# Surface gravity of a Killing horizon

1. Surface gravity is defined on the Killing horizon by $\xi^\mu \nabla_\nu \xi^\nu = \kappa \xi^\nu$ for the Killing vector $\xi$. Why can we interpret this as the force required at infinity to hold a unit mass at rest on the horizon?

2. Is there an obvious reason why $\nabla^\mu \xi^2$ is orthogonal to the horizon? Clearly, I can expand this and get $\nabla^\mu \xi^2 = - 2 \xi^\rho \nabla_\rho \xi^\mu =-2 \kappa \xi^\mu$ which is orthogonal since $\xi$ is orthogonal and $\kappa$ is just a constant. But, I want to know if there is an obvious, intuitive way to see this without doing any calculations (the author says it's "obvious") or if the person that wrote the notes was expecting me to do the above calculation to see this?

Number (2) follows from the fact that for any scalar function $\Phi(x^{\nu})$, the gradient $\partial_{\mu} \Phi$ is orthogonal to the level set $\Phi(x^{\nu})=const$.

Intuitively, this should hold since $\Phi$ doesn't change in directions tangent to the hypersurface. A proof follows from the chain rule. Let $\gamma^\mu (s)$ be a curve which lies entirely in the hypersurface $H$ defined by $\Phi(x^{\nu})=c$. Then since $\Phi(\gamma^{\mu}(t))=c$ , $$\frac{d\Phi(\gamma^{\mu}(t))}{dt} = \frac{\partial \Phi}{\partial \gamma^\mu} \frac{d\gamma^{\mu}}{dt} = \partial_\mu \Phi \; \frac{d\gamma^{\mu}}{dt} = 0 .$$ The derivative $\frac{d\gamma^{\mu}}{dt}$ of an arbitrary curve in $H$ gives an arbitrary tangent vector to $H$, so $\partial_\mu \Phi$ is normal to $H$. This is a property of differential geometry, and has nothing to do with null vectors or horizons.

In particular, then, the magnitude $\xi^2$ of the Killing vector field is a scalar function on spacetime, with the constant value $\xi^2=0$ on the horizon. So the gradient $\nabla_\mu \xi^2=\partial_\mu \xi^2$ must be normal to the horizon.

As for (1), take a look at Wald eqn 12.5.18 and problem 6.4.

First time giving an answer... hope it helps!

• This is a nice-looking first answer. Welcome!
– rob
Nov 17, 2016 at 6:46

1) Going to Rindler coordinates, in the near horizon limit of the schwarzschild black hole, you get that the surface gravity is precisely the constant acceleration of the Rindler observer. Indeed this is the equivalence principle at work: gravitation "=" acceleration.

For a static observer you can do an explicit calculation by defining the four-acceleration as $a^{\mu}=U^{\sigma} \nabla_{\sigma} U^{\mu}$, where $U^{\mu}$ is the velocity, and taking the time-translation Killing field $ξ^{\mu}=\sqrt{-ξ^2} U^{\mu}$. See Carroll's book pag. 246 for reference.

2)By definition the Killing vector is null on the horizon, so $ξ^2=0$ on the horizon. Moreover this means that is normal to himself and to the horizon.

• Thanks. That's a nice reference. 1, On p247 of his book, I can't derive eqns 6.15 and 6.16 though. Do you understand how he gets them? 2, I understand that $\xi^2=0$ on the horizon but why does this mean $\nabla^\mu \xi^2 =0$ also? The $\mu$ index isn't restricted to the horizon surface and whilst $\xi$ won't change along the horizon, it would be expected to change if I move perpendicular to it. Jan 28, 2016 at 10:12
• the way I understood this is $\nabla^\mu \xi^2$ is not zero. But since $\xi^2$ is zero, we can write $t^\mu \nabla_\mu \xi^2=0$. Where t is a tangent vector at Killing Horizon. This would imply $\nabla_\mu \xi^2= c. \xi_\mu$. Hence we can parameterize c and choose it to be zero for affine parameterization.
– Ari
Sep 10, 2016 at 4:36
• About 1), maybe I can add one more small piece to your answer. The killing vector is related to the four-velocity because it is readily to show that it is associated with an observer who marginally hangs on at the horizon without falling into the blackhole. In other words, his worldline is marginally time-like (light-like in practice) with $r=2M$. Apr 29, 2020 at 2:58