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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}

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