van.der.mark" target="_blank">general-bounces+martin.van.der.mark</a>=<a href="mailto:philips.com@lists.natureoflightandparticles.org" target="_blank">philips.com@lists.natureoflightandparticles.org</a>]
<b>On Behalf Of </b>Andrew Meulenberg<br>
<b>Sent:</b> woensdag 6 mei 2015 19:46<span class=""><br>
<b>To:</b> Nature of Light and Particles - General Discussion<br>
<b>Subject:</b> Re: [General] Quantisation of classical electromagnetism</span></span></font></p><font size="1">
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<p class="MsoNormal" style="margin-bottom:12pt"><font size="1"><span style="color:red">Dear John W, Martin, et al.,</span></font></p>
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<p class="MsoNormal" style="margin-bottom:12pt"><span style="color:red">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.</span></p>
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<p class="MsoNormal" style="margin-bottom:12pt"><span style="color:red">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,
<u>I </u>had been wrong and this smaller radius is probably correct.</span></p>
</span></font></div><p class="MsoNormal"><font size="1"><span style="color:red">some comments below:</span></font></p><font size="1">
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</font><p class="MsoNormal"><font size="1">On Wed, May 6, 2015 at 3:28 PM, Mark, Martin van der <</font><a href="mailto:martin.van.der.mark@philips.com" target="_blank">martin.van.der.mark@philips.com</a><font size="1">> wrote:</font></p><font size="1">
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</font><p class="MsoNormal"><span style="font-family:"Calibri","sans-serif";color:rgb(31,73,125)">Andrew, John, all</span></p><font size="1">
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</font><p class="MsoNormal"><span style="font-family:"Calibri","sans-serif";color:rgb(31,73,125)">John W is quite right as well, just a small remark on the hydrogen atom.</span></p><font size="1">
</font><p class="MsoNormal"><span style="font-family:"Calibri","sans-serif";color:rgb(31,73,125)">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).</span></p><font size="1">
</font><p class="MsoNormal"><span style="font-family:"Calibri","sans-serif";color:rgb(31,73,125)">For the structure of the atom there are three conditions, one of electromagnetic, and two of inertial
nature.</span></p><font size="1">
</font><p class="MsoNormal"><span style="font-family:"Calibri","sans-serif";color:rgb(31,73,125)">1) The coulomb potential runs to minus infinity, that is very deep. It comes from the charge of proton
and electron.</span></p><font size="1">
</font><p class="MsoNormal"><span style="font-family:"Calibri","sans-serif";color:rgb(31,73,125)">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),</span></p><font size="1">
</font><p class="MsoNormal"><span style="font-family:"Calibri","sans-serif";color:rgb(31,73,125)">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.</span></p><font size="1">
</font><p class="MsoNormal"><font size="1"><span style="font-family:"Calibri","sans-serif";color:rgb(31,73,125)">Cheers, Martin</span></font></p>
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<p class="MsoNormal"><span style="color:red">I also have three basic conditions:</span><u></u><u></u></p>
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<span style="color:red">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.</span><u></u><u></u></li><li class="MsoNormal">
<span style="color:red">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.</span><u></u><u></u></li></ul>
</span><p class="MsoNormal"><span style="font-size:11pt;font-family:"Calibri","sans-serif";color:rgb(119,147,60)">I do not really understand
what you mean by the double resonance. </span></p></div></div></div></div></div></div></div></div></div></div></blockquote><div><span style="color:rgb(180,95,6)"><br> <span style="font-size:11pt;font-family:"Calibri","sans-serif"">The levels identified with the classical description agree with those of the mechanical (QM) system.</span></span><span style="font-size:11pt;font-family:"Calibri","sans-serif";color:rgb(119,147,60)"></span></div><blockquote class="gmail_quote" style="margin:0px 0px 0px 0.8ex;border-left:1px solid rgb(204,204,204);padding-left:1ex"><div link="blue" vlink="purple" lang="EN-US"><div><div><div><div><div><div><div><div><div><span class="">
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<span style="color:red">the ground state is established by the requirement of a photon to have an angular momentum of hbar.</span><u></u><u></u></li></ul>
</span><p class="MsoNormal"><span style="font-size:11pt;font-family:"Calibri","sans-serif";color:rgb(119,147,60)">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!!!!! <br></span></p></div></div></div></div></div></div></div></div></div></div></blockquote><div><br><span style="color:rgb(180,95,6)"><span style="font-size:11pt;font-family:"Calibri","sans-serif"">[Exactly, therefore it cannot radiate a photon except to a system w ang. mom. = hbar. No levels below gnd state have that value.]</span></span><br> </div><blockquote class="gmail_quote" style="margin:0px 0px 0px 0.8ex;border-left:1px solid rgb(204,204,204);padding-left:1ex"><div link="blue" vlink="purple" lang="EN-US"><div><div><div><div><div><div><div><div><div><p class="MsoNormal"><span style="font-size:11pt;font-family:"Calibri","sans-serif";color:rgb(119,147,60)"></span></p>
<p class="MsoNormal"><span style="font-size:11pt;font-family:"Calibri","sans-serif";color:rgb(119,147,60)">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.</span></p></div></div></div></div></div></div></div></div></div></div></blockquote><div><span style="color:rgb(180,95,6)"><br></span></div><div><span style="color:rgb(180,95,6)">There is no reason <u>given</u> 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)</span> <span style="color:rgb(180,95,6)">This is the basis of Mill's hydrino states. The limiting factor is the photon.</span><br></div><div> </div><blockquote class="gmail_quote" style="margin:0px 0px 0px 0.8ex;border-left:1px solid rgb(204,204,204);padding-left:1ex"><div link="blue" vlink="purple" lang="EN-US"><div><div><div><div><div><div><div><div><div><p class="MsoNormal"><span style="font-size:11pt;font-family:"Calibri","sans-serif";color:rgb(119,147,60)"><u></u><u></u></span></p>
<p class="MsoNormal"><span style="font-size:11pt;font-family:"Calibri","sans-serif";color:rgb(119,147,60)">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…</span></p></div></div></div></div></div></div></div></div></div></div></blockquote><div><br></div><div><span style="color:rgb(180,95,6)">The solution assuming an infinite mass nucleus (hence no deBroglie wavelength & no resonance) still produces discrete levels. Therefore that issue cannot be causal.</span><br></div><div> </div><blockquote class="gmail_quote" style="margin:0px 0px 0px 0.8ex;border-left:1px solid rgb(204,204,204);padding-left:1ex"><div link="blue" vlink="purple" lang="EN-US"><div><div><div><div><div><div><div><div><div><p class="MsoNormal"><span style="font-size:11pt;font-family:"Calibri","sans-serif";color:rgb(119,147,60)"><u></u><u></u></span></p>
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<p class="MsoNormal"><span style="color:red">Looking at your conditions produced other thoughts.</span><u></u><u></u></p>
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<span style="color:red">The statement that "the coulomb potential runs to minus infinity" is a mathematician, not a physicist talking.
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</span><p class="MsoNormal"><span style="font-size:11pt;font-family:"Calibri","sans-serif";color:rgb(119,147,60)">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.</span></p></div></div></div></div></div></div></div></div></div></div></blockquote><div><span style="color:rgb(180,95,6)"><br></span></div><div><span style="color:rgb(180,95,6)">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.</span><br></div><div> </div><blockquote class="gmail_quote" style="margin:0px 0px 0px 0.8ex;border-left:1px solid rgb(204,204,204);padding-left:1ex"><div link="blue" vlink="purple" lang="EN-US"><div><div><div><div><div><div><div><div><div><p class="MsoNormal"><span style="font-size:11pt;font-family:"Calibri","sans-serif";color:rgb(119,147,60)"><u></u><u></u></span></p>
<p class="MsoNormal"><span style="font-size:11pt;font-family:"Calibri","sans-serif";color:rgb(119,147,60)">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)</span></p></div></div></div></div></div></div></div></div></div></div></blockquote><div><br></div><div><span style="color:rgb(180,95,6)">Do you consider the proton charge radius to be its field-energy density or the major extent of its <b>Coulomb potential</b> (e.g., the electron Compton radius)?</span> <span style="color:rgb(180,95,6)">We haven't defined charge yet have we? [Am I being the mathematician now for insisting on a valid definition?]</span><br></div><div> </div><blockquote class="gmail_quote" style="margin:0px 0px 0px 0.8ex;border-left:1px solid rgb(204,204,204);padding-left:1ex"><div link="blue" vlink="purple" lang="EN-US"><div><div><div><div><div><div><div><div><div><p class="MsoNormal"><span style="font-size:11pt;font-family:"Calibri","sans-serif";color:rgb(31,73,125)"><u></u><u></u></span></p><span class="">
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<span style="color:red">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. </span><span style="color:rgb(31,73,125)"><u></u><u></u></span></p>
</span><p class="MsoNormal"><span style="font-size:11pt;font-family:"Calibri","sans-serif";color:rgb(119,147,60)">Yes it can exceed its
RESTmass , and will be precisely gamma m0, see above, not exceeding its relativistic mass, of course.</span></p></div></div></div></div></div></div></div></div></div></div></blockquote><div><span style="color:rgb(180,95,6)"><br></span></div><div><span style="color:rgb(180,95,6)">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.</span><br><br></div><blockquote class="gmail_quote" style="margin:0px 0px 0px 0.8ex;border-left:1px solid rgb(204,204,204);padding-left:1ex"><div link="blue" vlink="purple" lang="EN-US"><div><div><div><div><div><div><div><div><div><p class="MsoNormal"><span style="font-size:11pt;font-family:"Calibri","sans-serif";color:rgb(119,147,60)"><u></u><u></u></span></p><span class="">
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<span style="color:red">This, like relativity, makes a big difference in some fields of physics.</span><u></u><u></u></p>
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<span style="color:red">The source of the wave nature of the electron is never defined in QM. Is it the 'hidden variable"?
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</span><p class="MsoNormal"><span style="font-size:11pt;font-family:"Calibri","sans-serif";color:rgb(119,147,60)">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.<u></u><u></u></span></p><span class="">
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<span style="color:red">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.</span><u></u><u></u></p>
</span><p class="MsoNormal"><span style="font-size:11pt;font-family:"Calibri","sans-serif";color:rgb(119,147,60)">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<u></u><u></u></span></p>
<p class="MsoNormal"><span style="font-size:11pt;font-family:"Calibri","sans-serif";color:rgb(119,147,60)">More beer required here
too</span></p></div></div></div></div></div></div></div></div></div></div></blockquote><div><br></div><div><span style="color:rgb(180,95,6)">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 <u><b>all</b></u> directions. Uniquely, it can have an infinity of angular momenta. That is why it can have spin 1/2 in <u><b>any</b></u> 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?)</span><br></div><blockquote class="gmail_quote" style="margin:0px 0px 0px 0.8ex;border-left:1px solid rgb(204,204,204);padding-left:1ex"><div link="blue" vlink="purple" lang="EN-US"><div><div><div><div><div><div><div><div><div><p class="MsoNormal"><span style="font-size:11pt;font-family:"Calibri","sans-serif";color:rgb(119,147,60)"><u></u><u></u></span></p><span class="">
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<font size="1"><span style="color:red">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.</span></font></li></ol>
<p><span style="color:red">More below:</span><u></u><u></u></p>
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<p class="MsoNormal"><font size="1"><b><span style="font-family:"Tahoma","sans-serif"">From:</span></b><span style="font-family:"Tahoma","sans-serif""> General [mailto:<a href="mailto:general-bounces%2Bmartin.van.der.mark" target="_blank">general-bounces+martin.van.der.mark</a>=<a href="mailto:philips.com@lists.natureoflightandparticles.org" target="_blank">philips.com@lists.natureoflightandparticles.org</a>]
<b>On Behalf Of </b>John Williamson<br>
<b>Sent:</b> woensdag 6 mei 2015 11:12</span></font></p><font size="1">
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<font size="1"><b>To:</b> Nature of Light and Particles - General Discussion<br></font>
<font size="1"><b>Cc:</b> Nick Bailey; Kyran Williamson; Michael Wright; Manohar .; Ariane Mandray<br></font>
<font size="1"><b>Subject:</b> Re: [General] Quantisation of classical electromagnetism</font></p>
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<p class="MsoNormal"><font size="1"><span style="font-family:"Tahoma","sans-serif";color:black">Hihi,<br>
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A lot of questions there Andrew.<br>
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All quantised means is "countable".</span></font></p>
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<p class="MsoNormal"><font size="1"><span style="color:red">QM is certainly putting a lot more weight to the word than that. Pointing out resonances has a physical meaning that can be useful.</span></font></p>
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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.
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</font><p class="MsoNormal"><font size="1"><span style="color:red">I would say that this looks at effect, not cause or definition.</span></font></p>
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<p class="MsoNormal"><font size="1"><span style="font-family:"Tahoma","sans-serif";color:black">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!</span></font></p>
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<font size="2">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!</font></span></p>
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<p class="MsoNormal"><span style="color:red">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. </span><span style="color:rgb(31,73,125)"><u></u><u></u></span></p>
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</span><p class="MsoNormal"><span style="font-size:11pt;font-family:"Calibri","sans-serif";color:rgb(119,147,60)">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)<u></u><u></u></span></p>
<p class="MsoNormal"><span style="font-size:11pt;font-family:"Calibri","sans-serif";color:rgb(119,147,60)">Oh and KE = -PE/2, PE is negative!!!</span></p></div></div></div></div></div></div></div></div></div></div></blockquote><div><br></div><div><span style="color:rgb(180,95,6)">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?</span><br></div><blockquote class="gmail_quote" style="margin:0px 0px 0px 0.8ex;border-left:1px solid rgb(204,204,204);padding-left:1ex"><div link="blue" vlink="purple" lang="EN-US"><div><div><div><div><div><div><div><div><div><p class="MsoNormal"><span style="font-size:11pt;font-family:"Calibri","sans-serif";color:rgb(119,147,60)"><u></u><u></u></span></p><span class="">
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<p class="MsoNormal"><span style="color:red">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.</span><u></u><u></u></p>
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<font size="1">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!</font></span></p><font size="1">
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<p class="MsoNormal"><span style="color:red">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.)
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</font></span><p class="MsoNormal"><font size="2"><span style="font-family:"Calibri","sans-serif";color:rgb(119,147,60)">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.</span></font></p></div></div></div></div></div></div></div></div></div></div></blockquote><div><font size="2"><span style="color:rgb(180,95,6)"><br></span></font></div><div><font size="1"><font size="2"><span style="color:rgb(180,95,6)">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?</span> </font></font></div><blockquote class="gmail_quote" style="margin:0px 0px 0px 0.8ex;border-left:1px solid rgb(204,204,204);padding-left:1ex"><div link="blue" vlink="purple" lang="EN-US"><div><div><div><div><div><div><div><div><div><p class="MsoNormal"><font size="1"><span style="font-family:"Calibri","sans-serif";color:rgb(119,147,60)"></span></font></p>
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Martin is, as usual, right in (pretty much) everything he says. Especially in that it is very important!<br>
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Regards, John W.</span></p>
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<p class="MsoNormal" style="margin-bottom:12pt"><b><span style="font-family:"Tahoma","sans-serif";color:black">From:</span></b><span style="font-family:"Tahoma","sans-serif";color:black"> General [general-bounces+john.williamson=<a href="mailto:glasgow.ac.uk@lists.natureoflightandparticles.org" target="_blank">glasgow.ac.uk@lists.natureoflightandparticles.org</a>]
on behalf of Mark, Martin van der [<a href="mailto:martin.van.der.mark@philips.com" target="_blank">martin.van.der.mark@philips.com</a>]<br>
<b>Sent:</b> Wednesday, May 06, 2015 8:48 AM<br>
<b>To:</b> Nature of Light and Particles - General Discussion<br>
<b>Cc:</b> Nick Bailey; Kyran Williamson; Michael Wright; Manohar .; Ariane Mandray<br>
<b>Subject:</b> Re: [General] Quantisation of classical electromagnetism</span></p>
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</font><p class="MsoNormal"><span style="font-family:"Calibri","sans-serif";color:rgb(31,73,125)">Dear Andrew,</span></p><font size="1">
</font><p class="MsoNormal"><span style="font-family:"Calibri","sans-serif";color:rgb(31,73,125)">I have good answers to most of your questions, but have no time right now to write them down,<br>
we must come back to this, it is very important indeed.</span></p><font size="1">
</font><p class="MsoNormal"><font size="1"><span style="font-family:"Calibri","sans-serif";color:rgb(31,73,125)">In any case it comes down to the following:</span></font></p>
</span><p><span style="font-size:11pt;font-family:Symbol;color:rgb(31,73,125)">·</span><span style="font-size:7pt;color:rgb(31,73,125)">
</span><span style="font-size:11pt;font-family:"Calibri","sans-serif";color:rgb(31,73,125)">Quantization comes from any wave equation with imposed boundary conditions.
</span><span style="font-size:11pt;font-family:"Calibri","sans-serif";color:red">[if you can establish standing waves?]</span><span style="font-size:11pt;font-family:"Calibri","sans-serif";color:rgb(31,73,125)">
</span><span style="font-size:11pt;font-family:"Calibri","sans-serif";color:rgb(119,147,60)">stationary waves may do already</span><u></u><u></u></p><span class="">
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</span><span style="font-size:11pt;font-family:"Calibri","sans-serif";color:rgb(31,73,125)">Uncertainty is no more than what the Fourier limit tells you.
</span><span style="font-size:11pt;font-family:"Calibri","sans-serif";color:red">[agreed]</span><u></u><u></u></p>
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</span><span style="font-size:11pt;font-family:"Calibri","sans-serif";color:rgb(31,73,125)">Copenhagen interpretation is Copenhagen mystification: although it is not very wrong at the simple level, it takes away any possibility for improvement by dogma.</span><span style="font-size:11pt;font-family:"Calibri","sans-serif";color:red">[agreed]</span><u></u><u></u></p>
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</span><span style="font-size:11pt;font-family:"Calibri","sans-serif";color:rgb(31,73,125)">Wave/particle dualism is the consequence of special relativity, see Louis de Broglie.</span><span style="font-size:11pt;font-family:"Calibri","sans-serif";color:red"> [do
you have a particular reference? I have not seen this statement before.]</span><span style="font-size:11pt;font-family:"Calibri","sans-serif";color:rgb(31,73,125)">
</span><span style="font-size:11pt;font-family:"Calibri","sans-serif";color:rgb(119,147,60)">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.</span></p></div></div></div></div></div></div></div></div></blockquote></div></div></div></div></div></div></div></div></div></blockquote><div><span style="color:rgb(180,95,6)">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.<br><br>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.<br><br></span><div style="margin-left:40px"><span style="color:rgb(180,95,6)">"This is exactly BOHR’s formula that he deduced from the theorem mentioned above<br>and which again can be regarded as a phase wave resonance condition for an electron in<br></span></div><div style="margin-left:40px"><span style="color:rgb(180,95,6)">orbit about a proton."<br><br></span></div><span style="color:rgb(180,95,6)">If this is also considered resonance, rather than just strict mechanics, then the energy levels have 3 resonances in coincidence.<br></span></div><div><br></div><div>Andrew<br></div><blockquote class="gmail_quote" style="margin:0px 0px 0px 0.8ex;border-left:1px solid rgb(204,204,204);padding-left:1ex"><div link="blue" vlink="purple" lang="EN-US"><div><div><div><div><div><div><div><div><div><div class="h5"><blockquote style="border-width:medium medium medium 1pt;border-style:none none none solid;border-color:-moz-use-text-color -moz-use-text-color -moz-use-text-color rgb(204,204,204);padding:0cm 0cm 0cm 6pt;margin-left:4.8pt;margin-right:0cm"><div><div><div><div><div><div><div><div><div><div><div><div><div><div><div><div><div><div><div><div><div><p class="MsoNormal"><br></p></div></div></div></div></div></div></div></div></div></div></div></div></div></div></div></div></div></div></div></div></div></blockquote></div></div></div></div></div></div></div></div></div></div></div>
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