Physics Asked on December 19, 2020
In a relativistic context, the energy of a "single" photon, thought of as a massless particle with an on-shell condition $p^2=0$ or $E^2=vec{p}^2$, depends on the frame. In other words, performing a boost preserves the light-like property of its four-momentum but it doesn’t preserve the actual components.
If we consider a scenario at $T=0$, so ignoring any thermal effects, let us imagine we have a bath of photons following some distribution due to some production mechanism. Will the photon number density observed depend on the frame of the observer? Would that mean that different observers see a different number of photons?
What about the case at $Tneq 0$? Will having a temperature break Lorentz invariance?
In any inertial frame, the number of photons is frame-independent. Just suppose you have an ideal photon detector that scoops up all the photons, clicking each time. The number of clicks is frame-independent, so the number of photons is too. Since volume is frame-dependent, that means the number density of photons is frame-dependent as well.
Subtleties appear when we consider non-inertial frames, in which case the essential problem is that detectors with relative acceleration have different definitions of the vacuum state, and hence different definitions of photons, which are excitations about it. Thus, the detectors don't even agree on the number of photons. This is the Unruh effect.
Correct answer by knzhou on December 19, 2020
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