Blog Post 28, Worksheet 9.1, Problem 1

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 \(\rho (t)\). 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?

We know from Newton's second law that the net force acting on an object is equal to its mass times it acceleration: \[F = m\dot{v}\] We know that the force is coming from gravity, so we have: \[F = -G\frac{Mm}{R^2} = m\dot{v}\] The small ms (mass of a particle) cancel, and we are left with: \[\dot{v} = \frac{-GM}{R^2}\]
(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 and call their sum C. Convince yourself the equation you've written down has units of energy per unit mass. 

From above, we have: \[\dot{v} = \frac{-GM}{R^2}\] Multiplying both sides by v, we get: \[\dot{v}v = \frac{-GM}{R^2}v\] We know that \(\dot{v} = \frac{dv}{dt}\) and \(v = \frac{dR}{dt}\). Substituting these in, we get: \[\frac{dV}{dt}v = \frac{-GM}{R^2}\frac{dR}{dt}\] \(\frac{1}{dt}\) cancels on both sides, and we are left with: \[v dv = \frac{-GM}{R^2} dR\] Now we can integrate both sides: \[\int v dv = \int \frac{-GM}{R^2} dR\] \[\frac{v^2}{2} = \frac{GM}{R} + C\] We know that \(v = \dot{R}^2\), so we can rewrite our equation as: \[\frac{1}{2}\dot{R}^2 - \frac{GM}{R} = C\]
(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 \((\frac{\dot{R}}{R})^2\), where \(\dot{R}\) is equal to \(\frac{dR}{dt}\).

We know that the total mass, M, is equal to \(\rho V\), where V is the total volume. As we are considering a sphere, \(V = \frac{4}{3}πR^3\). Plug this into our equation for C, and we get:\[ \frac{1}{2} \dot{R}^2 - \frac{G(\rho\frac{4}{3}πR^3)}{R} = C\] \[\frac{1}{2}\dot{R}^2- \frac{4G\rhoπR^2}{3} = C\] Rearranging the equation, we get: \[ (\frac{\dot{R}}{R})^2 = \frac{8G\rhoπ}{3} + \frac{2C}{R^2}\]
(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). 

Replacing R with a(t)r in our above equation, we get: \[ (\frac{\dot{R}}{a(t)r})^2 = \frac{8G\rhoπ}{3} + \frac{2C}{(a(t)r)^2}\] We know that \(\dot{R} = \frac{dR}{dt} = \frac{d(a(t)r)}{dt} = \dot{a}(t)r\), so we can substitute \(\dot{a}(t)r\) for \(\dot{R}\): \[ (\frac{\dot{a}(t)r}{a(t)r})^2 = \frac{8G\rhoπ}{3} + \frac{2C}{(a(t)r)^2}\]
(e) Rewrite the above expression so that \((\frac{\dot{a}}{a})^2) appears alone on the left side of the equation.

Both \(r^2\)s cancel on the left side of the equation, leaving us with: \[ (\frac{\dot{a}}{a})^2 = \frac{8G\rhoπ}{3} + \frac{2C}{(a(t)r)^2}\]

(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 \(2C/r^2 = -kc^2\) where k is the "curvature" parameter, you will get your first Friedmann equation. The Friedmann equation tells us about how the shell expands or contracts; in other words, it tells us about the Hubble expansion (or contraction) rate of the universe. 
Plugging in  \(H(t) = \frac{\dot{a}}{a}\) into our above equation, we get: \[ (H(t))^2 = \frac{8G\rhoπ}{3} + \frac{2C}{a(t)^2r^2}\] The constant \(2C/r^2 = -kc^2\) can be substituted in the term on the far right, leaving us with: \[ (H(t))^2 = \frac{8G\rhoπ}{3} + \frac{kc^2}{a(t)^2}\]
(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 \(\frac{\ddot{a}}{a} = \frac{-4π}{3}G\rho\), which is known as the second Friedmann equation. 

From part a, we know that \[\dot{v} = \frac{-GM}{R^2}\] We also know that \[M = \rho V = \rho (\frac{4}{3}πR^3)\] Plugging this in for M, we get: \[\dot{v} = -\frac{4}{3}G\rho πR = -\frac{4}{3}G\rho π a(t)r\] We can replace \(\dot{v}\) with \(\frac{d^2R}{dt^2}\). However, we know that \(\frac{d^2R}{dt^2} = \frac{d^2(a(t)r)}{dt^2} = \ddot{a}(t)r\). So we're left with: \[\ddot{a}(t)r = -\frac{4}{3}G\rho π a(t)r\]An r cancels on both sides, and we divide both sides by a(t) to get: \[\frac{\ddot{a}}{a} = \frac{-4Gπ\rho}{3}\]