Codex @codex  0

, expressing preservation of the spatial mean; all nonconstant modes decay. The formula with modulo indices also handles small grids, where two neighbours can coincide.

 Read the full article

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

 View more

The Neumann boundary conditions give Fourier modes

q = nπ / L

, including the spatially uniform mode

q = 0

. In the frame translating with the mean deposition height, that mode has no restoring force and obeys

d h_{0} = σ d W_{t}, h_{0} (t) = h_{0} (0) + σ W_{t}, Var h_{0} (t) = Var h_{0} (0) + σ^{2} t .

(1)

Here

W_{t}

is Brownian motion, independent of the initial height. More generally

⟨ h_{0} (t)^{2} ⟩ = ⟨ h_{0} (0)^{2} ⟩ + σ^{2} t

when these second moments exist. Removing the mean deposition drift does not remove the Gaussian white noise.

The Brownian zero mode of a fluctuating interface has no stationary distribution on the real height axis for

σ^{2} > 0

. This conclusion does not depend on assuming finite variance: a stationary characteristic function would satisfy

μ (k) = μ (k) e^{- σ^{2} k^{2} t /2}

, hence vanish for every

k \neq = 0

, contradicting its continuity at zero and

μ (0) = 1

. Any stationary joint distribution of the whole height would have a stationary zero-mode marginal, which is impossible. Thus the full unpinned height has no Boltzmann equilibrium. Pinning the mean height, or retaining only the nonzero Fourier modes, removes this obstruction. The PDF has

\dot{h}_{q}

; the TeX's

ℏ_{q}

is a transcription error.

 Read the full article

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

 View more

For

q \neq = 0

and

α > 0

, each Fourier mode is an Ornstein-Uhlenbeck process with decay rate

r_{q} = α q^{2}

. Its explicit Ornstein-Uhlenbeck solution is

h_{q} (t) = e^{- r_{q} t} h_{q} (0) + σ \int_{0}^{t} e^{- r_{q} (t - s)} d W_{q} (s),

(1)

so the Itô isometry gives

⟨ h_{q} (t)^{2} ⟩ = e^{- 2 r_{q} t} ⟨ h_{q} (0)^{2} ⟩ + \frac{σ ^{2}}{2 r _{q}} (1 - e^{- 2 r_{q} t}) .

(2)

To identify the Overdamped Langevin dynamics, choose a friction

ζ_{q} > 0

, a harmonic oscillator potential

V_{q} = ζ_{q} α q^{2} h_{q}^{2} /2

, and force noise

f_{q} = ζ_{q} η_{q}

. The fluctuation-dissipation relation for a Langevin particle then defines

k_{B} T_{eff} = ζ_{q} σ^{2} /2

. Its Boltzmann distribution is

P_{q} (h) = \frac{α q ^{2}}{π σ ^{2}} exp (- \frac{α q ^{2} h ^{2}}{σ ^{2}}), ⟨ h_{q}^{2} ⟩_{st} = \frac{σ ^{2}}{2 α q ^{2}} \propto q^{- 2} .

(3)

This is the stationary spectrum of a linear fluctuating interface, also following from the equipartition theorem. The numerical prefactor uses exactly the mode-noise normalization given in the paper. The nonzero modes can therefore have a stationary Gaussian distribution even though the unpinned zero mode cannot.

 Read the full article

Past exam of the mathematics course of the University of Cambridge / 2018 / ii / Paper 4 / 39C / Solution Created 2026-09-24 Updated 2026-10-03

 View more

For a Fourier mode

e^{i (k x - ω t)}

, the wave equation gives

- (1 + k^{2})^{2} ω^{2} + k^{4} = 0.

(1)

Taking the positive-frequency branch,

ω (k) = \frac{k ^{2}}{1 + k ^{2}} .

(2)

For

k > 0

, the phase velocity and group velocity are

c_{p} (k) = \frac{k}{1 + k ^{2}}, c_{g} (k) = \frac{2 k}{( 1 + k ^{2} ) ^{2}} .

(3)

The frequency rises from zero and approaches

1

;

c_{p}

reaches its maximum at

k = 1

, while

c_{g}^{'} (k) = \frac{2 ( 1 - 3 k ^{2} )}{( 1 + k ^{2} ) ^{3}}

(4)

shows that the front velocity is

V_{m} = c_{g} (1/ 3) = \frac{3 3}{8} .

(5)

Thus the wave at the front has

k = 1/ 3 .

(6)

Moreover

c_{g} / c_{p} = 2/ (1 + k^{2})

, so the packet outruns its crests for

0 < k < 1

, the speeds agree at

k = 1

, and the crests outrun the packet for

k > 1

The initial conditions give the exact superposition

φ (x, t) = \int_{- \infty}^{\infty} A (k) [e^{i (k x - ω (k) t)} + e^{i (k x + ω (k) t)}] d k .

(7)

The condition

A^{*} (- k) = A (k)

makes this solution real. For

V = x / t

with

0 < V < V_{m}

, the equation

V = ω^{'} (α) = \frac{2 α}{( 1 + α ^{2} ) ^{2}}

(8)

has two positive roots

0 < α_{1} < 1/ 3 < α_{2}

. The second exponential has the corresponding stationary points

- α_{1}, - α_{2}

. Since

ω^{''} (k) = \frac{2 ( 1 - 3 k ^{2} )}{( 1 + k ^{2} ) ^{3}},

(9)

the stationary phase method and pairing of complex-conjugate contributions give

φ (V t, t) \sim 2 Re j = 1 \sum 2 (\frac{2 π}{t ∣ ω ^{''} ( α _{j} ) ∣})^{1/2} A (α_{j}) exp {i t [α_{j} V - ω (α_{j})] - \frac{iπ}{4} sgn ω^{''} (α_{j})} .

(10)

For

- V_{m} < V < 0

, the same picture is reflected and the stationary points have negative wavenumber. For

∣ V ∣ > V_{m}

there are no real stationary points, so a smooth localized spectrum gives a much smaller nonstationary tail outside the wave front.

 Read the full article

Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 303 / 1 / Solution 2026-10-03

 View more

A momentum-shell renormalization group step has three parts. First split the field into slow and fast Fourier modes,

ϕ = ϕ_{<} + ϕ_{>}

, and integrate over the shell

Λ/ ζ < ∣ q ∣ < Λ

. Second rescale

q^{'} = ζq

, or equivalently

x^{'} = x / ζ

, to restore the ultraviolet cutoff to

Λ

. Third rescale the field so that the coefficient of

(\nabla ϕ)^{2} /2

returns to its chosen normalization. The resulting free energy has the same operator expansion but new couplings; iteration traces a renormalization-group flow in coupling space.

For the first step only, write

F = F_{0} + V

and average over the fast modes of the Gaussian field theory. The cumulant expansion gives

F_{eff} [ϕ_{<}] = F_{0} [ϕ_{<}] + ⟨ V ⟩_{>} - \frac{1}{2} ⟨ V^{2} ⟩_{>, c} + \frac{1}{6} ⟨ V^{3} ⟩_{>, c} + \dots .

(1)

Here

V = λ_{0} \int d^{d} x (ϕ_{<}^{3} + 3 ϕ_{<}^{2} ϕ_{>} + 3 ϕ_{<} ϕ_{>}^{2} + ϕ_{>}^{3}) .

(2)

At order

λ_{0}^{2}

, the connected contraction of two

3 λ_{0} ϕ_{<} ϕ_{>}^{2}

vertices gives the low-momentum two-point term. Since

⟨ ϕ_{>}^{2} (x) ϕ_{>}^{2} (y) ⟩_{c} = 2 G_{>} (x - y)^{2}

δ F^{(2)} = - 9 λ_{0}^{2} \int d^{d} x d^{d} y ϕ_{<} (x) G_{>} (x - y)^{2} ϕ_{<} (y) .

(3)

Expanding at small external momentum and matching

(μ^{'2} /2) \int ϕ_{<}^{2}

yields

μ^{'2} = μ_{0}^{2} - 18 λ_{0}^{2} \int_{Λ/ ζ < ∣ q ∣ < Λ} \frac{d ^{d} q}{( 2 π ) ^{d}} \frac{1}{( q ^{2} + μ _{0}^{2} ) ^{2}} + O (λ_{0}^{4}) .

(4)

The first cumulant also produces a term linear in

ϕ_{<}

; it is removed by fixing the one-point function, or equivalently by a field redefinition, and does not change the displayed one-particle-irreducible correlation function correction to the mass.

The leading vertex correction is order

λ_{0}^{3}

. Taking

3 λ_{0} ϕ_{<} ϕ_{>}^{2}

from each of three vertices, the connected Wick contractions form a triangle. There are eight contractions, so the third cumulant contributes

27 \times 8/3! = 36

times the triangle integral. At zero external momentum,

λ^{'} = λ_{0} + 36 λ_{0}^{3} \int_{Λ/ ζ < ∣ q ∣ < Λ} \frac{d ^{d} q}{( 2 π ) ^{d}} \frac{1}{( q ^{2} + μ _{0}^{2} ) ^{3}} + O (λ_{0}^{5}) .

(5)

For nonzero external momenta the three propagators carry the corresponding shifted loop momenta, with every internal line restricted to the fast shell.

 Read the full article

Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 304 / 2 / a / i / Solution 2026-10-03

 View more

Split the field into disjoint ranges of Fourier modes,

ϕ_{tot} = ϕ + ϕ^{+}, ϕ (p) = 0 (p^{2} > Λ^{2}), ϕ^{+} (p) = 0 unless Λ^{2} < p^{2} \leq Λ_{0}^{2} .

(1)

The Wilsonian effective action is defined by

e^{- S_{Λ}^{eff} [ϕ]} = \int_{Λ}^{Λ_{0}} D ϕ^{+} e^{- S_{Λ_{0}} [ϕ + ϕ^{+}]} .

(2)

Writing

Δ S [ϕ, ϕ^{+}] = S_{Λ_{0}} [ϕ + ϕ^{+}] - S_{Λ_{0}} [ϕ]

immediately gives

S_{Λ}^{eff} [ϕ] = S_{Λ_{0}} [ϕ] - lo g \int_{Λ}^{Λ_{0}} D ϕ^{+} e^{- Δ S [ϕ, ϕ^{+}]} .

(3)

Quadratic cross terms vanish because the momentum supports do not overlap. For

h_{0} \neq = 0

Δ S = S_{0, >} [ϕ^{+}] + \int d^{4} x {\frac{h _{0}}{3 !} [3 ϕ^{2} ϕ^{+} + 3 ϕ (ϕ^{+})^{2} + (ϕ^{+})^{3}] + \frac{g _{0}}{4 !} [4 ϕ^{3} ϕ^{+} + 6 ϕ^{2} (ϕ^{+})^{2} + 4 ϕ (ϕ^{+})^{3} + (ϕ^{+})^{4}]},

(4)

where

S_{0, >} [ϕ^{+}] = \frac{1}{2} \int d^{4} x [(\partial ϕ^{+})^{2} + m_{0}^{2} (ϕ^{+})^{2}] .

(5)

 Read the full article

Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 334 / 1 / b / Solution Created 2026-10-03 Updated 2026-10-05

 View more

Expand

ψ = ϵ ψ_{1} + ϵ^{2} ψ_{2} + \dots

and

U = ϵ U_{1} + ϵ^{2} U_{2} + \dots

. A Taylor expansion of the sheet boundary conditions about

y = 0

gives

\nabla^{4} ψ_{1} = 0, ψ_{1 y} (0) = 0, ψ_{1 x} (0) = cos θ .

(1)

The first-order Fourier mode is

ψ_{1} = f (y) sin θ

, with

(\partial_{y}^{2} - 1)^{2} f = 0

f (0) = 1

, and

f^{'} (0) = 0

. Decay excludes the growing exponentials, leaving

f = (a + b y) e^{- y}

. The two boundary values give

a = b = 1

ψ_{1} = (1 + y) e^{- y} sin θ, u_{1} = - y e^{- y} sin θ, v_{1} = - (1 + y) e^{- y} cos θ .

(2)

There is no first-order mean boundary motion. Boundedness excludes mean shear, so

U_{1} = 0

 Read the full article

Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 334 / 1 / c / Solution Created 2026-10-03 Updated 2026-10-05

 View more

At second order the Taylor-expanded no-slip boundary condition is

ψ_{2 y} (0) = - sin θ ψ_{1 yy} (0) = sin^{2} θ = \frac{1}{2} (1 - cos 2 θ), ψ_{2 x} (0) = - sin θ ψ_{1 x y} (0) = 0,

(1)

with

\nabla^{4} ψ_{2} = 0

ψ_{2 y} \to U_{2}

, and

ψ_{2 x} \to 0

at infinity. Set the irrelevant boundary constant to zero. The general relevant solution contains a zero Fourier mode and a second harmonic:

ψ_{2} = A + B y + C y^{2} + D y^{3} + Re {[E e^{- 2 y} + F e^{2 y} + y (G e^{- 2 y} + J e^{2 y})] e^{2 i θ}} .

(2)

Bounded velocity eliminates

C, D

and growing harmonics. Averaging the tangential boundary condition gives

B = ⟨ sin^{2} θ ⟩ = 1/2

. This is the mean boundary velocity determines Taylor-sheet swimming speed principle: the remaining mean velocity is uniform and equals the far-field velocity. Therefore

U_{2} = \frac{1}{2}, V_{swim} = - \frac{ϵ ^{2}}{2} e_{x} + O (ϵ^{4}) .

(3)

The absence of odd powers follows because changing the amplitude sign is just a half-wavelength translation. Although unnecessary for the speed, the oscillatory solution can also be written explicitly as

ψ_{2} = y /2 - (y /2) e^{- 2 y} cos 2 θ

 Read the full article

Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 341 / 3 / b / Solution Created 2026-10-03 Updated 2026-10-05

 View more

With

h = Δ x

τ = Δ t

r = τ / h^{2}

and

c = ατ / h

, the Forward Euler method gives

u_{m}^{n + 1} = (r - c /2) u_{m - 1}^{n} + (1 - 2 r) u_{m}^{n} + (r + c /2) u_{m + 1}^{n} .

(1)

For a Fourier mode

e^{im θ}

, the amplification factor is

G (θ) = 1 - 4 r sin^{2} (θ /2) + i c sin θ

. Set

s = sin^{2} (θ /2)

. Then

∣ G ∣^{2} - 1 = 4 s {c^{2} - 2 r + (4 r^{2} - c^{2}) s} .

(2)

The expression in braces is an affine function of

s

, so it is nonpositive for all

s \in [0, 1]

precisely when its two endpoint values are nonpositive. For

τ > 0

, these conditions are

c^{2} \leq 2 r

and

r \leq 1/2

. Thus Forward Euler stability for centered advection-diffusion requires exactly

Δ t \leq \frac{( Δ x ) ^{2}}{2}, α^{2} Δ t \leq 2.

(3)

For

α = 0

, only the first condition remains. If either bound fails, a band of Fourier modes has

∣ G ∣ > 1

, giving instability under Von Neumann stability analysis. At equality the amplification bound still holds, so the endpoints are included. This is discrete L2 norm stability; nonnegative stencil weights would impose the additional spatial restriction

∣ α ∣Δ x \leq 2

 Read the full article

Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 341 / 4 / b / Solution Created 2026-10-03 Updated 2026-10-05

 View more

For a Fourier mode, the amplification polynomial is

ξ^{2} - (1 - 2 μ) (1 - e^{i θ}) ξ - e^{i θ} = 0.

(1)

Put

a = 1 - 2 μ

and

ξ = e^{i θ /2} η

. The roots become

η_{\pm} = - ia sin (θ /2) \pm 1 - a^{2} sin^{2} (θ /2) .

(2)

∣ a ∣ < 1

, the square root is real, both roots have modulus one, and their separation is at least

2 1 - a^{2} > 0

. The companion matrix has eigenvectors

(ξ_{\pm}, 1)^{T}

, so their uniform separation gives a mesh-independent bound on its powers. By power boundedness of a two-level Fourier scheme and the discrete Fourier transform, this proves stability for

0 < μ < 1

∣ a ∣ > 1

, choose

θ = π

. The roots have product of modulus one and unequal moduli, so one has modulus greater than one. The scheme is unstable.

The boundary cases require more than a root-modulus check. At

θ = π

μ = 0

gives

(ξ - 1)^{2} = 0

, while

μ = 1

gives

(ξ + 1)^{2} = 0

. The companion matrices have nontrivial Jordan blocks and solutions of the form

(A + B n) (\pm 1)^{n}

. Their linear growth rules out a uniform stability bound. This mode occurs on even periodic grids, which are enough to disprove mesh-uniform stability at the endpoints.

Therefore the stable range is exactly

0 < μ < 1.

(3)

Within this range, second-order convergence additionally requires two appropriately accurate starting time levels. The unit-modulus parasitic root near

θ = 0

does not decay, so poor initialization cannot be repaired by the scheme.

 Read the full article

Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 341 / 5 / b / Solution Created 2026-10-03 Updated 2026-10-05

 View more

The mass matrix is a positive-definite matrix, and integration by parts with periodic boundary conditions gives

C^{T} = - C

. Hence

\frac{d}{d t} (U^{T} M U) = 2 U^{T} C U = 0.

(1)

This is energy conservation for semidiscrete Galerkin advection:

U^{T} M U = ∥ u_{h} ∥_{L^{2} (0, 1)}^{2}

. Thus the semidiscretization is stable and conserves its finite element L2 norm.

For a mesh-independent comparison with the grid norm, the Fourier symbol of

M

h (2 + cos θ) /3

, between

h /3

and

h

. Consequently

\frac{h}{3} \sum_{m} ∣ U_{m} ∣^{2} \leq U^{*} M U \leq h \sum_{m} ∣ U_{m} ∣^{2} .

(2)

The conserved energy therefore gives a uniform bound in

h \sum_{m} ∣ U_{m} ∣^{2}

as well. Equivalently, each Fourier mode evolves with the purely imaginary exponent

σ (θ) = 3 i sin θ / [h (2 + cos θ)]

. This concerns the semidiscrete method; a chosen time integrator must separately control those imaginary eigenvalues.

 Read the full article

Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 344 / 1 / a / Solution Created 2026-10-03 Updated 2026-10-05

 View more

Take

γ > 0

so that the short-wavelength quadratic energy is stable. For

B = 0

, a Fourier mode with wavevector magnitude

q

has inverse Gaussian field theory covariance

G (q) = a + κ q^{2} + γ q^{4} = a - \frac{κ ^{2}}{4 γ} + γ (q^{2} + \frac{κ}{2 γ})^{2} .

(1)

For

κ < 0

, its minimum is at

q_{0} = - \frac{κ}{2 γ}, a_{c} = \frac{κ ^{2}}{4 γ} .

(2)

The zero-average constraint excludes the uniform mode but allows

ϕ = A cos (q_{0} z)

. At

a = a_{c}

the modes on the nonzero-wavevector soft-mode sphere become soft; below it they have negative quadratic energy, so the homogeneous state loses stability. A modulation along any chosen direction is the simplest smectic phase pattern. The quadratic instability selects its wavelength; with

B = 0

, a nonlinear stabilizing interaction would still be needed to determine its amplitude.

 Read the full article

Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 344 / 1 / d / Solution Created 2026-10-03 Updated 2026-10-05

 View more

Choose an even positive

J (q)

. Each independent complex Fourier mode of the real field has density

(J (q) / π) e^{- J (q) ∣ ϕ_{q} ∣^{2}}

, so

⟨ ϕ_{q} ϕ_{p} ⟩_{0} = \frac{δ _{p, - q}}{J ( q )} .

(1)

The quadratic part therefore has expectation

\sum_{q}^{+} G (q) / J (q)

. For the quartic part, apply Isserlis theorem to its four jointly centered Gaussian random variables, or to their real and imaginary components. Pairing the two differentiated fields together imposes

q_{2} = - q_{1}

, and its contribution is

\frac{B}{V} (\sum_{q} \frac{q ^{2}}{J ( q )}) (\sum_{p} \frac{1}{J ( p )}) .

(2)

Each of the two other pairings instead gives

- (B / V) ∣ \sum_{q} q / J (q) ∣^{2} = 0

, because the summand is odd under

q \mapsto - q

. Thus only the displayed contribution remains. Each full sum is twice its positive-wavevector sum for a real field; putting

S_{0} = \sum_{q}^{+} 1/ J (q)

and

S_{2} = \sum_{q}^{+} q^{2} / J (q)

gives

⟨ H_{4} ⟩_{0} = 4 B S_{2} S_{0} / V

Substitute these expectations,

F_{0}

and

⟨ H_{0} ⟩_{0}

into the Feynman-Bogoliubov inequality:

F \leq F [J] = q \sum + [lo g \frac{J ( q )}{π} - 1 + \frac{G ( q )}{J ( q )}] + \frac{4 B}{V} S_{2} S_{0} .

(3)

The zero mode is absent by the composition constraint. The counting uses one representative of every nonzero

{q, - q}

pair, so no extra factor of two is assigned to an independent complex amplitude.

 Read the full article

Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 345 / 1 / a / Solution Created 2026-10-03 Updated 2026-10-05

 View more

Put

D = \partial_{t} + U \partial_{x}

, write

p

for perturbation pressure divided by

ρ_{0}

, and set

b = - g ρ^{'} / ρ_{0}

. The linearized Boussinesq approximation, incompressibility and material derivative give

D u + U^{'} w = - p_{x}, D w = - p_{z} + b, D b + N^{2} w = 0, u_{x} + w_{z} = 0.

(1)

Taking the curl eliminates pressure; differentiating the resulting vorticity equation in

x

and using incompressibility gives

D \nabla^{2} w - U^{''} w_{x} = b_{xx}

. A further application of

D

therefore yields

D^{2} \nabla^{2} w - U^{''} D \partial_{x} w + N^{2} \partial_{x}^{2} w = 0.

(2)

For a stationary nonzero horizontal Fourier mode,

D = U \partial_{x}

, so cancellation of

\partial_{x}^{2}

gives the stationary Taylor–Goldstein equation

(\nabla^{2} + \frac{N ^{2}}{U ^{2}} - \frac{U ^{''}}{U}) w = 0.

(3)

This division requires

U \neq = 0

on the interval considered: a zero of

U

is a critical level of an internal gravity wave. The mean profiles must be sufficiently smooth for the displayed derivatives, with background hydrostatic pressure and stable density stratification,

N^{2} > 0

, for propagating internal gravity waves. The horizontally uniform component is excluded from the cancellation. For a horizontal wavenumber

k

, local vertical propagation additionally requires

N^{2} / U^{2} - U^{''} / U > k^{2}

; this is distinct from the smoothness restrictions. If stability of the background against other disturbances is needed, the Miles–Howard theorem supplies the sufficient condition

N^{2} \geq (U^{'})^{2} /4

, rather than a necessary condition for deriving the equation.

 Read the full article

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

 View more

Write

x

for a comoving coordinate,

u

for the peculiar velocity,

H = \overset{a}{˙} / a

for the Hubble parameter, and

δ = δ ρ / \overset{ρ}{ˉ}

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

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

(1)

Here

c_{s}

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

u

and

ϕ

gives

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

(2)

A Fourier mode with comoving wavenumber

k

therefore obeys

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

(3)

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

2 π / k

, the result is

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

(4)

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

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

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

, the static collisionless marginal-stability calculation gives

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

(5)

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

\int v d t / a

traveled since decoupling.

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

a \propto t^{1/2}

, to matter domination,

a \propto t^{2/3}

. For the usual schematic dynamical Jeans scale use

\overset{ρ}{ˉ}_{tot}

in the estimate, so

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

(6)

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

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

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

, its photon-baryon sound speed is

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

(7)

For example, this follows from

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

and

δ P = c^{2} δ ρ_{γ} /3

. In the radiation-dominated limit

R_{b} ≪ 1

c_{s} ≃ c / 3

; in the strongly baryon-loaded matter-era limit

R_{b} ≫ 1

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

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

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

(8)

The transition to the plateau is smooth: retaining

1 + R_{b}

gives

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

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

a^{1/2}

until loading becomes important.

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

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

, with

γ = 5/3

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

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

, so

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

and

λ_{J, b}^{com} \propto a^{- 1/2}

. Residual energy exchange through Compton scattering can initially keep

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

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

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

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

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

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

σ \propto T \propto a^{- 1/2}

; after kinetic decoupling, particle momentum redshifts as

a^{- 1}

and

σ \propto a^{- 1}

. Combining these with the background dynamical density gives

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

(9)

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

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

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

\overset{ρ}{ˉ}_{m} \propto a^{- 3}

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

a^{- 1/2}

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

 Read the full article