# The equation of motion for a scalar field in curved spacetime in terms of the covariant derivative

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 $$\nabla_\mu A_\nu=\partial_\mu A_\nu-\Gamma_{\mu\nu}^{\rho}A_\rho.$$

• source? where did you read that those two expressions are equivalent? have you done any research before posting this here? Thanks. – AccidentalFourierTransform Aug 11 at 15:34
• @AccidentalFourierTransform Source: Answer here: physics.stackexchange.com/q/74779 and yes, I tried to derive 2 from 1. – mithusengupta123 Aug 11 at 16:00

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.