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

1. Jul 11, 2008

### Mentz114

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,

$$\frac{\nu_1}{\nu_2} = \frac{(g_{mn}u^m k^n)_1}{(g_{mn}u^m k^n)_2}$$ -------- (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

$$k_{m;n}k^n = 0$$ --------- (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,

$$\left(1 - \frac{2M}{r}\right)^{-\frac{1}{2}}, \left(1 - \frac{2M}{r}\right)^{\frac{1}{2}}, 0, 0$$

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

Now we can calculate $(g_{mn}u^m k^n)_i$ which gives,

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

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. Jul 15, 2008

### Mentz114

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,

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

gives the correct SR result for M=0, namely

$$g_{mn}u^m k^n = u^1 - u^0 = \gamma\beta - \gamma = \gamma(\beta - 1)$$.

The sign is different, but that disappears when we take the ratio.