I was reading about pair-instability supernovae, which is when gamma rays produced in the core of a large star become energetic enough to produce electron-positron pairs, which causes a loss of radiation pressure.

Am I correct in assuming that if a gamma ray of sufficient energy were traveling through free space it would oscillate between being a gamma ray and being a (perhaps virtual) electron-positron pair? Would the time it spends in the latter form cause it to take longer to get from point A to point B than expected from the speed of light?

According to this question Is pair production only with $\gamma$ photons? it requires another body nearby such as an atomic nucleus to exchange momentum with. OK - so suppose the gamma ray is traveling through space, but there are a few atoms per cubic meter of ionized gas. Would the effect occur then? Is the result different than the speed of light through the medium or the effect through scattering? (I.e. difference between actually bouncing off atoms vs. passing near atoms.)

  • $\begingroup$ It is said that gamma's produced inside the sun's core take about 1 million years to emerge at the surface, but that's more of a figure of speech. It's not the same gamma that comes out, but thermal radiation that was always in very close equilibrium with the surrounding matter, and that's the key. Radiation, in equilibrium, doesn't "travel", at all, it's basically stuck stochastically in one place. Only if there is a thermal gradient do we get a heat flow. $\endgroup$
    – CuriousOne
    Jun 15, 2016 at 20:16
  • $\begingroup$ A photon needs an interaction for pair production, otherwise conservation of energy and conservation of momentum cannot be both conserved. Therefore, a photon traveling alone cannot produce a particle pair. $\endgroup$
    – Peter R
    Jun 15, 2016 at 20:17
  • $\begingroup$ In the context of QFT, you're asking about the one-loop mass renormalization of the photon. It can be shown that there is none, i.e. the gamma ray will travel at $c$ even with this taken into account. $\endgroup$
    – knzhou
    Jun 15, 2016 at 20:50

1 Answer 1


As said in the comments, a photon by itself cannot produce the pair — it is easy to see by conservation of energy and momentum. However, those oscillations between photon and pair that you mention are indeed present in quantum mechanics. In simple terms it can be understood as a consequence of Heisenberg uncertainty principle: for short periods of time the energy may be not conserved. It is called vacuum polarization or photon self-energy. But full treatment of this in Quantum-Electrodynamics (QED) leads to the conclusion that hypothetical free photon does not acquire mass due to this interaction in the vacuum — hence, photons of all energies move at the speed of light. In the end, there is experimental evidence.

A photon in the medium is a completely different story. It does not make much sense to talk about propagation of a single photon in a dense medium — it is immediately scattered and/or absorbed. Instead, one has to consider an effective state (a quasiparticle) that acts as the photon, but has slightly different medium-dependent parameters (e.g., small mass and different interaction strength). Although the quantum-mechanical treatment exists as well, this is much easier to understand in terms of Maxwell electrodynamics. There, a propagating wave in the medium as well changes its properties and moves slower than the speed of light.

(I.e. difference between actually bouncing off atoms vs. passing near atoms.)

There is a big difference. Quantum-mechanically, passing near atoms means nothing — only act of interaction can influence the particle. But quantum-mechanical interaction happens much more often than you think. What is meant by "passing near atoms" is a long-range interaction between the atom and photon that allows it to satisfy energy/momentum conservation and produce particles.


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