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Year 1 Sem 2/LC Introduction to Particle Physics and Cosmology/Lecture 09

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Lecture 9: Cosmology III: Friedmann Equation
Mon 02 Mar 2026 12:00

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Equation of Motion for the Universe

Goal: To derive an equation of motion for the dynamics of the Universe.

Assumptions:

  • We will assume Newtonian Gravity.
  • We will assume The Cosmological Principle holds on the scales we’re considering.
  • We will assume uniform expansion.
  • We will assume no dissipation. This means that total energy for a particle of mass $m$ will be conserved.
  • As a result of these assumptions, we wil be using comoving coordinates.

We’ll be using a non-vectorised form of the comoving coordinates:

$$\boxed{r(t) = a(t) x}$$

Assume at some time $t_1$ we have some spherical volume of radius $r(t_1)$, containing mass $M$. As time progresses and the universe expands, this radius will increase. This is given by some generic $r(t)$ and volume $V(t)$

As a result of our assumptions, we assume that we have uniform mass density given by $\rho(t)$ for some $t$:

$$\rho(t) = \frac{M}{V(t)}$$

Consider a particle of mass $m$ on the boundary of our initial $t_1$ universe. This particle will have some gravitational potential energy, and some kinetic energy as it moves on expansion:

$$\text{total energy } = \text{potential energy } + \text{ kinetic energy}$$
$$U = V + T$$

For potential energy:

$$V = \frac{-G Mm}{r(t)} = - \frac{4 \pi}{3} Gm \rho (t) r^2(t) \qquad \text{ assuming a spherical universe.}$$

For kinetic energy:

$$T = \frac{1}{2} m v^2 = \frac{1}{2}m \left(\frac{dr(t)}{dt}\right)^2 = \frac{1}{2} m \dot{r}^2$$

Hence for total energy:

$$U = T+V =\frac{1}{2} m \dot{r}^2 - \frac{4 \pi}{3} Gm \rho (t) r^2(t)$$

In doing this, we’ve had to implicitly assume that we can treat the mass $M$ distributed throughout the sphere as if it were a point mass at the centre of the sphere. This gives us an ordinary differential equation in terms of $r(t)$.

Comoving Coordinates

We now want to transform this into comoving coordinates.

$$\dot{r(t)} = \frac{d}{dt} \left(a(t) x\right) = \dot{a}(t) x$$
$$\implies U = \frac{-4 \pi}{3} Gm \rho(t) a^2(t) x^2 + \frac{1}{2}m \dot{a}^2(t) x^2$$
$$\frac{2 U}{ma^2(t) x^2} = \left(\frac{\dot{a}(t)}{a(t)}\right)^2 - \frac{8 \pi G}{3} \rho(t)$$

Let $\frac{2U}{mx^2} \equiv - kc^2$, hence1:

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

This is called the Friedmann Equation. This is now an ODE in terms of scale factor and not in terms of radius. Before we can solve it however, we need to deal with the $\rho(t)$ term.

Considering the second term, we have:

$$[k] = \frac{1}{L^2}$$

This $k$ has different implications on the geometry of the universe, and we consider 3 cases:

  • $k=0$: A flat universe.
  • $k>0$: A positive-curvature (closed) universe with spherical geometry, analogous to the surface of a sphere.
  • $k<0$: A negative-curvature (open) universe with hyperbolic geometry, analogous to the surface of a saddle.

Current observations suggest that our universe is very close to the $k=0$ case, so we will disregard the second term in first year cosmology.

The Fluid Equation

In order to solve the Friedmann Equation, we need to build something to describe $\rho (t)$, called the “fluid equation”. We will assume the following:

  • We will assume that the universe is a perfect/ideal fluid. This means that it is characterised by a density $\rho$ and an isotropic pressure $p$.
  • We will assume that the 1st Law of Thermodynamics holds:
    $$dE = - p dV + TdS$$
  • We will assume that the process is adiabatic, $dS = 0$.

Considering a comoving unit volume, where:

$$x = 1 \implies r(t) = a(t)$$

Hence:

$$V(t) = \frac{4 \pi}{3} a^3(t)$$

The internal energy of our fluid is given by:

$$E = mc^2 = \frac{4 \pi}{3} \rho(t) a^3(t) c^2$$
$$\frac{dE}{dt} = \frac{4 \pi}{3} c^2 \frac{d}{dt} \left(\rho(t) a^3(t)\right)$$
$$= \frac{4 \pi c^2}{3} \left(\dot{\rho}(t) a^3(t) +\rho(t) 3a^2(t) \dot{a}(t)\right)$$

And:

$$V(t) = \frac{4 \pi}{3} a^3(t) \implies \frac{dV}{dt} = 4 \pi a^2(t) \dot{a}(t)$$

Hence:

$$\frac{4 \pi c^2}{3} \left(\dot{\rho}(t) a^3(t) +\rho(t) 3a^2(t) \dot{a}(t)\right) + p 4 \pi a^2(t) \dot{a}(t) = 0$$
$$\dot{\rho}(t) + 3 \frac{\dot{a}(t)}{a(t)} \left(\rho (t) + \frac{p}{c^2}\right) = 0$$

This is now much easier to solve!

Footnotes

1.

This seems very random, but all we are doing is setting a constant equal to another constant, so it does work!

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