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I know this is a classical system, and thus not compliant with the quantum nature of real atoms. But please bare with me.

I have heard this before: the orbiting electron should radiate in the classical Bohr atomic model (point-charge electron and point charge proton orbiting common center of mass).

When searching around, I found it in wikipedia and also in a very popular question and answer here.

Yet I can't see how. I have been checking the literature:

Using Lorentz Force where we account for both the electrical and magnetic fields $\textbf{F}_L = e\textbf{E} + e\textbf{v} \times \textbf{B}$ you find it radiates only if you fix the position of the proton.

If you are committed to be accurate, and work on the center of mass, then magnetic field terms cancel and only electric fields remain. This then leads to classical movement in central field where momentum conservation prevents the electron from falling into the nucleus.

Now for a more serious consideration,

Using Pointing Theorem and then trying to find the radiated energy coming out of the system, with a bit more work, I find that there is nothing coming out. Which is consistent with above, but inconsistent with calculations of radiation coming from an equivalent system: rotating dipole

Where is the mistake?

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    $\begingroup$ I think that your assumption of a fixed center of mass prohibits radiation, but the center of mass changes continually due to radiation. Have a look at the classical energy loss of the bohr atom here physics.princeton.edu/~mcdonald/examples/orbitdecay.pdf $\endgroup$ – anna v Sep 29 '15 at 15:58
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    $\begingroup$ The whole point of Bohr's atom was that continuous radiation was disallowed by the requirement to always occupy one of the predefined orbits. This was an ad hoc assumption with no fancy working behind it. $\endgroup$ – dmckee Sep 29 '15 at 18:14
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    $\begingroup$ @annav, thanks for your link but it has the same issues I discussed: it assumes the proton is fixed, and assumes that the radiation comes from the electron moving in a central electric field. I uses for this the Larmor formula. Also the center of mass can be fixed because is an isolated system, and the center of mass reference frame should be inertial. My question is more fundamental, if we don't fix the proton how do we say the system should radiate? $\endgroup$ – rmhleo Sep 29 '15 at 21:01
  • $\begingroup$ @dmckee Sommerfeld did fix the orbits later, in a way that no radiation is allowed, by defining the action integral to be an integer times Plank's constant. But I am considering Bohr's proposal, without this additional constraint, precisely because it doesn't seem to be needed for stability. $\endgroup$ – rmhleo Sep 29 '15 at 21:05
  • $\begingroup$ To keep the same center of mass for a radiating system one has to do the kinematics including all those leaving photons. Just the proton and the electron will no longer be in a center of mass once one photon leaves . As the stability of hydrogen cannot depend on the center of mass, showing that the system is not stable in one frame , in this case proton at rest, is enough. $\endgroup$ – anna v Sep 30 '15 at 3:13
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I will turn my comment into an answer:

One is dealing with classical physics in this question, at the level of the Bohr model for the atom special relativity is unknown.

An inertial frame can be defined as a frame where the laws of physics have the same mathematical form and measurements in one inertial frame can be converted to measurements in another by a simple transformation (the Galilean transformation).

This means that predictions for physical observables made in one inertial frame hold for all inertial frames, so one uses the simplest frame. In the frame where the proton is at rest and the electron is turning around it like a planet , classical electrodynamics predicts that there will be electromagnetic radiation. Transforming this prediction to the rest mass system of the initial conditions of (proton+electron) will still show in the time progress electromagnetic radiation. The original center of mass can only be defined if the momentum of the departing radiation is included for the Galilean transformation.

In the quote with no link that you give :

This then leads to classical movement in central field where momentum conservation prevents the electron from falling into the nucleus.

The discrepancy is in this statement: Momentum conservation does not prevent the electron to radiate in an electric field. The radiation takes away momentum. Again ignoring that the center of mass system will change.

I cannot check the Poynting vector comment because the maths are not shown.

But is is enough that the laws of physics are the same in all inertial frames to show that the assumed center of mass frame is not stable either, so some mistake will be there also.

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  • $\begingroup$ The frame where the proton is at rest, is not an inertial frame. Unless you provide a force that counters that of the electron at any moment, which is the case I'm avoiding. $\endgroup$ – rmhleo Sep 30 '15 at 4:03
  • $\begingroup$ I will include the Poynting vector calculations, that might be where the problem lies. $\endgroup$ – rmhleo Sep 30 '15 at 4:08
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    $\begingroup$ Are you saying that the earth at rest is not an inertial frame for the moon earth system? Please reconsider. $\endgroup$ – anna v Sep 30 '15 at 4:09
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    $\begingroup$ Mathematics is mathematics. The proton has a rest frame and instantaneously there is an electron with momentum p. then at t+deta(t) the electron has a momentum p-delta(p) the proton at rest. The center of mass of the two has changed by this loss of momentum but the rest frame of the proton has not. $\endgroup$ – anna v Sep 30 '15 at 5:12
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    $\begingroup$ In the framework where the electron is at rest the same holds, it will be the proton which will be losing delta(p). The fall of the electron comes from the sequential losses of momentum, when it loses all momentum it will fall down and keep losing potential energy in to radiation. ( or the proton on the electron in the electron rest frame). What other theory is needed? $\endgroup$ – anna v Sep 30 '15 at 5:25

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