# Time derivative of time-translation Killing vector

I'm working with the spherically symmetric, static black hole metric. In the problem I'm working on, I'm told that $K$ is the time-translation Killing vector, $\frac{\partial}{\partial t}$ or $K = (1, 0, 0, 0)$. I'm also told that $$\kappa = -\frac{1}{2} (\nabla^a K^b)(\nabla_a K_b) |_{rH}$$ ... where $rH$ is the black hole horizon.

In taking the covariant derivative of $K$, we have a $\nabla_t K_t$ term. Is the term above...

• $\nabla_t K_t = \partial_t K_t$ because $K_t$ is not a vector, it's just a component of a vector, and the covariant derivative of a function is just the partial of said function?
• The $\nabla_t K_t$ component of $\nabla_t K$? Which would be $\partial_t K_t - \Gamma_{t~t}^t K_t$? (In this case, the Christoffel symbol is actually zero, but it may not be in other cases).

Futhermore, is $\partial_t K_t = \frac{\partial ^2}{\partial t^2}$ or is $\partial_t K_t = \partial_t(1) = 0$?

• I can see where your confusion comes from. You may want to read my answer here, which can be summarized as "$\nabla_t K_t$ is understood to be shorthand for something more like $(\nabla \tilde{K})_{tt}$."
– user10851
Dec 5, 2014 at 8:43

You should always work with $$\nabla_\rho K_\sigma=\partial_\rho K_\sigma-\Gamma^\mu_{\rho\sigma} K_\mu$$ even if $K_\mu$ is a constant vector. Here $\partial_t K_t=0$ but you must consider the Christoffel symbols. For example: $$\nabla_t K_t=\partial_t K_t-\Gamma^\mu_{tt} K_\mu=-\Gamma^t_{tt} K_t$$ After that see what $\Gamma^t_{tt}$ is.
• We are going to take a covariant derivative of a vector and that's why we don't use $\nabla_\mu f(x) = \partial_\mu f(x)$. By $\nabla_\rho K_\sigma$, we mean the components of covariant derivative of a vector. Dec 5, 2014 at 8:36