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In my general relativity textbook (Carroll), he says that "the geodesic equation (together with metric compatibility) implies that the quantity

$\epsilon =-g_{\mu\nu}\frac{dx^\mu}{d\lambda}\frac{dx^\nu}{d\lambda}$

is constant along the path. For any trajectory we can choose the parameter $\lambda$ such that $\epsilon$ is a constant; we are simply noting that this is compatible with affine parameterization along a geodesic."

Maybe I'm missing something really obvious, but where does the conservation of this quantity come from?

theorist_wannabe
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1 Answers1

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Let $V^{\mu}=\frac{dx^\mu}{d\lambda}$. Then, $\epsilon=-V_\mu V^\mu$. To see the conservation of this quantity along the geodesic we have to look at the covariant derivative along the curve, that is $$V^\nu\nabla_\nu\epsilon=-V^\nu\nabla_\nu(V_\mu V^\mu)=-V^\nu V^\mu\nabla_\nu V_\mu -V^\nu V_\mu\nabla_\nu V^\mu.$$ But metric compatibility implies that $\nabla_\nu V_\mu=g_{\mu \sigma}\nabla_\nu V^\sigma$. Also using the geodesic equation $V^\nu \nabla_\nu V^\mu=0$, we find $$V^\nu\nabla_\nu\epsilon=0,$$ which shows that $\epsilon$ is a conserved quantity.

Mateus Sampaio
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