# Various values of $w$ in the equation $p=w\rho$

In cosmology, while deriving how the energy density dilutes with the scale factor one assumes $p=w\rho$ with $w=0$ for the matter, $w=-1$ for vacuum energy and $w=1/3$ for the radiation. In know the energy density of radiation varies as $p=\frac{1}{3}\rho$. But I do not understand why is it taken to be zero for matter and $-1$ for vacuum energy (This implies vacuum energy doesn't dilute with expansion. But if vaccum energy is the zero point energy of the fields I don't see a why it shouldn't dilute.)? I was reading Kolb and Turner's book but didn't get any answer there.

The Friedmann equations assume that all our matter fields are perfect fluids. Those, in the rest frame of the fluid, assume the form of the stress-energy tensor $$T^{\mu\nu}_{fluid} = \begin{pmatrix} \rho & 0 & 0 & 0 \\ 0&P&0&0 \\ 0&0&P&0 \\ 0&0&0&P \end{pmatrix}$$ where $\rho$ is the total energy density including both the energy coming from the rest mass of the particles and the internal energy coming from random microscopic motion velocities. A good model for matter at finite temperature, with $k_B T \ll mc^2$ to linear order, is the ideal gas non-relativistic in its rest frame $$p = n k_B T,\; \rho = mn + \frac{3}{2}n k_B T$$ where $n$ is the number density of particles in the rest frame of the fluid and $m$ is the particle mass. Now you see that for this gas there is really no relation $p=w\rho$ with $w$ a constant factor.

We may, however, still compute that under the assumption of adiabatic expansion (entropy per particle constant), the $\sim n k_B T$ term will scale with $(a/a_0)^{-5}$ whereas the $mn$ term will scale with $(a/a_0)^{-3}$.

A common approximation in cosmology is to do the "quick and dirty" approximation where even the $\sim n k_B T$ terms are completely neglected with respect to $\sim mn$.

This makes the computation tidy and tractable, verifiable to a certain degree by hand, but I cannot advocate for this approximation beyond that. Naturally, this leads to a trivial $p = w \rho$ with $w=0$.

As for vacuum energy, this is simply the cosmological constant which shows up in Einstein equations as $\Lambda g_{\mu \nu}$ on the "left-hand-side" (the gravitational part). This can be somehow linked with vacuum energy, but it can be simply understood as a constant of nature determining the properties of gravity in our universe. If we move it to the "right-hand side" of matter in the Einstein equations and understand it as vacuum energy, we get a stress-energy tensor in the frame of the cosmological fluid $$-\Lambda g^{\mu \nu} = T^{\mu\nu}_{vac} = \pmatrix{\Lambda&0&0&0\\0&-\Lambda&0&0\\ 0&0&-\Lambda&0\\0&0&0&-\Lambda}$$ If, then, you purely formally understand this stress-energy tensor as the stress-energy tensor of a perfect fluid as shown above, you get $P_\Lambda = - \rho_\Lambda$ and $w$ is simply $-1$.

If you would like to dig deeper, the reason why vacuum energy is allowed to behave this way comes from the construction of the Einstein equations which I have briefly described here.

Dark matter is assumed to be a pressureless fluid, only affected by gravity, therefore $p_{\rm dm} = 0$, or $w_{\rm dm} = 0$. In general, Friedmann equations coupled with the First Law of thermodynamics implies that the density of a component with EoS $p = w \rho$ should scale as

$$\rho \sim a^{-3(1 + w)}$$

From this, you can see again that for regular matter $w = 0$, so that $\rho\sim a^{-3}$ as expected. If you take $w = -1$ you see that $\rho\sim 1$, that means that the density does not change over time. One possibility for such a fluid is vacuum, indeed, the more cubic meters of empty universe you create, the more vacuum there is

• Vacuum energy is related to dark energy as far as I know and not dark matter. My question was why $w=0$ for regular matter and $w=-1$ for vacuum energy. What do you mean by $\rho\sim 1$? From $w=-1$, $\rho=$constant.
– SRS
Commented May 14, 2017 at 15:38
• My question is exactly that how do we know that vacuum energy doesn't dilute with expansion?
– SRS
Commented May 14, 2017 at 15:39
• @SRS Sorry for the confusion. I tried to justify that for matter (including dark matter) $w = 0$. Whereas if you set $w = -1$ you get something like vacuum Commented May 14, 2017 at 15:40
• @SRS Exactly because it is vacuum. Intuitively, the more universe you have, the more vacuum you have, so the density remains constant Commented May 14, 2017 at 15:44
• As far as I know vaccum energy is contributed by the zero point energies of the quantum fields. And I see no reason why that shouldn't dilute. Vaccum energy does couple to gravity and therefore it's not 'Nothing'.
– SRS
Commented May 14, 2017 at 15:47

Considering the FRW metric for the one can derive the energy conservation law:

$$\dot{\rho}=-3H(\rho+p)$$

where $\rho$ is energy density (not energy but energy density) and p is the pressure.

Equation of state of the matter:

Energy density of matter is given by (taking $k_{B}=1$)

$$\rho=nm+\frac{3}{2}nT$$

whereas the pressure $p=nT$. For a non-relativistic particle i.e., matter, $m\gg T$. So, $$\frac{p}{\rho}\approx\frac{T}{m}\approx0$$

Hence, w=0 for the matter

Equation of state of vacuum energy density:

Vacuum energy density is a constant energy density in the Einstein's equation.So, $\rho_{vacuum}=\Lambda$ remains constant.This gives, $\dot{\Lambda}=0$. Now, using the energy conservation law

This implies,

$$-3H(\Lambda+p)=0$$

which gives $p=-\Lambda$. This means, w=-1 for vacuum energy density

Presently, the vacuum energy density ( Dark energy) dominates the universe. And the proportion of vacuum energy density increases in the future as the temperature of universe is dropping due to expansion.