Mass density of radiation in Friedmann equation?

Question

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

In the case of matter, pressure $P=0$,

In the case of radiation, pressure $P=(1/3)ρ$,

In the case of cosmological constant, pressure $P=-ρ$.

When this goes into the Friedmann equation, does it have the following form?

Considering only the case of radiation term

$$\rho + 3P = 0 + 3(\frac{1}{3}{\rho _r}) = {\rho _r}$$

Is it this form?

The mass density of radiation = 0, and the pressure term of radiation $P=(1/3)ρ$

$$\rho + 3P = 0 + 3(\frac{1}{3}{\rho _r}) = {\rho _r}$$

Is this correct?

When we say that the energy density(or mass density) of radiation is ${\rho _r}$, which form is it in Friedmann equation?
$\rho + 3P = {\rho _r} + 3(\frac{1}{3}{\rho _r}) = 2{\rho _r}$ or
$\rho + 3P = 0 + 3(\frac{1}{3}{\rho _r}) = {\rho _r}$

Answer 1

The $\rho$ being used here is in all cases energy density, not mass density, so it's not true that $\rho = 0$ for radiation. It's not about using units with $c=1$; $E = mc^2$ is about rest energy and rest mass, and while radiation has neither, it still has energy. If $p = \rho/3$, then $\rho + 3p$ is simply $2\rho$.

Something that might be confusing is that when talking about matter, we do use $\rho$ to refer to mass density. That's because we're talking about nonrelativistic matter, for which most of its energy is the rest energy, proportional to the mass. Its kinetic energy is negligible, so $\rho \approx \rho_m c^2$. But the $\rho$ appearing in the Friedmann equations is still energy, not mass.