2.13: S15 Pressure and Energy Density Evolution SOLUTIONS
Exercise 15.1.1
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The equation is
\[a\frac{d\rho}{da} = -3(P/c^2+\rho) \nonumber\]
Plugging in \(\rho \propto a^n\) we get \(n\rho = -3(P/c^2 + \rho)\). Solving for \(P\) for \(n=-3, -4, 0\) we find \(P=0\), \(P=\rho c^2/3\), and \(P = -\rho c^2\) respectively.
Exercise 15.2.1
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We have \(H^2 = 8\pi G \rho/3 - k/a^2\), \( \Omega_i \equiv \rho_{i,0}/\rho_c\), and the critical density today, \(\rho_c\) defined indirectly via \(H_0^2 = 8\pi G \rho_c/3\). Recall that \(\rho\) in the Friedmann equation is the total density so \(\rho = \Sigma_i \rho_i\).
Let's take the Friedmann equation, evaluated today (so \(H_0^2 = 8\pi G \rho_0/3 - k\) and divide each term by either \(H_0^2\) or \( 8\pi G\rho_c/3\). We can divide by either because they are equal. We get
\[1 = \Sigma_i \rho_{i,0}/\rho_c - k/H_0^2 = \Sigma_i \Omega_i + \Omega_k\]
if we also use the given definition of \(\Omega_k \equiv -k/H_0^2\).