Assume I have a closed system of a stationary proton and a photon with sufficient energy for electron pair production to take place (say, $1.122\,\mathrm{MeV}$), where all the laws of physics apply.enter image description here Taking the rest energy of a proton to be $938.272\,\mathrm{MeV}$, the total energy of this system is $939.394\,\mathrm{MeV}$.

Now, when my photon undergoes pair production, I have an electron-positron pair with a combined K.E. of $0.1\,\mathrm{MeV}$ and a proton. The total energy is still the same as before.

enter image description here

However, the proton and electron attract each other, thus increasing the kinetic energy of both particles and the system as a whole. Supposing that the collision between the two particles is elastic, the energy of the system has now increased permanently.

When two charged particles attract or repel each other, the gain in kinetic energy is a result of the loss of potential energy. In this case, however, the photon didn't have any potential energy.

So where does this energy come from?

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    $\begingroup$ Wouldn't the proton have potential energy due to its interaction with the newly created positron and the electron? $\endgroup$ Aug 22, 2017 at 2:45

2 Answers 2


Pair production is a quantum mechanical phenomenon, and electrons and photon and protons are quantum mechanical entities and their interactions follow the laws of nature at the quantum mechanical level.

pair production

It is not simply the photon that turns into a particle pair, because this cannot happen: At the center of mass of the pair the momentum is zero, whereas a gamma can never have zero momentum. This answers your "In this case, however, the photon didn't have any potential energy."

A charged field is necessary, in this case provided by the Z of the nucleus, in your case a proton. So what happens is that two electromagnetic vertices enter: the photon vertex generating a real positron and a virtual electron which virtual electron scatters with a virtual gamma off the field of the nucleus.

Although potentials enter in the simple quantum mechanical equations they are not useful in the field theoretical representation in which calculations using Feynman diagrams are done. And the above is a lowest order diagram. The potential between the receding electron and positron and the proton will be represented by a summation of virtual photon exchanges with the proton, if one wanted to go to the trouble of it. But as each exchange introduces a vertex that depresses the crossection by 1/137, the effect is very small. The energy, potential and kinetic, in the interaction is provided by the incoming gamma energy .


EDIT: I interpreted this question in a semi-classical context, where photons and pair production exist, but the particles in question are sufficiently separated and localized to be described by classical electromagnetism. In particular, I assumed that the proton was too far away to meaningfully interact with the incoming photon(s).

First, one photon cannot decay into an electron and a positron. You need two photons for that - otherwise you can't conserve both energy and momentum (see above disclaimer - not true if the proton is sufficiently close by).

Secondly, before the pair production, the photon had no potential energy; after the pair production, the electron and positron have (initially) precisely the same potential energies due to their interaction with the proton, just with opposite signs, so they cancel.

Because we're being semi-classical, we can't talk about the electron+positron potential energy until they've separated somewhat, but whenever we do, that energy comes at the expense of their initial kinetic energy. In other words,

$$ E_{\gamma 1}+E_{\gamma 2} = KE_{p}+KE_{e-}+KE_{e+}+2m_e c^2 + PE_{p,e-}+PE_{p,e+}+PE_{e-,e+}$$

The mass-energy of the proton never comes into play, so we can ignore that.

We've also ignored the radiation due to the accelerating charged particles, but the concept is still the same - just tack that on to the right hand side.

  • $\begingroup$ Your first sentence is true in free-space but is incorrect in the presence of a heavy charge (a high Z nucleus is better than a proton, but the OP's scenario is OK within some set of parameters). This process is one half of the goings on in an electromagnetic shower. Other than that you have the core of a good answer. $\endgroup$ Aug 22, 2017 at 2:49
  • $\begingroup$ @dmckee Fair point. I interpreted this question in the context of classical electromagnetism + a sprinkling of quantum effects - edited to reflect that. $\endgroup$
    – J. Murray
    Aug 22, 2017 at 3:27
  • $\begingroup$ @dmckee: the presence of the proton can only allow the photon to decay if there is some virtual particle exchange between them. On its own the photon would not produce two particles that are on-shell, because it would violate momemtum conservation. Same situation in an electromagnetic shower. $\endgroup$ Aug 22, 2017 at 3:50

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