Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Frequency shift of light between observers in Schwarzschild space-time

  1. Jul 11, 2008 #1


    User Avatar
    Gold Member

    Considering the simple case of two observers O1 and O2 lying on the same radius at positions r=r1 and r=r2 respectively.

    Using a result from Stephani(1) I work out that the ratio of frequencies of light sent radially between these observers is given by this ratio, numerator and denominator evaluated at the points r=r1 and r=r2 respectively,

    [tex]\frac{\nu_1}{\nu_2} = \frac{(g_{mn}u^m k^n)_1}{(g_{mn}u^m k^n)_2}[/tex] -------- (1)

    where u(i)n is the 4-velocity of O1 and O2 and kn is a null vector defined by the photon trajectory ( up to a constant which will cancel out) so that

    [tex]k_{m;n}k^n = 0[/tex] --------- (2)

    which is the geodesic condition for a (transverse ?) plane wave.

    We need to find the null vector kn. Because only k0 and k1 are non-zero for a radial photon it is not difficult to solve for k from equation (2), up to a factor, which is all we need. I got the components of kn,

    [tex]\left(1 - \frac{2M}{r}\right)^{-\frac{1}{2}}, \left(1 - \frac{2M}{r}\right)^{\frac{1}{2}}, 0, 0[/tex]

    which satisfy equ (1) and also gmnkmkn = 0.

    Now we can calculate [itex](g_{mn}u^m k^n)_i[/itex] which gives,

    [tex](g_{mn}u^m k^n)_i = u_i^1\left(1-\frac{2M}{r_i}\right)^{-\frac{1}{2}} - u_i^0\left(1-\frac{2M}{r_i}\right)^{\frac{1}{2}}[/tex].

    Which looks as if it could be right. If the observers are at rest wrt to each other, then we can write ui0 = 1 and ui1 = 0, which reduces equation (1) to the usual gravitational frequency shift.

    For u to be a proper 4-vector gmnumun = 1 ( speed of light), so it should be possible to generalise this result a bit more.

    If it's correct. This must have been worked out somewhere.

    (1) 'General Relativity', Stephani, (Cambridge, 1993).
  2. jcsd
  3. Jul 15, 2008 #2


    User Avatar
    Gold Member

    I haven't had much time for this and resources on the web for geometric optics seem limited. I did notice that my formula derived above,

    [tex](g_{mn}u^m k^n)_i = u_i^1\left(1-\frac{2M}{r_i}\right)^{-\frac{1}{2}} - u_i^0\left(1-\frac{2M}{r_i}\right)^{\frac{1}{2}}[/tex]

    gives the correct SR result for M=0, namely

    [tex]g_{mn}u^m k^n = u^1 - u^0 = \gamma\beta - \gamma = \gamma(\beta - 1)[/tex].

    The sign is different, but that disappears when we take the ratio.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook