University Physics IV · Frontiers of Modern Physics · 15.6
Scale Factor & the Friedmann Equation
Cosmology has one dynamical unknown: a(t), the factor every proper distance gets multiplied by. Learn to read it straight out of a redshift, to get its equation of motion from a Newtonian energy argument, and to see precisely where that argument stops being enough.
Build the model
Connect the measurement to the mechanism.
Put the whole expansion into one number. Fix every galaxy at a comoving coordinate that never changes and let each proper separation be r = a(t)x, so recession is the ruler stretching rather than motion through space; differentiating gives Hubble's law with H = ȧ/a, and light emitted when the factor was a arrives with every wavelength multiplied by 1/a, which is the entire content of 1 + z = 1/a. The equation of motion for a follows from Newton alone: put a test galaxy on the surface of a uniform sphere, use the shell theorem to ignore everything outside it, and conserve energy.
Kinetic minus gravitational is constant, and dividing through by the sphere's size leaves H² = 8πGρ/3 − kc²/a², where the integration constant k is that sphere's total energy in disguise. Setting k = 0 defines the critical density ρc = 3H²/(8πG) — 8.5 × 10⁻²⁷ kg m⁻³, five hydrogen atoms per cubic metre — and dividing every density by it turns geometry into an addition problem. What it costs is pressure: relativity's companion equation carries ρ + 3p/c², so radiation gravitates twice as hard as its mass alone and a negative-pressure component pushes, while the curvature and Λ terms are imported rather than derived.
The Newtonian route reaches the right equation for a matter universe, and the wrong reason for an accelerating one.
- Simple definition
- The scale factor a(t) is the dimensionless number every proper distance in the universe is multiplied by, set to 1 today; the Friedmann equation is its equation of motion, fixing (ȧ/a)² from the total density.
- Example
- Light leaving a galaxy when a = 0.25 arrives with every wavelength four times longer, so 1 + z = 4 and z = 3: the 121.6 nm Lyman-α line of hydrogen is measured at 486.4 nm.
A galaxy 100 Mpc away recedes at 6740 km s⁻¹ without moving through space at all.
x is a fixed comoving separation in Mpc; a is dimensionless with a(t₀) = 1; H in s⁻¹
A measured redshift reads off how big the universe was. No velocity is needed or implied.
z dimensionless; a(tₑₘᵢₜ) is the scale factor at the moment the light left
Turns a density into an expansion rate. With Λ absent, k > 0 recollapses, k = 0 is flat, k < 0 expands forever.
ρ is the total density in kg m⁻³; both sides are s⁻²; k carries the sphere's total energy
Geometry becomes arithmetic: the Ω values sum to 1 when space is flat, above 1 closed, below 1 open.
H₀ = 67.4 km s⁻¹ Mpc⁻¹ gives ρc = 8.5 × 10⁻²⁷ kg m⁻³, five hydrogen atoms per cubic metre
One line gives H at any epoch, and whichever term is steepest names the era you are looking at.
Ωᵣ ≈ 9 × 10⁻⁵, Ωₘ ≈ 0.315, ΩΛ ≈ 0.685, and Ωₖ = 1 − ΣΩᵢ ≈ 0
An age follows from H alone: 2/(3H₀) is 9.7 Gyr, too young for the oldest globular clusters.
Valid for k = 0, Λ = 0 and one dominant component, where Ht is a pure number
Comoving coordinates put the expansion in the ruler
Label each galaxy by a comoving coordinate x that never changes, and let the proper separation be r(t) = a(t)x with a = 1 today. Differentiate once: ṙ = ȧx = (ȧ/a)r = Hr. Hubble's law is not an empirical coincidence but a theorem about this picture, and because it is linear in r it looks identical from every galaxy — there is no centre to find. With H₀ = 67.4 km s⁻¹ Mpc⁻¹ = 2.18 × 10⁻¹⁸ s⁻¹, a galaxy 100 Mpc away recedes at 6740 km s⁻¹, and one at 1000 Mpc at 67 400 km s⁻¹. Push far enough and ṙ passes c, at the Hubble radius c/H₀ = 4450 Mpc, about 14.5 billion light years; that is permitted because nothing is moving through space. Note also what does not expand: a system already bound by its own gravity or by electromagnetism has decoupled from the flow, so the Solar System, the Galaxy and the Local Group keep their sizes while a grows around them.
A redshift measures the scale factor, not a velocity
A wavelength is a length, so it is stretched by exactly the same factor as everything else between emission and arrival: λobs/λₑₘᵢₜ = a(t₀)/a(tₑₘᵢₜ) = 1/a. That single line is 1 + z = 1/a. A quasar at z = 3 emitted its light when a = 0.250, and the 121.6 nm Lyman-α line reaches us at 486.4 nm, in the visible. The same factor rescales a thermal spectrum, since T ∝ 1/a: the microwave background we measure at 2.725 K was released at z = 1100, when a = 9.1 × 10⁻⁴ and the temperature was 2.725 × 1101 = 3000 K — just hot enough to keep hydrogen ionised, which is why the universe turns transparent there and not somewhere else. Note what the measurement does and does not give. It gives a at the moment of emission, directly. It gives neither a distance nor a lookback time; converting z into either requires integrating the whole expansion history.
Derive the Friedmann equation by conserving energy on a sphere
Fill space uniformly with dust of density ρ, draw a sphere of proper radius r = ax about any point you like, and place a test mass m on its surface. The shell theorem says only the interior mass pulls, M = (4/3)πr³ρ, and homogeneity means the choice of centre cannot matter. Conserve energy: ½mṙ² − GMm/r = E, a constant. Substitute r = ax, ṙ = ȧx and M: ½mx²ȧ² − (4π/3)Gρa²x²m = E. Now divide throughout by ½ma²x², which is what removes the arbitrary x: (ȧ/a)² − (8π/3)Gρ = 2E/(mx²a²). The right-hand side falls as a⁻², so write it as −kc²/a² and you have H² = 8πGρ/3 − kc²/a². The sign of E is the whole story: E < 0 is a gravitationally bound sphere that must turn round (k > 0), E = 0 is the marginal case (k = 0), and E > 0 escapes (k < 0).
Critical density turns geometry into an addition problem
Set k = 0 and solve the Friedmann equation for the density that makes it so: ρc = 3H²/(8πG). Today, with H₀ = 67.4 km s⁻¹ Mpc⁻¹ = 2.184 × 10⁻¹⁸ s⁻¹, that is ρc = 3(2.184 × 10⁻¹⁸)²/(8π × 6.674 × 10⁻¹¹) = 8.53 × 10⁻²⁷ kg m⁻³, roughly five hydrogen atoms per cubic metre — a harder vacuum than any laboratory reaches. Now measure every density against it, Ωᵢ = ρᵢ/ρc, and define Ωₖ = −kc²/(a²H²). Dividing the Friedmann equation by H² then makes it read ΣΩᵢ + Ωₖ = 1 exactly. The measured budget is Ωₘ = 0.315, ΩΛ = 0.685, Ωᵣ ≈ 9 × 10⁻⁵ and |Ωₖ| < 0.005. That flatness is an observation the equation accommodates, not a result it predicts.
Which term wins, and the power laws that follow
Each component dilutes at its own rate. Matter goes as a⁻³, pure volume. Radiation goes as a⁻⁴: the same volume factor, plus one more because every photon's energy falls as 1/a with its wavelength. Curvature enters as a⁻² and Λ as a constant, giving H²/H₀² = Ωᵣ a⁻⁴ + Ωₘ a⁻³ + Ωₖ a⁻² + ΩΛ. Run it backwards and the steepest term takes over: radiation overtakes matter at a = Ωᵣ/Ωₘ = 2.9 × 10⁻⁴, or z ≈ 3400, while matter last equalled Λ at a = (Ωₘ/ΩΛ)¹⁄³ = 0.77, z ≈ 0.30. When one component dominates a flat model the equation integrates in a line: ȧ² ∝ a⁻¹ gives a ∝ t²⁄³ and H = 2/(3t), while ȧ² ∝ a⁻² gives a ∝ t¹⁄² and H = 1/(2t). Since Ht is then a pure number, an age follows from H alone — and 1/H₀ = 14.5 Gyr makes a flat matter-only universe just 9.7 Gyr old, younger than its own oldest stars.
What the Newtonian sphere borrows and cannot supply
Three debts are worth naming. First, pressure. General relativity's companion equation is ä/a = −(4πG/3)(ρ + 3p/c²) + Λc²/3, and an energy balance for dust carries only ρ. So radiation, with p = ρc²/3, gravitates twice as strongly as its mass alone, and a component with p = −ρc² pushes outward — neither effect can emerge from Newtonian bookkeeping. Second, the a⁻⁴ dilution of radiation actually comes from the relativistic fluid equation ρ̇ = −3H(ρ + p/c²); Newtonian mass conservation would only ever give a⁻³. Third, k arrived here as an energy constant, and calling it the curvature of space, like admitting Λ at all, is an import from the full theory. Even the shell theorem needs Birkhoff's theorem to be legitimate in a relativistic setting. Treat the derivation as a mnemonic with its proof kept elsewhere: exact for pressureless matter, and silent about almost everything the last thirty years of cosmology has been about.
Change one variable at a time
Make the relationship visible.
Drag the epoch left from today: matter overtakes ΩΛ near a = 0.77, and radiation overtakes matter near a = 3 × 10⁻⁴. Then raise Ωₘ to 1.0 and the matter-Λ crossing slides off the right-hand edge, so a universe that dense never gets a dark-energy era at all.
SCALE FACTOR a0.3162
REDSHIFT z2.2
H / H₀3.26
Ωₘ + ΩΛ + Ωᵣ1.000
Live interpretationSCALE FACTOR a: 0.3162. REDSHIFT z: 2.2. H / H₀: 3.26. Ωₘ + ΩΛ + Ωᵣ: 1.000
Catch the common trap
Explain before calculating.
Light from a distant galaxy arrives with 1 + z = 5.0. Which statement about that measurement is correct?
Choose an answer to test the model.
Practice & worked examples
Reason from the model, then test the result.
EasyA quasar is observed at redshift z = 3.00. Find the scale factor when its light was emitted, and the observed wavelengths of its 121.6 nm Lyman-α line and its 656.3 nm Hα line.
- 1 + z = 4.00, and 1 + z = 1/a, so a = 1/4.00 = 0.250: every proper separation then was a quarter of what it is now.
- A wavelength is a length, so it is stretched by the same factor: λobs = λₑₘᵢₜ(1 + z) = 121.6 nm × 4.00 = 486.4 nm — emitted in the ultraviolet, received as blue-green.
- The same factor acts on every feature of the spectrum: 656.3 nm × 4.00 = 2625 nm, so Hα lands in the infrared.
- Nothing here needed a velocity. The factor 4.00 is simply how much a has grown since the light left, and because it multiplies every wavelength alike the pattern of lines is untouched, which is how the object stays identifiable.
Answera = 0.250; Lyman-α is observed at 486.4 nm and Hα at 2625 nm (2.63 μm).
MediumTake H₀ = 67.4 km s⁻¹ Mpc⁻¹. Find the critical density today, express it as hydrogen atoms per cubic metre, and use Ωb = 0.049 to find the mean baryon density. Use 1 Mpc = 3.086 × 10²² m, G = 6.674 × 10⁻¹¹ N m² kg⁻² and mH = 1.674 × 10⁻²⁷ kg.
- Put H₀ in SI. The metres cancel between the km and the Mpc, leaving an inverse time: H₀ = 6.74 × 10⁴ m s⁻¹ ÷ 3.086 × 10²² m = 2.184 × 10⁻¹⁸ s⁻¹.
- Numerator: 3H₀² = 3 × (2.184 × 10⁻¹⁸)² = 1.431 × 10⁻³⁵ s⁻². Denominator: 8πG = 8π × 6.674 × 10⁻¹¹ = 1.677 × 10⁻⁹ m³ kg⁻¹ s⁻².
- ρc = 3H₀²/(8πG) = 1.431 × 10⁻³⁵ ÷ 1.677 × 10⁻⁹ = 8.53 × 10⁻²⁷ kg m⁻³.
- As atoms: 8.53 × 10⁻²⁷ ÷ 1.674 × 10⁻²⁷ = 5.1 hydrogen atoms per cubic metre. The critical density is a far better vacuum than any pump can reach.
- Baryons are only Ωb of that: ρb = 0.049 × 8.53 × 10⁻²⁷ = 4.18 × 10⁻²⁸ kg m⁻³, which is 0.25 atoms per cubic metre — one hydrogen atom in every four cubic metres, averaged over the universe.
Answerρc = 8.53 × 10⁻²⁷ kg m⁻³, about 5.1 H atoms m⁻³; ρb = 4.18 × 10⁻²⁸ kg m⁻³, about one atom per 4 m³.
HardA flat model has Ωₘ = 0.315, ΩΛ = 0.685 and H₀ = 67.4 km s⁻¹ Mpc⁻¹; ignore radiation. (a) Find H at z = 1. (b) Find the redshift at which the matter and Λ densities were equal. (c) Find the redshift at which the expansion began to accelerate, and say which of the three the Newtonian derivation cannot reach.
- Flat means Ωₖ = 0, so H²/H₀² = Ωₘ(1 + z)³ + ΩΛ. At z = 1: 0.315 × 8 + 0.685 = 2.520 + 0.685 = 3.205.
- H(z = 1) = √3.205 × H₀ = 1.790 × 67.4 = 121 km s⁻¹ Mpc⁻¹. The universe was expanding 1.79 times faster then, and it was matter, not Λ, that dominated the sum.
- Equal densities need Ωₘ(1 + z)³ = ΩΛ, so (1 + z)³ = 0.685/0.315 = 2.175, giving 1 + z = 1.296 and z = 0.30. Λ has led the density budget only since then.
- Acceleration is a different question and needs the relativistic equation ä/a = −(4πG/3)(ρ + 3p/c²). Matter has p ≈ 0; Λ has p = −ρΛc², so its 3p/c² term is −3ρΛ. Setting ρₘ − 2ρΛ = 0 gives Ωₘ(1 + z)³ = 2ΩΛ.
- So (1 + z)³ = 2 × 0.685/0.315 = 4.349, 1 + z = 1.632 and z = 0.63. Acceleration began well before Λ became the larger density, because its negative pressure counts three times over.
- Parts (a) and (b) are Friedmann bookkeeping and the Newtonian sphere delivers them. Part (c) it cannot: an energy balance for dust has no pressure term at all, so nothing in it can say when ä changes sign.
AnswerH(z = 1) = 121 km s⁻¹ Mpc⁻¹; matter and Λ had equal density at z = 0.30; acceleration began at z = 0.63. Only the first two follow from the Newtonian route.