Friedmann Equation

Friedmann Equation#

The Friedmann equation applies the Friedmann metric,

d s^{2} = - d t^{2} + a (t)^{2} [d r^{2} + S_{k} (r)^{2} d Ω^{2}]

to Einstein’s field equation,

G_{μ ν} = \frac{8 π G}{c^{4}} T_{μ ν}

The resultant is an equation that describes how the Hubble parameter varies over time $H^{2} (t)$ . You may interpret this as how the universe expands over time. The Friedmann equation is an implicit function of time more explicitly it depends on the energy density $ρ$ of the universe.

Often times, we use the definition of the Hubble’s parameter,

H (t) \equiv \frac{\dot{a}}{a}

Mass Density Form :

{(\frac{\dot{a}}{a})}^{2} = - \frac{4 π G}{3} (ρ + \frac{3 ρ}{c^{2}}) + \frac{Λ c^{2}}{3}

Energy Density Form : $ ${(\frac{\dot{a}}{a})}^{2} = \frac{8 π G}{3 c^{2}} μ - \frac{κ c^{2}}{R_{0}^{2} a^{2}}$ $

Natural Unit Form : $ ${(\frac{\dot{a}}{a})}^{2} = \frac{8 π G}{3} μ - \frac{k}{a^{2}}$ $

$μ$ : Energy density of the universe
$k$ : Curvature constant where $k \equiv κ / R_{0}^{2}$

We will be using the natural unit form since it’s natural.

Hubble’s Constant#

Immediately one can infer the Hubble’s constant from the Friedmann equation setting $t = t_{0}$ which is today.

\begin{array}{r} \begin{array}{c} {(\frac{{\dot{a}}_{0}}{a_{0}})}^{2} = H_{0}^{2} \\ H_{0}^{2} = \frac{8 π G}{3} μ_{0} - \frac{k}{a^{2}} \end{array} \end{array}

Non-Relativistic Form#

The Friedmann equation can be thought of as an expanding finite sphere of mass density $ρ (t)$

{(\frac{\dot{a}}{a})}^{2} = \frac{8 π G}{3} ρ (t) + \frac{2 C}{r_{s}^{2}} \frac{1}{a (t)^{2}}

$C$ : Constant that affects the expansion rate
$r_{s}$ co-moving radius

Proof#

Density Parameter Form#

{(\frac{\dot{a}}{a})}^{2} = H_{0}^{2} [\frac{Ω_{r, 0}}{a^{4}} + \frac{Ω_{m, 0}}{a^{4}} + Ω_{Λ, 0} + \frac{1 - Ω_{0}}{a^{2}}]

Fluid Equation#

The fluid equation is the equation of motion for the energy density $ρ$ as a function of time. It is given by an adiabatic expansion of the universe.,

\begin{array}{r} \begin{array}{c} \dot{μ} + 3 \frac{\dot{a}}{a} (μ + P) = 0 \\ \dot{μ} = - 3 \frac{\dot{a}}{a} (μ + P) \\ \dot{μ} \frac{a}{\dot{a}} = - 3 (μ + P) \end{array} \end{array}

Proof - Newtonian#

You may prove exactly, the fluid equation for the adiabatic expansion of a sphere with total internal energy $E = V (t) μ (t)$ where $V (t)$ is,

V (t) = \frac{4 π}{3} r_{s}^{3} a (t)^{3}

Acceleration Equation#

Knowing the Friedmann equation and fluid equation we can take the derivative to get,

\frac{\ddot{a}}{a} = - \frac{4 π G}{3 c^{2}} (μ + 3 P)

The condition of acceleration is to make $μ + 3 P < 0$ so $P < - \frac{1}{3} μ$ . Unfortunately we do not know how $P$ relates to $μ$ which both share the same units.

Equation of State#

We assume that pressure $P$ and energy density $μ$ are linearly proportional such that,

\begin{array}{r} P = w μ w = {\begin{cases} 0 & matter \\ 1 / 3 & radiation \\ - 1 & dark energy \end{cases} \end{array}

$w$ : A constant relating to the type of energy. Notice that from the acceleration equation, the universe accelerates for $w < - 1 / 3$ so it motivates that the unknown energy (dark energy) is negative. Why we use exactly $w = - 1$ for dark energy is explained in the cosmological constant

For a universe of several energy components, $μ$ is a weighted sum of those energies,

P = \sum_{i} w_{i} μ_{i}

Now let’s apply each type of energy density to the fluid equation:

\begin{array}{r} \begin{array}{c} {\dot{μ}}_{i} + 3 \frac{\dot{a}}{a} (μ_{i} + P_{i}) = 0 \\ \frac{d μ_{i}}{μ_{i}} = - 3 (1 + w_{i}) \frac{d a}{a} \end{array} \end{array}

μ_{i} (a) = μ_{i, 0} a^{- 3 (1 + w_{i})}

This gives the useful proportional relations,

\begin{array}{r} μ_{m} \propto a^{- 3} \\ μ_{r} \propto a^{- 4} \\ μ_{Λ} = μ_{Λ, 0} \end{array}

Cosmological Constant - Incorrect Version#

The cosmological constant $Λ$ started out as a correction to Einstein’s field equation for a universe full of matter and radiation. In Einstein’s years, scientist did not believe that the universe was either expanding or contracting thus conclude that a universe of matter and radiation must be a stable universe. Unfortunately, Einstein equation does not agree so Einstein added the cosmological constant such that the Friedmann equation changes to:

{(\frac{\dot{a}}{a})}^{2} = \frac{8 π G}{3} μ - \frac{k}{a^{2}} + \frac{λ}{3}

The acceleration equation is then,

\frac{\ddot{a}}{a} = - \frac{4 π G}{3} (μ + 3 P) + \frac{Λ}{3}

Notice that at a stable universe where $\ddot{a} (t) = 0$ , when $P = 0 \cdot μ + \frac{1}{3} μ$ and

\begin{array}{r} Λ \equiv 4 π G μ μ_{Λ} \equiv \frac{1}{8 π G} Λ \end{array}

$μ_{Λ}$ : Cosmological constant component of energy density (note: $μ \neq μ_{Λ}$ ), independent of time.

Let’s plug in $Λ$ into the Friedmann equation for a static universe $\dot{a} = 0$ and $a = 1$ ,

\begin{array}{r} \begin{array}{c} \frac{2 Λ}{3} - \frac{k}{a^{2}} + \frac{Λ}{3} \\ k = Λ \end{array} \end{array}

Since $Λ > 0$ , this indicates that the curvature $k > 0$ therefore a open but static universe. This sounds incorrect and it is; in the next section we will explore the more acceptable use of the cosmological constant.

Motivation on the Incorrectness : Consider even a slight increase in $Λ$ then $\dot{a} > 0$ which causes $μ_{m} \to 0$ and by the Friedmann equation this feedbacks to $\dot{a} \to 0$ even more. We got a non-linear expanding universe

Cosmological Constant - Vacuum Energy#

The most accepted “correct” way to interpret the cosmological constant is as vacuum energy. Yes, vacuum has energy which many may first have been introduced as the Heinsenberg uncertainty principle,

Δ E Δ t \leq h