In a previous post I wrote about the phenomenon of gravitational red-shift in stationary space-times. Of particular importance is the following fact:

Given a time-like Killing vector field $X$ (which we take to represent a preferred stationary background family of observers), the red-shift factor of observer $A$ relative to signals sent by observer $B$ is given by $H = \sqrt{\dfrac{g(X,X)|_B}{g(X,X)|_A}}$. That is, if observer $B$ sends a signal of frequency $f$, observer $A$ will see it as having frequency $Hf$.

We can also consider the infinitesimal version of the frequency shift. Taking the derivative of $H$ (calculate it when the observers $A,B$ are infinitesimally close to each other) we see that $\nabla H = -\nabla \log (-g(X,X))$.

Now suppose we are at a *Killing horizon*. That is, suppose we have a null hypersurface $\mathfrak{h}$ and a Killing vector field $X$ such that (a) $X$ is time-like on one side of the hypersurface (b) $X$ degenerates to a null vector field along $\mathfrak{h}$ and is in fact tangent to its null generator. Therefore $\nabla_XX$ is parallel to $X$ along the horizon. Write $\nabla_XX = \kappa X$. Using that $X$ is Killing, we have that $\kappa X = \nabla( - g(X,X))$ along the horizon. Comparing this with the paragraph above (and using Taylor's theorem for smooth functions), we see now the connection between $\kappa$, the *surface gravity* of a stationary black-hole, and the red-shift effect for observers extremely close to the event horizon: if $\kappa > 0$, for observers very close to the event horizon, but sits to the past of the black hole (i.e. outside the black hole), this tells us that the out-ward gradient of $H$ is negative, and so we expect a red-shift.