I had a question about https://physics.stackexchange.com/questions/320426/what-prevents-us-to-build-a-model-where-in-the-atom-the-positive-and-negative-el?noredirect=1#comment720010_320426. The question is about the contribution of the electric fields of charged particles to the EM radiation during the approach of the electron to the nucleus.

Let me explain this thought. Having an electron in infinity from a positive charged hydrogen ion both constituents exert a force on each other. From this starting position it has to be possible to calculate the top speed with which the electron approaches the nucleus. Furthermore it should be possible to compare

  • the EM radiation from this interaction until the electron reaches the lowest bonded state in the hydrogen atom
  • with the EM radiation of an electron of the same speed, coming in rest under the influence of an external magnetic field (under the influence of the Lorentz force).

Is there a difference in the value (the amount) of the emitted EM radiations from electrons with same kinetic energy

  • reaching the lowest bonded state in a hydrogen atom and
  • coming to standstill in a magnetic field?

In short: Both electrons have the same amount of potential energy (in the case of the hydrogen atom) and kinetic energy (in the case of the magnetic field) and came into standstill in both cases. Are the emitted EM radiation of the same amount?

  • $\begingroup$ I don't think an electron can come to rest in a hydrogen atom but if it could I think it would be accelerated differently from Photons emitted from the hydrogen's proton and other electron. On the other hand an electron being accelerated by electromagnetic field is interacting with a uniform field of photons. $\endgroup$ – Bill Alsept Mar 22 '17 at 6:40
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    $\begingroup$ What makes you think that an electron can come to rest under the action of a magnetic field? Magnetic fields don't perform work on charged particles. $\endgroup$ – Emilio Pisanty Mar 22 '17 at 11:22
  • $\begingroup$ @Emilio Under the conditions of Lorentz force charges moves in spiral path until exhausting in standstill;-) $\endgroup$ – HolgerFiedler Mar 22 '17 at 13:16
  • $\begingroup$ @EmilioPisanty what about charged particles in a synchrotron accelerator ? Large magnets are used to accelerate the particles. $\endgroup$ – Bill Alsept Mar 22 '17 at 19:52
  • $\begingroup$ @BillAlsept If you have a specific question about that disparate topic, please ask it separately. But if you can't distinguish between a static and an oscillating field, the answer is unlikely to go beyond "see your preferred introductory EM textbook". $\endgroup$ – Emilio Pisanty Mar 22 '17 at 20:18

Yes, the radiation in those two situations would be drastically different. For concreteness, I'll consider the following two situations:

  1. An electron is dropped in as a wavepacket with mean velocity $\mathbf v$ and a small spatial extent $\sigma$ a large distance away from a proton at rest, with some nonzero impact parameter, in such a way that the total energy is negative (so the electron is bound and only a negligible fraction of the electron population will be scattered to infinity) but arbitrarily close to zero.

  2. An electron is suddenly injected in a wavepacket of spatial extent $\sigma$ and central velocity $\mathbf v$ into a region with a uniform, static magnetic field $\mathbf B$ orthogonal to $\mathbf v$, with no additional external fields. Since this is an accelerated charge, it will radiate, and therefore slowly spiral down to rest at some location close to the center of its initial circular orbit.

In situation 1, the electron will radiate its energy through all the possible radiative decay pathways, i.e. all the possible dipole-allowed steps down the Bohr-model states, until it reaches the ground state. You will observe a bunch of different photon energies, all of them of the form $\hbar\omega= \mathrm{Ry}\left(\frac{1}{m^2}-\frac{1}{n^2}\right)$, starting small (since the spacings in the Rydberg region are small) and growing until you get to the Lyman-alpha line.

In situation 2, the electron will radiate exclusively at the cyclotron frequency at that magnetic field, $$ \omega_c = \frac{eB}{m_e}.$$ (This frequency gets its name from the fact that cyclotrons operate at that frequency, but it is a much more general concept.) From a classical perspective, the electron will follow circular orbits, with the crucial characteristic that the frequency of the orbit is independent of the orbit's size or the electron's velocity or energy. This means that the emitted radiation will always be at that frequency.

From a more quantum perspective, the electron has been prepared in a superposition of excited Landau levels, and through coupling to the electromagnetic vacuum it will experience radiative decay down that ladder. Mirroring the classical constant-frequency property, the Landau ladder is harmonic and equispaced, i.e. all the levels are separated by $\hbar\omega_c$, so you can only radiate at multiples of that frequency (and, moreover, dipole radiation is only possible between adjacent levels).

As to why you think this is useful, or a meaningful comparison, I'm still completely in the dark. And, just to not let this slide, the statement that

From this starting position it has to be possible to calculate the top speed with which the electron approaches the nucleus.

is completely meaningless - it completely relies on an underlying trajectory, which is meaningless in quantum mechanics. Starting off with counterfactual statements will not get you very far.

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  • $\begingroup$ Emilio, thanks for the first part about Bohr- and Rydberg-steps. Could you please rethink the second part. The Lorentz force is accompanied by photon emission. And since you can use a permanent magnet and this magnet definitely will not lose strength, the magnetic field doesn't contribute to the emission losses of the electron. The electron emits EM radiation an loses kinetic energy by this. It comes to rest. $\endgroup$ – HolgerFiedler Mar 22 '17 at 19:51
  • $\begingroup$ @HolgerFiedler What are you talking about? That's precisely what I'm describing. $\endgroup$ – Emilio Pisanty Mar 22 '17 at 20:17
  • $\begingroup$ Emilio, in cyclotrons "the particles are held to a spiral trajectory by a static magnetic field and accelerated by a rapidly varying (radio frequency) electric field" Wikipedia. That isn't what I described. I want to know the behavior of simple a moving electron in a homogeneous magnetic field. $\endgroup$ – HolgerFiedler Mar 23 '17 at 4:25
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    $\begingroup$ @HolgerFiedler sigh, it's almost like you're misreading everything on purpose. This is elementary material and forms part of the required background to your question to begin with. I find it very frustrating that you repeatedly try to make brash assertions about high-flying topics and yet seem to be unable to make the most basic of conceptual leaps. Did you read the Wikipedia article I linked to? I'm talking about exactly the same situation that you described. $\endgroup$ – Emilio Pisanty Mar 23 '17 at 11:03

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