# Question about $3p/c^2$ in Friedmann equation

The Friedmann Equation:

$$\frac{\ddot{a}}{a} = -\frac{4\pi G}{3}\left(\rho + \frac{3p}{c^2}\right) + \frac{\Lambda c^2}{3}$$

I want to know, where does the $3p/c^2$ come from. I know it is something like density of energy or something like that but I don't know where it comes from.

• Did you read the Wikipedia article? The $p$ stands for pressure, while $\rho$ is the energy density. – pela Jul 13 '17 at 21:07
• It sounds like you're asking about the origin and significance of that specific term. To really understand that requires tracing through the derivation of the equation. Is that what you're after? – zephyr Jul 14 '17 at 14:36
• @zephyr I want to find how $\rho_{rad}$ is equal to $3p/c^2$. – titansarus Jul 14 '17 at 18:58

There are two Friedman equations.

• One related to the energy-density component in the Einstein field equations.
• One related to the pressure component in the Einstein field equations.

The Einstein field equations are the following ten equations (16 equations for 4x4 tensor components but only ten of them are independent due to symmetry)

$$G_{\mu\nu} + \Lambda g_{\mu\nu} = \kappa T_{\mu\nu} = \frac{8 \pi G}{c^4} T_{\mu\nu}$$

where $G_{\mu\nu} = R_{\mu\nu} - \frac{1}{2} g_{\mu\nu}R$ is the Einstein tensor, $g_{\mu\nu}$ the metric tensor, $\Lambda$ is the cosmological constant, $\kappa = \frac{8 \pi G}{c^4}$ is the Einstein constant, $T_{\mu\nu}$ is the energy-momentum tensor

Friedman was actually assuming that only the matter/energy density component is non-zero $T_{44}= \epsilon = \rho c^2$. He assumed "Die Materie ist inkohärent" (ie non interacting dust with $p=0$) by which the components $T_{ii}$ are non-zero for $i \neq 4$. The other components (energy flux and momentum) are set equal to zero based on the assumption that the motion of energy and matter is considered to be negligibly small.

LeMaître kept the pressure in place arguing that the pressure associated with the kinetic energy of matter might be negligible but this may not be the case for radiation pressure. So he used also $T_{11} = T_{22} = T_{33} = -p$

Using the Robertson Walker metric metric $d s^2 = dt^2 - a(t)^2 \left[ \frac{dr^2}{1-k r^2}+r^2d\Omega^2\right]$ the equations become expressions in terms or the scale factor $a(t)$:

$$\begin{array}{rcrcl} G_{00} &=& c^{-2}\frac{3k}{a^2} + 3c^{-2} \frac{\dot a^2}{a^2} &=& \frac{8 \pi G}{c^4} \rho c^2 - \Lambda \\ -G_{11} = -G_{22} = -G_{33} &=& c^{-2}\frac{k}{a^2} + c^{-2}\frac{\dot a^2}{a^2} + 2 c^{-2} \frac{\ddot a}{a} &=& -\frac{8 \pi G}{c^4} p + \Lambda \end{array}$$

If you subtract the second equations from the first equations

$$G_{00}+G_{11}+G_{22}+G_{33} = -6 c^{-2} \frac{\ddot a}{a} = \frac{8 \pi G}{c^4} (3p+ \rho c^2) + 2\Lambda$$

and rearrange some terms, then you could also write

$$\frac{\ddot a}{a} = -\frac{4 \pi G}{3} \left( \rho + \frac{3p}{c^2} \right) + \frac{ \Lambda c^2}{3}$$

That was a lovely answer. Since in a comment, titansarus asked for the connection between 3p (I'll take c=1, it's just a constant) and the radiation energy density, I will add that it is easy to show that pressure is always E/3 for any isotropic relativistic gas (where E is kinetic energy density and "relativistic" means rest energy is unimportant, so each particle has its kinetic energy equal to the magnitude of its momentum). Pressure is momentum flux per area, so form a ratio of that to the kinetic energy density, for all particles in a given energy bin, integrating over all directions. If that comes out 1/3, you have your answer.

The numerator of that ratio is the integral over direction cosine of the particle density times its momentum times its speed (here 1) times the square of the direction cosine (that's what gives momentum flux, one cosine comes from the normal component of the momentum, and the other from the need to transport it across the boundary via the speed normal to the boundary), and the denominator is exactly the same integral except without the square of the direction cosine (because it's energy density, so has neither of those two reasons to include the cosine). Hence, the ratio is just the integral over the direction cosine of the cosine squared, divided by the integral over the direction cosine. That's 1/3, simple as that.