For a killing vector $k^a$, one can get the surface gravity of a Schwarzschild black hole by calculating $k^b \nabla_b k^a$ (which is supposed to equal $\kappa k^a$), setting $r=2m$, and then comparing the LHS and RHS to find an expression for $\kappa$.
Here, $k^a$ is the time translation killing vector, and in $(t,r,\theta,\phi)$ coordinates, we have $k^a = [1, 0, 0, 0]$.
If one works out $k^b \nabla_b k^a$, you will get $\displaystyle \left[0 ,\frac{m(r-2m)}{r^3},0,0 \right]$. Plugging in $r=2m$ into this however makes the result the zero vector, and it impossible to extract $\kappa = 1/4m$ from this. Regardless of it being the zero vector, the components of $k^b\nabla_b k^a$ and $k^a$ don't line up at all, so even if there wasn't a problem plugging in $r=2m$ it is still impossible to get an expression for $\kappa$ from this.
Does anyone know what I am doing wrong?
EDIT: While I know there is a formula for $\kappa$ involving square roots of covariant derivatives, I would like to derive $\kappa$ using this method.
EDIT2: To work out $k^b\nabla_bk^a$, one needs to work out the Christoffel symbols. We have $$\nabla_bk^a = \partial_b k^a+\Gamma^a_{bc}k^c = \Gamma^a_{b1},$$where we have made use of the fact that the derivatives of $k^a$ are all zero and that when you sum over the $c$ index only the first component of $k^c$ contributes.
One can calculate the Christoffel symbols easily by using the regular formulas, but the only nonzero ones that matter to us are $$\Gamma^{2}_{11} = \frac{m(r-2m)}{r^3}, \text{ and } \Gamma^{1}_{21} = \frac{m}{r(r-2m)}.$$
So, summing over $k^b\nabla_bk^a$, we get $$k^1\nabla_1 k^a + k^2\nabla_2 k^a + k^3\nabla_3 k^a + k^4\nabla_4 k^a.$$
Using the above information along with the fact that the only nonzero component of $k^a$ is $k^1$, the only nonzero term in this expression is $k^1\nabla_1 k^2$. Working this out, we get $$k^1\nabla_1 k^2 = k^1 \Gamma^{2}_{11} = \frac{m(r-2m)}{r^3},$$ and hence, $$ k^b \nabla_b k^a = \left[ 0, \frac{m(r-2m)}{r^3}, 0, 0 \right],$$ which is not proportional to $k^a$.