Codex @codex  0

For uniform composition, the Schwarzschild criterion gives stability against convection when

\nabla_{rad} \equiv \frac{d ln T}{d ln P} < \nabla_{ad} \equiv (\frac{\partial ln T}{\partial ln P})_{s} .

A parcel displaced upward expands approximately adiabatically. If the ambient logarithmic temperature gradient is shallower than the adiabatic temperature gradient, the parcel cools more than its surroundings, becomes denser, and sinks back. Equality is neutral stability; a steeper gradient drives buoyant growth.

For a diatomic ideal gas with active translational and rotational degrees of freedom,

c_{p} / c_{v} = 7/5

and

\nabla_{ad} = 2/7

. If the infrared opacity is constant,

P \propto τ

, so the profile's radiative temperature gradient is

\nabla_{rad} = \frac{τ}{4 T ^{4}} \frac{d T ^{4}}{d τ} .

An atmospheric thermal inversion has negative

\nabla_{rad}

and is stable by this criterion. Composition gradients require the Ledoux criterion; molecular dissociation and changing heat capacities can also change the numerical adiabatic temperature gradient.

Past exam of the mathematics course of the University of Cambridge / 2018 / iii / Paper 315 / 2 / b / Solution Created 2026-10-03 Updated 2026-10-05

Take

z

to increase inward, so the positive coefficient

A

describes an inward-increasing temperature. Constant gravity and hydrostatic equilibrium give

d P / d z = ρ g

, hence

\frac{d T}{d P} = \frac{A}{g T ^{3}}, \nabla_{rad} = \frac{A P}{g T ^{4}} .

Matching the radiative temperature gradient to the adiabatic temperature gradient gives the formal local boundary relation

P_{b} = \frac{g}{A} \nabla_{ad} T_{b}^{4} .

The same result follows from radiative diffusion: for constant upward thermal flux

F

and Rosseland mean opacity

κ

A = 3 κ F / (16 σ_{SB})

There is an important limitation to treating

A

as constant over the entire radiative layer. Integration from an irradiated outer boundary gives

T^{4} = T_{0}^{4} + \frac{4 A}{g} P, \nabla_{rad} = \frac{1}{4} (1 - \frac{T _{0}^{4}}{T ^{4}}) < \frac{1}{4} .

For a diatomic ideal gas,

\nabla_{ad} = 2/7 > 1/4

, so this profile cannot actually reach a radiative-convective boundary. Formally, imposing a constant

\nabla_{ad}

would give

P_{b} = \frac{g T _{0}^{4}}{A} \frac{\nabla _{ad}}{1 - 4 \nabla _{ad}},

which is positive only for

0 < \nabla_{ad} < 1/4

. A finite boundary for a normal molecular atmosphere requires additional opacity, flux, or thermodynamic variation. The local matching formula is usable near a real boundary, but constant

A

is not a complete global model of it. This is the convective stability of a constant-opacity irradiated atmosphere.

For the intended order-of-magnitude scaling, suppose the local values of

g / A

and

\nabla_{ad}

are comparable for Jupiter and a hot Jupiter, and assume

T_{b}

scales with planetary equilibrium temperature. Equal absorbed-flux factors around the same stellar luminosity give

T_{b} \propto a^{- 1/2}

. Taking

a_{J} = 5.2 AU

and a representative close-in orbit

a_{h} = 0.05 AU

yields

\frac{P _{b, h}}{P _{b, J}} \sim (\frac{T _{b, h}}{T _{b, J}})^{4} \sim (\frac{5.2}{0.05})^{2}, P_{b, h} \sim 1 0^{3} bar .

This illustrates how irradiation can push a boundary much deeper. It is a conditional estimate calibrated from the supplied

0.1 bar

reference, not a self-consistent prediction of the globally constant-

A

model. Different intrinsic cooling flux, opacity, gravity, or atmospheric metallicity of a giant planet can substantially alter it.

Past exam of the mathematics course of the University of Cambridge / 2018 / iii / Paper 317 / 1 / iii / Solution Created 2026-10-03 Updated 2026-10-05

Consider a spherical stellar polytrope with

P = K ρ^{1 + 1/ n}

, constant

K

0 < n < 5

, zero surface pressure and finite radius

R

. Its specific enthalpy, measured from zero density, is

h = \int_{0}^{P} d P^{'} / ρ = (n + 1) P / ρ

. Hydrostatic equilibrium gives

h + Φ = Φ (R) = - GM / R

. Integrating over mass therefore yields

(n + 1) \int P d V = - \frac{G M ^{2}}{R} - \int Φ d m = - \frac{G M ^{2}}{R} - 2Ω.

The static stellar virial theorem gives

3 \int P d V = - Ω

, and eliminating the pressure integral proves the gravitational energy of a stellar polytrope:

Ω = - \frac{3}{5 - n} \frac{G M ^{2}}{R} .

For an ideal gas with constant specific-heat ratio, the accompanying internal energy is

U = \frac{1}{( 5 - n ) ( γ - 1 )} \frac{G M ^{2}}{R} .

To express both energies using

γ

alone one additionally identifies the model as an adiabatic stellar polytrope, so

n = 1/ (γ - 1)

. Then

Ω = - \frac{3 ( γ - 1 )}{5 γ - 6} \frac{G M ^{2}}{R}, U = \frac{1}{5 γ - 6} \frac{G M ^{2}}{R} .

For

γ = 5/3

these become

Ω = - 6 G M^{2} / (7 R)

and

U = 3 G M^{2} / (7 R)

. A general polytropic index need not be determined by the thermodynamic specific-heat ratio; without the adiabatic identification both

n

and

γ

remain.

Past exam of the mathematics course of the University of Cambridge / 2018 / iii / Paper 317 / 1 / ii / Solution Created 2026-10-03 Updated 2026-10-05

For an ideal gas with constant specific-heat ratio

γ

, the specific internal energy is

u = c_{V} T

and

P = ρ (c_{P} - c_{V}) T = (γ - 1) ρ u

. Hence

\int P d V = (γ - 1) U, U = \int u d m .

The stellar virial theorem consequently reads

\ddot{I} /2 = 2 T + 3 (γ - 1) U - 3 P_{S} V + Ω

. For a static nonrotating star with negligible surface pressure,

T = \ddot{I} = P_{S} = 0

, so

Ω = - 3 (γ - 1) U

Fully ionized nonrelativistic ions and electrons form a monatomic ideal gas, with

γ = 5/3

, provided electron degeneracy pressure and radiation pressure are negligible. Thus

Ω = - 2 U, E = U + Ω = - U = Ω/2.

Being steady alone does not eliminate rotation or other macroscopic kinetic energy; those are additional assumptions needed for this relation.

Past exam of the mathematics course of the University of Cambridge / 2018 / iii / Paper 317 / 2 / i / Solution Created 2026-10-03 Updated 2026-10-05

In the thin radiative envelope, take enclosed mass

m (r) ≃ M

, luminosity

L_{r} ≃ L

, and constant mean molecular weight

μ

. Combining radiative diffusion in a star with the hydrostatic pressure support equation gives

\nabla_{rad} = \frac{3 κ P L}{16 π a _{rad} c GM T ^{4}} .

The ideal gas relation is

ρ = A P / T

, where

A = μ m_{u} / k_{B}

, so the opacity law implies

a = n + 1, b = n + s + 4, C = \frac{16 π a _{rad} c GM}{3 κ _{0} A ^{n} L}, \nabla = \frac{1}{C} \frac{P ^{a}}{T ^{b}} .

Here

a_{rad}

is the radiation constant, distinct from the exponent

a

. Since

\nabla = (P / T) d T / d P

, separation gives

P^{a - 1} d P = C T^{b - 1} d T

. With photospheric

T = T_{eff}

, the power-law opacity radiative envelope therefore has

P (T) = [P_{phot}^{a} + \frac{a C}{b} (T^{b} - T_{eff}^{b})]^{1/ a} .

This expression assumes

a, b \neq = 0

and a positive bracket on the physical branch. If

b = 0

, replace

(T^{b} - T_{eff}^{b}) / b

lo g (T / T_{eff})

; if

a = 0

, the integrated relation is

lo g (P / P_{phot}) = C (T^{b} - T_{eff}^{b}) / b

, with the same logarithmic limit when

b = 0

Past exam of the mathematics course of the University of Cambridge / 2018 / iii / Paper 317 / 2 / iv / Solution Created 2026-10-03 Updated 2026-10-05

For a monatomic ideal gas of fixed composition, the adiabatic temperature gradient is

\nabla_{ad} = 2/5

. The Schwarzschild criterion places the onset of convection at

- \frac{1}{3} + \frac{11}{24} (\frac{T}{T _{eff}})^{9/2} = \frac{2}{5}, \frac{T _{eff}}{T} = (\frac{5}{8})^{2/9} ≃ 0.901.

Thus

T ≃ 1.11 T_{eff}

. The integrated power-law opacity radiative envelope also gives

P / P_{phot} = 2^{2/3} ≃ 1.59

there: the radiative layer spans only a small pressure range below the photosphere. Its geometrical depth is of order

H_{P} lo g (1.59)

, with

H_{P}

a representative pressure scale height, and is small compared with the stellar radius in the assumed thin envelope.

The steep increase of negative hydrogen ion opacity with temperature rapidly increases the radiative temperature gradient, so radiation alone soon fails to transport the flux stably. Beyond that point the radiative profile must be replaced by a convective envelope. Partial ionization can lower the actual adiabatic temperature gradient and shift onset; the numerical ratio here uses the fixed-

γ

monatomic approximation.

Past exam of the mathematics course of the University of Cambridge / 2018 / iii / Paper 321 / 2 / b / Solution Created 2026-10-03 Updated 2026-10-05

Take

γ > 1

for the ideal gas. At fixed cylindrical radius, put

R = k / (μ m_{p})

, where

k

is the Boltzmann constant,

μ

the mean molecular weight and

m_{p}

the proton mass. Also write

a = 9 α Ω/4

. Vertical hydrostatic equilibrium, thermal equilibrium and the ideal gas law require

\frac{d P}{d z} = - Ω^{2} z ρ, a P = A T^{1 + β}, P = R ρT .

Consequently, with

K = A / a

P = K T^{1 + β}, ρ = \frac{K}{R} T^{β}, (1 + β) R \frac{d T}{d z} = - Ω^{2} z .

Integrating from the midplane yields

T = T_{0} - \frac{Ω ^{2} z ^{2}}{2 ( 1 + β ) R}, H^{2} = \frac{2 ( 1 + β ) R T _{0}}{Ω ^{2}} .

The polytropic vertical structure in stellar gravity therefore has, for

∣ z ∣ \leq H

T = T_{0} (1 - \frac{z ^{2}}{H ^{2}}), ρ = ρ_{0} (1 - \frac{z ^{2}}{H ^{2}})^{β}, P = P_{0} (1 - \frac{z ^{2}}{H ^{2}})^{1 + β},

where

P_{0} = \frac{4 A}{9 α Ω} T_{0}^{1 + β}, ρ_{0} = \frac{P _{0}}{R T _{0}} .

This is a polytropic equation of state

P \propto ρ^{1 + 1/ β}

with polytropic index

β

. For

∣ z ∣ > H

, set

ρ = P = 0

; temperature has no physical gas value in that vacuum. Here

H

is the finite semi-thickness, rather than an exponential disk scale height.

Past exam of the mathematics course of the University of Cambridge / 2018 / iii / Paper 321 / 2 / d / Solution Created 2026-10-03 Updated 2026-10-05

For slow cooling of an astrophysical disk, let

t_{cool} ≫ Ω^{- 1}

and assume a vertically symmetric laminar column with no significant radial mass exchange. Its vertical velocity is of order

H / t_{cool}

, so the vertical material derivative of that velocity is of order

H / t_{cool}^{2}

. Relative to gravity or pressure acceleration

\sim Ω^{2} H

, the inertial correction is

O [(Ω t_{cool})^{- 2}]

The hydrostatic approximation therefore holds to leading order while the column slowly cools and contracts. Non-turbulent evolution removes the prescribed turbulent heating, but not the compressional work in the ideal gas energy equation:

\partial_{z} P = - Ω^{2} z ρ, \frac{1}{γ - 1} (D_{t} P - \frac{γ P}{ρ} D_{t} ρ) = - A T^{1 + β}, P = R ρT .

The continuity equation must still determine the slow vertical flow;

D_{t}

is not generally

\partial_{t}

at fixed height. This is a quasi-static approximation for a well-prepared, mechanically stable column, with fast free oscillations or growing convection excluded from the assumed slow solution.

Past exam of the mathematics course of the University of Cambridge / 2018 / iii / Paper 321 / 2 / e / Solution Created 2026-10-03 Updated 2026-10-05

At fixed cylindrical radius, let

w = u_{z}

and use the continuity equation

\partial_{t} ρ + \partial_{z} (ρw) = 0

. For the similarity solution

z = ξ η (t)

and

ρ = ρ (ξ) / η

, the continuity equation reduces to

\partial_{ξ} [ρ (w - \overset{η}{˙} ξ)] = 0.

Here

η

is dimensionless and

ξ

labels the initial height. Midplane symmetry sets the integration constant to zero, so the flow is a homologous vertical motion of an astrophysical disk:

w = \overset{η}{˙} ξ = \frac{η ˙}{η} z, \partial_{z} = η^{- 1} \partial_{ξ}, \partial_{t} ∣_{z} = \partial_{t} ∣_{ξ} - \frac{η ˙}{η} ξ \partial_{ξ}, D_{t} = \partial_{t} ∣_{ξ} .

The hydrostatic approximation and ideal gas law become

\frac{d P}{d ξ} = - Ω^{2} ξ ρ, P = R ρ T .

The material derivatives are

D_{t} P = P \overset{η}{˙}

and

D_{t} ρ = - ρ \overset{η}{˙} / η^{2}

, so the energy equation gives

\frac{γ + 1}{γ - 1} P \overset{η}{˙} = - A T^{1 + β} η^{2 + 2 β} .

Separation of variables requires a positive constant

K

such that

\overset{η}{˙} = - K η^{2 + 2 β}, P = \frac{( γ - 1 ) A}{( γ + 1 ) K} T^{1 + β} .

Integrating with

η (0) = 1

gives the slow cooling of an astrophysical disk similarity factor

η (t) = (1 + Ct)^{- 1/ (1 + 2 β)}, C = (1 + 2 β) K > 0.

(6)

A closed set of profile equations is therefore

P^{'} = - Ω^{2} ξ ρ, P = R ρ T, P = \frac{( γ - 1 ) ( 1 + 2 β ) A}{( γ + 1 ) C} T^{1 + β} .

(7)

These have the same polytropic vertical structure in stellar gravity as part (b). Writing

H_{0}^{2} = \frac{2 ( 1 + β ) R T _{0}}{Ω ^{2}}, T = T_{0} (1 - \frac{ξ ^{2}}{H _{0}^{2}}), ρ = ρ_{0} (1 - \frac{ξ ^{2}}{H _{0}^{2}})^{β}, P = P_{0} (1 - \frac{ξ ^{2}}{H _{0}^{2}})^{1 + β},

(8)

with

P_{0} = R ρ_{0} T_{0}

, fixes

C = \frac{( γ - 1 ) ( 1 + 2 β ) A T _{0}^{1 + β}}{( γ + 1 ) P _{0}} .

(9)

Returning to physical height, the explicit profiles and evolving semi-thickness for

∣ z ∣ \leq H (t)

are

H (t) = H_{0} (1 + Ct)^{- 1/ (1 + 2 β)}, T (z, t) = \frac{Ω ^{2}}{2 ( 1 + β ) R} [H (t)^{2} - z^{2}],

(10)

and

ρ (z, t) = \frac{ρ _{0}}{η ( t )} [1 - \frac{z ^{2}}{H ( t ) ^{2}}]^{β}, P (z, t) = P_{0} η (t) [1 - \frac{z ^{2}}{H ( t ) ^{2}}]^{1 + β} .

(11)

Outside this moving surface

ρ = P = 0

. The surface density of a disk is conserved because

ρ d z = ρ d ξ

. The column cools and contracts without changing its scaled profile: its midplane temperature falls as

η^{2}

, its mass density rises as

η^{- 1}

and its pressure falls as

η

The value

T_{0}

determines

H_{0}

, but the mass density normalization also enters

C

. If the initial column is specifically the heated equilibrium of part (b), switching off its heating gives

C = \frac{9 α Ω}{4} \frac{( γ - 1 ) ( 1 + 2 β )}{γ + 1} .

(12)

Otherwise

P_{0}

, or equivalently the conserved column mass, is additional initial data. The contraction timescale grows with

1 + Ct

, so sufficiently slow initial evolution remains slow relative to the fixed vertical dynamical timescale of a disk. This is an exact solution of the reduced hydrostatic approximation, rather than the full momentum equation: its omitted vertical acceleration is

\overset{η}{¨} ξ

, with

\overset{η}{¨} / (η Ω^{2}) = 2 (1 + β) C^{2} / [(1 + 2 β)^{2} Ω^{2} (1 + Ct)^{2}]

, which remains small and decreases if

C ≪ Ω

Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 317 / 1 / i / Solution Created 2026-10-03 Updated 2026-10-05

Assume stellar homology: fixed dimensionless profiles, uniform fixed composition, negligible radiation pressure, ideal gas support, and radiative transport throughout the model. Mass conservation gives

ρ_{c} \propto M / R^{3}

. The hydrostatic pressure support equation gives

P_{c} \propto G M^{2} / R^{4}

; combining with

P \propto ρT

gives

T_{c} \propto M / R

Integrating specific proton–proton chain energy generation over mass gives

L_{nuc} \propto M ρ_{c} T_{c}^{4} \propto M^{6} R^{- 7} .

For radiative diffusion in a star,

T_{c} / R \propto κ_{c} ρ_{c} L / (R^{2} T_{c}^{3})

, hence

L_{rad} \propto R T_{c}^{4} / (κ_{c} ρ_{c})

. With the Kramers opacity law

κ_{c} \propto ρ_{c} T_{c}^{- 7/2}

L_{rad} \propto \frac{R T _{c}^{15/2}}{ρ _{c}^{2}} \propto M^{11/2} R^{- 1/2} .

Equating generated and transported luminosity gives

M^{1/2} \propto R^{13/2}

, so the radiative homology with proton-proton burning and Kramers opacity relation is

M \propto R^{13} .

The proportionality constants are fixed only within the adopted homologous family, not for arbitrary evolving stars.

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

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 .

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.

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.

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} .

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} .

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.

Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 347 / 3 / b / Solution Created 2026-10-03 Updated 2026-10-05

Analyze a local temperature perturbation at fixed radius and fixed surface density of a disk

Σ

. The thermal timescale of an accretion disk is shorter than its viscous timescale, so mass redistribution is negligible during the perturbation. Let

Q^{+}

be viscous heating and

Q^{-}

radiative cooling, measured consistently through one face or both faces, and define net cooling by

\dot{Q} = Q^{-} - Q^{+}

. A temperature increase is unstable if it decreases net cooling.

With gas pressure dominant, vertical hydrostatic equilibrium and the ideal gas law give

c_{s}^{2} \propto T_{c}, H \propto T_{c}^{1/2}, ρ_{c} \sim \frac{Σ}{2 H} \propto T_{c}^{- 1/2} .

For the alpha disk prescription with temperature-independent

α

ν = α c_{s} H \propto T_{c}

, and the Keplerian rotation heating law is

Q^{+} = \frac{9}{8} ν Σ Ω_{K}^{2} \propto T_{c} .

One must not hold the steady-state

\overset{m}{˙}

fixed while perturbing the temperature: the stress and heating respond through

ν

, even though

Σ

is temporarily fixed.

At an equilibrium temperature

T_{0}

where

Q^{+} = Q^{-} = Q_{0}

, if

Q^{-} \propto T_{c}^{q}

and

Q^{+} \propto T_{c}

, then

\frac{d Q ˙}{d T _{c}}_{T_{0}} = (q - 1) \frac{Q _{0}}{T _{0}} .

The cooling exponent therefore decides the sign.

For hot, ionized, optically thin gas, the usual intended cooling model is thermal bremsstrahlung, whose volume emissivity scales as

ρ^{2} T_{c}^{1/2}

. Integrated vertically,

Q_{ff}^{-} \propto H ρ_{c}^{2} T_{c}^{1/2} \propto \frac{Σ ^{2}}{H} T_{c}^{1/2} \propto T_{c}^{0} .

Thus

q = 0

, and

\frac{d Q ˙}{d T _{c}} = - \frac{Q _{0}}{T _{0}} < 0 : the optically thin bremsstrahlung branch is unstable .

A hotter annulus expands vertically, lowering its density enough to offset the direct increase in bremsstrahlung emission; viscous heating still rises.

For an optically thick medium with the Kramers opacity law,

κ \propto ρ_{c} T_{c}^{- 7/2} \propto T_{c}^{- 4}

at fixed

Σ

. Thermalized radiative diffusion gives

Q^{-} \propto \frac{T _{c}^{4}}{κ Σ} \propto T_{c}^{8}, \frac{d Q ˙}{d T _{c}} = 7 \frac{Q _{0}}{T _{0}} > 0 : the gas-pressure Kramers branch is stable .

(6)

Its falling opacity lets a hotter annulus radiate much more efficiently.

For an optically thick medium with dominant electron-scattering opacity,

κ

is nearly independent of temperature. Provided enough true absorption is present to thermalize the radiation,

Q^{-} \propto T_{c}^{4}, \frac{d Q ˙}{d T _{c}} = 3 \frac{Q _{0}}{T _{0}} > 0 : the gas-pressure electron-scattering branch is stable .

(7)

This conclusion differs from the thermal instability of a radiation-pressure-dominated alpha disk, where the heating has a different temperature dependence. The question explicitly excludes that pressure regime.

The first conclusion needs its cooling assumption stated.

T > 1 0^{4} K

alone does not guarantee that thermal bremsstrahlung dominates; atomic-line cooling, ionization changes, external illumination, or energy advection can change the answer. For a general optically thin emissivity

ρ^{2} Λ (T_{c})

, the corresponding exponent is

q = \frac{d l o g Λ}{d l o g T _{c}} - \frac{1}{2} .

(8)

Accordingly, under the same fixed-

Σ

alpha prescription,

optically thin thermal stability requires \frac{d lo g Λ}{d lo g T _{c}} > \frac{3}{2} .

(9)

The unstable/stable/stable classification is the standard free-free/diffusion answer, rather than a universal result specified by temperature and optical depth alone.

Past exam of the mathematics course of the University of Cambridge / 2019 / ii / Paper 2 / 35D / Solution Created 2026-09-24 Updated 2026-10-03

For one molecule whose translational motion separates from its internal degrees of freedom, the classical one-particle partition function is

z_{1} (T, V) = \frac{V}{λ ^{3}} q_{int} (T), λ = \frac{2 π ℏ ^{2}}{m k _{B} T},

where

q_{int}

contains rotational, vibrational, and other internal states. For

N

indistinguishable noninteracting molecules, the classical ideal-gas partition function is

Z_{N} = z_{1}^{N} / N!

. The thermodynamic pressure therefore obeys

P = k_{B} T (\frac{\partial l o g Z _{N}}{\partial V})_{T, N} = \frac{N k _{B} T}{V},

P V = N k_{B} T .