According to classical electromagnetic theory, accelerated charges should emit radiation and lose energy. The reason given in my book why atoms don't emit radiation (say, when the atom moves along a circle) is because the atom is neutral. I can understand how this works for a neutral particle like a neutron but the atom has constituent charges within in. How can the "presence" of an opposite charge nearby stop what seems to be an intrinsic process independent of the surroundings? Do the electrons and protons emit radiation that destructively interferes or something of that sort?

  • 4
    $\begingroup$ What silly book is this? $\endgroup$ – BMS Apr 17 '14 at 5:58
  • $\begingroup$ possible duplicate of Where did Schrödinger solve the radiating problem of Bohr's model? $\endgroup$ – kleingordon Apr 17 '14 at 6:03
  • $\begingroup$ I don"t think its a duplicate. The question you refer to says that once we consider the electron as a wave as well, it cannot be considered as accelerating and hence does not emit radiation. I mean to ask if I whirl an atom in a circle, clearly the charges are accelerating, why doesn"t(if it indeed doesn"t) the atom emit radiation. $\endgroup$ – user42991 Apr 17 '14 at 6:14
  • $\begingroup$ @user42991: Remember neutron is composed of charged quarks, so there is no difference in considering atom or neutron. You seem to differentiate atom and neutron as non elementary charged and elementary charged respectively. $\endgroup$ – Immortal Player Apr 17 '14 at 6:16
  • $\begingroup$ Sorry, my bad. Anyway, what is the answer to the question? $\endgroup$ – user42991 Apr 17 '14 at 6:26

Your book is wrong. "Atomic Bremsstrahlung" is a thing, and occurs when an neutral atom has a dipole moment and is accelerated somehow.

As a practical matter, situations in which something as massive as an atom is accelerated up to a sizeable fraction of its rest-mass, while at the same time not being ionized by the forces involved are pretty few and far between, so the phenomena doesn't come up that often.

But there's nothing, quantum or classically, about being part of a larger neutral system that prevents a charged particle from radiating if the whole system is accelerating.


According to classical electromagnetic theory...

I think this is the key assumption you are building your question on, and for atoms/nucleons/electrons/everything smaller this assumption just doesn't hold true. All these objects have to be described with quantum mechanics, so there is no trajectory of a localized charge or something alike - all you're left with is a probability density to find a localized charge at a certain point in space-time. This probability-density might change over time, but there is no classical acceleration, there is not even a particle...

  • $\begingroup$ I thought Newtonian mechanics was a special case of quantum mechanics. So even if the precise location of the particle cannot be determined on a large scale can"t it be said to approximately move in a circle? Also, what explains thermal radiation then in the quantum mechanical sense? Classically it is due to accelerating charges. $\endgroup$ – user42991 Apr 17 '14 at 6:50
  • $\begingroup$ Newtonian mechanics is not a special case, it's better to look at it as a big average hiding all the tiny quantum effects/fluctuations. But please get rid of the idea of "something moving" - there is NOTHING, particles are only a model to describe reality, and that model is false here $\endgroup$ – Benedikt Apr 17 '14 at 7:07
  • $\begingroup$ According the quantum mechanical books I have read, most recently "In Search of Schrodinger"s cat", Newtonian mechanics is a special case. Many devices(say, the aeroplane) have been built taking Newtonian mechanics into consideration and they work. Newtonian mechanics is a good approximation. Granted, the uncertainty in the position and or velocity of the atom orbiting will be great, owing to its low mass, but that does not mean there is no acceleration. There will be uncertainty in the acceleration. As far as most people are concerned detection in experiments= reality. $\endgroup$ – user42991 Apr 19 '14 at 3:30
  • $\begingroup$ Maybe that's just an argument about words - a special case is a subset of a theory, taking some assumptions that make the calculation easier, but still using the same equations and the same theory. For example in gas dynamics, you often consider the special case of a ideal gas of point like hard interacting particles - they are not point like but the error is often negligible. With quantum mechanics and newtonian it's different - you'll never describe the plane as a quantum object, it's a system of myriads of quantum objects. All effects summed up result in the newtonian laws. Its a superset. $\endgroup$ – Benedikt Apr 22 '14 at 11:15

Down to quantum size as atomic scale, particle idea does not work anymore. Instead in quantum mechanics wave function and eigenstate are basic concepts. Physical quantities familiar in classical theory are operators and their expectation or average value is what one observes in the classical sense. To tell if an electron is moving it's the electron density distribution that one should look at.

When you say an atom does not emit EM radiation, you assume it sits in the ground state, the lowest eigenstate. The electron density of an eigenstate is invariant as time flies. Interpreted classically, as you like, the electron in the atom is sitting there doing nothing, so no radiation occurs. On the other hand, atoms do radiate in many situations. That's because electrons jump between eigenstates. Now the electron density is changing with time, meaning classically they are accelerating and so emitting photons.

As for free electrons, such as in an electron beam, their wave function (a wave packet) is not confined. Therefore they don't have discrete eigenstates to stay in. They are traveling all the time, more similar to the classical world. When the velocity (still the expectation value) is not constant, radiation will happen.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.