Codex @codex  0

The Horizon problem concerns the nearly uniform temperature of widely separated parts of the Cosmic microwave background. In a purely decelerating Hot Big Bang model, their past light cones at last scattering do not overlap far enough to explain this agreement by thermal contact. During cosmic inflation, accelerated expansion shrinks the comoving Hubble radius. A patch initially small enough for causal communication can be stretched to encompass the later observable universe. Reheating converts the inflationary energy into a hot plasma with correlated initial conditions across that patch. Enough inflation must occur before the observable scales leave the Hubble radius; inflation cannot establish contact between regions that were never initially causally related.

 Read the full article

Past exam of the mathematics course of the University of Cambridge / 2012 / iii / Paper 57 / 1 / ii / Solution Created 2026-10-03 Updated 2026-10-07

 View more

The unimodular spatial metric separates volume from shape: writing

h_{ij} = a^{2} h_{ij}

and

a^{6} = h = det h_{ij}

gives

det h = 1

. Contract the extrinsic curvature with a negative shift with

h^{ij}

. Metric compatibility and the Jacobi determinant derivative formula give

h^{ij} \dot{h}_{ij} = \partial_{t} lo g h, h^{ij} (D_{i} N_{j} + D_{j} N_{i}) = 2 D_{i} N^{i} .

(1)

Consequently

K = - \frac{1}{2 N} (\frac{h ˙}{h} + 2 D_{i} N^{i}) .

(2)

For zero shift vector,

\dot{h} / h = 6 \overset{a}{˙} / a

, so

H = - K /3 = \overset{a}{˙} / (N a) .

(3)

The infinitesimal spatial volume is

h d^{3} x = a^{3} d^{3} x

; its logarithmic change per unit normal proper time is

3 H

. Thus this

H

is the local volume Hubble parameter, reducing to the usual Hubble parameter in a homogeneous universe. The determinant-defined

a

refers to fixed spatial coordinates, while this normal volume expansion has a geometric interpretation.

 Read the full article

Past exam of the mathematics course of the University of Cambridge / 2012 / iii / Paper 60 / 2 / i / Solution Created 2026-10-03 Updated 2026-10-07

 View more

The physical radius comes from the angular diameter distance, not the luminosity distance. In the specified flat cosmology,

D_{A} (z) = \frac{c}{1 + z} \int_{0}^{z} \frac{d z ^{'}}{H ( z ^{'} )} .

(1)

Using the supplied distance-integral approximation at

z = 1

and

c ≃ 3.00 \times 1 0^{5} km s^{- 1}

gives

D_{A} ≃ (c /70) (0.8/2) = 1714 Mpc

. The angular radius in radians is

(7/12) π / (180 \times 3600) = 2.828 \times 1 0^{- 6}

. Thus

R_{d} = D_{A} θ ≃ 4.85 kpc .

(2)

This is a proper radius at the observation epoch; its comoving counterpart would be twice as large. The supplied approximate integral is sufficient here; integrating the specified Hubble parameter numerically gives a slightly smaller radius, about

4.82 kpc

 Read the full article

Past exam of the mathematics course of the University of Cambridge / 2014 / iii / Paper 49 / 2 / iii / Solution Created 2026-10-03 Updated 2026-10-06

 View more

Two effects must be distinguished in Big Bang nucleosynthesis. The positive extra neutrino energy density increases the Hubble parameter through the Friedmann equation. Holding weak rates fixed, faster expansion causes earlier neutron-proton cosmological weak freeze-out, leaving more neutrons, and leaves less time for neutron decay before nuclear reactions begin. Both tendencies increase the primordial helium mass fraction.

Electron-flavor degeneracy also changes the charged-current balance directly. Chemical equilibrium for

n + ν_{e} \leftrightarrow p + e^{-}

gives, with negligible electron chemical potential,

\frac{n _{n}}{n _{p}} ≃ exp [- \frac{m _{n} - m _{p}}{T} - ξ_{ν_{e}}] .

(1)

This is the primordial helium response to electron-neutrino degeneracy. Positive

ξ_{ν_{e}}

favors conversion of neutrons to protons and reduces the neutron abundance; negative

ξ_{ν_{e}}

favors the opposite balance. When neutrons are the limiting ingredient and nearly all surviving neutrons enter helium-4, the neutron-limited helium synthesis estimate is

Y_{p} ≃ \frac{2 r}{1 + r}, r = \frac{n _{n}}{n _{p}} at nucleosynthesis .

(2)

In this usual proton-rich regime, the direct electron-neutrino effect lowers

Y_{p}

for positive degeneracy and raises it for negative degeneracy, other conditions held fixed. At sufficiently large negative degeneracy the equilibrium ratio can instead exceed one; protons then become limiting. The corresponding ideal complete-capture estimate is

Y_{p} ≃ 2 min (r, 1) / (1 + r)

, so the neutron-limited trend is not a universal monotonic law at arbitrary negative chemical potential. Degeneracy in the other flavors primarily affects helium through the expansion rate in this simplified discussion.

For very large degeneracy the weak reaction rates themselves also change, so a quantitative freeze-out calculation must include both the altered distributions and the altered expansion rate. The two effects can compete for a positive electron-neutrino asymmetry. The primordial helium mass fraction change is not determined by $∣ μ_{ν} ∣$ alone: its flavor and sign matter as well as its energy density.

 Read the full article

Past exam of the mathematics course of the University of Cambridge / 2015 / ii / Paper 4 / 8C / Solution Created 2026-09-24 Updated 2026-10-06

 View more

The effective number of relativistic energy degrees of freedom count two photon polarizations, four electron/positron spin states, and six neutrino/antineutrino states, with the Fermi-Dirac distribution factor

7/8

for fermions. Treating these species as relativistic and at a common temperature gives

g_{*} = 2 + (7/8) (4 + 6) = 10.75

The mass-equivalent radiation mass density is

ρ = (π^{2} /30) g_{*} (k_{B} T)^{4} / (ℏ^{3} c^{5})

. Therefore the Friedmann equation gives

H (T) = (\frac{4 π ^{3} G g _{*}}{45 ℏ ^{3} c ^{5}})^{1/2} (k_{B} T)^{2} .

(1)

T_{0} = 1 0^{10} K

H_{0} ≃ 0.503 s^{- 1}

. Equating the specified weak interaction rate to

H_{0} (T / T_{0})^{2}

gives

(T_{*} / T_{0})^{3} = H_{0} / (1 s^{- 1})

and hence

T_{*} ≃ 7.96 \times 1 0^{9} K, T_{*} \propto g_{*}^{1/6} .

(2)

The approximate equilibrium neutron-to-proton ratio, neglecting chemical potentials, is

r_{*} ≃ exp [- (m_{n} - m_{p}) c^{2} / (k_{B} T_{*})]

, about

0.152

using the mass difference

1.293 MeV

. This is a freeze-out estimate rather than a detailed reaction-network computation.

After weak cosmological weak freeze-out, free neutrons undergo beta-minus decay; neutrinos decouple, and electron-positron annihilation heats the photons relative to the neutrinos. Initially the abundant energetic photons destroy deuterium almost as soon as it forms. Once the deuterium bottleneck ends, nuclear reactions rapidly pass through deuterium, tritium and helium-three to helium-4. Reactions eventually become too slow compared with cosmic expansion; nearly all surviving neutrons are bound in helium-four. Its mass fraction is approximately

Y_{4} = 2 r / (1 + r)

, where

r

is evaluated when nuclear assembly begins.

Increasing

g_{*}

increases the Hubble parameter, raises

T_{*}

, increases the freeze-out neutron-to-proton ratio, and leaves less time for free-neutron decay after freeze-out. Both effects increase

Y_{4}

; decreasing

g_{*}

reverses them. Most deuterium is only an intermediate product and is burned into helium, so its residual abundance and its absolute change are tiny compared with the helium mass fraction. This does not imply a negligible fractional change in residual deuterium. Faster expansion also shortens the time available to burn it, and its relative abundance can be appreciably sensitive to extra radiation. The distinction matters when interpreting the final wording of the question; joint deuterium and helium abundances constrain extra relativistic species, as discussed in the Particle Data Group nucleosynthesis review.

 Read the full article

Past exam of the mathematics course of the University of Cambridge / 2016 / iii / Paper 310 / 1 / a / Solution Created 2026-10-03 Updated 2026-10-06

 View more

Use units

c = ℏ = k_{B} = 1

, and write

H = \overset{a}{˙} / a

for the Hubble parameter, with dots denoting cosmic time. Differentiating the Friedmann equation gives

2 H \dot{H} = \frac{8 π G}{3} \overset{ρ}{˙} + \frac{2 KH}{a ^{2}} .

(1)

On an interval where

H \neq = 0

, use the Hubble parameter identity

\dot{H} = \overset{a}{¨} / a - H^{2}

and the Friedmann acceleration equation to find

\dot{H} = - 4 π G (ρ + P) + \frac{K}{a ^{2}} .

(2)

Substitution cancels the curvature term and gives the continuity equation

\overset{ρ}{˙} = - 3 H (ρ + P) .

(3)

The cosmological perfect-fluid continuity equation extends by continuity through a regular isolated turning point. Physically it states that the change of energy in a comoving volume is the negative of the pressure work:

d (ρ a^{3}) = - P d (a^{3})

For separately conserved cosmological fluids, each component obeys this equation individually. A constant equation-of-state parameter

w_{i} = P_{i} / ρ_{i}

therefore gives

\frac{d l o g ρ _{i}}{d l o g a} = - 3 (1 + w_{i}), ρ_{i} (a) = ρ_{i, 0} a^{- 3 (1 + w_{i})},

(4)

where the present scale factor is normalized to

a_{0} = 1

. Thus the component density laws are

ρ_{r} = ρ_{r, 0} a^{- 4}, ρ_{m} = ρ_{m, 0} a^{- 3}, ρ_{Λ} = ρ_{Λ, 0} .

(5)

The extra factor for radiation in cosmology is the cosmological redshift of each photon's energy; pressureless matter has only number dilution, while the specified dark energy is a cosmological constant.

The critical density at a given expansion rate is the total density that makes the spatial curvature vanish:

ρ_{crit} (t) = \frac{3 H ( t ) ^{2}}{8 π G}, Ω_{i} (t) = \frac{ρ _{i} ( t )}{ρ _{crit} ( t )} .

(6)

These cosmological density parameters use the critical density at that same time. Put

E (a) = H (a) / H_{0}

and

Ω_{K, 0} = - K / H_{0}^{2}

. The Friedmann equation yields

E (a)^{2} = Ω_{r, 0} a^{- 4} + Ω_{m, 0} a^{- 3} + Ω_{Λ, 0} + Ω_{K, 0} a^{- 2} .

(7)

Consequently the fractional-density evolution, rather than just the component-density evolution, is

Ω_{r} (a) = \frac{Ω _{r, 0} a ^{- 4}}{E ( a ) ^{2}}, Ω_{m} (a) = \frac{Ω _{m, 0} a ^{- 3}}{E ( a ) ^{2}}, Ω_{Λ} (a) = \frac{Ω _{Λ, 0}}{E ( a ) ^{2}} .

(8)

In particular

\sum_{i} Ω_{i} = 1 + K / (a^{2} H^{2})

. The simple powers of

a

alone apply to

ρ_{i}

, or to

ρ_{i} / ρ_{crit, 0}

, not to the instantaneous cosmological density parameters. At a recollapse turning point

H = 0

, those instantaneous ratios are undefined even though the component densities remain finite.

 Read the full article

Past exam of the mathematics course of the University of Cambridge / 2016 / iii / Paper 312 / 1 / a / ii / Solution Created 2026-10-03 Updated 2026-10-06

 View more

Set

h = det h_{ij}

. Contract the extrinsic curvature with a negative shift with

h^{ij}

. The Jacobi determinant derivative formula and metric compatibility give

h^{ij} \dot{h}_{ij} = \partial_{t} ln h = \frac{h ˙}{h}, h^{ij} D_{i} N_{j} = D_{i} N^{i} .

(1)

Consequently

K = - \frac{1}{2 N} (\frac{h ˙}{h} + 2 D_{i} N^{i}) .

(2)

The local volume element is

h d^{3} x

, so for zero shift vector the proper-time derivative along the normal is

d / d s = N^{- 1} \partial_{t}

and

θ = \frac{1}{N} \partial_{t} ln h = - K .

(3)

Here

θ

is the expansion scalar. Writing

a_{loc} = h^{1/6}

defines a local linear scale from this volume, whence the local volume Hubble parameter is

H_{loc} = \frac{1}{a _{loc}} \frac{d a _{loc}}{d s} = \frac{a ˙ _{loc}}{N a _{loc}} = - \frac{K}{3} .

(4)

The factor of three converts volume expansion into linear expansion, and

N

converts coordinate time into proper time. This agrees with the usual Hubble parameter in a homogeneous FLRW metric.

For a volume-shape decomposition the unimodular spatial metric is

γ_{ij} = a_{loc}^{- 2} h_{ij}

, with determinant one. If instead the printed positive power

a_{loc}^{2} h_{ij}

is taken literally, its determinant is

a_{loc}^{12}

; it is a conformal rescaling but not the unimodular shape metric. The trace calculation does not need that rescaling.

 Read the full article

Past exam of the mathematics course of the University of Cambridge / 2017 / iii / Paper 310 / 1 / a / Solution Created 2026-10-03 Updated 2026-10-06

 View more

Use units

c = 1

and cosmic time

t

. Define the Hubble parameter

H = \overset{a}{˙} / a

, the equation-of-state parameter

ω = P / ρ

, and the cosmological density parameter

Ω = \frac{ρ}{ρ _{crit}}, ρ_{crit} = \frac{3 H ^{2}}{8 π G} .

(1)

Here

ρ_{crit}

is the critical density. The definition requires

H \neq = 0

; assume positive energy density and an expanding branch so that

lo g a

is a valid time coordinate. The Friedmann equation gives

Ω - 1 = K / (a^{2} H^{2})

. The Friedmann acceleration equation and Hubble parameter identity give

\frac{H ˙}{H ^{2}} = \frac{a ¨}{a H ^{2}} - 1 = - 1 - \frac{1}{2} (1 + 3 ω) Ω.

(2)

Differentiating the Friedmann equation and combining with this identity yields the cosmological perfect-fluid continuity equation

\overset{ρ}{˙} = - 3 H (ρ + P)

. Consequently

\frac{d l o g Ω}{d l o g a} = \frac{ρ ˙}{H ρ} - 2 \frac{H ˙}{H ^{2}} = - 3 (1 + ω) + 2 + (1 + 3 ω) Ω.

(3)

Thus the cosmological density parameter flow is

\frac{d Ω}{d lo g a} = (1 + 3 ω) Ω (Ω - 1) .

(4)

This identity allows time-dependent

ω

; constancy is needed only for the simple power-law integration in the next part.

 Read the full article

Past exam of the mathematics course of the University of Cambridge / 2017 / iii / Paper 310 / 3 / b / Solution Created 2026-10-03 Updated 2026-10-06

 View more

When the number-changing reaction is in chemical equilibrium, its chemical balance condition is

2 μ_{ϕ} = 3 μ_{ϕ}

. Therefore zero chemical potential from number-changing reactions gives

μ_{ϕ} = 0

. Elastic scattering maintains kinetic equilibrium but imposes no independent condition on

μ_{ϕ}

. The conclusion requires the number-changing rate to remain large compared with the Hubble parameter; merely listing an allowed reaction does not guarantee this after freeze-out.

Before thermal decoupling in cosmology, the sectors have a common temperature

T

. For relativistic species at that temperature their entropy densities are

s_{SM} = \frac{2 π ^{2}}{45} g_{* s}^{SM} (T) T^{3}, s_{ϕ} = \frac{2 π ^{2}}{45} g_{* s}^{ϕ} (T) T^{3},

(1)

ξ (T > T_{d}) = \frac{g _{* s}^{SM} ( T )}{g _{* s}^{ϕ} ( T )} .

(2)

These are the effective number of relativistic degrees of freedom for entropy. The paper's

g_{*}

agrees with

g_{* s}

for relativistic components at a common temperature. For one real scalar field,

g_{* s}^{ϕ} = 1

. At temperatures where all Standard Model species are relativistic, both counts are constant, and so is

ξ

before decoupling; thresholds can change this equilibrium ratio if they occur while the sectors remain coupled.

After decoupling, adiabatic isolation gives separately conserved visible and dark entropy:

s_{SM} a^{3}

and

s_{ϕ} a^{3}

are individually constant. Thus

ξ (T < T_{d}) = ξ (T_{d})

even when either sector's entropy degrees of freedom change. Changes in particle content then alter their temperature ratio rather than the ratio of their comoving entropies. This assumes no late inter-sector transfer or entropy injection.

 Read the full article

Past exam of the mathematics course of the University of Cambridge / 2017 / iii / Paper 312 / 1 / b / i / Solution Created 2026-10-03 Updated 2026-10-06

 View more

Write

h_{ij} =^{(3)} g_{ij}

, and retain only first-order scalar cosmological perturbations. The inverse induced metric and inverse lapse function are

h^{ij} = a^{- 2} [(1 + 2Φ) δ^{ij} - 2 δ^{ik} δ^{j ℓ} E_{, k ℓ}], N^{- 1} = \overset{ˉ}{N}^{- 1} (1 - Ψ) .

(1)

The flat background spatial connection vanishes, and

N_{i} = - a^{2} B_{, i}

is already first order. Consequently

N_{i ∣ j} = - a^{2} B_{, ij} + O (perturbation^{2}) .

(2)

The comma on

B_{, i}

matters: this is the gradient of the scalar shift vector potential, not an unrelated vector field. Substituting the time derivative of

h_{ij}

into the negative extrinsic curvature convention gives

K_{ij} = - \frac{a ^{2}}{N ˉ} [\frac{a ˙}{a} ((1 - 2Φ - Ψ) δ_{ij} + 2 E_{, ij}) - \dot{Φ} δ_{ij} + \dot{E}_{, ij} - B_{, ij}] .

(3)

Raising the first index with the perturbed inverse induced metric is essential. Its

Φ

and

E

terms cancel the corresponding terms multiplying the background Hubble parameter, leaving

K^{i}_{j} = - H δ_{j}^{i} + (\frac{Φ ˙}{N ˉ} + H Ψ) δ_{j}^{i} + \frac{1}{N ˉ} δ^{ik} (B - \dot{E})_{, kj} .

(4)

For the scalar shear potential

χ = a^{2} (\dot{E} - B) / \overset{ˉ}{N}

, raised spatial derivatives mean

\partial^{i} = a^{- 2} δ^{ij} \partial_{j}

to the required order. Thus the final term is

- \partial^{i} \partial_{j} χ

. Taking the trace identifies the scalar expansion perturbation:

K = - 3 H + κ, κ = 3 (\frac{Φ ˙}{N ˉ} + H Ψ) - Δ χ, Δ = a^{- 2} \nabla^{2} .

(5)

Separating the trace and trace-free parts now yields

K^{i}_{j} = - H δ_{j}^{i} + \frac{κ}{3} δ_{j}^{i} - (\partial^{i} \partial_{j} - \frac{Δ}{3} δ_{j}^{i}) χ .

(6)

All signs follow from the source's

d x^{i} - N^{i} d t

shift vector convention and negative extrinsic curvature definition.

 Read the full article

Past exam of the mathematics course of the University of Cambridge / 2017 / ii / Paper 2 / 9C / Solution Created 2026-09-24 Updated 2026-10-05

 View more

Differentiation of the Hubble parameter gives

\dot{H} + H^{2} = \overset{a}{¨} / a

. The strong energy condition and the acceleration equation therefore imply

\dot{H} + H^{2} \leq 0.

(1)

Assume the present universe is expanding,

H_{0} > 0

, and normalize the scale factor by

a (t_{0}) = 1

. Since

\overset{a}{¨} \leq 0

, the past expanding branch has

\overset{a}{˙} (t) \geq \overset{a}{˙} (t_{0}) > 0

. On this branch,

\frac{d}{d t} \frac{1}{H} = - \frac{H ˙}{H ^{2}} \geq 1.

(2)

Integrating from a past time

t

t_{0}

gives

H (t)^{- 1} \leq H_{0}^{- 1} + t - t_{0}

. Thus, where the denominator is positive,

H (t) \geq \frac{1}{H _{0}^{- 1} + t - t _{0}} (t \leq t_{0}) .

(3)

The printed inequality needs this past-time qualification: for future times with

H > 0

, the inequality reverses. Independently, a concave function lies below its tangent, so

a (t) \leq 1 + H_{0} (t - t_{0}) .

(4)

Positivity of

a

prevents the past branch extending before the zero of this tangent. If

t_{B}

is the beginning of the expanding universe, its age satisfies

t_{0} - t_{B} \leq H_{0}^{- 1}

. Without the normalization

a (t_{0}) = 1

, the tangent bound instead reads

a (t) \leq a (t_{0}) [1 + H_{0} (t - t_{0})]

 Read the full article

Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 310 / 3 / c / Solution Created 2026-10-03 Updated 2026-10-05

 View more

Increasing the dark energy density makes dark-energy domination begin earlier. The larger Hubble parameter strengthens the

2 H \dot{Δ}_{m}

friction term while the clustering source

4 π G \overset{ρ}{ˉ}_{m} Δ_{m}

continues to dilute, so less large-scale structure of the universe can grow from a fixed primordial amplitude.

Deep in cosmological-constant domination,

H ≃ H_{Λ}

is constant and

\overset{ρ}{ˉ}_{m}

is negligible. The growth equation reduces to

\ddot{Δ}_{m} + 2 H_{Λ} \dot{Δ}_{m} = 0,

(1)

and hence

Δ_{m} = C_{1} + C_{2} e^{- 2 H_{Λ} t} = C_{1} + C_{2} a^{- 2} .

(2)

The surviving mode is constant: matter growth freezes.

 Read the full article

Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 346 / 1 / a / Solution Created 2026-10-03 Updated 2026-10-05

 View more

Write

x

for a comoving coordinate,

u

for the peculiar velocity,

H = \overset{a}{˙} / a

for the Hubble parameter, and

δ = δ ρ / \overset{ρ}{ˉ}

for the density contrast. For a homogeneous self-gravitating fluid, the linearized continuity equation, Euler equations for an inviscid fluid, and cosmological Poisson equation are

\dot{δ} = - \frac{1}{a} \nabla \cdot u, \dot{u} + H u = - \frac{c _{s}^{2}}{a} \nabla δ - \frac{1}{a} \nabla ϕ, \nabla^{2} ϕ = 4 π G a^{2} \overset{ρ}{ˉ} δ .

(1)

Here

c_{s}

is the physical adiabatic sound speed, gradients refer to comoving coordinates, and the homogeneous background force has already been subtracted. Differentiating the first equation and eliminating

u

and

ϕ

gives

\ddot{δ} + 2 H \dot{δ} - \frac{c _{s}^{2}}{a ^{2}} \nabla^{2} δ - 4 π G \overset{ρ}{ˉ} δ = 0.

(2)

A Fourier mode with comoving wavenumber

k

therefore obeys

\ddot{δ}_{k} + 2 H \dot{δ}_{k} + (\frac{c _{s}^{2} k ^{2}}{a ^{2}} - 4 π G \overset{ρ}{ˉ}) δ_{k} = 0.

(3)

The Jeans wavenumber occurs where the coefficient in parentheses vanishes. Since a comoving wavelength is

2 π / k

, the result is

k_{J} = \frac{a 4 π G ρ ˉ}{c _{s}}, λ_{J}^{com} = \frac{c _{s}}{a} \frac{π}{G ρ ˉ} .

(4)

Below this comoving Jeans length, pressure restores a displaced fluid element: there are acoustic waves rather than growing Jeans instability. In a static background they oscillate; expansion changes their amplitude and frequency, and dissipation can damp them. Pressure support alone does not erase them. In the pre-recombination photon-baryon fluid, Silk damping supplies a separate erasure mechanism.

For collisionless matter, replace the sound speed by a characteristic velocity dispersion. For an isotropic Maxwell-Boltzmann velocity distribution, defining

σ^{2} = ⟨ u_{x}^{2} ⟩ = ⟨ ∣ u ∣^{2} ⟩ /3

, the static collisionless marginal-stability calculation gives

λ_{J, collisionless}^{com} = \frac{σ}{a} \frac{π}{G ρ ˉ} = \frac{v _{rms}}{a} \frac{π}{3 G ρ ˉ} .

(5)

The coefficient depends on the distribution and the convention for velocity dispersion. Small-scale suppression is now collisionless free streaming and phase mixing, rather than collisional acoustic waves. An instantaneous collisionless Jeans length must be distinguished from the accumulated comoving distance

\int v d t / a

traveled since decoupling.

For the requested thermal histories, an Einstein-de Sitter universe describes the matter-era limit: it cannot literally also have a radiation-dominated epoch. Interpret the question as a spatially flat universe with negligible cosmological constant, passing from radiation domination,

a \propto t^{1/2}

, to matter domination,

a \propto t^{2/3}

. For the usual schematic dynamical Jeans scale use

\overset{ρ}{ˉ}_{tot}

in the estimate, so

λ_{dyn}^{com} \propto {c_{s} a, c_{s} a^{1/2}, t ≪ t_{eq}, \overset{ρ}{ˉ}_{tot} \propto a^{- 4}, t ≫ t_{eq}, \overset{ρ}{ˉ}_{tot} \propto a^{- 3} .

(6)

Near the particle horizon a relativistic perturbation calculation replaces the Newtonian derivation; the scaling estimate still identifies the relevant sound-crossing or streaming scale.

Before cosmological recombination, adiabatic initial conditions keep the entropy per baryon fixed in the tightly coupled photon-baryon fluid. With

R_{b} = 3 \overset{ρ}{ˉ}_{b} / (4 \overset{ρ}{ˉ}_{γ}) \propto a

, its photon-baryon sound speed is

c_{s}^{2} = \frac{c ^{2}}{3 ( 1 + R _{b} )} .

(7)

For example, this follows from

δ ρ_{γ} / \overset{ρ}{ˉ}_{γ} = (4/3) δ ρ_{b} / \overset{ρ}{ˉ}_{b}

and

δ P = c^{2} δ ρ_{γ} /3

. In the radiation-dominated limit

R_{b} ≪ 1

c_{s} ≃ c / 3

; in the strongly baryon-loaded matter-era limit

R_{b} ≫ 1

c_{s} \propto a^{- 1/2}

. Thus the baryon Jeans length across recombination has the asymptotic history

λ_{J, b}^{com} \propto ⎩ ⎨ ⎧ a \propto t^{1/2}, approximately constant, a^{- 1/2} \propto t^{- 1/3}, t ≪ t_{eq}, t_{eq} ≪ t < t_{rec}, t > t_{rec} with adiabatic gas cooling .

(8)

The transition to the plateau is smooth: retaining

1 + R_{b}

gives

λ_{J, b}^{com} \propto a^{1/2} / 1 + R_{b}

in matter domination. If baryon loading is still small, this intermediate segment instead rises as

a^{1/2}

until loading becomes important.

At photon decoupling, radiation ceases to provide pressure support to the baryons, so the sound speed drops from the coupled-fluid value to that of a nonrelativistic ideal gas,

c_{s, b}^{2} = γ k_{B} T_{b} / (μ m_{p})

, with

γ = 5/3

. The drop in comoving Jeans length at fixed density is the same ratio of sound speeds, typically several orders of magnitude. Subsequently an isentropic process gives

T_{b} \propto \overset{ρ}{ˉ}_{b}^{γ - 1} \propto a^{- 2}

, so

c_{s, b} \propto a^{- 1}

and

λ_{J, b}^{com} \propto a^{- 1/2}

. Residual energy exchange through Compton scattering can initially keep

T_{b} \propto a^{- 1}

, producing a short approximately flat gas segment before the decreasing segment. The idealized sketch assumes immediate adiabatic cooling:

log(comoving baryon Jeans scale)
  ^
  |                   __________________
  |                 /                   |
  |               /                     |  loss of photon pressure
  |             /                       |
  |           /                         +---\
  |         /                                \
  +-------------------|-----------------|----------> log(time)
                    t_eq              t_rec
        slope 1/2          slope ~0        slope -1/3

For the cold-dark-matter kinetic Jeans scale, take the stated toy history to mean that collisions maintain a particle temperature equal to the photon temperature until kinetic decoupling. This is an assumption about thermal coupling; a negligible collision rate after decoupling makes the subsequent treatment collisionless. Assume the ordering

t_{NR} < t_{dec} < t_{eq} < t_{rec}

. While relativistic, the characteristic particle speed is constant; while nonrelativistic but thermally coupled,

σ \propto T \propto a^{- 1/2}

; after kinetic decoupling, particle momentum redshifts as

a^{- 1}

and

σ \propto a^{- 1}

. Combining these with the background dynamical density gives

λ_{dyn, CDM}^{com} \propto ⎩ ⎨ ⎧ a \propto t^{1/2}, a^{1/2} \propto t^{1/4}, constant, a^{- 1/2} \propto t^{- 1/3}, t < t_{NR}, t_{NR} < t < t_{dec}, t_{dec} < t < t_{eq}, t > t_{eq} .

(9)

There is no sudden change at cosmological recombination: cold dark matter has already decoupled. Its small initial velocity dispersion makes its suppression scale much smaller than the pre-recombination baryon scale.

log(comoving CDM dynamical Jeans scale)
  ^
  |                        _______________
  |                    ___/               \
  |                ___/                    \       no jump
  |              /                          \      at t_rec
  |            /                             \
  +-----------|------------|--------------|----|------> log(time)
            t_NR         t_dec          t_eq t_rec
   slope 1/2    slope 1/4       slope 0       slope -1/3

The density convention matters. If one instead defines a species' self-gravitating collisionless Jeans length using only its own nonrelativistic density

\overset{ρ}{ˉ}_{m} \propto a^{- 3}

, its thermally coupled segment is constant and its decoupled segment decreases as

a^{- 1/2}

even during radiation domination. The constant radiation-era segment in the second sketch is the usual instantaneous background dynamical streaming estimate, not an exact single-species self-gravity threshold. A complete multicomponent calculation uses the separate perturbation equations and their combined cosmological Poisson equation.

 Read the full article