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202211221415 Exercise 29.4.1

Use this fact, that the difference between a vacuum energy density \rho_v and \rho_m or \rho_r is that \rho_v does not change as the universe expands (though it may change for other physical reasons), the basic Friedman equation Eq. (29.13):

\displaystyle{\dot{a}^2-\frac{8\pi G}{3}(\rho_m+\rho_r+\rho_v)a^2=K}

with K=0, and the assumption of vacuum dominance (\rho_m=\rho_r\approx 0) to show that Eq. (29.9):

a(t)=a(t_s)\exp \Big(\sqrt{\frac{8}{3}\pi G\rho_v}(t-t_s)\Big).

This form of the equation is valuable because it no longer refers to the present state of the universe.

Extracted from Thomas A. Moore. (2013). A General Relativity Workbook.

Roughwork.

This solution is not mine. It was found on the Internet some years ago, to whose author(s) I lost references.

\begin{aligned} 0 & = \dot{a}^2-\frac{8\pi G}{3}\rho_va^2\\ \bigg(\frac{\mathrm{d}a}{\mathrm{d}t}\bigg)^2 & = \frac{8\pi G}{3}\rho_va^2 \\ \frac{1}{a}\frac{\mathrm{d}a}{\mathrm{d}t} & = \sqrt{\frac{8\pi G}{3}\rho_v} \\ \ln \big( a(t)\big) & = \bigg(\sqrt{\frac{8\pi G}{3}\rho_v}\bigg)\cdot t + C \\ a(t)& =a(t_s)\exp \bigg(\sqrt{\frac{8}{3}\pi G\rho_v}(t-t_s)\bigg) \\ \end{aligned}