# 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*}\]