Spontaneous pair production?

Question

So I've been looking into particle-antiparticle pair production from a gamma ray and don't understand one thing.

Let's say I have a 1,1 MeV photon and it hits a nucleus - electron-positron pair with some momentum will be created and the nucleus will probably get some momentum as well because of the impact.

But why does the photon need the nucleus at all? Why can't it just fly through space and suddenly, with some probability, change into a electron-positron pair with momentum? I see that the momentum of the system wouldn't be conserved but I don't really understand how the nucleus helps it.

Thank you a lot!

Answer 1

look at energy-momentum conservation:

$$p_\gamma = p_1+p_2$$

the photon has invariant mass 0 wheras the electron and positron have mass $m_e$

$$p_\gamma^2 = (p_1+p_2)^2 = p_1^2+p_2^2+2p_1\cdot p_2$$ $$0 = 2m_e^2 + 2p_1\cdot p_2$$ $$-m_e^2= p_1\cdot p_2 = E_1E_2-|\vec{p_1}||\vec{p_2}|cos\theta > E_1E_2-|\vec{p_1}||\vec{p_2}| = E_1E_2(1-\beta_1\beta_2) > 0$$

The betas cannot be greater than one. So the right hand side always stays positive. The nucleon helps. because it changes the initial state to one with a invariant mass greater than zero.

Answer 2

Pair production is not the same as decay of a particle. A particle can decay into two components according to its decay probability without needing an extra interaction. A lamda in its rest frame will decay into a proton and a pion, for example, within a predictable decay time .

There is no rest frame for the photon since its mass is 0 and it is always travelling with the velocity of light. If it were to decay spontaneously into an electron positron pair, they do have a rest mass and a rest frame, and their invariant mass would be at least 2*m_e, which should have been the mass of the photon. A contradiction.

It can interact though with the fields of other particles . How does the photon interact? The interaction probabilities can be calculated given the charges of the target particles, the easiest way using Feynman diagrams. One can envisage a photon as sequentially turning into virtual loops of e+e- . One of the virtual electrons interacts with the field of a real charged particle exchanging enough energy and momentum so that both e+ and e- become real while energy and momentum are conserved in a three body interaction.

The nucleus helps by ensuring the momentum in the final state (e+e_Nucleus) to be the same as the one in the initial state (photon nucleus).

Answer 3

If you look at a photon in a vacuum, there is no preferred inertial frame of reference. And the energy of the photon is dependent upon which frame is chosen. So some frames have more than the needed energy, others less. It would be really awkward to have pair production in reference frames that lack the needed energy.

Answer 4

I think that you can see a process like this...

photon + photon -> electron + positron

without a nucleus. It is an endothermic collision (only gamma rays) and surely its rate grows rapidly above the threshold.