# Do composite particles that are electrically neutral but have charged constituents radiate?

For example an electron radiates when accelerated. So does a positron. But is the radiation emitted by accelerated positronium the sum of the radiation emitted by each separately? If not, why not? If so, does this provide a way of testing whether a given neutral particle is composite? For example, does a neutron bremsstrahlung when decelerated?

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It depends on how fast you accelerate the particle. For slow accelerations, no, for fast accelerations (where you see the constituents) yes. Neutrons have a magnetic moment by the way, so the answer is just yes. But I suppose you are interested in accelerating an Helium atom. –  Ron Maimon Jun 22 '12 at 1:13

In the view of classical electrodynamics, electron-positron pair would definitely emit radiation when accelerated. But in quantum mechanics, positronium at the singlet state (spin anti-parallel) with 0 orbital angular momentum will be perfectly spherically symmetric in its rest frame. In other words, it will have no net charge, net dipole moment, net quadruple moment, .... So it will not radiate. Triplet state and excited states may, however.

Neutron will surely emit radiation when accelerated, although very small, because it has non-zero magnetic dipole moment. That is not called bremsstrahlung however.

And this would not be a good way to probe whether a particle is composite or not. The higher-order radiation is small and hard to detect, and it is much more straightforward to simply measure dipole or quadruple moments.

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I'm curious how quantum mechanics conspires to cause the singlet state to not radiate. I suppose the obvious conclusion is that the probability amplitude for a charge to radiate is opposite in sign to the amplitude for a state of opposite charge (but which is otherwise identical). This is interesting, but perhaps I could have guessed that the probability amplitude would be proportional to the signed charge. OK, I'm sort of thinking out loud here, but I think this makes sense to me now. Thanks! –  user1247 Jun 22 '12 at 9:14
@user1247: This is exactly right--- the amplitude for photon emission is proportional to the charge. But the conspiracy only works for slow inverse accelerations compared to the radius of the particle (the inverse acceleration and the radius have the same units when c=1). At larger accelerations you see the substructure. –  Ron Maimon Jun 23 '12 at 8:58