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What methods have been used to rigorously prove that classical electrodynamics does not admit a robustly stable atom?

The conclusion is often stated and I am aware of the standard responses such as the sub-second collapse of the classical version of the Bohr atom. I am also aware of static stable configurations that are not stable under perturbations (the robust condition). This question is asking about broader techniques and techniques that are rigorous and not approximations. It is not a question about quantum mechanics and I have a moderately strong academic background in theoretical mechanics.

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  • $\begingroup$ If you're aware of the reasons why classical methods don't work, what "broader" methods are you talking about ? $\endgroup$ – StephenG Jul 12 '18 at 12:48
  • $\begingroup$ It's hardly an "approximation" to say that, classically, a moving electron in any orbit will decay rapidly. What techniques would not demonstrate this if the classical model applied? ,You can't pick and choose the classical bits you like and not the rest.....no wait....Bohr did that :) $\endgroup$ – user198207 Jul 12 '18 at 13:38
  • $\begingroup$ Terminology comment: instead of "robust stable" or "stable under perturbations", the common term is just "stable". A configuration that is maintained in time unless perturbation destroys it, is called "unstable equilibrium". Also, history comment: the argument about the collapse is due to Bohr and directed at Rutherford's model (not "classical version of the Bohr model"). $\endgroup$ – Ján Lalinský Jul 12 '18 at 15:10
  • $\begingroup$ I am not sure that Stephen G is aware of the reasons himself. There is a difference between being convinced of an answer because of being told, and having a proof. And regarding Study^3, the approximation is that the radiation from the electron was approximated by Bohr. Hence leaving room for doubt. Bohr simply proved using approximate methods that one model of the atom collapses classically. The broader idea is to prove that all classical models would collapse. Please see Jan Lalinsky's answer below, which does answer the question and has given me a direction to take. $\endgroup$ – Ponder Stibbons Jul 12 '18 at 21:08
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The argument about collapse of the atom in Rutherford's model is originally, I think, due to Niels Bohr. It is, originally, not "rigorous", if that word means a mathematical proof.

Bohr's himself formulated the argument in this way: (see N. Bohr, On the constitution of atoms and molecules, Philos. Mag. 26,1 (1913); http://web.ihep.su/dbserv/compas/src/bohr13/eng.pdf)

,,Let us now, however, take the effect of the energy radiation into account, calculated in the ordinary way from the acceleration of the electron. In this case the electron will no longer describe stationary orbits. $W$ will continuously increase, and the electron will approach the nucleus describing orbits of smaller and smaller dimensions, and with greater and greater frequency; the electron on the average gaining in kinetic energy at the same time as the whole system loses energy. This process will go on until the dimensions of the orbit are the same order of magnitude as the dimensions of the electron or those of the nucleus. A simple calculation shows that the energy radiated out during the process considered will be enormously great compared with that radiated out by ordinary molecular processes. ``

Calculating energy radiated using Larmor's formula is easy, but making this into mathematical proof of collapse is much harder. One would have to use strictly specified model of motion, for example, some specific differential equations, analyze them and find that typical evolution of initial circular orbit leads to a collapse. This is not easy to do, because the equations of motion for charged particle interacting with another charged particle, even if they are points, are very complicated to solve.

There are, however, some numerical calculations of this "two-body problem of electrodynamics":

J. L. Synge, On the electromagnetic two–body problem., Proc. Roy. Soc. A 177 118–39

Synge assumes purely retarded interactions, with no self forces. He starts with circular orbits of both electron and proton and he finds out that if both are allowed to move (the proton is not fixed), the orbit radii decrease in time, but much more slowly than the Larmor formula would suggest.

Both Bohr and Synge assumed that the fields are purely retarded just as in macroscopic physics, and they concluded the collapse. On the other hand, on the microscopic level, the fields may not be purely retarded but each particle field may contain some part of advanced field. L. Page and others showed this would suppress loss of energy by radiation from the atom:

L. Page, Advanced Potentials and their Application to Atomic Models, Phys. Rev. 24, 296 (1924)

Later people also realized that Rutherford's model ignores other EM forces acting on the electron and proton; for example those of other distant charged particles, or "background radiation". These forces are usually assumed to be very weak, but in fact they may compensate for the damping effect arising from the retarded character of mutual electron-proton forces. So, even if the Synge model leads to collapse for isolated system, in reality this collapse may be prevented due to presence of the background radiation, which there is always some.

There is some numerical evidence this may work, see

D.C. Cole, Yi Zou, Quantum mechanical ground state of hydrogen obtained from classical electrodynamics, Phys. Lett. A 317, p. 14-20 (2003) see https://doi.org/10.1016/j.physleta.2003.08.022

T.H. Boyer, Comments on Cole and Zou's Calculation of the Hydrogen Ground State in Classical Physics Found Phys Lett (2003) 16,p. 613,

see https://doi.org/10.1023/B:FOPL.0000012787.05764.4d

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  • $\begingroup$ Thanks, this is the type of answer I was looking for. I want to upvote or whatever it is on this forum. But, I don't see the button. How do I acknowledge that this answer was a good one? $\endgroup$ – Ponder Stibbons Jul 12 '18 at 21:09
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    $\begingroup$ Look for gray arrows on the left-hand side and a number between them. $\endgroup$ – Ján Lalinský Jul 12 '18 at 21:43
  • $\begingroup$ I did click on the gray arrows - but it said that I did not have enough reputation to make a difference. Sorry. I will work on that. I appreciate that you understood what the question was. The bit that was most useful to me was the paper by Cole and Yue, it got me on a new aspect of the problem. $\endgroup$ – Ponder Stibbons Jul 13 '18 at 10:43
  • $\begingroup$ Hi Jan - I got up to 19 reputation, so I have upvoted your answer. Thanks again. Ponder. $\endgroup$ – Ponder Stibbons Jul 17 '18 at 12:48

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