This question on binary black hole solutions, led to me think about the similar question from the perspective of what we know about the Hydrogen atom.

Prior to quantum mechanics, it was not understood what led to the stability of the Hydrogen atom against collapse of the electron orbits due to Bremsstrahlung, i.e. the emission of radiation due to the fact that it is in a non-inertial (accelerated) frame of reference. Bohr and Sommerfeld came up with a somewhat ad-hoc procedure - the first ever instance of quantization - according to which in the quantum theory only those classical orbits are allowed the value of whose action is quantized in units of $\hbar$.

$$ \int_i p dq = 2 \pi n_i $$

where the integral is over the $i^{th}$ (closed) classical orbit.

Now what I'm thinking of next has probably been thought of before but I haven't done a literature review to find out.

Classically, we expect the accelerating electron to radiate resulting in the catastrophic collapse of its orbit. However, in a complete description we must also take the proton into consideration. It is also a charged object and as is well known from the two-body, inverse square law, central force problem (see eg. Goldstein) both the proton and electron orbit each other. Therefore the proton, being a charged object, must also radiate if we don't ignore its (accelerated) motion around the electron. An observer sitting at a distance $d \gg r$, where $r$ is the mean size of the two-body system, will measure radiation which is a superposition of that coming from both the electron and the proton.

The question is this: a). What is the phase difference between the two contributions ($\mathbf{E}_e$ and $\mathbf{E}_p$) to the net electric field $\mathbf{E}$ as seen by this observer?, b). What is the value of the total energy $E$ emitted by the electron-proton system, given by the integral of the Poynting vector across a closed surface enclosing the system, as seen by this observer?

[ My motivation is to see if we can learn more about the Bohr-Sommerfeld quantization condition by considering the classical electrodynamics of the full electron-proton system. The quantity $E$ will depend on the size and shape of the classical orbits of the two charged objects or more simply of their mean separation $E:=E(r)$. As we vary $r$ from $0$ to some value $r_{max} \ll d$ we would expect $E(r)$ to oscillate and have local minima for some classical orbits. If these classical minima occur for orbits which satisfy the Bohr-Sommerfeld condition then we would have established a connection between the full classical problem and its quantization. ]


4 Answers 4


Although a point charge does radiate when accelerated, this isn't necessarily true of all charge distritutions, where the radiation fields of all the charge elements can cancel in some cases.

This nonradiation condition has been known since 1910 when Paul Ehrenfest published a paper:

Ungleichförmige Elektrizitätsbewegungen ohne Magnet- und Strahlungsfeld: Phys. Z. 11 (1910), 708-709)

For more information, have a look at:

Nonradiation condition http://en.wikipedia.org/wiki/Nonradiation_condition

Invisibility Physics: Acceleration without radiation, part I http://skullsinthestars.com/2008/04/19/invisibility-physics-acceleration-without-radiation-part-i/

Recent classic papers are:

Goedecke, G. H. (1964). "Classically Radiationless Motions and Possible Implications for Quantum Theory". Physical Review 135: B281–B288. doi:10.1103/PhysRev.135.B281

Haus, H. A. (1986). "On the radiation from point charges". American Journal of Physics 54: 1126

  • $\begingroup$ Thanks @John. That is the gist of the my question. I'm looking at the papers you've mentioned. $\endgroup$
    – user346
    Jan 23, 2011 at 10:08

A better starting place for this study might be looking at the quantum mechanics of positronium, which is a bound state of an electron and a positron. In this case, both particles have equal masses, and therefore, would radiate equally.

In practice, positronium isn't qualitatively very different from the hydrogen atom--the reduced mass of the electron is obviously very different, and it is necessary to apply a correction for the fact that both charges are moving, but you get the same basic energy level structure that you do for the hydrogen atom.


I would guess that the contribution of the proton would be fairly negligible, because of the difference in mass. The classical radiated field amplitude will depend on the magnitude of the acceleration, and while the proton and electron both experience the same force, the proton is 1836 times heavier, so the acceleration is reduced by a factor of 1836. Which means that the field amplitude will presumably be reduced by the same factor, in which case the phase relation isn't that important, because even if they were perfectly out of phase, the reduction of the total field due to the proton's contribution would be minimal.

  • $\begingroup$ Nice answer @Chad. $\endgroup$
    – user346
    Jan 23, 2011 at 0:06
  • $\begingroup$ I did think of this aspect of the problem. I'm hoping there could be some other consideration which I might have overlooked. Of course, if there isn't, then this is an open-and-shut case. $\endgroup$
    – user346
    Jan 23, 2011 at 0:21

The problem you have posed is very interesting.I made a similar trial some time ago. If you like you can go to www.anconanotizie.it where i published a note on the possibility to explain the stability of the hidrogen atom on a classical basis. The crucial point, according to my opinion, is the following: the proton is a very strong source of magnetic field mainly due to its spin. So the electron will be submitted to the Lorentz force which presumably will oppose to the Coulomb force.If the electron try to collapse on the proton a strong repulsive force will be established between the two particles explained by the induction electromagnetic law of Faraday-Lenz. So it's possibile that will exist motion conditions that prevent the system from radiation just because the system should not collapse.

Ronchini Riccardo


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