I want to show that $$\Gamma ^{\mu}_{\mu \nu}=\partial _\nu (\ln \sqrt{|g|}) .$$ (Here $|g|$ denotes the determinant of the metric.)

Working out the left hand side:\begin{align} \Gamma ^{\mu}_{\mu \nu} &= \dfrac{1}{2} g^{\mu \rho}(\partial _{\mu} g_{\nu \rho } + \partial _{\nu} g_{\rho \mu} - \partial _{\rho} g_{\mu \nu})\\ &= \dfrac{1}{2} g^{\mu \rho} \partial _{\nu} g_{\rho \mu} \end{align} where I've used the symmetry of the metric.

The right hand side yields \begin{align} \partial _\nu (\ln \sqrt{|g|})&= \dfrac{1}{2|g|}\partial _\nu |g|\\ &= \dfrac{1}{2}|g^{-1}|\partial _\nu |g| \end{align}

A definition for the determinant of the metric is \begin{align} \tilde{\epsilon}^{\mu \nu \rho \sigma}|g|= \tilde{\epsilon}_{\mu' \nu' \rho' \sigma'} g^{\mu' \mu}g^{\nu' \nu}g^{\rho' \rho}g^{\sigma' \sigma} \end{align} where $\tilde{\epsilon}^{\mu \nu \rho \sigma}$ is the Levi-Cevita symbol.

My questions:

  • Is what I've done so far correct?

  • How can I match the result $\dfrac{1}{2} g^{\mu \rho} \partial _{\nu} g_{\rho \mu}$ with $\dfrac{1}{2}|g^{-1}|\partial _\nu |g|$? I don't see how to incorporate the definition of the determinant...


I am not sure for the correctness of your equation with Levi-Civita symbol.

Instead, I recommend using the Jacobi's formula for derivative of matrix determinant. Taking the equation from corollary part of wikipedia's page (as more suitable for our needs) and after subsituting some symbols (to avoid confusion in application of this to GR) we have $$ \frac{d}{d\lambda} \det[ g_{\mu\nu}] = \frac{d}{d\lambda} \det[ g(\lambda)] = \det[ g(\lambda)] \, \mathrm{tr} \left[g(\lambda)^{-1} \, \frac{d}{d\lambda} g(\lambda)\right] = \det[ g(\lambda)] \,\, g^{\mu\nu} \frac{d}{d\lambda} g_{\nu\mu}. $$

Since variable $\lambda$ could be anything, including coordinate variable $x^{\nu}$, it means that the right hand side of your original equation could be written as $$ \partial_\nu\left(\ln\sqrt{|g|}\right)= \frac12 |g|^{-1} \partial_\nu{|g|} =\frac12 g^{\mu\rho}\partial _\nu g_{\rho\mu}, $$ which is what you already obtained for the left hand side. Note: obviously, $g$ does not change sign, so the $|\dots|$ does not affect differentiation.

  • $\begingroup$ Thank you, this answers my question! I still have a remark about your note: why is it obvious that g does not change sign? $\endgroup$ – Anne O'Nyme Oct 20 '13 at 9:04
  • $\begingroup$ @user52125: Since the sign change of $g$ would mean that the metric signature changed. Also, if $g$ passes zero it would mean that metric is degenerate at some point. $\endgroup$ – user23660 Oct 20 '13 at 9:22

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