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? 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).
\begin{align*}
F_g &= m\dot{v}\\
-\frac{Gm^2}{R^2} &= m\dot{v}\\
\dot{v} &= -\frac{Gm}{R^2}
\end{align*}
(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 $\frac{dR}{dt}$, cancel $dt$, and integrate both sides of your equation, This is an indefinite integral, so you will have a constant integration; combine these integration constants call their sum $C$. You should arrive at the following equation:
\begin{align*}
\frac{1}{2}\dot{R}^2 - \frac{GM}{R} = C
\end{align*}
Convince yourself the equation you've written down has units of energy per unit mass .
\begin{align*}
\dot{v} &= -\frac{Gm}{R^2}\\
\dot{v}v &= -\frac{Gm}{R^2}v\\
\dot{v}\frac{dR}{dt}&= -\frac{Gm}{R^2}\frac{dR}{dt}\\
\dot{v}dR &= -\frac{Gm}{R^2}dR\\
\int \ddot{Rdr} &= \int -\frac{Gm}{R^2}dR\\
\frac{1}{2}\dot{R}^2 + C_1 &= \frac{Gm}{R} + C_2\\
C &= \frac{1}{2}\dot{R}^2 - \frac{Gm}{R}
\end{align*}
(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 $\dot{R}$ is equal to $\frac{dR}{dt}$.
The total mass, $M$ in terms of density is:
\begin{align*}
M = \frac{4}{3}\pi R^3 \rho(t)
\end{align*}
We can plug in this mass, $M$, to the equation in part (b) as follows:
\begin{align*}
C &= \frac{1}{2}\dot{R}^2 - \frac{Gm}{R}\\
C &= \frac{1}{2}\dot{R}^2 - \frac{G\left(\frac{4}{3}\pi R^3 \rho(t)\right)}{R}\\
2C &= \dot{R}^2 - G\left(\frac{8}{3}\pi R^2 \rho(t)\right)\\
\dot{R}^2 &= 2C + G\left(\frac{8}{3}\pi R^2 \rho(t)\right)\\
\left(\frac{\dot{R}}{R}\right)^2 &= \frac{2C}{R^2} + \frac{8 G \pi \rho(t)}{3}
\end{align*}
(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 , $a(t)$.
We know that $R = a(t)r$. We also need to know what $\dot{R}$ is. In order to do that, we need to find the derivative of $R = a(t)r$, which is $\dot{R} = \dot{a}(t)r$. We can substitute these values for $R$ and $\dot{R}$ in the equation we derived in part (c) as follows:
\begin{align*}
\left(\frac{\dot{a}(t)r}{a(t)r}\right)^2 &= \frac{2C}{(a(t)r)^2} + \frac{8 G \pi \rho(t)}{3}
\end{align*}
(e) Rewrite the above expression so that $\left(\frac{\dot{a}}{a}\right)^2$ appears alone on the left side of the equation.
Rewriting the above equation in terms of $\left(\frac{\dot{a}}{a}\right)^2$, we get:
\begin{align*}
\left(\frac{\dot{a}}{a}\right)^2 &= \frac{2C}{a^2 r^2} + \frac{8 G \pi \rho(t)}{3}
\end{align*}
(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 expands or contracts; in other words, it tells us about the Hubble expansion (or contraction) rate of the universe.
Substituting $H(t) = \frac{\dot{a}}{a}$ and $\frac{2C}{r^2} = -kc^2$ to the equation in part (e), we get the first Friedmann Equation:
\begin{align*}
H^2(t) &= \frac{8 G \pi \rho(t)}{3} - \frac{kc^2}{a^2}
\end{align*}
(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 $\frac{\dot{a}}{a} = -\frac{4 \pi}{3}Gp$, which is known as the second Friedmann equation.
The more complete second Friedmann equation 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 follow the same basic steps as we did to get the first Friedmann equation, and start with the equation we derived in part (a):
\begin{align*}
\dot{v} &= -\frac{Gm}{R^2}
\end{align*}
We know that acceleration, represented as the first derivative of velocity, $\dot{v}$, can also be represented by the second derivative of distance, as $\ddot{r}$. Since we know that $R = a(t)r$, we can say that the second derivative of a distance can be represented as, $\ddot{r} = \ddot{a}(t)r$. Therefore :
\begin{align*}
\dot{v} = \ddot{r} = \ddot{a}(t)r
\end{align*}
This can be substituted in the equation from part (a) to get the equation:
\begin{align*}
\ddot{a}(t)r &= -\frac{Gm}{\left(a(t)r\right)^2}\\
\frac{\ddot{a}(t)r}{a(t)} &= -\frac{Gm}{r^2}\\
\frac{\ddot{a}}{a} &= -\frac{Gm}{r^3}\\
\end{align*}
Finally, we can substitute the mass for the mass density, $m = \frac{4}{3}\pi r^3 \rho(t)$ to derive the Second Friedmann Equation:
\begin{align*}
\frac{\ddot{a}}{a} &= -\frac{Gm}{r^3}\\
\frac{\ddot{a}}{a} &= -\frac{G\left(\frac{4}{3}\pi r^3 \rho(t)\right)}{r^3}\\
\frac{\ddot{a}}{a} &= -\frac{4}{3} G \pi \rho(t)
\end{align*}
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