We are bounded in a nutshell of Infinite space: Just how stretched out can the universe get?

We are bounded in a nutshell of Infinite space: Blog Post #36, Worksheet # 12.1, Problem #1 & #2d: Just how stretched out can the universe get?

1. Linear perturbation theory. In this and the next exercise we study how small fluctuations in the initial condition of the universe evolve with time, using some basic fluid dynamics. In the early universe, the matter/radiation distribution of the universe is very homogeneous and isotropic. At any given time, let us denote the average density of the universe as \(\bar{\rho}(t)\). Nonetheless, there are some tiny fluctuations and not everywhere exactly the same. So let us define the density at comoving position r and time t as \(\rho (x,t)\) and the relative density contrast as \[\delta(r,t) = \frac{\rho(r,t) - \bar{\rho}(t)}{ \bar{\rho}(t)}\]. In this exercise we focus on the linear theory, namely, the density contrast in the problem remains small enough so we only need consider terms linear in \(\delta\). We assume that cold dark matter, which behaves like dust (that is, it is pressureless) dominates the content of the universe at the early epoch. The absence of pressure simplifies the fluid dynamics equations used to characterize the problem.

(a) In the linear theory, it turns out that the fluid equations simplify such that the density contrast \(\delta\) satisfies the following second-order differential equation \[\frac{d^2\delta}{dt^2} + \frac{2\dot{a}}{a} \frac{d\delta}{dt} = 4\pi G \bar{\rho}\delta,\] where \(a(t)\) is the scale factor of the universe. Notice that remarkably in the linear theory this equation does not contain spatial derivatives. Show that this means that the spatial shape of the density fluctuations is frozen in comoving coordinates, only their amplitude changes. Namely this means that we can factorize \[\delta(x,t) = D(t)\tilde{\delta}(x)\] , where \(\tilde{\delta}(x)\) is arbitrary and independent of time, and \(D(t)\) is a function of time and valid for all x. \(D(t)\) is not arbitrary and must satisfy a differential equation. Derive this differential equation.

(b) Now let us consider a matter dominated flat universe, so that \(\bar{\rho}(t) = a^{-3} \rho_{c,0}\) where \(\rho_{c,0}\) is the critical density today, \(3H_0 ^2/8\pi G\) as in Worksheet 11.1 (aside: such a universe sometimes is called the Einstein-de Sitter model). Recall that the behavior of the scale factor of this universe can be written \(a(t) = (3H_0 t /2)^{2/3}\) , which you learned in previous worksheets, and solve the differential equation for \(D(t)\). Hint: you can use the ansatz \(D(t) \propto t^q\) and plug it into the equation that you derived above; and you will end up with a quadratic equation for q. There are two solutions for q, and the general solution for D is a linear combination of two components: One gives you a growing function in t, denoting it as \(D_+ (t)\); another decreasing function in t, denoting it as \(D_- (t)\).

(c) Explain why the \(D)_+\) component is generically the dominant one in structure formation, and show that in the Einstein-de Sitter model, \(D_+ (t) \propto a(t)\).

(a) The key for this problem is solving to eliminate the x, which would make the differential equation true for all space and would change for a specific time. Thus, starting off with the original equation \[\frac{d^2\delta}{dt^2} + \frac{2\dot{a}}{a} \frac{d\delta}{dt} = 4\pi G \bar{\rho}\delta,\] and the equivalency of the space-time factor with its specific parts separated into time and space: \[\delta(x,t) = D(t)\tilde{\delta}(x)\] we just have to plug into the first equation and take the time derivative to solve: \[ \ddot{D}(t)\tilde{\delta}(x) + \frac{2\dot{a}}{a} \dot{D}(t)\tilde{\delta}(x)= 4\pi G \bar{\rho} D(t)\tilde{\delta}(x),\] and since we can show with this that the space dimension is not affected, the common factor on both sides can be eliminated: \[ \ddot{D}(t) + \frac{2\dot{a}}{a} \dot{D}(t) = 4\pi G \bar{\rho} D(t),\] and we thus have a differential equation valid for all x.

(b) First off, we need to establish the correct expressions for the scale factor and the critical density: \[ a(t) = \left(\frac{3H_0 t}{2}\right)^{2/3},\] and if we were to take the first derivative: \[ \dot{a}=\frac{2}{3} \left(\frac{3H_0 t}{2}\right)^{-1/3} \frac{3h_0}{2} \] \[\dot{a} =H_0 \left(\frac{3H_0 t}{2}\right)^{-1/3} ,\] and then divided by the original definition of the scale factor: \[\frac{\dot{a}}{a}= \frac{ H_0 \left(\frac{3H_0 t}{2}\right)^{-1/3}}{\left(\frac{3H_0 t}{2}\right)^{2/3}}\] \[\frac{\dot{a}}{a} = \frac{H_0}{\frac{3H_0 t}{2}}, \] we now have: \[\frac{\dot{a}}{a} = \frac{2}{3t}.\]

As for the critical density: \[\bar{\rho}(t) = a^{-3} \rho_{c,0},\] so the value of a can simply be put into the density expression: \[\bar{\rho}(t) = \left[\left(\frac{3H_0 t}{2}\right)^{2/3}\right]^{-3} \rho_{c,0}\] \[ \rho_{c,0} = 3H_0 ^2/8\pi G\] \[\bar{\rho}(t) = \frac{4}{9 H_0 ^2 t^2} \frac{3H_0 ^2 }{8\pi G} \] \[\bar{\rho}(t) = \frac{1}{6\pi G t^2}.\]

Taking the differential equation we solved for all space, we can place the values we have just found like: \[ \ddot{D}(t) + \frac{2\dot{a}}{a} \dot{D}(t) = 4\pi G \bar{\rho} D(t),\] \[ \ddot{D}(t) + \frac{2\dot{a}}{a} \dot{D}(t) = 4\pi G D(t) \frac{1}{6\pi G t^2},\] \[ \ddot{D}(t) + 2\left(\frac{2}{3t}\right) \dot{D}(t) = \frac{2 D(t)}{3 t^2},\] \[ \ddot{D}(t) + \frac{4}{3t} \dot{D}(t) - \frac{2}{3t^2} D(t) = 0 .\] And now that the equation has been simplified as much as it can be, we add in some squiggle math, knowing an important relationship between density with respect to time and time to a variable q:\[D(t) \propto t^q\] and taking the necessary derivatives, we have: \[ q(q-1) t^{q-2} + \frac{4}{3t} q t^{q-1}- \frac{2}{3t^2} t^q \sim 0 .\] \[ q(q-1) t^{q-2} + \frac{4}{3t} q t^{q-1}- \frac{2}{3t^2} t^q \sim 0 .\] \[ (q^2 - q) t^{q-2} + \frac{4}{3} q t^{q-2}- \frac{2}{3} t^{q-2} \sim 0 .\] So after simplifying the equation a bit, we can take out a couple of common factor and be left with a solvable polynomial: \[ t^{q-2} [(q^2 - q) + \frac{4}{3} q - \frac{2}{3t}] \sim 0 .\] \[ \frac{t^{q-2}}{3} [ 3q^2 + q - 2 \sim 0 ,\] and now we can use the quadratic formula to solve for q \[ q = \frac{-1 + \sqrt{1-4(3)(-2)}}{2(3)}\] \[ q = \frac{2}{3} ~,~ -1\] with these two values for q, they correspond to the increasing and decreasing functions the problem talks about, as: \[D_+ (t) \propto t^{2/3} ~,~ D_- (t) \propto \frac{1}{t} ,\] which means that a combination of these two expressions describes the entirety of the density with respect to time function: \[D(t) \approx D_+ (t) + D_- (t) .\]

(c) Because of the nature of \( D_+ (t) \) and its value we now know, it is clearly the dominant factor in establishing the development of density I the universe over time, for its growth is much more sustained and clear than that of \( D_- (t) \), which actually tends towards 0. However, the \(D_- (t)\) part of the function once was the dominant figure, at very low/small t, which makes it an important factor to consider. Furthermore, the \(D_+ (t)\) model closely resembles (it is actually identical to) the description of the expansion of the universe in a matter dominated universe, as we saw in a previous post (http://ay17-rcordova.blogspot.com/2015/11/cosmology-101-part-2.html), which indicates how a flat universe expands is identical to the scale factor for a matter dominated universe.

2. Spherical collapse. Gravitational instability makes initial small density contrasts grow in time. When the density perturbation grows large enough, the linear theory, such as the one presented in the above exercise, breaks down. Generically speaking, non-linear and non-perturbative evolution of the density contrast have to be dealt with in numerical calculations. We will look at some amazingly numerical results later in this worksheet. However, in some very special situations, analytical treatment is possible and provide some insights to some important natures of gravitational collapse. In this exercise we study such an example.

(d) Plot r as a function of t for all three cases (i.e. use y-axis for r and x-axis for t), and show that in the closed case, the particle turns around and collapse; in the open case, the particle keeps expanding with some asymptotically positive velocity; and in the flat case, the particle reaches an infinite radius but with a velocity that approaches zero.

(d)Taking key facts from the other parts of this problem, we know that for a closed universe, the equations for the distance from the origin r and the time that defines this are: \[r =A(1-\cos\eta),\] \[t = B(\eta - \sin\eta), ~~(0 \leq \eta \leq 2\pi),\] whereas for an open universe, the equations are: \[r = A(\cosh\eta - 1),\] \[t = B(\sinh\eta - \eta), ~~(0\leq \eta \leq \infty). \] Finally, for a flat universe, the equations become: \[ r = A\eta^2 / 2,\] \[t = B\eta^3 / 6, ~~ (0\leq \eta \leq \infty),\] where one can be expressed in terms of the other as: \[t = B\eta^3 / 6\] \[\eta = \frac{\sqrt[3]{6t}}{B}\] \[ r = A\eta^2 / 2\] \[ r = \frac{A \left(\frac{\sqrt[3]{6t}}{B}\right)^2}{ 2},\] \[ r = \frac{A\left(\frac{6t}{B}\right)^{2/3}}{2}.\] Knowing these values, limits and expressions, we can plot these into Python and create a good model to describe how particles act in these descriptions of the universe.

First, we just have to create the environment for the program:

import numpy as np

import matplotlib

import matplotlib.pyplot as plt

Next, we start describing each of the r and t and the \(\eta\) limits, for each system, while setting the

constants to 1 since these are only scale factor and do not directly influence the overall shape the graph takes:

#closed

n1=np.arange(0,2.*np.pi, 0.1)

r1=1.*(1.-np.cos(n1))

t1=1.*(n1-np.sin(n1))

plt.plot(t1,r1,label="Closed")

We do this process again with the open universe, first describing the limits (scaled here so they all fit in one graph at the end) and then the equations with the constants once again set to 1.

#open

n2=np.arange(0,2.7,0.1)

r2=1.*(np.cosh(n2)-1.)

t2=1.*(np.sinh(n2)-n2)

plt.plot(t2,r2,label="Open")

And here, for the flat case, we place the solved equation for r in terms of t we derived earlier and give t the limits as we did in the open universe description.

#flat

t3=np.arange(0,2.*np.pi,0.1)

r3= (((6.*t3)**(2./3.))/2.)

plt.plot(t3,r3,label="Flat")

Now with the final programs to define the plot with labels and legends:

plt.legend(loc=2)

plt.xlabel('Scaled Time')

plt.ylabel('Scaled Growth of the Radius of the Universe')

plt.show()

We have:

Full code used: