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Four-Momentum of Photon / Doppler shift in General Relativity

  1. Dec 6, 2011 #1
    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][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.

  2. jcsd
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