Year 1 Sem 2/LC Introduction to Particle Physics and Cosmology/Lecture 11

Lecture 11: Cosmology V: Cosmological Models
Mon 09 Mar 2026 12:00

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Cosmological Models

Flat, Matter Only Universe Continued

We have the Friedmann equation:

$$\left(\frac{\dot{a}}{a}\right)^2 = \frac{8 \pi G}{3} \rho(t)$$

And using our equation of state for a matter only universe:

$$= \frac{8 \pi G}{3} \frac{\rho_0}{a^3}$$

$$\implies \dot{a}^2 = \frac{8 \pi G}{3} \rho_0 \frac{1}{a}$$

$$\implies \dot{a} = \sqrt{\frac{8 \pi G}{3} \rho_0} \sqrt{\frac{1}{a}}$$

And hence to get to $a(t)$ we can separate variables:

$$\int \sqrt{a} da = \sqrt{\frac{8 \pi G}{3} \rho_0} \int dt$$

$$\frac{2}{3}a^{3/2} + C = \sqrt{\frac{8 \pi G \rho_0}{3}} t$$

We need a boundary condition to determine the constant. We again say that the scale factor at $t=0$ is zero.

$$a(t = 0) = 0 \implies C = 0$$

$$a(t) = (6 \pi G \rho_0)^{1/3} t^{2/3}$$

And it is common to rewrite this using $a(t_0) \equiv a_0 = 1$. This gives us (by dividing through by $a_0$, as dividing by one changes nothing):

$$a(t) = \left(\frac{t}{t_0}\right)^{2/3}$$

This causes power law growth for all $t$.

Flat, Radiation Only Universe

Now, we keep $k=0$ but the equation of state is now $p = \rho c^2/3$. The fluid equation now gives us:

$$\dot{\rho} + \frac{3 \dot{a}}{a} \left(\rho + \frac{\rho c^2}{3 c^2}\right) = 0$$

$$\dot{\rho} + \frac{\dot{a}}{a} \left(4 \rho\right) = 0$$

This is almost directly equivalent to the matter only case, except the factor of 4:

$$\frac{1}{a^4} \frac{d}{dt} \left(\rho a^4\right) = 0$$

Hence:

$$\rho(t) = \frac{\rho_0}{a^4(t)}$$

This extra factor of $1/a$ causes an apparent loss of energy. This is due to redshift as the universe expands.

We can express $\rho$ as energy per unit volume:

$$\rho = \frac{E}{V} = \frac{hv}{V} = \frac{hc}{V \lambda}$$

This gives us the same $1/a^3$ due to volume dilution, but also the extra $1/a$ due to the $1/\lambda$ term as the wavelength of photons becomes stretched out. As before, we want to solve the Friedmann equation, so we substitute in our new $\rho$:

$$\left(\frac{\dot{a}}{a}\right)^2 = \frac{8 \pi G}{3} \frac{\rho_0}{a^4}$$

Following the same steps we get:

$$a(t) = \left(\frac{t}{t_0}\right)^{1/2}$$

This is still power law growth, but where the universe grows more slowly compared to the matter-only case.

Mixture Model

These two models are unrealistic for a true universe that contains both matter and radiation. We can form a more general $\rho(t)$ which is a sum of the two species.

We still assume a flat $k = 0$ universe. This gives us:

$$\left(\frac{\dot{a}}{a}\right)^2 = \frac{8 \pi G}{3} \left(\rho_m(t) + \rho_r(t)\right)$$

We can do this by considering limits, i.e. what happens when:

$r_m \ll r_r$
$r_m \gg r_r$

These are universes that are dominated by one species or the other, but not solely.

Case I: $\rho_r \gg \rho_m$

Since $\rho(t) = \rho_m(t) + \rho_r(t)$, in this limit it approximates to just the radiation component. When this happens, $a \propto \sqrt{t}$. In this phase where radiation dominates, we have:

$$\rho_r(t) = \frac{\rho_{r, 0}}{a^4(t)} \implies \rho_r(t) \propto \frac{1}{t^2}$$

And hence:

$$\rho_m(t) = \frac{\rho_{m, 0}}{a^3(t)} \implies \rho_m(t) \propto \frac{1}{t^{3/2}}$$

Case II: $\rho_m \gg \rho_r$

In this limit, the matter component dominates, and we have $a \propto t^{2/3}$. In this phase where matter dominates, we have:

$$\rho_m(t) = \frac{\rho_{m, 0}}{a^3(t)} \implies \rho_m(t) \propto \frac{1}{t^2}$$

And:

$$\rho_r(t) \propto \frac{1}{t^{8/3}}$$

Point of Equipartition

At some point in this evolution form a radiation to matter dominated world, there will be some crossover point where $\rho_m = \rho_r$. This is called the point of equipartition.

Curvature - Non-Flat Matter-Only Universe

We’ve already seen what happens with the $k=0$ case of a flat universe. When $k<0$ we have a hyperbolic geometry and $k>0$ spherical geometry (and continue to assume that we are only dealing with a matter only universe)

$$\left(\frac{\dot{a}}{a}\right)^2 = \frac{8 \pi G}{3} \rho_m(t) - \frac{kc^2}{a^2(t)}$$

Crucially, is the unbound nature of the expansion wrt time still true for these new geometries?

If $k<0$ (Hyperbolic):

Here, the negative sign of $k$ causes the final second term to be positive. This means that the expansion is still unbound. As:

$$H^2(t) = \left(\frac{\dot{a}}{a}\right)^2 > 0$$

This causes a larger magnitude of $H^2(t)$ hence the expansion is at a more rapid rate.

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