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Gravitational instability is the growth of a perturbation because its increased self-gravity overcomes the opposing support or restoring response. For a uniform nonexpanding gas, linearization gives the Jeans instability dispersion relation

ω^{2} = c_{s}^{2} k^{2} - 4 π Gρ

: sufficiently long wavelengths grow rather than oscillate. In an expanding universe the background and Hubble friction modify the growth rate; a pressureless growing density mode can eventually become nonlinear. Rotation, magnetic stress, pressure and geometry can all alter the stability threshold in other systems.

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Linear growth equation Created 2026-09-29 Updated 2026-10-07

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For pressureless matter on subhorizon scales, the linear growth factor obeys

\ddot{D} + 2 H \dot{D} - 4 π G \overset{ρ}{ˉ}_{m} D = 0.

(1)

The second term is dilution by cosmic expansion, while the final term drives Jeans instability.

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Past exam of the mathematics course of the University of Cambridge / 2001 / iii / Paper 35 / 1 / Solution Created 2026-10-03 Updated 2026-10-14

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Write

τ = t - t_{0}

, let

R

be the scale factor,

H = \dot{R} / R

the Hubble parameter, and

μ = Λ/3

the asymptotic expansion rate for the positive cosmological constant. The continuity equation for a homogeneous Hubble flow gives

ρ R^{3} = ρ_{0}

. Isotropy and the Poisson equation for Newtonian gravity give a relative acceleration proportional to displacement from any chosen origin:

g = - \frac{4 π Gρ - Λ}{3} r .

(1)

This uses the local relative gravitational acceleration, rather than trying to sum an absolutely convergent potential over an infinite universe. A fluid element with constant comoving coordinate

x

has

r = R x

. The spatially uniform pressure has no gradient, so the Euler equations for an inviscid fluid give

\ddot{R} = - \frac{4 π G ρ _{0}}{3 R ^{2}} + \frac{Λ}{3} R .

(2)

Multiply by

2 \dot{R}

and integrate. With

A = 8 π G ρ_{0} /3

, the resulting first integral is

\dot{R}^{2} = A / R + μ^{2} R^{2} - κ .

(3)

For an expanding branch that reaches arbitrarily large

R

\dot{R} / R \to μ

; more precisely,

H = μ + O (R^{- 2})

and the correction to

ln R - μ t

is integrable. Therefore

R \sim C e^{μ t}

. Positive

Λ

alone does not make every initially expanding branch escape: sufficiently large positive

κ

can give a turning point or a limiting static radius. Under the stated vacuum-dominance inequalities at

R = 1

, however,

A / R

and

∣ κ ∣

remain small compared with

μ^{2} R^{2}

for the future expanding branch, and

R ≃ e^{μτ}

is the required approximation.

To derive the linear cosmological density perturbation equation, write

δ = ρ^{'} / ρ

and

v = H r + u

, where

u

is the peculiar velocity. At fixed

x = r / R

, the linearized continuity equation, Euler equations for an inviscid fluid and perturbation Poisson equation are

\dot{δ} + R^{- 1} \nabla_{x} \cdot u = 0, \dot{u} + H u = - R^{- 1} \nabla_{x} (c_{s}^{2} δ + ϕ), \nabla_{x}^{2} ϕ = 4 π G R^{2} ρ δ .

(4)

Taking a divergence of the second equation and differentiating the first eliminates the peculiar velocity. For a Fourier mode of comoving wavenumber

k

, this gives

\overset{ε}{¨} + 2 H \overset{ε}{˙} + (\frac{c _{s}^{2} k ^{2}}{R ^{2}} - 4 π Gρ) ε = 0.

(5)

The adiabatic sound speed for the four-thirds polytropic equation of state is

c_{s}^{2} = d p / d ρ \propto ρ^{1/3} = ρ_{0}^{1/3} / R

, hence

c_{s}^{2} = c_{0}^{2} / R

. Both the restoring pressure term and the gravitational term are proportional to

R^{- 3}

. Thus, in the exponential approximation,

\overset{ε}{¨} + 2 μ \overset{ε}{˙} + η^{2} e^{- 3 μτ} ε = 0, η^{2} = c_{0}^{2} k^{2} - 4 π G ρ_{0} .

(6)

The sign of

η^{2}

is the Jeans instability distinction, but exponential expansion makes the total subsequent amplification finite.

Remove Hubble friction by setting

ε = e^{- μτ} y

. The standard-form equation is

y^{''} + Q (τ) y = 0, Q = η^{2} e^{- 3 μτ} - μ^{2} .

(7)

Where

Q > 0

and

∣ Q^{'} ∣ ≪ Q^{3/2}

, the WKB approximation gives

ε ≃ e^{- μτ} Q^{- 1/4} [A cos (\int^{τ} Q d s) + B sin (\int^{τ} Q d s)] .

(8)

Where

Q < 0

, put

q = - Q

; away from a turning point the corresponding WKB approximation is

ε ≃ e^{- μτ} q^{- 1/2} [A e^{\int^{τ} q d s} + B e^{- \int^{τ} q d s}], ∣ q^{'} ∣ ≪ q^{2} .

(9)

For

η^{2} = - s^{2} < 0

q = (μ^{2} + s^{2} e^{- 3 μτ})^{1/2}

never vanishes. The exponential integral can be evaluated explicitly:

\int_{0}^{τ} q (t) d t = \frac{2}{3 μ} [Ψ (q (0)) - Ψ (q (τ))], Ψ (q) = q + \frac{μ}{2} ln \frac{q - μ}{q + μ} .

(10)

Differentiate this expression, using

q^{'} = - 3 μ (q^{2} - μ^{2}) / (2 q)

, to verify it. Early on, if coefficients change little over the interval considered, the two local exponents for

ε

are

- μ \pm μ^{2} + s^{2}

: one increases and one decreases. At late times

q \to μ

and the displayed integral is

μτ + O (1)

. The nominally growing WKB approximation branch of

y

therefore becomes a constant branch of

ε

, while the other branch decays like

e^{- 2 μτ}

. The WKB approximation is not automatically accurate at all intermediate times; its stated derivative criterion must be checked.

For

η^{2} > 0

, the early frozen-coefficient exponents are

- μ \pm μ^{2} - η^{2}

. There are damped oscillations if

η^{2} > μ^{2}

, a repeated-root solution

(A + B τ) e^{- μτ}

at equality, and two real decaying local branches if

0 < η^{2} < μ^{2}

. In the oscillatory case

Q

vanishes at

τ_{t} = ln (η^{2} / μ^{2}) / (3 μ)

: the ordinary WKB approximation breaks down there, and a turning-point connection is needed. After that transition the solutions again approach a constant or decay. A positive pressure-minus-gravity restoring term does not imply indefinitely continuing oscillation; its physical frequency is redshifted away.

The limiting conclusion does not depend on the accuracy of WKB approximation. With

x = 2∣ η ∣ e^{- 3 μτ /2} / (3 μ)

, the equation for

y

x^{2} y_{xx} + x y_{x} + (sgn (η^{2}) x^{2} - \frac{4}{9}) y = 0.

(11)

For

η^{2} > 0

its basis is

J_{2/3} (x), Y_{2/3} (x)

, the Bessel functions of the first kind and Bessel functions of the second kind. For

η^{2} < 0

use

I_{2/3} (x), K_{2/3} (x)

, the Modified Bessel functions of the first kind and modified Bessel functions of the second kind. Their small-

x

branches are proportional to

x^{2/3}

and

x^{- 2/3}

. Multiplication by

e^{- μτ}

gives the decaying and constant branches respectively. At

η^{2} = 0

the exact solution is

A + B e^{- 2 μτ}

. Thus Bessel density modes during exponential expansion prove

ε (τ) ⟶ ε_{\infty} for every finite initial pair and every real η^{2}, Λ > 0.

(12)

The same conclusion holds in the full four-thirds perturbation equation on an exponentially escaping background:

(R^{2} \overset{ε}{˙})^{'} = - η^{2} R^{- 1} ε

. Its integrated equation has an absolutely integrable Volterra kernel, bounded by a constant times

e^{- 3 μτ}

. The Gronwall inequality first bounds

ε

; then

R^{- 1} ε

is integrable and

\overset{ε}{˙} = O (R^{- 2})

is integrable. A finite limiting amplitude follows without setting

H

exactly constant. This argument requires positive

Λ

and the escaping branch; it is not a claim about

Λ = 0

or recollapse.

The requested strict increase of that limit needs an initial-velocity qualification. For

η^{2} = - s^{2} < 0

(e^{2 μτ} \overset{ε}{˙})^{'} = s^{2} e^{- μτ} ε .

(13)

ε (0) > 0

and

\overset{ε}{˙} (0) \geq 0

, this identity prevents a first loss of positivity and makes the amplitude strictly increase for

τ > 0

. Integrating it twice gives

ε_{\infty} = ε (0) + \frac{ε ˙ ( 0 )}{2 μ} + \frac{s ^{2}}{2 μ} \int_{0}^{\infty} e^{- 3 μτ} ε (τ) d τ > ε (0) .

(14)

This is monotone Jeans growth with positive cosmological constant. Without the extra initial condition the printed assertion is false:

ε = e^{- μτ} I_{2/3} (2 s e^{- 3 μτ /2} / (3 μ))

is positive at

τ = 0

and tends to zero. An arbitrarily small multiple remains an infinitesimal linear cosmological density perturbation, so linearity does not remove the counterexample.

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Past exam of the mathematics course of the University of Cambridge / 2002 / iii / Paper 41 / 1 / Solution Created 2026-10-03 Updated 2026-10-14

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Integrating the given expansion equation gives

R^{3/2} = t / t_{0}

, with the integration constant fixed by

R (t_{0}) = 1

. Thus the scale factor and Hubble parameter are

R = (t / t_{0})^{2/3}

and

H = 2/ (3 t)

. The Newtonian derivation of the flat Friedmann equation gives

H^{2} = 8 π Gρ /3

, so

ρ = \frac{1}{6 π G t ^{2}}, c^{2} = γ K ρ^{γ - 1} = c^{2} (t_{0}) (t / t_{0})^{- 2 (γ - 1)} .

(1)

Use the comoving coordinate

q = x / R

and the density contrast

δ = ρ_{1} / ρ

. Linearized mass conservation, the peculiar-velocity Euler equation for fluid motion and the Poisson equation for the peculiar gravitational potential are

\dot{δ} = - R^{- 1} \nabla_{q} \cdot v, \dot{v} + H v = - R^{- 1} \nabla_{q} (c^{2} δ + Φ), \nabla_{q}^{2} Φ = 4 π G R^{2} ρ δ .

(2)

Differentiate the first equation, substitute the second and use the third. A Fourier component of comoving wavenumber

k

obeys

\overset{ε}{¨} + 2 H \overset{ε}{˙} + (\frac{c ^{2} k ^{2}}{R ^{2}} - 4 π Gρ) ε = 0.

(3)

Since

c^{2} k^{2} / R^{2} = k^{2} c^{2} (t_{0}) t_{0}^{2 γ - 2/3} t^{- 2 γ + 2/3}

, the normalization is

Λ = k c (t_{0}) t_{0}^{γ - 1/3}

, as printed in the PDF. The resulting linear cosmological density perturbation equation is

\overset{ε}{¨} + \frac{4}{3 t} \overset{ε}{˙} + (\frac{Λ ^{2}}{t ^{2 γ - 2/3}} - \frac{2}{3 t ^{2}}) ε = 0.

(4)

To reduce it to a Bessel equation, put

ν = γ - 4/3

z = Λ/ (ν t^{ν})

and

ε = t^{a} y (z)

. As

\overset{z}{˙} = - ν z / t

, multiplying the substituted equation by

t^{2 - a}

gives

ν^{2} z^{2} y^{''} + ν (ν - 2 a - 1/3) z y^{'} + [ν^{2} z^{2} + a^{2} + a /3 - 2/3] y = 0.

(5)

Taking

a = - 1/6

makes the first two coefficients proportional to those in the Bessel equation, and the remaining constant is

- 25/36

. Hence

ν = γ - \frac{4}{3} \in (0, 1/3), λ = \frac{5}{6 ν} = \frac{5}{6 γ - 8} .

(6)

For noninteger

λ

, the Bessel density modes of a polytropic expanding universe are

ε = t^{- 1/6} [A J_{- λ} (z) + B J_{λ} (z)] .

(7)

There is an exceptional case within the allowed interval: if

λ = n

is an integer,

J_{- n} = (- 1)^{n} J_{n}

, so these are not two independent solutions. The full solution then uses

t^{- 1/6} [A Y_{n} (z) + B J_{n} (z)]

, with the Bessel function of the second kind providing the missing branch. For example

γ = 29/18

gives

λ = 3

. A basis using

J_{λ}, Y_{λ}

is valid for every allowed exponent.

For early times,

z \to \infty

. The large-argument asymptotic expansion of the Bessel function of the first kind has oscillatory phase

z - πa /2 - π /4

and envelope

(2/ (π z))^{1/2}

for order

a

; the Bessel function of the second kind has the same envelope. Thus

∣ ε ∣_{envelope} \propto t^{- 1/6} z^{- 1/2} \propto t^{ν /2 - 1/6} = t^{(3 γ - 5) /6} .

(8)

This decreases because

3 γ < 5

. At late times

z \to 0

, the leading powers are

J_{\pm λ} (z) \propto z^{\pm λ}

for noninteger order, giving

t^{- 1}

and

t^{2/3}

. The

Y_{n}

branch gives the same growing power in the integer case. Therefore a generic perturbation grows as $t^{2/3}$ ; an initial condition with zero growing-mode coefficient instead has the decaying

t^{- 1}

mode.

To find the growth scale including its dependence on

ν

, set

ε = t^{2/3} h (t)

. The equation becomes

h^{''} + \frac{8}{3 t} h^{'} + Λ^{2} t^{- 2 ν - 2} h = 0.

(9)

Substitute

h = 1 + a Λ^{2} t^{- 2 ν} + \dots

. The leading coefficient equation is

- 2 ν (5/3 - 2 ν) a + 1 = 0

. The pressure correction to a growing polytropic cosmological mode is consequently

ε_{+} = C t^{2/3} [1 + \frac{Λ ^{2} t ^{- 2 ν}}{2 ν ( 5/3 - 2 ν )} + \dots] .

(10)

Since

c k / R = Λ t^{- ν - 1}

and

t^{- 2} = 6 π Gρ

, the correction becomes small when

(c k / R)^{2} t^{2}

is small compared with

2 ν (5/3 - 2 ν)

. The factor

2 (5/3 - 2 ν)

is between two and

10/3

, giving the stated order-of-magnitude onset condition

\frac{c k}{R} ≲ 6 π ν Gρ .

(11)

For a well-developed asymptotic growing mode, replace

≲

≪

. This is a Jeans instability interpretation: the physical acoustic frequency must be sufficiently low compared with the gravitational expansion scale, or equivalently the physical wavelength must be sufficiently long for self-gravity to overcome pressure support. It is an approximate condition for the dust-like growing branch, not an exact initial-condition-independent instant at which any perturbation begins increasing. The instantaneous sign change of the coefficient in the original equation instead occurs at

c^{2} k^{2} / R^{2} = 4 π Gρ

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Past exam of the mathematics course of the University of Cambridge / 2003 / iii / Paper 64 / 1 / Solution Created 2026-10-03 Updated 2026-10-14

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Write

H = \dot{R} / R

. A fluid element in the homologous spherical flow obeys

\dot{r} = H r

, so

r / R

remains constant. Also

\nabla \cdot u_{0} = 3 H

. Along every element, the continuity equation therefore gives

D ρ_{0} / D t = - 3 H ρ_{0}

. Every element starts with the same mass density, and hence has the same mass density at subsequent times:

ρ_{0} (t) = ρ_{00} (\frac{R _{0}}{R ( t )})^{3} .

(1)

This proves preservation of uniformity, rather than assuming it when conserving the total mass.

By the shell theorem, only the mass interior to radius

r

contributes to its gravitational acceleration. That mass is

4 π ρ_{0} r^{3} /3

. The same result follows directly from the Poisson equation for Newtonian gravity: spherical symmetry gives

(r^{2} Φ_{r})_{r} = 4 π G ρ_{0} r^{2}

, and regularity at the centre sets the integration constant to zero. Hence

Φ_{r} = 4 π G ρ_{0} r /3

and

g = - \frac{4 π G ρ _{0}}{3} r .

(2)

The material derivative of the background velocity is

(\dot{H} + H^{2}) r = (\ddot{R} / R) r

. There is no pressure force, and the momentum equation consequently reduces to

R^{2} \ddot{R} = - K, K = \frac{4 π G}{3} ρ_{00} R_{0}^{3} = GM .

(3)

Multiplication by

\dot{R}

and integration, using collapse from rest, gives

\frac{1}{2} \dot{R}^{2} = K (\frac{1}{R} - \frac{1}{R _{0}}) .

(4)

Choose the negative square root for contraction. Put

C^{2} = 2 K / R_{0}

and

R = R_{0} cos^{2} (ϕ /2)

. For

0 < ϕ < π

, the energy integral then gives

\dot{R} = - C tan (ϕ /2)

, while

d R / d ϕ = - R_{0} sin ϕ /2

. Their ratio is

\frac{d t}{d ϕ} = \frac{R _{0}}{2 C} (1 + cos ϕ) = \frac{R}{C} .

(5)

Integrating from the initial turning point proves the cycloid parametrization of pressureless homologous spherical collapse:

R = \frac{R _{0}}{2} (1 + cos ϕ), Ct = \frac{R _{0}}{2} (ϕ + sin ϕ), C^{2} = \frac{8 π G ρ _{00} R _{0}^{2}}{3} .

(6)

It covers the entire contracting branch, with complete collapse at

ϕ = π

and

t_{c} = 3 π / (32 G ρ_{00})

. At

R = 0

the uniform dust configuration is singular, so the smooth calculation is restricted to

t < t_{c}

Each Fourier mode among the comoving perturbations of a homologously collapsing cloud has its phase fixed on each fluid element. In fact, for

k = q / R

(\partial_{t} + u_{0} \cdot \nabla) e^{i k \cdot r} = i (\dot{k} + H k) \cdot r e^{i k \cdot r} = 0.

(7)

Its physical wavelength contracts in proportion to

R

, while its wavelength in the comoving coordinate

r / R

stays fixed. This is the physical reason to use a time-dependent wavenumber; holding the physical wavenumber fixed would not follow the contracting pattern. Boundary effects are being excluded in this bulk Fourier mode calculation.

Let

ρ^{'} = ρ_{1} (t) e^{i k \cdot r}

and let the perturbed velocity and Newtonian gravitational potential have amplitudes

u_{1} (t)

and

Φ_{1} (t)

. Linearizing the continuity equation, the uniform background eliminates

u_{1} \cdot \nabla ρ_{0}

. The advected phase eliminates its own time derivative, leaving

\overset{ρ}{˙}_{1} + 3 \frac{R ˙}{R} ρ_{1} + \frac{i ρ _{0}}{R} q \cdot u_{1} = 0.

(8)

For the density contrast

δ = ρ_{1} / ρ_{0}

, use

\overset{ρ}{˙}_{0} = - 3 H ρ_{0}

to obtain

\dot{δ} = - i k \cdot u_{1}

. The linear momentum equation and Poisson equation for Newtonian gravity are

\dot{u}_{1} + H u_{1} = - i k Φ_{1}, - k^{2} Φ_{1} = 4 π G ρ_{0} δ .

(9)

Differentiating the density contrast equation must include

\dot{k} = - H k

\ddot{δ} = - i (\dot{k} \cdot u_{1} + k \cdot \dot{u}_{1}) = 2 i H k \cdot u_{1} - k^{2} Φ_{1} = - 2 H \dot{δ} + 4 π G ρ_{0} δ .

(10)

Thus the compressive Fourier mode with

k \neq = 0

satisfies

\ddot{δ} + 2 \frac{R ˙}{R} \dot{δ} - 4 π G ρ_{0} δ = 0.

(11)

The pressureless approximation removes the usual acoustic restoring term of Jeans instability. A transverse velocity perturbation instead obeys

\dot{u}_{1 ⊥} + H u_{1 ⊥} = 0

and does not drive this density contrast.

Since

d / d t = (C / R) d / d ϕ

, the density contrast equation becomes

δ_{ϕϕ} - tan (ϕ /2) δ_{ϕ} - \frac{3}{1 + c o s ϕ} δ = 0.

(12)

For a direct verification, put

s = tan (ϕ /2)

. As

d / d ϕ = (1 + s^{2}) d / (2 d s)

, the first-derivative terms cancel, reducing the equation to

δ_{ss} - \frac{6}{1 + s ^{2}} δ = 0.

(13)

The function

δ_{g} = s (1 + s^{2}) /2

has second derivative

3 s

and satisfies this equation. Transforming back gives the required growing branch:

δ_{g} = \frac{sin ϕ}{( 1 + cos ϕ ) ^{2}} .

(14)

To establish what the other solution does, use reduction of order, not just the growth of this example. The coefficient of

δ_{ϕ}

d lo g R / d ϕ

, so an independent solution is proportional to

δ_{g} \int d ϕ / (R δ_{g}^{2})

. Choose the integration endpoint at complete collapse to select the decaying branch:

δ_{d} = δ_{g} \int_{ϕ}^{π} \frac{( 1 + cos s ) ^{3}}{sin ^{2} s} d s = δ_{g} [4 cot (ϕ /2) + sin ϕ - 3 (π - ϕ)] .

(15)

For example, evaluating the integral with

a = tan (s /2)

gives integrand

4/ [a^{2} (1 + a^{2})^{2}]

and antiderivative

- 4/ a - 6 arctan a - 2 a / (1 + a^{2})

. Its endpoint value gives the displayed expression. Its Wronskian with

δ_{g}

- (1 + cos ϕ)^{- 1}

, so the two solutions are independent on

0 < ϕ < π

For

ϵ = π - ϕ \to 0

R \sim R_{0} ϵ^{2} /4

δ_{g} \sim 4 ϵ^{- 3}

and the integral in

δ_{d}

ϵ^{5} /40 + O (ϵ^{7})

. Consequently the two density modes of a collapsing pressureless sphere have

δ_{g} \propto R^{- 3/2}, δ_{d} \propto R .

(16)

Equivalently, with

τ = t_{c} - t

, the local powers are

τ^{- 1}

and

τ^{2/3}

. There is one independent growing branch: every solution is

A δ_{g} + B δ_{d}

, and only the branch with

A = 0

decays at collapse. The uniqueness is up to normalization and addition of the decaying solution; it does not mean that only one literal function can grow. A fixed nonzero initial perturbation will eventually violate

∣ δ ∣ ≪ 1

, so the divergent linear result is not a nonlinear solution through collapse.

Figure 1.
Homogeneous collapse from rest and the growing and decaying fractional-density modes
. **Details**.

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Past exam of the mathematics course of the University of Cambridge / 2007 / iii / Paper 64 / 2 / ii / Solution Created 2026-10-03 Updated 2026-10-14

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Let

w = c_{s}^{2}

be constant and small, and write

θ = i k \cdot v

. The supplied synchronous gauge equations imply

δ^{'} = - (1 + w) (θ + h^{'} /2), θ^{'} + (1 - 3 w) H θ - \frac{w}{1 + w} k^{2} δ = 0.

(1)

Differentiating the first and eliminating

θ

gives

δ^{''} + (1 - 3 w) H δ^{'} + w k^{2} δ = - \frac{1 + w}{2} [h^{''} + (1 - 3 w) H h^{'}] .

(2)

For the dominant single component, the Einstein equation gives

h^{''} + H h^{'} = - 3 H^{2} (1 + 3 w) δ

. Therefore the exact algebraic intermediate relation is

δ^{''} + (1 - 3 w) H δ^{'} + [w k^{2} - \frac{3}{2} (1 + w) (1 + 3 w) H^{2}] δ = \frac{3}{2} w (1 + w) H h^{'} .

(3)

The requested nonrelativistic density equation in synchronous gauge follows to leading nonrelativistic order: discard pressure corrections to the background, Hubble damping, and metric tensor terms while retaining

c_{s}^{2} k^{2} δ

, which can be comparable to gravity at the Jeans wavenumber. Since

w ≪ 1

, that scale lies well inside the Hubble radius, so keeping its enhanced gradient term is consistent. The resulting equation is

δ^{''} + H δ^{'} + (c_{s}^{2} k^{2} - 4 π G a^{2} \overset{ρ}{ˉ}_{m}) δ = 0,

(4)

where the Friedmann equation gave

3 H^{2} /2 = 4 π G a^{2} \overset{ρ}{ˉ}_{m}

. Now

δ^{'} = a \dot{δ}

and

δ^{''} = a^{2} (\ddot{δ} + H \dot{δ})

, so

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

(5)

This is the specified approximation, not an exact finite-

w

relativistic identity. The pressure-gradient term stabilizes short wavelengths, while gravity drives the long-wavelength Jeans instability.

Now apply the stipulated equation to

P = K ρ^{4/3}

with

K > 0

. In the pressure-negligible matter background,

a \propto t^{2/3}, \overset{ρ}{ˉ}_{m} = \frac{1}{6 π G t ^{2}}, c_{s}^{2} = \frac{4}{3} K \overset{ρ}{ˉ}_{m}^{1/3} \propto t^{- 2/3} \propto a^{- 1} .

(6)

Thus

c_{s}^{2} k^{2} / a^{2}

is proportional to

t^{- 2}

, exactly like the gravitational term. Define

k_{J}^{2} = \frac{4 π G ρ ˉ _{m} a ^{2}}{c _{s}^{2}}, q = \frac{k ^{2}}{k _{J}^{2}} .

(7)

The comoving Jeans wavenumber and

q

are constant in this era. The equation becomes the Euler-Cauchy equation

\ddot{δ} + \frac{4}{3 t} \dot{δ} + \frac{2 ( q - 1 )}{3 t ^{2}} δ = 0.

(8)

Setting

δ \propto t^{p}

gives

p^{2} + p /3 + 2 (q - 1) /3 = 0

. For distinct real roots the full pair of modes is

δ (t) = A (\frac{t}{t _{*}})^{p_{+}} + B (\frac{t}{t _{*}})^{p_{-}}, p_{\pm} = \frac{- 1 \pm 25 - 24 q}{6} .

(9)

These are the Jeans modes of a four-thirds polytropic cosmological fluid. The physical Jeans length is

λ_{J} = \frac{2 πa}{k _{J}} = c_{s} \frac{π}{G ρ ˉ _{m}}, q = (\frac{λ _{J}}{λ})^{2}, λ = \frac{2 πa}{k} .

(10)

Here

λ_{J} \propto a

, so its comoving value is constant. For

λ > λ_{J}

q < 1

and

p_{+} > 0

, giving genuine growth; the other solution decays. In the long-wavelength pressure-free limit,

p_{+} = 2/3

and

p_{-} = - 1

, the familiar matter-era modes. At

λ = λ_{J}

, the two powers are

0

and

- 1/3

, so the leading mode is constant rather than growing.

For

1 < q < 25/24

both powers are negative: pressure prevents growth, although the modes still decay without oscillation. At

q = 25/24

they coincide and the complete solution is

δ = t^{- 1/6} [A + B ln (t / t_{*})] .

(11)

For

q > 25/24

they become a complex-conjugate pair; a real basis is

δ = t^{- 1/6} [A cos (ω ln (t / t_{*})) + B sin (ω ln (t / t_{*}))], ω = \frac{24 q - 25}{6} .

(12)

These short-wavelength acoustic oscillations have a decaying envelope. The onset of instability is

q = 1

, while the threshold for oscillatory time dependence is slightly larger,

q = 25/24

; expansion damping explains why these two thresholds are not identical.

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Past exam of the mathematics course of the University of Cambridge / 2017 / iii / Paper 320 / 3 / a / Solution Created 2026-10-03 Updated 2026-10-06

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On a length scale

ℓ

, pressure communicates support over a sound-crossing time

ℓ / v_{s}

, whereas self-gravity acts over a time of order

(G ρ_{0})^{- 1/2}

. When the former is longer, gravity can amplify an overdensity before pressure smooths it. Thus Jeans instability occurs at scales of order

ℓ ≳ v_{s} / G ρ_{0}

; the precise numerical factor depends on whether scale means wavelength or inverse wavenumber.

Take a uniform static background, small perturbations, an inviscid barotropic fluid, and

δ p = v_{s}^{2} δ ρ

with

v_{s}^{2} = (d p / d ρ)_{0} > 0

. An infinite Newtonian homogeneous density cannot itself have zero background gravitational acceleration while satisfying the unmodified Poisson equation for Newtonian gravity. Specify the Jeans swindle: subtract the homogeneous field, or regard an external prescription as balancing it, and apply Poisson's equation only to perturbations. The linearized continuity equation and Euler equations for an inviscid fluid are

\partial_{t} δ ρ + ρ_{0} \nabla \cdot δ v = 0, \partial_{t} δ v = - \frac{v _{s}^{2}}{ρ _{0}} \nabla δ ρ - \nabla δ ϕ, \nabla^{2} δ ϕ = 4 π G δ ρ .

(1)

Taking a time derivative of the first equation and substituting the second yields

\partial_{t}^{2} δ ρ = v_{s}^{2} \nabla^{2} δ ρ + 4 π G ρ_{0} δ ρ

. A plane wave

e^{i (k \cdot x - ω t)}

therefore gives

ω^{2} = v_{s}^{2} k^{2} - 4 π G ρ_{0}, k_{J} = \frac{4 π G ρ _{0}}{v _{s}} .

(2)

For

0 < k < k_{J}

there is exponential growth at rate

4 π G ρ_{0} - v_{s}^{2} k^{2}

. At

k_{J}

the density mode is neutral to this linear order, and above it there are acoustic oscillations. The critical wavelength is

2 π / k_{J} = π v_{s} / G ρ_{0}

. Transverse velocity perturbations decouple as zero-frequency vortical modes in this idealization.

For the attractive Yukawa potential, assume a real screening parameter

α \geq 0

. Its Green function obeys

(\nabla^{2} - α^{2}) (- G m e^{- α r} / r) = 4 π G m δ^{(3)} (r)

, including the delta source. Thus

δ ϕ_{k} = - 4 π G δ ρ_{k} / (k^{2} + α^{2})

. The same fluid elimination gives Jeans instability with Yukawa gravity:

ω^{2} = k^{2} [v_{s}^{2} - \frac{4 π G ρ _{0}}{k ^{2} + α ^{2}}] .

(3)

There exists a growing nonzero-wavenumber mode precisely when

4 π G ρ_{0} > v_{s}^{2} α^{2}, 0 < k < k_{J}^{2} - α^{2} .

(4)

Equality gives no exponentially growing nonzero mode. If screening exceeds this threshold, the finite-range gravitational response is too weak to overcome pressure at any wavelength. Otherwise all sufficiently long but nonconstant waves remain unstable, with a longer minimum unstable wavelength than in Newtonian gravity. For

α > 0

the long-wave gravitational contribution also scales as

k^{2}

, so the growth rate tends to zero as

k \to 0

; screening does not produce a second positive lower cutoff in wavenumber. A homogeneous screened background can have the constant gravitational potential

- 4 π G ρ_{0} / α^{2}

, with zero force, so no Newtonian background subtraction is needed for nonzero screening. The limits

k \to 0

and

α \to 0

are nonuniform. An exact

k = 0

density change is not one of the growing finite-wavelength modes. In the pressureless edge case

v_{s} = 0

, every finite

k > 0

is unstable for positive density and finite screening.

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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 / 2020 / ii / Paper 3 / 14D / i / Solution Created 2026-09-24 Updated 2026-09-29

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Expand the density contrast in Fourier modes,

δ (x, t) = \int δ_{k} (t) e^{i k \cdot x} d^{3} k

. In the static universe each mode satisfies

\ddot{δ}_{k} + c_{s}^{2} (k^{2} - k_{J}^{2}) δ_{k} = 0.

(1)

For

k > k_{J}

there are sound waves

δ_{k} = A_{k} e^{i ω_{k} t} + B_{k} e^{- i ω_{k} t}, ω_{k} = c_{s} k^{2} - k_{J}^{2} .

(2)

For

k < k_{J}

, Jeans instability gives growing and decaying exponentials with rate

c_{s} k_{J}^{2} - k^{2}

. At the Jeans wavenumber the two solutions are constant and linear in time.

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Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 347 / 1 / a / Solution 2026-09-28

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Let the mean particle mass be

μ m_{p}

. For a monatomic ideal gas with

γ = 5/3

P = \frac{ρ _{0} k _{B} T}{μ m _{p}}, c_{s}^{2} = \frac{γ k _{B} T}{μ m _{p}} .

(1)

A uniform sphere of mass

M = (4 π /3) ρ_{0} r^{3}

has Newtonian gravitational energy

W = - \frac{3 G M ^{2}}{5 r} .

(2)

At the virial threshold, the pressure term

3 \int P d V = 3 M k_{B} T / (μ m_{p})

balances

∣ W ∣

. Therefore

M = \frac{5 k _{B} T}{G μ m _{p}} r .

(3)

Eliminating

r

gives the Jeans mass

M_{J} = (\frac{5 k _{B} T}{G μ m _{p}})^{3/2} (\frac{3}{4 π ρ _{0}})^{1/2} .

(4)

The numerical coefficient depends on the convention used to identify a finite cloud with a Jeans mode. For example, assigning the mass inside a sphere of radius half the standard Jeans instability wavelength gives

M_{J} = \frac{π ^{5/2}}{6} \frac{c _{s}^{3}}{G ^{3/2} ρ _{0}^{1/2}} .

(5)

Both conventions have the physically invariant scaling

M_{J} \propto T^{3/2} ρ_{0}^{- 1/2} .

(6)

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Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 310 / 4 / c / Solution Created 2026-09-24 Updated 2026-09-25

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For a barotropic equation of state,

δ P = c_{s}^{2} δ ρ = c_{s}^{2} \overset{ρ}{ˉ} δ

. A spatial Fourier mode of comoving wavenumber

k

therefore satisfies

\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.

(1)

The Jeans wavenumber is defined by equality of the pressure and self-gravity terms:

k_{J} = \frac{a 4 π G ρ ˉ}{c _{s}} .

(2)

For

k ≫ k_{J}

, pressure dominates and produces acoustic oscillations whose amplitude is affected by Hubble friction. For

k ≪ k_{J}

, self-gravity dominates and a growing mode develops: this is Jeans instability.

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Pressure correction to a growing polytropic cosmological mode 2026-10-14

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Put

δ = t^{2/3} h (t)

in the Bessel density modes of a polytropic expanding universe equation. It becomes

h^{''} + 8 h^{'} / (3 t) + Λ^{2} t^{- 2 ν - 2} h = 0

. Substituting

h = 1 + a Λ^{2} t^{- 2 ν} + \dots

gives

a = [2 ν (5/3 - 2 ν)]^{- 1}

. Since

c k / R = Λ t^{- ν - 1}

, the fractional pressure correction is small when

(c k / R)^{2} t^{2} ≪ 2 ν (5/3 - 2 ν)

. For

0 < ν < 1/3

, this is the order-of-magnitude condition

c k / R ≪ 6 π ν Gρ

. It describes approach to the dust-like growing branch, not an exact universal time at which every initial perturbation starts increasing. The instantaneous Jeans instability balance instead compares

(c k / R)^{2}

with

4 π Gρ

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