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The equation of motion for a scalar field in curved spacetime $$\frac{\partial\mathcal{L}}{\partial\phi}=\frac{1}{\sqrt{-g}}\partial_{\mu}\left[\sqrt{-g}\frac{\partial\mathcal{L}}{\partial\left(\partial_{\mu}\phi\right)}\right]\tag{1}$$ can be written in terms of the covariant derivative as $$\frac{\partial\mathcal{L}}{\partial\phi}=\nabla_{\mu}\left[\frac{\partial\mathcal{L}}{\partial\left(\partial_{\mu}\phi\right)}\right].\tag{2}$$

Here $\mathcal{L}$ is a Lagrangian scalar function. How is Eq.$(2)$ obtained from Eq.$(1)$? The action of a covariant on a vector field $A_\mu$ is given by $$\nabla_\mu A_\nu=\partial_\mu A_\nu-\Gamma_{\mu\nu}^{\rho}A_\rho.$$

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  • $\begingroup$ source? where did you read that those two expressions are equivalent? have you done any research before posting this here? Thanks. $\endgroup$ – AccidentalFourierTransform Aug 11 at 15:34
  • $\begingroup$ @AccidentalFourierTransform Source: Answer here: physics.stackexchange.com/q/74779 and yes, I tried to derive 2 from 1. $\endgroup$ – mithusengupta123 Aug 11 at 16:00
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I'll give hints instead of a complete solution, in case this ends up being categorized as homework.

I'll write $|g|$ instead of $-g$, because the former works for either sign convention (mostly-plus or mostly-minus). The quantity $$ V^\mu := \frac{\partial {\cal L}}{\partial(\partial_\mu\phi)} \tag{1} $$ is a vector (I mean, the components of a vector), so the question is how to derive $$ \frac{1}{\sqrt{|g|}}\partial_\mu\left(\sqrt{|g|} V^\mu\right) =\nabla_\mu V^\mu \tag{2} $$ for a vector $V$. Use the definition of $\nabla_\mu$ to see that (2) can also be written $$ \frac{1}{\sqrt{|g|}} V^\mu\partial_\mu\sqrt{|g|} = \Gamma^{\alpha}_{\alpha\mu}V^\mu. \tag{3} $$ The left-hand side of (3) involves the quantity $$ \frac{1}{\sqrt{|g|}}\partial_\mu\sqrt{|g|} = \frac{1}{2}\partial_\mu\log|g|. \tag{4} $$ We can think of $g$ as a square matrix, and $|g|$ is its determinant. The matrix is invertible, so $|g|$ is the product of its eigenvalues $\lambda_n$. Rewrite the right-hand side of (4) in terms of $\lambda_n$, and recall that if $A,B$ are diagonalizable square matrices, then the trace of their product is equal to the sum of the products of their eigenvalues. Use this write the right-hand side of (4) in terms of the matrices $g^{-1}$ and $\partial_\mu g$. Written in terms of indices, this becomes $$ \frac{1}{\sqrt{|g|}}\partial_\mu\sqrt{|g|} =\frac{1}{2}g^{\alpha\beta}\partial_\mu g_{\alpha\beta}. \tag{5} $$ We can use this to rewrite the left-hand side of (3). Now write the $\Gamma^{\alpha}_{\alpha\mu}$ on the right-hand side of (3) in terms of the metric and its derivatives to see that the identity (3) is indeed true.

Reference: Chapter 3 in Ward (1984), General Relativity, University of Chicago Press. Equation (3.4.10) in that book is equation (3) above.

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