# 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?

-

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; $$\frac{D}{d\lambda}\left[g_{\mu\nu}(x(\lambda))\right] = 0$$

-
Ok, thanks. I have made an attempt to do the substitution - see my edited question. Can I assume in most simple problems that the time derivative of the metric is zero? – rgvcorley Apr 20 '13 at 8:36
Note that the overdots you write should be derivatives with respect to affine parameter in general, not necessarily time. It's much safer not to make any such assumptions, even for simple problems. – joshphysics Apr 21 '13 at 22:47

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:

$$\frac{D}{d\lambda } U_\sigma = U^\alpha \nabla_\alpha U_\sigma = \dot{U}_\sigma - g^{\alpha \beta }\Gamma^\mu_{\alpha \sigma} U_\beta U_\mu=0$$

-
Are you certain that your last equation is correct? If one defines $\dot U_\sigma = \frac{d}{d\lambda} U_\sigma$ then it is not true in general that $\dot U_\sigma = g_{\sigma\mu}\dot U^\mu$. – joshphysics Apr 20 '13 at 6:18
It's true as long as $\frac{d}{d \lambda} g_{\mu \nu} = 0$, which is true for most choices of $\lambda$ when considering a timelike geodesic (which is usually what one's interested in) when there's a timeline Killing vector. In physical scenarios where there's no t-symmetry, i.e. the FLRW metric, one's generally not interested in test particles anyway. – jld Apr 20 '13 at 6:37
I'm quite skeptical. Notice that $\frac{d}{d\lambda} g_{\mu\nu}(x(\lambda)) = \dot x^\sigma(\lambda)\partial_\sigma g_{\mu\nu}(x(\lambda))$ and it is not true in general that $\partial_\sigma g_{\mu\nu}=0$. It is, however, true that $\frac{D}{d\lambda}g_{\mu\nu}(x(\lambda)) = \dot x^\sigma(\lambda)\nabla_\sigma g_{\mu\nu}(x(\lambda)) = 0$ for any metric-compatible connection, namely one for which $\nabla_\sigma g_{\mu\nu} = 0$. – joshphysics Apr 20 '13 at 7:39
Even if there is a timelike killing vector, it's not clear to me how the partial $\partial_\sigma g_{\mu\nu}$ of the metric vanishes for all $\sigma$, and even if this is somehow true, I don't think the original question intended the class of metrics and/or geodesics to be so restricted. – joshphysics Apr 20 '13 at 7:47
I also think that this answer is misleading. $\lambda$ has nothing to do with space-time coordinates, so there is nothing to do with Killing vectors. As I get it, $\lambda$ is the proper time, otherwise the whole equation will be different (it is not invariant under reparametrizations, and this nice form works only for affine parametrization). – Peter Kravchuk Apr 20 '13 at 17:11

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.

-