## Saturday, November 7, 2015

### 29 | Worksheet 9.1, Problem 2: General Relativity Modification to Newtonian Friedmann Equation

In the previous post, we derived the Friedmann equation in a matter-only universe using 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 full Friedmann equations follow from Einstein's general relativity (GR). Analogous to the equations that we derived previously, the full Friedmann equations express the expansion/contraction rate of the scale factor of the universe in terms of the properties of the content in the universe, such as the density, pressure, and cosmological constant. We will directly quote the equations below and study some important consequences.

The first Friedmann equation (1):
$\left(\frac{\dot{a}}{a}\right)^2 = \frac{8\pi}{3}G \rho - \frac{kc^2}{a^2} + \frac{\Lambda}{3}$
The second Friedmann equation (2):
$\frac{\ddot{a}}{a} = - \frac{4\pi G}{3c^2} (\rho c^2 + 3P) + \frac{\Lambda}{3}$
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 universes, 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.
$\dot{\rho}c^2 = -3 \frac{\dot{a}}{a}(\rho c^2 + P)$
To derive this equation, first multiply $$a^2$$ on both sides of (1) and then take the time derivative on both sides; plug (2) into your expression to eliminate $$\ddot{a}$$.

The first Friedmann equation is given by:
$\frac{\dot{a}^2}{a^2} = \frac{8\pi}{3} G \rho - \frac{kc^2}{a^2} + \frac{\Lambda}{3}$
If we multiply both sides by $$a^2$$, we get:
$\dot{a}^2 = \frac{8\pi}{3}G \rho a^2 - kc^2 + \frac{\Lambda a^2}{3}$
Now we can take the derivative with respect time time. In this expression, both $$a$$ and $$\rho$$ are functions of time. Therefore, the time derivative is:
$2\dot{a}\ddot{a} = \frac{8\pi}{3} G a^2 \dot{\rho} + \frac{8\pi}{3}\rho (2a\dot{a}) + \frac{\Lambda}{3}2a\dot{a}$
Notice the $$-kc^2$$ term drops out, as this has no time dependence. If we simplify and solve for $$\frac{\ddot{a}}{a}$$, we get:
$\frac{\ddot{a}}{a} = \frac{4\pi G \dot{\rho} a}{3\dot{a}} + \frac{8\pi}{3} G \rho + \frac{\Lambda}{3}$
The second Friedmann equation tells us:
$\frac{\ddot{a}}{a} = -\frac{4\pi G}{3c^2} (\rho c^2 + 3P) + \frac{\Lambda}{3}$
Since we now have two expressions for $$\ddot{a}/a$$, we can set them equal:
$\frac{4\pi G \dot{\rho} a}{3\dot{a}} + \frac{8\pi G \rho}{3} + \frac{\Lambda}{3} = - \frac{4\pi G}{3c^2}(\rho c^2 + 3P) + \frac{\Lambda}{3}$
Algebraic simplifcation and rearrangement to solve for $$\dot{\rho} c^2$$ gives:
$\boxed{\dot{\rho}c^2 = - \frac{3\dot{a}}{a}(\rho c^2 + P)}$

Now we can use these equations to derive some fun consequences of different kinds of universe, some of which describe our own universe.

For simplicity, in the exercise below, let us always set $$k = 0$$, namely consider a flat universe. Luckily for us, state-of-the-art observations suggest that our universe is likely flat.

(b) Cold matter dominated universe. 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 matter $$\rho$$ as a function of the scale factor of the universe $$a$$. You can leave this equation in terms of $$\rho$$, $$\rho_0$$, $$a$$, and $$a_0$$, where $$\rho_0$$ and $$a_0$$ are current values of the mass density and scale factor.

The third Friedmann equation says:
$\dot{\rho}c^2 = -\frac{3\dot{a}}{a} (\rho c^2 + P)$
In this cold universe, $$P = 0$$. This simplifies our expression:
$\dot{\rho} = -3\rho \frac{\dot{a}}{a}$
Dividing both sides by $$\rho$$ gives:
$\frac{\dot{\rho}}{\rho} = -3 \frac{\dot{a}}{a}$
Now we can integrate both sides. We will use the limits provided in the problem:  $$\rho$$, $$\rho_0$$, $$a$$, and $$a_0$$.
$\int_{\rho_0}^{\rho} \frac{\dot{\rho'}}{\rho'} = -3 \int_{a_0}^{a} \frac{\dot{a'}}{a'}$
The evaluated integral gives:
$\ln{\rho} - \ln{\rho_0} = -3(\ln{a} - \ln{a_0})$
Using log rules, we can write this as:
$\ln{\frac{\rho}{\rho_0}} = -3\ln{\frac{a}{a_0}}$
Exponentiating both sides, we get:
$\boxed{\frac{\rho}{\rho_0} = \left(\frac{a}{a_0}\right)^{-3}}$
We know that $$a_0$$ is the scale factor of the universe in the present day. By definition, $$a_0 =1$$, so we could rewrite this expression as:
$\boxed{\rho = \rho_0 a^{-3}}$

The result you got has the following simple interpretation. The cold matter behaves like "cosmological dust" and it is pressureless (not to be confused with warm/hot dust in the interstellar medium!). As the universe expands, the mass of each dust particle is fixed, but the number density of the dust is diluted - inversely proportionally to the volume.

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 the 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 differential equation, recall that at time $$t=0$$ (the Big Bang), $$a=0$$. At time $$t=t_0$$ (present day), $$a = a_0 =1$$.

The first Friedmann equation is:
$\frac{\dot{a}^2}{a^2} = \frac{8\pi}{3} G \rho - \frac{kc^2}{a^2} + \frac{\Lambda}{3}$
However, we will consider the universe flat, so $$k = 0$$. Also, we are considering a cold universe, such that $$\Lambda = 0$$. Therefore, our expression becomes:
$\left(\frac{\dot{a}}{a}\right)^2 = \frac{8\pi}{3}G \rho$
We just showed that $$\rho = \rho_0 a^{-3}$$. If we plug this in for $$\rho$$, we get:
$\left(\frac{\dot{a}}{a}\right)^2 = \frac{8\pi G \rho_0}{3a^3}$
If we solve for $$\dot{a}$$, we arrive at:
$\dot{a} = \left(\frac{8\pi G \rho_0}{3a}\right)^{\frac{1}{2}}$
Let's rewrite $$\dot{a}$$ as $$\mathrm{d}a/\mathrm{d}t$$:
$\frac{\mathrm{d}a}{\mathrm{d}t} = \left(\frac{8\pi G \rho_0}{3a}\right)^{\frac{1}{2}}$
Rearranging a few terms gives:
$a^{\frac{1}{2}}\mathrm{d}a = \left(\frac{8\pi G \rho_0}{3}\right)^{\frac{1}{2}} \mathrm{d}t$
We're only interested in a proportionality between $$a$$ and $$t$$, so we can drop the constants:
$a^{\frac{1}{2}}\mathrm{d}a \propto \mathrm{d}t$
Now we can integrate both sides:
$\int_{0}^{a} a'^{\frac{1}{2}}\mathrm{d}a' \propto \int_{0}^{t} \mathrm{d}t'$
$a^{\frac{3}{2}} \propto t$
So,
$\boxed{a \propto t^{\frac{2}{3}}}$

(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 the scale factor.

$\dot{\rho}c^2 = -\frac{3\dot{a}}{a} (\rho c^2 + P)$
In this radiation dominated universe, we will still say that $$\Lambda =0$$, but we now are concerned with a pressure term. We are told that for radiation, $$P = \frac{1}{3}\rho c^2$$. The third Friedmann equation becomes:
$\dot{\rho} = -4 \rho \frac{\dot{a}}{a}$
Like we did before, let's divide by $$\rho$$:
$\frac{\dot{\rho}}{\rho} = -4 \frac{\dot{a}}{a}$
Following the same integration steps we used above, this gives us the expression:
$\frac{\rho}{\rho_0} = \left(\frac{a}{a_0}\right)^{-4}$
Or, since $$a_0 = 1$$,
$\boxed{\rho = \rho_0 a^{-4}}$

The result also has a simple interpretation. Imaging the radiation being a collection of photons. Similar to the matter case, the number density of photons is diluted, inversely proportional to the volume. In contrast to the dust particle, each photon can be thought of as a wave. The wavelength of a photon is also stretched as the universe expands, proportional to the scale factor of the universe. According to quantum mechanics, the energy of each photon is inversely proportional to its wavelength: $$E = h \nu$$, unlike the dust case where each particle has a fixed energy. So, in an expanding universe, the energy of each photon is decreasing inversely proportional to the scale factor. Check that this is consistent with the result you got.

Our result is, indeed consistent with this energy consideration. We can see from the above equation that the mass density is inversely proportional to the scale factor. We also previously showed that the scale factor is directly proportional to the wavelength of the radiation. Since wavelength and frequency are inversely proportional by $$\nu = c/\lambda$$, then frequency must be inversely proportional to the scale factor. Furthermore, since energy is directly proportional to frequency, energy must also be inversely proportional to the scale factor.

Again, using the relation between $$\rho$$ and $$a$$ and the first Friedmann equation, show that $$a(t) \propto t^{1/2}$$ for the radiation-only universe.

Recall that our modified version of the first Friedmann equation is:
$\left(\frac{\dot{a}}{a}\right)^2 = \frac{8\pi}{3}G \rho$
As we did before, we can plug in the expression we found for $$\rho$$, which gives:
$\left(\frac{\dot{a}}{a}\right)^2 = \frac{8\pi G \rho_0}{3a^4}$
Following the exact same steps as we did for (b), we arrive at the following integral:
$\int a \mathrm{d}a \propto \int \mathrm{d}t$
So,
$a^2 \propto t$
Therefore, we conclude that:
$\boxed{a \propto t^{\frac{1}{2}}}$

(d) Cosmological constant/dark energy dominated universe: Imagine a universe dominated by the cosmological-constant-like term. Namiely 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 righ-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?

The third Friedmann equation tells us:
$\dot{\rho} c^2 = -3 \frac{\dot{a}}{a} (\rho c^2 + P)$
In this model of the universe, $$\rho = 0$$ and $$P=0$$. Therefore,
$\dot{\rho} = 0$
This suggest that the energy density of the dark energy does not 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?

When we allow $$P =0$$ and $$\rho=0$$, the first Friedmann equation is simply:
$\left(\frac{\dot{a}}{a}\right)^2 = \frac{\Lambda}{3}$
This means that the Hubble parameter of this universe is:
$\boxed{H = \left(\frac{\Lambda}{3}\right)^{\frac{1}{2}}}$
We can rewrite the first Friedmann equation as:
$\frac{\mathrm{d}a}{\mathrm{d}t} = \left(\frac{\Lambda}{3}\right)^{\frac{1}{2}}a$
Rearranging and dropping constants, we are left with:
$\frac{\mathrm{d}a}{a} \propto \mathrm{d}t$
Integrating both sides gives:
$\int \frac{\mathrm{d}a}{a} \propto \int \mathrm{d}t$
$\ln{a} \propto t$
Therefore, we arrive at the relationship that:
$\boxed{a \propto \mathrm{e}^t}$

(e) Suppose the energy density of a universe at its very early time is dominated by half matter and half radiation. (This is 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?

In the questions, above we derived relationships between the scale factor of the universe and the denisty. In a matter-dominated universe:
$\rho_m \propto a^{-3}$
$\rho_r \propto a^{-4}$
By these relations, we can see that the density of radiation decays quicker than the density of matter. Therefore, radiation will be "diluted" more rapidly. As a result, in this half matter/half radiation universe, matter will be the dominant component.

(f) Suppose the energy density of a universe is dominated by similar amounts of matter and dark energy. (This is the case for our universe today. Today our universe is roughly 68% in dark energy and 32% in matter, including 28% dark matter and 5% usual matter, which is why it is acceleratedly expanding today.) 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?

As mentioned above, a matter-dominated universe can be described by:
$\rho_m \propto a^{-3}$
However, we showed that the density of dark energy did not have a time dependence. Therefore, as the universe continues to expand, dark energy will become the dominant component. With this knowledge, we can consider the scale factor-time relation we found for a dark energy universe:
$a \propto \mathrm{e}^t$
This suggests that as we become a dark energy-dominated universe, the size of the universe will expand exponentially!

#### 1 comment:

1. Very good explanations in e/f/
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