# Can we regard dark matter as a special fluid rather than individual particles?

If we assume that dark matter is an $N$-body system of sizeless and collisionless particles which only interact via newtonian gravity, do these huge quantity of particles manifest some collective behaviors which can be described with simple equations?

For example, gas is consist of innumerable billiard balls, but we can use some equations such as state equation of ideal gas $PV=nRT$, heat equation and Navier-Stokes equations to describe the behavior collectively rather than tracing the motion of particles individually. Is this the case of dark matter?

• I don't know that the EOS for DM has been yet resolved, so that's probably one issue. Sep 28, 2015 at 14:38

Considering dark matter as a perfect fluid is useful for understanding cosmological evolution via the Friedmann acceleration equation $$3\frac{\ddot{a}}{a}=\Lambda - 4\pi G(\rho + 3p).$$ (note that this is a general relativistic equation, so not strictly Newtonian). Cosmologists use the equation of state parameter $w$ to relate the pressure of a perfect fluid $p$ to its energy density $\rho$: $$p = w\rho.$$ Knowing $w$ allows us to plug into the Friedmann equations to solve for the time dependence of the scale factor $a(t)$. We also learn from these equations how fluids with different $w$ parameters are diluted as space expands. For example, non-relativistic matter (such as cold dark matter) has an equation of state parameter $w\approx 0$, so that $\rho\propto a^{-3} =V^{-1}$. This is an intuitive answer, as the density of the matter decreases inversely proportional to the volume. On the other hand, ultra-relativistic matter (e.g. photons) has an equation of state parameter $\omega \approx \frac{1}{3}$, so that the energy density of radiation scales as $a^{-4}$, where the additional factor of $1/a$ comes from the fact that the expansion of space redshifts radiation and decreases its energy by a factor of $1/a$. A cosmological constant has an equation of state parameter $w=-1$, implying that a positive dark energy density creates negative pressure. We also see that $\rho_{\Lambda}$ has no $a$ dependence, hence the name Cosmological Constant.