<p>Yes. Just convert the mass to its energy equivalent via $c^2$. </p>

<p>$7\times10^{-30}$ g/cc is entirely equivalent to $7\times 10^{-27}$ kg/m$^3$ or $6.3\times 10^{-10}$ J/m$^3$ or 3.94 GeV/m$^3$ or 3.94 keV/cc.</p>

<p>It is quoted as a mass density so that its importance in terms of the dynamics of the universe, can be compared with that of normal and dark matter. In fact in most applications it is more normal to quote and compare in terms of energy density, since that takes account of the fact that matter has kinetic energy as well as rest-mass energy and more neatly includes radiation energy into the comparison.</p>

CC BY-SA 3.0
2017-03-18T10:53:05.890Z
2017-03-18T10:53:05.890Z
ProfRob
<p>This is a case of using a sort of <a href="https://en.wikipedia.org/wiki/Natural_units#Systems_of_natural_units" rel="nofollow noreferrer">natural units</a> (similar to <a href="https://en.wikipedia.org/wiki/Geometrized_unit_system" rel="nofollow noreferrer">geometerized units</a>), which are often convenient in relativity (special and general). Take, for instance, the <a href="https://en.wikipedia.org/wiki/Minkowski_space#Minkowski_metric" rel="nofollow noreferrer">Minkowski metric</a> of special relativity. In dimensional units, it is
$$\mathrm{d}s^2=-c^2\mathrm{d}t^2+\mathrm{d}x^2+\mathrm{d}y^2+\mathrm{d}z^2$$
Setting $c=1$ makes it a bit more convenient, and gives time and space the same units - which is more intuitive, if you think of time as another dimension.</p>

<p>You also brought up $E=mc^2$. This is a special case of the <a href="https://en.wikipedia.org/wiki/Energy%E2%80%93momentum_relation" rel="nofollow noreferrer">energy-momentum relation</a>:
$$E^2=p^2c^2+m_0^2c^4$$
Setting $c=1$, we have the much more convenient form
$$E=p^2+m_0^2$$
This comes up again and again. Add to this the <a href="https://en.wikipedia.org/wiki/Energy%E2%80%93momentum_relation" rel="nofollow noreferrer">mass-energy equivalence</a> that is also important in relativity (especially the right-hand side of the Einstein field equations), and it simply makes sense to treat energy and mass as the same thing. In fact, it becomes perhaps even more useful than in special relativity, because we also often set $G=1$ - and that comes up a <em>lot</em> in general relativity.</p>

<p>Now, the units of density under this system are $\text{Length}^{-2}$, and the density above is clearly not in those units. I'm guessing that this just isn't a case of setting $G=1$ - and that makes sense, because in cosmology and astronomy, many types of densities are measured in g/cm<sup>3</sup>: dark matter, interstellar gas and dust, and sometimes dark energy. Keeping it in $\text{Mass}\cdot\text{Length}^{-3}$ makes it easy to compare these, working in a system that is more commonly used.</p>

<p>In a nutshell, then: Cosmologists sometimes like $c=1$ (and sometimes $G=1$) because it makes a lot of equations a lot easier to write out. Here, it becomes really easy to compare mass densities and energy densities.</p>

CC BY-SA 3.0
2017-03-17T15:15:33.813Z
So... how much would it be in $J/m^3$ ?
SF.
CC BY-SA 3.0
2017-03-17T17:59:07.053Z
@SF. If $\rho_m$ is the mass density, the energy density is just $\rho_mc^2$ - where $c$ is of course in the correct units.
HDE 226868
CC BY-SA 3.0
2017-03-18T11:25:56.370Z
So, that would be $6.3×10^{−16}  J/cm^3$, right?
SF.
CC BY-SA 3.0
2017-03-18T13:25:20.670Z
@SF. Yes, that's correct.
HDE 226868
CC BY-SA 3.0
2017-03-17T14:06:08.670Z
2017-03-17T14:06:08.670Z
HDE 226868
<p>Aside from cosmology,  the advantage of reporting density in $\frac{g}{cc}$ is that density and <a href="https://en.wikipedia.org/wiki/Specific_gravity" rel="nofollow noreferrer">specific gravity </a>  are identical in this case (unless some wiseguy doesn't use water as the reference material).</p>

CC BY-SA 3.0
2017-03-17T16:18:24.927Z
uh, but that's *energy* density. How would that comparison work?
SF.
CC BY-SA 3.0
2017-03-17T16:42:19.750Z
You might have to explain further. I don't see how a normalized expression of energy density in units of $g/cm^3$ is equivalent to the dimensionless concept of specific gravity. I don't see how even a standard concept of density is equivalent to specific gravity.
zephyr
CC BY-SA 3.0
2017-03-17T18:26:42.117Z
@zephyr: the latter case is numerical equivalence with unit mismatch, for metric. Water density is very close to 1g/cc or 1kg/l, and its specific gravity is 1. Balsa, at specific gravity of 0.2 will be 0.2g/cc. But yeah, the energy density equivalence? I don't quite see it here.
SF.
CC BY-SA 3.0
2017-03-17T19:04:20.950Z
@SF. I do see that that would be true (although I'd argue that Carl's use of the word "identical" is a bit too strong since the density of water at STP is 0.9999720 g/cc - a better description might be nearly identical). I just think Carl missed the mark on this answer really.
zephyr
CC BY-SA 3.0
2017-03-17T15:45:46.283Z
2017-03-17T15:45:46.283Z
Carl Witthoft
<p>According to <a href="https://en.wikipedia.org/wiki/Dark_energy" rel="noreferrer">Wikipedia article</a>,</p>

<blockquote>
  <p>The density of dark energy ($~ 7 × 10^{−30} g/cm^3$) is very low, much less than the density of ordinary matter or dark matter within galaxies. However, it comes to dominate the mass–energy of the universe because it is uniform across space.</p>
</blockquote>

<p>Other sources I found also give the density either as percentage of mass of the universe, or just as mass density like above, $kg/m^3$ or similar.</p>

<p>In about all other applications and contexts of usage of energy density, I'd find <a href="https://en.wikipedia.org/wiki/Energy_density" rel="noreferrer">MJ/L</a>, or other units of <em>energy per volume</em>. Why is energy of density given as <em>mass per volume</em>, and how do I convert it to more familiar units? Using $E=mc^2$ ?</p>

CC BY-SA 3.0
2017-03-18T19:56:45.470Z
$1 g/cm^3$ is the density of water. It lets the reader compare it to something familiar.
Keith Thompson
CC BY-SA 3.0
2017-03-20T06:55:31.120Z
One *could* argue that, as it does not in fact exist*, it doesn't matter what units you use :-).
(* we just don't all know it yet).
Russell McMahon
<dark-energy>
Why is dark energy density given in grams per cm$^3?$
SF.
# Why is dark energy density given in grams per cm$^3?$

(Asked by: username_0 on 2017-03-17)

According to [Wikipedia article](https://en.wikipedia.org/wiki/Dark_energy),

> The density of dark energy ($~ 7 × 10^{−30} g/cm^3$) is very low, much less than the density of ordinary matter or dark matter within galaxies. However, it comes to dominate the mass–energy of the universe because it is uniform across space.

Other sources I found also give the density either as percentage of mass of the universe, or just as mass density like above, $kg/m^3$ or similar.

In about all other applications and contexts of usage of energy density, I'd find [MJ/L](https://en.wikipedia.org/wiki/Energy_density), or other units of *energy per volume*. Why is energy of density given as *mass per volume*, and how do I convert it to more familiar units? Using $E=mc^2$ ?

**Tags:** dark-energy
> **Comments:**
> **username_1**: $1 g/cm^3$ is the density of water. It lets the reader compare it to something familiar.  
> **username_2**: One *could* argue that, as it does not in fact exist*, it doesn't matter what units you use :-). (* we just don't all know it yet).  

---

## 3 answers

### Answer by username_3 on 2017-03-18. Score: 3 | Accepted answer

Yes. Just convert the mass to its energy equivalent via $c^2$.

$7\times10^{-30}$ g/cc is entirely equivalent to $7\times 10^{-27}$ kg/m$^3$ or $6.3\times 10^{-10}$ J/m$^3$ or 3.94 GeV/m$^3$ or 3.94 keV/cc.

It is quoted as a mass density so that its importance in terms of the dynamics of the universe, can be compared with that of normal and dark matter. In fact in most applications it is more normal to quote and compare in terms of energy density, since that takes account of the fact that matter has kinetic energy as well as rest-mass energy and more neatly includes radiation energy into the comparison.


---

### Answer by username_4 on 2017-03-17. Score: 4

This is a case of using a sort of [natural units](https://en.wikipedia.org/wiki/Natural_units#Systems_of_natural_units) (similar to [geometerized units](https://en.wikipedia.org/wiki/Geometrized_unit_system)), which are often convenient in relativity (special and general). Take, for instance, the [Minkowski metric](https://en.wikipedia.org/wiki/Minkowski_space#Minkowski_metric) of special relativity. In dimensional units, it is
$$\mathrm{d}s^2=-c^2\mathrm{d}t^2+\mathrm{d}x^2+\mathrm{d}y^2+\mathrm{d}z^2$$
Setting $c=1$ makes it a bit more convenient, and gives time and space the same units - which is more intuitive, if you think of time as another dimension.

You also brought up $E=mc^2$. This is a special case of the [energy-momentum relation](https://en.wikipedia.org/wiki/Energy%E2%80%93momentum_relation):
$$E^2=p^2c^2+m\_0^2c^4$$
Setting $c=1$, we have the much more convenient form
$$E=p^2+m\_0^2$$
This comes up again and again. Add to this the [mass-energy equivalence](https://en.wikipedia.org/wiki/Energy%E2%80%93momentum_relation) that is also important in relativity (especially the right-hand side of the Einstein field equations), and it simply makes sense to treat energy and mass as the same thing. In fact, it becomes perhaps even more useful than in special relativity, because we also often set $G=1$ - and that comes up a *lot* in general relativity.

Now, the units of density under this system are $\text{Length}^{-2}$, and the density above is clearly not in those units. I'm guessing that this just isn't a case of setting $G=1$ - and that makes sense, because in cosmology and astronomy, many types of densities are measured in g/cm3: dark matter, interstellar gas and dust, and sometimes dark energy. Keeping it in $\text{Mass}\cdot\text{Length}^{-3}$ makes it easy to compare these, working in a system that is more commonly used.

In a nutshell, then: Cosmologists sometimes like $c=1$ (and sometimes $G=1$) because it makes a lot of equations a lot easier to write out. Here, it becomes really easy to compare mass densities and energy densities.

> **Comments:**
> **username_0**: So... how much would it be in $J/m^3$ ?  
> **username_4**: @SF. If $\rho_m$ is the mass density, the energy density is just $\rho_mc^2$ - where $c$ is of course in the correct units.  
> **username_0**: So, that would be $6.3×10^{−16}  J/cm^3$, right?  
> **username_4**: @SF. Yes, that's correct.  

---

### Answer by username_5 on 2017-03-17. Score: 0

Aside from cosmology, the advantage of reporting density in $\frac{g}{cc}$ is that density and [specific gravity](https://en.wikipedia.org/wiki/Specific_gravity)  are identical in this case (unless some wiseguy doesn't use water as the reference material).

> **Comments:**
> **username_0**: uh, but that's *energy* density. How would that comparison work?  
> **username_6**: You might have to explain further. I don't see how a normalized expression of energy density in units of $g/cm^3$ is equivalent to the dimensionless concept of specific gravity. I don't see how even a standard concept of density is equivalent to specific gravity.  
> **username_0**: @zephyr: the latter case is numerical equivalence with unit mismatch, for metric. Water density is very close to 1g/cc or 1kg/l, and its specific gravity is 1. Balsa, at specific gravity of 0.2 will be 0.2g/cc. But yeah, the energy density equivalence? I don't quite see it here.  
> **username_6**: @SF. I do see that that would be true (although I'd argue that Carl's use of the word "identical" is a bit too strong since the density of water at STP is 0.9999720 g/cc - a better description might be nearly identical). I just think Carl missed the mark on this answer really.  

---

