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Let $f$ be a function on a spacetime $(M, g_{ab})$ whose gradient, $k_a = \nabla_a f$, ie everywhere null, i.e., $k_ak^a = 0$ throughout $M$. Does it necessarily follow that the integral curves of $k^a$ are null geodesics?

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1 Answer 1

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The condition that $k$ is a gradient implies that it is curl-free; i.e. $\nabla_i k_j = \nabla_j k_i$. Let $\gamma$ be an integral curve of $k^i$; i.e. $\dot \gamma^i(t) = k^i(\gamma(t))$. Then replacing $\dot \gamma$ with $k$ in the covariant acceleration and applying the curl-free condition we get $$\ddot \gamma^i = \dot \gamma^j \nabla_j \dot \gamma^i = k^j \nabla_j k^i = k^j \nabla^i k_j.$$

Now we're basically done - just recognize the RHS as $\frac12 \nabla^i (k^j k_j)$, at which point the fact that $k$ is null ($k^j k_j = 0$) tells us that $\ddot \gamma = 0$; i.e. $\gamma$ is a geodesic.

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