# Prove that $U^\sigma \nabla_\sigma U^\mu = g^{\mu\nu}\nabla_\nu \ln V$

Let $$U^\mu$$ be a vector in 4-dimensional Minkowski space with norm $$-1$$ and $$K^\mu = V(x)U^\mu$$ a vector proportional to it. We can write $$V(x) = \sqrt{-K_\nu K^\nu}$$.

(This setup comes from physics where $$U^\mu$$ is a four-velocity, $$K^\mu$$ is a normalized time-like Killing vector for an observer at infinity and $$V(x)$$ is called the redshift factor.)

Then, define $$a^\mu$$ (the four-acceleration) by $$U^\sigma \nabla_\sigma U^\mu$$, where $$\nabla$$ is the Christoffel connection.

According to Sean Carroll, Spacetime and Geometry, p. 247, $$a_\mu = \nabla_\mu\ln V$$. Why?

Attempt:

\begin{align}\nabla_\mu\ln V &= \frac 1{2V^2}\nabla_\mu\left(-K_\nu K^\nu\right)\\ &=-\frac 1{2V^2}\left((\nabla_\mu K_\nu)K^\nu + K_\nu\nabla_\mu K^\nu\right)\\ &=-\frac 1{2V^2}\left((\nabla_\mu g_{\rho\nu}K^\rho)K^\nu + K_\nu\nabla_\mu K^\nu\right)\\ &=-\frac 1{2V^2}\left(K_\rho\nabla_\mu K^\rho + K_\nu\nabla_\mu K^\nu\right)\\ &=-\frac {K_\nu\nabla_\mu K^\nu}{V^2}\\ &= -\frac 1{V}U_\nu\nabla_\mu\left(VU^\nu\right)\\ &= -U_\nu\nabla_\mu U^\nu - \frac 1 VU_\nu U^\nu\nabla_\mu V\end{align}