A helpful proof in contracting the Christoffel symbol?

Question

Out of all of my time learning General relativity, this is the one identity that I cannot get around. $$ \Gamma_{\alpha \beta}^{\alpha} = \partial_{\beta}\ln\sqrt{-g} \tag{1}$$ where $g$ is the determinant of the metric tensor $g_{\alpha \beta}$.

With the Christoffel symbol, we start by contracting

$$ \begin{align} \Gamma_{\alpha \beta}^{\alpha} &= \frac{1}{2} g^{\alpha\gamma} (\partial_{\alpha} g_{\beta\gamma} + \partial_{\beta} g_{\alpha\gamma} - \partial_{\gamma} g_{\alpha\beta} ) \\ &= \frac{1}{2} g^{\alpha\alpha} ( \partial_{\beta} g_{\alpha\alpha}) \\ &= \frac{1}{2g_{\alpha\alpha}} ( \partial_{\beta} g_{\alpha\alpha}) \end{align}\tag{2}$$

where I took $\gamma \rightarrow \alpha$ and $g^{\alpha\alpha} = 1/g_{\alpha\alpha}$.

The next steps to take now, I have no clue. MTW gives a hint by saying to use the results from some exercise, which are,

$$\det A = \det||A^{\lambda}_{\ \ \rho}|| = \tilde{\epsilon}^{\alpha\beta\gamma\delta}A^{0}_{\ \ \alpha}A^{1}_{\ \ \beta}A^{2}_{\ \ \gamma}A^{3}_{\ \ \delta} $$

$$(A^{-1})^{\mu}_{\ \ \alpha}(\det A) = \frac{1}{3!}\delta_{\alpha\beta\gamma\delta}^{\mu\nu\rho\sigma} A^{\beta}_{\ \ \nu} A^{\gamma}_{\ \ \rho}A^{\delta}_{\ \ \sigma} $$

$$ \mathbf{d}\ln|\det A| = \mathrm{trace}(A^{-1}\mathbf{d}A) ,\tag{3}$$ where $\mathbf{d}A$ is the matrix $||\mathbf{d}A^{\alpha}_{\ \ \mu}||$ whose entries are one-forms.

I fail to reason why the metric turns into the determinant from what I have done and then becomes the result at the top.

Answer 1

Recall the matrix identity $$\tag{1}\log\det M=\operatorname{tr}\log M.$$ If $M=M(\lambda)$ is differentiable in $\lambda$, then $$\tag{2}\frac{d}{d\lambda}\log\det M=\operatorname{tr}\left(M^{-1}\frac{d}{d\lambda} M\right).$$ The proof of $(1)$ for symmetric matrices follows from the usual formulae for the trace and determinant in terms of eigenvalues$^{1}$.

As for the Christoffels, we have $$\Gamma^i{}_{ij}=\frac{1}{2}g^{ik}(\partial_i g_{jk}+\partial_j g_{ik}-\partial_k g_{ij})=\frac{1}{2}g^{ik}\partial_j g_{ik}=\frac{1}{2}\operatorname{tr}(g^{-1} \partial_j g).$$ The last equality is just what the contraction of indices means for the (symmetric!) matrix $g=(g_{ij})$, and there is an error in the indices in OP's post. Now, using $(2)$ we have $$\Gamma^i{}_{ij}=\frac{1}{2}\partial_j\log \det g.$$ This can be brought into the form $$\Gamma^i{}_{ij}=\partial_j \log\sqrt{|\det g|}$$ by the usual rules of calculus.

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$^{1}$ For symmetric matrices, such as $g$, it is easy because $g$ can be diagonalized. For other matrices you might need a Jordan normal form to compute $\log M$.

Answer 2

The original and the most general definition of determinant is given by Gauss . For the determinant of metric tensor we write \begin{eqnarray} g&:=& \frac{1}{4!}\varepsilon^{{\alpha\beta}{\gamma\delta}}\varepsilon^{{\mu\nu}{\rho\sigma}}g_{\alpha\mu}g_{\beta\nu}g_{\gamma\rho}g_{\delta\sigma}.\\ \therefore \delta g &=& \frac{1}{3!}\varepsilon^{{\alpha\beta}{\gamma\delta}}\varepsilon^{{\mu\nu}{\rho\sigma}}g_{\alpha\mu}g_{\beta\nu}g_{\gamma\rho}\delta g_{\delta\sigma},\\ &=&\frac{-g}{3!}\epsilon^{{\alpha\beta}{\gamma\delta}}\epsilon^{{\mu\nu}{\rho\sigma}} g_{\alpha\mu}g_{\beta\nu}g_{\gamma\rho}\delta g_{\delta\sigma},\\ &=&\frac{-g}{3!}\epsilon^{{\alpha\beta}{\gamma\delta}}\epsilon_{{\alpha\beta}\gamma}{}^\sigma \delta g_{\delta\sigma},\\ &=&g \, g^{\delta\sigma}\delta g_{\delta\sigma}. \end{eqnarray} (Equivalently, $$ \delta(\ln |\det g|)=Tr (g^{-1}\delta g)= Tr ( \delta \ln g )$$) By using this result we the have $$ \frac{1}{g} \partial_\beta g =g^{\delta\sigma}\partial_\beta g_{\delta\sigma} $$

Note: You may need some basic manipulation of these quantities, $\varepsilon$ is Levi-Civita symbol, $\epsilon$ is Levi-Civita tensor.

$$\epsilon^{{\alpha\beta}{\gamma\delta}} = -\frac{1}{\sqrt{-g}}\varepsilon^{{\alpha\beta}{\gamma\delta}}$$ $$\epsilon_{{\alpha\beta}{\gamma\delta}} = \sqrt{-g}\varepsilon_{{\alpha\beta}{\gamma\delta}}$$

$$\varepsilon_{{\alpha\beta}{\gamma\delta}}=\varepsilon^{{\alpha\beta}{\gamma\delta}}= \delta^{[\alpha}_0 \delta^\beta_1 \delta^\gamma_2 \delta^{\delta]}_3$$

$$\varepsilon^{{\alpha\beta}{\gamma\delta}}\varepsilon_{{\mu\nu}{\rho\sigma}} =-\epsilon^{{\alpha\beta}{\gamma\delta}}\epsilon_{{\mu\nu}{\rho\sigma}}= 4! \delta^{[\alpha}_\mu \delta^\beta_\nu \delta^\gamma_\rho \delta^{\delta]}_\sigma \equiv \delta^{\alpha\beta\gamma\delta}_{\mu\nu\rho\sigma}\equiv \left| \begin{matrix} \delta^\alpha_\mu & \delta^\alpha_\nu & \delta^\alpha_\rho & \delta^\alpha_\sigma \\ \delta^\beta_\mu & \delta^\beta_\nu & \delta^\beta_\rho & \delta^\beta_\sigma \\ \delta^\gamma_\mu & \delta^\gamma_\nu & \delta^\gamma_\rho & \delta^\gamma_\sigma \\ \delta^\delta_\mu & \delta^\delta_\nu & \delta^\delta_\rho & \delta^\delta_\sigma \end{matrix} \right|$$

$$\epsilon^{{\alpha\beta}{\gamma\delta}}\epsilon_{{\alpha\nu}{\rho\sigma}} = -\left| \begin{matrix} \delta^\alpha_\alpha & \delta^\alpha_\nu & \delta^\alpha_\rho & \delta^\alpha_\sigma \\ \delta^\beta_\alpha & \delta^\beta_\nu & \delta^\beta_\rho & \delta^\beta_\sigma \\ \delta^\gamma_\alpha & \delta^\gamma_\nu & \delta^\gamma_\rho & \delta^\gamma_\sigma \\ \delta^\delta_\alpha & \delta^\delta_\nu & \delta^\delta_\rho & \delta^\delta_\sigma \end{matrix} \right| =-\left| \begin{matrix} 1 & 0 & 0 & 0 \\ 0 & \delta^\beta_\nu & \delta^\beta_\rho & \delta^\beta_\sigma \\ 0 & \delta^\gamma_\nu & \delta^\gamma_\rho & \delta^\gamma_\sigma \\ 0 & \delta^\delta_\nu & \delta^\delta_\rho & \delta^\delta_\sigma \end{matrix} \right| =-\left| \begin{matrix} \delta^\beta_\nu & \delta^\beta_\rho & \delta^\beta_\sigma \\ \delta^\gamma_\nu & \delta^\gamma_\rho & \delta^\gamma_\sigma \\ \delta^\delta_\nu & \delta^\delta_\rho & \delta^\delta_\sigma \end{matrix} \right|$$

$$\epsilon^{{\alpha\beta}{\gamma\delta}}\epsilon_{{\alpha\beta}{\rho\sigma}} = -\left| \begin{matrix} \delta^\beta_\beta & \delta^\beta_\rho & \delta^\beta_\sigma \\ \delta^\gamma_\beta & \delta^\gamma_\rho & \delta^\gamma_\sigma \\ \delta^\delta_\beta & \delta^\delta_\rho & \delta^\delta_\sigma \end{matrix} \right|=-\left| \begin{matrix} 2 & 0 & 0 \\ 0 & \delta^\gamma_\rho & \delta^\gamma_\sigma \\ 0 & \delta^\delta_\rho & \delta^\delta_\sigma \end{matrix} \right|=-2 \left| \begin{matrix} \delta^\gamma_\rho & \delta^\gamma_\sigma \\ \delta^\delta_\rho & \delta^\delta_\sigma \end{matrix} \right| $$

$$\epsilon^{{\alpha\beta}{\gamma\delta}}\epsilon_{{\alpha\beta}{\gamma\sigma}} =-2 \left| \begin{matrix} \delta^\gamma_\gamma & \delta^\gamma_\sigma \\ \delta^\delta_\gamma & \delta^\delta_\sigma \end{matrix} \right| =-2 \left| \begin{matrix} 3 & 0 \\ 0 & \delta^\delta_\sigma \end{matrix} \right| = -3! \delta^\delta_\sigma $$

$$\epsilon^{{\alpha\beta}{\gamma\delta}}\epsilon_{{\alpha\beta}{\gamma\delta}} =-3! \delta^\delta_\delta = -4!$$

Note2: \begin{eqnarray} g^{-1}&:=& \frac{1}{4!}\varepsilon_{{\alpha\beta}{\gamma\delta}}\varepsilon_{{\mu\nu}{\rho\sigma}}g^{\alpha\mu}g^{\beta\nu}g^{\gamma\rho}g^{\delta\sigma}.\\ \therefore \delta g^{-1} &=& \frac{1}{3!}\varepsilon_{{\alpha\beta}{\gamma\delta}}\varepsilon_{{\mu\nu}{\rho\sigma}}g^{\alpha\mu}g^{\beta\nu}g^{\gamma\rho}\delta g^{\delta\sigma},\\ &=&\frac{-g^{-1}}{3!}\epsilon_{{\alpha\beta}{\gamma\delta}}\epsilon_{{\mu\nu}{\rho\sigma}} g^{\alpha\mu}g^{\beta\nu}g^{\gamma\rho}\delta g^{\delta\sigma},\\ &=&\frac{-g^{-1}}{3!}\epsilon_{{\alpha\beta}{\gamma\delta}}\epsilon^{{\alpha\beta}\gamma}{}_\sigma \delta g^{\delta\sigma},\\ &=&g^{-1} \, g_{\delta\sigma}\delta g^{\delta\sigma}.\\ \therefore -g^{-2}\delta g &=&g^{-1} \, g_{\delta\sigma}\delta g^{\delta\sigma}.\\ \delta g &=& -g\,g_{\delta\sigma}\delta g^{\delta\sigma}. \end{eqnarray} By comparing this $\delta g$ to the first result, we have $$\delta g_{\alpha\beta}=- g_{\alpha\mu}g_{\beta\nu} \delta g^{\mu\nu}$$

Answer 3

Here's a direct/theoretical way to arrive at the formula for $\Gamma^{\nu}_{\mu \nu}$ in terms of the square root of the determinant of the metric from first principles. Note $\sqrt{-g}$ appears in covariant four-dimensional integrals, so you can expect the formula to be derived using integrals.

Using $$\nabla g_{\mu \nu} = \nabla g = 0 \ \ \ , \ \ \ \nabla^2 = g^{\mu \nu} \nabla_{\mu} \nabla_{\nu} \ \ , $$ we have (ignoring integration by parts terms assumed to vanish) $$\int d^4 x \sqrt{-g} (\nabla^2 \phi) \phi = - \int d^4 x \sqrt{-g} g^{\mu \nu} \nabla_{\mu} \phi \nabla_{\nu} \phi = - \int d^4 x \sqrt{-g} g^{\mu \nu} \partial_{\mu} \phi \partial_{\nu} \phi = \int d^4 x \sqrt{-g} \frac{1}{\sqrt{-g}} \partial_{\mu} (\sqrt{-g} g^{\mu \nu} \partial_{\nu} \phi) \phi $$ giving the curvilinear form of the Laplacian $$\nabla^2 \phi = \frac{1}{\sqrt{-g}} \partial_{\mu} (\sqrt{-g} g^{\mu \nu} \partial_{\nu} \phi) $$ so that on expanding it directly $$\nabla_{\mu} (\nabla^{\mu} \phi) = \partial_{\mu} (\partial^{\mu} \phi) + \Gamma^{\mu}_{\mu \nu} (\partial^{\nu} \phi) = \frac{1}{\sqrt{-g}} \partial_{\mu} (\sqrt{-g} g^{\mu \nu} \partial_{\nu} \phi) = \partial_{\mu} \partial^{\mu} \phi + (\frac{1}{\sqrt{-g}} \partial_{\nu} \sqrt{-g}) \partial^{\nu} \phi$$ we find the following expression for the contracted Christoffel symbols $$\Gamma^{\mu}_{\mu \nu} = \frac{1}{\sqrt{-g}} \partial_{\nu} \sqrt{-g} = \partial_{\nu} \ln \sqrt{-g} .$$