# Gravitational redshift in a general stationary metric

Suppose you have a general metric $g_{\mu \nu}(t,r,\theta,\phi)$ which don't depend explicitly on $t$ coordinate, i.e a stationary metric. Light travels along a geodesic from A (at which the frequency is $\omega_0$) to B. Along the geodesic the metric and the Riemann tensors completely regular (Riemann tensor not zero).

What is the frequency of the light observed at B? (You may assume the metric is diagonal, but I prefer a more general solution)

• Note that the observed frequency changes because you and your clocks go slower when you're lower. – John Duffield Jul 26 '15 at 15:59

Cinsider an observer who is stationary at point $A$. Because they are stationary $dr = d\theta = d\phi = 0$ and the metric becomes:

$$ds = \sqrt{g_{00}(r_A, \theta_A, \phi_A)} dt$$

And likewise for an observer at point $B$:

$$ds = \sqrt{g_{00}(r_B, \theta_B, \phi_B)} dt$$

The relative change in the frequency of the light is simply the relative time dilation between the two observers, so:

$$\frac{\omega_B}{\omega_A} = \sqrt{\frac{g_{00}(r_B, \theta_B, \phi_B)}{g_{00}(r_A, \theta_A, \phi_A) }}$$

• I think you accidently missed the sqrt sign. Other than that your answer seems to be correct, though it took me some time to convince myself that for 2 signals sent from stationary observer with time dilation of $\Delta \tau=\sqrt{-g_{tt}}\Delta t$, we get that $\Delta t_A=\Delta t_B$ because of the metric's independence on $t$ coordinate. – Alexander Jul 26 '15 at 15:36
• It's interesting that the formula can give a finite, believable result, even in cases where a stationary observer is unphysical (e.g. inside the ergosphere). – user10851 Jul 26 '15 at 16:18