(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 \[\frac{d^2\delta}{dt^2}+\frac{2\dot{a}}{a}\frac{d\delta}{dt} = 4πG\bar{\rho}\delta.\] 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 is comoving coordinates, only their amplitude changes. Namely this means that we can factorize \[\delta(x, t) = D(t)\bar{\delta}(x),\] where \(\bar{\delta}(x)\) 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.
We are given the equation in the form \[\frac{d^2\delta}{dt^2}+\frac{2\dot{a}}{a}\frac{d\delta}{dt} = 4πG\bar{\rho}\delta.\] We want to show that the equation for \(\delta\) can be written in the form \(\delta(x, t) = D(t)\bar{\delta}(x),\). To show this, we can use this equation as a substitution for \(\delta\) in above differential equation: \[\ddot{D}(t)\bar{\delta}(x) + \frac{2\dot{a}}{a}(\dot{D}(t)(\bar{\delta}(x)) = 4πG\bar{\rho}D(t)\bar{\delta}(x)\] This simplifies to: \[\ddot{D}(t) + \frac{2\dot{a}}{a}\dot{D}(t) = 4πG\bar{\rho}D(t),\] which is in the same form as our first equation.
(b) Now let us consider a matter dominated flat universe, so that \(\bar{\rho} = a^{-3}\rho_{c,0}\), where \(\rho_{c,0}\) is the critical density today, \(2H_0^2/8πG\) as in Worksheet 11.1 (aside: such a universe sometimes is called the Einstein-de Sitter model). Recall that the behaviour of the scale factor of this universe can be written \(a(t) = (3H_0t/2)^{2/3}\), 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 is \(D_+(t)\); another decreasing function in t, denoting it as \(D\_(t)\).
We are given that \(a(t) = (3H_0t/2)^{2/3}\), and thus we know that \(\dot{a}(t) = H_0 (\frac{3H_0t}{2})^{-1/3}\). We combine these to get the quantity: \[\frac{\dot{a}}{a} = \frac{H_0 (\frac{3H_0t}{2})^{-1/3}}{(3H_0t/2)^{2/3}}\] \[\frac{\dot{a}}{a} = \frac{2}{3t}\] We are also given that \(\bar{\rho} = a^{-3}\rho_{c,0}\). We can substitute in our expression for a above to get a new expression for \(\bar{\rho}\): \[\bar{\rho} = [3H_0t/2)^{2/3}]^{-3}(2H_0^2/8πG) = \frac{1}{6πGt^2}\] Now we can plug these two expressions (for \(\bar{\rho}\) and for \(\frac{\dot{a}}{a}\)) into our differential equation from part (a): \[\ddot{D}(t) + 2(\frac{2}{3t})\dot{D}(t) = \frac{4πGD(t)}{6πGt^2}\] \[\ddot{D}(t) + \frac{4}{3t}\dot{D}(t) = \frac{2}{3t}D(t)\] We can now use the substitution suggested in the hint above, \(D(t) \propto t^q\) to make a substitution for \(D(t), \dot{D}(t)\), and \(\ddot{D}(t)\): \[q(q=1)t^{q-2} + \frac{4}{3t}qt^{q-1} - \frac{2}{33t^2}t^q = 0\] This simplifies to: \[\frac{1}{3}t^{q-2}[3q62 + q - 2] = 0\] We can use the quadratic equation to solve for our two possible answers for q, \(q = \frac{2}{3}\) and \(q = -1\). We plug these bag into the ansatz given in the question to get: \[D_+(t) \propto t^{2/3}\] \[D\_(t)\propto \frac{1}{t}\] (c) Explain why the \(D_+\) component is generically the dominant one in structure formation, and show that in the Einstein-de Sitter model, \(D_+(t) \propto a(t).\)
As t increases over time, \(D\_(t)\) will get closer and closer to 0, as \(D\_(t)\propto \frac{1}{t}\), and thus the \(D_+\) component is generically the dominant one in structure formation.
In the Einstein-de Sitter model, we know that \(a(t) = (3H_0t/2)^{2/3}\), and thus \(a(t) \propto t^{2/3}\). Thus we have: \[D_+(t) \propto t^{2/3}\] \[a(t) \propto t^{2/3}\] \[D_+(t) \propto a(t)\]
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.
(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 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.
Closed case
\[r = A(1 - \text{cos }\eta)\] \[t = B(\eta - \text{sin }\eta)\]
In this plot, t is on the x-axis, and r(t) is on the y-axis. (Scaled to not include the multiplication factors of A and B). As the time increases, the particle at first moves outward some distance r, and then turns around and returns to a distance of r = 0.
Open case
\[r = A(\text{cosh }\eta - 1 \] \[t = B(\text{sinh }\eta - \eta)\]
The following two plots show the open case with with relatively large and small limits for \(\eta\), respectively, with t on the x-axis, and r(t) on the y-axis:
As time increases, the distance r from the center of the sphere continues to increase. The velocity of the particle is always positive, and thus the particle keeps expanding with some asymptotically positive velocity.
Flat case
\[r = A\eta^2/2\] \[t = B\eta^3/6\]
In this case, as time increases, while the distance of the particle continues to increase, the velocity of the particle decreases. Thus, the particle reaches an infinite radius but with a velocity that approaches zero.
Well done!
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