7.5: The Friedmann Equation

Up until now, we haven't talked about how to determine the scale factor function $a(t)$. In 1922, Alexander Friedmann combined the FLRW metric with the Einstein Field equations and discovered a way to determine $a(t)$. We now call it the Friedmann Equation

$$H^2(t)=\frac{8\pi\rho_\text{tot}(t)}{3}-\frac{K}{a^2(t)}\label{eq:FriedmannEquation}$$

where the Hubble Parameter $H(t)$ is defined as

$$H(t)\equiv \frac{\dot{a}(t)}{a(t)}\label{eq:HubbleParameter}$$

and where $\rho_\text{tot}$ is the mass-energy density of the universe. The constant $K$ determines whether the geometry of the universe is spherical ( $K>0$), flat ( $K=0$), or saddle-like ( $K<0$).

Note

The "dot" over the $a(t)$ in Equation $\ref{eq:HubbleParameter}$ is a shorthand notation for a derivative with respect to time.

By setting $K=0$, we can determine an expression for the critical density $\rho_\text{crit}(t)$. A density greater than the critical density will produce a universe with a spherical geometry, and anything less will produce a universe with a saddle-like geometry.

Definition: Critical Density

The critical density is the density of the stuff in the universe that would produce a universe with flat geometry.

Setting $K=0$ in Equation $\ref{eq:FriedmannEquation}$ in solving for $\rho (t)$ yields

$$\rho_\text{crit}(t)=\frac{3H^2(t)}{8\pi}\label{eq:rhoCrit}.$$

As you can see, the critical density changes with time. The critical density today is written as

$$\rho_\text{crit,0}=\frac{3H_0^2}{8\pi}\label{eq:rhoCritToday}.$$

Exercise $\PageIndex{1}$

Use the conversion 1 pc = $3.1\times 10^{16}$ m and 1 kg = $7.42\times 10^{-28}$ m to determine the value of the critical density today in $\frac{\text{kg}}{\text{m}^3}$. How many hydrogen atoms per cubic meter is that equivalent to?

Answer

Let's start by converting the Hubble constant.

$$H_0=70\frac{\text{km/s}}{\text{Mpc}}\times \frac{1000\text{ m}}{1\text{ km}}\times \frac{1\text{ Mpc}}{10^6\text{ pc}}\times \frac{1\text{ pc}}{3.1\times 10^{16}\text{ m}}=2.26\times 10^{-18}\frac{1}{\text{s}} \nonumber$$

Then we square it.

$$H_0^2=5.1\times 10^{-36}\frac{1}{\text{s}^2}\nonumber$$

Now we need to figure out how to get this in units of $\frac{\text{kg}}{\text{m}^3}$. We can start by using the speed of light to convert seconds to meters.

$$H_0^2=5.1\times 10^{-36}\frac{1}{\text{s}^2}\times \left (\frac{1\text{ s}}{3\times 10^8\text{ m}}\right)^2=5.67\times 10^{-53}\frac{1}{\text{m}^2}\nonumber$$

Now we can write $\frac{1}{\text{m}^2}$ as $\frac{\text{m}}{\text{m}^3}$ and convert the top to kilograms.

$$H_0^2=5.67\times 10^{-53}\frac{\text{m}}{\text{m}^3}\times \frac{1\text{ kg}}{7.42\times 10^{-28}\text{ m}}=7.6\times 10^{-26}\frac{\text{kg}}{\text{m}^3} \nonumber$$

Therefore the critical density is

$$\rho_\text{crit,0}=\frac{3}{8\pi}\left(7.6\times 10^{-26}\frac{\text{kg}}{\text{m}^3}\right)=9.1\times 10^{-27}\frac{\text{kg}}{\text{m}^3}. \nonumber$$

The mass of a hydrogen atom is about $1.67\times 10^{-27}\text{ kg}$, so the critical density is equivalent to a little over five hydrogen atoms per cubic meter of space. That is a very tiny number, especially considering the fact that the density of everything around us is much bigger than that. On the other hand, space is really big. As we will find out later, it appears as the universe as a whole has a density extremely close to the critical density. That goes to show you just how much empty space there is in the universe.