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. This comparison assumes the same potential-energy structure coefficient; it is the usual top-hat model rather than an exact claim about every halo profile.

Halving the radius multiplies the density by eight. During the interval to

t_{coll} = 2 t_{m a x}

, the background density falls by four. Thus

Δ_{vir} \equiv \frac{ρ _{vir}}{ρ ˉ ( t _{coll} )} = 8 \times 4 \times \frac{9 π ^{2}}{16} = 18 π^{2} ≃ 178.

(6)

The formal pressureless trajectory reaches zero radius; the physical virialized object instead has a finite radius. Its density is evaluated at the same collapse epoch, not at the instantaneous singular solution's density.

For a physical estimated virial radius and enclosed mass,

ρ_{h} = 3 M / (4 π r_{vir}^{3})

. Equating this to

Δ_{vir} \overset{ρ}{ˉ}_{m, 0} (1 + z_{f})^{3}

gives a halo density estimate of formation redshift:

1 + z_{f} ≃ [\frac{3 M}{4 π r _{vir}^{3} Δ _{vir} ρ ˉ _{m, 0}}]^{1/3} .

(7)

In the pure Einstein-de Sitter universe,

\overset{ρ}{ˉ}_{m, 0}

equals the present critical density. The inference assumes the measured density retains the collapse-epoch normalization. Later accretion, mergers and a changing virial-radius convention can change it, so it is a model-dependent formation estimate rather than a unique historical date.

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Past exam of the mathematics course of the University of Cambridge / 2016 / iii / Paper 310 / 1 / b / Solution Created 2026-10-03 Updated 2026-10-14

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Insert the component-density laws into the Friedmann equation, using

ρ_{i, 0} = 3 H_{0}^{2} Ω_{i, 0} / (8 π G)

and

a_{0} = 1

. Multiplying by

a^{2}

gives

\overset{a}{˙}^{2} = H_{0}^{2} (\frac{Ω _{r, 0}}{a ^{2}} + \frac{Ω _{m, 0}}{a} + Ω_{Λ, 0} a^{2}) - K .

(1)

The radiation in cosmology and pressureless matter contributions add; the printed first formula has lost the plus sign between them. In the Friedmann acceleration equation,

ρ + 3 P = 2 ρ_{r} + ρ_{m} - 2 ρ_{Λ}

, so

\overset{a}{¨} = - H_{0}^{2} (\frac{Ω _{r, 0}}{a ^{3}} + \frac{Ω _{m, 0}}{2 a ^{2}} - Ω_{Λ, 0} a) .

(2)

Evaluating the first equation today gives

K = H_{0}^{2} (Ω_{r, 0} + Ω_{m, 0} + Ω_{Λ, 0} - 1)

Now remove dark energy and assume a positive pressureless matter or radiation in cosmology density, starting on an expanding branch. Let

f (a) = H_{0}^{2} (Ω_{r, 0} / a^{2} + Ω_{m, 0} / a)

. It decreases from infinity to zero, and

\overset{a}{˙}^{2} = f (a) - K

, while

\overset{a}{¨} < 0

. This immediately distinguishes all three cases of spatial curvature of an FLRW universe.

Closed, $K > 0$ . There is a unique maximum size $a_{m a x}$ with $f (a_{m a x}) = K$ . Expansion from a Big Bang slows to $\overset{a}{˙} = 0$ and then becomes contraction, ending in a Big Crunch. The turning point is reached in finite time because $f (a) - K$ has a simple zero there. The acceleration remains negative at the maximum, so it is a recollapse rather than a static state.
Open, $K < 0$ . The expanding solution never stops because $\overset{a}{˙}^{2} = f (a) + ∣ K ∣ > 0$ . It expands forever, eventually approaching a curvature-dominated universe with $\overset{a}{˙} \to ∣ K ∣$ and $a (t) \sim ∣ K ∣ t$ . Locally its late-time behavior approaches the empty Milne universe.
Spatially flat, $K = 0$ . Expansion also continues forever, but $\overset{a}{˙} \to 0$ instead of approaching a nonzero constant. With pressureless matter present, the late-time solution has $a \propto t^{2/3}$ , the Einstein-de Sitter universe behavior; a flat solution containing only radiation in cosmology has $a \propto t^{1/2}$ . With both components, an early radiation domination era gives way to matter domination.

The early small-

a

solution has

a \propto t^{1/2}

if radiation in cosmology is present, and

a \propto t^{2/3}

for pressureless matter alone. The figure isolates the effect of curvature in three matter-only examples, rather than comparing models with the same present density parameters.

Figure 1.
Matter-only scale-factor evolution: closed recollapse, flat power-law expansion, and open expansion tending to a constant speed.
**Details**.

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Past exam of the mathematics course of the University of Cambridge / 2018 / iii / Paper 312 / 2 / i / Solution Created 2026-10-03 Updated 2026-10-05

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The background Einstein-de Sitter universe satisfies

a^{'} = H a

and

H^{'} = - H^{2} /2

. Introduce

F_{n}

and

G_{n}

for the spatial convolutions in the Fourier transform representation with kernels

F_{n}

and

G_{n}

, respectively, so

δ^{(n)} = a^{n} F_{n}

and

θ^{(n)} = - H a^{n} G_{n}

. Differentiation gives

(δ^{(n)})^{'} = n H a^{n} F_{n}, (θ^{(n)})^{'} = - (n - \frac{1}{2}) H^{2} a^{n} G_{n} .

(1)

At first order the cosmological continuity equation yields

F_{1} - G_{1} = 0

, while the cosmological Euler equation gives

3 F_{1} /2 - 3 G_{1} /2 = 0

. The growing-mode normalization therefore gives

G_{1} = F_{1} = 1

At second order, take the unsymmetrized order of the inputs printed in the paper. Matching the alpha mode-coupling kernel and beta mode-coupling kernel terms gives

2 F_{2} - G_{2} = α (q_{1}, q_{2}), \frac{5}{2} G_{2} - \frac{3}{2} F_{2} = β (q_{1}, q_{2}) .

(2)

Hence

F_{2} = \frac{5}{7} α + \frac{2}{7} β, G_{2} = \frac{3}{7} α + \frac{4}{7} β .

(3)

With the paper's unsymmetrized convention

α = β = (q_{1} + q_{2}) / q_{1}

, one possible unsymmetrized representation is

F_{2} (q_{1}, q_{2}) = G_{2} (q_{1}, q_{2}) = \frac{q _{1} + q _{2}}{q _{1}} .

(4)

Only the symmetric part enters a convolution of two identical linear fields. Symmetrizing gives

F_{2, s} = G_{2, s} = \frac{1}{2} [\frac{q _{1} + q _{2}}{q _{1}} + \frac{q _{1} + q _{2}}{q _{2}}] = \frac{( q _{1} + q _{2} ) ^{2}}{2 q _{1} q _{2}} .

(5)

This is the one-dimensional cosmological density kernel. The physical symmetric one-dimensional Euler kernel is also

(q_{1} + q_{2})^{2} / (2 q_{1} q_{2})

: symmetrizing the printed

β

reproduces it, so its apparently different unsymmetrized form is not a contradiction. Unsymmetrized kernels themselves are not unique, since antisymmetric parts integrate to zero. Zero input momenta require the usual separate treatment of the homogeneous background.

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Past exam of the mathematics course of the University of Cambridge / 2018 / ii / Paper 1 / 9B / Solution Created 2026-09-24 Updated 2026-10-14

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Let

H = a^{'} / a = \overset{a}{˙}

, where a prime denotes differentiation with respect to conformal time

d τ = d t / a

. Multiplying the Friedmann equation by

a^{2}

gives

H^{2} + k c^{2} = \frac{8 π G}{3} ρ a^{2} .

(1)

The critical density is the value obtained when

k = 0

, so

Ω = \frac{ρ}{ρ _{crit}} = \frac{8 π Gρ a ^{2}}{3 H ^{2}}, \frac{k c ^{2}}{H ^{2}} = Ω - 1 .

(2)

Since

H^{'} = a \overset{a}{¨}

, the pressure-free Raychaudhuri equation gives

2 H^{'} = - \frac{8 π G}{3} ρ a^{2} = - (H^{2} + k c^{2}),

(3)

2 H^{'} + H^{2} + k c^{2} = 0.

(4)

Differentiating

Ω = 1 + k c^{2} / H^{2}

and substituting this equation yields

Ω^{'} = H Ω (Ω - 1) .

(5)

For an expanding universe

H > 0

. The Einstein-de Sitter universe

Ω = 1

is an unstable constant solution: curves starting with

0 < Ω < 1

decrease towards

0

, while curves starting with

Ω > 1

rise away from

1

and diverge as the closed model reaches maximum expansion. This is the flatness problem.

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Past exam of the mathematics course of the University of Cambridge / 2018 / ii / Paper 2 / 9B / b / Solution Created 2026-09-24 Updated 2026-10-14

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The physical particle horizon is

d_{H} (t) = a (t) \int_{0}^{t} \frac{c d t ^{'}}{a ( t ^{'} )} .

(1)

For an Einstein-de Sitter universe,

a (t) \propto t^{2/3}

, so

d_{H} (t_{0}) = t_{0}^{2/3} c \int_{0}^{t_{0}} t^{' - 2/3} d t^{'} = 3 c t_{0} .

(2)

Since

H_{0} = 2/ (3 t_{0})

d_{H} (t_{0}) = \frac{2 c}{H _{0}} .

(3)

The horizon problem is that widely separated regions of the observed universe have nearly identical conditions even though, in a decelerating standard Big Bang history, their past light cones do not overlap early enough to establish causal thermal equilibrium.

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Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 310 / 1 / c / Solution Created 2026-10-03 Updated 2026-10-05

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Normalize the present scale factor to

a (t_{0}) = 1

. In a flat universe containing radiation in cosmology, pressureless matter, and a cosmological constant, the Friedmann equation is

H (a) = H_{0} Ω_{r, 0} a^{- 4} + Ω_{m, 0} a^{- 3} + Ω_{Λ, 0},

(1)

where

Ω_{r, 0} + Ω_{m, 0} + Ω_{Λ, 0} = 1

. Since

d t = d a / (a H)

, the age of an FLRW universe is

t_{0} = \frac{1}{H _{0}} \int_{0}^{1} \frac{d a}{a Ω _{r, 0} a ^{- 4} + Ω _{m, 0} a ^{- 3} + Ω _{Λ, 0}} .

(2)

For a matter-only flat Einstein-de Sitter universe,

t_{0} = \frac{2}{3 H _{0}} .

(3)

Taking the present value

H_{0} ≃ 70 km s^{- 1} Mp c^{- 1}

, for which the Hubble time is

H_{0}^{- 1} ≃ 14.0 Gyr

, gives

t_{0} ≃ 9.3 Gyr .

(4)

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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 346 / 3 / Solution Created 2026-10-03 Updated 2026-10-14

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Assume nonrelativistic pressureless matter, an initial irrotational vector field of growing-mode linear cosmological density perturbations, a homogeneous expanding background, and no shell crossing. All gradients below refer to initial comoving coordinates. Let

ρ_{0} = \overset{ρ}{ˉ}_{m} (t) a^{3} (t) = \overset{ρ}{ˉ}_{m} (t_{i})

, with

a (t_{i}) = 1

, and normalize the linear growth factor by

D (t_{i}) = 1

. The paper calls

D

a growth rate, but it is the dimensionless growth factor;

\dot{D}

is its time derivative.

Writing

x_{i}

for the actual position at

t_{i}

, the Zeldovich approximation is, to first order,

x (t) = x_{i} + [D (t) - 1] s (x_{i}), \dot{x} = \dot{D} s (x_{i}) .

(1)

This explicitly gives

x (t_{i}) = x_{i}

. The usual notation

x = q + D s (q)

instead uses the uniform reference Lagrangian coordinate

q

, with

x_{i} = q + s (q)

. The two labels can be interchanged inside a first-order displacement, but not in the unperturbed position without the initial displacement correction.

To determine

s

, mass conservation between the initial and current positions gives

1 + δ (t) = \frac{1 + δ _{i}}{d e t [ 1 + ( D - 1 ) \nabla s ]} = 1 + δ_{i} - (D - 1) \nabla \cdot s + O (δ_{i}^{2}) .

(2)

Demanding

δ (t) = D (t) δ_{i}

gives

\nabla \cdot s = - δ_{i}

. The irrotational vector field assumption makes

s

a gradient; the initial cosmological Poisson equation is

\nabla^{2} Φ_{i} = 4 π G ρ_{0} δ_{i}

. With the same boundary conditions for these potential equations, and after removing an irrelevant uniform translation,

s = - \frac{\nabla Φ _{i}}{4 π G ρ _{0}}, \dot{x} = - \frac{D ˙}{4 π G ρ ˉ _{m} a ^{3}} \nabla Φ_{i} .

(3)

The evolution of

D

follows from the linearized peculiar-motion equation

\ddot{x} + 2 H \dot{x} = - \nablaΦ/ a^{2}

. Since cosmological Poisson equation gives

Φ (t) = (D / a) Φ_{i}

to this order, substitution yields the linear growth equation

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

(4)

Thus the approximation extrapolates the linear growing displacement along a fixed initial direction, even as its density mapping becomes nonlinear.

The early spin of a dark matter halo comes from an external gravitational torque on its nonspherical initial mass region. A uniform external acceleration moves its centre of mass without spinning it; the spatially varying tidal tensor exerts different forces on different parts. The tidal torque theory requires misalignment of the region's shape and the surrounding tidal field. A local irrotational vector field of velocity does not imply zero integrated angular momentum for a nonspherical region.

Put

y = x_{i} - \hat{x}_{i}

and use the physical peculiar velocity

v = a \dot{x}

. The physical lever arm is

a (x - \hat{x})

, while a leading-order mass element is

ρ_{0} d^{3} x_{i}

. Substituting the velocity formula into the supplied definition of angular momentum about the centre of mass gives

J = - \frac{a ^{2} D ˙}{4 π G} \int_{V_{L}} y \times \nabla Φ_{i} d^{3} x_{i} .

(5)

Terms from the perturbed mass measure are higher order. Subtracting the barycentre's velocity also changes nothing because

\int_{V_{L}} y d^{3} x_{i} = 0

at this order.

Apply the divergence theorem componentwise:

ϵ_{jk l} \int_{V_{L}} y_{k} \partial_{l} Φ_{i} d^{3} x_{i} = ϵ_{jk l} \int_{Σ_{L}} Φ_{i} y_{k} d S_{l} - ϵ_{jk l} \int_{V_{L}} Φ_{i} δ_{k l} d^{3} x_{i} = ϵ_{jk l} \int_{Σ_{L}} Φ_{i} y_{k} d S_{l} .

(6)

The antisymmetry of the Levi-Civita symbol kills the last term. Consequently, for the ordinary oriented comoving surface element

d S

J = - \frac{a ^{2} D ˙}{4 π G} \int_{Σ_{L}} Φ_{i} y \times d S .

(7)

The original PDF prints a different coefficient,

- \dot{D} / (4 π G \overset{ρ}{ˉ}_{m} a^{2})

. That coefficient does not follow from its own mass measure and velocity equation and has the wrong dimensions for total angular momentum. The corrected coefficient above also gives the time dependence consistent with the paper's later tensor expression; this is a source typo, not a TeX transcription issue.

For a spherical

V_{L}

centered on its barycentre, the outward normal is parallel to

y

, so

y \times d S = 0

pointwise. Hence

J = 0 for a spherical initial region .

(8)

For an equipotential surface,

Φ_{i} = Φ_{*}

Σ_{L}

, and

ϵ_{jk l} \int_{Σ_{L}} y_{k} d S_{l} = ϵ_{jk l} \int_{V_{L}} \partial_{l} y_{k} d^{3} x_{i} = ϵ_{jk l} δ_{k l} Vol (V_{L}) = 0.

(9)

Therefore

J = 0 if the initial boundary is an equipotential .

(10)

The second conclusion holds for any boundary shape at this order, not just a sphere.

To extract the leading tidal torque, expand the initial peculiar gravitational potential about the barycentre using its Taylor series:

Φ_{i} (\hat{x}_{i} + y) = Φ_{0} + g_{k} y_{k} + \frac{1}{2} H_{km} y_{k} y_{m} + \dots, H_{km} = \partial_{k} \partial_{m} Φ_{i} (\hat{x}_{i}) .

(11)

The constant has no force;

g

has zero integrated torque because the first mass moment vanishes. With the mass second-moment tensor

I_{km} = ρ_{0} \int_{V_{L}} y_{k} y_{m} d^{3} x_{i},

(12)

we obtain

J_{j} = - \frac{a ^{2} D ˙}{4 π G ρ _{0}} ϵ_{jk l} H_{l m} I_{km} = \frac{D ˙}{4 π G ρ ˉ _{m} a} ϵ_{jk l} H_{km} I_{m l} .

(13)

The second equality swaps the indices

k, l

and uses the symmetry of both tensors. Define the initial acceleration tidal tensor by

T_{km} = - \partial_{k} \partial_{m} Φ_{i} (\hat{x}_{i}) = - H_{km} .

(14)

Then the formula in the question has exactly the stated sign:

J_{j} = - \frac{D ˙}{4 π G ρ ˉ _{m} a} ϵ_{jk l} T_{km} I_{m l} .

(15)

If one defines

T

as the positive Hessian matrix of

Φ_{i}

instead, the displayed contraction has a plus sign. Defining the convention is essential. Also, the paper's

I

is a mass second-moment tensor; the mechanical inertia tensor is

tr (I) 1 - I

Only the anisotropic parts contribute: a multiple of the identity produces zero contraction with the Levi-Civita symbol. In axes where

I = diag (I_{1}, I_{2}, I_{3})

, for example,

J_{1} = \frac{D ˙}{4 π G ρ ˉ _{m} a} T_{23} (I_{2} - I_{3}) .

(16)

Thus a spherical region or tensors sharing principal axes have zero leading torque; unequal principal moments together with off-diagonal tidal components create spin. The Taylor series truncation assumes that higher spatial derivatives of the tidal field are sufficiently small across the region.

In an Einstein-de Sitter universe,

a \propto t^{2/3}

D \propto a

, and

\overset{ρ}{ˉ}_{m} \propto a^{- 3}

. The initial

I

and

T

are time independent, so

J \propto a^{2} \dot{D} \propto t^{4/3} t^{- 1/3} \propto t .

(17)

This is early-time growth for fixed initial matter, before collapse invalidates the extrapolation. It does not predict indefinite linear spin growth for a virialized halo.

Two reasons for only approximate agreement with simulations are that nonlinear collapse and shell crossing change the trajectories and the tidal field, invalidating the first-order displacement and fixed leading tidal tensor; and that real halos undergo dark-matter halo mergers, anisotropic accretion, and exchange of material and angular momentum, so the halo identified at a later time need not be the same isolated collection of initial particles. These processes can change both the magnitude and the direction of the spin.

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Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 310 / 1 / b / Solution 2026-09-28

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The Friedmann acceleration equation and

H = \overset{a}{˙} / a

give the deceleration parameter

q = - \frac{a ¨}{a H ^{2}} = \frac{4 π G}{3 H ^{2}} \sum_{i} ρ_{i} (1 + 3 w_{i}) = \frac{1}{2} i \sum Ω_{i} (z) (1 + 3 w_{i}) .

(1)

An Einstein-de Sitter universe contains only pressureless matter, so

q_{0} = 1/2

and cannot explain the observed negative value. For the stated Lambda-CDM model,

q_{0} = \frac{1}{2} [0.3 + 0.7 (1 - 3)] = - 0.55,

(2)

so its cosmological constant produces the required accelerating universe.

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Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 310 / 1 / c / Solution 2026-09-28

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The age of an FLRW universe follows from

d t = d a / (a H)

. Neglecting radiation and spatial curvature gives

t_{0} = \frac{1}{H _{0}} \int_{0}^{1} \frac{d a}{a Ω _{m, 0} a ^{- 3} + Ω _{Λ, 0}} = \frac{1}{H _{0}} \int_{0}^{1} \frac{a ^{1/2} d a}{Ω _{m, 0} + Ω _{Λ, 0} a ^{3}} .

(1)

For an Einstein-de Sitter universe,

Ω_{m, 0} = 1

and

Ω_{Λ, 0} = 0

, so

t_{0} = H_{0}^{- 1} \int_{0}^{1} a^{1/2} d a = \frac{2}{3} H_{0}^{- 1} .

(2)

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Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 310 / 1 / d / Solution 2026-09-28

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The measured Hubble time would give the Einstein-de Sitter age

t_{0} = \frac{2}{3} (14 Gyr) ≃ 9.3 Gyr,

(1)

which is less than the measured

12 Gyr

age of stars that must themselves be younger than the universe. The measurements are therefore incompatible with an Einstein-de Sitter universe. For the stated spatially flat Lambda-CDM model, direct evaluation of the preceding integral gives

H_{0} t_{0} = \frac{2}{3 0.7} arsinh \frac{0.7}{0.3} ≃ 0.964,

(2)

and hence

t_{0} ≃ 13.5 Gyr

. The period of accelerating expansion caused by the cosmological constant therefore allows an age consistent with the stellar lower bound.

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Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 346 / 1 / Solution 2026-09-29

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Write physical position as

r = a (t) x

and physical velocity as

\dot{r} = H r + v, v = a \dot{x} .

(1)

Subtract the homogeneous expanding background from the pressureless Euler equation. To linear order in peculiar velocity and density contrast, the convective term is negligible and

\dot{v} + H v = - \frac{1}{a} \nablaΦ,

(2)

\frac{d}{d t} (a v) = - \nablaΦ .

(3)

Here

\nabla

differentiates with respect to comoving position and

Φ

is the peculiar gravitational potential.

The linear continuity and Poisson equations are

\dot{δ} + \frac{1}{a} \nabla \cdot v = 0, \nabla^{2} Φ = 4 π G a^{2} \overset{ρ}{ˉ}_{m} δ .

(4)

For the growing mode

δ (x, t) = D (a) δ_{i} (x)

, matter conservation gives

\overset{ρ}{ˉ}_{m} \propto a^{- 3}

, so the potential scales as

Φ (x, t) = \frac{D ( a )}{a} Φ_{i} (x)

(5)

after absorbing the fiducial normalization into

Φ_{i}

. Integrating the linear Euler equation with a negligible decaying mode gives

v = - \frac{\nabla Φ _{i}}{a} \int^{t} \frac{D ( a )}{a} d t^{'} .

(6)

Taking the divergence of Euler and using continuity and Poisson yields the linear growth equation

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

(7)

In the fiducial normalization used in the question this can be rearranged as

\frac{D ( a )}{a} = \frac{1}{4 π G ρ ˉ _{m, i}} \frac{d}{d t} (a^{2} \frac{d D}{d t}) .

(8)

The cosmological Lagrangian displacement obeys

\dot{ψ} = v / a

. Combining the last two relations and choosing the growing displacement gives the Zeldovich approximation

ψ (t) = - \frac{D ( a )}{4 π G ρ ˉ _{m, i}} \nabla Φ_{i} .

(9)

Define

b (t) = \frac{D ( a )}{4 π G ρ ˉ _{m, i}} .

(10)

Then

x = x_{i} - b \nabla Φ_{i}

and

v = - a \dot{b} \nabla Φ_{i}

. Keeping the lowest nonvanishing order in displacement in the halo angular momentum gives

J = - \overset{ρ}{ˉ}_{m} a^{5} \dot{b} \int_{V_{L}} d^{3} x_{i} (x_{i} - \overset{ˉ}{x}_{i}) \times \nabla Φ_{i} .

(11)

The angular momentum vanishes for a spherical Lagrangian patch, and also whenever the patch's inertia principal axes align with the local tidal-field principal axes. It likewise vanishes in a locally isotropic tidal field.

Taylor-expand

\partial_{j} Φ_{i} (x_{i}) = \partial_{j} Φ_{i} (\overset{ˉ}{x}_{i}) + T_{j l} (x_{i} - \overset{x}{ˉ}_{i})_{l} + \dots, T_{j l} = \partial_{j} \partial_{l} Φ_{i} ∣_{\overset{ˉ}{x}_{i}} .

(12)

The constant-gradient term vanishes by the barycentre definition. With

I_{l k} = \int_{V_{L}} d^{3} x_{i} (x_{i} - \overset{x}{ˉ}_{i})_{l} (x_{i} - \overset{x}{ˉ}_{i})_{k} a^{3} \overset{ρ}{ˉ}_{m},

(13)

one obtains the tidal torque theory result

J_{i} = - a^{2} \dot{b} ϵ_{ijk} T_{j l} I_{l k} .

(14)

I

is the protohalo inertia tensor and

T

is the local tidal, or gravitational Hessian, tensor. Their eigenframe misalignment produces the torque.

In an Einstein-de Sitter universe,

D \propto a \propto t^{2/3}

. Hence

\dot{b} \propto \overset{a}{˙} \propto t^{- 1/3}

and

∣ J ∣ \propto a^{2} \dot{b} \propto t .

(15)

This linear estimate is normally stopped near turnaround. Simulated haloes subsequently gain angular momentum through nonlinear torques, anisotropic accretion, and mergers, including the orbital angular momentum of infalling subhaloes, so the early linear estimate underpredicts the final value.

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Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 312 / 4 / i / Solution 2026-09-28

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The displayed kernel is the growing-mode result of standard perturbation theory in cosmology. Its assumptions are a Newtonian, weak-field, subhorizon treatment of pressureless cold matter; a single-stream, irrotational velocity field; and negligible velocity-dispersion tensor of collisionless matter, so

σ_{ij} = 0

. The background is the Einstein-de Sitter universe, which makes the growing mode proportional to

a

and permits the separated powers

a^{n} F_{n}

. The density contrast is assumed perturbatively small and the decaying solutions are discarded. Gaussian initial conditions are not needed to derive

F_{2}

, but they are needed for the loop contractions in the later parts.

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Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 346 / 1 / a / Solution Created 2026-09-28 Updated 2026-10-05

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For an adiabatic cosmological perturbation, the baryon and photon fractional density perturbations obey

δ_{b} = 3 δ_{γ} /4

. Since the pressure of the photon-baryon fluid is supplied by the photons,

d P = \frac{1}{3} d \overset{ρ}{ˉ}_{γ}, d \overset{ρ}{ˉ} = d \overset{ρ}{ˉ}_{γ} + d \overset{ρ}{ˉ}_{b} = (1 + \frac{3 ρ ˉ _{b}}{4 ρ ˉ _{γ}}) d \overset{ρ}{ˉ}_{γ} .

(1)

The photon-baryon sound speed is therefore

c_{s} = c \frac{d P}{d ρ ˉ} = \frac{c}{3} (1 + \frac{3 ρ ˉ _{b}}{4 ρ ˉ _{γ}})^{- 1/2} .

(2)

Use

\overset{ρ}{ˉ}_{b} \propto a^{- 3}

and

\overset{ρ}{ˉ}_{γ} \propto a^{- 4}

in the comoving Jeans length

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

(3)

Well before matter-radiation equality, photon inertia dominates,

c_{s} ≃ c / 3

, and

λ_{J, com} \propto a^{1/2} \propto t^{1/4} .

(4)

Once baryon loading dominates while tight coupling still holds,

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

, so

λ_{J, com}

is approximately constant. At cosmological recombination, photon pressure support disappears and the baryonic Jeans scale drops sharply. For a subsequently adiabatic monatomic gas,

T_{b} \propto a^{- 2}

and

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

, giving

λ_{J, com} \propto a^{- 1/2} \propto t^{- 1/3}

(5)

in an Einstein-de Sitter universe. The requested graph therefore rises as

t^{1/4}

, flattens before recombination, jumps downward there, and then decreases as

t^{- 1/3}

For collisionless matter, the same instantaneous estimate uses its one-dimensional velocity dispersion

σ_{v}

instead of

c_{s}

; more precisely, suppression is described by collisionless free streaming. Thus

λ_{J, com} \sim \frac{σ _{v}}{a} \frac{π}{G ρ ˉ _{cdm}} .

(6)

While the particles are relativistic,

σ_{v} ≃ c

and

λ_{J, com} \propto a^{1/2}

. Once nonrelativistic but thermally coupled to radiation,

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

, so

σ_{v} \propto a^{- 1/2}

and the scale is constant. After kinetic decoupling, momentum redshifts as

a^{- 1}

, so

σ_{v} \propto a^{- 1}

and

λ_{J, com} \propto a^{- 1/2}

. The second graph joins these three power laws at

t_{NR}

and

t_{dec}

; unlike the baryonic graph, its final decline begins at dark-matter kinetic decoupling rather than recombination.

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Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 346 / 3 / Solution 2026-09-28

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The definition of the virial radius of a dark-matter halo immediately gives the virial mass of a dark-matter halo

M_{v} = \frac{4 π}{3} v ρ_{c}^{0} r_{v}^{3} .

(1)

In an Einstein-de Sitter universe, spherical collapse gives

v = 18 π^{2} ≃ 178

, conventionally rounded to about

200

For the Navarro--Frenk--White profile, put

x = r / r_{s} = cs

. Direct integration gives

M (r) = 4 π ρ_{c}^{0} δ_{char} r_{s}^{3} [lo g (1 + x) - \frac{x}{1 + x}] .

(2)

Using

r_{s} = r_{v} / c

and

δ_{char} = v c^{3} g (c) /3

therefore yields

\frac{M ( s )}{M _{v}} = g (c) [lo g (1 + cs) - \frac{cs}{1 + cs}] .

(3)

At small radius the bracket is

(cs)^{2} /2 + O ((cs)^{3})

, so

M (r) \propto r^{2}

, consistent with the central

ρ \propto r^{- 1}

cusp. At large radius it is

lo g (cs) - 1 + o (1)

, so the mass diverges logarithmically unless the halo is truncated.

The primordial free streaming of warm dark matter erases small-scale density fluctuations and lowers the central phase-space density, tending to replace the smallest, earliest cold-dark-matter cusps by shallower central profiles or cores.

Since

V_{v}^{2} = G M_{v} / r_{v}

, the spherical potential that vanishes at infinity is

Φ (s) = - g (c) V_{v}^{2} \frac{l o g ( 1 + cs )}{s} .

(4)

Indeed,

r d Φ/ d r = GM (r) / r

, and normalization at

s = 1

requires

g (c) = [lo g (1 + c) - \frac{c}{1 + c}]^{- 1} .

(5)

The circular speed is consequently

V^{2} (s) = V_{v}^{2} \frac{g ( c )}{s} [lo g (1 + cs) - \frac{cs}{1 + cs}] .

(6)

It rises as

s^{1/2}

near the centre, peaks at

r ≃ 2.16 r_{s}

, and then declines approximately as

lo g r / r

. Increasing the concentration of a dark-matter halo moves the peak inward in units of

r_{v}

and raises it relative to

V_{v}

. Thus an NFW curve can be fairly broad but is not exactly a flat galaxy rotation curve; stellar and gas contributions matter when comparing with an observed galaxy rotation curve.

Lower-mass haloes typically collapse earlier, when the cosmic background density is larger. Their characteristic inner densities are consequently larger relative to the present virial density, producing the mass-concentration relation of dark-matter haloes in which concentration decreases weakly with mass.

For a mass smaller by

1024 = 2^{10}

M_{v} ≃ 9.8 \times 1 0^{8} M_{⊙}, r_{v} = 200102 4^{- 1/3} kpc ≃ 19.8 kpc,

(7)

and

c = 10102 4^{0.1} = 20

. Hence

r_{s} = r_{v} / c ≃ 0.99 kpc .

(8)

A halo of roughly

1 0^{9} M_{⊙}

and kiloparsec scale naturally hosts a dwarf galaxy, possibly an extremely faint one if star formation is inefficient.

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Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 346 / 4 / Solution Created 2026-09-28 Updated 2026-09-29

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Write proper position as

r = a (t) x

and proper velocity as

u = \dot{r} = H r + v, v = a \dot{x} .

(1)

Substitute this decomposition into the proper-coordinate Euler equations for an inviscid fluid, use

\nabla_{r} = a^{- 1} \nabla_{x}

, and subtract the homogeneous-background acceleration. With

ρ = \overset{ρ}{ˉ} (1 + δ)

, one obtains the comoving peculiar-velocity equation

\frac{\partial v}{\partial t} + \frac{a ˙}{a} v + \frac{1}{a} (v \cdot \nabla) v = - \frac{1}{a} \nablaΦ - \frac{1}{a ρ ˉ ( 1 + δ )} \nabla P .

(2)

The peculiar gravitational potential

Φ

is the total Newtonian potential with the potential of the exactly homogeneous expanding background subtracted. Its gradient therefore generates only accelerations relative to the Hubble flow.

For a pressureless fluid, linearization discards the quadratic advection term, leaving