1
$\begingroup$

I know that the following is true: $$V^{\mu}_{~~~~~;\nu} = \frac{\partial V^{\mu}}{\partial x^{\nu}} +\Gamma^{\mu}_{~\sigma\nu} V^{\sigma}.$$

Also, by definition, we have that $V_{\rho} = g_{\rho\nu}V^{\nu}$.

I would like to show that $$V_{\rho;\nu} = \frac{\partial V_{\rho}}{\partial x^{\nu}} - \Gamma^{\sigma}_{~\rho\nu} V_{\sigma}$$

I have attempted to do this by taking $(g_{\rho\nu}V^{\nu})_{;\mu}$ and using the Leibniz rule and the fact that $g_{\rho\nu;\mu}=0$ but I am missing the minus sign!

$\endgroup$
0
$\begingroup$

$\def\m{\mu} \def\n{\nu} \def\r{\rho} \def\s{\sigma} \def\t{\tau} \def\G{\Gamma}$While the method in @DanielC's answer is often the one followed it is possible, straightforward, and instructive to approach this in the way you have outlined in your question statement. We have \begin{align*} V_{\r;\n} &= (g_{\r\m}V^\m)_{;\n} \\ &= \underbrace{g_{\r\m;\n}}_0 V^\m + g_{\r\m}V^\m_{;\n} \\ &= g_{\r\m}(V^\m_{,\n} + \G^\m_{\ \s\n}V^\s) \\ &= g_{\r\s}V^\s_{,\n} + \G_{\r\s\n}V^\s \\ &= (g_{\r\s}V^\s)_{,\n} - g_{\r\s,\n}V^\s + \G_{\r\s\n}V^\s \\ &= V_{\r,\n} - (g_{\r\s,\n} - \G_{\r\s\n})V^\s \\ &= \ldots \\ &= V_{\r,\n} - \G^\t_{\ \r\n}V_\t. \end{align*} In the omitted steps one will need to use the well-known identity $$g_{\r\s,\n} = \G_{\s\r\n} + \G_{\r\n\s},$$ where $\G_{\r\s\t} = g_{\m\r} \G^\m_{\ \s\t}$.

$\endgroup$
1
$\begingroup$

This is not the way to do it. You need to compute $$\nabla_\mu \left( V^\alpha V_\alpha\right) $$ using the Leibniz rule and the fact that there is a scalar inside the bracket.

$\endgroup$
  • $\begingroup$ Hi Daniel, thanks for your answer! So we can say that the covariant derivative you stated is equal to 0? $\endgroup$ – user10000654 May 31 '18 at 6:19
  • $\begingroup$ After using this method, I now have a '-' sign in front of $V_{\rho ,\nu}$. Could you expand your answer a bit please to help me see what I am missing? $\endgroup$ – user10000654 May 31 '18 at 6:33
  • $\begingroup$ The covariant derivative of a scalar is equal to the regular partial derivative of that scalar. So you need to apply the Leibniz rule again and subtract alike terms. $\endgroup$ – DanielC May 31 '18 at 6:39
  • $\begingroup$ I have just noticed that, sorry for the confusion! I also need to use that permutation of indices for Christoffel symbols are antisymmetric, right? $\endgroup$ – user10000654 May 31 '18 at 6:42
  • $\begingroup$ Well, the indices of the Christoffel symbols are symmetric, if the metric is covariantly constant. $\endgroup$ – DanielC May 31 '18 at 7:09

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.