GR: Show that the directional covariant derivative of the velocity yields the geodesic eqation

Question

I want to show that the directional covariant derivative of the velocity vector yields the geodesic equation but I get stuck at a partial derivative. To make it clear what I want to show, the statement is:

Suppose $\gamma$ is a path on a manifold $M$ and $\varphi$ is a coordinate map. Let $x^\mu (\lambda) = (\varphi \circ \gamma)^\mu(\lambda)$ be the component functions of the path in $\mathbb{R}^n$, then $$ 0 = \frac{D}{d\lambda} \frac{dx^\mu}{d \lambda}= \frac{d^2x^\mu}{d \lambda^2} + \Gamma^{\mu}_{\rho \sigma} \frac{d x^\rho}{d\lambda}\frac{d x^\sigma}{d\lambda}. $$ This is my attempt. I get stuck at the end: Suppose $\frac{dx^\mu}{d \lambda} = U^\mu$, then unwinding definitions $$ 0 = \frac{D}{d\lambda} \frac{dx^\nu}{d \lambda}= \frac{D}{d\lambda} U^\nu = \frac{d x^\mu}{d \lambda} \nabla_\mu \left( U^\nu \right) = \frac{d x^\mu}{d \lambda} \partial_\mu U^\nu + \frac{d x^\mu}{d \lambda} \Gamma^\nu_{\mu \rho} U^\rho = \frac{d x^\mu}{d \lambda} \partial_\mu U^\nu + \Gamma^\nu_{\mu \rho} \frac{d x^\mu}{d \lambda}\frac{d x^\rho}{d \lambda} $$ It remains to show that $$ \frac{d x^\mu}{d \lambda} \partial_\mu U^\nu = \frac{d^2x^\mu}{d \lambda^2} $$ Using the definition $\partial_\mu f = \frac{\partial}{\partial x^\mu}(f \circ \phi^{-1})$, I cannot see that it is correct: $$ \frac{d x^\mu}{d \lambda} \partial_\mu U^\nu = \frac{d x^\mu}{d \lambda} \partial_\mu \frac{d x^\nu}{d \lambda} = \frac{d x^\mu}{d \lambda} \frac{\partial}{\partial(\phi \circ \gamma)^\mu} \left(\frac{d (\phi \circ \gamma)^\nu}{d \lambda} \circ \phi^{-1} \right) = ? $$ I am confused. Any help? It would be great if you could be precise about what you mean about each step.

Answer 1

You've stumbled upon a nontrivial issue, the issue being that $\dot\gamma$ is not a vector field on $M$, so there's no way for $\partial_\mu \dot x^\nu$ to actually make sense without extra work. This is treated very carefully in classical mathematics texts such as Helgason and do Carmo, but I don't think physicists really care. The precise statement you're looking for can be found on page 50 of do Carmo, Riemannian Geometry:

If $V$ is a vector field along a curve $c:I\to M$, and if $V$ is the restriction of a vector field $Y$ on $M$ ($Y\circ c=V$), then $$\frac{\mathrm DV}{\mathrm d\lambda }=\nabla_{\dot c}Y.$$

Let's check this. Writing $V=v^\mu \partial_\mu$, we have $$\tag{$*$}\frac{\mathrm DV^\mu}{\mathrm d\lambda }=\frac{\mathrm dv^\mu}{\mathrm d\lambda }\partial_\mu+v^\nu\frac{\mathrm d x^\rho}{\mathrm d\lambda}\Gamma^\mu{}_{\nu\rho},$$ where $\lambda\mapsto x^\mu$ is the coordinate representation of the curve $c$. On the other hand, $$\nabla_{\dot c}Y^\mu=\frac{\mathrm dx^\nu}{\mathrm d\lambda}\partial_\nu Y^\mu+Y^\nu\frac{\mathrm dx^\rho}{\mathrm d\lambda}\Gamma^\mu{}_{\nu\rho}.$$ Now, we are evaluating this along $c(t)$, so $Y^\nu=v^\nu$. Furthermore, we have $v^\mu=Y^\mu(c(t))$, so we can use the chain rule: $$\frac{\mathrm dv^\mu}{\mathrm d\lambda }=\frac{\mathrm d Y^\mu(c(t))}{\mathrm d\lambda }=\frac{\mathrm d x^\nu}{\mathrm d\lambda }(\partial_\nu Y^\mu)(c(t)).$$ This is nothing more than the chain rule. Thus we see the two equations are equivalent.

So the issue is that $\partial_\mu U^\nu$ doesn't make sense! $U^\nu$ is not a function of $x$. You first have to extend it to a vector field for things to make sense. You can do this via a partition of unity.