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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.

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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.

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    $\begingroup$ A simpler version of this argument is the following. Go into the center of mass frame of the final state. In that frame, the momentum of the photon in the initial state must have been zero. But then that photon had zero energy, so conservation of energy has been violated. $\endgroup$
    – user4552
    Commented Aug 14, 2011 at 22:56
  • $\begingroup$ Photon really has invariant mass $m=0$ but how come you equall its momentum $p_\gamma=0$? Momentum of a photon isn't $p_\gamma = m v_\gamma =0$ it is $$p_\gamma = \frac{h}{\lambda}$$ and $\lambda \neq 0$. So how come you state that $p_\gamma=0$? I really don't understand. $\endgroup$
    – 71GA
    Commented Jan 13, 2013 at 12:57
  • $\begingroup$ he's saying that since photon momentum cannot be zero it's a violation. $\endgroup$
    – user18764
    Commented Mar 7, 2014 at 17:39
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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 lambda 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).

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    $\begingroup$ The second paragraph is sufficient. $\endgroup$
    – user4552
    Commented Aug 14, 2011 at 22:54
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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.

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Actually, the photon does not need the nucleus to change into a electron-positron pair. According to current theories, it does that all the time. The alectron-positron pair converged again into a photon after a very short tine.

I am not aware if this effect has been experimentally proven yet. The observable effect should be that light (a strong LASER beam) can scatter light (anotehr string LASER beam), because the temporarily exisiting pair's electric field interacts with other pairs' electric fields. With your regular lab LASER, the effect is far too small to be observed.

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  • $\begingroup$ This answer is actually correct except that you are talking about two or more photons interacting with each other. (Presumably by exchange of a charged particle.) Yes it is theoretically possible to have photon-photon scattering, and indeed it has not been experimentally observed. This is still a matter of open discussion among research scientists. $\endgroup$
    – user3728501
    Commented Jun 18, 2022 at 11:41
  • $\begingroup$ eg: If you shine two lasers into each other, the beams of light are not observed to scatter, so the interaction probability must be low. Whereas if you shine two proton beams into each other, they can be observed to interact (provided the beam density is high enough, etc, etc the point is obviously nuanced) $\endgroup$
    – user3728501
    Commented Jun 18, 2022 at 11:42
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Because a photon is a packet of electromagnetic energy, so it must be interacted by the same field or any of electric or magnetic field. As nucleus has a certain electric field by which photon's field actually interacts and the result of this interaction is pair production-to conserve energy momentum and charge.

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