Through substituting for values of $\rho$ and $k$, I have:

$$H^2=\left(\frac{\dot{a}}{a}\right)^2=\frac{8\pi G C}{3a^4} + \frac{\Lambda c^2}{3}$$

$a=a(t)$, and $a(t=0)=0$. Note that $C$ is a constant of integration, which I calculated in the derivation of $\rho=Ca^{-6}$. introduced from the substitution to eliminate $\rho$ - it is not the same as $c$.

Further to this,

$$\frac{\ddot{a}}{a}=\frac{-16 \pi GC}{3a^6}+\frac{\Lambda c^2}{3}$$

However, this universe is dominated by a fluid where $p=\rho c^2$. Hence;

$$H^2=(\frac{\dot{a}}{a})^2=\frac{8\pi G C}{3a^4}$$ and $$\frac{\ddot{a}}{a}=\frac{-16 \pi GC}{3a^6}$$

I'm not sure if it would be possible to integrate the second of those two, and substitute the result into the first to get an equation in $a$? I'm also not sure that those two equations even hold.

I'm not really sure as to how to get the 'temporal dependence of $a$, and can't manage to integrate or solve it any other way myself. I've left out everything done beforehand, though a lot of it is available in the chat or my other questions.


closed as off-topic by David Z Jul 20 '16 at 11:28

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  • 1
    $\begingroup$ you need to use the other Friedmann equation as well to truly nail down the temporal dependence of $a$. You need an equation with $\ddot a$. Using that, you can see that, at late times, $a(t)=e^{Ht}$. At early times, it gets a bit messy $\endgroup$ – Jim Jul 20 '16 at 11:30
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    $\begingroup$ Hi all. I've discussed this question extensively in the H-Bar, and have done alot of work to get to this point, mainly with John Rennie. It's not a homework question, and I believe that the amount of work put into it so far, as well as prior discussion merits it being re-opened. Please consider this! Thanks. $\endgroup$ – Noah P Jul 20 '16 at 11:36
  • $\begingroup$ Jim, I'm not sure what I'm exactly trying to find - what form am I looking for? $\endgroup$ – Noah P Jul 20 '16 at 11:37
  • $\begingroup$ Actually, if you allow $t=0$ to represent the present, you'll find that $a(t)=e^{Ht}$ is a good general approximation $\endgroup$ – Jim Jul 20 '16 at 11:37
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    $\begingroup$ Hi Noah P. If you haven't already done so, please take a minute to read the definition of when to use the homework-and-exercises tag, and the Phys.SE policy for homework-like problems. $\endgroup$ – Qmechanic Jul 20 '16 at 13:10