I understand that energy of photons is defined by their wavelength/frequency. This frequency (and so energy) will be different for different observers: observer moving towards the photon will see higher frequency than observer moving away from the photon. So basically, the photon might look more energetic from one reference frame and less energetic from some other. It means that photon's energy is based on the observer?

When there are two highly-energetic photons, pair production of electron/positron might take place. How can this actually happen when the energy is based on the observer? It would mean that some observer might see pair production and another one wouldn't see it at all?

  • $\begingroup$ In a frame of reference where one photon has more energy, the other has less; and the particles produced will also have different energies depending on the observer. It all adds up. $\endgroup$ – Floris Apr 29 '16 at 22:32
  • $\begingroup$ But what about a reference frame, where the photons does not have enough energy to create particles? $\endgroup$ – Vojtěch Apr 30 '16 at 5:30
  • $\begingroup$ You need conservation of energy and momentum for pair production. This necessitates the photons traveling opposite directions with a sum energy exceeding 1022 keV regardless of any other momentum they may have. And that won't change with reference frame. Can you draw an example where you think the math would fail? $\endgroup$ – Floris Apr 30 '16 at 11:56
  • $\begingroup$ Ah, okay, i did not realize the photons need to trave opposite directions, i thought they might travel along each other in the same direction. $\endgroup$ – Vojtěch Apr 30 '16 at 12:10
  • $\begingroup$ There's a bit of confusion here. You do not need a pair of photons. A single photon of sufficient energy near a nucleus will do quite nicely. en.wikipedia.org/wiki/Pair_production $\endgroup$ – dolphus333 May 3 '16 at 20:49

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