# Perfect fluids in cosmology?

In cosmology, it is often assumed that the equation of state of a cosmological fluid is of the form $p=w\rho$. Why is this? Is it the equation of a perfect fluid?

Why does $w=0$ for matter $1/3$ for radiation and $-1$ for a cosmological constant?

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I take it wikipedia doesn't help? – Emilio Pisanty Apr 23 '13 at 17:37
@EmilioPisanty It doesn't help much. – user12345 Apr 23 '13 at 17:41
I think this is really two separate questions. – Ben Crowell Apr 23 '13 at 19:27
@BenCrowell I wouldn't know. – user12345 Apr 23 '13 at 20:59
@user12345 John Rennie answered the main thrust of the question well, but I'll just say something about the first bit. $p=w\rho$ is not the equation for a perfect fluid, it is the barotropic equation of state. A perfect fluid is a fluid that has no viscosity. You can actually prove that for a homogenous and isotropic universe the energy-momentum tensor must be that of a uniform perfect fluid, but the equation of state is determined by other physics, as John Rennie discussed. – Michael Brown Apr 24 '13 at 13:03

The Wikipedia article on the equation of state discusses this.

## Matter

Consider non-relativistic matter first, because this is easy. We write the pressure as:

$$p = \rho_m \space RT = \rho_m \space C^2$$

where $\rho_m$ is the mass density, $RT$ is the thermal energy of the matter (or whatever) in your fluid, and $C$ is some sort of characteristic velocity. The justification for replacing $RT$ by $C^2$ is that kinetic energy is proportional to the square of velocity.

$\omega$ is given by:

$$\omega = \frac{p}{\rho_e}$$

where $\rho_e$ is the energy density $\approx \rho_mc^2$, and substituting this into the above equation we get:

$$\omega = \frac{\rho_m C^2}{\rho_m c^2} = \frac{C^2}{c^2}$$

and for non-relativistic matter $C \ll c$ so $\omega \approx 0$.

Now consider radiation, which is harder because we can't relate $p$ to a mass density. I don't know of an intuitive way to show this, but this article goes into the details of calculating radiation pressure (it's complicated!!), and the final result turns out to be:

$$p = \frac{1}{3} \rho_e$$

and obviously we immediately get $\omega$ = 1/3.

## Cosmological constant

Start with the Friedmann equation with a non-zero cosmological constant, $\Delta$:

$$3\frac{\ddot{a}}{a} = \Delta - 4\pi G(\rho + 3p)$$

we want to treat $\Delta$ as an effective pressure and energy density so we define:

$$p_{eff} = p - \frac{\Delta}{8 \pi G}$$

$$\rho_{eff} = \rho + \frac{\Delta}{8 \pi G}$$

We do this because now the Friedmann equation simplifies to:

$$3\frac{\ddot{a}}{a} = - 4\pi G(\rho_{eff} + 3p_{eff})$$

as you can see by just substituting for $p_{eff}$ and $\rho_{eff}$ to get back the original Friedmann equation. So now we get an effective value for $\omega$:

$$\omega_{eff} = \frac{p_{eff}}{\rho_{eff}} = \frac{p - \frac{\Delta}{8 \pi G}}{\rho + \frac{\Delta}{8 \pi G}}$$

and if we assume that the cosmological constant is dominant, i.e. $p \approx 0$ and $\rho \approx 0$, we get the value of $\omega$ for just dark energy is -1.

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Sorry for taking ages to respond. Your matter answer is really intuitive! The radiation one is good enough for me (i.e. 'it comes from the BB spectrum). However, the cosmo const.(CC) part makes little sense to me. +1 overall though. I think the $w=-1$ of the CC is what you get if you ask 'what is the vacuum energy density'. Perfect fluid: $T_{\mu\nu} = (\rho + p/c^{2})u_{\mu}u_{\nu} - pg_{\mu\nu}$. If $p=-\rho$ then $T_{\mu\nu}$ depends on $g_{\mu\nu}$ only, so it's a vacuum property... and we just call it the CC. – user12345 Apr 25 '13 at 14:47