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

$${\displaystyle p=w\rho c^{2},}p=w\rho c^2$$, where ${\displaystyle p}$ is the pressure, ${\displaystyle \rho }$ is the mass density of the fluid in the comoving frame and $w$ is some constant.

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

$${\displaystyle a(t)=a_{0}\,t^{\frac {2}{3(w+1)}}} a(t)=a_0\,t^{\frac{2}{3(w+1)}}$$ 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 $-1$?

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  • $\begingroup$ I don't know if the math is displaying properly... I cut-and-pasted the 'Useful solutions' section from the Wikipedia article on 'Friedmann equations'.... $\endgroup$ – Kurt Hikes Oct 15 '20 at 23:12
  • $\begingroup$ Edited to move the maths into maths mode. Could probably be tidied up furthetr $\endgroup$ – Steve Linton Oct 16 '20 at 7:43
  • $\begingroup$ 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. $\endgroup$ – ProfRob Oct 16 '20 at 7:48
  • $\begingroup$ Please, fix bizarre repetition in the line going after “with equation of state”. $\endgroup$ – Incnis Mrsi Dec 22 '20 at 18:06
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You answered your question actually. You have said

Note that this solution is not valid for domination of the cosmological constant, which corresponds to an w=−1

then you are saying

But if 'w' is negative-one then 'a' is proportional to tt

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

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