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Physics LibreTexts

S13. The Evolution of Mass-Energy Density and a First Glance at the Contents of the Cosmos - SOLUTIONS

  • Page ID
    7786
  • Exercise 13.1.1

    Answer

    If the particles are not being created or destroyed then the number, \(N\) in some comoving volume is fixed. But the volume in the comoving volume is increasing as \(a^3\). The number density \(n = N/V\) is thus proportional to \(a^{-3}\).

    Exercise 13.2.1

    Answer

    Matter Case:

    \[\begin{equation*}
      \begin{aligned}
            \dot a^2 = a^2 \frac{8\pi G \rho_0}{3}a^{-3} = \frac{H_0^2}{a}
      \end{aligned}
    \end{equation*}\]

    Since a is increasing with time, \(H_0^2/a\) decreases with time, which means \(\dot a^2\) decreases with time.

    \(\Longrightarrow\) the magnitude of \(\dot a\) decreases with time, and since \(\dot a > 0\) that means \(\dot a\) decreases with time.

    More mathematically:

    \[\begin{equation*}
      \begin{aligned}
            \dot a = \frac{H_0}{a^{1/2}} \quad \Longrightarrow \quad \ddot a = -\frac{H_0}{2a^{3/2}}\dot a
      \end{aligned}
    \end{equation*}\]

    Since \(\dot a > 0\), it implies \(\ddot a < 0\).

    Radiation Case:

    \[\begin{equation*}
      \begin{aligned}
            \dot a^2 = a^2 \frac{8\pi G \rho_0}{3}a^{-4}
      \end{aligned}
    \end{equation*}\]

    \[\begin{equation*}
      \begin{aligned}
            \dot a = \frac{H_0}{a}
      \end{aligned}
    \end{equation*}\]

    \[\begin{equation*}
      \begin{aligned}
            \ddot a = -\frac{H_0}{a^2}\dot a
      \end{aligned}
    \end{equation*}\]

    Since \(\dot a > 0\) we see \(\ddot a < 0\).

    Exercise 13.2.2

    Answer

    \[\begin{equation*}
      \begin{aligned}
            \frac{\dot a^2}{a^2} = \frac{8\pi G \rho_0}{3}a^0 \quad \Longrightarrow \quad \dot a = H_0a
      \end{aligned}
    \end{equation*}\]

    Therefore, \(\ddot a = H_0\dot a > 0\).