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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 / ii / Paper 3 / 13C / a / Solution Created 2026-09-24 Updated 2026-10-03

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Define the Multivariate Fokker-Planck probability current by

J_{i} = A_{i} P - \frac{1}{2} \frac{\partial}{\partial x _{j}} (B_{ij} P) .

(1)

Then

- \frac{\partial J _{i}}{\partial x _{i}} = - \frac{\partial}{\partial x _{i}} (A_{i} P) + \frac{1}{2} \frac{\partial ^{2}}{\partial x _{i} \partial x _{j}} (B_{ij} P),

(2)

so the Fokker-Planck equation is exactly the continuity equation

\partial_{t} P + \nabla \cdot J = 0

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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 2 / 9B / a / Solution Created 2026-09-24 Updated 2026-10-03

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Differentiate the Friedmann equation

H^{2} = 8 π Gρ /3

and use the continuity equation

\overset{ρ}{˙} = - 3 H (ρ + P)

2 H \dot{H} = \frac{8 π G}{3} \overset{ρ}{˙} = - 8 π G H (ρ + P),

(1)

\dot{H} = - 4 π G (ρ + P)

. Since

H = \overset{a}{˙} / a

\frac{a ¨}{a} = \dot{H} + H^{2} = - 4 π G (ρ + P) + \frac{8 π G}{3} ρ .

(2)

Hence the Raychaudhuri equation is

\frac{a ¨}{a} = - \frac{4 π G}{3} (ρ + 3 P) .

(3)

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Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 329 / 1 / Solution 2026-09-29

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In a cross-section normal to the cylinder axes, measure

y

from the line of contact. The two circular boundaries have the parabolic approximation

h (y) = \frac{y ^{2}}{a} + O (\frac{y ^{4}}{a ^{3}})

(1)

for their separation. The contact point of the meniscus is at

y = w + o (w)

, so the leading cross-sectional area is

A (w) = \int_{0}^{w} \frac{y ^{2}}{a} d y = \frac{w ^{3}}{3 a} .

(2)

The area of the small meniscus cap is

O (w^{4} / a^{2})

and is lower order. At its upper end the gap has width

h (w) = w^{2} / a

. A tangent semicircle therefore has radius

w^{2} / (2 a)

and curvature

κ = \frac{2 a}{w ^{2}} .

(3)

The Young–Laplace equation makes the liquid pressure, relative to the nearly uniform gas pressure,

p = - γκ = - \frac{2 γa}{w ^{2}}, p_{z} = \frac{4 γa}{w ^{3}} w_{z} .

(4)

At a fixed

y

, lubrication theory gives a planar Poiseuille flow through a gap of width

h (y)

, with axial flux per unit

y

d Q = - \frac{h ( y ) ^{3}}{12 μ} p_{z} d y .

(5)

Integrating across the cusp,

Q = - \frac{p _{z}}{12 μ} \int_{0}^{w} \frac{y ^{6}}{a ^{3}} d y = - \frac{p _{z} w ^{7}}{84 μ a ^{3}} = - \frac{γ}{21 μ a ^{2}} w^{4} w_{z} .

(6)

The continuity equation

A_{t} + Q_{z} = 0

now gives

\frac{\partial w ^{3}}{\partial t} = \frac{γ}{7 μ a} \frac{\partial}{\partial z} (w^{4} \frac{\partial w}{\partial z}) .

(7)

Set

q = w^{3}

. Since

w^{4} w_{z} = \frac{1}{5} (q^{5/3})_{z}

, this is the porous medium equation

q_{t} = K (q^{5/3})_{zz}, K = \frac{γ}{35 μ a} .

(8)

Conservation of the fixed volume

V = \int_{R} A d z = \frac{1}{3 a} \int_{R} q d z

(9)

and dimensional analysis give the self-similar solution

q = t^{- 3/8} F (z / t^{3/8})

. One integration of the resulting ordinary differential equation yields the compactly supported Barenblatt solution

F (η) = (C - B η^{2})_{+}^{3/2}, B = \frac{21 μ a}{8 γ} .

(10)

Equivalently,

w (z, t) = t^{- 1/8} (C - B \frac{z ^{2}}{t ^{3/4}})_{+}^{1/2} .

(11)

Its tip is

z_{N} = (C / B)^{1/2} t^{3/8}

. Using

\int_{- 1}^{1} (1 - s^{2})^{3/2} d s = \frac{3 π}{8}

(12)

in the volume constraint gives

C^{2} = 8 aV B / π

. Therefore

z_{N} (t) = (\frac{8 Va}{π})^{1/4} (\frac{8 γ t}{21 μ a})^{3/8} .

(13)

For vertical cylinders at equilibrium, hydrostatic pressure gives

p - p_{atm} = - ρ g z

. Balancing this with the capillary pressure

- 2 γa / w^{2}

gives the large-height profile

w (z) \sim (\frac{2 γa}{ρ g z})^{1/2} .

(14)

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Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 347 / 3 / a / Solution 2026-09-29

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Vertically integrating the continuity equation and imposing a steady axisymmetric flow gives

\frac{1}{R} \frac{d}{d R} (R Σ u_{R}) = 0.

(1)

Thus the inward-positive mass accretion rate is

\overset{m}{˙} = - 2 π R Σ u_{R} = constant .

(2)

The zero-torque condition at

R_{ISCO}

gives the standard Shakura--Sunyaev thin disk relation

ν Σ = \frac{m ˙}{3 π} [1 - (\frac{R _{ISCO}}{R})^{1/2}] .

(3)

Combining this relation with mass conservation, or substituting it into the stated drift equation, yields

u_{R} = - \frac{3 ν}{2 R} [1 - (\frac{R _{ISCO}}{R})^{1/2}]^{- 1} .

(4)

Far outside the inner edge this reduces to

u_{R} ≃ - 3 ν / (2 R)

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Past exam of the mathematics course of the University of Cambridge / 2022 / ib / Paper 3 / 6B / b / Solution Created 2026-09-24 Updated 2026-09-25

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The probability density and probability current are

ρ = ∣ ψ ∣^{2}, J = \frac{ℏ}{2 mi} (ψ^{*} \frac{\partial ψ}{\partial x} - ψ \frac{\partial ψ ^{*}}{\partial x}) .

(1)

For a stationary state,

ψ (x, t) = χ (x) e^{- i Et /ℏ}

, so

ρ = ∣ χ ∣^{2}

is independent of time. The continuity equation then gives

\partial J / \partial x = 0

. Hence

J is independent of x .

(2)

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Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 329 / 3 / Solution 2026-09-28

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The nearly occluding sphere in a cylindrical tube has gap thickness

H = O (ϵ a)

and axial length

L = O (a H) = O (a ϵ)

. In the gap, lubrication theory therefore gives

p_{x} = O (\frac{μU}{ϵ ^{2} a ^{2}}), p = O (\frac{μU}{a ϵ ^{3/2}}), σ_{x y} = O (\frac{μU}{ϵ a}) .

(1)

Far ahead of and behind the sphere, the length and width scales are both

a

, so

p_{x} = O (\frac{μU}{a ^{2}}), σ_{r x} = O (\frac{μU}{a}) .

(2)

In the sphere frame the tube wall at

y = 0

moves with velocity

- U

and the sphere surface at

y = h

is stationary. The local Couette-Poiseuille flow in a thin gap is

u = - U + \frac{U}{h} y + \frac{p _{x}}{2 μ} y (y - h) .

(3)

Its axial volume flux per unit circumferential width is

q = - \frac{h ^{3}}{12 μ} p_{x} - \frac{U h}{2} .

(4)

Differentiating the profile and eliminating

p_{x}

gives the two wall stresses

\frac{σ _{x y}}{μ}_{y = 0} = \frac{4 U}{h} + \frac{6 q}{h ^{2}}, \frac{σ _{x y}}{μ}_{y = h} = - \frac{2 U}{h} - \frac{6 q}{h ^{2}} .

(5)

The continuity equation integrated across the gap says that changes of

q

along

x

are balanced by circumferential flux divergence. Circumferential variations occur on scale

a

, much longer than the axial scale

a ϵ

, so

q = q (θ)

is independent of

x

at leading order.

Put

h = h_{0} (1 + ξ^{2})

with

x = 2 a h_{0} ξ

. The leading

O (ϵ^{- 3/2})

pressure jump must vanish in the global balance from part ii. Equivalently, the pressure recovery condition in lubrication flow gives

0 = \int_{- \infty}^{\infty} p_{x} d x = - 12 μ q \int h^{- 3} d x - 6 μU \int h^{- 2} d x .

(6)

Using

I_{2} = π /2

and

I_{3} = 3 π /8

\frac{\int h ^{- 2} d x}{\int h ^{- 3} d x} = \frac{4}{3} h_{0}, q = - \frac{2}{3} U h_{0} (θ) .

(7)

The total flux through the gap is consequently

Q = a \int_{0}^{2 π} q d θ = - \frac{4}{3} π ϵ U a^{2} .

(8)

Far from the sphere, translation of the tube contributes

- π a^{2} U

, while Hagen-Poiseuille flow contributes

π a^{4} p_{x} / (8 μ)

. Equating these fluxes gives

p_{x} \sim \frac{8 μU}{a ^{2}} (1 - \frac{4}{3} ϵ) .

(9)

At the next order, substitute

q = - 2 U h_{0} /3

into the tube-wall shear and integrate through the gap:

\int_{- \infty}^{\infty} σ_{x y} (x, 0) d x = 4 μU \int (\frac{1}{h} - \frac{h _{0}}{h ^{2}}) d x = 2 π μU \frac{2 a}{h _{0}} .

(10)

The balance in part ii then gives the leading pressure drop

Δ p = \frac{2}{ϵ} \frac{2 μU}{a} \int_{0}^{2 π} \frac{d θ}{1 + λ cos θ} .

(11)

Finally, the shear on the sphere integrates to

\int_{- \infty}^{\infty} σ_{x y} (x, h) d x = μU \int (- \frac{2}{h} + \frac{4 h _{0}}{h ^{2}}) d x = 0,

(12)

because

I_{1} = π

and

I_{2} = π /2

. Thus the narrow gap exerts no net leading axial shear force on the sphere even though its local shear is nonzero; the leading hydrodynamic force transfer to the sphere is through the large lubrication pressure acting on its gently sloping surface.

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Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 344 / 1 / a / Solution 2026-09-28

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For a binary fluid mixture, the scalar compositional order parameter

ϕ (r)

may be taken as the local concentration difference between its two molecular species. Its spatial integral is fixed by their separately fixed total amounts:

\int_{V} ϕ (r) d r = V \overset{ˉ}{ϕ} .

(1)

Locally,

ϕ

can change only through transport and therefore obeys a continuity equation. By contrast,

p

is the local polar order parameter measuring the mean tail-to-head orientation of the surfactant molecules. Individual molecules can rotate in place, so the integral of

p

need not be conserved.

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Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 347 / 2 / a / Solution 2026-09-28

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The standard Shakura--Sunyaev thin disk assumes: a steady state; axial symmetry; a geometrically thin disk

H / R ≪ 1

; Newtonian, nearly Keplerian circular motion outside the inner edge; subsonic radial drift

∣ u_{R} ∣ ≪ u_{ϕ}

; vertical hydrostatic equilibrium; negligible disk self-gravity; an optically thick, locally thermal spectrum; local radiative balance between viscous heating and cooling; and a local alpha-viscosity stress with a zero-torque inner boundary near the innermost stable circular orbit. These assumptions also exclude dynamically dominant winds and large-scale external torques from the standard solution.

Start from the continuity equation

\partial_{t} ρ + \nabla \cdot (ρ u) = 0.

(1)

Integrate vertically, use axial symmetry and steadiness, and define the surface density of a disk

Σ = \int ρ d z

. Then

\frac{1}{R} \frac{d}{d R} (R Σ u_{R}) = 0,

(2)

so the inward-positive accretion rate is

\overset{m}{˙} = - 2 π R Σ u_{R} = constant .

(3)

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Past exam of the mathematics course of the University of Cambridge / 2022 / ii / Paper 1 / 15A / a / Solution Created 2026-09-24 Updated 2026-09-25

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The homogeneous part of the continuity equation gives

\dot{\overset{n}{ˉ}} + 3 H \overset{n}{ˉ} = 0

. At first order,

\dot{δ n} + 3 Hδ n + \frac{n ˉ}{a} \nabla \cdot δ v = 0.

(1)

For the density contrast

δ = δ n / \overset{n}{ˉ}

, the background equation cancels the expansion terms, leaving

\dot{δ} + \frac{1}{a} \nabla \cdot δ v = 0 .

(2)

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Past exam of the mathematics course of the University of Cambridge / 2022 / ii / Paper 3 / 36B / c / Solution Created 2026-09-24 Updated 2026-09-25

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The electric dipole moment of a localized charge density is

p (t) = \int_{R^{3}} x ρ (x, t) d^{3} x .

(1)

Differentiate and use the continuity equation

\overset{ρ}{˙} + \nabla \cdot J = 0

\overset{p}{˙}_{i} = - \int x_{i} \partial_{j} J_{j} d^{3} x = \int J_{i} d^{3} x - \int_{\partial R^{3}} x_{i} J_{j} n_{j} d S .

(2)

Localization makes the boundary integral vanish. Component by component,

\dot{p} (t) = \int_{R^{3}} J (x, t) d^{3} x .

(3)

This is the time derivative of the electric dipole moment.

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Past exam of the mathematics course of the University of Cambridge / 2023 / ib / Paper 2 / 15D / b / ii / Solution Created 2026-09-24 Updated 2026-10-03

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For a plane wave

A e^{i (k x - Et /ℏ)}

, the probability current is

J = \frac{ℏ k}{m} ∣ A ∣^{2} .

(1)

The incident amplitude is one and the transmitted amplitude is

T

, so

\frac{J _{tr}}{J _{in}} = ∣ T ∣^{2} = sech^{2} (2 ka) .

(2)

This is the transmission probability for quantum tunnelling. In the stationary state,

\partial ρ / \partial t = 0

, so the continuity equation makes the net current independent of position. The reflected current therefore supplies the remainder:

\frac{∣ J _{refl} ∣}{J _{in}} = 1 - sech^{2} (2 ka) = tanh^{2} (2 ka) .

(3)

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Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 312 / 3 / ii / Solution Created 2026-09-28 Updated 2026-10-03

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Write

μ = k \cdot p

and use the Legendre polynomial recurrence relation. The definition in the question is inverted by

Θ (μ) = \sum_{ℓ \geq 0} (2 ℓ + 1) (- i)^{ℓ} Θ_{ℓ} P_{ℓ} (μ) .

(1)

For Newtonian-gauge potentials the scanned geodesic equation is the standard identity

\frac{d l o g ϵ}{d η} = - \frac{d Ψ}{d η} + Φ^{'} + Ψ^{'} = Φ^{'} - p \cdot \nablaΨ.

(2)

Its monopole and dipole are

S_{0} = Φ^{'}

and

S_{1} = k Ψ/3

. Projecting the photon Boltzmann hierarchy onto

P_{0}

and

P_{1}

therefore gives

Θ_{0}^{'} + k Θ_{1} = Φ^{'},

(3)

Θ_{1}^{'} = \frac{k}{3} (Θ_{0} + Ψ - 2 Θ_{2}) .

(4)

The first is the photon continuity equation; the second is the photon Euler equation, with the photon quadrupole providing the anisotropic-stress term. Indeed, with photon density contrast

δ_{γ} = 4 Θ_{0}

and velocity divergence

θ_{γ} = 3 k Θ_{1}

δ_{γ}^{'} = - \frac{4}{3} θ_{γ} + 4 Φ^{'}, θ_{γ}^{'} = k^{2} (δ_{γ} /4 + Ψ - 2 Θ_{2}) .

(5)

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Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 314 / 4 / b / Solution 2026-09-28

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For

γ = 5/3

, the shock data give

ρ_{ps} = 4 ρ_{0}

and

u_{ps} = 3 V_{s} /4

. Since

V_{s} = R / (3 t)

, the prescribed homologous spherical flow is

u (r, t) = \frac{r}{4 t} .

(1)

Put

ξ = r / R (t)

and

ρ = ρ_{0} D (ξ)

. The spherical continuity equation becomes

- \frac{ξ}{3} D^{'} + \frac{1}{4} (3 D + ξ D^{'}) = 0,

(2)

D \propto ξ^{9}

. Matching the post-shock density gives

ρ (r, t) = 4 ρ_{0} ξ^{9} .

(3)

The material acceleration is

\frac{D u}{D t} = - \frac{3 r}{16 t ^{2}} .

(4)

The radial Euler equations for an inviscid fluid thus give

d p / d r = 3 ρ r / (16 t^{2})

. Integrating from the centre to the shock,

p_{ps} - p (0, t) = \frac{3}{16 t ^{2}} \int_{0}^{R} ρ r d r = \frac{3 ρ _{0} R ^{2}}{44 t ^{2}} .

(5)

Since

p_{ps} = ρ_{0} R^{2} / (12 t^{2})

p (0, t) = \frac{ρ _{0} R ^{2}}{66 t ^{2}} = \frac{2}{11} p_{ps} .

(6)

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Past exam of the mathematics course of the University of Cambridge / 2023 / ii / Paper 3 / 38C / c / Solution Created 2026-09-24 Updated 2026-09-25

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Integrating the incompressible continuity equation

u_{x} + v_{y} = 0

across the moving gap, with

v (x, 0) = 0

and

v (x, h) = - V

, gives the Reynolds lubrication equation

\frac{d Q}{d x} = V .

(1)

Hence

Q = V x + C

(2)

for a constant

C

, and the flux formula from part (b) gives

p_{x} = \frac{6 μU}{h ^{2}} - \frac{12 μ ( V x + C )}{h ^{3}} .

(3)

Define the spatial mean

⟨ f ⟩ = \frac{1}{L} \int_{0}^{L} f (x) d x .

(4)

Because both ends meet fluid at pressure

p_{0}

, the pressure recovery condition in lubrication flow is

0 = p (L) - p (0) = \int_{0}^{L} p_{x} d x .

(5)

It determines

C = \frac{U}{2} \frac{⟨ h ^{- 2} ⟩}{⟨ h ^{- 3} ⟩} - V \frac{⟨ x h ^{- 3} ⟩}{⟨ h ^{- 3} ⟩} .

(6)

Therefore the requested pressure gradient in a translating and squeezing finite gap is

p_{x} = \frac{6 μU}{h ^{2}} - \frac{12 μ}{h ^{3}} [V x + \frac{U}{2} \frac{⟨ h ^{- 2} ⟩}{⟨ h ^{- 3} ⟩} - V \frac{⟨ x h ^{- 3} ⟩}{⟨ h ^{- 3} ⟩}] .

(7)

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Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 321 / 1 / b / i / Solution 2026-09-25

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With

G = 0

and

T = 2 π v_{m} h^{'} Σ

, the evolution equation reduces to the radial continuity equation

\partial_{t} Σ = - \frac{1}{r} \partial_{r} (r v_{m} Σ),

(1)

or, for the mass per unit radius

M_{r} = 2 π r Σ

\partial_{t} M_{r} + \partial_{r} (v_{m} M_{r}) = 0.

(2)

The outward mass flux is

F_{M} = 2 π r v_{m} Σ

. If

\dot{M} > 0

denotes the inward accretion rate, steady mass conservation requires

F_{M} = - \dot{M}

. Hence

2 π r v_{m} Σ = - \dot{M}, Σ (r) = - \frac{M ˙}{2 π r v _{m} ( r )} .

(3)

This is positive because the magnetic drift velocity satisfies

v_{m} < 0

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Photon continuity equation 2026-10-06

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In Newtonian gauge in cosmology, the Free-streaming photon Boltzmann equation is

\dot{Θ} + e \cdot \nablaΘ = \dot{ϕ} - e \cdot \nabla ψ

. Angular averaging and the photon angular temperature moments give the displayed continuity equation. Its dipole yields the photon Euler equation, with the sign of the anisotropic stress term determined by the stress convention.

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Pressure-based acoustic analogy with a potential body force 2026-10-06

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Eliminating momentum between the continuity equation and conservation of momentum gives

ρ_{tt} = \partial_{i} \partial_{j} (ρ u_{i} u_{j} - σ_{ij}) + \nabla^{2} (p + χ)

when the force density is

- \nabla χ

. For

q = p - (P - χ)

and time-independent reference fields,

(c_{0}^{- 2} \partial_{t}^{2} - \nabla^{2}) q = \partial_{i} \partial_{j} W_{ij} + Q_{tt}, W_{ij} = ρ u_{i} u_{j} - σ_{ij}, Q = q / c_{0}^{2} - (ρ - ρ_{0}) .

(1)

No energy equation is needed for this identity. The arbitrary reference fields should match the stationary medium where the outgoing sound is evaluated.

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