Cosmology: Blog 28, Worksheet 9.1, Problem 1

A Matter-only Model of the Universe in Newtonian Approach 

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 ρ(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).

Because this acceleration is due to gravity, we know \[ ma  = F_g \] \[ m\dot{v} = - \frac{GMm}{R^2} \] \[ \dot{v} = - \frac{GM}{R^2} \]

(b) To derive an energy equation, it is a common trick to multiply both sides of your acceleration equation by v. You should arrive at the following equation: \[ \frac{1}{2} \dot{R}^2 - \frac{GM}{R} = C \]

Using what we found in a: \[ \frac{dR}{dt} \dot{v} = - \frac{GM}{R^2} \frac{dR}{dt} \] Which simplifies to \[v \: dv = - \frac{GM}{R^2} dR \] The integral of this is \[ \frac{1}{2} v^2 = \frac{GM}{R} + C \] Where C is a constant. Substituting \( v = \dot{R} \) we have: \[ \frac{1}{2} \dot{R}^2 - \frac{GM}{R} = C \]

(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 \( \left( \frac{ \dot{R}}{R} \right)^2 \), where \(  \dot{R} = \frac{dR}{dt}  \). \[ M = \rho V = \frac{4}{3} \pi R^3 \rho \] Plugging this into our expression from part B gives us:  \[ \frac{1}{2} \dot{R}^2 - \frac{G \left( \frac{4}{3} \pi R^3 \rho \right)}{R} = C \] Rearranging this makes it: \[ \left( \frac{\dot{R}}{R} \right)^2 = \frac{2C}{R^2} + \frac{8 \pi G}{3} \rho (t) \]

(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). \[  \frac{\dot{R}}{R} =  \frac{\dot{a}(t) r}{a(t) r} =  \frac{\dot{a}}{a} \] So our expression is  \[ \left( \frac{\dot{a}}{a} \right)^2 = \frac{2C}{R^2} + \frac{8 \pi G}{3} \rho (t) \]

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

(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 \( \frac{2C}{r} = \kappa c^2 \) 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. \[ \left( \frac{\dot{a}}{a} \right)^2 = \frac{8 \pi}{3} G \rho - \frac{ \kappa c^2 }{a^2} \]

(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 a: \( \frac{\ddot{a}}{a}  = - \frac{4 \pi G}{3c^2} \rho \), 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.
\[ \dot{v} = - \frac{GM}{R^2} = -\frac{4}{3} \pi G \rho (t) a(t) r \] We know \( \dot{v} = \ddot{R} \) so \(  \dot{v} = \ddot{a}r \) and \[\ddot{a}r = -\frac{4}{3} \pi G \rho (t) a(t) r \]  \[ \frac{\ddot{a}}{a}  = - \frac{4 \pi G}{3c^2} ( \rho c^2 + 3P) + \frac{ \Lambda }{3} \]