Re: Redshift and Planck's Equation

The photon is always emitted at the same energy in the rest frame of the
original atom. However, if you are moving with respect to that atom, you will
measure the energies of both the atom and the emitted photon differently.
Exactly how you measure them requires relativity to analyze, but the effect is
there nonrelativistically as well.

For example, if I am in a train moving away from you, and I throw a ball out the
window towards you, I would measure it as having a certain amount of energy in
my own rest frame. You would measure it as having less energy, because you would
measure its speed as being less than I would. And if the train is moving towards
you, the opposite happens - you see the ball as having more energy than I see
it as having.

So if the atom is moving away from you, the photon will appear to you to have
less energy; if it is moving toward you it will appear to have more. The speed
of the photon doesn't change, of course, but its frequency (and therefore
energy and "color") do. In either case, though, energy is conserved during the
total reaction according to both observers (you and the one on the star); you
just measure the energies of the individual objects involved differently.

Here's another way to think about this problem: Suppose you were initially at
rest with respect to the atoms which are emitting photons of a single energy.
Now you start moving away from them. What has happened to the energy of the
photons? Nothing. But something has happened to you, and you now measure the
energies of the photons differently. Is the energy of the photons conserved?
Yes (although you need to take your own motion into account to determine this,
if your motion is changing while you are measuring the photons).

A reference that I like for special relativity is Spacetime Physics, by
Taylor and Wheeler. (Unfortunately, I don't know whether it is still in
print.)