6
$\begingroup$

I have found this question here Can a free particle absorb/emit photons?, along with other resources that show a free particle cannot emit a photon (in a vacuum).

Now, I am 90% sure it does, but does this result hold generally, ie not just photons/ photons but in a medium?

If you go in the rest frame no matter what it emits energy is not conserved. The reason I am questioning this is that an exam question said 'if it can emit it find the angle relative to the motion of the electron'. Seems odd that they would give away marks for just repeating that it can't.

$\endgroup$
  • 1
    $\begingroup$ How about a free neutron decaying to a proton, electron & antineutrino? $\endgroup$ – PM 2Ring Dec 31 '18 at 13:06
  • 1
    $\begingroup$ But then it changes rest mass, it doesn't just emit things $\endgroup$ – Toby Peterken Dec 31 '18 at 13:11
10
$\begingroup$

True and true - there is no contradiction here.

The complete answer is contained in the answer to linked question:

This is because energy and momentum are not both conserved if a free charged particle (say, an electron) emits a photon

and the comment to that answer:

Note that if the particle has internal structure, this argument can fail. For example, atomic nuclei emit gamma rays. This is because they have more than one internal state with different energies.

Applying these two principles we arrive at:

  1. A single free electron cannot emit a photon
  2. A single free hydrogen atom with electron in an excited state can emit a photon by spontaneous emission
  3. A single free nucleus can emit a photon again by relaxation of its internal structure.

Note: In this answer I have only considered the dynamical restrictions involved in emitting a single photon. As pointed out in the comments there are many more complex decay processes of either elementary (e.g. tau) or composite (e.g. neutron) particles which may result in the emission of a photon as one of many decay products.

$\endgroup$
  • 2
    $\begingroup$ What about lepton decay, say the Tau particle decaying into an electron and a few neutrinos? I think it's not that much about elementarity but rather about the existence of a final state which in particular has a lower total mass than the initial state. edit: However, the argument makes sense if you modify it a little: An elementary particle can only decay into different particles which would not fit into the "emission" picture. In order to emit something, it needs to be a compound particle such that some interaction energy is involved which can be used to ensure energy conservation. $\endgroup$ – Photon Dec 31 '18 at 14:15
  • 3
    $\begingroup$ @Photon OK yes there is more to it: the original linked question referred to a single particle emitting a single photon without changing. The two quotes I gave are still correct. $\endgroup$ – Bruce Greetham Dec 31 '18 at 14:25
  • $\begingroup$ Yes, energy-momentum (4-momentum) should be conserved, that's why I wrote "in particular". My only point was that it is necessary to differentiate more clearly between a decay and a particle emission. But maybe your answer does so sufficiently, I might have read it not carefully enough and identified decay and particle emission which is clearly wrong here. My bad, sorry for the possible confusion! $\endgroup$ – Photon Dec 31 '18 at 14:42
  • 1
    $\begingroup$ @Photon I think your comment was very pertinent - I learnt from it thanks. $\endgroup$ – Bruce Greetham Dec 31 '18 at 14:55
  • 1
    $\begingroup$ @rob thanks: I have tried to edit to capture the main points of your and Photon's feedback - please reply if I can clarify further (without turning answer into a complete survey of particle physics!) $\endgroup$ – Bruce Greetham Dec 31 '18 at 18:19
1
$\begingroup$

If you go in the rest frame no matter what it emits energy is not conserved.

But note that this argument fails if the parent particle is massless, since then there is no such rest frame. A gluon can decay into two gluons. See Decay of massless particles .

$\endgroup$

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.