0
$\begingroup$

Consider the question whether a photon can decay, in the free space, into an electron and a positron - a pair production event. Such an event I don't think is possible, because both energy and momentum cannot be conserved (please correct me if I am wrong). When trying to show this I worked totally in the center of mass frame of the decay products. We then get:

Conservation of energy: $$E_\gamma=E_{e^-}+E_{e^+}$$ and conservation of momentum: $${E_\gamma \over c}=0$$ I would then have concluded that since in the center of mass frame we have $E_{e^-}=E_{e^+}$ that $E_{e^-}=0$ and that this is not possible.

I, however, think that this analysis is wrong since it totally ignores the potential energy between the positron and the electron due to their opposite charges. If this was included we could indeed have $E_{e^-}=0$ and then a photon could 'decay' into a positron and an electron in empty space.

So which analysis is right? Also how would I incorporate potential energy into this situation?

$\endgroup$
4
  • $\begingroup$ related: physics.stackexchange.com/q/22916/58382, physics.stackexchange.com/q/13513/58382 and links therein $\endgroup$
    – glS
    Commented Jan 15, 2015 at 9:44
  • $\begingroup$ The lowest energy $e^+ e^-$ bound state has energy $2m_e - 6.8\, \text{eV} > 0$, so even taking this into account the decay is not possible. $\endgroup$ Commented Jan 15, 2015 at 10:02
  • $\begingroup$ @RobinEkman Yes but we could define that potential to be as negative as we want, simply by changing the point we detonate to have 0 potential. I could define the potential to have the value e.g. $-2m_e$ at the point of decay meaning that the total energy would be 0. $\endgroup$ Commented Jan 15, 2015 at 10:19
  • $\begingroup$ @RobinEkman : and how big is the energy of a gamma ray? $\endgroup$
    – Sofia
    Commented Jan 15, 2015 at 11:07

2 Answers 2

1
$\begingroup$

Your analyses is not wrong, the photon is a stable particle. The most immediate argument is that linear momentum cannot be conserved. In the center-of-mass frame of the produced pair, the total linear momentum is zero, while the photon should have a linear momentum since there is no frame in which the photon is at rest.

$\endgroup$
0
$\begingroup$

Sorry but your question has a little error. The photon cannot decay into a pair in empty space. It has to be a quantized spectrum of energy levels present for this process to happen. This quantized energy level configuration can be provided with the presence of an atom (where the threshold energy for the process is $2mc^{2}$) or an electron (where the threshold energy for the process is $4mc^{2}$). In the case of the presence of an atom the recoil energy of the atom is very little due to his large mass compared with the $e-p$ but in the case of the presence of an electron his recoil momentum cannot be neglected and in this case we speak of an "triplet production" because we have a $e-p$ pair and a recoil electron.

$\endgroup$
1
  • $\begingroup$ to your attention, Joseph says that the photon decay is impossible, but his formulation of the question was unclear. So, I suggest that you take that into account in your answer. $\endgroup$
    – Sofia
    Commented Jan 15, 2015 at 11:17

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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