Acceleration in stationary spacetime How would I show the following equation for the acceleration vector
$$ a^\mu = u^\nu \nabla_\nu u^\mu = g^{\mu\nu}\nabla_\nu \ln V $$
for an observer instantaneously 'at rest,' where $u^\mu = dx^\mu/d\tau$ and $V^2 = -\xi_\mu \xi^\mu$ ($\xi^\mu$ is a time-like killing vector field). I guess we're assuming that we have a stationary spacetime metric.
I'm looking at Sean Carroll's Spacetime and Geometry, page 274 equation 6.15. They leave out the derivation for this. So $V$ is the redshift factor.
 A: Never mind, I got it. For the instruction of others, here it is:
Using four velocity $u^\mu u_\mu=-1$ and then $\nabla_\nu(u^\mu u_\mu)=0$
$u^\mu\nabla_\nu u_\mu + u_\mu\nabla_\nu u^\mu = 0 ⇒ 2u_\mu\nabla_\nu u^\mu = 0 ⇒ u_\mu\nabla_\nu u^\mu = 0.$
Using this, the Killing equation, $\nabla_\mu K_\nu + \nabla_\nu K_\mu = 0$ and $K_\alpha = V(x)u_\alpha$, we have
$$
\begin{align*}
&(\nabla_\mu V)u_\nu + V(\nabla_\mu u_\nu) + (\nabla_\nu V)u_\mu + V(\nabla_\nu u_\mu) &=& 0 \\
\implies &(\nabla_\mu V)u^\nu u_\nu + V u^\nu(\nabla_\mu u_\nu) + (\nabla_\nu V)u^\nu u_\mu + V u^\nu(\nabla_\nu u_\mu) &=& 0 \\
\implies &−\nabla_\mu V + u^\nu u_\mu(\nabla_\nu V) + V u^\nu(\nabla_\nu u_\mu) &=& 0 \\
\implies &−u^\mu \nabla_\mu V + u^\nu u^\mu u_\mu(\nabla_\nu V) + V u^\nu u^\mu(\nabla_\nu u_\mu) &=& 0  \\
\implies &−u^\mu \nabla_\mu V − u^\nu \nabla_\nu V &=& 0 \\
\implies &u^\mu \nabla_\mu V &=& 0 \\
\implies &−\nabla_\mu V + Vu ^\nu (\nabla_\nu u_\mu) &=& 0
\end{align*}
$$
So $$a_\mu = (\nabla_\mu V)/V = \nabla_\mu lnV.$$
Thanks, paper by Ying-Ming Huang.
