Blog Post 29, Worksheet 9.1, Problem 2

2. GR modification to Newtonian Friedmann Equation: 

In Question 1, you have derived the Friedmann Equation in a matter-only universe in the Newtonian approach. That is, you now have an equation that describes the rate of change of the size of the universe, should the universe be made of matter (this includes stars, gas, and dark matter) and nothing else. Of course, the universe is not quite so simple. In this question we'll introduce the full Friedmann equation which describes a universe that contains matter, radiation and/or dark energy. We will also see some correction terms to the Newtonian derivation. 

(a) The first Friedmann equation: \[(\frac{\dot{a}}{a})^2 = \frac{8π}{3}G\rho - \frac{kc^2}{a^2} + \frac{\Lambda}{3}\] The second Friedmann equation: \[\frac{\ddot{a}}{a} = -\frac{4πG}{3c^2}(\rho c^2 + 3P) + \frac{\Lambda}{3}\]
In these equations, \(\rho\) and P are the density and pressure of the content, respectively. k is the curvature parameter; k = -1, 0, 1 for open, flat, and closed universe, respectively. \(\Lambda\) is the cosmological constant. Note that in GR, not only density but also pressure are the sources of energy. 

Starting from these two equations, derive the third Friedmann equation, which governs the way average density in the universe changes with time: \[\dot{\rho}c^2 = -3\frac{\dot{a}}{a}(\rho c^2 + P)\] First, we multiply both sides of the first Friedmann equation by \(a^2\): \[\dot{a}^2 = \frac{8πG\rho}{3}a^2-kc^2+\frac{\Lambda}{3}a^2\] Next we take the time derivative of each side. This gives us: \[2\dot{a} \ddot{a} = \frac{8πG}{3}a^2 \dot{\rho} + \frac{16}{3}πG\rho a \dot{a} + \frac{2\Lambda}{3}a\dot{a}\] \[\ddot{a} = \frac{4πGa^2\dot{\rho}}{3\dot{a}} + \frac{8πG\rho a}{3} + \frac{\Lambda}{3}a\] Now we can plug this expression for \(\ddot{a}\) into the second Friedmann equation: \[\frac{4πGa\dot{\rho}}{3\dot{a}} + \frac{8πG\rho}{3} + \frac{\Lambda}{3} = -\frac{4πG}{3c^2}(\rho c^2 + 3P) + \frac{\Lambda}{3}\] Several terms cancel, and we are left with: \[\frac{\dot{\rho}a}{\dot{a}} + 2\rho = -\rho - \frac{3P}{c^2}\] We can now rearrange the equation to give us the third Friedmann equation: \[\dot{\rho}c^2 = -3\frac{\dot{a}}{a}(\rho c^2 + P)\]
(b) If the matter is cold, its pressure P = 0, and the cosmological constant \(\Lambda = 0\). Use the third Friedmann equation to derive the evolution of the density of the matter \(\rho\) as a function of the scale factor of the universe a. You can leave this equation int terms of \(\rho, \rho_0, a, \text{ and } a_0\), where \(\rho_0\) and \(a_0\) are current values of the mass density and scale factor. 

In the third Friedmann equation, P becomes zero, leaving us with: \[\dot{\rho}c^2 = -3\frac{\dot{a}}{a}(\rho c^2)\] A \(c^2\) cancels on both sides, giving us \[\frac{\dot{\rho}}{\rho} = -3\frac{\dot{a}}{a}\] Now we can integrate: \[\int_{\rho_0}^{\rho} \frac{\dot{\rho}}{\rho} = -3\int_{a_0}^{a} \frac{\dot{a}}{a}\] We know that \(\dot{\rho} = \frac{d\rho}{dt}\), and \(\dot{a} = \frac{da}{dt}\). If we substitute both of these in, the dts cancel, and we have: \[\int_{\rho_0}^{\rho} \frac{1}{\rho}d\rho = -3\int_{a_0}^{a} \frac{1}{a}da\] \[\text{ln}(\frac{\rho}{\rho_0}) = -3\text{ln}(\frac{a}{a_0})\] If we raise e to the power of both sides, we get: \[\frac{\rho}{\rho_0} = e^{-3\text{ln}(\frac{a}{a_0})} = (e^{\text{ln}(\frac{a}{a_0})})^{-3} = (\frac{a}{a_0})^{-3}\] This gives us: \[\frac{\rho}{\rho_0} = (\frac{a}{a_0})^{-3} = (\frac{a_0}{a})^3\] \[\rho = \frac{\rho_0 a_0^3}{a^3}\]
Using the relation between \(\rho\) and a that you just derived and the first Friedmann equation, derive the differential equation for the scale factor a for the matter dominated universe. Solve this differential equation to show that \(a(t) \propto t^{2/3}\). This is the characteristic expansion history of the universe if it is dominated by matter. (Hint: When solving this final differential equation, recall that at time t = 0, a = 0 (the Big Bang). At time \(t = t_0, a = a_0 = 1\) (present day).) 

From above, we know that \(\rho \propto a^{-3}\). The first Friedmann equation tells us that \[\dot{a}^2 = \frac{8π}{3}G\rho a^2 - kc^2 + \frac{\Lambda a^2}{3}\] We set k = 0 and \(\Lambda\) = 0, as per parts a and b, giving us: \[\dot{a}^2 = \frac{8π}{3}G\rho a^2\] We can plug in \(\rho = \frac{\rho_0 a_0^3}{a^3}\) to give us: \[\dot{a}^2 = \frac{8πG\rho_0 a_0^3}{3a}\] Take the square root of both sides to get: \[\dot{a} = (\frac{8πG\rho_0 a_0^3}{3})^{1/2}a^{-1/2}\] \[\dot{a} \propto a^{-1/2}\] We know that \(\dot{a} = \frac{da}{dt}\), so we can separate the variables to get: \[\frac{da}{dt} \propto a^{-1/2}\] \[a^{1/2}da \propto dt\] Now we integrate: \[\int a^{1/2}da \propto \int dt\] This gives us \(a^{3/2} \propto t\), or \(a \propto t^{2/3}\).

(c) Radiation dominated universe. Let us repeat the above exercise for a universe filled with radiation only. For radiation, \(P = \frac{1}{3}\rho c^2\) and \(\Lambda = 0\). Again, use the third Friedmann equation to see how the density of the radiation changes as a function of scale factor. 

We repeat the process above, except we replace P = 0 with \(P = \frac{1}{3}\rho c^2\). Friedmann's third equation becomes: \[\dot{\rho}c^2 = -3(\frac{\dot{a}}{a})(\frac{4}{3}\rho c^2)\] This simplifies to: \[\frac{\dot{\rho}}{\rho} = -4 \frac{\dot{a}}{a}\] Once again we substitute for \(\dot{\rho}\) and \(\dot{a}\) and integrate to get \[\rho = \frac{a_0 \rho_0}{a^4}\] \[\rho \propto a^{-4}\]
Again using the relation between \(\rho\) and a and the first Friedmann equation to show that \(a(t) \propto t^{1/2}\).

Plugging this relation into the first Friedmann equation, we get that \[\dot{a}^2 \propto \rho a^2\] \[\dot{a}^2 \propto a^{-2}\] \[\dot{a} \propto a^{-1}\] Substitute \(\dot{a} = \frac{da}{da}\): \[\frac{da}{dt} \propto a^{-1}\] \[ \int a da \propto \int dt\] \[a^2 \propto t\] \[a \propto t^{1/2}\]

(d) Cosmological constant/dark energy dominated universe. Imagine a universe dominated by the cosmological-constant-like term. Namely in the Friedmann equation, we can set \(\rho = 0\) and P = 0 and only keep \(\Lambda\) nonzero. 
As a digression, notice that we said "cosmological-constant-like" term. This is because the effect of the cosmological constant may be mimicked by a special content of the universe which has a negative pressure \(P = \rho c^2\). Check that the effect of this content on the right-hand-side of the third Friedmann equation is exactly like that of the cosmological constant. To be general we call this content the dark energy. How does the energy density of the dark energy change in time? 

Show that the scale factor of the cosmological-constant-dominated universe expands exponentially in time. What is the Hubble parameter of this universe?

This is the same process again as in (c) and (d), with P = 0, \(\rho\) = 0, \(\Lambda \neq 0\), which yields \[a \propto e^t\] The energy density of the dark energy decreases exponentially in time. The Hubble parameter is thus: \[H(t) = \frac{\dot{a}}{a} = \frac{e^t}{e^t} = 1\]
(e) Suppose the energy density of a universe at its very early time is dominated by half matter and half radiation. (This is in fact the case for our universe 13.7 billion years ago and only 60 thousand years after the Big Bang). As the universe keeps expanding, which content, radiation or matter, will become the dominant component? Why?

The scale factor for matter (\(a \propto t^{2/3}\)) is larger than the scale factor for radiation (\(a \propto t^{1/2}\)). As a result, as the universe keeps expanding, matter will become the dominant component.


(f) Suppose the energy density of a universe is dominated by similar amount of matter and dark energy. As the universe keeps expanding, which content, matter or the dark energy, will become the dominant component? Why? What is the fate of our universe?

Similar to above, the scale factor for dark energy (\(a \propto e^t\)) grows at a faster rate than the scale factor for matter (\(a \propto t^{2/3}\)), and thus as the universe keeps expanding, dark energy will become the dominant component.