Past exam of the mathematics course of the University of Cambridge / 2025 / ii / Paper 1 / 9E (Cosmology)

Table of contents
- a 9E
  - Solution a
- b 9E
  - Solution b

a

Solution

Successive radial wave crests travel the same fixed comoving distance, so

\int_{t_{e}}^{t_{0}} \frac{d t}{a ( t )} = \int_{t_{e} + δ t_{e}}^{t_{0} + δ t_{0}} \frac{d t}{a ( t )} .

To first order,

δ t_{0} / a (t_{0}) = δ t_{e} / a (t_{e})

. Since wavelength is proportional to the period,

1 + z = \frac{λ _{0}}{λ _{e}} = \frac{δ t _{0}}{δ t _{e}} = \frac{a ( t _{0} )}{a ( t _{e} )} .

b

Solution

Each photon loses a factor

(1 + z)^{- 1}

in energy, and cosmological time dilation reduces the photon arrival rate by another factor

(1 + z)^{- 1}

. Hence

L_{received} = \frac{L}{( 1 + z ) ^{2}},

so

n = 2

. The physical area of the sphere today is

4 πa (t_{0})^{2} x^{2}

F = \frac{L}{4 πa ( t _{0} ) ^{2} x ^{2} ( 1 + z ) ^{2}} .

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