## Saturday, December 5, 2015

### 36 | Worksheet 12.1, Problems 1 and 2(d): Large Scale Structure

1. Linear perturbation theory: In this and the next exercise, we study how small fluctuations in the initial conditions 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 $$\vec{x}$$ and time $$t$$ as $$\rho(\vec{x}, t)$$ and the relative density contrast as
$\delta(\vec{x},t) \equiv \frac{\rho(\vec{x},t) - \bar{\rho}(t)}{\bar{\rho}(t)}$
In this exercise, we focus on the linear theory, namely, the density contrast in the problem remains small enough so we only need to 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
$\delta(\vec{x},t) = D(t)\tilde{\delta}$
where $$\tilde{\delta}(\vec{x})$$ is arbitrary and independent of time, and $$D(t)$$ is a function of time and valid for all $$\vec{x}$$. $$D(t)$$ is not arbitrary and must satisfy a differential equation. Derive this differential equation.

To do this, we need to show that the position-dependent term $$\tilde{\delta}$$ is an arbitrary term in determining the density contrast. We show that the density contrast is given by:
$\delta(\vec{x},t) = D(t) \tilde{\delta}(\vec{x})$
The differential equation is:
$\frac{d^2\delta}{dt^2} + \frac{2\dot{a}}{a}\frac{d\delta}{dt} = 4\pi G \bar{\rho} \delta$
Let's plug our expression for $$\delta$$ into this differential equation:
$\frac{d^2}{dt^2}\left(D(t)\tilde{\delta}(\vec{x})\right) + \frac{2\dot{a}}{a} \frac{d}{dt}\left(D(t) \tilde{\delta}(\vec{x})\right) = 4\pi G \bar{\rho} \left(D(t) \tilde{\delta}(\vec{x})\right)$
This gives us
$D''(t) \tilde{\delta}(\vec{x}) + \frac{2\dot{a}}{a} D'(t) \tilde{\delta}(\vec{x}) = 4\pi G \bar{\rho} D(t)\tilde{\delta}(\vec{x})$
We can see that all of the $$\tilde{\delta}(\vec{x})$$ terms cancel, giving us:
$D''(t) + \frac{2\dot{a}}{a} D'(t) = 4\pi G \bar{\rho} D(t)$
Since all of the position-dependent terms cancel out, we can see that the $$\tilde{\delta}(\vec{x})$$ term is arbitrary and the equation is dependent entirely on time.

(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, $$3H_0^2/8\pi G$$. Recall that the behavior of the scale factor of this universe can be written $$a(t) = (3H_0 t/2)^{2/3}$$. Solve the differential equation for $$D(t)$$. 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 $$D_-(t)$$.

We know that
$a = \left(\frac{3H_0 t}{2}\right)^{\frac{2}{3}}$
So, the time derivative of $$a$$ is:
$\dot{a} = \frac{2}{3} \left(\frac{3H_0}{2}\right)^{\frac{2}{3}} t^{-\frac{1}{3}}$
Furthermore, we know that:
$\bar{\rho} = a^{-3} \frac{3H_0^2}{8\pi G}$
If we plug these values into our differential equation:
$D''(t) + \frac{2\dot{a}}{a}D'(t) = 4\pi G \bar{\rho} D(t)$
and simplify, we get:
$D''(t) + \frac{4}{3t}D'(t) = \frac{2}{3t^2}D(t)$
Now we need to solve this differential equation. We will start by saying:
$D = kt^q$
The first- and second-order time derivatives of $$D$$ are,
$D'(t) = kqt^{q-1}$
$D''(t) = kq(q-1)t^{q-2}$
If we plug this into our differential equation, we get:
$kq(q-1)t^{q-2} + \frac{4}{3t}kqt^{q-1} = \frac{2}{3t^2}kt^q$
Note that by simplifying, we get:
$kq(q-1)t^{q-2} + \frac{4}{3}kqt^{q-2} = \frac{2}{3}kt^{q-2}$
Now we can cancel all $$k$$ and $$t$$, giving:
$q^2 + \frac{1}{3}q - \frac{2}{3} = 0$
If we factor this, we get find that
$q = -1,\frac{2}{3}$
As a result, we find that:
$\boxed{D_+ \propto t^{2/3} \, \, \, \, \, \, \, \, D_- \propto t^{-1}}$

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

In order for structure to form, by definition, the density contrast must increase. Since $$D(t)$$ represents the density contrast, the component that leads to an increased density contrast with time must be the dominant component of structure formation. Since $$D_+$$ represents an increasing density contrast, this is the dominant component of structure formation.

Earlier, we've shown that in a matter-dominated universe,
$a(t) \propto t^{2/3}$
And here, we have shown that
$D_+(t) \propto t^{2/3}$
Therefore,
$\boxed{D_+(t) \propto a(t)}$

2(d) 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.

Consider a spherical patch of uniform over-density. Let's study the motion of a particle in terms of its distance $$r$$ from the center of the sphere as a function of time $$t$$. Recall that from Newtonian dynamics, we have derived that this function satisfies the following equation:
$\frac{1}{2}\left(\frac{dr}{dt}\right)^2 -\frac{GM}{r} = C$
Where $$C$$ is a constant. It turns out that this differential equation can be solved analytically for constant $$C$$.

The parametric equation that solves this differential equation for the closed case is:
$r = A(1-\cos \eta)$
$t = B(\eta - \sin \eta)$
where $$\eta$$ is a function of $$t$$. $$A$$ and $$B$$ are positive constants satisfying the relationship $$A^3 = GMB^2$$.

For the open case, the solutions to this differential equation are:
$r = A(\cosh \eta - 1)$
$t = B(\sinh \eta - \eta)$

And finally, for the flat case, the solutions are:
$r = \frac{A}{2} \eta^2$
$t = \frac{B}{6} \eta^3$

Plot $$r$$ as a function of $$t$$ for all three cases, and show that in the closed case, the particle turns around and collapses; 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.

Here are the plots:
As you can see, the particle turns around and collapses with time in the closed case.

In the open case, the radius of the sphere appears to increase forever in what appears to be approximately linear with time.

Finally, the flat case also exhibits and infinite increase in the radius of the sphere, but the rate at which this distance increases steadily decreases over time. In other words, the velocity approaches zero with time.