# Mass density of radiation in Friedmann equation?

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• darkdark10
In summary: P_{ii}=\frac13\epsilon##, so that ##\epsilon+3P## is indeed 4 times the pressure tensor's trace. The "momentum density" does exist in the hydrodynamical sense, but it is not an independent variable, it is not an observable. The same is true for the "mass density". It is not an independent variable. So it is not a valid question to ask what form it takes when entering the equation. The equation \rho + \frac{{3P}}{{{c
darkdark10
$$\frac{1}{R}\frac{{{d^2}R}}{{d{t^2}}} = - \frac{{4\pi G}}{3}\left[ {{\rho _m} + {\rho _{rad}} + {\rho _\Lambda } + \frac{{3({P_m} + {P_{rad}} + {P_\Lambda })}}{{{c^2}}}} \right]$$
In the Friedmann equation, ρ is the mass density.
===
https://en.wikipedia.org/wiki/Friedmann_equations
They were first derived by Alexander Friedmann in 1922 from Einstein's field equations of gravitation for the Friedmann–Lemaître–Robertson–Walker metric and a perfect fluid with a given mass density ρ and pressure p.
~~~
ρ and p are the volumetric mass density (and not the volumetric energy density) and the pressure,
===

In the case of matter, pressure P=0,
In the case of radiation, pressure P=(1/3)ρc^2,
In the case of cosmological constant, pressure P=-ρc^2.

When we call the energy densities of matter, radiation, and dark energy (ρ_m)c^2, (ρ_rad)c^2, (ρ_lambda)c^2, what form does this take when entering the equation?
$${\rho _m} + \frac{{3{P_m}}}{{{c^2}}} = {\rho _m} + 0 = {\rho _m}$$
$${\rho _\Lambda } + \frac{{3{P_\Lambda }}}{{{c^2}}} = {\rho _\Lambda } + \frac{{3( - {\rho _\Lambda }{c^2})}}{{{c^2}}} = - 2{\rho _\Lambda }$$

$${\rho _{rad}} + \frac{{3{P_{rad}}}}{{{c^2}}} = {\rho _{rad}} + \frac{{3(\frac{1}{3}{\rho _{rad}}{c^2})}}{{{c^2}}} = 2{\rho _{rad}}$$
or
$${\rho _{rad}} + \frac{{3{P_{rad}}}}{{{c^2}}} = 0 + \frac{{3(\frac{1}{3}{\rho _{rad}}{c^2})}}{{{c^2}}} = {\rho _{rad}}$$

Last edited:
darkdark10 said:
In the Friedmann equation, ρ is the mass density.
No. It's the energy density in mass density units. "Mass density" doesn't even make sense as a separate concept for radiation.

darkdark10 said:
When we call the energy densities of matter, radiation, and dark energy (ρ_m)c^2, (ρ_rad)c^2, (ρ_lambda)c^2, what form does this take when entering the equation?
What you just wrote. All you're doing is changing from mass density units to energy density units. The conversion factor between those is ##c^2##.

vanhees71
PeterDonis said:
No. It's the energy density in mass density units. "Mass density" doesn't even make sense as a separate concept for radiation.

What you just wrote. All you're doing is changing from mass density units to energy density units. The conversion factor between those is ##c^2##.

We know that mass density = energy density/c^2. The question is, when entering the expression (ρ+3P/c^2), which form is correct? When looking at the radiation term,
$$\rho + \frac{{3P}}{{{c^2}}} = {\rho _{rad}} + \frac{{3(\frac{1}{3}{\rho _{rad}}{c^2})}}{{{c^2}}} = 2{\rho _{rad}}$$
or
$$\rho + \frac{{3P}}{{{c^2}}} = 0 + \frac{{3(\frac{1}{3}{\rho _{rad}}{c^2})}}{{{c^2}}} = {\rho _{rad}}$$

darkdark10 said:
We know that mass density = energy density/c^2.
That depends on what you mean by "mass density". The only meaning that works with the Friedmann equation is to consider "mass density" as the same as energy density, just in different units.

darkdark10 said:
The question is, when entering the expression (ρ+3P/c^2), which form is correct?

vanhees71
PeterDonis said:
That depends on what you mean by "mass density". The only meaning that works with the Friedmann equation is to consider "mass density" as the same as energy density, just in different units.I already gave you the answer.
The only meaning that works with the Friedmann equation is to consider "mass density" as the same as energy density, just in different units.
$$\rho + \frac{{3P}}{{{c^2}}} = {\rho _{rad}} + \frac{{3(\frac{1}{3}{\rho _{rad}}{c^2})}}{{{c^2}}} = 2{\rho _{rad}}$$
Your argument leads to the claim that when matter and radiation have the same energy density ρ0, the radiation exerts twice the gravitational force than the matter.

Is this a valid argument?

darkdark10 said:
Your argument leads to the claim that when matter and radiation have the same energy density ρ0, the radiation exerts twice the gravitational force than the matter.

Is this a valid argument?
Yes. More precisely, radiation of a given energy density causes the expansion of the universe to decelerate twice as fast as matter (more precisely cold matter, assumed to have negligible pressure) of the same energy density.

darkdark10 said:
We know that mass density = energy density/c^2. The question is, when entering the expression (ρ+3P/c^2), which form is correct? When looking at the radiation term,
$$\rho + \frac{{3P}}{{{c^2}}} = {\rho _{rad}} + \frac{{3(\frac{1}{3}{\rho _{rad}}{c^2})}}{{{c^2}}} = 2{\rho _{rad}}$$
or
$$\rho + \frac{{3P}}{{{c^2}}} = 0 + \frac{{3(\frac{1}{3}{\rho _{rad}}{c^2})}}{{{c^2}}} = {\rho _{rad}}$$
At least at this point it should be clear, how confusing the concept of "relativistic mass" is. Unfortunately Einstein introduced it in his 1905 paper, before the formulation of special-relativistic point mechanics was fully understood, which to my knowledge has been achieved first in a paper by Planck in 1906. Note that Einstein himself was not in favor of the "relativistic mass" idea for very long. He rather adviced people early on to use mass in its only well-defined meaning, which is "invariant mass". In GR there's not even a chance to make sense of a "relativistic mass". Inertia is related to energy, and the source of the gravitational field is the energy-momentum-stress tensor of "matter and radiation". Nowhere enters a separate quantity "relativistic mass (density)".

The trace of the energy-momentum-stress tensor of the electromagnetic field is 0. This is due to conformal invariance of the free Maxwell equations. For thermal radiation it implies that ##\epsilon=3P## (where ##\epsilon## is the energy density and ##P## the pressure).

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