# Euler-Lagrange equation in curved spacetime

The action of a field $$\phi^\mu$$ in flat $$n$$-dimensional spacetime is $$S = \int \text{d}^n x \mathscr{L}(\phi^\mu(x),\partial_\alpha \phi^\mu(x))$$ From an infinitesimal variation of field configuration, and discarding a boundary term, we have E-L equations $$\frac{\partial \mathscr{L}}{\partial \phi^\mu} - \partial_\alpha \frac{\partial \mathscr{L}}{\partial (\partial_\alpha \phi^\mu)}$$ In curved spacetime the action should now be $$S = \int \text{d}^n x \sqrt{|g|} \mathscr{L}(\phi^\mu(x),\nabla_\alpha \phi^\mu(x))$$ but the covariant derivative and the element $$\sqrt{|g|}$$ make it hard to isolate a surface term. I tried various way round, but I can't obtain the expected, seemingly covariant, E-L equation $$\frac{\partial \mathscr{L}}{\partial \phi^\mu} - \nabla_\alpha \frac{\partial \mathscr{L}}{\partial (\nabla_\alpha \phi^\mu)}$$ even because this very formula may be wrong; I don't know in fact what is the actual form of E-L equation to be found.

I've looked a lot on this site and various sources, but I can't find a step-by-step demonstration of E-L equations in curved spacetime and a clear final statement, so what is the path to follow that brings to the result?

• I discuss this in Sec. 1.3 here. The change-$\partial$s-to-$\nabla$s trick works if $\phi$ is independent of the metric tensor. It's therefore insufficient for the treatment of gravity in general relativity.
– J.G.
Commented Aug 19, 2022 at 19:40
• @J.G.: It seems to me that the OP is only concerned with variation of a field in a fixed curved background, so no variations of the metric are required here and the $\partial \to \nabla$ rule works fine. But it's a good point to keep in mind if they want to extend things. Commented Aug 19, 2022 at 20:50
• @MichaelSeifert My sentiments exactly. +1 for your answer, by the way; it basically makes the same points I would have, which is why I only left a comment.
– J.G.
Commented Aug 19, 2022 at 20:53
• @J.G. Thank you and compliments for your work Commented Aug 20, 2022 at 10:07

Let us consider the action $$S = \int_\mathcal{M} \text{d}^n x \sqrt{|g|} \mathscr{L}(\phi^\mu(x),\nabla_\alpha \phi^\mu(x))$$ where $$\mathcal{M}$$ is a region of spacetime with a fixed background metric $$g$$, and vary $$\phi^\mu$$. To do this, we take $$\phi^\mu$$ and replace it with $$\phi^\mu + \delta \phi^\mu$$, and then discard any terms that are higher order than $$\mathcal{O}(\delta \phi^\mu)$$ to find $$\delta S$$. The requirement that $$\delta S = 0$$ for arbitrary $$\delta \phi^\mu$$ will then allow us to find the Euler-Lagrange equations.

The variation of $$\delta \phi^\mu$$ yields $$S + \delta S = \int_\mathcal{M} \text{d}^n x \sqrt{|g|} \mathscr{L}(\phi^\mu(x)+ \delta \phi^\mu,\nabla_\alpha \phi^\mu(x) + \nabla_\alpha \delta \phi^\mu)$$ and so the first-order variation $$\delta S$$ can be seen to be $$\delta S = \int_\mathcal{M} \text{d}^n x \sqrt{|g|} \left[ \frac{\partial \mathscr{L}}{\partial \phi^\mu} \delta \phi^\mu + \frac{\partial \mathscr{L}}{\partial (\nabla_\alpha \phi^\mu)} \nabla_\alpha \delta \phi^\mu \right]$$

We now need to integrate by parts, as we would in flat spacetime. In the context of a curved background metric, Gauss's theorem is $$\int_\mathcal{M} \text{d}^n x \sqrt{|g|} \nabla_\alpha A^\alpha = \oint_{\partial \mathcal{M}} \text{d}^{n-1} x \sqrt{|h|} n_\alpha A^\alpha$$ where $$h_{\mu \nu}$$ is the induced metric on the boundary $$\partial \mathcal{M}$$ and $$n_\alpha$$ is the unit normal on $$\partial \mathcal{M}$$.1 (See Appendix B of Wald's General Relativity or Chapter 3 of Poisson's A Relativist's Toolkit for a derivation.) In the present case, this means that we have $$\delta S = \int_\mathcal{M} \text{d}^n x \sqrt{|g|} \left[ \frac{\partial \mathscr{L}}{\partial \phi^\mu} - \nabla_\alpha \frac{\partial \mathscr{L}}{\partial (\nabla_\alpha \phi^\mu)} \right] \delta \phi^\mu + \oint_{\partial \mathcal{M}} \text{d}^{n-1} x \sqrt{|h|} n_\alpha \frac{\partial \mathscr{L}}{\partial (\nabla_\alpha \phi^\mu)} \delta \phi^\mu.$$ If we are considering field variations $$\delta \phi^\mu$$ that vanish on the boundary but are otherwise arbitrary,2 then the boundary term vanishes and the quantity in brackets in the bulk integrand must vanish, leaving us with the result $$\frac{\partial \mathscr{L}}{\partial \phi^\mu} - \nabla_\alpha \frac{\partial \mathscr{L}}{\partial (\nabla_\alpha \phi^\mu)} = 0$$ as expected.

1 This implicitly assumes that $$\partial \mathcal{M}$$ is not a null hypersurface. See Wald and/or Poisson for details on how such cases can be treated.

2 Alternately, if one wishes to consider a wider class of variations, one can instead modify the action by adding an explicit surface term which cancels this term. This turns out to sometimes be helpful if the Lagrangian contains higher-order derivatives. The best-known example of this is the Gibbons-Hawking-York boundary term, an addition to the Einstein-Hilbert action that has various nice properties that the "plain-vanilla" Einstein-Hilbert action does not.

• Very clear, excellent explanation, thank you. I still have a doubt, because seems to me that you implicitly state $\nabla_\alpha (AB) = \nabla_\alpha A\, B + A \nabla_\alpha B$, but that's not true. Commented Aug 20, 2022 at 10:01
• Moreover for a scalar field $\nabla_\alpha=\partial_\alpha$, but this would give incorrect results Commented Aug 20, 2022 at 10:54
• @RobTan: (1) The covariant derivative does satisfy the Liebniz product rule. It's constructed in such a way that for any tensors $A_{\dots}$ & $B_{\dots}$, we have $\nabla_a (A_{\dots} B_{\dots}) = (\nabla_a A_{\dots}) B_{\dots} + A_{\dots} \nabla_a (B_{\dots})$. See, for example, these notes. ... Commented Aug 20, 2022 at 13:58
• (2) From your comment on the other answer, I think you're implicitly assuming that for a scalar field, all covariant derivatives of scalar fields are equal to the corresponding expressions with partial derivatives. But that's not true; it's only true for the first derivative. So while we do have $\nabla_\mu \phi = \partial_\mu \phi$, this expression is a one-form field, and so $\nabla_\nu (\nabla_\mu \phi) \neq \partial_\nu (\nabla_\mu \phi) = \partial_\nu (\partial_\mu \phi)$. In particular, this means that the EOM for a massless Klein-Gordon field in curved spacetime is ... Commented Aug 20, 2022 at 14:15
• $$\nabla_\mu \left( g^{\mu \nu} \nabla_\nu \phi\right) = 0$$and not $$\partial_\mu \left( g^{\mu \nu} \partial_\nu \phi\right) = 0$$as you seem to assume it should be; the left-hand sides of these equations are not equivalent. Commented Aug 20, 2022 at 14:18

I’ll consider the case of a scalar field.

You just need to apply the usual method of deriving the Euler-Lagrange equations by the variational principle. You get:

$$\frac{\partial \mathcal L}{\partial \phi}-\frac{1}{\sqrt g}\partial_k\left(\sqrt g \frac{\partial\mathcal L}{\partial \phi_{,k}}\right)=0$$

You can prove that this equation is covariant. This is because $$\frac{\partial\mathcal L}{\partial \phi_{,k}}e_k$$ is a well defined vector field and the second term is its divergence. You can express it in terms of the covariant derivative in order to make the covariance more explicit. This is done by the following general formula of the divergence for any vector field $$v$$: $$\frac{1}{\sqrt g}\partial_k(\sqrt g v^k)=(\nabla_k v)^k$$ using the identity: $$\Gamma_{ij}^j=\frac{1}{\sqrt g}\partial_i \sqrt g$$

You can therefore rewrite the equation as: $$\frac{\partial \mathcal L}{\partial \phi}-\nabla \cdot \frac{\partial\mathcal L}{\partial\nabla\phi}=0$$

Hope this helps.

• Thank you very much. I'm still confused though, because basing on the previous answer I expect E-L equations to be the same of the flat spacetime situation for a scalar field, since a scalar field does not depend on coordinate basis vectors and its covariant variation is simply a derivative, meaning $\nabla_\alpha = \partial_\alpha$. Your equation is instead different: how? Commented Aug 20, 2022 at 10:06
• I don’t quite understand your discrepancy. There is no reference to curvature in the equations, so the equation is the same whether the metric describes a flat manifold or not.
– LPZ
Commented Aug 20, 2022 at 13:54
• I was confusing the derivative of the scalar field with the derivative of $\partial \mathscr{L}/\partial \phi_{,k}$, that is not a scalar field, but now I think I understand, I should try myself Commented Aug 20, 2022 at 14:23
• Oh ok. Yeah you must be careful, a good sanity check is to look at the indices which must always match. Btw $\frac{\partial\mathcal L}{\partial \phi_{,k}}$ are the coefficients of a vector and transform accordingly when you change coordinates.
– LPZ
Commented Aug 20, 2022 at 17:17