In my working, I found it hard to solve this action

$$S = \int d^4x \sqrt{-g} \partial_\mu\phi \partial_\nu\phi G^{\mu\nu} $$

My goal is to find variation to metric:

$$ \delta S = \int d^4x (\delta{\sqrt{-g}) \partial_\mu\phi \partial_\nu\phi G^{\mu\nu}} + \int d^4x {\sqrt{-g}(\delta[{ \partial_\mu\phi \partial_\nu\phi G^{\mu\nu}}}]) $$

I can solve the first line of equations above, but I don't have any idea to solve variation Einstein tensor.

Can you help me to solve that or give me some hints, please?

  • $\begingroup$ Where has this problem arisen? The action you're considering isn't a scalar, which doesn't seem correct. As for $\delta G^{\mu \nu}$, you can just write the Einstein tensor in terms of the Ricci scalar & Ricci tensor and do the usual calculations. $\endgroup$
    – Eletie
    Commented Mar 4, 2021 at 20:19
  • $\begingroup$ I am sorry, my bad. I simply change the actual action. I have edited the action, is that correct form? $\endgroup$
    – paseo cas
    Commented Mar 4, 2021 at 20:33
  • 2
    $\begingroup$ Have you used the variational method to derive things like EFE from the Einstein-Hilbert action? If so, this should all be familiar. If you vary with respect to the field $\phi$ you'll get a set of equations, and similarly if you vary with respect to the metric $g_{\mu \nu}$ you'll get a set of equations. To compute the metric variation of the Einstein tensor you want to write it in terms of the metric, which is just $G^{\mu \nu} = R^{\mu \nu} - \frac{1}{2}g^{\mu \nu}R$. The variation of these individual terms is then standard, but will take a while to write out. $\endgroup$
    – Eletie
    Commented Mar 4, 2021 at 20:44
  • $\begingroup$ Take a look at the other answers for the method, e.g. physics.stackexchange.com/questions/93157/… It should also be available in most GR notes online. $\endgroup$
    – Eletie
    Commented Mar 4, 2021 at 20:46

1 Answer 1


This calculation is tedious.

You will need:

$$\delta R_{ab} = \nabla_{b}(\delta \Gamma^{c}_{ac}) - \nabla_{c}(\delta \Gamma^{c}_{ab})$$

$$\delta \Gamma^{c}_{ab} = \cfrac{1}{2}g^{cd}\Big(\nabla_{b}\delta g_{da} + \nabla_{a}\delta g_{db} - \nabla_{c} \delta g_{ab}\Big)$$

$$G_{ab} = R_{ab} - \cfrac{1}{2}g_{ab}R$$

First of all use the expression for the variation of the Christoffel symbols to obtain the variation of the Ricci tensor as covariant derivatives of variations of the metric tensor.

Then for the terms:

$$\delta R_{ab}\partial^{a}\phi \partial^{b}\phi \hspace{1.0cm} g_{ab}\delta(R)\partial^{a}\phi\partial^{b}\phi$$

you have to perform integration by parts to cancel total derivative terms. (see my answers here: Derivation of $f(R)$ field equations, problem with integration by parts , Metric field equations for the Jordan-Brans-Dicke action)

The calculation is not hard once you understand what you have to do, but it is going to take some time. I strongly encourage you to derive the field equations by hand, the satisfaction once you get the right answer does really worth the struggle.

  • 1
    $\begingroup$ They will also need to integrate by parts for the $g_{ab}R$ term too, which again involves using the $\delta R_{ab}$ identity, so I definitely agree the calculation takes a while - but as you mention, satisfying too! $\endgroup$
    – Eletie
    Commented Mar 4, 2021 at 20:58
  • 1
    $\begingroup$ You're right! Thank you for the comment, i'm going to edit the answer. $\endgroup$
    – Noone
    Commented Mar 4, 2021 at 21:00
  • $\begingroup$ Thank you, I will try to derive it $\endgroup$
    – paseo cas
    Commented Mar 5, 2021 at 9:02

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