akaSmith said:
I'm not transferring into the photon's rest frame. I'm using SR's predicted time rate, as seen by an observer in their rest frame, not the photon's frame. Does the prediction work or not?
First of all you should not think about photons yet. First you should learn about classical electrodynamics, which is a relativistic field theory and that's why Einstein discovered special relativity while thinking about electrodynamics of moving bodies or more formally about how to transform from one inertial frame of reference to another such that Maxwell's equations are form-invariant, as it must be for any theory.
At the same time the electromagnetic field is somewhat special, because it is a massless vector field. This means that the free electromagnetic field, i.e., a field in a vacuum is a wave field with a frequency ##\omega## and a wave number ##\vec{k}## related by ##\omega=c |\vec{k}|##, where ##c## is the phase velocity, which is the speed of light. Any free massless field has this phase velocity, and it's a special velocity in the theory of relativity, because it's the "limiting speed", i.e., two inertial reference frames can have at most the relative speed ##c##.
Particularly there are no "photons" at rest. It's anyway wrong to think about photons as particles. They are far from being particles already by the fact that there's no proper definition of its position. That's another specialty of the masslessness of the electromagnetic field. It's better think about photons as representing an electromagnetic field with the smallest possible intensity possible for a given frequency. Technically speaking it is a socalled Fock state of the quantized electromagnetic field. The energy of a photon is given by ##\hbar \omega## and its momentum ##\vec{q}=\hbar \vec{k}##.
If you take the analogy with massless particles, it's clear that there cannot be a rest frame of a massless particle, because of the energy-momentum relation ##E^2/c^2=\vec{p}^2##. That means that the speed is ##|\vec{p} c^2/E=c## wrt. to the inertial reference frame we observe the particle in (i.e., our rest frame). Now there's no frame of reference, which moves with a velocity ##c## or larger, i.e., we cannot transform into any frame, where the particle is at rest. It moves with ##c## in any reference frame.
What changes, however, is the frequency as well as the wave vector (or the energy and the momentum of the corresponding photon) when looking at the em. wave from a different frame of reference. A natural specification of the frequency and wave vector is the rest frame of the source of this wave, but you can of course transform to any frame of reference, where the light source moves and the frequency and wave vector change when looking at the wave from a frame, where the light source moves. That's the Doppler effect (i.e., if the light source moves towards you, the frequency is large than when you look at it frome the frame where the light source is at rest). But the change is always such that ##\omega^2-c^2 \vec{k}^2=0## in all frames (or equivalently for the corresponding photon ##E^2-c^2 \vec{p}^2=0##, which tells you that the mass (more precisely the invariant mass, which is the only mass we use in physics nowadays) is always 0. This again implies that in any frame the phase velocity of an em. wave in vacuo is ##c##, the universal limiting speed of special relativity.
