Skip to main content
Physics LibreTexts

S11. The Friedmann Equation - SOLUTIONS

Box \(\PageIndex{1}\)

Assume the universe is filled with a fluid with mass density \(\rho\), that is flowing in a manner consistent with Hubble's law. Consider a test particle of mass \(m\) a distance \(a(t)\ell\) away from the origin of the coordinate system, that moves along with the fluid; i.e., all of its motion relative to the origin is due to the changing of the scale factor. We take the origin of the coordinate system to be at rest.

Exercise 11.1.1: Express the test particle's kinetic energy as a function of \(\dot a, \ell\) and \(m\)

Solution \(\PageIndex{1.1}\)

K.E. is \(\frac{1}{2}mv^2\), where \(v = \dot a \ell\), so the test particle's kinetic energy is

\[\begin{equation*}
  \begin{aligned}
        \frac{1}{2}m\dot a^2 \ell ^2
  \end{aligned}
\end{equation*}\]

 

Exercise 11.1.2: Calculate the test particle's potential energy. Do so by considering just the mass interior to a sphere centered on the origin, with radius \(a(t)\ell\), as is justified for spherical mass distributions in Newtonian mechanics.

Solution \(\PageIndex{1.2}\)

P.E. is \(-\frac{GM(<d)m}{d}\), where by \(M(<d)\) we mean the mass contained in the sphere of radius \(d\), so \(M(<d) = \frac{4}{3}\pi d^3 \rho.\)

Therefore the test particle's potential energy is

\[\begin{equation*}
  \begin{aligned}
        -G \frac{4}{3} \pi \rho d^2 m
  \end{aligned}
\end{equation*}\]

 

Exercise 11.1.3: Add these two together and set them equal to a constant. Call the constant \(\kappa\).

Solution \(\PageIndex{1.3}\)

\[\begin{equation*}
  \begin{aligned}
        \frac{1}{2}m\dot a^2 \ell ^2 - G \frac{4}{3} \pi \rho d^2 m = \kappa
  \end{aligned}
\end{equation*}\]

 

Exercise 11.1.4: Manipulate your equation from Exercise 11.1.3 to get:

\[\begin{equation*}
  \begin{aligned}
        \left(\frac{\dot a}{a}\right)^2 = \frac{8\pi G \rho}{3} + \frac{2\kappa}{ml^2} \times \frac{1}{a^2}
  \end{aligned}
\end{equation*}\]

Solution \(\PageIndex{1.4}\)

Recall that \(d = a \ell \; \Longrightarrow \; \ell = \frac{d}{a}\), substituting this in and rearranging our equation we get

\[\begin{equation*}
  \begin{aligned}
        \frac{1}{2}md^2 \frac{\dot a^2}{a^2}  - \frac{1}{2}md^2 \frac{8\pi G\rho}{3} = \kappa
  \end{aligned}
\end{equation*}\]

dividing through by \(\frac{1}{2}md^2\) gives

\[\begin{equation*}
  \begin{aligned}
        \frac{\dot a^2}{a^2}  - \frac{8\pi G\rho}{3} = \frac{2\kappa}{m}\frac{1}{d^2}
  \end{aligned}
\end{equation*}\]

Then we substitute back in \(d = a \ell\) and solve for \(\big(\frac{\dot a}{a}\big)^2\):

\[\begin{equation*}
  \begin{aligned}
        \left(\frac{\dot a}{a}\right)^2 = \frac{8\pi G \rho}{3} + \frac{2\kappa}{ml^2} \times \frac{1}{a^2}
  \end{aligned}
\end{equation*}\]