Timeline for Surface gravity of a Killing horizon

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Apr 29, 2020 at 2:58	comment	added	gamebm		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 $r=2M$.
Sep 10, 2016 at 4:36	comment	added	Ari		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.
Jan 28, 2016 at 10:12	comment	added	user11128		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 27, 2016 at 20:58	history	answered	Rexcirus	CC BY-SA 3.0