# Derivation of $f(R)$ field equations, problem with integration by parts

I am following the derivation of the field equations on the the Wikipedia page for $$f(R)$$ gravity.

But I do not understand the following step: $$\delta S = \int \frac{1}{2\kappa} \sqrt{-g} \left(\frac{\partial f}{\partial R} (R_{\mu\nu} \delta g^{\mu\nu}+g_{\mu\nu}\Box \delta g^{\mu\nu}-\nabla_\mu \nabla_\nu \delta g^{\mu\nu}) -\frac{1}{2} g_{\mu\nu} \delta g^{\mu\nu} f(R) \right)$$ the wiki article says, the next step is to integrate the second and third terms by parts to yield: $$\delta S = \int \frac{1}{2\kappa} \sqrt{-g}\delta g^{\mu\nu} \left(\frac{\partial f}{\partial R} R_{\mu\nu}-\frac{1}{2}g_{\mu\nu} f(R)+[g_{\mu\nu}\Box -\nabla_\mu \nabla_\nu] \frac{\partial f}{\partial R} \right)\, \mathrm{d}^4x$$ In other words, integrating by parts should yield: $$\int \sqrt{-g} \left(\frac{\partial f}{\partial R} (g_{\mu\nu}\Box \delta g^{\mu\nu}-\nabla_\mu \nabla_\nu \delta g^{\mu\nu}) \right)\, d^4x$$ $$= \int \sqrt{-g}\delta g^{\mu\nu} \left([g_{\mu\nu}\Box -\nabla_\mu \nabla_\nu] \frac{\partial f}{\partial R} \right) \mathrm{d}^4x$$ From there getting the usual f(R) field equations is trivial. What I'm confused by is how to integrate by parts to get that.

I have tried many different ways the one I think is most correct is: assuming $$g_{\mu \nu} \Box$$ and $$\nabla_\mu \nabla_\nu$$ are differential operators then $$u' = g_{\mu \nu} \Box \delta g^{\mu\nu}$$ and $$v = f'$$, similarly with the $$\nabla_\mu \nabla_\nu$$ so using the formula for integration by parts: $$\int u'v = uv -\int uv'$$ I get: $$\int \sqrt{-g} \left(f' (g_{\mu\nu}\Box \delta g^{\mu\nu}-\nabla_\mu \nabla_\nu \delta g^{\mu\nu}) \right)\, d^4x$$ $$= -\int \sqrt{-g}\delta g^{\mu\nu} \left([g_{\mu\nu}\Box -\nabla_\mu \nabla_\nu] f' \right) \mathrm{d}^4x$$ because the $$uv$$ term will disappear.

So can any one explain to me why I have the minus sign and Wikipedia doesn't? Is it ok to use $$g_{\mu \nu} \Box$$ as a differential operator? I have tried other ways such as writing $$\Box$$ explicitly and using integration by parts twice but I also couldn't get the correct answer as i end up with terms such as $$\nabla_\nu \nabla_\mu$$ which cant be correct.

There is a similar post on physics forums on this step but it does not answer my question and is now closed.

• Or, you could just use the Palatini formalism, to avoid the nasty second derivatives: relativity.livingreviews.org/… Aug 2, 2014 at 23:28
• Everything is explained in this article on Scholarpedia here. Also, consider also this paper for derivation here. If you still need the proof, I can write down all the steps. Jun 26, 2016 at 15:19

You need to use recursive integration by parts to deal with the second derivatives.

See, for example, equation (18) here.

The last term in (18) has a multiplying factor of $(-1)^n$, where $n$ is the order of the derivative. In your case, $n=2$ and the minus sign vanishes.

The box operator $$\Box = g^{αβ}\nabla_{α}\nabla_{β} = \nabla^{α}\nabla_{α}$$ and the two covariant derivatives $$\nabla_{μ}\nabla_{ν}$$ act on the variation of the inverse metric tensor. You want $$δg^{μν}$$ to be a multiplying factor so you must integrate by parts twice (because you have two derivatives) to make the derivatives act on $$\cfrac{df}{dR}$$. The box term will be:

$$f'(R)\nabla^{α}\nabla_{α}(δg^{μν})$$

From Leibniz rule we know that: $$\nabla^{α}(f'(R)\nabla_{α}δg^{μν}) = \nabla^{α}f'(R)\nabla_{α}δg^{μν} + f'(R)\nabla^{α}\nabla_{α}δg^{μν}$$

The left hand side term is a total derivative so it is zero which means:

$$f'(R)\nabla^{α}\nabla_{α}δg^{μν} = - \nabla^{α}f'(R)\nabla_{α}δg^{μν}$$

Applying Leibniz rule again:

$$\nabla_{α}(-\nabla^{α}f'(R)δg^{μν}) = - \nabla_{α}\nabla^{α}f'(R)δg^{μν} - \nabla_{α}δg^{μν}\nabla^{α}f'(R)$$

The left hand side term is a total derivative.

$$- \nabla_{α}δg^{μν}\nabla^{α}f'(R) = \nabla_{α}\nabla^{α}f'(R)δg^{μν} \Rightarrow$$

$$f'(R)\nabla^{α}\nabla_{α}δg^{μν} = \nabla_{α}\nabla^{α}f'(R)δg^{μν}$$.

You have to do the same with the $$f'(R)\nabla_{μ}\nabla_{ν}δg^{μν}$$ term.

The box term also contains $$g_{μν}$$ but metric is compatible so:

$$\nabla^{α}\nabla_{α}(g_{μν}f'(R)δg^{μν}) = \nabla^{α}\nabla_{α}(f'(R)δg^{μν})g_{μν}$$

EDIT: Another derivation can be found here: https://arxiv.org/abs/1002.0617