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When getting the perturbed Ricci scalar in a Modified Gravity theory of the form $\mathcal{L}_{gr}=F\left(\phi,R\right)R$, $\phi$ being a scalar field, it is easy to arrive at an expression of it in terms of $\delta \phi$ and $\delta F$, simply by doing:

$ \delta R = \frac{1}{F_{,R}}\delta F-\frac{F_{,\phi}}{F_{,R}}\delta \phi. $

However, suppose $F=F\left(\phi\right)$. How would you get a similar expression of $\delta R$ in terms of either $\delta \phi$ or $\delta F$? I know one can obtain $\delta R$ by choosing a metric, perturbing it and getting all the corresponding geometrical quantities, but I want to obtain the dependence specified above.

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You can do it by writing the Einstein equation of motion, taking the trace, obtaining $R=R(\phi)$ and then computing the variation of $R$. Let me show it to you.

Start from the action

$$ S=\int d^4x \frac{1}{2}F(\phi) R $$

varying the action with respect to the metric you obtain the Einstein equation

$$ G_{\mu\nu} = \frac{1}{F(\phi)} [\nabla_\mu \nabla_\nu - g_{\mu\nu}\Box]F(\phi) $$

With $ G_{\mu \nu} = R_{\mu \nu} - \frac{1}{2} g_{\mu \nu} R$

Taking the trace you obtain

$$ -R = -\frac{3}{F(\phi) } \Box F(\phi) \rightarrow R = \frac{3}{F(\phi) } \Box F(\phi) $$

Now you perform the variaton of the R.H.S and obtain what you want

$$ \delta R = 3\delta \bigg(\frac{1}{F(\phi)}\bigg) \Box F(\phi) + 3 \frac{1}{F(\phi)} \delta (\Box F(\phi)) $$

The variation of the first term is of course $$ \delta \bigg(\frac{1}{F(\phi)}\bigg) = - \frac{1}{F^2(\phi)}{F,}_{\phi}\, \delta \phi $$

Where ${F,}_{\phi}$ means derivative of $F$ with respect to $\phi$

The variation of the box is not so trivial, in fact it is

$$ \delta (\Box F(\phi)) = \delta \bigg[\, \frac{1}{\sqrt{-g}} \partial_\mu \big(\sqrt{-g}g^{\mu\nu} \partial_\nu F(\phi)\big)\, \bigg] $$

Where you have to perform the variation term by term, it's a pretty lengthy and boring computation.

One usually chooses the metric and a gauge in perturbation theory to compute this term. In the case of a flat Friedmann Robertson Walker metric, in the synchronous gauge you have for example $$ \delta (\Box F(\phi)) = -\delta \ddot{F} - 3 H \delta \dot{F} + \frac{1}{a^2} \partial^2 \delta F - \frac{1}{2} \dot{h} \dot{F} $$

Where $h$ is the trace of the perturbation to the metric, defined in the synchronous gauge in the following way:

$$ H_{\alpha\beta} = a^2 h_{\alpha\beta} = \begin{cases} 0, & \mbox{for } \alpha = 0 \mbox{ and/or} \, \beta = 0 \\ a^2 h_{ij}, & \mbox{ otherwise} \end{cases} $$

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  • $\begingroup$ thank you very much for your detailed answer. It seems one cannot avoid having metric perturbation terms as in the $F(\phi,R)$ case, doesn't it? (I say it because of the $\dot h$ one). $\endgroup$ – J.J Oct 19 at 17:48
  • $\begingroup$ @J.J Sorry for my late answer. Yes one can apparently avoid those terms, but maybe using the Klein-Gordon equation too in a model of modifed gravity you could get rid of those terms, with that lagrangian doesn't seem possible to me, but writing a proper modifed gravity lagrangian with the kinetic term eccetera one could maybe simplify those, but I think it varies from case to case $\endgroup$ – AnOrAn Oct 27 at 14:13

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