Open, closed and flat universes

The sign of k decides the fate of a matter-filled universe. The dividing line is the critical density — about five protons per cubic metre — and a flat universe grows as t to the 2/3.

Follows MIT 8.286 lecture 7MIT 8.286 Lecture 7: The Dynamics of Homogeneous Expansion, Part IIIOn MIT OpenCourseWare: the lecture and its readings

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The Friedmann equation8.286 · Lecture 7

Full reading path (10 pages)

Vectors (8.01)
Position and velocity in one dimension (8.01)
Acceleration in one dimension (8.01)
Motion in two dimensions and projectiles (8.01)
Newton's laws of motion (8.01)
The universal law of gravitation (8.01)
The homogeneous expanding universe (8.286)
Why a static universe cannot work (8.286)
The expanding sphere and the equation for a(t) (8.286)
The Friedmann equation (8.286)

On this page · 5 sections

Three kinds of universe
The critical density
Our universe
The flat universe grows as t^(2/3)
Summary

The Friedmann equation contains one constant, $k$ , or equivalently the conserved quantity $E = -kc^2/2$ . In the second half of Lecture 7 Guth shows that its sign decides the whole future of a universe filled with ordinary (pressureless) matter.

Three kinds of universe#

Write the conserved quantity as an equation for the expansion speed:

Expansion speed

\dot a^2 = 2E + \frac{8\pi}{3}\,\frac{G\rho_i}{a}

In words

a dot squared equals 2E plus 8 pi over 3 G rho sub i over a

Source: 8.286 Lecture 7 transcript

The second term is always positive and shrinks as the universe grows. Everything depends on $E$ :

$E > 0$ ( $k < 0$ ), open universe. $\dot a^2$ never drops below $2E$ , so the universe expands forever, never slower than a minimum rate.
$E < 0$ ( $k > 0$ ), closed universe. The system is gravitationally bound. $\dot a$ falls to zero when the universe reaches its maximum size, then the expansion reverses and the universe collapses toward $a = 0$ — the Big Crunch.
$E = 0$ ( $k = 0$ ), critical or flat universe. The borderline: the universe expands forever, but ever more slowly, with $\dot a \to 0$ .

Maximum size of a closed universe

a_{\max} = -\frac{4\pi G\rho_i}{3E} \qquad (E < 0)

In words

a sub max equals minus 4 pi G rho sub i over 3E (E less than 0)

Source: 8.286 Lecture 7 transcript

The names open, closed and flat describe geometry, which comes later in the course. For now they label how the universe evolves in time.

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ȧ² = 2E + 1/a for the three cases (units where 8πGρᵢ/3 = 1), computed. The closed curve reaches zero at a_max = 5, where the universe turns around (it never gets larger); the flat one approaches zero only as a → ∞.

The critical density#

The flat case marks a definite density. Set $k = 0$ in the Friedmann equation, $H^2 = \tfrac{8\pi}{3}G\rho$ , and solve for $\rho$ :

Critical density

\rho_c = \frac{3H^2}{8\pi G}

In words

rho sub c equals 3H squared over 8 pi G

Source: 8.286 Lecture 7 transcript

A universe denser than this is closed, a less dense one open. It is convenient to measure density in units of $\rho_c$ :

Density parameter

\Omega \equiv \frac{\rho}{\rho_c}: \qquad \Omega > 1\ \text{closed}, \quad \Omega = 1\ \text{flat}, \quad \Omega < 1\ \text{open}

In words

omega is defined as rho over rho sub c : omega greater than 1 closed , omega equals 1 flat , omega less than 1 open

Source: 8.286 Lecture 7 transcript

Writing today's Hubble constant as $H_0 = 100\,h_0$ km/s/Mpc, the critical density is

Critical density in numbers

\rho_c = 1.88\,h_0^2 \times 10^{-29}\ \text{g/cm}^3

In words

rho sub c equals 1.88 h sub 0 squared times 10 to the power minus 29 g/cm cubed

Source: 8.286 Lecture 7 transcript

With $h_0 = 0.67$ (the Planck value Guth uses) that is $8.4 \times 10^{-30}$ g/cm³, about five protons per cubic metre — some $10^{-29}$ times the density of water, a far better vacuum than anything made in a laboratory.

Worked example

How big does a closed universe get?

At some initial time a matter-only universe has $\Omega_i = 2$ . Taking $a_i = 1$ , how large does it grow before it recollapses? What if $\Omega_i = 1.01$ ?

Express E with Ω

With $a_i = 1$ and $\dot a_i = H_i$ : $E = \tfrac12 H_i^2 - \tfrac{4\pi}{3}G\rho_i$ . Since $\tfrac{4\pi}{3}G\rho_c = \tfrac12 H_i^2$ , this is $E = \tfrac12 H_i^2(1 - \Omega_i)$ , negative for $\Omega_i > 1$ .

Maximum size

$a_{\max} = -\dfrac{4\pi G\rho_i}{3E} = \dfrac{\tfrac12 H_i^2\,\Omega_i}{\tfrac12 H_i^2(\Omega_i - 1)} = \dfrac{\Omega_i}{\Omega_i - 1}$ .

Numbers

$\Omega_i = 2$ gives $a_{\max} = 2$ : the universe only doubles in size. $\Omega_i = 1.01$ gives $a_{\max} = 101$ . The closer $\Omega$ starts to 1, the longer the universe behaves almost exactly like the flat case.

Our universe#

What does the real universe do? In 2013 Guth summarised the measurements as:

the total density is within about half a percent of critical, $\Omega_\text{total} \approx 1$ ;
matter (ordinary and dark) makes up only about 30% of it, $\Omega_\text{matter} \approx 0.3$ ;
the rest, about 70%, is dark energy, discovered in 1998.

So today the pressureless equations on this page do not describe our universe well. They apply to the long era of matter domination, from roughly 50,000 years after the Big Bang until a few billion years ago, when dark energy took over. Before 1998, inflation theorists expected $\Omega = 1$ while observers kept finding 0.2–0.3 in matter; both turned out to be partly right.

The flat universe grows as t^(2/3)#

The flat case is simple enough to solve right away. With $E = 0$ , $\dot a^2 = \text{const}/a$ , so

Separating the variables

\frac{da}{dt} = \frac{\text{const}}{a^{1/2}} \;\Longrightarrow\; a^{1/2}\,da = \text{const}\;dt \;\Longrightarrow\; \tfrac23\,a^{3/2} = \text{const}\cdot t + c'

In words

da over dt equals const over a to the power 1/2 which gives a to the power 1/2 da equals const dt which gives 2 over 3 a to the power 3/2 equals const times t plus c prime

Source: 8.286 Lecture 7 transcript

The zero of cosmic time is ours to choose. Choosing $t = 0$ at the instant when $a = 0$ — the Big Bang — makes $c' = 0$ , and

Flat, matter-dominated universe

a(t) \propto t^{2/3}

In words

a(t) is proportional to t to the power 2/3

Source: 8.286 Lecture 7 transcript

Loading graph…

Computed growth of the scale factor, for two universes with the same size and expansion rate today. Gravity makes the flat universe decelerate, so it reached today's size sooner: its age is 2/3 of 1/H₀.

Guth works out the consequence for the age of the universe in Lecture 8:

Worked example

The age of a flat matter-only universe

For $a \propto t^{2/3}$ , find the Hubble rate as a function of time. If today $H_0 = 67.4$ km/s/Mpc, how old would such a universe be?

Hubble rate

$H = \dot a/a = \dfrac{\tfrac23 t^{-1/3}}{t^{2/3}} = \dfrac{2}{3t}$ , so the age is $t_0 = \dfrac{2}{3H_0}$ .

Numbers

$1/H_0 = (97.8/67.4)\times10^{10}$ yr $= 14.5$ billion years, so $t_0 = \tfrac23 \times 14.5 = 9.7$ billion years.

A problem

Some stars appear to be older than that. A flat universe of matter alone is too young — one of the clues that something else, dark energy, is in the universe. Later lectures come back to this tension between the age of the universe and the ages of the oldest stars.

Summary#

For pressureless matter, $E > 0$ means an open universe (forever expanding), $E < 0$ a closed one (maximum size, then Big Crunch), $E = 0$ the flat borderline.
The dividing density is $\rho_c = 3H^2/(8\pi G) \approx 1.88\,h_0^2\times10^{-29}$ g/cm³, about five protons per cubic metre; $\Omega = \rho/\rho_c$ .
Today $\Omega_\text{total} \approx 1$ , but only about 30% is matter; about 70% is dark energy.
A flat matter-dominated universe grows as $a \propto t^{2/3}$ and is $2/(3H_0)$ old.

Symbols on this page

Show the 22 symbols used here

$\pi$ pi: The ratio of a circle's circumference to its diameter, π ≈ 3.14159. Say it: “pi”.
$\rho$ rho: Density: mass (or energy divided by c²) per unit volume. ρ_c is the critical density, ρ̄ the mean density. Say it: “rho”.
$\Omega$ capital omega: The density parameter Ω = ρ/ρ_c. Subscripts pick one ingredient: Ω_m matter, Ω_vac (or Ω_Λ) vacuum energy, Ω_r radiation, Ω_k curvature. Say it: “omega”.
$\approx$ approximately equal: Is approximately equal to. Say it: “is approximately”.
$\propto$ proportional to: Is proportional to: doubling one doubles the other. Say it: “is proportional to”.
$\equiv$ identical to: Is defined as (or is identically equal to). Say it: “is defined as”.
$\to$ arrow: Goes to (tends to), or becomes. Say it: “goes to”.
$\Rightarrow$ implies: Implies: what follows is a consequence of what came before. Say it: “which gives”.
$\times$ times: Times (multiplication). Between two vectors: the cross product. Say it: “times”.
$\cdot$ dot: Times (multiplication). Between two vectors: the dot product. Say it: “times”.
$\max$ maximum: Maximum: the largest value. Say it: “max”.
$\dot a$ dot: Rate of change with time: ȧ = da/dt. Say it: “dot”.
$a$ a: The scale factor a(t): how much the universe has expanded. Physical distance = a × comoving distance (Guth's units: metres per notch). Say it: “a”.
$H$ H: The Hubble parameter H = ȧ/a: the expansion rate of the universe at a given time. Say it: “H”.
$G$ G: Newton's gravitational constant, G ≈ 6.674 × 10⁻¹¹ m³ kg⁻¹ s⁻². Say it: “G”.
$c$ c: The speed of light; c′ (with a prime) is a constant of integration. Say it: “c”.
$k$ k: The curvature constant: k > 0 closed, k = 0 flat, k < 0 open (Guth's units: 1/notch²). Say it: “k”.
$H_{0}$ Hubble constant: The Hubble constant: today's value of H, about 67–73 km/s/Mpc. Say it: “H nought”.
$h_{0}$ little h: H₀ in units of 100 km/s/Mpc: h₀ ≈ 0.67–0.7. Say it: “h nought”.
$t_{0}$ today: Today: the present time, about 13.8 billion years after the Big Bang in the real universe (simple models on some pages give other ages). Say it: “t nought”.
$\rho_{c}$ critical density: The critical density, 3H²/(8πG): the density of a flat universe (about 9 × 10⁻²⁷ kg/m³ today). Say it: “rho c”.
$\rho_{i}$ initial density: The density when a = 1, so that ρ = ρᵢ/a³ for matter. Say it: “rho i”.

Tap or click a symbol in a formula to see the same explanation. All symbols: symbols and notation.

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A matter-only universe has $E < 0$ (so $k > 0$ ). What happens?
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