# What is the 'scale factor' equation for a dark-matter dominated universe?

The Friedmann equations can be solved exactly in presence of a perfect fluid with equation of state

$${\displaystyle p=w\rho c^{2}} \qquad p=w\rho c^2$$

where $${\displaystyle p}$$ is the pressure, $${\displaystyle \rho }$$ is the mass density of the fluid in the co-moving frame and $$w$$ is some constant.

In spatially flat case ($$k = 0$$), the solution for the scale factor is

$${\displaystyle a(t)=a_{0}\,t^{\frac {2}{3(w+1)}}}$$ where $${\displaystyle a_{0}}$$ is some integration constant to be fixed by the choice of initial conditions. This family of solutions labelled by $${\displaystyle w}$$ is extremely important for cosmology. E.g. $${\displaystyle w=0}$$ describes a matter-dominated universe, where the pressure is negligible with respect to the mass density. From the generic solution one easily sees that in a matter-dominated universe the scale factor goes as

$${\displaystyle a(t)\propto t^{2/3}}$$ matter-dominated Another important example is the case of a radiation-dominated universe, i.e., when $${\displaystyle w=1/3}$$. This leads to

$${\displaystyle a(t)\propto t^{1/2}}$$ radiation dominated Note that this solution is not valid for domination of the cosmological constant, which corresponds to an $${\displaystyle w=-1}$$. In this case the energy density is constant and the scale factor grows exponentially.

So, '$$a$$' is proportional to $$t^{2/3}$$ or $$t^{1/2}$$ for matter- or radiation-dominated universes, respectively... But if '$$w$$' is negative-one then '$$a$$' is proportional to $$t^t$$? I mean, what is the exponent in this 'exponential growth' phase where the '$$w$$' 'constant' is $$-1$$?

• Exponential means $a \propto \exp(t/t_0)$. Your question title mentions dark matter, but the body is asking about dark energy? They aren't the same thing. – ProfRob Oct 16 '20 at 7:48

The $$w=-1$$ does not apply to $$a(t) \propto t^{2/3(1+w)}$$