Given a lagrangian of a form: \begin{equation}\mathcal{L}=f(\phi,\partial_{\mu}\phi\partial^{\mu}\phi)\end{equation} where $f$ is a function, I need to derive pressure and density in a FLRW universe with $g^{\mu}_{\nu}=\delta^{\mu}_{\nu}$.

My approach is using: \begin{equation}T_{\mu\nu}=-\frac{2}{\sqrt{-g}}\frac{\partial}{\partial g^{\mu\nu}}(\sqrt{-g}\mathcal{L})\end{equation} \begin{equation}=g_{\mu\nu}\mathcal{L}-2\frac{\partial\mathcal{L}}{\partial g^{\mu\nu}}.\end{equation}

And finally, \begin{equation}\rho=T^0_{ 0}\end{equation} \begin{equation}P=T^i_{ i}\end{equation}.

The problem I am facing right now is how to explicitly use the form of lagrangian to simplify the energy momentum tensor. Can anyone please help me?

Edit I: Just to clarify, $\phi$ depends only on $t$ and is independent of $x^i$.

Edit II: Okay, I solved it. Here's the way to do it: \begin{equation}T_{\mu\nu}=g_{\mu\nu}\mathcal{L}-2\frac{\partial\mathcal{L}}{\partial X}\partial_{\mu}\phi\partial_{\nu}\phi.\end{equation} \begin{equation}\rho=T^0_{ 0}=\mathcal{L}-2\frac{\partial\mathcal{L}}{\partial X}\dot{\phi}^2\end{equation} \begin{equation}P=T^i_{ i}=\mathcal{L}\end{equation}.

where $X=\partial_{\mu}\phi\partial_{\nu}\phi$.

  • $\begingroup$ You have assume that $T_{\mu\nu}$ is isotropic. However, this is not always true for arbitrary functional form $f$ of $\phi$ and $\partial_{\mu} \phi$. $\endgroup$
    – mastrok
    Commented Jul 2, 2014 at 4:50
  • $\begingroup$ I have just edited the original question. $T_{\mu\nu}$ is isotropic because $phi$ only depends on $t$, and f is a function of $\phi$ and $\dot{\phi}$. Sorry for the confusion. $\endgroup$
    – titanium
    Commented Jul 2, 2014 at 5:30
  • 1
    $\begingroup$ btw, the upper/lower split indices on the metric are always $g^\mu_\nu = \delta^\mu_\nu$ by definition (no negative signs, no matter your sign convention) $\endgroup$
    – user10851
    Commented Jul 2, 2014 at 5:50
  • $\begingroup$ I guess you have a sign problem since a stable field $\phi=const$ should give negative $w=p/\rho$. $\endgroup$
    – mastrok
    Commented Jul 2, 2014 at 12:36
  • $\begingroup$ Also, if your metric is Minkowski, $T_{\mu\nu}$ has to be zero. $\endgroup$ Commented Aug 1, 2014 at 17:07

1 Answer 1


I give a general derivation even if it is an-isotropic. As it is your homework, I wont give all the detial derivations.

Notation: $\eta_{\mu\nu}$ is mostly positive, let $K=g_{\mu\nu}\partial^{\mu}\phi\partial^{\nu}\phi $

$$ S_{\phi} =\int d^4x \sqrt{-g}{\cal L}(\phi,K) \\ T_{\phi}^{\mu\nu}=\frac{2}{\sqrt{-g}}\frac{\delta{\cal L}}{\delta g_{\mu\nu}} = -g^{\mu\nu}{\cal L} + \frac{\partial {\cal L}}{\partial K}g^{\mu\alpha}g^{\nu\beta}\partial_{\alpha}\phi\partial_{\beta}\phi $$

Now, we can compare it with the energy momentum tensor for perfect fluid, $$ T^{\mu\nu} = pg^{\mu\nu} +(\rho+p)u^{\mu}u^{\nu}, \quad u^{\mu}u_{\mu}=-1$$ The quantities can be identified, $$p = -{\cal L} \\ \rho = \mbox{you can figure it out} \\ u^{\mu} = -\frac{\partial^{\mu}\phi}{\sqrt{-K}}$$

Thus, the rest is simple as you just substitute a time dependent $\phi(t)$.


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