Considering the simple case of two observers O(adsbygoogle = window.adsbygoogle || []).push({}); _{1}and O_{2}lying on the same radius at positions r=r_{1}and r=r_{2}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=r_{1}and r=r_{2}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 O_{1}and O_{2}and k^{n}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 k^{n}. Because only k^{0}and k^{1}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 k^{n},

[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 g_{mn}k^{m}k^{n}= 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 u_{i}^{0}= 1 and u_{i}^{1}= 0, which reduces equation (1) to the usual gravitational frequency shift.

For u to be a proper 4-vector g_{mn}u^{m}u^{n}= 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).

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# Frequency shift of light between observers in Schwarzschild space-time

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