The Friedmann equation and the critical density

Builds on: Sections 13.1, 13.2 and S4
Reading time: 40 min
Exercises: 4
Olympiad problems: 2

The Universe is astonishingly empty. If all the matter and energy in it were spread out evenly, each cubic metre would hold the mass of about five and a half hydrogen atoms. A very good laboratory vacuum still holds billions of molecules in each cubic metre. This tiny number, the critical density, is the density that separates a universe that keeps expanding from one that eventually falls back. The Universe turns out to have very nearly this density. In this section we find it from the same reasoning about energy and escape that Section 3.2 used for a single rocket.

13.3.1An expanding ball of galaxies

Before you readMake a guess first

About how many hydrogen atoms in each cubic metre would give the Universe its critical density?

Take any point of the Universe and draw a large ball round it, so large that the matter inside is spread almost evenly. The matter outside, arranged evenly all round, pulls equally in every direction and has no overall effect on the ball, much as the pull of a hollow shell on a body inside it is zero. A galaxy on the edge of the ball therefore feels only the pull of the mass inside, as if all of it were at the centre. As the Universe expands, the ball grows with it, its radius in proportion to the scale factor, $R = a r_{0}$ , and the galaxy's energy stays the same, exactly as for a rocket climbing away from Earth.

ResultThe Friedmann equation(13.5)

H^{2} = {(\frac{\dot{a}}{a})}^{2} = \frac{8 π G}{3} ρ - \frac{k c^{2}}{a^{2}}

$H$	the Hubble parameter $\dot{a} / a$ , the rate of expansion at any time, equal to $H_{0}$ today
$\dot{a}$	the rate at which the scale factor grows
$ρ$	the density of the Universe at that time, including the mass equivalent $ε / c^{2}$ of any energy density $ε$
$k$	the curvature constant, positive, zero or negative

Alexander Friedmann found this equation from general relativity in 1922. The Newtonian argument gives the same form, with

k

standing for the energy of the expanding ball. In general relativity

k

describes the curvature of space, and the density must include every form of energy, as Section 13.4 explains.

Show the derivation

A galaxy of mass $m$ on the edge of a ball of radius $R$ and density $ρ$ feels the pull of the mass $M = \frac{4}{3} π R^{3} ρ$ inside it. Its energy is constant,

\frac{1}{2} m {\dot{R}}^{2} - \frac{G M m}{R} = constant .

Dividing by $\frac{1}{2} m R^{2}$ and putting in $M$ gives

{(\frac{\dot{R}}{R})}^{2} = \frac{8 π G}{3} ρ + \frac{constant}{R^{2}} .

With $R = a r_{0}$ , the ratio $\dot{R} / R$ equals $\dot{a} / a$ , and the last term goes as $1 / a^{2}$ . Writing that term as $- k c^{2} / a^{2}$ gives the Friedmann equation.

13.3.2The critical density

When $k = 0$ the ball has exactly zero energy, like a rocket moving at precisely the escape speed. Every galaxy moves away at just the speed needed to escape the pull of the matter inside its ball, and the expansion slows for ever without ever stopping. The density for which this happens at a given rate of expansion is the critical density, and the density of the Universe is usually given as a fraction of it, the density parameter $Ω$ .

ResultThe critical density and the density parameter(13.6)

\begin{aligned} ρ_{c} & = \frac{3 H^{2}}{8 π G} \\ Ω & = \frac{ρ}{ρ_{c}}, Ω - 1 = \frac{k c^{2}}{a^{2} H^{2}} \end{aligned}

$ρ_{c}$	the critical density at the time when the Hubble parameter is $H$
$Ω$	the density parameter, the ratio of the actual density to the critical density

Today, with

H_{0} = 70

km/s/Mpc

= 2.27 \times 10^{- 18}

s⁻¹,

ρ_{c} = 9.20 \times 10^{- 27}

kg/m³, or

1.36 \times 10^{11}

solar masses in each cubic megaparsec. The second line is the Friedmann equation divided by

H^{2}

. A density above the critical value means

k

is positive, and below it negative.

Quick checkThe critical density for another Hubble constant

What is the critical density if $H_{0} = 67.8$ km/s/Mpc? Take 1 Mpc $= 3.086 \times 10^{22}$ m and $G = 6.674 \times 10^{- 11}$ N m² kg⁻².

× 10⁻²⁷ kg/m³

The sign of $k$ fixes the shape of space. With $k = 0$ space is flat, and the angles of a very large triangle add up to 180°, as in ordinary geometry. With $k$ positive space is closed, curving back on itself like the surface of a sphere, and with $k$ negative it is open, curving the other way. In a universe that holds only matter, the sign of $k$ also decides its fate. A closed universe stops expanding and falls back, while flat and open ones expand for ever. Section 13.4 shows how dark energy changes this.

Measurements of the cosmic microwave background (Section 13.7) show that space is flat to within about a per cent, so the total density is very close to the critical density. Ordinary matter, the atoms of stars, gas and planets, gives $Ω_{b} \approx 0.05$ , about 0.28 hydrogen atoms in each cubic metre. Dark matter, which pulls by gravity but gives out no light, adds about 0.25, and dark energy makes up the remaining 0.70.

Example 13.4A universe with matter alone

Suppose the Universe held only matter, with $Ω_{m} = 0.3$ and $H_{0} = 70$ km/s/Mpc. What is its density, and is it open, flat or closed?

Solution

By (13.6), $ρ = 0.3 \times 9.20 \times 10^{- 27} = 2.76 \times 10^{- 27}$ kg/m³. Since $Ω - 1 = - 0.7$ is negative, $k$ is negative.

AnswerIts density is

2.76 \times 10^{- 27}

kg/m³, and it would be open and expand for ever. Our Universe has this density of matter but is flat, because dark energy makes up the difference, 0.7 of the critical density.

Example 13.5Every galaxy at the escape speedfinish it

In a flat universe of matter alone with $H_{0} = 70$ km/s/Mpc, a galaxy is 10 Mpc from us. How fast does it move away, and what is the escape speed from the ball of radius 10 Mpc round us?

The start

The galaxy moves away at $H_{0} d = 70 \times 10 = 700$ km/s. The ball holds $M = \frac{4}{3} π d^{3} ρ_{c}$ .

Your stepNow find the escape speed

\sqrt{2 G M / d}

, using

ρ_{c} = 3 H_{0}^{2} / 8 π G

km/s

Common errorsSection 13.3

The critical density depends on the square of the Hubble constant, so it must be found with $H$ in s⁻¹. Leaving $H_{0}$ in km/s/Mpc gives a number that is wrong by a factor of about $10^{39}$ .
A flat universe is not one with no matter. Flat means that the total density equals the critical density, and the matter, the radiation and the dark energy together add up to it.
The density parameter changes with time, since both $ρ$ and $H$ change. Today's values carry the subscript 0, as $Ω_{m, 0}$ , when the difference matters.
In a universe with dark energy, flat space does not mean that the expansion slows for ever. The fate depends on the dark energy as well as on the curvature.

ExercisesSection 13.34 questions

13.3.1
If the Hubble constant were twice as large, by what factor would the critical density change?
times
13.3.2
A survey finds a mean density of matter of $2.0 \times 10^{- 27}$ kg/m³. What is $Ω_{m}$ for $H_{0} = 70$ km/s/Mpc?
13.3.3
What is the critical density for $H_{0} = 100$ km/s/Mpc?
× 10⁻²⁶ kg/m³
13.3.4
A universe holds only matter, at twice the critical density. What happens to it?

Olympiad problemsPast IOAA problems, restated2 problems

2009Neutrinos as dark matter10 points

Adapted from IOAA 2009 (Tehran), theory problem 11. We have restated it in our own words and kept the original data. The solution was last checked on 21 October 2026.

Assume that neutrinos have a mass of $m_{ν} = 10^{- 5} m_{e}$ , where $m_{e} = 9.109 \times 10^{- 31}$ kg. How many neutrinos would each cubic metre need to hold to make up the dark matter of the Universe? Assume that the Universe is flat, that 25% of its mass is dark matter, and that $H_{0} = 75$ km/s/Mpc. A flat universe has zero total energy in the sense of Section 13.3.

I solved it without the solution

AnswerAbout $2.90 \times 10^{8}$ neutrinos in each cubic metre. The neutrinos left from the Big Bang number about $3.4 \times 10^{8}$ in each cubic metre, enough in number, but their masses are known to be far below $10^{- 5} m_{e}$ , so they cannot be most of the dark matter.

2013Matter and motion in our neighbourhood50 points

Adapted from IOAA 2013 (Volos), theory problem 1. We have restated it in our own words and kept the original data. The solution was last checked on 21 October 2026.

In a uniform universe the density parameter of matter, ordinary and dark together, is $Ω_{m} = ρ_{m} / ρ_{c} = 0.32$ . Take $H_{0} = 67.8$ km/s/Mpc.

(1) Find the average density of matter in our neighbourhood. [16 points]

(2) Find the escape speed of a galaxy 100 Mpc away from the matter inside that distance, taking the matter density of (1). [14 points]

(3) Suppose that this galaxy moved on a circular orbit round the centre of our cluster of galaxies, with us at that centre. What would its angular speed on the sky be? [12 points]

(4) Two such galaxies start on the same line of sight but orbit at different radii. Could we ever see them separate? Answer yes or no. [8 points]

I solved it without the solution

Answer(1) $2.76 \times 10^{- 27}$ kg/m³. (2) 3835 km/s. (3) $8.8 \times 10^{- 19}$ rad/s, the same at every distance. (4) No. In a uniform sphere every circular orbit has the same period, so the two galaxies stay lined up.