[Physics] Metric determinant and its partial and covariant derivative

differential-geometrydifferentiationgeneral-relativitymetric-tensortensor-calculus

question : $\nabla_a \nabla_b \sqrt{g} \phi =\partial_a \sqrt{g} \partial_b \phi$ is true?

because $\nabla_a \sqrt{g}=0$ so we can write $\sqrt{g} \nabla_a \nabla_b \phi$ , but because the determinant of the metric does not transform like a scalar, we can not write partial derivative instead of covariant derivative.

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$$ g=\frac{1}{4!}\varepsilon^{abcd}\varepsilon^{efgh}g_{ae}g_{bf}g_{dg}g_{dh}\\ \therefore \quad \nabla_m g = \frac{1}{3!} \varepsilon^{abcd}\varepsilon^{efgh}g_{ae}g_{bf}g_{dg}\nabla_m g_{dh}\\=0\;. $$ Note that $\varepsilon^{abcd}$ is Levi-Civita symbol, it is constant. It is tensor density of weight 0.5 and make $g$ be a tensor density of weight 1.

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General Relativity – Variation of Modified Einstein-Hilbert Action Explained

Find the answer to question 1 below. Question 2. is weird since $\nabla_\mu g_{\alpha\beta} = 0$ (when the connection is metric compatible) as mentioned by @Trimok. In any case, the variation of the action can be derived using the method described below.

We start with the BD action $$ S = \frac{1}{16\pi}\int d^4 x \sqrt{-g} \left[ \phi R - \frac{\omega}{\phi} g^{\mu\nu} \partial_\mu \phi \partial_\nu \phi \right] + S_M $$ where $S_M$ is the matter action. To determine Einstein's field equations, we vary the action w.r.t to the metric. We will use the formulae (ref. wikipedia) \begin{equation} \begin{split} \delta R = R_{\mu\nu} \delta g^{\mu\nu} + \nabla_\sigma \left( g^{\mu\nu} \delta \Gamma^\sigma_{\mu\nu} - g^{\mu\sigma} \delta \Gamma^\rho_{\rho\mu} \right) \end{split} \end{equation} The variation of the Christoffel tensor is \begin{equation} \begin{split} \delta \Gamma^\lambda_{\mu\nu} & = \delta g^{\lambda\rho} g_{\rho\alpha} \Gamma^\alpha_{\mu\nu} + \frac{1}{2} g^{\lambda\rho} \left( \partial_\mu \delta g_{\nu\rho} + \partial_\nu \delta g_{\mu\rho} - \partial_\rho \delta g_{\mu\nu} \right) \\ & = \frac{1}{2} g^{\lambda\rho} \left( \nabla_\mu \delta g_{\nu\rho} + \nabla_\nu \delta g_{\mu\rho} - \nabla_\rho \delta g_{\mu\nu} \right) \\ & = - \frac{1}{2} \left( g_{\nu\alpha} \nabla_\mu \delta g^{\alpha\lambda} + g_{\mu\alpha} \nabla_\nu \delta g^{\alpha\lambda} - g_{\mu\alpha} g_{\nu\beta} \nabla^\lambda \delta g^{\alpha\beta} \right) \end{split} \end{equation} where we used $\delta g_{\mu\nu} = - g_{\mu\alpha} g_{\nu\beta} \delta g^{\alpha\beta}$. This implies \begin{equation} \begin{split} g^{\mu\nu} \delta \Gamma^\sigma_{\mu\nu} &= -\nabla_\alpha \delta g^{\alpha\sigma} + \frac{1}{2} g_{\alpha\beta}\nabla^\sigma\delta g^{\alpha\beta} \\ g^{\mu\sigma} \delta \Gamma^\lambda_{\lambda\mu} &= - \frac{1}{2}g_{\alpha\beta} \nabla^\sigma \delta g^{\alpha\beta} \end{split} \end{equation} which implies \begin{equation} \begin{split} \delta R = R_{\mu\nu} \delta g^{\mu\nu} - \nabla_\mu \nabla_\nu \delta g^{\mu\nu} + g_{\mu\nu} \nabla^2 \delta g^{\mu\nu} \end{split} \end{equation} Finally, from 1, we also have $$ \delta \sqrt{-g} = - \frac{1}{2} \sqrt{-g} g_{\mu\nu} \delta g^{\mu\nu} $$ Finally, we are ready to compute the variation of the action. We have \begin{equation} \begin{split} \delta S &= \frac{1}{16\pi}\int d^4 x \delta \sqrt{-g} \left[ \phi R - \frac{\omega}{\phi} g^{\mu\nu} \partial_\mu \phi \partial_\nu \phi \right] \\ &~~~~~~~~~~~~ + \frac{1}{16\pi}\int d^4 x \sqrt{-g} \left[ \phi \delta R - \frac{\omega}{\phi} \delta g^{\mu\nu} \partial_\mu \phi \partial_\nu \phi \right] + \delta S_M \\ &= - \frac{1}{32\pi}\int d^4 x \sqrt{-g} g_{\mu\nu} \left[ \phi R - \frac{\omega}{\phi} g^{\alpha\beta} \partial_\alpha \phi \partial_\beta \phi \right] \delta g^{\mu\nu} + \int d^4 x \frac{ \delta S_M}{\delta g^{\mu\nu}} \delta g^{\mu\nu} \\ &~~~~~~~~~~~~ + \frac{1}{16\pi}\int d^4 x \sqrt{-g} \left[ \left( \phi R_{\mu\nu} - \nabla_\mu \nabla_\nu \phi + g_{\mu\nu} \nabla^2 \phi \right) - \frac{\omega}{\phi} \partial_\mu \phi \partial_\nu \phi \right] \delta g^{\mu\nu} \end{split} \end{equation} Requiring that the variation of the action vanish (to leading order in $\delta g^{\mu\nu}$) gives \begin{equation} \begin{split} G_{\mu\nu} &= - \frac{16\pi}{\phi \sqrt{-g}} \frac{ \delta S_M}{\delta g^{\mu\nu}} + \frac{\omega}{\phi^2} \left[ \partial_\mu \phi \partial_\nu \phi - \frac{1}{2} g_{\mu\nu} \partial_\alpha \phi \partial^\alpha \phi \right] +\frac{1}{\phi} \left[ \nabla_\mu \nabla_\nu \phi - g_{\mu\nu} \nabla^2 \phi \right] \end{split} \end{equation} Recall that the stress-tensor is defined as \begin{equation} \begin{split} T_{\mu\nu} = - \frac{2}{\sqrt{-g}} \frac{ \delta S_M}{\delta g^{\mu\nu}} \end{split} \end{equation} Thus \begin{equation} \begin{split} G_{\mu\nu} &= \frac{8\pi}{\phi} T_{\mu\nu} + \frac{\omega}{\phi^2} \left[ \partial_\mu \phi \partial_\nu \phi - \frac{1}{2} g_{\mu\nu} \partial_\alpha \phi \partial^\alpha \phi \right] +\frac{1}{\phi} \left[ \nabla_\mu \nabla_\nu \phi - g_{\mu\nu} \nabla^2 \phi \right] \end{split} \end{equation} which is the Brans-Dicke equation.

General Relativity – Why Is the Covariant Derivative of the Determinant of the Metric Zero?

Comments to the post (v2):

Note that $\sqrt{|g|}$ transforms as a density rather than a scalar under general coordinate transformations. In particular, the covariant derivative of $\sqrt{|g|}$ does not necessarily coincide with the partial derivative of $\sqrt{|g|}$.
Here is a heuristic explanation using local coordinates. The Levi-Civita connection is compatible with the metric $g_{\mu\nu}\mathrm{d}x^{\mu}\odot \mathrm{d}x^{\nu}$. That a connection $\nabla$ is compatible with a metric means that $\nabla_{\lambda}g_{\mu\nu}=0$. Using linearity and Leibniz rule, the covariant derivative $\nabla_{\lambda}$ then annihilates any sufficiently nice function $f(g_{00},g_{01}, \ldots)$ of the metric. In particular, the square root of the determinant $\sqrt{|g|}$, so $\nabla_{\lambda}\sqrt{|g|}=0$.

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General Relativity – Variation of Modified Einstein-Hilbert Action Explained

General Relativity – Why Is the Covariant Derivative of the Determinant of the Metric Zero?

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