# Calculating surface gravity

I have some trouble with understanding how surface gravity/Killing horizon equation works, for example in following form:

$$κ^2=-\frac 12 (\nabla^aK^b)(\nabla_aK_b)$$ with Killing vector $K$.

I'm confused with 1st bracket where covariant derivative have upper index- I can contract it with metric to get $g^{ac}\nabla_c$ but then I have 'problematic' term $K^b$. For example for time-like Killing vector $K^b=(1,0,0,0)$, I can get $0$ and then whole equation is $0$ (with $K^b=(\partial_0,0,0,0$) case is even more complicated) and of course it isn't true when dealing with static space-times (like RN or Schwarzschild's with time-like killing vector field).

For this metric (adv. Eddington-Finkelstein) in static case (where functions depend only on $r$):

$$\mathrm ds^2=-A^2(v,r)\Delta(v,r)\mathrm dv^2+2A(v,r)\mathrm dv\,\mathrm dr+r^2\mathrm d\Omega^2$$

with $(\Delta=1-2m(v,r)/r$) and time-like, normalized Killing vector with components $(1,0,0,0)$. I'm getting zeros (covariant derivative acting on $(1,0,0,0)$ gives me 0, covariant acting on $(\partial_0,0,0,0)$ gives higher order terms, but I should get (for positive root):

$$\kappa=\frac {A(r_H)}{2m(r_H)}(1-2m'(r_H))$$ where $r_h$ is the Killing horizon and $m'=\partial_rm$

How can I compute surface gravity using this expression?

In a metric independent of the time coordinate, e.g. Schwarzschild, the covariant derivative of the time Killing vector $$K^\mu = (1, 0, 0, 0)$$ is not zero. In fact the covariant derivative of a vector is defined as $$\nabla_\mu A^\nu = \partial_\mu A^\nu + \Gamma^\nu _{\mu \lambda} A^\lambda$$. The $$\Gamma's$$ not necessarily are nil.
Moreover in the question the definition $$K^\mu = (\partial_t, 0, 0, 0)$$ is inconsistent. Instead should be $$K^\mu = \partial_t = (1, 0, 0, 0)$$.