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2.13: S15 Pressure and Energy Density Evolution SOLUTIONS

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Nov 8, 2022
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- 1.37: Redshifts
- Back Matter

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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 $\Omega_k \equiv -k/H_0^2$ .

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Exercise 15.1.1

Exercise 15.2.1

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