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According to Bohr's atomic model, the electrons do not radiate energy when they revolve in the predefined orbits. But according to Maxwell's theory of electromagnetism, an accelerated charge produces electromagnetic waves and loses energy.

In Bohr's atomic model of an atom the electrons are in a circular motion so they are essentially accelerating thus they should lose energy due to the reasons mentioned above and eventually collapse into the nucleus. But it doesn't happen, so isn't it a violation of the EM theory?

Edit: The question is related to consensus of Bohr and Maxwell rather than Rutherford and Bohr.

think__tech
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Yes. Bohr's model is not really a theory at all -- it's just the observation that you get the right atomic spectra if you simply assume the particle may only have angular momentum $L = n \hbar$ and that the $n = 0$ state is stable. No reason is given for this, so of course it contradicts Maxwell's equations.

Bohr was simply trying to find the simplest assumptions that would fit the data. The real explanation is quantum mechanics, which of course is not compatible with classical electromagnetism. Roughly speaking, in quantum mechanics the configuration of the electron in a given orbital is a standing wave. Unlike classical particles, standing waves have a lowest possible frequency, which musicians know as the fundamental. When the electron is in this state, it can't lose any more energy, so it doesn't radiate.

knzhou
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    So, does wave mechanics violate Maxwell's theory? Does not the electron still accelerate? – Archisman Panigrahi Oct 04 '18 at 13:24
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    @ArchismanPanigrahi You can try to have quantum matter (which is what I assume you mean by wave mechanics) along with a classical electromagnetic field. That was what was initially tried historically, and it works fine in certain regimes, but ultimately it doesn't make sense -- you can't have a sensible theory if only some things are quantized. The right theory is quantum electrodynamics, where the electromagnetic field is quantum as well, and that's very far from Maxwell's theory. – knzhou Oct 04 '18 at 13:27
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    @ArchismanPanigrahi In any case, you shouldn't speak of whether the electron is accelerating in quantum theory, because it's simply not a particle with a definite position. An electron in its ground state is in a stationary state analogous with a standing wave. It's not "going" anywhere. – knzhou Oct 04 '18 at 13:30
  • @knzhou As far as I know, Bohr's model was proposed as a replacement to Rutherford's model. And one of Rutherford's model's flaw was that it couldn't explain why doesn't the electron collapse into the nucleus despite it radiating energy. So how come was Bohr's model accepted over Rutherford's model? PS: This was by far one of the best explanations I've got! But I just wanted to clear it completely, I hope its ok... – think__tech Oct 04 '18 at 13:51
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    @think__tech It wasn't accepted as an explanation, it just fit the data better. Back then nobody knew at all how an atom worked. The Bohr model is an example of the "old quantum theory", a family of ad hoc assumptions that kind of worked, which were eventually explained and replaced by quantum mechanics as we know it. – knzhou Oct 04 '18 at 13:58
  • @Archisman Panigrahi: It "violates" Maxwell's theory in the same way that Einstein's relativity violates Newton's laws of gravity & motion. Both of those work quite well, if you don't look at things that are too small, too massive, or going too fast. – jamesqf Oct 04 '18 at 23:47
  • Just because it's a standing wave doesn't mean there's no acceleration. Apparently you can define an acceleration operator in QM. If this operator has non-zero expectation in the electron ground state shouldn't we still expect radiation? Or is this prevented by conservation of energy and lepton number (there is no lower energy state of an electron and we can't make the electron disappear)? But can this acceleration operator be used to predict the radiation from an excited state? – Alex Oct 08 '18 at 23:19
  • @Alex I think radiation is more complicated than just looking at $\langle \hat{a} \rangle$. However, in any case, $\langle \hat{a} \rangle$ vanishes in any stationary state. – knzhou Oct 09 '18 at 08:59
  • Why does it vanish? The link I gave seems to suggest the operator is $\hat{a} = -\nabla U/m$. Maybe it vanishes on a spherically symmetric state, but I don't think it will on a generic stationary state. – Alex Oct 09 '18 at 16:31
  • @Alex It's because $\langle \hat{v} \rangle$ vanishes for a stationary state; the probability isn't moving at all. So its derivative vanishes too. Go check for a few simple cases if you don't believe me! – knzhou Oct 09 '18 at 16:43
  • $\hat{v} = \hat{P}/m$. Are you saying all stationary states have zero expected momentum? That's not true. Maybe you have an unusual definition of "stationary state"? I always encountered it to mean an eigenstate of the Hamiltonian. – Alex Oct 09 '18 at 20:22
  • @Alex Indeed, it's true. Every stationary (bound) state has precisely zero expected momentum. That includes every orbital of the hydrogen atom. Again, please go check it if you don't believe me, or show me a counterexample! – knzhou Oct 09 '18 at 20:25
  • I think you're confusing three concepts: stationary, bound, and symmetric. They all mean different things. If the potential is symmetric then energy eigenstates are either even or odd and the expected momentum for an even or odd state is zero. If the potential is not symmetric then the energy eigenstates can have finite expected momentum -- e.g. the lowest energy eigenstate of the half-harmonic oscillator. – Alex Oct 09 '18 at 21:42
  • @Alex No, I know basic quantum mechanics. There are non-bound stationary states with nonzero $\langle \hat{p} \rangle$, such as the simple $e^{ikx}$, but they are irrelevant to this question, which is about electrons bound to atoms. Bound stationary states have zero $\langle \hat{p} \rangle$. Also, I never brought up symmetry, and on top of that your example is wrong. Just actually compute $\langle \hat{p} \rangle$ for the state you're talking about, and you'll see it vanishes. – knzhou Oct 09 '18 at 21:48
  • Ok, I guess my question is then: how can you show the expectation of the momentum is zero for any bound energy eigenstate, regardless of the symmetry of the potential? – Alex Oct 09 '18 at 22:12
  • Incidentally, I think Ehrenfest's theorem sheds some light on my first question. For a state that does not depend on time (e.g. an energy eigenstate) the expectation of the acceleration operator is zero. I guess the reason why real hydrogen atoms radiate from their excited states might be that the usual $n=2$ state, say, actually isn't an energy eigenstate because of the coupling with the electromagnetic field. And so $\frac{\partial}{\partial t} \langle \hat{P}\rangle$ might not be identically zero for a real hydrogen atom. – Alex Oct 09 '18 at 22:23
  • @Alex Well, I still can't think of any reason why the radiation rate should have anything to do with the derivative of $\langle \hat{p} \rangle$. Again it's simply more complicated than that. – knzhou Oct 10 '18 at 08:32
  • I was thinking about think__tech's original reasoning that accelerating charges radiate (e.g. as described by the Larmor formula). If there's a time derivative of $\langle \hat{P} \rangle$ maybe that can be interpreted as an acceleration. – Alex Oct 10 '18 at 14:27