I've seen in other questions that pair production is not possible in vacuum, i.e. it needs a nucleus or something massive in order for conservation of momentum to hold. I've seen the derivation of that, where you end up with energy of the photon not even being larger than $2m_e c^2$ of the electron and positron. I think I got that covered.

But how does the equations/math look if I were to include the nucleus ? Do I just add a nucleus on the left and right side of the equation ($p_\gamma + p_\text{nuc} = p_\text{nuc'} + p_e + p_p$), and then just go to town with that, or is there something tricky about it ? Because after having be searching for it on Google, there really doesn't seem to be any "proof" of how this is done. So either is really easy, or not as easy as I think it might be.

So yes, what is the right way to do it ?


A single photon cannot decay into anything with mass because the impulse-energy relations for massive and mass-less particles differ by the rest mass, so you cannot conserve both impulse and energy. If you add a particle to absorb impulse and energy, you add enough freedom through new variables that this system of 4 equations (energy and impulse in three spatial directions) has solutions, which exist if your photon has enough energy and the particle to absorb impulse and energy is sufficiently heavy (and you can trade-off between enough energy and enough mass to some extent).

The impulse vector equation you wrote is sensible and correct: Prior total impulse is final total impulse, with that of the nucleus changed. You can simplify by assuming your nucleus is at rest initially (possibly even if it flies inside a particle accelerator by doing the problem in its rest frame). Do not forget to look at energy conservation next.


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.