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<div>Hi andrew, thanks for the reply.</div>
<div>Some quick remarks/answers.</div>
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<div>Yes, the relativistic models are the somerfeld model and the solutions to the dirac equation. </div>
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<div>By the fact that space is 3d and by the fundamental meaning of Gauss's law, any charged potato has a potential that approaches Coulomb's law at large distance. At short distance the structure of the thing may be determined by stronger forces. There is
an exception: the 3 leptons. Hence another measure of size is needed for those, and it is found on a theoretical basis by calculating the total energy in the coulomb field down to the putative radius of the particle, assuming that all energy (=mass) is of
electromagnetic origin and matches the particle's rest mass</div>
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<div>For the rest, i do not know where you want to go, really, so beer is the best option, and i am looking forward to any occasion for it that may come!</div>
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<div>Cheers<br>
<br>
Verstuurd vanaf mijn iPhone</div>
<div><br>
Op 12 mei 2015 om 20:08 heeft Andrew Meulenberg <<a href="mailto:mules333@gmail.com">mules333@gmail.com</a>> het volgende geschreven:<br>
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<div>Dear Martin,<br>
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Unless we are willing and able to argue constructively, we don't really have a 'critical mass' needed to carry a project forward. We need to be able to hone our ideas and their presentation. We might even be able to change our minds.<br>
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response below (in black):<br>
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<div class="gmail_quote">On Tue, May 12, 2015 at 2:00 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>
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<p class="MsoNormal"><span style="font-size:11pt;font-family:"Calibri","sans-serif";color:rgb(112,48,160)">Dear Andrew, I have been away for a few days.</span></p>
<p class="MsoNormal"><span style="font-size:11pt;font-family:"Calibri","sans-serif";color:rgb(112,48,160)">In your previous replies you tried to challenge me to be precise, and have questioned the correctness of my statements. That is very good.</span></p>
<p class="MsoNormal"><span style="font-size:11pt;font-family:"Calibri","sans-serif";color:rgb(112,48,160)"> </span></p>
<p class="MsoNormal"><span style="font-size:11pt;font-family:"Calibri","sans-serif";color:rgb(112,48,160)">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.
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<p class="MsoNormal"><span style="font-size:11pt;font-family:"Calibri","sans-serif";color:rgb(112,48,160)"> </span></p>
<p class="MsoNormal"><span style="font-size:11pt;font-family:"Calibri","sans-serif";color:rgb(112,48,160)">I hope no one is color blind, yet another color will be used: purple! (just in case some haze troubles the mind…)</span></p>
<p class="MsoNormal"><span style="font-size:11pt;font-family:"Calibri","sans-serif";color:rgb(112,48,160)"> </span></p>
<p class="MsoNormal"><span style="font-size:11pt;font-family:"Calibri","sans-serif";color:rgb(112,48,160)">So: purple (Martin) responds to
</span><span style="font-size:11pt;font-family:"Calibri","sans-serif";color:rgb(152,72,7)">brown (Andrew) responds to
</span><span style="font-size:11pt;font-family:"Calibri","sans-serif";color:rgb(79,98,40)">green (Martin) responds to
</span><span style="font-size:11pt;font-family:"Calibri","sans-serif";color:red">red (Andrew)</span><span style="font-size:11pt;font-family:"Calibri","sans-serif";color:rgb(112,48,160)"></span></p>
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<p class="MsoNormal"><span style="font-size:11pt;font-family:"Calibri","sans-serif";color:rgb(112,48,160)"> </span></p>
<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="font-size:10pt;font-family:"Arial","sans-serif";color:navy"></span><span style="font-size:11pt;font-family:"Calibri","sans-serif";color:navy"></span></p>
<p class="MsoNormal"><span style="font-size:10pt;font-family:"Arial","sans-serif";color:navy"><span dir="ltr"><span dir="ltr"><span><span><span dir="ltr"><span dir="ltr"><span><span></span><span></span></span></span></span></span><span></span></span></span></span></span><span style="font-size:11pt;font-family:"Calibri","sans-serif";color:rgb(31,73,125)"></span></p>
<p class="MsoNormal"><font size="1"><span style="font-family:"Calibri","sans-serif";color:navy"> </span></font></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>Andrew Meulenberg<br>
<b>Sent:</b> donderdag 7 mei 2015 7:40</span></font></p>
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<div><font size="1"><br>
</font><font size="1"><b>To:</b> Nature of Light and Particles - General Discussion<br>
</font><font size="1"><b>Subject:</b> Re: [General] Quantisation of classical electromagnetism</font></div>
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<p class="MsoNormal" style="margin-bottom:12pt"><font size="1">Dear Martin,</font></p>
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<p class="MsoNormal"><font size="1">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.</font></p>
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<p class="MsoNormal"><font size="1">On Thu, May 7, 2015 at 3:50 AM, Mark, Martin van der <<a href="mailto:martin.van.der.mark@philips.com" target="_blank">martin.van.der.mark@philips.com</a>> wrote:</font></p>
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<p class="MsoNormal"><font size="1"><span style="font-family:"Calibri","sans-serif";color:rgb(119,147,60)">Andrew, thanks, please see below, in green</span><span></span></font></p>
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<p class="MsoNormal"><font size="1"><span style="font-family:"Calibri","sans-serif";color:rgb(31,73,125)" lang="NL"> </span><span style="font-family:"Arial","sans-serif";color:navy" lang="DE">Dr. Martin B. van der Mark</span><span lang="NL"></span></font></p>
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<p class="MsoNormal"><font size="1"><span style="font-family:"Arial","sans-serif";color:navy">Principal Scientist, Minimally Invasive Healthcare</span></font></p>
<p class="MsoNormal"><b><span style="font-size:11pt;font-family:"Calibri","sans-serif";color:rgb(31,73,125)"><u></u> <u></u></span></b></p>
<p class="MsoNormal"><font size="1"><b><span style="font-family:"Tahoma","sans-serif"">From:</span></b></font><span style="font-size:10pt;font-family:"Tahoma","sans-serif""><font size="1"> 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</font><br>
<font size="1"><b>To:</b> Nature of Light and Particles - General Discussion<br>
<b>Subject:</b> Re: [General] Quantisation of classical electromagnetism</font></span><font size="1"><span></span></font></p>
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<p class="MsoNormal"><font size="1"><span> <span style="color:red">Dear John W, Martin, et al.,</span></span><span></span></font></p>
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<p class="MsoNormal" style="margin-bottom:12pt"><font size="1"><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></font></p>
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<p class="MsoNormal" style="margin-bottom:12pt"><font size="1"><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></font></p>
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<p class="MsoNormal"><span style="font-size:7.5pt;color:red">some comments below:</span><u></u><u></u></p>
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<p class="MsoNormal"><span style="font-size:7.5pt"> On Wed, May 6, 2015 at 3:28 PM, Mark, Martin van der <</span><a href="mailto:martin.van.der.mark@philips.com" target="_blank">martin.van.der.mark@philips.com</a><span style="font-size:7.5pt">> wrote:</span><u></u><u></u></p>
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<p class="MsoNormal"><font size="1"><span> </span><span style="font-family:"Calibri","sans-serif";color:rgb(31,73,125)">Andrew, John, all</span></font></p>
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<p class="MsoNormal"><font size="1"><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></font></p>
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<p class="MsoNormal"><font size="1"><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></font></p>
<p class="MsoNormal"><font size="1"><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></font></p>
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<p class="MsoNormal"><font size="1"><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></font></p>
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<p class="MsoNormal"><font size="1"><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></font></p>
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<p class="MsoNormal"><font size="1"><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></font></p>
<p class="MsoNormal"><font size="1"><span style="font-family:"Calibri","sans-serif";color:rgb(31,73,125)">Cheers, Martin</span><span></span></font></p>
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<p class="MsoNormal"><font size="1"> <span style="color:red">I also have three basic conditions:</span></font></p>
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<li class="MsoNormal"><font size="1"><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></font></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>
<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.
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<p class="MsoNormal"><span style="color:rgb(180,95,6)"> </span><span style="font-size:11pt;font-family:"Calibri","sans-serif";color:rgb(180,95,6)">The levels identified with the classical description agree with those of the mechanical (QM) system.</span><span style="font-size:11pt;font-family:"Calibri","sans-serif""><u></u><u></u></span></p>
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<p class="MsoNormal"><span style="font-size:11pt;font-family:"Calibri","sans-serif";color:rgb(112,48,160)">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.</span></p>
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<div>The classical description, based on conservation of energy and momentum, should include the photon energy, E(gamma).<br>
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<li> TE = KE + PE + E(gamma)</li></ul>
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<div>E(gamma) is not included in the standard QM derivation of the H atom. (It should be in the time-dependent Schrodinger equation. However, that is not what is generally taught for the H atom. I'm sure that it is at some level. Did you have it taught to you?
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<li class="MsoNormal"><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>
<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!!!!!
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<span style="font-size:11pt;font-family:"Calibri","sans-serif";color:rgb(180,95,6)">[Exactly, therefore it cannot radiate a photon except to a system w ang. mom. = hbar. No levels below gnd state have that value.]</span><br>
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<p class="MsoNormal"><span></span><span style="color:rgb(112,48,160)">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).<u></u><u></u></span></p>
<p class="MsoNormal"><span style="font-size:11pt;font-family:"Calibri","sans-serif";color:rgb(112,48,160)">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. </span></p>
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<div>We may be talking at crossed purposes here. Let me try to straighten it out. (We still may have to talk, but this will help me to focus and remember.) Forgive me if I, for brevity, make flat statements as if they are known facts. And please correct me
when I am wrong.<br>
<ol>
<li>To go below the 'ground state', by photonic emission, it is necessary for this 'final' (ground) state to become an emitter.<br>
</li><li>To get to the ground state, an electron must radiate a photon (which has ang mom of hbar). The final state of the system includes that photon. If we renormalize the H + photon system by subtracting the photon to get the bare atom, in the ground state, then
the final state has zero ang mom and any lower state, if one existed, would also have zero ang mom.
</li><li>To go to any lower state (by the same process) requires a second photon. If neither the 'final' nor any lower state has sufficient ang mom to form a 2nd photon, no photonic transition can occur (unless we have other electrons or atoms close enough for longitudinal
photons to exchange energy - the Mill's process?). Zero-zero photonic transitions are highly forbidden. [Note that my original comment (in red) stated<span><span style="color:red"> "... the requirement of a photon to have an angular momentum of hbar."<span style="color:rgb(0,0,0)">
The photon, <u>not</u> the bare ground state, has ang mom of hbar.</span><br>
</span></span></li><li>The ground state has an isotropic electron distribution. The H gnd state angular momentum of zero (to +/- hbar/2, from the HUP) allows its electron to have a linear orbit, through the nucleus, and the full range of elliptical orbits (including all circular
orbits of +/- hbar/2) with different axes. QM provides the probability distribution of an electron in all of those orbits.</li><li>The radial orbit and the circular orbit are both (extreme) solutions and have the same conditions of the eigenvalue.</li><li>My statement was about the transition from the ground state to a lower state, not the transition to the ground state.<br>
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<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><u></u><u></u></p>
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<p class="MsoNormal"><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></p>
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<p class="MsoNormal"><span style="font-size:11pt;font-family:"Calibri","sans-serif";color:rgb(112,48,160)">See previous answer and understand that the reason is implicitly given there already. A mode of a resonator or waveguide, a quantum state, these are 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!</span></p>
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<div>I may like that resonance answer. However, in the atomic system there are only losses by photonic emission (true?), so
<u>any</u> orbit that cannot radiate has an infinite Q value. The HUP indicates that the resonance may be very broad. The precision of the energy level (which is an average value?) is extremely high because it is a value determined from an infinite number of
'measurements' (the time independent solution). What is the standard deviation of individual orbit measurements?<br>
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<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 is heavy and slow, but both have
the same momentum! Hence they have the same de Broglie wavelength…</span> <br>
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<p class="MsoNormal"><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><u></u><u></u></p>
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<p class="MsoNormal"><span style="font-size:11pt;font-family:"Calibri","sans-serif";color:rgb(112,48,160)">Bullshit, sorry, but here you should really know better since:<u></u><u></u></span></p>
<p><u></u><span style="font-size:11pt;font-family:"Calibri","sans-serif";color:rgb(112,48,160)"><span>1)<span style="font:7pt "Times New Roman"">
</span></span></span><u></u><span style="font-size:11pt;font-family:"Calibri","sans-serif";color:rgb(112,48,160)">I have told you already you are referring to the BORN APPROXIMATION<u></u><u></u></span></p>
<p><u></u><span style="font-size:11pt;font-family:"Calibri","sans-serif";color:rgb(112,48,160)"><span>2)<span style="font:7pt "Times New Roman"">
</span></span></span><u></u><span style="font-size:11pt;font-family:"Calibri","sans-serif";color:rgb(112,48,160)">It is in every basic textbook, second year physics<u></u><u></u></span></p>
<p><u></u><span style="font-size:11pt;font-family:"Calibri","sans-serif";color:rgb(112,48,160)"><span>3)<span style="font:7pt "Times New Roman"">
</span></span></span><u></u><span style="font-size:11pt;font-family:"Calibri","sans-serif";color:rgb(112,48,160)">The so-called reduced mass gives a slightly different set of levels, the correct levels. Remember the Rydberg constant? And its slightly different
numbers?<u></u><u></u></span></p>
<p class="MsoNormal"><span style="font-size:11pt;font-family:"Calibri","sans-serif";color:rgb(112,48,160)">Just do your homework, and go to Wikipedia, that will be good enough.<u></u><u></u></span></p>
<p class="MsoNormal"><span style="font-size:11pt;font-family:"Calibri","sans-serif";color:rgb(112,48,160)"><u></u> <u></u></span></p>
<p class="MsoNormal"><span style="font-size:11pt;font-family:"Calibri","sans-serif";color:rgb(112,48,160)">Here another remark regarding your earlier question, whether I was talking physics or mathematics. (see next question below)<u></u><u></u></span></p>
<p class="MsoNormal"><span style="font-size:11pt;font-family:"Calibri","sans-serif";color:rgb(112,48,160)">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.<u></u><u></u></span></p>
<p class="MsoNormal"><span style="font-size:11pt;font-family:"Calibri","sans-serif";color:rgb(112,48,160)">Sorry I just hate mediocrity.</span></p>
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<div>I think that you missed my point. I was objecting to your assumption of a resonance between the nuclear and electron deBroglie wavelengths. The mathematical levels and deBroglie wavelengths exist whether the nuclear mass is finite or infinite. The values
do change, but not much.<br>
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<div>However, the basis of my objection was ill-founded and you naturally responded to a different thought. I should have thought about the conservation laws as being resonances, rather than assuming that they were not.
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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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<li class="MsoNormal"><span style="color:red">The statement that "the coulomb potential runs to minus infinity" is a mathematician, not a physicist talking.
</span><u></u><u></u></li></ol>
<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>
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<div>I am responding to 60 years of mathematical physicists rejecting the anomalous solution of the Dirac equations because of the consequences of a 'singularity' at r = 0. Physicists should never go there because there is no such thing as a point charge. I
do not believe that having point-like behavior includes the singularity. <br>
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<p class="MsoNormal"><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><u></u><u></u></p>
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<p class="MsoNormal"><span style="font-size:11pt;font-family:"Calibri","sans-serif";color:rgb(112,48,160)">The Coulomb potential is BY DEFINITION a 1/r potential associated with a POINT CHARGE.<u></u><u></u></span></p>
<p class="MsoNormal"><span style="font-size:11pt;font-family:"Calibri","sans-serif";color:rgb(112,48,160)">It therefore runs, by definition, to minus infinity.</span></p>
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The Coulomb potential, including r=0, is a non-physical entity. It is a mathematical model - with limitations.<br>
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<p class="MsoNormal"><span style="font-size:11pt;font-family:"Calibri","sans-serif";color:rgb(112,48,160)">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.<u></u><u></u></span></p>
<p class="MsoNormal"><span style="font-size:11pt;font-family:"Calibri","sans-serif";color:rgb(112,48,160)">I know you want to make that point, and I fully agree. But please do not misuse the argument or definition of things.
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If the Coulomb 'law' were supplemented with an r > 0 statement, perhaps it would not be mistaken and misused by the mathematical physicists. I have to be careful not to misuse existing definitions - even if they are flawed.<br>
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<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>
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<p class="MsoNormal"><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></p>
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<p class="MsoNormal"><span style="font-size:11pt;font-family:"Calibri","sans-serif";color:rgb(112,48,160)">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.</span></p>
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<div>I find it interesting that the electron radius is defined on a theoretical basis; while the proton charge radius is based on an experimental process.<br>
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<p class="MsoNormal" style="margin-left:36pt"><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></p>
<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>
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<p class="MsoNormal" style="margin-bottom:12pt"><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></p>
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<p class="MsoNormal" style="margin-left:36pt"><span style="color:red">This, like relativity, makes a big difference in some fields of physics.</span></p>
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<li class="MsoNormal"><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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<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.</span></p>
<p class="MsoNormal" style="margin-left:36pt"><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></p>
<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</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>
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<p class="MsoNormal"><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
<b><u>all</u></b> directions. Uniquely, it can have an infinity of angular momenta. That is why it can have spin 1/2 in
<b><u>any</u></b> 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></p>
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<p class="MsoNormal"><span style="font-family:"Calibri","sans-serif";color:rgb(112,48,160)">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.</span></p>
<p class="MsoNormal"><span style="font-family:"Calibri","sans-serif";color:rgb(112,48,160)">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!</span></p>
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<div>I thought that I had gotten that '97 paper and at least looked it over. However, I can't find it on my computer now. I'll have to download it again tomorrow. I'll also have to examine the EM fields as an inviscid fluid. Can such fluid waves then pass thru
each other? EM waves can (but not always).<br>
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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></li></ol>
<p><span style="color:red">More below:</span></p>
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<p class="MsoNormal"><b><span style="font-size:7.5pt;font-family:"Tahoma","sans-serif"">From:</span></b><span style="font-size:7.5pt;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></p>
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<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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<p class="MsoNormal"><span style="font-size:7.5pt;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></p>
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<p class="MsoNormal"><span style="font-size:7.5pt;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></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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<p class="MsoNormal"><span style="font-size:7.5pt;color:red">I would say that this looks at effect, not cause or definition.</span></p>
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<p class="MsoNormal"><span style="font-size:7.5pt;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></p>
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</span><span style="font-family:"Tahoma","sans-serif";color:black">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!</span></font></p>
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<p class="MsoNormal"><font size="1"><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.
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<p class="MsoNormal"><font size="1"><span style="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)</span></font></p>
<p class="MsoNormal"><font size="1"><span style="font-family:"Calibri","sans-serif";color:rgb(119,147,60)">Oh and KE = -PE/2, PE is negative!!!</span></font></p>
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<p class="MsoNormal"><span style="color:rgb(180,95,6)">In<font size="1">troducing 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?</font></span></p>
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<p class="MsoNormal"><span style="font-family:"Calibri","sans-serif";color:rgb(112,48,160)">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.</span><span style="font-family:"Calibri","sans-serif";color:rgb(31,73,125)"> </span></p>
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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></p>
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</span></font><font size="1"><span style="font-family:"Tahoma","sans-serif";color:black">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!</span></font><span style="font-size:18pt"></span></p>
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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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<p class="MsoNormal"><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><span style="font-size:16pt"></span></p>
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<p class="MsoNormal"><span style="font-size:16pt"> </span><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> </p>
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<p class="MsoNormal"><span style="font-family:"Calibri","sans-serif";color:rgb(112,48,160)">The electron is a wave and a particle at the same time, right?<br>
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<div>I consider an electron to be<span style="color:rgb(0,0,0)"> <span><span style="font-size:11pt;font-family:"Calibri","sans-serif"">a particle consisting of a self-wrapped photon and with an 'external'</span></span><span><span style="font-size:11pt;font-family:"Calibri","sans-serif"">
wave intimately associated with it. The wave is not extended (although there are associated evanescent waves that are); I believe it to be one of orientation.<br>
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A photon is the reverse, it is a wave that is confined (limited in extent, but not wrapped) to act as a (sometimes long) particle.<br>
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<div><span style="color:rgb(0,0,0)"><span><span style="font-size:11pt;font-family:"Calibri","sans-serif"">A photon could be multiply 'wrapped' about a proton, an electron cannot be wrapped about a proton.<br>
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<div><span style="color:rgb(0,0,0)"><span><span style="font-size:11pt;font-family:"Calibri","sans-serif"">Best regards;<br>
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<div><span style="color:rgb(0,0,0)"><span><span style="font-size:11pt;font-family:"Calibri","sans-serif"">Andrew<br>
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</span><span style="color:red">More below:</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!</font><br>
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<font size="1">Regards, John W.</font></span></p>
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<p class="MsoNormal" style="margin-bottom:12pt"><font size="1"><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></font></p>
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<p class="MsoNormal"><font size="1"><span style="font-family:"Calibri","sans-serif";color:rgb(31,73,125)">Dear Andrew,</span></font></p>
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<p class="MsoNormal"><font size="1"><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></font></p>
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<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>
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<p><font size="1"><span style="font-family:Symbol;color:rgb(31,73,125)">·</span><span style="color:rgb(31,73,125)">
</span><span style="font-family:"Calibri","sans-serif";color:rgb(31,73,125)">Quantization comes from any wave equation with imposed boundary conditions.
</span><span style="font-family:"Calibri","sans-serif";color:red">[if you can establish standing waves?]</span><span style="font-family:"Calibri","sans-serif";color:rgb(31,73,125)">
</span><span style="font-family:"Calibri","sans-serif";color:rgb(119,147,60)">stationary waves may do already</span></font></p>
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</span><span style="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-family:"Calibri","sans-serif";color:red">[agreed]</span></font></p>
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</span><span style="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-family:"Calibri","sans-serif";color:red">[agreed]</span></font></p>
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</span><span style="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-family:"Calibri","sans-serif";color:red"> [do you have a particular reference?
I have not seen this statement before.]</span><span style="font-family:"Calibri","sans-serif";color:rgb(31,73,125)">
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<p><span style="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>
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<p class="MsoNormal" style="margin-bottom:12pt"><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.</span><span style="color:rgb(31,73,125)"></span></p>
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<p class="MsoNormal" style="margin-bottom:12pt"><font size="1"><span style="font-family:"Calibri","sans-serif";color:rgb(112,48,160)">It is not your fault, it is mentioned rarely. But since I have, now you know.</span></font></p>
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<p class="MsoNormal" style="margin-bottom:12pt"><font size="1"><span style="color:rgb(180,95,6)">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.</span></font></p>
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<p class="MsoNormal"><font size="1"><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</span></font></p>
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<p class="MsoNormal"><font size="1"><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.</span></font></p>
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<p class="MsoNormal"><font size="1"><span style="font-family:"Cambria","serif";color:rgb(112,48,160)">Dear Andrew I hope to have cleared up a few things, surely we should talk about them some more,</span></font></p>
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<p class="MsoNormal"><font size="1"><span style="font-family:"Cambria","serif";color:rgb(112,48,160)">Very best regards, Martin</span></font></p>
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