Matter dominated universe
"Leonard Susskind solves the expansion equation for universes with zero total energy, and then adds a non-zero total energy term, which leads to an exploration of matter versus radiation dominated universes. Originally presented in the Stanford Continuing Studies Program."\(^{[1]}\)
Figure 1: As you can see from the graph above, for the first roughly 10,000 years most of the energy density in the universe was due to the presence of radiation. During the time interval from when the universe was 10,000 years old to when it was several billion years old, the energy density of the universe was dominated by matter. In our present epoch, the energy density of the universe is dominated by dark energy (or vacuum energy). This graph has profound implications. As time progresses, dark energy will become more and more dominant and will remain the dominant source of energy in the universe; since dark energy causes everything in the universe to expand, the universe will continue to expand without ever stopping. This graph, essentially, implies the ultimate fate of the universe. The universe will continue to expand more and more until it becomes dark, empty, and lifeless.
Suppose that we lived in a Universe where there are only particles of mass \(m\) dispersed homogeneously and isotopically throughout space that are all at rest in the \(R\)-frame with coordinates \(x^i\). (Recall that this just means that none of the particles move along the spatial dimensions \(x^i\); all of their “motion” through spacetime is through the \(x^0\)-dimension or the time dimension). Even though the space may be expanding and the distances between the galaxies may be changing, their coordinates do not change because the coordinates \(x^i\) are “embedded” and “attached to” the galaxies. Since all of the galaxies of masses \(m\) are at rest in the coordinate system, the only form of energy present is the rest energy \(E=mc^2=m\) and potential energy \(PE=-MmG/D\) where \(M=ρ(4πD^3/3)\). For any mass \(m\) with coordinates \(x^i\) at a distance \(D\) away from the origin at \(x^i=0\), we can set \(PE=0\) at \(x^i\) and call that point the reference point. (By analogy, if a rock on the Earth’s surface is a distance \(r\) away from the Earth’s center with a potential energy of \(PE=-M_{Earth}m_{rock}G/r_{Earth}\), we can set \(PE=0\) and call that location where the rock is the reference point.) Thus, for any galaxy of mass \(m\) at a distance \(D\) from the origin, the only form of energy that galaxy has that we have to worry about is its rest energy \(E=mc^2=m\).
At some arbitrary point \(x^i\) where a galaxy may be, there will also be galaxies surrounding that point each with energy \(E=mc^2=m\). Imagine that we draw a box of volume \(V=a^3\) (so the edges are separated by \(∆x^1=1\), \(∆x^2=1\), and \(∆x^3=1\) ) centered around the point at \(x^i\). Suppose there are an \(N\) number of galaxies in that box. As the box expands or contracts (due to the expansion/contraction of space), the number of galaxies \(N\) in the box stays the same. The energy density \(ρ_M\) at the point \(x^i\) is defined as the ratio of the total energy \(E_{total}\) in the box and the volume \(V=a^3\) of the “unit box.” To find \(E_{total}\), we just take the sum of the energies of each of the galaxies (recall that the energy of each galaxy of mass \(m\) is \(E=mc^2=m\)) to get \(E_total=NE=Nm\). The energy density is just
$$ρ_M=\frac{E_{total}}{V}=\frac{NE}{a^3}=\frac{Nm}{a^3}=\frac{k}{a^3}.$$
Thus,
$$ρ_M∝\frac{1}{a^3}.$$
When we plug \(ρ_M\) into the FRW equation, we see that the FRW equation is a differential equation whose solution is the scaling factor \(a(t)\). Plugging in \(ρ_M\), we have
$$\frac{(da/dt)^2}{a^2}=\frac{8}{3}ρ_MπG=\frac{8MπG}{3a^3}.\tag{1}$$
(Note that \(Nm\) is just the total mass within the “unit cube." If we let \(M\) denote the total mass within this unit cube, then \(M=Nm\).) We will solve this differential equation for \(a(t)\) using the “guess and check” method in just a moment, but first I will just tell you what the solution is (it is \(a(t)=Ct^{2/3}\)) and make a few remarks about its physical interpretation. Basically, it’s just telling us that \(a\) is growing as the \(2/3^{rd}\) power of time; that is to say, the distances \(D=a(t)∆r\) are growing with time since \(a(t)\) is growing (recall that the “coordinate separation” \(∆r\) between galaxies stays the same). Just to reemphasize this point, it is the energy density \(T^{00}=ρ_M\) which determines the geometry \(G^{00}\) of the \(x^0\)-dimension (or time-dimension) of spacetime; it is this curvature which “tells” everything how to move through space (and time). In the equation
$$G^{00}=\frac{(da/dt)^2}{a^2}=\frac{8ρ_MπG}{3}=\frac{8πG}{3}T^{00},$$
the right-hand side of the EFE’s is where we specified how mass and energy is distributed throughout space as \(ρ_M\); this distribution of mass and energy tells spacetime to curve and become distorted in the particular way specified by \(G^{00}=\frac{(da/dt)^2}{a^2}\). This curved spacetime geometry is unusual in that the space doesn’t curve and is expanding while the time-dimension is curved.
But in a universe filled with only radiation, the distribution of energy is different and in this universe the energy density at each point is \(T^{00}=ρ_r\). The energy distribution in this Universe tells spacetime to become curved and distorted in the particular way specified by \(G^{00}=\frac{(da/dt)^2}{a^2}\) where \(\frac{(da/dt)^2}{a^2}\)(the square of Hubble's constant) is something completely different. (This is because \(a(t)\) is completely different in a universe filled with radiation. In this Universe, \(a(t)=Ct^{1/2}\).) In a Universe filled with only radiation, \(a(t)\) grows more slowly with time and, therefore, space is expanding more slowly with time.
Solving for the solution to the FRW equation in a matter dominated universe using the “guess and check” method
We guess that \(a(t)=Ct^p\). To solve this differential equation, we must find values for \(C\) and \(p\) (and, hence, we must find an \(a(t)\) ) such that the left-hand side equals the right-hand side. First, let’s plug \(a(t)=Ct^p\) into the right-hand side of the differential equation to obtain \((da/dt)^2/a^2=8MπG/3C^3t^{3p}\). Let’s now find \((da/dt)^2/a^2\) and then make the substitution:
$$\frac{da}{dt}=Cpt^{p-1}⇒\frac{da/dt}{a}=\frac{Cpt^{p-1}}{Ct^p}=pt^{-1}=\frac{p}{t}$$
$$\frac{(da/dt)^2}{a^2}=\frac{p^2}{t^2}=\frac{8MπG}{3C^3t^{3p}}$$
To solve this differential equation, we want the left-hand side to equal the right-hand side. The only way the left-hand side can equal the right-hand side is if both sides have a \(\frac{1}{t^2}\) term; another way of saying this is that \(p\) must equal 2/3. Then, when \(p=2/3\), the differential equation simplifies to
$$\frac{(2/3)^2}{t^2}=\frac{8MπG}{3C^3t^{3(2/3)}}=\frac{8MπG}{3C^3t^2}.$$
Dividing by \((2/3)^2\)and multiplying by \(t^2\) on both sides we get,
$$1=\frac{6πGM}{C^3}⇒C^3=6πGM⇒C=(6πGM)^{1/3}.$$
Thus \(p=2/3\) and \(C=(6πGM)^{1/3}\). The solution to the differential equation is
$$a(t)=(6πGM)^{1/3}t^{2/3}.$$
For fun, we might imagine that perhaps we lived in some other Universe where the distribution of mass and energy looked different and the energy density at each point is \(ρ\). Suppose that the energy density \(ρ\) was of a particular form such that when we plugged it into the FRW equation to solve for \(a(t)\), we obtained \(a(t)=\frac{1}{t}\). If we plotted \(\frac{1}{t}\) vs. \(t\), we would see that this particular \(a(t)\) corresponds to a Universe which is contracting. Each point in space is specified by coordinates \(x^i\). The distance between each point in space is shrinking with \(t\) because \(a(t)\) is shrinking with \(t\).
More general model
Instead of considering the two special cases of having two different “unit cubes,” one filled with static galaxies (modeled as particles) which remain at rest in the box and the other filled with photons (interestingly, wave packets such as photons can also be modeled as particles) which are “buzzing around” very rapidly in the box, we can consider these as two special cases of a more general model. In this more general model, we consider a box filled with particles which can be “moving around” at different speeds. They can be stationary (as they are in our model for the matter dominant Universe), they can be moving at the speed of light, or they can have slower speeds which fall somewhere in between those two extremes. In this more general model, we can think of the particles as having a temperature \(T\). The fact that \(T=0\) in the matter dominant case does not mean the galaxies and their stars are cold; it just means they are not moving around in the box. Now imagine that the sides of the box are walls which the particles can bounce off of. This allows us to define the pressure \(P\) that the particles exert on the walls of the box which is \(P=\frac{ρ}{w}\). Using the conservation of energy (\(dE=-PdV\)), we can derive the relationship
$$ρ=\frac{C}{a^{3(w+1)}}.\tag{2}$$
This equation describes how \(ρ\) is related to \(a(t)\) based on the value of \(w\).
This article is licensed under a CC BY-NC-SA 4.0 license.
References
1. Leonard Susskind. "Matter and radiation dominated universes". theoreticalminimum.com.