# Spontaneous pair production?

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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This question is related, but not identical: physics.stackexchange.com/q/12488 – Ben Crowell Aug 14 '11 at 22:57
for par production to happen, it needs a high electrical field. So that's the reason why pair production happens near nucleus...en.wikipedia.org/wiki/Virtual_particle#Pair_production – Vineet Menon Sep 14 '11 at 9:48
The process you describe violates the energy-momentum conservation. Two photons, however, can produce a pair of fermions in the vacuum. – Solenodon Paradoxus Dec 26 '15 at 11:19

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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The second paragraph is sufficient. – Ben Crowell Aug 14 '11 at 22:54

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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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.

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Look at the diagram in Wiki en.wikipedia.org/wiki/Two-photon_physics . The interaction with the nucleus happens through one of the photons being virtual. This has a much higher crossection since the size of the nucleus is of the order of a fermi, and a photon can have a chance of hitting its field. A photon meeting another photon in a beam beam interaction has very low crossection, since the photon is practically pointlike ( there is a cloud of virtual vacuum pairs accompanying each photon, but the scattering probability is low). – anna v Nov 13 '12 at 7:53

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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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. – Ben Crowell Aug 14 '11 at 22:56
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. – 71GA Jan 13 '13 at 12:57
he's saying that since photon momentum cannot be zero it's a violation. – user18764 Mar 7 '14 at 17:39

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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I'm interested in how this meshes with (for instance) the GZK limit on cosmic ray energies, above which cosmic ray protons will interact with the microwave background. – Richard Terrett Aug 14 '11 at 14:20
It would be nice to mention the effect of electron-positron attraction (potential energy contribution) into all that ;-) – Vladimir Kalitvianski Aug 14 '11 at 18:02
@Richard Terrett: The GZK limit is a limit relative to the CMB's rest frame. – Ben Crowell Aug 15 '11 at 2:23