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In the narrow polar flux tube, the magnetic field lines are almost radial and the transverse width is much smaller than the radial scale. The Lorentz force density has no component along the magnetic field, so the longitudinal magnetically channelled accretion is hydrodynamic. Let

v (r) > 0

denote inward speed. Mass conservation, the Euler equations for an inviscid fluid, and the polytropic equation of state give

ρ v A = \dot{M}_{tube}, v \frac{d v}{d r} = - \frac{1}{ρ} \frac{d p}{d r} - \frac{G M _{⋆}}{r ^{2}}, p = K ρ^{γ} .

(1)

Using the adiabatic sound speed

c^{2} = γ K ρ^{γ - 1}

and

A \propto r^{3}

\frac{ρ ^{'}}{ρ} = - \frac{v ^{'}}{v} - \frac{3}{r}, (v - \frac{c ^{2}}{v}) v^{'} = \frac{3 c ^{2}}{r} - \frac{G M _{⋆}}{r ^{2}} .

(2)

The Bernoulli equation is

\frac{v ^{2}}{2} + \frac{c ^{2}}{γ - 1} - \frac{G M _{⋆}}{r} = \frac{c _{0}^{2}}{γ - 1},

(3)

where the right-hand side comes from matching to a nearly stationary reservoir with negligible gravitational potential. The fixed magnetic field provides transverse confinement; neglecting magnetic forces in this one-dimensional equation concerns their longitudinal component.

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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 / iii / Paper 347 / 2 / a / Solution Created 2026-10-03 Updated 2026-10-05

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Take

u (r) > 0

to be the inward radial speed and select the regular critical solution for a gas reservoir of density

ρ_{\infty}

and adiabatic sound speed

c_{\infty}

. Other mathematically subsonic steady solutions need a different inner boundary condition. For the polytropic equation of state, let

γ = 1 + 1/ n

, so

c_{s}^{2} = \frac{d p}{d ρ} = \frac{n + 1}{n} K ρ^{1/ n}, h = \int \frac{d p}{ρ} = (n + 1) K ρ^{1/ n} = n c_{s}^{2} .

(1)

Mass conservation and the Bernoulli equation give

\dot{M} = 4 π r^{2} ρ u, \frac{u ^{2}}{2} + n c_{s}^{2} - \frac{G M _{BH}}{r} = n c_{\infty}^{2} .

(2)

The gas is at rest at infinity, so its kinetic and gravitational terms vanish there.

Differentiate these two conserved quantities, using

d lo g ρ / d r = - 2/ r - u^{'} / u

. Eliminating the density derivative yields the polytropic Bondi accretion equation

(u - \frac{c _{s}^{2}}{u}) \frac{d u}{d r} = \frac{2 c _{s}^{2}}{r} - \frac{G M _{BH}}{r ^{2}} .

(3)

At a regular sonic point, both sides vanish:

u_{c} = c_{c}, r_{c} = \frac{G M _{BH}}{2 c _{c}^{2}} .

(4)

The Bernoulli equation at that point reduces to

n c_{\infty}^{2} = (n - 3/2) c_{c}^{2}

. For

n > 3/2

, equivalently

1 < γ < 5/3

c_{c}^{2} = \frac{2 n}{2 n - 3} c_{\infty}^{2} = \frac{2 c _{\infty}^{2}}{5 - 3 γ}, r_{c} = \frac{2 n - 3}{4 n} \frac{G M _{BH}}{c _{\infty}^{2}} .

(5)

Since

c_{s}^{2} \propto ρ^{1/ n}

, the critical density is

ρ_{c} = ρ_{\infty} [2 n / (2 n - 3)]^{n}

. Evaluate the conserved mass accretion rate at the sonic point:

\dot{M}_{B} = \frac{π G ^{2} M _{BH}^{2} ρ _{\infty}}{c _{\infty}^{3}} (\frac{2 n}{2 n - 3})^{n - 3/2} = 4 πλ (γ) \frac{G ^{2} M _{BH}^{2} ρ _{\infty}}{c _{\infty}^{3}},

(6)

where

λ (γ) = \frac{1}{4} (\frac{2}{5 - 3 γ})^{(5 - 3 γ) / [2 (γ - 1)]} .

(7)

The isothermal limit

γ \to 1

gives

λ = e^{3/2} /4

, agreeing with Isothermal Bondi accretion.

For

γ = 5/3

n = 3/2

and the Newtonian critical radius tends to zero while

c_{c}

diverges. The limiting critical solution approaches Mach number one only as

r \to 0

; it has no finite-radius Newtonian sonic transition. This is a failure of the Newtonian inner approximation, not an obstruction to accretion onto a black hole. A physical Schwarzschild black hole has a finite event horizon, and the regular relativistic inflow becomes supersonic outside it.

The singular-looking factor in the rate has a finite limit. Writing

ϵ = n - 3/2

, its logarithm is

ϵ lo g [n / ϵ] \to 0

, so the critical Bondi accretion rate for gamma equals five thirds is

\dot{M}_{B, 5/3} = \frac{π G ^{2} M _{BH}^{2} ρ _{\infty}}{c _{\infty}^{3}} (λ = 1/4) .

(8)

This is also the leading cold-reservoir limit of relativistic spherical accretion; an exact rate for relativistically hot gas requires the relativistic equations, not this Newtonian expression.

For completeness, the finite relativistic sonic point can be seen without assigning it an arbitrary inner radius. Let

b = c_{s}^{2} / c^{2}

denote the relativistic squared sound speed and

g = γ - 1

. For an isentropic relativistic gas with

p = K ρ_{rest}^{γ}

, its dimensionless specific enthalpy is

h = g / (g - b)

. Conservation of rest-mass flux and relativistic Bernoulli energy gives the critical relations

r_{c} = \frac{G M _{BH}}{c ^{2}} \frac{1 + 3 b _{c}}{2 b _{c}}, (g - b_{c}) 1 + 3 b_{c} = g - b_{\infty} .

(9)

For

g = 2/3

and

c_{\infty} ≪ c

, expansion gives

b_{\infty} ≃ (9/4) b_{c}^{2}

, hence

r_{c} ≃ \frac{3 G M _{BH}}{4 c c _{\infty}} > \frac{2 G M _{BH}}{c ^{2}} .

(10)

Here the asymptotic relativistic and Newtonian sound speeds agree to leading order. This finite-radius transition explains why the gas can reach the horizon despite the Newtonian

r_{c} = 0

result.

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Past exam of the mathematics course of the University of Cambridge / 2020 / ii / Paper 1 / 40B / a / Solution Created 2026-09-24 Updated 2026-09-29

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Write the density and pressure as

ρ = ρ_{0} + ρ^{'}

and

p = p_{0} + p^{'}

, and let

u

be the small fluid velocity. The linear homentropic acoustic equations are

ρ_{t}^{'} + ρ_{0} \nabla \cdot u = 0, ρ_{0} u_{t} = - \nabla p^{'}, p^{'} = c_{0}^{2} ρ^{'},

(1)

where the adiabatic sound speed is

c_{0}^{2} = (\frac{\partial p}{\partial ρ})_{s} .

(2)

If the flow is irrotational, write

u = \nabla ϕ

. The momentum equation integrates spatially to

p^{'} = - ρ_{0} ϕ_{t},

(3)

a purely time-dependent integration term having been absorbed into

ϕ

. Substitution into continuity gives the wave equation

ϕ_{tt} - c_{0}^{2} \nabla^{2} ϕ = 0,

(4)

so the wave speed is

c_{0}

To obtain energy conservation, dot the momentum equation with

u

and use the product rule:

\frac{\partial}{\partial t} (\frac{1}{2} ρ_{0} ∣ u ∣^{2}) = - \nabla \cdot (p^{'} u) + p^{'} \nabla \cdot u .

(5)

The continuity and pressure-density relations give

p^{'} \nabla \cdot u = - \frac{p ^{'} p _{t}^{'}}{ρ _{0} c _{0}^{2}} = - \frac{\partial}{\partial t} (\frac{p ^{'2}}{2 ρ _{0} c _{0}^{2}}) .

(6)

Therefore the acoustic energy conservation law is

\frac{\partial E}{\partial t} + \nabla \cdot I = 0,

(7)

with acoustic energy density and intensity

E = \frac{1}{2} ρ_{0} ∣ u ∣^{2} + \frac{p ^{'2}}{2 ρ _{0} c _{0}^{2}}, I = p^{'} u .

(8)

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Past exam of the mathematics course of the University of Cambridge / 2020 / ii / Paper 4 / 39B / b / i / Solution Created 2026-09-24 Updated 2026-09-29

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For

p (ρ) = A - \frac{B ^{2}}{ρ},

(1)

the adiabatic sound speed is

c^{2} = \frac{d p}{d ρ} = \frac{B ^{2}}{ρ ^{2}}, c = \frac{B}{ρ} .

(2)

Since

\int (c / ρ) d ρ = - B / ρ = - c

, choose invariants that vanish in the equilibrium state

(u, c) = (0, c_{0})

R_{+} = u - c + c_{0}, R_{-} = u + c - c_{0} .

(3)

Consequently

u + c = c_{0} + R_{-}, u - c = - c_{0} + R_{+} .

(4)

In a right-moving simple wave with

R_{-} = 0

, the

R_{+}

characteristic speed is therefore

u + c = c_{0}

, independent of amplitude. The transport equation becomes

(\partial_{t} + c_{0} \partial_{x}) R_{+} = 0

, so the whole velocity profile translates rigidly at the equilibrium sound speed

c_{0}

without nonlinear distortion.

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Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 321 / 2 / a / ii / Solution 2026-09-25

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Put

K (r) = e^{s (r)}

. The perfect gas relation in the question becomes the polytropic equation of state

P = K (r) ρ^{γ} .

(1)

At fixed

r

, vertical hydrostatic balance is

γ K ρ^{γ - 1} \frac{\partial ρ}{\partial z} = - ρ Ω_{K}^{2} z .

(2)

After one integration,

ρ^{γ - 1} = ρ_{0}^{γ - 1} - \frac{γ - 1}{2 γ K} Ω_{K}^{2} z^{2} .

(3)

Defining the midplane adiabatic sound speed by

c_{s 0}^{2} = γ P_{0} / ρ_{0} = γ K ρ_{0}^{γ - 1}

gives

ρ = ρ_{0} (1 - \frac{z ^{2}}{H ^{2}})^{m}, m = \frac{1}{γ - 1}, H^{2} = \frac{2 c _{s 0}^{2}}{( γ - 1 ) Ω _{K}^{2}} .

(4)

This is a vertically truncated polytropic atmosphere; the physical perfect-gas case has

γ > 1

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Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 314 / 1 / b / Solution Created 2026-09-24 Updated 2026-09-25

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Let

u > 0

be the inward radial speed. Steady spherical Bondi accretion conserves mass:

\dot{M} = 4 π r^{2} ρ u .

(1)

For

γ = 5/3

p = K ρ^{5/3}

and the adiabatic sound speed satisfies

v_{s}^{2} = γ p / ρ \propto ρ^{2/3}

. Matching the reservoir therefore gives

ρ = ρ_{0} (\frac{v _{s}}{v _{s 0}})^{3} .

(2)

The Bernoulli integral is

\frac{u ^{2}}{2} + \frac{v _{s}^{2}}{γ - 1} - \frac{GM}{r} = \frac{3 v _{s 0}^{2}}{2} .

(3)

Writing the Mach number as

M = u / v_{s}

and eliminating

v_{s}

with mass conservation gives

r v_{s}^{2} = (\frac{M ˙ v _{s 0}^{3}}{4 π ρ _{0}})^{1/2} M^{- 1/2} .

(4)

Multiplication of the Bernoulli equation by

r /2

now yields

A \dot{M}^{1/2} (M^{3/2} + 3 M^{- 1/2}) = GM + B r,

(5)

where

A = (\frac{v _{s 0}^{3}}{16 π ρ _{0}})^{1/2}, B = \frac{3}{2} v_{s 0}^{2} .

(6)

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Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 314 / 3 / a / Solution Created 2026-09-24 Updated 2026-09-25

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The adiabatic sound speed and Alfvén speed are

v_{s}^{2} = (\frac{\partial p}{\partial ρ})_{s} = \frac{γ p}{ρ}, v_{A} = \frac{B}{μ _{0} ρ}, v_{A} = \frac{B}{μ _{0} ρ} .

(1)

The entropy is held fixed in the sound-speed derivative.

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Plane-parallel stellar f-mode 2026-10-14

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The plane-parallel Cowling displacement-pressure equations admit this stellar oscillation for every compatible equilibrium density and adiabatic sound speed profile. In the upward-displacement convention its horizontal amplitude is

i k ξ / k

, its Eulerian pressure amplitude is

ρ g ξ

, and its Lagrangian density and pressure perturbations vanish. Its displacement decays into the star but grows upward without bound, so a pure linear continuation to infinite height is not physical.

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Polytropic flow Created 2026-10-06 Updated 2026-10-07

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A polytropic flow uses a polytropic equation of state. When

K

and the adiabatic exponent are constant, a smooth streamline has specific enthalpy

h = γ K ρ^{γ - 1} / (γ - 1)

and adiabatic sound speed

c_{s}^{2} = γ K ρ^{γ - 1}

. Spatially varying entropy can make

K

depend on streamline; the constant-

K

spherical wind is a special case.

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For

γ > 1

and positive mass density and pressure, let

ξ = x / t

in one-dimensional adiabatic compressible flow. The smooth equations are

(u - ξ) ρ^{'} + ρ u^{'} = 0

(u - ξ) u^{'} + p^{'} / ρ = 0

and

(u - ξ) p^{'} + γ p u^{'} = 0

. Away from a constant branch, their coefficient determinant forces

ξ = u + σ c

with

σ = \pm 1

and adiabatic sound speed

c = γ p / ρ

. The entropy equation gives

p = K ρ^{γ}

, and combining continuity with

c^{'} / c = (γ - 1) ρ^{'} / (2 ρ)

gives

u^{'} = \frac{2}{γ + 1}, c^{'} = σ \frac{γ - 1}{γ + 1}, u - \frac{2 σ c}{γ - 1} = J .

(1)

Thus

u = [2 ξ + (γ - 1) J] / (γ + 1)

c = σ (γ - 1) (ξ - J) / (γ + 1)

ρ = (c^{2} / (γ K))^{1/ (γ - 1)}

and

p = K ρ^{γ}

, on intervals where

c > 0

. The specific entropy is constant. Together with uniform branches, these exhaust smooth positive-gas self-similar solutions; a contact discontinuity is not a smooth fan.

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Spherically symmetric adiabatic flow 2026-10-06

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A smooth radial compressible flow with a fixed adiabatic exponent conserves