I'm reading TASI Lectures on Inflation(https://arxiv.org/pdf/0907.5424.pdf). On page 20, it says

... also called the Friedmann Equations $$\tag{21} \boxed{H^2=\left(\frac{\dot{a}}{a}\right)^2=\frac13\rho-\frac{k}{a^2}} $$ ...
... Eqn.(24) may be integrated to give $$\tag{26} \rho\propto a^{-3(1+w)} $$ Together with the Friedmann Equation(21) this leads to the time evolution of the scale factor $$\tag{27} a(t)\propto\cases{t^{2/3(1+w)}\quad w\ne-1,\\e^{Ht}\quad w=-1,} $$

I've tried to get eq(27) using eq(21) and eq(26) but failed. So could you please show me how to derive this? Besides, it seems not to have $t^{2/3(1+w)}\sim e^{Ht}$ when $w\to-1$. Why is it this case? Does this mean $w$ is in fact not a continious parameter?


1 Answer 1


Let's take $k=0$ (flat universe).

If $w=-1$, then according to (26), $\rho$ is constant, so $H^2=$constant. Therefore: $$\big(\frac{\dot{a}}{a}\big)^2=H^2 \rightarrow \dot{a}=H a \rightarrow a(t)\sim e^{H t}$$

If $w\ne -1$, then $\dot{a}^2\sim a^2 \times a^{-3(1+w)}$ which gives $$\dot{a}\sim a^{-\frac{1}{2}(1+3w)} \rightarrow a^{\frac{1}{2}(1+3w)}da\sim dt \rightarrow a^{\frac{3}{2}(1+w)}=t \rightarrow a\sim t^{\frac{2}{3}(1+w)}$$.

  • $\begingroup$ Thank you very much! Now I know how to get $a(t)$. But I'm still confused about why there isn't $t^{\frac{2}{3}(1+w)}\sim e^{Ht}$ when $w\to -1$. Could you please explain more about what's the physics in it? $\endgroup$
    – Photon
    Jan 7 at 8:45

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