<p>Since relativistic physics almost always takes units of $c = 1$, there is no difference between units of energy density and mass density. Conceptually, however, the gravitational charge is energy, not mass.</p>

<p>In general relativity, a perfect fluid has no viscuous shear stresses and no heat flux, and can be described by the stress-energy tensor $T^{\mu\nu} = (\rho + p)u^\mu u^\nu + pg^{\mu\nu}$, where $\rho$ is density, $p$ is pressure, and $u^\mu$ is the fluid four-velocity. In the case of vanishing pressure, the result is called a <em>dust</em>, and is appropriate for modeling cold dark matter, galaxies on the cosmological scale, etc. On the other hand, a <em>radiation fluid</em> has $\rho = 3p$, and is appropriate for modeling incoherent electromagnetic radiation, or a fluid made of strongly relativistic particles (e.g., hot dark matter), and so forth.</p>

<p>Thus, in the specific context of cosmology, the <em>matter-dominated</em> regime has $p\sim 0$ and <em>radiation-dominated</em> regime has $p\sim\rho/3$. Note that in the latter case, the fluid particles have zero (or at least negligible) mass. In-between, we can talk about a fluid of particles that are not strongly relativistic, and hence have some mass.</p>

<p>On the other hand, the standard model of dark energy is a cosmological constant, which is equivalent to a perfect fluid $p = -\rho$. It does not fit into the above scale; it is not matter in the above sense, nor is radiation, nor anything in-between. Furthermore, it has the peculiar property of being exactly the same in every local inertial frame. This frame-independence enables us to think of the cosmological constant as an <em>intrinsic</em> energy density of spacetime, but it also makes it inappropriate to think of it as a collection of fluid particles that have some rest frame.</p>

<p>For your purposes, this important because $E = mc^2$ is only valid in a rest frame.</p>

CC BY-SA 3.0
2014-10-15T12:06:00.747Z
So the formula says that mass equals energy, but the concepts dont fit the formula? Something tells me i need to read your post every 3 hours for a few days, and it will somehow sink in, but right now it hasnt yet
usethedeathstar
CC BY-SA 3.0
2014-10-15T16:33:44.920Z
You can apply the formula to find a mass density if you like, but the point was only that unlike the previous cases, you can't decompose dark energy into a collection of massive particles zipping about.
Stan Liou
CC BY-SA 3.0
2014-10-15T08:49:59.327Z
2014-10-15T08:49:59.327Z
Stan Liou
<p>We call dark matter dark matter, because of the way it acts on galaxies, but why do we look at dark energy as an energy? With $E = m c^2$, can't dark energy be mass as well? Why dont we look at dark energy as a mass, in the same way as we look at dark matter as being a missing mass (as halo around galaxies)? </p>

<dark-energy>
why do we look at dark energy as an energy?
usethedeathstar
# why do we look at dark energy as an energy?

(Asked by: username_0 on 2014-10-15)

We call dark matter dark matter, because of the way it acts on galaxies, but why do we look at dark energy as an energy? With $E = m c^2$, can't dark energy be mass as well? Why dont we look at dark energy as a mass, in the same way as we look at dark matter as being a missing mass (as halo around galaxies)?

**Tags:** dark-energy


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## 1 answers

### Answer by username_1 on 2014-10-15. Score: 2 | Accepted answer

Since relativistic physics almost always takes units of $c = 1$, there is no difference between units of energy density and mass density. Conceptually, however, the gravitational charge is energy, not mass.

In general relativity, a perfect fluid has no viscuous shear stresses and no heat flux, and can be described by the stress-energy tensor $T^{\mu\nu} = (\rho + p)u^\mu u^\nu + pg^{\mu\nu}$, where $\rho$ is density, $p$ is pressure, and $u^\mu$ is the fluid four-velocity. In the case of vanishing pressure, the result is called a *dust*, and is appropriate for modeling cold dark matter, galaxies on the cosmological scale, etc. On the other hand, a *radiation fluid* has $\rho = 3p$, and is appropriate for modeling incoherent electromagnetic radiation, or a fluid made of strongly relativistic particles (e.g., hot dark matter), and so forth.

Thus, in the specific context of cosmology, the *matter-dominated* regime has $p\sim 0$ and *radiation-dominated* regime has $p\sim\rho/3$. Note that in the latter case, the fluid particles have zero (or at least negligible) mass. In-between, we can talk about a fluid of particles that are not strongly relativistic, and hence have some mass.

On the other hand, the standard model of dark energy is a cosmological constant, which is equivalent to a perfect fluid $p = -\rho$. It does not fit into the above scale; it is not matter in the above sense, nor is radiation, nor anything in-between. Furthermore, it has the peculiar property of being exactly the same in every local inertial frame. This frame-independence enables us to think of the cosmological constant as an *intrinsic* energy density of spacetime, but it also makes it inappropriate to think of it as a collection of fluid particles that have some rest frame.

For your purposes, this important because $E = mc^2$ is only valid in a rest frame.

> **Comments:**
> **username_0**: So the formula says that mass equals energy, but the concepts dont fit the formula? Something tells me i need to read your post every 3 hours for a few days, and it will somehow sink in, but right now it hasnt yet  
> **username_1**: You can apply the formula to find a mass density if you like, but the point was only that unlike the previous cases, you can't decompose dark energy into a collection of massive particles zipping about.  

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