1
$\begingroup$

I'm studying General Relativity, and I'm currently facing the questions of the metric tensor, the $\Gamma$ connection, and the fact that $\Gamma$ is not a tensor. I'm also reading about the fact that the derivative of a tensor is, in general, not a tensor and in addition to that I'm also reading about the covariant derivative, and here is the question: we have defined the connection in terms of the metric tensor and its derivatives:

$$\Gamma^{\lambda}_{\mu\nu} = \frac{1}{2}\ g^{\lambda\alpha}\left(\frac{\partial g_{\mu\alpha}}{\partial x^{\nu}} + \frac{\partial g_{\nu\alpha}}{\partial x^{\mu}} - \frac{\partial g_{\mu\nu}}{\partial x^{\alpha}}\right)$$

and it's all ok. The covariant derivative has been defined too, and now the professor "proved" that since the normal derivative of a tensor is not a tensor, the covariant derivative will. As a test, he, again, "proved" to run the covariant derivative on the metric tensor, getting:

$$\mathsf{D}_{\lambda} g_{\mu\nu} = \frac{\partial g_{\mu\nu}}{\partial x^{\lambda}} - \Gamma^{\rho}_{\mu\lambda}g_{\rho\nu} - \Gamma^{\rho}_{\nu\lambda}g_{\rho\mu}$$

Ok again, until here. Then "after a little algebra" we find that:

$$\mathsf{D}_{\lambda} g_{\mu\nu} = 0$$

But the "little algebra" is the missing part. Now I would like to prove it, but I'm stuck on the calculation and I'm going to tell you how and why.

First of all, let me tell you that I started with substituting the $\Gamma$, getting the long expression:

\begin{equation*} \begin{split} \mathsf{D}_{\lambda} g_{\mu\nu} & = \frac{\partial g_{\mu\nu}}{\partial x^{\lambda}} - \bigg\{\frac{1}{2}\ g^{\rho k}\left(\frac{\partial g_{\mu k}}{\partial x^{\lambda}} + \frac{\partial g_{\lambda k}}{\partial x^{\mu}} - \frac{\partial g_{\mu\lambda}}{\partial x^{k}}\right)\bigg\}g_{\rho\nu} \\\\ & - \bigg\{\frac{1}{2}\ g^{\rho \alpha}\left(\frac{\partial g_{\nu \alpha}}{\partial x^{\lambda}} + \frac{\partial g_{\lambda \alpha}}{\partial x^{\nu}} - \frac{\partial g_{\lambda\nu}}{\partial x^{\alpha}}\right)\bigg\}g_{\rho\nu} \end{split} \end{equation*} \ \ Then I continued to develop the calculation but here is the problem : whilst developing the calculation, I met lots of terms of this form (for example the very fist one):

$$\frac{1}{2} g^{\rho k}\frac{\partial g_{\mu k}}{\partial x^{\lambda}}\ g_{\rho\nu}$$

There are other five terms like this, with different indexes of course, but I don't know how to proceed. Tensor calculus when it comes to manipulation of index is somehow "strange". I never know what to do because nobody taught us anything about if not the basics. And when the professor says "after little algebra" I would like to see that little algebra at least once in a lifetime, right?

So, any hint / explanation / something for this? My problem is how to treat objects like that. There are too many indexes, and I need to really understand how to proceed and work. Thank you for the time!

$\endgroup$
2
$\begingroup$

This is a nice exercise and it is not a difficult one if one knows how to deal with it and I will flesh out @mike stones answer (which is correct) a bit:

  1. You have a typo in your fourth equation: the last metric should read $g_{\rho\mu}$.
  2. You are almost there if you put everything in correctly. The last expression is really your only problem here so lets focus on that. But before that allow me to introduce the compact notation $\frac{\partial g_{\alpha\beta}}{\partial x^\gamma}\equiv g_{\alpha\beta,\gamma}$.

$$ g^{\rho\kappa}g_{\mu\kappa,\lambda}g_{\rho\nu}=g^{\rho\kappa}g_{\rho\nu}g_{\mu\kappa,\lambda} $$

We are working only with components when we use this compact notation and Einstein's sum convention, which means we can allways change the order in our summands as we please. Now we have one contracted indice $\rho$ here and we sum over contracted indices and as @mike stones pointed out

$$g^{\rho\kappa}g_{\rho\nu}=g^{\kappa\rho}g_{\rho\nu}=\delta^\kappa_{~~\nu}.$$

This is a fundamental relation between the metic and its inverse; it is bassically the defining property of the inverse metric. In matrix notation that would read $(g^{-1})(g)=diag(1,1,1,1)$.

If you use that in your two expressions you can eliminate one of the indices which are in the Kronecker-Delta since all cases where $\kappa \ne \nu$ are $0$ anyway. If you do that you should get out that the covariant derivative of the metric vanishes.

Allways remember that this are just sums over the contracted indices, all objects are just components and to check if you have the right amount of contracted and non contracted indices in your equations. Using the symmetries of $g$ and higher objects as well as relabeling indices in different summands will become necessary for more advanced computations.

$\endgroup$
  • $\begingroup$ Great answer, really clear and clean! Thank you. Unfortunately, professor never spent much time in explaining tensor calculus. I thought about delta identities, but as I wrote in the comment below, I wasn't sure I could move the metric tensor from the right to the left. Thank you! I'm going to do it now :D $\endgroup$ – Les Adieux Sep 15 '16 at 19:31
  • 1
    $\begingroup$ Yeah some times making things like this clear and going over it in detail comes a bit short in the lectures. So following and trying to understand the steps or really show the identities mentioned with "it can be show/one can show" helps a lot with understanding the math/mechanics going on. $\endgroup$ – N0va Sep 15 '16 at 19:42
1
$\begingroup$

Use $g^{\rho k} g_{\rho \nu}= \delta^k_\nu$ to get $$ g^{\rho k} \frac{\partial g_{\mu k}}{\partial x^\lambda} g_{\rho\nu}= \delta^k_\nu \frac{\partial g_{\mu k}}{\partial x^\lambda}= \frac{\partial g_{\mu \nu}}{\partial x^\lambda} $$ and so on....

$\endgroup$
  • $\begingroup$ You indirectly solved another hidden doubt I had, namely: so I can actually move the metric from the right to the left (or vice versa) without any problem! That simplify the problem a lot! $\endgroup$ – Les Adieux Sep 15 '16 at 19:08
  • 1
    $\begingroup$ @Entropy The elements of the metric are just numbers, which commute, so you can definitely switch their order :) $\endgroup$ – drglove Sep 15 '16 at 19:22

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.