# Energy variance, and pair production

1. Apr 19, 2010

### Matterwave

Ok, so here's a question. The energy of say photons is frame-dependent. Photons are blue-shifted or red-shifted depending on my velocity towards or away from the source. However, what happens when I apply this to pair-production? For example, if my photons are energetic enough, they may create a positron-electron pair. If my photons are NOT energetic enough they don't create the pair.

However, since the energy is frame-dependent, it would seem to suggest that whether a pair is created or not is frame-dependent. The existence of particles, however, should not be frame dependent. Paradox! Obviously my logic is faulty somewhere.

Please resolve this for me. Thanks. =)

2. Apr 19, 2010

### Staff: Mentor

A photon all by itself cannot convert into an electron-positron pair and conserve both energy and momentum. There has to be another object in the picture, typically an atomic nucleus which is normally (almost) at rest. In a reference frame in which the photon has a small energy, the nucleus has a lot of kinetic energy which can be contributed towards creating the pair.

3. Apr 19, 2010

### Matterwave

What if I replace the atomic nucleus with another photon? Are you saying 1 photon will be blue-shifted and one will be red-shifted or some such?

4. Apr 20, 2010

### Phrak

Assume this occurs as the reverse of electron-positron annihilation where two photons are generated.

Take the annihilation event and work it backwards. In the center-of-mass frame of two leptons, the photons will travel in opposite directions with equal wavelength so that momentum is conserved, as jtbell elucidated.

If the center of mass has a non zero velocity, the wavelengths will not be equal.