Past exam of the mathematics course of the University of Cambridge / 2003 / iii / Paper 66 / 3 / a

Solution

H^{2} = A a^{- 3} + B a^{- 2} + C, A = H_{0}^{2} Ω_{m}, B = - k, C = Λ/3.

Differentiating on the expanding branch gives

\dot{H} = - \frac{3}{2} A a^{- 3} - B a^{- 2}, \ddot{H} = H (\frac{9}{2} A a^{- 3} + 2 B a^{- 2}) .

Consequently

\ddot{H} + 2 H \dot{H} - \frac{3}{2} A a^{- 3} H = H [(\frac{9}{2} - 3 - \frac{3}{2}) A a^{- 3} + (2 - 2) B a^{- 2}] = 0.

Λ

and curvature contributions to the background. It does not extend without change to a radiation background or arbitrary dynamical dark energy.

a \propto t^{2/3}

H = 2/ (3 t)

and

4 π G \overset{ρ}{ˉ} = 2/ (3 t^{2})

\ddot{D} + \frac{4}{3 t} \dot{D} - \frac{2}{3 t ^{2}} D = 0.

Substitute

D = t^{p}

p (p - 1) + 4 p /3 - 2/3 = 0

, with roots

2/3

and

- 1

. Therefore

D (t) = C_{+} t^{2/3} + C_{-} t^{- 1}, D_{+} \propto a, D_{-} \propto H \propto a^{- 3/2} .

The Hubble-parameter solution is the decaying mode in the Einstein-de Sitter universe.

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