Blog #28: WS9.1, #1

1. In this exercise, we will derive the first and second Friedmann equations of a homogenous, 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's a function of time. 
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 (homogenous 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 \(\dot{v}\) to avoid confusion with the scale factor a. 

Since F=ma, we can also say that \(F=m\dot{v}\). The only force in this system is gravity, so we can equate this with the gravitational force.
\(\frac{-GMm}{R^2}=m\dot{v}\)
\(\dot{v}=\frac{-GM}{R^2}\)

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

\(\dot{v}\frac{dR}{dt}=\frac{-GM}{R^2}\frac{dR}{dt}\)
We can replace \(\dot{v}\) with \(\ddot{R}\) since velocity is the time derivative of the radius:
\(\int\ddot{R}d\dot{R}=-GM\int\frac{1}{R^2}dR\)
\(\frac{1}{2}\dot{R}^2+C=\frac{GM}{R}+C\)
\(\frac{1}{2}\dot{R}^2-\frac{GM}{R}=C\)

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 \(\frac{dR}{dt}\).

The mass of the sphere can be expressed as \(M=\frac{4}{3}\pi R^3\rho(t)\)
\(\frac{1}{2}\dot{R}^2-\frac{4G\pi R^2\rho(t)}{3}=C\)
\((\frac{\dot{R}}{R})^2=\frac{8G\pi R^2\rho(t)}{3}+\frac{2C}{R^2}\)

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 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). 

If \(R=a(t)r\), then \(\dot{R}=\dot{a}(t)r\).
\((\frac{\dot{a}(t)}{a(t)})^2=\frac{8G\pi\rho(t)}{3}+\frac{2C}{a^2(t)r^2}\)

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

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

f) Derive the first Friedmann Equation: 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 the 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. 

\(H^2(t)=\frac{8G\pi\rho(t)}{3}-\frac{kc^2}{a^2(t)}\)

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. 

\(\dot{v}=\frac{-GM}{R}=\frac{-4}{3}G\pi R\rho(t)\)
Since \(\dot{v}=\ddot{R}\) and \(\dot{R}=\dot{a}(t)r\), then \(\dot{v}=\ddot{a}(t)\).
\(\ddot{a}(t)=\frac{-4}{3}G\pi\rho(t)a(t)\)
\(\frac{\ddot{a}}{a}=\frac{-4}{3}G\pi\rho(t)\)

The more complete second Friedmann equation actually has another term involving the pressure following from Einstein's general relative, 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.