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, 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

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Past exam of the mathematics course of the University of Cambridge / 2018 / iii / Paper 344 / 1 / c / Solution Created 2026-10-03 Updated 2026-10-05

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At thermal equilibrium, microscopic reversibility equates the probabilities of a path and its reversed path when both include their Boltzmann distribution initial weights. Denote their endpoint states by

z_{1}, z_{2}

, including velocity if needed. Since the Hamiltonian is even under time reversal in classical mechanics,

ρ_{eq} (z_{1}) P_{F} = ρ_{eq} (z_{2}) P_{B}, ρ_{eq} (z) \propto e^{- β H (z)}, β = \frac{1}{k _{B} T} .

(1)

This detailed balance condition and the energy balance for an autonomous Lagrangian yield

\frac{P _{F}}{P _{B}} = e^{- β (H_{2} - H_{1})}, \frac{2 ζ}{σ ^{2}} = β, σ^{2} = 2 ζ k_{B} T .

(2)

This is the fluctuation-dissipation relation for a Langevin particle: the strength of Gaussian white noise is fixed by the damping and temperature, with

k_{B}

the Boltzmann constant.

For an equilibrium coarse-grained variable, the unresolved microscopic states contribute entropy; their statistical weight is encoded in the Helmholtz free energy, rather than in a single microscopic energy. Relative to the same reference measure,

ρ_{eq} (x) \propto e^{- βF (x)}

, so microscopic reversibility becomes

\frac{P _{F}}{P _{B}} = e^{- β (F_{2} - F_{1})} .

(3)

This extension assumes an equilibrium coarse-grained description with reversible path statistics; externally driven dynamics need not obey this relation.

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Past exam of the mathematics course of the University of Cambridge / 2018 / ii / Paper 3 / 36A / b / ii / Solution Created 2026-09-24 Updated 2026-10-03

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At fixed temperature, the excited population is largest as

μ ↑ 0

, hence

N_{ϵ}^{m a x} (T) = \frac{M}{e ^{ϵ / (k T)} - 1} .

(1)

N > N_{ϵ}^{m a x}

, the excited levels cannot accommodate all particles. The excess

N - N_{ϵ}^{m a x}

occupies the zero-energy level macroscopically: this is Bose-Einstein condensation.

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Past exam of the mathematics course of the University of Cambridge / 2018 / ii / Paper 4 / 35A / a / Solution Created 2026-09-24 Updated 2026-10-03

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Expanding the trace in the spin basis gives

Tr (M^{N}) = \sum_{s_{1}, \dots, s_{N}} M_{s_{1} s_{2}} M_{s_{2} s_{3}} \dots M_{s_{N} s_{1}} = \sum_{{s_{i}}} e^{- β H} = Z,

(1)

so the transfer matrix for the one-dimensional Ising model reproduces the partition function.

In the ordered basis

s = + 1, - 1

M = (e^{β (J + B)} e^{- β J} e^{- β J} e^{β (J - B)}) .

(2)

Its eigenvalues are

λ_{\pm} = e^{β J} cosh (βB) \pm e^{2 β J} sinh^{2} (βB) + e^{- 2 β J} .

(3)

Since

λ_{+} > ∣ λ_{-} ∣

at every positive temperature, the thermodynamic limit gives the Helmholtz free energy per spin

f = - k T lo g λ_{+} .

(4)

This is an analytic function of

T

and

B

for

T > 0

, so the one-dimensional short-range Ising model has no finite-temperature phase transition.

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Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 303 / 2 / a / Solution 2026-10-03

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Write

m_{i} = μ_{i}^{2}

and

r^{2} = ϕ_{1}^{2} + ϕ_{2}^{2}

. The uniform Landau free energy is

V (ϕ_{1}, ϕ_{2}) = \frac{1}{2} m_{1} ϕ_{1}^{2} + \frac{1}{2} m_{2} ϕ_{2}^{2} + g r^{4},

(1)

so stationarity requires

ϕ_{i} (m_{i} + 4 g r^{2}) = 0.

(2)

For

m_{1}, m_{2} > 0

, the unique ground state is

(0, 0)

and the

Z_{2} \times Z_{2}

discrete symmetry is unbroken. If

m_{1} < 0

and

∣ m_{1} ∣ > ∣ m_{2} ∣

, then

(ϕ_{1}, ϕ_{2}) = (\pm \frac{- m _{1}}{4 g}, 0),

(3)

which breaks the first

Z_{2}

and leaves the second intact. This includes the case in which both masses are negative, because

m_{1}

is then the more negative one. The remaining possible ordered case under the stated inequality is

m_{1} > 0 > m_{2}

, for which

(ϕ_{1}, ϕ_{2}) = (0, \pm \frac{- m _{2}}{4 g}),

(4)

and only the second

Z_{2}

is broken. The Hessian matrix in each ordered state is positive because the uncondensed direction has squared mass

m_{k} - m_{j} > 0

, where

m_{j}

is the condensed, more negative mass.

Near either continuous transition the nonzero order parameter is proportional to

(- m_{i})^{1/2}

. Hence the mean-field critical exponents are

β_{1} = β_{2} = \frac{1}{2} .

(5)

The lower critical dimension is the dimension at or below which fluctuations destroy the proposed finite-temperature ordered phase. Here the broken symmetry is discrete, so

d_{l} = 1

. In one dimension a domain wall interpolating between the two signs has finite energy, whereas its possible position gives an entropy growing as

lo g L

. Domain walls therefore occur with nonzero density at every positive temperature and split the system into domains of finite typical length. Thus there is no finite-temperature ordered phase in one dimension.

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Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 303 / 3 / i / Solution 2026-10-03

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At a critical point, a local field operator

O

has scaling dimension

Δ_{O}

O (x) \mapsto b^{- Δ_{O}} O (x / b)

under coarse-graining by

b

. The coupling

u

\int d^{d} x u O

has renormalization-group eigenvalue

y_{u} = d - Δ_{O} .

(1)

A perturbation is a relevant operator when

y_{u} > 0

, an irrelevant operator when

y_{u} < 0

, and a marginal operator when

y_{u} = 0

; nonlinear terms in its renormalization-group beta function decide the fate of a marginal coupling. Irrelevant microscopic interactions decay under coarse-graining, so many distinct systems approach the same fixed point and share one universality class.

Near criticality the singular free energy density is of order one per correlation volume:

f_{s} \sim ξ^{- d} \sim t^{ν d} .

(2)

The heat capacity contains two derivatives with respect to temperature, so

c_{s} \sim \frac{d ^{2} f _{s}}{d t ^{2}} \sim t^{ν d - 2} = t^{- α} .

(3)

Consequently the hyperscaling relation is

α = 2 - d ν .

(4)

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Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 311 / 4 / b / Solution Created 2026-10-03 Updated 2026-10-05

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Classically, the laws of black-hole mechanics have the same form as the laws of thermodynamics, but this could have been only a formal analogy. Hawking radiation makes it physical: a stationary black hole emits approximately thermal radiation with

T_{H} = \frac{ℏ κ}{2 π k _{B} c},

(1)

so surface gravity is proportional to genuine temperature.

The correspondence is:

constancy of $κ$ on a stationary horizon becomes constancy of equilibrium temperature;
the First law of black-hole mechanics

d M = \frac{κ}{8 π G} d A + Ω_{H} dJ + Φ_{H} d Q

(2)

matches the first law of thermodynamics and fixes the Bekenstein-Hawking entropy

S_{B H} = \frac{k _{B} c ^{3} A}{4 G ℏ};

(3)

Hawking's area theorem becomes a second law, replaced in the quantum theory by the generalized second law for $S_{B H} + S_{outside}$ ;
the impossibility of reaching $κ = 0$ by a finite process parallels the unattainability of absolute zero.

A real Hawking temperature therefore identifies horizon area as thermodynamic entropy and turns the geometric laws into thermodynamic laws.

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Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 314 / 1 / b / Solution Created 2026-10-03 Updated 2026-10-05

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The calculation in part (a) leaves the cross-helicity conservation law source

Q_{h} = T B \cdot \nabla s .

(1)

At positive temperature, this vanishes exactly when the magnetic field is tangent to constant-specific entropy surfaces:

B \cdot \nabla s = 0

. A homentropic flow is a sufficient special case; uniform entropy throughout space is not necessary. Mere advection of specific entropy,

Ds / D t = 0

, constrains its variation along the velocity rather than along the magnetic field, so it does not by itself eliminate this source. To conserve total cross-helicity in a volume, the net boundary flux must also vanish.

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Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 315 / 2 / a / Solution Created 2026-10-03 Updated 2026-10-05

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Assume local thermodynamic equilibrium, negligible scattering of thermal radiation, and a plane-parallel atmosphere. With inward optical depth

τ_{λ}

and outward ray cosine

μ

, the radiative transfer equation is

μ \frac{d I _{λ}}{d τ _{λ}} = I_{λ} - B_{λ} [T (τ_{λ})] .

(1)

For a deep atmosphere, its formal solution of the radiative transfer equation is

I_{λ} (0, μ) = \int_{0}^{\infty} B_{λ} [T (t)] e^{- t / μ} \frac{d t}{μ} .

(2)

The weighting samples

τ_{λ}

of order unity. The Eddington-Barbier relation makes this precise when the source function is nearly linear:

I_{λ} (0, μ) ≃ B_{λ} [T (τ_{λ} = μ)]

A molecular band with greater opacity reaches unit optical depth higher than the nearby continuum. If temperature decreases upward, that band samples cooler gas and appears in absorption. If an atmospheric thermal inversion makes the upper gas hotter, the band appears in emission. If both depths have the same temperature, an opaque isothermal atmosphere has

I_{λ} = B_{λ} (T),

(3)

and no opacity-dependent features. This last conclusion assumes all wavelengths are opaque or the lower boundary emits the same Planck function; an optically thin isothermal layer over a different background need not be featureless.

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Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 315 / 2 / c / Solution Created 2026-10-03 Updated 2026-10-05

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Assume all three planets retain approximately their original hydrogen-helium bulk composition. Stellar encounters change orbital energy and irradiation, not automatically the planet's elemental abundances. Their atmospheric structures can become approximately stationary long before their interiors finish cooling.

For the unperturbed Jupiter-like planet at

a ≃ 5.2 au

, the zero-albedo globally averaged planetary equilibrium temperature is about

122 K

. Stellar light heats the outer atmosphere, while internal cooling supports a deeper temperature gradient and convection. At cooler pressures, chemical equilibrium favors methane and ammonia; condensate clouds can include ammonia at high levels and water deeper down. Disequilibrium chemistry in an exoplanet atmosphere can preserve carbon monoxide or other species from deeper layers.

For the inward-migrated planet at

a ≃ 0.052 au

, the irradiation-only planetary equilibrium temperature is ten times higher, about

1220 K

, because

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

. Its irradiated planetary atmosphere has a heated radiative exterior, potentially strong day-night differences, and a deep radiative-convective boundary. At suitable pressures, chemical equilibrium increasingly favors carbon monoxide over methane; water remains important, while alkali absorption and high-temperature condensates can matter. An atmospheric thermal inversion depends on absorbers, clouds, and irradiation and is not guaranteed simply by migration.

The ejected object is a rogue planet. Its irradiation-based planetary equilibrium temperature becomes very small, but its actual emitting temperature is set primarily by internal cooling and Kelvin-Helmholtz contraction. After only several million years it can remain warm and self-luminous, with deep convection and an outward-cooling radiative atmosphere. Its photospheric chemistry and clouds depend on that cooling temperature; they cannot be inferred from the absence of a host star alone. None of these cases fixes an exact pressure-temperature profile without a cooling model, opacities, and atmospheric composition.

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Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 315 / 3 / c / Solution Created 2026-10-03 Updated 2026-10-05

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Exoplanet transit photometry discovers a giant planet through a periodic flux decrement and measures

R_{p} / R_{⋆} ≃ δ

. Stellar-radius estimates turn this into a planetary radius. The radial-velocity method discovers the stellar orbital reflex motion and constrains

M_{p} sin i

. Follow-up exoplanet transit photometry is required to measure a geometric radius when such a planet also transits; a radial-velocity detection alone gives no direct size. Alternatively, exoplanet direct imaging finds young luminous giants, whose sizes are inferred less directly from luminosity, temperature, distance, and a planetary mass-radius relation or atmosphere model.

The two broad explanations for hot-Jupiter radius inflation are retention of primordial heat and addition of new interior power.

Delayed cooling of an inflated giant planet can arise from enhanced atmospheric opacity, which slows radiative escape; an irradiation-maintained radiative blanket, which insulates the convective interior; or compositional stratification and inefficient layered convection, which inhibit the outward transport of heat. These alter the rate of Kelvin-Helmholtz contraction.

Heating of an inflated giant planet can arise from tidal heating maintained by eccentricity or obliquity; Joule heating of currents driven by atmospheric winds through a magnetic field; or downward transport and dissipation of atmospheric mechanical energy generated by irradiation. To affect radius, the energy must be deposited at a depth and rate that changes the interior cooling balance. Simply absorbing starlight high in the atmosphere does not automatically supply deep heating. These are proposed mechanisms with different efficiencies, not six universally established contributions in every inflated planet.

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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

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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.

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Past exam of the mathematics course of the University of Cambridge / 2019 / ii / Paper 3 / 35D / Solution Created 2026-09-24 Updated 2026-10-03

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The chemical potential is the change in internal energy when one particle is added at fixed entropy and volume:

μ = (\frac{\partial E}{\partial N})_{S, V} .

(1)

Equivalently,

μ = (\partial F / \partial N)_{T, V}

for the Helmholtz free energy.

Let a microstate

s

have energy

E_{s}

and particle number

N_{s}

. Maximizing the Gibbs entropy subject to normalization and fixed mean values of

E

and

N

is the maximum-entropy derivation of equilibrium ensembles. With Lagrange multipliers

α

β

, and

- β μ

, vary

- \sum_{s} p_{s} lo g p_{s} - α (\sum_{s} p_{s} - 1) - β \sum_{s} p_{s} (E_{s} - μ N_{s}) .

(2)

The stationarity equation is

- lo g p_{s} - 1 - α - β (E_{s} - μ N_{s}) = 0

. Normalization therefore gives the grand canonical ensemble

p_{s} = \frac{e ^{- β (E_{s} - μ N_{s})}}{Z}, Z = s \sum e^{- β (E_{s} - μ N_{s})}, β = \frac{1}{k _{B} T} .

(3)

Here

Z

is the grand canonical partition function and

k_{B}

is the Boltzmann constant.

For one fermionic quantum state of energy

ε

, the Pauli exclusion principle permits occupation numbers only

n = 0, 1

. Its two grand-canonical weights are

1

and

e^{- β (ε - μ)}

, so its mean occupation is the Fermi-Dirac distribution

n_{F} (ε) = \frac{e ^{- β (ε - μ)}}{1 + e ^{- β (ε - μ)}} = \frac{1}{e ^{β (ε - μ)} + 1} .

(4)

For a free nonrelativistic particle,

ε = ℏ^{2} k^{2} / (2 m)

. In a region of area

A

, the number of wave-vector states in the annulus from

k

k + d k

, including the two spin angular momentum states, is

2 \frac{A}{( 2 π ) ^{2}} 2 πk d k .

(5)

Since

k d k = (m / ℏ^{2}) d ε

, the two-dimensional free-electron density of states is constant:

g (ε) = \frac{A m}{π ℏ ^{2}} .

(6)

We now use units in which

k_{B} = 1

, as in the question. At zero temperature, the Fermi-Dirac distribution is a step function, and hence

N = g \int_{0}^{ε_{F}} d ε = g ε_{F} .

(7)

At positive temperature the same fixed particle number satisfies the exact relation

N = g \int_{0}^{\infty} \frac{d ε}{e ^{(ε - μ) / T} + 1} = g T lo g (1 + e^{μ / T}) .

(8)

Thus

μ = T lo g (e^{ε_{F} / T} - 1) = ε_{F} + T lo g (1 - e^{- ε_{F} / T}) = ε_{F} + O (T e^{- ε_{F} / T}) .

(9)

This is the low-temperature particle-number cancellation for constant density of states. Therefore, for

T ≪ ε_{F}

, the mean number of particles in the energy interval

[ε, ε + d ε]

g (ε) n_{F} (ε) d ε ≃ \frac{N}{ε _{F}} \frac{d ε}{e ^{(ε - ε_{F}) / T} + 1} .

(10)

Finally, compare the internal energy with its zero-temperature value. The thermally excited particles above

ε_{F}

and holes below

ε_{F}

have equal leading particle numbers, so their terms proportional to

ε_{F}

cancel. With

x = ∣ ε - ε_{F} ∣/ T

U (T) - U (0) = 2 g T^{2} \int_{0}^{\infty} \frac{x d x}{e ^{x} + 1} = \frac{π ^{2}}{6} g T^{2},

(11)

where the integral follows from the fermion-to-boson thermal integral ratio. Differentiation gives the linear low-temperature heat capacity of a two-dimensional Fermi gas:

C_{A} = (\frac{\partial U}{\partial T})_{N, A} = \frac{π ^{2}}{3} g T = \frac{π ^{2} N}{3 ε _{F}} T .

(12)

The requested power law is therefore linear in $T$ , with exponent one.

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Past exam of the mathematics course of the University of Cambridge / 2022 / ii / Paper 3 / 35A / a / Solution Created 2026-09-24 Updated 2026-09-25

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Bosons have symmetric many-particle states, so any number may occupy the same one-particle state. At low temperature, an ideal Bose gas therefore places a macroscopic population in its lowest one-particle state; below the critical temperature this is Bose-Einstein condensation.

Fermions have antisymmetric many-particle states and obey the Pauli exclusion principle, so each one-particle state can hold at most one fermion of each available internal label. At zero temperature, a noninteracting Fermi gas fills distinct states in increasing order of energy up to the Fermi energy. Its many-particle ground state is thus the filled Fermi sea.

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Past exam of the mathematics course of the University of Cambridge / 2023 / ii / Paper 2 / 37A / f / Solution Created 2026-09-24 Updated 2026-09-25

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The condition

L ≪ na

implies

∣ τ a ∣ ≪ 1

. The small-argument expansion of the hyperbolic tangent gives

tanh x = x + O (x^{3})

, so

\frac{L}{na} ≃ τ a .

(1)

Using

f = k_{B} T τ

yields the entropic Hooke law for a one-dimensional chain

f ≃ \frac{k _{B} T}{n a ^{2}} L .

(2)

Thus the effective spring constant is

k_{B} T / (n a^{2})

At fixed tension, the exact extension is

L = na tanh (\frac{f a}{k _{B} T}) .

(3)

Increasing the temperature decreases the positive argument of the hyperbolic tangent, so the chain contracts. In the small-extension regime this becomes