# Pressure and density using a general Lagrangian

Given a lagrangian of a form: $$\mathcal{L}=f(\phi,\partial_{\mu}\phi\partial^{\mu}\phi)$$ 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: $$T_{\mu\nu}=-\frac{2}{\sqrt{-g}}\frac{\partial}{\partial g^{\mu\nu}}(\sqrt{-g}\mathcal{L})$$ $$=g_{\mu\nu}\mathcal{L}-2\frac{\partial\mathcal{L}}{\partial g^{\mu\nu}}.$$

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

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: $$T_{\mu\nu}=g_{\mu\nu}\mathcal{L}-2\frac{\partial\mathcal{L}}{\partial X}\partial_{\mu}\phi\partial_{\nu}\phi.$$ $$\rho=T^0_{ 0}=\mathcal{L}-2\frac{\partial\mathcal{L}}{\partial X}\dot{\phi}^2$$ $$P=T^i_{ i}=\mathcal{L}$$.

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

• You have assume that $T_{\mu\nu}$ is isotropic. However, this is not always true for arbitrary functional form $f$ of $\phi$ and $\partial_{\mu} \phi$. Commented Jul 2, 2014 at 4:50
• 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. Commented Jul 2, 2014 at 5:30
• 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)
– user10851
Commented Jul 2, 2014 at 5:50
• I guess you have a sign problem since a stable field $\phi=const$ should give negative $w=p/\rho$. Commented Jul 2, 2014 at 12:36
• Also, if your metric is Minkowski, $T_{\mu\nu}$ has to be zero. Commented Aug 1, 2014 at 17:07

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 $\phi(t)$.