Cosmology 101 Part 1

We are bounded in a nutshell of Infinite space: Blog Post #28, Worksheet # 9.1, Problem #1: Cosmology 101 Part 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? 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).

(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 of integration; combine these integration constants call their sum C. You should arrive at the following equation  \[\frac{1}{2}\dot{R}^2 - \frac{GM}{R} = C\] Convince yourself the equation you’ve written down has units of energy per unit mass.

(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 \(\frac{dR}{dt}\)  .

(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, \(a(t)\) .

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

(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 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.

(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.

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.

(a) In any frame, there are basic equations that permit us to physically interpret the system. One is: \[ F = ma\] as per Newton’s second law of motion. This can be rewritten as \[F= m\dot{v}\] since the dot represents the derivative of the variable below it. Now with this, the force of gravity can be equated with it: \[F_g = \frac{GMM}{R^2} \] \[ m\dot{v}=- \frac{GMM}{R^2}\] and canceling out the appropriate terms, we’re left with \[\dot{v}=-\frac{GM}{R^2},\] the acceleration of the shell.

(b) Multiplying the above expression by v  on both sides, we have: \[v \cdot\dot{v}=-\frac{GM}{R^2} \cdot v ,\] which can be rewritten as: \[v \frac{dv}{dt}=-\frac{GM}{R^2} \frac{dR}{dt},\] and canceling out the dt’s and integrating both sides with respect to the correct limits, we have: \[\frac{v^2}{2} = \frac{GM}{R} + C ,  \] which is  \[\frac{1}{2}\dot{R}^2 - \frac{GM}{R} = C\] once rearranged.

(c) The total mass of this system can be expressed as the density multiplied by volume, such that: \[M = \rho V \] and V can also be rewritten for a spherical system as: \[ V = \frac{4}{3} \pi R^3 , \] and plugging both of these into the above equation, we get:  \[\frac{1}{2}\dot{R}^2 - \frac{GM}{R} = C, \] \[\frac{1}{2}\dot{R}^2 - \frac{G\rho\frac{4}{3} \pi R^3 }{R} = C, \] and by working with the algebra a bit, and taking into consideration that the C is a fixed value, we have: \[\frac{1}{2}\dot{R}^2  =G\rho\frac{4}{3} \pi R^2  + C, \]  \[\dot{R}^2  =G\rho\frac{8}{3} \pi R^2 + 2C, \] \[\left(\frac{\dot{R}}{R}\right)^2  =G\rho\frac{8}{3} \pi  + \frac{2C}{R^2}, \] which is what the question asked for.

(d) Now we simply rewrite the equation with a slight change to redefine R as something else. Here, \(R = a(t)r\), and as such: \[\left(\frac{\dot{a}(t)r}{a(t)r}\right)^2  =G\rho\frac{8}{3} \pi  + \frac{2C}{(a(t)r)^2}, \] what we do next is a bit more interesting.

(e) Now, we simplify the expression back into a more manageable system with the right side that we want: \[\left(\frac{\dot{a}}{a}\right)^2  =G\rho\frac{8}{3} \pi  + \frac{2C}{a(t)^2 r^2}. \]

(f) Next, it is simply a matter of following the instructions correctly, just substituting \( \frac{\dot{a}}{a}\) with H(t) and \(2C/r^2\)  with \( -kc^2\). So we have: \[H(t)^2  =\frac{8}{3}\pi G\rho - \frac{kc^2}{a^2}, \] which is the Newtonian mechanics version of the first Friedmann equation.

(g) For the Second Friedmann equation, it is just a matter of going back to the beginning of the previous derivation process we just did, and having: \[\dot{v}=-\frac{GM}{R^2},\] and at this moment plugging in the definition of mass we had established previously, \[ M =\rho \frac{4}{3} \pi R^3 , \] we know simplify down to: \[\dot{v}=-\frac{G\rho \frac{4}{3} \pi R^3 }{R^2},\]   \[\dot{v}= -\frac{4}{3} \pi G\rho R,\] and redefining R and a as we did beforehand, \[\ddot{a} (t) r = -\frac{4}{3} \pi G\rho a(t) r,\] and placing everything and simplifying a bit more, we get:  \[\frac{\ddot{a}}{a} = -\frac{4}{3} \pi G\rho,\] the other Newtonian mechanics version of the Friedmann equations.