Blog #28, Worksheet #9.1, Question 1, Matter-Only Universe

1) In this exercise, we will derive the first and second Friedmann equations of a homogeneous, isotropic and matter-only universe. We use the Newtonian approach. Consider a universe filled with matter which has a mass density \(\rho (t)\). Note that as the universe expands or contracts, the density of the matter changes with time, which is why it is a function of time t. Now consider a mass shell of radius R within this universe. The total mass of the matter enclosed by this shell is M. In the case we consider (homogeneous and isotropic universe), there is no shell crossing, so M is a constant.

(a) What is the acceleration of this shell? Express the acceleration as the time derivative of velocity, \(\dot{v}\) (pronounced v-dot) to avoid confusion with the scale factor a (which you learned about last week).

We start with the simple equation for force so that we can relate acceleration to constants in our system. We know that \[F = m \dot{v} = - \frac{GMm}{r^2}\] We then simply cancel m's and rearrange to solve \[\dot{v} = - \frac {GM}{r^2} \]

(b) To derive an energy equation, it is a common trick to multiply both sides of your acceleration equation by v. Turn your velocity, v, into dR/dt , cancel dt, and integrate both sides of your equation. This is an indefinite integral, so you will have a constant of integration; combine these integration constants call their sum C. 

Once we multiply both sides by v and cancel dt, we are left with \[\dot{v} dr = - \frac{GM}{r^2} dr \] and when we integrate both sides we get \[\frac{\dot{r}^2}{2} = \frac{GM}{r} + C\]Then we rearrange to get our desired final answer of \[\frac{\dot{r}^2}{2} - \frac{GM}{r} = C \]

(c) Express the total mass M using the mass density, and plug it into the above equation. Rearrange your equation to give an expression for \((\frac{\dot{R}}{R})^2\) where \(\dot{R}\) is equal to dR/dt.
We know that \[\rho = \frac{M}{\frac{4}{3} \pi r^3}\] which gives us \[M = \frac{4}{3} \pi r^3 \rho\]We then plug in to our equation from part b) and get \[\frac{\dot{r}}{2} - \frac{4}{3} \pi G \rho r^2 = C \] Finally we multiply through by 2 and divide by \(R^2\) to get the desired left side of the equation. Our final answer comes out to \[(\frac{\dot{r}}{r})^2 = \frac{8}{3} \pi G \rho + \frac{2C}{r^2}\]

(d) R is the physical radius of the sphere. It is often convenient to express R as R = a(t)r, where r is the comoving radius of the sphere. The comoving coordinate for a fixed shell remains constant in time. The time dependence of R is captured by the scale factor a(t). The comoving radius equals to the physical radius at the epoch when a(t) = 1. Rewrite your equation in terms of the comoving radius, R, and the scale factor, a(t).

(e) Rewrite the above expression so \((\frac{\dot{a}}{a})^2 appears alone on the left side of the equation.

This step only requires replacing the r's in part c) with a's to account for the comoving radius. 

\[(\frac{\dot{a}}{a})^2 = \frac{8}{3} \pi G \rho + \frac{2C}{(a(t)r)^2}\]

(f) Derive the first Friedmann Equation: From the previous worksheet, we know that \(H(t) = \frac{\dot{a}}{a} \). Plugging this relation into your above result and identifying the constant \(\frac{2C}{r^2} = -kc^2 \) where k is the “curvature” parameter, you will get the first Friedmann equation. The Friedmann equation tells us about how the shell of expands or contracts; in other words, it tells us about the Hubble expansion (or contracration) rate of the universe.

We know \[H(t) = \frac{\dot{a}}{a} \qquad and \qquad \frac{2C}{r^2} = -kc^2\]Plugging into our equation from part d) we get \[H(t) = \frac{8}{3} \pi G \rho - \frac{kc^2}{a^2}\]

(g) Derive the second Friedmann Equation: Now express the acceleration of the shell in terms of the density of the universe, and replace R with R = a(t)r. You should see that \(\frac{\ddot{a}}{a} = - \frac{4 \pi}{3} G \rho \) which is known as the second Friedmann equation.The more complete second Friedmann equation actually has another term involving the pressure following from Einstein’s general relativity (GR), which is not captured in the Newtonian derivation. If the matter is cold, its pressure is zero. Otherwise, if it is warm or hot, we will need to consider the effect of the pressure.

We begin with \[\dot{v} = \ddot{r} = \ddot{a}(t) r = - \frac{GM}{r^2}\]Then we substitute as we did earlier for density, replacing r with a(t)r and get \[\frac{-4 \pi G \rho a(t)r}{3} = \ddot{a}(t)r\]Finally we cancel r's and divide by a to get a final answer of \[-\frac{4}{3} \pi G \rho = \frac{\ddot{a}}{a}\]