I have the action

$$S=\int d^4x\sqrt{-g} \Big[\frac{1}{8}\phi^2R- \frac{1}{2}g^{\mu\nu}\partial_\mu\phi\partial_\nu\phi - \frac{1}{2}m^2\phi^2\Big]$$ where $\phi$ is a scalar field and $R$ is the scalar curvature (signature $-+++$)

I want to get the equations of motion and then substitute in the FRW metric.

Could someone show me how to vary this action?

Alternatively could I simply substitute the FRW metric into the above action and then calculate the Euler-Lagrange equations for the scale factor $a(t)$ and the scalar field $\phi$?

I guess in that case I would only get two equations rather than the three I would get by substituting the FRW metric into the full equations of motion.

  • $\begingroup$ Can you please give a reference where this theory is used? There exists no pure gravitational term it's interesting! $\endgroup$
    – Noone
    Oct 16, 2020 at 6:22
  • $\begingroup$ It's a form of the dilaton field from the low energy effective action of string theory. By describing a variable Planck mass it describes how all mass in the Universe scales. It's also the Brans-Dicke scalar squared with w=1 and a mass term. Here's a reference arxiv.org/abs/gr-qc/0604082 $\endgroup$ Oct 16, 2020 at 8:30
  • $\begingroup$ What I'm saying about the variable Planck mass is speculative. I want to apply the Lagrangian to the Universe by substituting the FRW metric in the equations of motion and using the hypothesis that $\phi=a(t)$. I want to show that the FRW expansion is actually an apparent effect caused by atomic masses increasing which leads to our rulers shrinking. $\endgroup$ Oct 16, 2020 at 8:52
  • $\begingroup$ What is the physical meaning of $\phi=a(t) $?? $\endgroup$
    – Noone
    Oct 16, 2020 at 11:09
  • $\begingroup$ My hypothesis is that the increasing mass scale $\phi$ is responsible for the apparent expansion of the Universe characterized by the scale factor $a(t)$. I use this hypothesis to find an equation for $a(t)$. $\endgroup$ Oct 16, 2020 at 12:53

1 Answer 1


You vary the action with brute force Variation for simplicity. I assume that you can handle the last two terms and the problem is the first one. Varying with respect to $\phi$ you'll obtain:

$$\Box \phi + \cfrac{1}{4}R\phi -m^2 \phi =0$$ Varying with respect to the metric you have to integrate by parts the first term.

$$\delta(\phi ^2 R) = \phi^2 \delta R = \phi^2 (g_{\mu\nu}\Box \delta g^{\mu\nu} - \nabla_{\mu} \nabla_{\nu}\delta g^{\mu\nu} + R_{\mu\nu}\delta g^{\mu\nu})$$

You want $\delta g^{\mu\nu} $ to be a multiplying factor so you integrate by parts twice and now the derivatives will act on $\phi^2$.

For the integration check my answer here: Derivation of $f(R)$ field equations, problem with integration by parts

Τhe final answer will be:

$$ \cfrac{1}{8}\left( g_{\mu\nu}\Box - \nabla_{\mu}\nabla_{\nu} + G_{\mu\nu}\right)\phi^2 - \cfrac{1}{2}\nabla_{\mu}\phi\nabla_{\nu}\phi + \cfrac{1}{4}g_{\mu\nu}\nabla^{\xi}\phi\nabla_{\xi}\phi + \cfrac{1}{4}g_{\mu\nu}m^2\phi^2=0$$ You will obtain $4$ Einstein's equations and one Klein-Gordon. Only two of them are independent though. You can of course plug in the metric and vary with respect to the scale factor. The result will be the same. If you want to vary the action by hand and then use a program to obtain the components of equations, varying with respect to the field surely contains less manipulations than varying with respect to the scale factor $a$.

  • $\begingroup$ I'm afraid this is too advanced for me. I would need someone to do the work for me! $\endgroup$ Oct 16, 2020 at 13:53
  • $\begingroup$ If I plugged in the FRW metric and varied the scale factor and $\phi$ would I get all the equations? $\endgroup$ Oct 16, 2020 at 13:54
  • $\begingroup$ What do you feel is advanced? If you don't want to do it by hand there's the xAct package. I strongly recommend to struggle computing everything by hand at least once, it will help you understand. $\endgroup$
    – Noone
    Oct 16, 2020 at 14:05
  • $\begingroup$ For $\phi =a(t) $ there's only one independent equation. $\endgroup$
    – Noone
    Oct 16, 2020 at 14:05
  • $\begingroup$ For the one degree of freedom metric of cosmology you will obtain one equation. You'll need several degrees of freedom to obtain the others. $\endgroup$
    – Noone
    Oct 16, 2020 at 14:09

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