I'm trying to derive the modified gravity EFT field equations and, from their 00 component, this Friedmann equation: \begin{equation} H^{2}+H\frac{\dot{\Omega}}{\Omega}=\frac{\kappa \rho_{m}+\Lambda+\Gamma}{3\Omega}\\ \end{equation} By varying this action: \begin{equation} S=\int d^{4}x \sqrt{g}\left(\frac{1}{2\kappa}(\Omega(t)R-2\Lambda(t)-\Gamma(t)\delta g^{00})+\mathcal{L}_{m}\right) ~. \end{equation} However, while I can get the correct right hand side, I cannot get the correct left hand side as I don't know what to do with the $\Omega$ term: \begin{equation} H^{2}+\frac{R}{3\Omega}\frac{\delta \Omega}{\delta g^{00}}=\frac{\kappa \rho_{m}+\Lambda+ \Gamma}{3\Omega} ~. \tag{1} \end{equation}


To start with I took the variance with respect to the metric tensor and set the whole thing equal to zero in accordance with the principle of least action. As with the standard derivation of the Einstein field equations, I then took the $\sqrt{g}$ outside the brackets and defined \begin{eqnarray} T_{\mu\nu}=\frac{-2}{\sqrt{g}}\frac{\delta \mathcal{L}_{m}\sqrt{g}}{\delta g^{\mu\nu}} ~, ~~~~~~~~~~ \frac{\delta R}{\delta g^{\mu\nu}}=R_{\mu\nu} ~, ~~~~~~~~~~ \frac{1}{\sqrt{g}}\frac{\delta \sqrt{g}}{\delta g^{\mu\nu}}=-\frac{1}{2}g_{\mu\nu} ~, \end{eqnarray} From which I can get to \begin{equation} R_{00}+\frac{1}{2}R+\frac{R}{\Omega}\frac{\delta \Omega}{\delta g^{00}}=\frac{\kappa \rho_{m}+\Lambda+ \Gamma}{\Omega} ~. \end{equation} where I have set the derivatives on the right hand side equal to zero and used $T_{00}=\rho_{m}$. I know the definitions of $R$ and $R_{00}$ in the FLRW metric are \begin{align} R_{00}&=-3\frac{\ddot{a}}{a}\\ R&=6\left(\frac{\ddot{a}}{a}+\frac{\dot{a}^{2}}{a^{2}}\right) \end{align}

where $H=\dot{a}/a$, giving Equation 1. I've experimented with using the chain rule on the $\Omega$ term as follows: \begin{equation} \frac{\delta \Omega}{\delta g^{00}}=\frac{\delta \Omega}{\delta R}\frac{\delta R}{\delta g^{00}}=\frac{\delta \Omega}{\delta R}R_{00}=\frac{\delta \Omega}{\delta t}\frac{\delta t}{\delta R}R_{00}=\dot{\Omega}\frac{R_{00}}{\dot{R}} ~, \end{equation}

However, no matter what I do from there I can't get the term to equal $H \frac{\dot{\Omega}}{\Omega}$. If anyone would be willing to comment upon what I've already done, and perhaps give some advice regarding how to proceed, I would very much appreciate it.

Thank you

  • 1
    $\begingroup$ Maybe I'm missing something, but don't you have some uncontracted indicies in equation 1? Shouldn't it read $\frac{\delta \Omega}{\delta g^{0 0}}$? $\endgroup$ – drglove Jun 22 '16 at 21:11
  • $\begingroup$ @drglove The OP addresses that at the end of the post by looking to try and reduce the derivative term to just its time component. It might be useful if the OP tags all the equations with numbers to make reference easier in comments / answers? $\endgroup$ – Alexander McFarlane Jun 23 '16 at 12:15
  • $\begingroup$ Thanks, I had been a little bit inconsistent; I was implicitly setting $\mu=\nu=0$ in some places (but still writing $\mu$ and $\nu$) and explicitly in others. I've edited it for consistency, and to remove a couple of typos. And Alex, I still need to end up with some $R_{\mu\nu}$ terms at the end; I actually had to insert one via the chain rule in order to get rid of the metric derivative. It becomes something workable in the form of $H$ once one sets $\mu=\nu=0$; I simply hadn't written it all out consistently. $\endgroup$ – Harlequin Jun 23 '16 at 12:30
  • $\begingroup$ I don't think it is by any coincidence that you have a $\frac{R}{3}$ factor in (1). Have you tried expressing $R_{00}$ in terms of $H$? Also not entirely sure how $\frac{1}{\Omega}$ comes into play? Also label all your equations to make it easier to comment $\endgroup$ – Alexander McFarlane Jun 23 '16 at 13:21

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