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I'm trying to get an equation for the energy density of matter of the universe $\rho(t)$, assuming the number of particles is conserved and rest mass energy is much greater than kinetic energy. $\rho(t)$ should be a function of the scale factor of the universe $a(t)$, mass of particles $m$, and the number density of the particles $n(t_0)$.

I'm not sure how to approach this since $\rho(t)$ is in units of $\frac{Mass}{Length^3}$, $m$ is in units of $\frac{Mass}{Particle}$, $n(t_0)$ is in units of $\frac{Particle}{Length^3}$, and $a(t)$ is in units of $Length$. I guess there could be a factor of $\frac{a(t)}{a(t_0)}$ or $\frac{a(t_0)}{a(t)}$, but I don't know where I'd get that from.

Any help?

EDIT! I forgot that $a(t)$ is dimensionless.

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Energy density $\rho \sim nmc^2$ if kinetic energy is unimportant (and you are ignoring dark energy and other forms of energy density).

But the number density of particles varies as $a^{-3}$.

Thus $$ \frac{\rho}{\rho_0} = a^{-3},$$ where $\rho_0 = n(0)mc^2$ is the present day energy density and the present day scale factor is 1.

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  • $\begingroup$ Thanks, I ended up figuring this out. But are you sure it's not $a^{-4}$? Because next I need to integrate the Friedmann Equation, and that is a lot simpler using $\rho \propto a^{-4}$. $\endgroup$
    – Spuds
    Commented Feb 20, 2017 at 22:31
  • $\begingroup$ @spuds I am sure I have answered your (cross-posted) question. Mass density (and energy density if only rest mass is important) scales as $a^{-3}$. $\endgroup$
    – ProfRob
    Commented Feb 20, 2017 at 22:37

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