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S15 Pressure and Energy Density Evolution SOLUTIONS

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    7882
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    Exercise 15.1.1

    Answer

    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

    Answer

    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 \(\Omega_k \equiv -k/H_0^2\). 

     


    This page titled S15 Pressure and Energy Density Evolution SOLUTIONS is shared under a CC BY 4.0 license and was authored, remixed, and/or curated by Lloyd Knox.