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

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

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  • $\begingroup$ $\rho$ is the energy density, not the mass density. $\endgroup$
    – Javier
    May 29, 2023 at 18:07
  • $\begingroup$ c = 1, so ρ is used as the mass density. By the way, even if you call it energy density, ρ+3P=0+3(ρ_r/3) does this hold? That is, the energy density of radiation = 0, and only the pressure term exists? $\endgroup$
    – D will
    May 30, 2023 at 1:58
  • $\begingroup$ If you have the Λ/3 in you equation you don't need to use the ρ and P for dark energy, and if you do you don't need the extra Λ/3 term. For the evolution of the different densities see here and the links therein. $\endgroup$
    – Yukterez
    May 30, 2023 at 3:06
  • $\begingroup$ Dear Yukterez, You are right about Λ/3 term. By the way, what I'm asking about is the radiation term. I'd love to hear your thoughts on this. $\endgroup$
    – D will
    May 30, 2023 at 3:40

1 Answer 1

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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.

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  • $\begingroup$ //The ρ being used here is in all cases energy density, not mass density, so it's not true that ρ=0 for radiation. If p=ρ/3, then ρ+3p is simply 2ρ.// Are you sure of the above claim? $\endgroup$
    – D will
    May 30, 2023 at 15:36
  • $\begingroup$ In Wikipedia's Friedmann equation(en.wikipedia.org/wiki/Friedmann_equations), there is an explanatory phrase: ρ and P are the volumetric mass density (and not the volumetric energy density) $\endgroup$
    – D will
    May 30, 2023 at 15:37
  • $\begingroup$ ρ+3P is the source of gravity. If your argument is correct, doesn't it lead to the argument that radiation with energy density ρ exerts a gravitational force of 2ρ? $\endgroup$
    – D will
    May 30, 2023 at 16:25
  • $\begingroup$ Yes, I am sure. I checked Weinberg's book, which clearly says that $\rho$ is the energy density. The Wikipedia article is confusing at best and wrong at worst. I think they're talking about units: my interpretation is that by "mass density" they just mean "energy density divided by $c^2$". $\endgroup$
    – Javier
    May 30, 2023 at 18:24
  • $\begingroup$ And in GR there isn't really a concept of the gravitational force exerted by something - and it certainly couldn't be $\rho$ or $2\rho$, which don't have units of force. But it is true that when it comes to radiation, pressure plays a much more important role than it does for matter. $\endgroup$
    – Javier
    May 30, 2023 at 18:26

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