# Defining Surface gravity of a black hole

For a Killing horizon associated with a Killing vector $$K$$, the surface gravity $$\kappa$$ can be computed by various methods, like $$\kappa^2 = - \frac{1}{2} \nabla^\mu K^\nu \nabla_\mu K_\nu \ .$$ However Killing vectors multiplying a constant is still a Killing vector, and $$c K$$ is of course also null on the Killing horizon.

I wonder what is the criteria to fix the constant $$c$$? For Schwarzschild and Reissner-Nordstrom, I guess one can impose $$|K|^2 \to -c^2$$ at spacial infinity, but this doesn't seem to be the case for the Kerr black hole, since $$|\partial_t + \Omega_H \partial_\varphi|^2 \to +\infty \ .$$

• Great question! Of course the answer is hidden in the difference. Jun 7, 2022 at 16:09
• However in a static, asymptotically free spacetime with time translation Killing vector $K=\partial_t$ one can set the normalisation $\lim_{r\rightarrow \infty}K^2 =-1$ and this fixes the surface gravity of any associated Killing horizon. In a stationary asymptotically flat spacetime the Killing horizon is associated with a linear combination of $K=\partial_t$ and $R=\partial_\phi$ and fixing the normalisation of $K$ fixes the linear combination and so also the surface gravity. Jun 7, 2022 at 17:03

You're absolutely right. Because the Killing vectors are defined to satisfy Killing's equation $$2\nabla_{(a} K_{b)} = 0 \ ,$$ a constant scaling is possible, and the value of $$\kappa$$ changes. You're also right that for the Schwarzschild metric (and for all static asymptotically flat spacetimes), one can fix the constant by demanding $$K^a K_a|_{\infty} =-1$$.
They then go on to discuss some other more general ways to define the surface gravity $$\kappa$$ in terms of different definitions of horizons/trapped surfaces. But the conclusion is really that, in general, there's no prescribed way of doing this (and depends on the context of the problem you're working with).