4 The Friedmann Equations

Our discussion previously has laid the groundwork for describing the expansion of the Universe. We can now turn to the dynamics of that expansion - how that expansion is sourced by the various components of the Universe.

The complete derivation of these equations follows from assuming the FLRW metric and a homogeneous stress-energy tensor and substituting these into Einstein’s equations. The calculation is straightforward, albeit tedious. However, since we will largely not need that machinery in this class, we will just state the resulting equations and then proceed.

It should be noted that in much of what follows, $c=1$.

4.1 Introducing the Friedmann Equations

The first Friedmann eqn is \[ \left(\frac{\dot{a}}{a}\right)^{2} = \frac{8 \pi G}{3} \rho(t) - \frac{k}{R_0^{2} a^{2}} \] where $\rho(t)$ is the time-dependent energy density of the Universe (including all components - matter, radiation), $k$ is the curvature constant that we defined and $R_{0}$ the radius of curvature today.

The second Friedmann eqn is \[ \frac{\ddot{a}}{a} = -\frac{4\pi G}{3}(\rho(t) + 3P(t)) \] where $P$ is the pressure of the matter component.

Finally, we have the continuity equation \[ \dot{\rho} + 3 H (\rho + P) = 0 \]

Non-independence

These three equations are not independent. To see this, start with the first Friedmann equation and take a time derivative \[ \begin{aligned} \frac{d}{dt} \left(\frac{\dot{a}}{a}\right)^{2} = 2 \left( \frac{\dot{a}}{a} \right) \left[ \frac{\ddot{a}}{a} - \frac{\dot{a}^{2}}{a^{2}}\right] \\ = -2 \left( \frac{\dot{a}}{a} \right) \left(\frac{12\pi G}{3} (\rho + P) - \frac{k}{R_0^{2} a^{2}} \right) \\ = \frac{8 \pi G}{3} \dot{\rho} + 2 \frac{k \dot{{a}}}{R_0^{2} a^{3}} \\ \end{aligned} \] which just yields the continuity equation.

Newtonian Derivation

While the complete derivation of the first Friedmann and continuity equations requires GR, we can gain intuition from a partial Newtonian derivation of these. We leave the second Friedmann equation without explicit derivation.

Friedmann I

Consider a sphere of pressureless matter of radius $R_{s}(t)$ and energy/mass density $\rho(t)$. The mass inside is \[ M(t) = \rho(t) \frac{4}{3} \pi R_{s}^{3}(t) \] The acceleration of this sphere is given by \[ \frac{d^{2}R_{s}}{dt^{2}} = - G \frac{M(t)}{R_{s}^{2}} \] Multiplying by $\dot{R}_{s}$ and integrating, we have \[ \frac{1}{2} \left( \frac{d}{dt} R_{s} \right)^{2} = \frac{GM}{R_{s}} + E \] where $E$ is the integration constant. Now, writing $R_{s} = a r_{s}$ where $r_{s}$ is the comoving radius (and therefore constant in time), we get \[ \frac{1}{2} r_{s}^2 \dot{a}^{2} = \frac{4}{3} \pi G \rho a^{2} + E \] or \[ \left( \frac{\dot{{a}}}{a} \right)^{2} = \frac{8}{3}\pi G \rho + \frac{2E}{r_{s}^{2}} \frac{1}{a^{2}} \] which clearly has the form of the Friedmann equation. Note that in the relativistic version, $\rho$ includes all sources of matter/energy. The integration constant here gets identified with the curvature:

$E>0$, $k=-1$, Universe expands forever
$E<0$, $k=1$, maximum size
$E=0$, $k=0$, stops in the far future. Note that these conclusions are in the absence of dark energy, which changes this discussion.

Why does this work at all? Birkhoff’s theorem helps here.

Continuity Equation

Start from the first law of thermodynamics \[ dQ = dE + P dV \] If we consider our previous sphere and adiabatic expansion ($dQ=0$), this gives us \[ \dot{E} + P \dot{V} = 0 \] Now \[ \begin{aligned} E = \rho V \\ \dot{E} = \dot{\rho} V + \rho \dot{V} \\ \dot{\rho} V + \dot{V}(\rho + P) = 0 \\ V = \frac{4}{3} \pi r_{s}^{3} a^{3} \implies \dot{V} = 3 \frac{\dot{a}}{a} V = 3 H V \\ \dot{\rho} + 3 H (\rho + P) = 0 \end{aligned} \]

4.2 The Critical Density

Consider the first Friedmann eqn
\[ \left(\frac{\dot{a}}{a}\right)^{2} = H^2 = \frac{8 \pi G}{3} \rho(t) - \frac{k}{R_0^{2} a^{2}} \] Dividing by $H^{2}$, we have \[ 1 = \frac{8 \pi G}{3 H^2} \rho(t) - \frac{k}{R_0^{2} a^{2} H^{2}} \] Notice that when \[ \rho = \frac{3H^{2}}{8 \pi G} \] the curvature $k$ is zero. We therefore define a critical density \[ \rho_\text{crit} = \frac{3H^{2}}{8 \pi G} \] Note that the value of the critical density depends on $H$ and therefore varies with time. The value today is \[ \begin{aligned} \rho _\text{crit} &= 1.877 \times 10^{-26} h^{2} \text{kg}\,\text{m}^{-3} \\ &= 2.775 \times 10^{11} h^{2} M_{\odot} \text{Mpc}^{-3} \\ &= 1.053 \times 10^{10} \text{eV}^{4} \end{aligned} \]

We can measure the density relative to the critical density \[ \Omega = \frac{\rho}{\rho _\text{crit}} \] Here, we’re keeping $\Omega$ to represent all energy density, but very soon, we will write down these $\Omega$’s as a function of the type - radiation, matter, dark energy.

Plugging back into the Friedmann equation, we find \[ 1 - \Omega(t) = -\frac{k}{R_{0}^{2}a^{2}H^{2}} \] Note that the sign of $1-\Omega$ is fixed and doesn’t change with time. In particular, if $k=0$, $\Omega(t)=1$.

# The Friedmann Equations Our discussion previously has laid the groundwork for describing the expansion of the Universe. We can now turn to the dynamics of that expansion - how that expansion is sourced by the various components of the Universe. The complete derivation of these equations follows from assuming the FLRW metric and a homogeneous stress-energy tensor and substituting these into Einstein's equations. The calculation is straightforward, albeit tedious. However, since we will largely not need that machinery in this class, we will just state the resulting equations and then proceed. It should be noted that in much of what follows, $c=1$. ## Introducing the Friedmann Equations The first Friedmann eqn is $$ \left(\frac{\dot{a}}{a}\right)^{2} = \frac{8 \pi G}{3} \rho(t) - \frac{k}{R_0^{2} a^{2}} $$ where $\rho(t)$ is the time-dependent energy density of the Universe (including all components - matter, radiation), $k$ is the curvature constant that we defined and $R_{0}$ the radius of curvature today. The second Friedmann eqn is $$ \frac{\ddot{a}}{a} = -\frac{4\pi G}{3}(\rho(t) + 3P(t)) $$ where $P$ is the pressure of the matter component. Finally, we have the continuity equation $$ \dot{\rho} + 3 H (\rho + P) = 0 $$ ### Non-independence These three equations are not independent. To see this, start with the first Friedmann equation and take a time derivative $$ \begin{aligned} \frac{d}{dt} \left(\frac{\dot{a}}{a}\right)^{2} = 2 \left( \frac{\dot{a}}{a} \right) \left[ \frac{\ddot{a}}{a} - \frac{\dot{a}^{2}}{a^{2}}\right] \\ = -2 \left( \frac{\dot{a}}{a} \right) \left(\frac{12\pi G}{3} (\rho + P) - \frac{k}{R_0^{2} a^{2}} \right) \\ = \frac{8 \pi G}{3} \dot{\rho} + 2 \frac{k \dot{{a}}}{R_0^{2} a^{3}} \\ \end{aligned} $$ which just yields the continuity equation. ### Newtonian Derivation While the complete derivation of the first Friedmann and continuity equations requires GR, we can gain intuition from a partial Newtonian derivation of these. We leave the second Friedmann equation without explicit derivation. **Friedmann I** Consider a sphere of pressureless matter of radius $R_{s}(t)$ and energy/mass density $\rho(t)$. The mass inside is $$ M(t) = \rho(t) \frac{4}{3} \pi R_{s}^{3}(t) $$ The acceleration of this sphere is given by $$ \frac{d^{2}R_{s}}{dt^{2}} = - G \frac{M(t)}{R_{s}^{2}} $$ Multiplying by $\dot{R}_{s}$ and integrating, we have $$ \frac{1}{2} \left( \frac{d}{dt} R_{s} \right)^{2} = \frac{GM}{R_{s}} + E $$ where $E$ is the integration constant. Now, writing $R_{s} = a r_{s}$ where $r_{s}$ is the comoving radius (and therefore constant in time), we get $$ \frac{1}{2} r_{s}^2 \dot{a}^{2} = \frac{4}{3} \pi G \rho a^{2} + E $$ or $$ \left( \frac{\dot{{a}}}{a} \right)^{2} = \frac{8}{3}\pi G \rho + \frac{2E}{r_{s}^{2}} \frac{1}{a^{2}} $$ which clearly has the form of the Friedmann equation. Note that in the relativistic version, $\rho$ includes all sources of matter/energy. The integration constant here gets identified with the curvature: - $E>0$, $k=-1$, Universe expands forever - $E<0$, $k=1$, maximum size - $E=0$, $k=0$, stops in the far future. Note that these conclusions are in the absence of dark energy, which changes this discussion. Why does this work at all? Birkhoff's theorem helps here. **Continuity Equation** Start from the first law of thermodynamics $$ dQ = dE + P dV $$ If we consider our previous sphere and adiabatic expansion ($dQ=0$), this gives us $$ \dot{E} + P \dot{V} = 0 $$ Now $$ \begin{aligned} E = \rho V \\ \dot{E} = \dot{\rho} V + \rho \dot{V} \\ \dot{\rho} V + \dot{V}(\rho + P) = 0 \\ V = \frac{4}{3} \pi r_{s}^{3} a^{3} \implies \dot{V} = 3 \frac{\dot{a}}{a} V = 3 H V \\ \dot{\rho} + 3 H (\rho + P) = 0 \end{aligned} $$ ## The Critical Density Consider the first Friedmann eqn $$ \left(\frac{\dot{a}}{a}\right)^{2} = H^2 = \frac{8 \pi G}{3} \rho(t) - \frac{k}{R_0^{2} a^{2}} $$ Dividing by $H^{2}$, we have $$ 1 = \frac{8 \pi G}{3 H^2} \rho(t) - \frac{k}{R_0^{2} a^{2} H^{2}} $$ Notice that when $$ \rho = \frac{3H^{2}}{8 \pi G} $$ the curvature $k$ is zero. We therefore define a **critical density** $$ \rho_\text{crit} = \frac{3H^{2}}{8 \pi G} $$ Note that the value of the critical density depends on $H$ and therefore varies with time. The value today is $$ \begin{aligned} \rho _\text{crit} &= 1.877 \times 10^{-26} h^{2} \text{kg}\,\text{m}^{-3} \\ &= 2.775 \times 10^{11} h^{2} M_{\odot} \text{Mpc}^{-3} \\ &= 1.053 \times 10^{10} \text{eV}^{4} \end{aligned} $$ We can measure the density relative to the critical density $$ \Omega = \frac{\rho}{\rho _\text{crit}} $$ Here, we're keeping $\Omega$ to represent all energy density, but very soon, we will write down these $\Omega$'s as a function of the type - radiation, matter, dark energy. Plugging back into the Friedmann equation, we find $$ 1 - \Omega(t) = -\frac{k}{R_{0}^{2}a^{2}H^{2}} $$ Note that the sign of $1-\Omega$ is fixed and doesn't change with time. In particular, if $k=0$, $\Omega(t)=1$.