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

<p>But the number <em>density</em> of particles varies as $a^{-3}$.</p>

<p>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.</p>

CC BY-SA 3.0
2017-02-20T22:31:20.693Z
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 $\rho \propto a^{-4}$.
Spuds
CC BY-SA 3.0
2017-02-20T22:37:31.807Z
@spuds I am sure I have answered your (cross-posted) question. Mass density (and energy density if only rest mass is important) scales as $a^{-3}$.
ProfRob
CC BY-SA 3.0
2017-02-20T22:13:12.707Z
2017-02-20T22:13:12.707Z
ProfRob
<p>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 $n(t_0)$.</p>

<p>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.</p>

<p>Any help?</p>

<p>EDIT! I forgot that $a(t)$ is dimensionless.</p>

CC BY-SA 3.0
2017-02-20T21:55:35.490Z
Are you familiar with the [Friedmann equations](https://en.wikipedia.org/wiki/Friedmann_equations)?
HDE 226868
<universe><cosmology><matter><density>
Finding equation for energy density of matter
Spuds
# Finding equation for energy density of matter

(Asked by: username_0 on 2017-02-20)

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.

**Tags:** universe, cosmology, matter, density
> **Comments:**
> **username_1**: Are you familiar with the [Friedmann equations](https://en.wikipedia.org/wiki/Friedmann_equations)?  

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## 1 answers

### Answer by username_2 on 2017-02-20. Score: 2

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.

> **Comments:**
> **username_0**: 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 $\rho \propto a^{-4}$.  
> **username_2**: @spuds I am sure I have answered your (cross-posted) question. Mass density (and energy density if only rest mass is important) scales as $a^{-3}$.  

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