Past exam of the mathematics course of the University of Cambridge / 2021 / ii / Paper 1 / 9B (Cosmology)

Table of contents
- a 9B
  - Solution a
- b 9B
  - Solution b
- c 9B
  - Solution c

a

Solution

In the homogeneous background,

v = 0

and the physical velocity is the Hubble flow

u = a H x = \overset{a}{˙} x .

The comoving coordinate

x

is held fixed by the time derivative in the fluid equation, so

\frac{\partial u}{\partial t} = \overset{a}{¨} x .

Moreover, spatial homogeneity gives

\nabla \overset{ˉ}{P} = 0

. The Euler equation therefore reduces to

\overset{ρ}{ˉ} a \overset{a}{¨} x = - \overset{ρ}{ˉ} \nabla \overset{ˉ}{Φ}, \nabla \overset{ˉ}{Φ} = - a \overset{a}{¨} x .

Taking the divergence in three spatial dimensions gives

\nabla^{2} \overset{ˉ}{Φ} = - 3 a \overset{a}{¨} .

Consistency with the Poisson equation,

\nabla^{2} \overset{ˉ}{Φ} = \frac{4 π G}{c ^{2}} \overset{ρ}{ˉ} a^{2},

then requires

\frac{a ¨}{a} = - \frac{4 π G}{3 c ^{2}} \overset{ρ}{ˉ} .

This is the pressureless, zero-cosmological-constant Raychaudhuri equation, obtained by the Newtonian fluid derivation of the Raychaudhuri equation.

b

Solution

The elementary Newtonian derivation of the Friedmann equation follows a test particle on the boundary of a finite expanding sphere. It therefore introduces a distinguished centre and can misleadingly suggest that all matter expands away from one origin.

The fluid derivation is local: the Euler equation and Poisson equation must be compatible in every comoving neighbourhood. It consequently describes a genuinely homogeneous infinite universe, in which every comoving observer sees the same Hubble flow, without selecting a physical centre. This is the improvement captured by the Newtonian fluid derivation of the Raychaudhuri equation.

c

Solution

Insert

ρ = \overset{ρ}{ˉ} + ϵ δ ρ, v = ϵ δ v

into the continuity equation. Its zeroth-order part is the homogeneous cosmological continuity equation

\dot{\overset{ρ}{ˉ}} + 3 H \overset{ρ}{ˉ} = 0.

Since

\overset{ρ}{ˉ}

is spatially homogeneous, the terms of first order in

ϵ

give

\frac{\partial δ ρ}{\partial t} + 3 Hδ ρ + \frac{ρ ˉ}{a} \nabla \cdot δ v = 0.

On the other hand, the quotient rule and the background equation imply

\frac{\partial}{\partial t} (\frac{δ ρ}{ρ ˉ}) = \frac{1}{ρ ˉ} \frac{\partial δ ρ}{\partial t} - \frac{δ ρ}{ρ ˉ ^{2}} \dot{\overset{ρ}{ˉ}} = \frac{1}{ρ ˉ} \frac{\partial δ ρ}{\partial t} + 3 H \frac{δ ρ}{ρ ˉ} .

Substitution of the first-order equation yields the linearized cosmological continuity equation

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

δ ρ / \overset{ρ}{ˉ}

is the density contrast.

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