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?


1 Answer 1


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.

  • $\begingroup$ 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

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