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I'm trying to understand a derivation / question for my revision. It has to do with the four momentum of a photon and doppler/redshifts.

1. The problem statement, all variables and given/known data

Show that ratio between emitted and received frequencies (for two observers E and R moving with velocities u and v respectively) is given by

[itex] \frac{\nu_E}{\nu_R} =\frac{\textbf{u} \bullet\textbf{k}}{\textbf{v}\bullet\textbf{k}}[/itex]

2. Relevant equations

My first issue is deriving the four-wavevector for a photon. I know this is related to the four-momentum by p = hbar * k. Which one do I derive first? Assuming I'm given the four-wavevector

[itex] k^a = \left( \omega / c, \vec{k}\right) [/itex]

Then I also proceed with the knowledge that scalar products of four-vectors are invariant under frame transformations

3. The attempt at a solution

Transform to emitter's frame (E). In this frame

[itex][u^{a}]=\left(c,0,0,0\right)[/itex]

and

[itex][k^{a}_E]=\left(\omega_E , \vec{k_E}\right)[/itex]

Taking the scalar product gives [itex]\textbf{u} \bullet \textbf{k}_{E} = \omega_{E}[/itex]

Performing similar steps gives

[itex]\textbf{u} \bullet \textbf{k}_{R} = \omega_{R}[/itex]

This bit I'm unclear on. [itex]\textbf{k}_E = \textbf{k}_R[/itex]

With this, we divide both scalar products giving the answer.

Elsewhere in my notes I have a comment saying that "metric is independent of 't', therefore [itex]\textbf{p}_0[/itex] at time of emission is the same as at the time of reception"

This is related so could someone comment on how this fits into the overall picture.

Thanks

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# Homework Help: Four-Momentum of Photon / Doppler shift in General Relativity

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