University Physics IV · Frontiers of Modern Physics · 15.7
Cosmological Evidence & the Dark Sector
Four measurements that have nothing to do with one another — a blackbody spectrum, a deuterium line, a flat rotation curve, a supernova magnitude — are made to close on a single budget. Learn to run that audit, and to keep the column that says how much strictly apart from the column that says what.
Build the model
Connect the measurement to the mechanism.
The evidence for the standard cosmology is not one observation but four that had no reason to agree. The microwave background is a blackbody to better than one part in 10⁴ of its peak, which forces a past dense and hot enough to thermalise radiation, and its first acoustic peak at ℓ ≈ 220 fixes the geometry as flat to about half a per cent. Big-bang nucleosynthesis then predicts, from nuclear rates and a single free number, both a helium mass fraction near 0.25 — which barely moves with that number, so it tests the framework rather than measuring it — and a deuterium abundance that falls steeply with it, so D/H = 2.5 × 10⁻⁵ pins the baryons at Ωb h² = 0.0224.
Rotation curves, cluster dynamics and gravitational lensing weigh all the matter and return Ωₘ ≈ 0.31, six times more than nucleosynthesis permits to be baryonic. Type Ia supernovae near z = 0.5 arrive about 0.25 magnitudes fainter than any decelerating model allows. Flatness then closes the sum, and the remainder, ΩΛ ≈ 0.69, is what is left after everything that can be weighed has been.
That is the cost of the method: every one of these is a gravitational or a thermal effect, so the budget fixes how much and never what — and the early-universe and late-universe routes to H₀ still disagree by some 8 per cent.
- Simple definition
- The cosmological density budget is the set of dimensionless parameters Ωᵢ = ρᵢ/ρc stating each component's share of the critical density — quantities fixed by measurement without any claim about what the components are made of.
- Example
- With H₀ = 67.4 km s⁻¹ Mpc⁻¹ the critical density is 8.5 × 10⁻²⁷ kg m⁻³, about five hydrogen atoms per cubic metre; the measured budget Ωb = 0.049, Ωc = 0.266, ΩΛ = 0.685 sums to 1.000, which is why the geometry comes out flat.
ρc = 1.88 × 10⁻²⁶ h² kg m⁻³, or 5.1 protons m⁻³ at h = 0.674.
H₀ in s⁻¹, G in m³ kg⁻¹ s⁻², ρ in kg m⁻³; h ≡ H₀/(100 km s⁻¹ Mpc⁻¹), so Ωᵢ is dimensionless
3.362 mK of dipole means v = 370 km s⁻¹ — 110 times the intrinsic 10⁻⁵ signal.
ΔT in K, v the observer's speed in m s⁻¹, θ measured from the direction of motion
1/5 at freeze-out, 1/7 after decay, so Yₚ = 0.25 — a clock, not a baryometer.
Δmc² = 1.293 MeV, kTf ≈ 0.8 MeV; Yₚ is a mass fraction, n/p a pure number
A 1% error in D/H costs only 0.6% in Ωb h², giving Ωb = 0.049 at h = 0.674.
D/H is a number ratio, measured as 2.53 × 10⁻⁵ in metal-poor quasar absorbers
v = 220 km s⁻¹ at r = 20 kpc encloses 2.3 × 10¹¹ M☉, far outside the visible disc.
v in m s⁻¹, r in m, M in kg; assumes circular orbits and rough spherical symmetry
ℓ₁ = 220.6 gives Ωₖ = 0.001 ± 0.002 — flat enough to close the budget by subtraction.
ℓ is the multipole of the angular power spectrum; Ωₜₒₜ and Ωₖ are dimensionless
A blackbody no astrophysical source can fake
The COBE-FIRAS spectrum fits a Planck curve at T₀ = 2.7255 ± 0.0006 K with residuals below one part in 10⁴ of the peak — the most exact blackbody ever measured, and nothing in the present universe can make one. Scattered starlight, warm dust and free-free emission all leave a spectrum with a shape, and a shape is what is absent here. So the spectrum by itself establishes a past that was optically thick and in thermal equilibrium, and dates the release of these photons to recombination near T = 3000 K, which the present 2.7255 K places at redshift 3000/2.7255 − 1 ≈ 1100. Notice what it does not do. It fixes a temperature and a thermal history, and the photon energy density that follows, Ωγ ≈ 5.4 × 10⁻⁵, is negligible in today's budget. Every density parameter in this lesson comes from something else.
Subtract your own motion before reading an anisotropy
The largest departure from uniformity on the sky is a pure dipole of amplitude 3.362 mK — one part in 810 of T₀, so about 10⁻³. It is kinematic: a moving observer sees ΔT/T₀ = (v/c) cos θ, so the amplitude is v/c and nothing more, and it would appear in a perfectly smooth universe. It gives 370 km s⁻¹ for the solar-system barycentre, and about 620 km s⁻¹ for the Local Group once the Sun's galactic orbit is removed. Fit that dipole, subtract it, and the residual sky has an rms near 30 μK, or 1.1 × 10⁻⁵. Two orders of magnitude separate the observer's velocity from the physics, which is why every published anisotropy map is a map of what is left after the dipole has been taken out.
The first acoustic peak is a ruler, and it reads flat
Before recombination the photons and baryons are one fluid: gravity compresses an overdensity, radiation pressure pushes back, and the modes that have completed exactly half an oscillation when the photons decouple show maximum contrast. That sets a physical length — the sound horizon, about 147 Mpc comoving — of known size at a known distance, so its angular size is a measurement of geometry, ℓ₁ ≈ 220/√Ωₜₒₜ. The observed ℓ₁ = 220.6 gives Ωₜₒₜ = 1.000 with Ωₖ = 0.001 ± 0.002. A second number falls out of the same spectrum: baryons weigh the fluid down, deepening odd compression peaks against even rarefaction peaks, and the ratio returns Ωb h² = 0.0224 independently of any nuclear physics. Flatness is the assumption that later lets ΩΛ be obtained by subtraction.
Helium is a clock; deuterium is the scale
Above about 0.8 MeV, weak interactions hold n/p at its equilibrium value exp(−Δmc²/kT) with Δmc² = 1.293 MeV. Freeze-out at 0.8 MeV leaves n/p = exp(−1.616) = 0.199, near 1/5; free neutrons then decay for the few minutes it takes deuterium to survive photodissociation, dropping the ratio to about 1/7, and essentially every surviving neutron is swept into ⁴He. So Yₚ = 2(1/7)/(1 + 1/7) = 0.25, a number governed by the expansion rate and the neutron lifetime and only logarithmically by the baryon density — a test of the framework, not a scale. Deuterium is its opposite: a fragile intermediate burned away faster the denser the baryons, with D/H ∝ (Ωb h²)^−1.6. Measure it where no star has processed it, in metal-poor absorbers along quasar sightlines, D/H = 2.53 × 10⁻⁵, and Ωb h² = 0.0224 follows — Ωb = 0.049 at h = 0.674.
Weigh the matter three ways; it beats the baryons six times
Rotation curves come first: v stays near 220 km s⁻¹ far beyond the optical disc, so M(<r) = v²r/G keeps rising linearly while the light has stopped, giving 2.3 × 10¹¹ M☉ inside 20 kpc. Cluster galaxies move far too fast for the mass their light implies, which is what Zwicky found in Coma in 1933. Weak lensing then maps the potential with no equilibrium assumption at all, and in the Bullet Cluster the lensing mass travels with the galaxies while the X-ray gas — which carries most of the baryonic mass — is left behind in the collision, separating a collisionless component from the baryons in a single image. The three methods converge on Ωₘ ≈ 0.315, while nucleosynthesis caps baryons at 0.049. About 84 per cent of the matter therefore takes part in neither nuclear nor electromagnetic processes.
Acceleration, the residual, and one number that disagrees
Type Ia supernovae, standardised through the width–luminosity relation, work as distance indicators to z ≈ 1. Near z = 0.5 they arrive roughly 0.25 magnitudes fainter than a decelerating Ωₘ = 1 model predicts — a flux ratio 10(−0.25/2.5) = 0.79, so about 12 per cent farther away than that model puts them. Combine flat geometry from the CMB with Ωₘ = 0.315 from dynamics and ΩΛ = 0.685 follows by subtraction; the supernovae then confirm the same value along an independent route. What does not agree is H₀. The CMB read through ΛCDM gives 67.4 ± 0.5 km s⁻¹ Mpc⁻¹ and the Cepheid-calibrated distance ladder gives 73.0 ± 1.0, five standard deviations apart. Because the CMB constrains Ω h² rather than Ω, that disagreement moves the whole budget: at h = 0.730 the same measured densities give Ωₘ = 0.268.
Change one variable at a time
Make the relationship visible.
Set Ωb h² = 0.0224 and Ωₘ h² = 0.143, then drag h from 0.674 to 0.730 — the same measured densities push Ωₘ down from 0.315 to 0.268 and swell the dark-energy residual to 0.732. That is the H₀ tension moving the whole budget.
Ωb BARYONS0.049
Ωc DARK MATTER0.265
ΩΛ DARK ENERGY0.685
NON-BARYONIC MATTER84 %
Live interpretationΩb BARYONS: 0.049. Ωc DARK MATTER: 0.265. ΩΛ DARK ENERGY: 0.685. NON-BARYONIC MATTER: 84 %
Catch the common trap
Explain before calculating.
Big-bang nucleosynthesis predicts a primordial helium mass fraction Yₚ ≈ 0.25 and a deuterium abundance D/H ≈ 2.5 × 10⁻⁵. Only one of the two is used to measure the baryon density. Which, and why?
Choose an answer to test the model.
Practice & worked examples
Reason from the model, then test the result.
EasyTake H₀ = 67.4 km s⁻¹ Mpc⁻¹ and G = 6.674 × 10⁻¹¹ m³ kg⁻¹ s⁻². Find the critical density, then use Ωb = 0.049 to find how many protons of ordinary matter the universe holds per cubic metre. Use 1 Mpc = 3.086 × 10²² m and mₚ = 1.673 × 10⁻²⁷ kg.
- Put H₀ into SI. H₀ = 67.4 × 10³ m s⁻¹ ÷ 3.086 × 10²² m = 2.184 × 10⁻¹⁸ s⁻¹. The megaparsec is the only unit doing any work here.
- Critical density: ρc = 3H₀²/(8πG) = 3 × (2.184 × 10⁻¹⁸)² ÷ (8π × 6.674 × 10⁻¹¹) = 1.431 × 10⁻³⁵ ÷ 1.677 × 10⁻⁹ = 8.53 × 10⁻²⁷ kg m⁻³.
- In protons that is 8.53 × 10⁻²⁷ ÷ 1.673 × 10⁻²⁷ = 5.1 m⁻³ — the whole critical density is five hydrogen atoms in a cubic metre.
- Ordinary matter is only the fraction Ωb of it: ρb = 0.049 × 8.53 × 10⁻²⁷ = 4.18 × 10⁻²⁸ kg m⁻³, and nb = 4.18 × 10⁻²⁸ ÷ 1.673 × 10⁻²⁷ = 0.25 m⁻³.
Answerρc = 8.5 × 10⁻²⁷ kg m⁻³, about 5.1 protons per cubic metre. Baryons supply 0.25 m⁻³ — one proton per four cubic metres, and that is the part of the budget we claim to understand.
MediumThe CMB monopole is T₀ = 2.7255 K and the largest anisotropy on the sky is a dipole of amplitude ΔT = 3.362 mK. Find the observer's speed relative to the frame in which the CMB is isotropic, and compare that dipole with the intrinsic fluctuations, whose rms is about 30 μK.
- The dipole is kinematic, not cosmological: to first order in v/c a moving observer sees ΔT/T₀ = (v/c) cos θ, so the amplitude of the pattern is v/c exactly.
- ΔT/T₀ = 3.362 × 10⁻³ K ÷ 2.7255 K = 1.2335 × 10⁻³, one part in 810.
- v = 1.2335 × 10⁻³ × 2.998 × 10⁵ km s⁻¹ = 370 km s⁻¹. That is the solar-system barycentre; removing the Sun's 220 km s⁻¹ galactic orbit and the Galaxy's motion leaves about 620 km s⁻¹ for the Local Group.
- Now the comparison: 3.362 × 10⁻³ K ÷ 30 × 10⁻⁶ K = 112. The signal carrying the cosmology is a hundredth of the signal carrying our own velocity, which is why the dipole is fitted and removed before any map is published.
Answerv ≈ 370 km s⁻¹, and the dipole is about 112 times the intrinsic anisotropy. The largest feature on the CMB sky is a fact about the observer, not about the universe.
HardPlanck reports the physical densities Ωb h² = 0.0224 and Ωₘ h² = 0.143, and the first acoustic peak fixes Ωₜₒₜ = 1.000. With h = 0.674 find Ωb, Ωₘ, Ωc and ΩΛ and the non-baryonic share of the matter. Then repeat with the distance-ladder value h = 0.730, and say what the H₀ tension does to the budget.
- h² = 0.674² = 0.4543. The CMB constrains Ω h², so every fraction below is a division by this number.
- Ωb = 0.0224 ÷ 0.4543 = 0.0493 and Ωₘ = 0.143 ÷ 0.4543 = 0.3148, so the non-baryonic matter is Ωc = 0.3148 − 0.0493 = 0.2655.
- Non-baryonic share of the matter: 0.2655 ÷ 0.3148 = 0.843, so 84 per cent of the matter is neither protons nor neutrons — and that conclusion needs the BBN cap on Ωb, not the dynamics alone.
- Flatness closes the sum: ΩΛ = 1.000 − 0.3148 = 0.685. Nothing in this calculation measured dark energy; it is the residual, which is why an error anywhere else lands on it.
- Repeat with h = 0.730, h² = 0.5329: Ωb = 0.0420, Ωₘ = 0.2683, ΩΛ = 0.732. The measured densities did not move; only the conversion did.
- An 8.3 per cent disagreement in H₀ therefore shifts Ωₘ by 15 per cent and ΩΛ by 7 per cent. The H₀ tension is not a quibble about a speed — it is a disagreement about the composition.
AnswerAt h = 0.674: Ωb = 0.049, Ωc = 0.266, ΩΛ = 0.685, with 84 per cent of matter non-baryonic. At h = 0.730 the same data give Ωb = 0.042, Ωₘ = 0.268, ΩΛ = 0.732.