I am following Carroll's book "Spacetime and Geometry" p. 245-247. I have the following expression for the surface gravity of a Killing horizon: $$\kappa^2=-\frac{1}{2}(\nabla_{\mu}K_{\nu})(\nabla^{\mu}K^{\nu})$$ where $K^{\mu}$ is the timelike Killing vector that is orthogonal to the horizon. If I have a static spacetime with the time translation Killing vector the four-velocity of a static particle is proportional to the Killing vector: $$K^{\mu}=V(x)U^{\mu}$$ where V is the redshift factor. Then, Carroll claims the following: $$\kappa = \sqrt{(\nabla_{\mu}V)(\nabla^{\mu}V)}$$ How can I prove that? I have attempted to substitute, use the Leibniz rule and cancel two terms by using the identity $U^{\mu}\nabla_{\nu}U_{\mu}=0$ (from conservation of the modulus of four-velocity), but I cannot go further.
1 Answer
Unfortunately, I do not know how to continue this approach, but I can suggest a different way. One can start by considering the expression $$\sqrt{\nabla_\mu V \nabla^\mu V}$$ By plugging in the expression for $V=\sqrt{- K_\alpha K^\alpha}$, one finds $$\sqrt{\nabla_\mu V \nabla^\mu V}=\frac{1}{2V}\sqrt{\nabla_\mu (K_\alpha K^\alpha) \nabla^\mu (K_\beta K^\beta)}$$ Using the product rule yields
$$ \frac{1}{V}\sqrt{K^\alpha\nabla_\mu K_\alpha K_\beta \nabla^\mu K^\beta} $$ By apllying Killing's equation, this yields
$$\frac{1}{V}\sqrt{-K^\alpha\nabla_\alpha K_\mu K_\beta \nabla^\mu K^\beta} $$ Here, one can use the definition of the surface gravity $K^\alpha \nabla_\alpha K^\beta = -\kappa K^\beta$. This leads to $$\frac{1}{V}\sqrt{\kappa K_\mu K_\beta \nabla^\mu K^\beta} $$ $K_\beta$ and $K_\mu$ can be exchanged and the formula for the surface gravity ccan be applied again: $$\frac{1}{V}\sqrt{\kappa^2 (-K_\beta K^\beta)} $$ Since $\sqrt{-K_\beta K^\beta}$ is $V$, the result is
$$\sqrt{\nabla_\mu V \nabla^\mu V}=\kappa $$
I hope this is helpful :)
