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.
WARNING: MATH AHEAD
(a) Find the shell's acceleration.
This is pretty simple. We will use Newton's gravity equation.
\[F_g=\frac{GMm}{R^2}\]
We know that \(\dot(v)=F/m\), so:
\[\dot{v}=\frac{GM}{R^2}\]
(b) Convert to energy.
We will use the quick trick of multiplying both sides by velocity and integrating.
\[\dot{v}v=\frac{GMv}{R^2}\]
\[vdv=\frac{GMdR}{R^2}\]
\[\frac{1}{2}v^2=\frac{GM}{R}+C\]
\[\frac{1}{2}\dot{R}^2-\frac{GM}{R}=C\]
(c) Express using mass density.
\[M=\frac{4}{3}\pi R^3 \rho\]
\[\frac{1}{2}\dot{R}^2-\frac{G(\frac{4}{3}\pi R^3 \rho)}{R}=C\]
\[\frac{1}{2}\dot{R}^2-\frac{G4\pi R^2 \rho}{3}=C\]
\[\left(\frac{\dot{R}}{R}\right)^2=\frac{8\pi G \rho}{3}+\frac{2C}{R^2}\]
(d,e) Express using R=a(t)r where r is the coming radius of the sphere.
\[\left(\frac{\dot{R}}{R}\right)^2=\frac{8\pi G \rho}{3}+\frac{2C}{R^2}\]
\[\left(\frac{\dot{a}}{a}\right)^2=\frac{8\pi G \rho}{3}+\frac{2C}{a^2 r^2}\]
(f) In the last worksheet we showed that \(H(t) = \frac{\dot{a}}{a}\). Express the First Friedmann Equation using this term. We can also turn \(2c/r^2\) into \(-kc^2\), a curvature term.
\[\left(\frac{\dot{a}}{a}\right)^2=\frac{8\pi G \rho}{3}+\frac{2C}{a^2 r^2}\]
\[\boxed{H^2=\frac{kc^2}{a^2}+\frac{8\pi G \rho}{3}}\]
(g) Derive the Second Friedmann Equation by expressing the acceleration of our initial shell in terms of the universal density.
From before we established:
\[\dot{v}=\frac{GM}{R^2}\]
\[\ddot{R}=\frac{4}{3}GM\pi R \rho\]
\[\boxed{\frac{\ddot{a}}{a}=-\frac{4}{3}\pi G \rho}\]
These Friedmann equations only apply to a universe that contains only matter without any pressure. The more extensive Friedmann equations come from General Relativity.
I worked with B. Brzycki, G. Grell, and N. James on this problem.
Super clear.
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