 Solution

ID: past-exam-of-the-astrophysics-course-of-the-university-of-oxford/2021/4/c/solution

d s^{2} = - c^{2} d t^{2} + a^{2} (t) [d χ^{2} + χ^{2} (d ϑ^{2} + sin^{2} ϑ d φ^{2})] .

The source's

- c d t^{2}

is missing a factor

c

, and its angular term must mean the unit-sphere line element

d Ω^{2}

, not the solid angle element

d Ω = sin ϑ d ϑ d φ

. For example,

c d t^{2}

d s^{2}

has dimensions length squared. The corrected convention above removes both ambiguities.

A radial light ray has

d s^{2} = 0

and

∣ d χ ∣ = c ∣ d t ∣/ a

. Thus

\frac{d χ}{d z} = \frac{c}{H ( z )}, χ (z) = \frac{c}{H _{0}} \int_{0}^{z} \frac{d u}{( 1 + u ) ^{3/2}} = \frac{2 c}{H _{0}} (1 - \frac{1}{1 + z}) .

A proper shell volume at emission is

d V_{proper} = a^{3} χ^{2} d χ d Ω

. Multiplying by

n (z) = n_{0} a^{- 3}

cancels the expansion factors, equivalently using the comoving volume element directly. Hence

\frac{d ^{2} N}{d Ω d z} = n_{0} χ^{2} \frac{d χ}{d z} = \frac{4 c ^{3} n _{0} ( 1 + z - 1 ) ^{2}}{H _{0}^{3} ( 1 + z ) ^{5/2}} .

This counts every galaxy in the shell per solid angle, rather than intersections weighted by individual galaxy areas. Near

z = 0

it reduces to

n_{0} (c / H_{0})^{3} z^{2}

, as expected for a Euclidean volume shell.

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