Surface Gravity of the Schwarzschild Blackhole

To work out the surface gravity $\kappa$ of the Schwarzschild Blackhole, we use the formula given in R. M. Wald (12.5.14):

(1)
\begin{align} \kappa^2 = -\frac{1}{2} \left ( \nabla^a \chi^b \right ) \left ( \nabla_a \chi_b \right ) \end{align}

Here, $\chi^a$ is the Killing field whose null surface is the horizon (in this case, $\partial_t$), and the expression is to be evaluated at the horizon.

The Killing Vector in our case is $\chi^a \sim (1, 0, 0, 0)$ (So, $\chi_a \sim \left ( 1-\frac{r_s}{r}, 0, 0, 0 \right )$ ). Just as a check, we will show that it satisfies Killing's equation. To start with, we must compute the Christoffel symbols:

(2)
\begin{align} \Gamma^t_{r t} = \frac{1}{2} \left ( 1 - \frac{r_s}{r} \right )^{-1} \frac{r_s}{r^2}, \end{align}
(3)
\begin{align} \Gamma^r_{t t} = \frac{1}{2} \left ( 1 - \frac{r_s}{r} \right ) \frac{r_s}{r^2}, \end{align}

Then, we note that:

(4)
\begin{align} \nabla^r \chi^t = - \frac{r_s}{2 r^2} = -\nabla^t \chi^r \end{align}

Thus verifying that $\chi^a$ is indeed a Killing vector. Note that it is also null on the event horizon, i.e. $\chi^a \chi_a = 1 \cdot \left (1 - \frac{r_s}{r} \right ) = 0$ when $r = r_s$. Thus, the event horizon is a Killing horizon.

Now, we note that the only terms in $\nabla^a \chi^b$ (or equivalently, in $\nabla_a \chi_b$) that are nonzero are $\nabla^r \chi^t$ and $\nabla^t \chi^r$, which we have already computed above. Note that $\nabla_t \chi_r = -\nabla^t \chi^r$ and $\nabla_r \chi_t = -\nabla^r \chi^t$ numerically, because $g_{tt} = -( g_{rr} )^{-1}$. Thus, we have

(5)
\begin{align} \kappa^2 = -\frac{1}{2} \left ( \nabla^t \chi^r \nabla_t \chi_r + \nabla^r \chi^t \nabla_r \chi_t \right ) = - \nabla^t \chi^r \nabla_t \chi_r, \end{align}

where we have used Killing's equation to equate the second term above with the first. Substituting in and evaluating the expression for $\kappa$ at the horizon gives us

(6)
\begin{align} \kappa^2 = \frac{1}{4 r_s^2 }. \end{align}
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