Energy versus rate of expansion YouTube

The matter-dominated universe

We can ask how energy density evolves over time.

Figure 9.3 - Density in co-moving coordinates

Figure 9.3 - Density in co-moving coordinates

Suppose, back in co-moving coordinates, that $M$ is the total amount of energy and matter in a cubic region of side $\Delta x$, then the density is the constant $$C = \frac{M}{\Delta x^3}$$

In physical coordinates, the volume of the region expands as $$\left(a(t) \Delta x\right)^3$$ whilst the mass (and energy) $M$ in the region remains fixed, thus the physical energy density is $$\rho(t) = \frac{M}{\left(a(t) \Delta x\right)^3} = \frac{M/\Delta x^3}{a(t)^3} = \frac{C}{a(t)^3} \rightarrow 0 \text{ as } t \rightarrow \infty$$

Thus, as expected, in physical coordinates, the energy density decreases as $\frac{1}{a(t)^3}$ over time.

We can compute $C$ if we have values for the density and scale factor at a given time. We can choose today say, which is the same as setting $t = T$, the age of the universe. Then, $$\rho(T) = \frac{C}{a(T)^3}$$ or $$C = \rho(T) a(T)^3$$ and the energy density becomes $$\rho(t) = \frac{\rho(T)a(T)^3}{a(t)^3}$$

The energy equation becomes $$\left(\frac{\dot{a}}{a(t)}\right)^2 = \frac{8 \pi G}{3} \frac{\rho(T)a(T)^3}{a(t)^3} = \frac{C'}{a(t)^3}$$ where $$C' = \frac{8 \pi G}{3} \rho(T) a(T)^3$$ is a constant. Multiplying both sides by $a(t)^2$ and taking the square root, we get $$\dot{a} = \sqrt{\frac{C'}{a}}$$

Separating variables and integrating, we get $$\int \sqrt{a} \mathrm{d}a = \int \sqrt{C'}\mathrm{d}t$$ or $$\frac{2}{3} a^{3/2} = {C'}^{1/2}t$$

Re-arranging, we recover the scale factor as a function of $t$, $$a(t) = \left(\frac{9}{4} C'\right)^{1/3} t^{2/3}$$

If we gather up the constants $$k = \left(\frac{9}{4} C'\right)^{1/3} = \left(6 \pi G \rho(T)\right)^{1/3} a(T)$$ then we see that the scale factor is proportional to $t^{2/3}$, that is, $$a(t) = k t^{2/3}$$

Further, $$H(t) = \frac{\dot{a}}{a} = \frac{\frac{2}{3} k t^{-1/3}}{k t^{2/3}} = \frac{2}{3} \frac{1}{t^{1/3} t^{2/3}}$$

Hence the Hubble constant changes according to the very simple formula, $$H(t) = \frac{2}{3t}$$ and since the time-dependence is now explicit, maybe we should start calling it the Hubble parameter.

Now we can use the correct Hubble parameter in the original argument, $V = HD$. Photons from a galaxy $X$ light-years away will have taken $X$ years to reach us, so we must use the Hubble parameter at time $t = T - X$, where $T$ is the age of the universe (now). That is, $$H = H\left(T - X\right) = \frac{2}{3\left(T - X\right)}$$

This model, called the matter-dominated universe, uses the assumption that there is a fixed amount of energy (and mass) in any co-moving volume, meaning that the physical density of energy in that volume decreases over time.

We can make different assumptions about the dominant form of energy in the universe, and see how these lead to different scale factors.

The radiation-dominated universe

Suppose, for argument's sake, we consider that the universe only contained photons, and no other type of matter or energy, which we could call a radiation-dominated universe. Since photons have no mass at all, it turns out that we must calculate the total energy (and mass), $M$, in a given volume, $\Delta x^3$, in a different way to the matter-dominated model.

The energy associated with a single photon is given by a famous equation in physics, $$E = \frac{h}{\lambda}$$ where $\lambda$ is called the wavelength associated with the photon.

We now assume the wavelengths of the photons all have the same fixed wavelength, as measured in the co-moving frame.

Nb. We discuss why this assumption might be true in the next section, but for now, note that we are not talking about photons in general, but those from what is called the sphere of last scattering.

Since the wavelength of a given photon is a fixed length in co-moving coordinates, it follows that, in physical coordinates, it must stretch out over time, $$\lambda(t) = a(t) \lambda$$ and so the energy of a single photon, in physical coordinates, must decrease over time $$E = \frac{h}{a(t) \lambda}$$

Suppose there are a fixed number, $N$, of photons in the our volume $\Delta x^3$, and they all have a common wavelength, $\lambda$ (as defined in the co-moving coordinates), then in physical coordinates then, the total energy is just $$M = \frac{Nh/\lambda}{a(t)}$$ where $\frac{Nh}{\lambda}$ is constant. Now when we calculate the energy density, we find that $$\rho(t) = \frac{M}{\left(a(t) \Delta x\right)^3} = \frac{Nh/\lambda}{a(t)} \frac{1}{\left(a(t) \Delta x\right)^3} = \frac{C}{a(t)^4}$$ where, again, $C = \frac{Nh}{\lambda \Delta x^3}$ is constant.

That is, the energy density in a radiation-dominated universe has one more factor of $a(t)$ in the denominator than it has in a matter-dominated universe.

The energy equation becomes $$\frac{\dot{a}}{a} = \frac{\sqrt{8 \pi G C}}{a^2}$$ and on separating out the variables and integrating, we get $$\frac{1}{2} a^2 = \int a \mathrm{d}a = \int \left(8 \pi G C\right)^{1/2} \mathrm{d}t = \left(8 \pi G C\right)^{1/2} t$$ or $$a(t) = k t^{1/2}$$ where $k = \left(32 \pi G C\right)^{1/4}$ is constant.

Thus we have now shown that, proportionately at least, the rate of expansion in a radiation-dominated universe ($a(t) \propto t^{1/2}$) is less than in a matter-dominated universe ($a(t) \propto t^{2/3}$).

There is good evidence to say that the early universe was radiation-dominated. To see why this might be true, note that as we go back in time and the universe gets smaller, whilst the energy associated with a photon will increase (due to its wavelength shortening), the mass of a given atom remains fixed and the total vacuum energy decreases (as we shall see next).

Hence, for a small enough universe, the energy from the photons - radiation - will dominate all other forms of mass/energy.

The vacuum-energy-dominated universe

Vacuum energy, also called dark energy, is defined (for our purposes) to be that component of the total energy in the universe that does not dilute over time, in that the density of vacuum energy remains fixed over time and space, $$\rho = \rho_\text{vac}$$

In other words, vacuum energy is proportional to the size of the universe, at any given time. In the early, very small, universe, it would be negligible, but as the universe expands, vacuum energy increases proportionately, and eventually dominates all other forms of mass and energy.

Plugging this into the energy equation, we get $$\frac{\dot{a}}{a} = \sqrt{8 \pi G \rho_\text{vac}}$$

On separating variables and integrating this time, we find $$log_e a = \int \frac{\mathrm{d}a}{a} = \int k \mathrm{d}t = kt$$ where $k = \sqrt{8 \pi G \rho_\text{vac}}$ is a constant, which leads to the solution, $$a(t) = e^{kt}$$

This means that in a vacuum-energy-dominated universe, the rate of expansion of the universe is exponential.

Current observations tell us that the universe is expanding exponentially. Roughly speaking, as a gauge, it will take approximately $10$ billion years for the universe to double in size.

This rate of expansion implies that approximately $70$% of all known matter and energy in the universe is vacuum energy.