Transforming an equation to the co-vector version Ok, this question is more a result of my lack of knowledge of how to manipulate equations involving index notation rather than about physics...
I have the geodesic equation with $U^\lambda\equiv\dot{x}^\lambda$:-
$$
\dot{U^\lambda} + \Gamma^\lambda_{\mu\nu} U^\mu U^\nu
$$
And I want to transform to the co-vector $U_\mu=g_{\mu\lambda}U^\lambda$.
Can I simply multiply each vector by $g_{\mu\nu}$? Like so:-
$$
g_{\mu\lambda}\dot{U^\lambda} + \Gamma^\lambda_{\mu\nu}g_{\mu\alpha}U^\alpha g_{\nu\beta}U^\beta
$$
Or do I need to use $g^{\sigma\nu}g_{\nu\mu} = \delta^\sigma_\mu$ to rewrite $U_\mu=g_{\mu\lambda}U^\lambda$ and then sub it in?
Edit: Here's my attempt at the sub in method
So using $g^{\lambda\mu}g_{\mu\lambda} = \delta^\lambda_\lambda$ to rewrite $U_\mu=g_{\mu\lambda}U^\lambda$ as $U^\lambda=g^{\lambda\mu}U_\mu$. (Is this even correct?). Then differentiate:-
$$
\dot{U}^\lambda=\dot{g}^{\lambda\mu}U_\mu +  g^{\lambda\mu}\dot{U}_\mu
$$
Can I assume that the differential of the metric wrt time is going to be zero? Obivously this is not going to be true in general since massive bodies move! But generally in simple problems would this be taken as true?
 A: You generally need to do the second thing where you sub in.  For an affinely parameterized geodesic $x(\lambda) = (x^\mu(\lambda))$ we have
$$
  x_\mu(\lambda)= g_{\mu\nu}(x(\lambda))x^\nu(\lambda)
$$
Denoting derivatives with respect to affine parameter by overdots, it follows that
\begin{align}
  \dot x_\mu(\lambda) 
&= \frac{d}{d\lambda}\left[g_{\mu\nu}(x(\lambda))\right]x^\nu(\lambda) + g_{\mu\nu}(x(\lambda))\dot x^\nu(\lambda)
\end{align}
Therefore, notice that you can only "simply multiply each vector by $g_{\mu\nu}$" if the first term in this expression vanishes, which is only true then the metric components are constant along the curve.  This is not in general the case, although it is the case, for example, in flat space coordinates where $g_{\mu\nu} = \delta_{\mu\nu}$.  What is generally true, however, is that the directional covariant derivative of the metric is zero along a given curve;
$$
  \frac{D}{d\lambda}\left[g_{\mu\nu}(x(\lambda))\right] = 0
$$
A: Just as in any equation, you can do whatever you want to it as long as you do the exact same thing to both sides. Multiplying one side of the equation by $g_{\mu \lambda}$ and multiplying the other side by $g_{\mu \alpha} g_{\nu \beta}$ doesn't make any sense. Also, I should point out that $\mu$ and $\nu$ are dummy indices, i.e. they're already being summed over, so you can't sum over them again.
What you should do is take the covariant derivative of $U_\sigma$ to get a covector version of the geodesic equation:
$$\frac{D}{d\lambda } U_\sigma = U^\alpha \nabla_\alpha U_\sigma = \dot{U}_\sigma - g^{\alpha \beta }\Gamma^\mu_{\alpha \sigma} U_\beta U_\mu=0$$
A: Recall the definition of the covariant derivative of vector $V$ along the direction $U$:
$$
(\nabla_U V)^\nu=\partial_U V^\nu+\Gamma^\nu_{\mu\lambda}V^\lambda U^\mu
$$
where $\partial_U$ is the ordinary directional derivative. If $V$ is defined on a curve $\gamma(t)$ with $U$ the tangent vector, then $\partial_U V=\partial_tV$.
You can see that the geodesics equation is
$$
\nabla_U U^\mu=0
$$
for $U$ the velocity vector.
So if you know what the covariant derivative is, you can lower the index and write for the covelocity $U_\mu$:
$$
0=\nabla_UU_\mu=\partial_tU_\mu-\Gamma_{\lambda\mu}^\nu U_\nu U^\lambda=\partial_tU_\mu-\Gamma_{\;\;\,\mu}^{\nu\lambda} U_\nu U_\lambda
$$
If for some reason you dont want to use covariant derivatives, then you have to do the $t$  derivative of metric via $\partial_t g_{\mu\nu}=U^\lambda\partial_\lambda g_{\mu\nu}$ and use the formulae for the Cristoffel symbols in order to derive the given expression.
