<p>The simplest and shortest answer to this question is that since our universe is flat (to within our error bars), its density must be that of the critical density - that's what it means to be flat! In that case the overall density of the universe, accounting for matter, dark matter, dark energy, etc. is (assuming $H_0=100\ \mathrm{km/s/Mpc}$)</p>

<p>$$\rho_c = 1.8788\times 10^{-26}\ \mathrm{kg/m^3} = 1.8788\times 10^{-29}\ \mathrm{g/cm^3}$$</p>

<p>If you want to be more precise about it and actually use measurements of our universe, you can go about calculating it by using the density parameters. By definition, the density parameter is defined as</p>

<p>$$\Omega \equiv \frac{\rho}{\rho_c}$$</p>

<p>where $\rho_c = 1.8788\times 10^{-26}h\ \mathrm{kg/m^3}$ (notice that little $h$ now included). Numerous missions such as WMAP and Planck have measured these density parameters for the variety of mass-energies in our universe and you can look them up and put them all together to get the total mean density of our universe. Overall, you should find that</p>

<p>$$\rho = (\Omega_m + \Omega_c + \Omega_\Lambda + ...)\rho_c$$</p>

<p>where the various $\Omega$s are defined for the various mass-energies in our universe. You can look up values for the density parameters from recent results such as from <a href="https://lambda.gsfc.nasa.gov/product/map/current/params/lcdm_wmap9.cfm" rel="nofollow noreferrer">WMAP</a> or <a href="https://arxiv.org/pdf/1502.01589.pdf" rel="nofollow noreferrer">Planck</a>. Just decide on your chosen model and pull out the $\Omega$ values $-$ or more properly the $\Omega h^2$ values if you want to use your own value of $H_0$. It's not too difficult to plug everything in. You'll pretty much always find though, that the sum of the density parameters very nearly equals one and to an order of magnitude you can say $\rho=\rho_c$.</p>

CC BY-SA 3.0
2018-01-31T15:40:25.297Z
2018-01-31T15:40:25.297Z
zephyr
<p>I know cosmic mean density for baryons is around $(2\ \mathrm{to}\ 5)\times10^{-31}\ \mathrm{g/cm^3}$, assuming that Hubble's Constant is $100\ \mathrm{km/s/Mpc}$. But what is the current cosmic mean density including all particles in the universe (dark matter etc.)?</p>

CC BY-SA 3.0
2018-01-31T14:23:04.337Z
We don't know what dark matter is nor do we know where it is, but you want us to estimate its particle density?   This question needs the <NOTFAIR> tag.
Carl Witthoft
CC BY-SA 3.0
2018-01-31T14:45:00.110Z
@CarlWitthoft And yet we've measured the density of dark matter with ever increasing precision for decades and from it proven the universe is flat!
zephyr
CC BY-SA 3.0
2018-01-31T19:47:59.913Z
@zephyr  OK -- I was thinking more of particle density instead of mass density.
Carl Witthoft
CC BY-SA 3.0
2018-01-31T21:41:07.423Z
@CarlWitthoft Oh yes, that would be tricky if you have no idea what the particles are.
zephyr
<cosmology>
What is the cosmic mean density?
bhjghjh
# What is the cosmic mean density?

(Asked by: username_0 on 2018-01-31)

I know cosmic mean density for baryons is around $(2\ \mathrm{to}\ 5)\times10^{-31}\ \mathrm{g/cm^3}$, assuming that Hubble's Constant is $100\ \mathrm{km/s/Mpc}$. But what is the current cosmic mean density including all particles in the universe (dark matter etc.)?

**Tags:** cosmology

> **Comments:**
> **username_1**: We don't know what dark matter is nor do we know where it is, but you want us to estimate its particle density?   This question needs the <NOTFAIR> tag.  
> **username_2**: @username_1 And yet we've measured the density of dark matter with ever increasing precision for decades and from it proven the universe is flat!  
> **username_1**: @username_2  OK -- I was thinking more of particle density instead of mass density.  
> **username_2**: @username_1 Oh yes, that would be tricky if you have no idea what the particles are.  

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## 1 answers

### Answer by username_2 on 2018-01-31. Score: 3 | Accepted answer

The simplest and shortest answer to this question is that since our universe is flat (to within our error bars), its density must be that of the critical density - that's what it means to be flat! In that case the overall density of the universe, accounting for matter, dark matter, dark energy, etc. is (assuming $H\_0=100\ \mathrm{km/s/Mpc}$)

$$\rho\_c = 1.8788\times 10^{-26}\ \mathrm{kg/m^3} = 1.8788\times 10^{-29}\ \mathrm{g/cm^3}$$

If you want to be more precise about it and actually use measurements of our universe, you can go about calculating it by using the density parameters. By definition, the density parameter is defined as

$$\Omega \equiv \frac{\rho}{\rho\_c}$$

where $\rho\_c = 1.8788\times 10^{-26}h\ \mathrm{kg/m^3}$ (notice that little $h$ now included). Numerous missions such as WMAP and Planck have measured these density parameters for the variety of mass-energies in our universe and you can look them up and put them all together to get the total mean density of our universe. Overall, you should find that

$$\rho = (\Omega\_m + \Omega\_c + \Omega\_\Lambda + ...)\rho\_c$$

where the various $\Omega$s are defined for the various mass-energies in our universe. You can look up values for the density parameters from recent results such as from [WMAP](https://lambda.gsfc.nasa.gov/product/map/current/params/lcdm_wmap9.cfm) or [Planck](https://arxiv.org/pdf/1502.01589.pdf). Just decide on your chosen model and pull out the $\Omega$ values $-$ or more properly the $\Omega h^2$ values if you want to use your own value of $H\_0$. It's not too difficult to plug everything in. You'll pretty much always find though, that the sum of the density parameters very nearly equals one and to an order of magnitude you can say $\rho=\rho\_c$.


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