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

Question

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.

Answer 1

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

Answer 2

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.

Answer 3

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.