# A simple calculation about surface gravity in classical GR

I am reading An Introduction to General Relativity Spacetime and Geometry by Sean Carroll, but simple calculations stop me.

At page 245, a formula for the surface gravity is given $$\kappa^2=-\frac{1}{2}(\nabla_\mu\chi_\nu)(\nabla^\mu\chi^\nu) \tag{6.9}$$ where $\kappa$ is a parameter called surface gravity, $\chi$ is the Killing vector with a Killing horizon.

Can you derive the above formula step by step using Killing's equation $\nabla_{(\mu}\chi_{\nu)}=0$ and the fact that $\chi_{[\mu}\nabla_\nu\chi_{\sigma]}=0$ ? The starting point is the geodesic equation : $\chi^\mu\nabla_\mu\chi^\nu=-\kappa\chi^\nu$.

I don't know why this hasn't been properly answered. Actually, a previous answer was quite confusing in my opinion. From the Killing equation you know that $$\nabla_{(a} \chi_{b)} =0$$ which means that this derivative is antisymmetric. Therefore, \begin{align} 0&=\chi _{[a} \nabla_b \chi_{c]} \\ &= \chi_a \nabla_b \chi_c + \chi_b \nabla_c \chi_a + \chi_c \nabla_a \chi_b\\ \nonumber \therefore \\ \chi_c \nabla_a \chi_b &= -\chi_a \nabla_b \chi_c + \chi_b \nabla_a \chi_c \end{align} where we have used again the Killing equation.

Contracting this with $$\nabla^a \chi^b$$ and using the geodesic equation for an "inaffine" parameter: $$\chi^a \nabla_a \chi^b = -\kappa \chi^b$$ gives you (after a couple of steps) $$(\nabla^a \chi^b) (\chi_c \nabla_a \chi_b) = - 2 \kappa^2 \chi_c$$ for all $$\chi_c$$ and thus, you obtain the result you were looking for.

• The other answer has the right approach but contains many errors in the calculation. Funnily enough, those errors miraculously cancel each other. BTW, the author claims that $\kappa$ is a constant over the horizon. Do you perhaps know why? Commented Feb 20, 2023 at 5:00

Starting with $\chi_{[a}\nabla_{b} \chi_{c]} = 0$, we can use the Killing equation ($\nabla_{b} \chi_{c} = - \nabla_{c} \chi_{b}$) to get,

$$$\chi_{c} \nabla_{a} \chi_{b} = -2 \chi_{a}\nabla_{b}\chi_{c} + 2 \chi_{b}\nabla_{a}\chi_{c}$$$

Contracting both sides with $\nabla^{a}\chi^{b}$, we get (need to use $\chi^{a}\chi_{a} = 0$ on the Killing horizon),

\begin{align} \chi_{c} (\nabla^{a} \chi^{b})(\nabla_{a} \chi_{b}) &= -2 (\chi_{a}\nabla^{a}\chi^{b})\nabla_{b}\chi_{c} \\ &= -2 \kappa \chi^{b}\nabla_{b}\chi_{c} \\ &= -2 \kappa^2 \chi_{c} \end{align}

Hence, we obtain,

$\kappa^{2} = -\frac{1}{2} (\nabla_{a}\chi_{b})(\nabla^{a}\chi^{b})$