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:
import numpy as np
import matplotlib
import matplotlib.pyplot as plt
#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")
#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")
#flat
t3=np.arange(0,2.*np.pi,0.1)
r3= (((6.*t3)**(2./3.))/2.)
plt.plot(t3,r3,label="Flat")
plt.legend(loc=2)
plt.xlabel('Scaled Time')
plt.ylabel('Scaled Growth of the
Radius of the Universe')
plt.show()
