2
$\begingroup$

Imagine a universe dominated by matter, but it is balanced with a cosmological constant $\Lambda=4\pi G\rho$ so the universe is static ($H=0$). However, what would happen if some of that matter turns into radiation?

I was trying to see the Friedmann Equations and found that $\dot{H}+H^2=-\frac{4\pi G}{3}(\rho+3p)+\frac{\Lambda}{3}$ so

\begin{align*} \dot{H}+H^2&=-4\pi G\rho/3-4\pi Gp+4\pi G\rho/3\\ &=-4\pi Gp\\ &=-4\pi G\rho w\\ &=-4\pi G\left(\frac{1}{3}\rho_{\text{rad}}+0\cdot\rho_{\text{matter}}\right)\\ &=-\frac{4\pi G}{3}\rho_{\text{rad}}\\ &=-\beta\\ \Rightarrow\ -\frac{\dot{H}}{H^2+\beta}&=1\\ \therefore\ H(t)&=-\sqrt\beta\tan(\sqrt\beta t) \end{align*}

Does it mean that this universe will expand and contract from time to time? Did I do something wrong with the steps?

$\endgroup$

1 Answer 1

1
$\begingroup$

You have the hubble constant, but you have to integrate that to get the cosmological parameter:

$$\begin{align} \frac{\dot a}{a} &= - \sqrt{\beta}\tan \left(\sqrt{\beta}t\right)\\ \ln a &= \ln a_{0} + \ln \cos \left(\sqrt{\beta}t\right)\\ a &= a_{0}cos\left(\sqrt{\beta}t\right)\\ \end{align}$$

so, you have a big bang and big crunch at $t = \mp \frac{\pi}{2\sqrt{\beta}}$

which is consistent with the behaviour of a closed cosmology.

$\endgroup$
1
  • $\begingroup$ Thanks! I must have made a mistake in the interpretation of the Friedmann Equations. However, I cannot see the problem. $\endgroup$ Commented Apr 29, 2022 at 19:50

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.