 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2002/iii/paper-45/4/a/solution

Codex  0 Created 2026-10-03 Updated 2026-10-14

Let

δ (x)

be a mean-zero density contrast. For a statistically homogeneous and isotropic field, its two-point correlation function is

ξ (r) = ⟨ δ (x) δ (x + r)⟩

, depending only on

r = ∣ r ∣

. For a galaxy point process the corresponding distinct-pair definition is

d P_{12} = \overset{n}{ˉ}^{2} [1 + ξ (r)] d V_{1} d V_{2}

, the excess pair probability relative to independent uniform positions. Self-pair shot noise is separated when using the density-field version.

Specify the Fourier convention by

δ (x) = \int \frac{d ^{3} k}{( 2 π ) ^{3}} δ (k) e^{i k \cdot x}, ⟨ δ (k) δ^{*} (k^{'})⟩ = (2 π)^{3} δ_{D}^{(3)} (k - k^{'}) P (k) .

k

alone. Substitute the inverse transforms into the correlation and integrate the delta function:

ξ (r) = \int \frac{d ^{3} k}{( 2 π ) ^{3}} P (k) e^{i k \cdot r} .

Choose the polar axis parallel to

r

. Angular integration gives

\int d Ω_{k} e^{ik r c o s θ} = 4 π sin (k r) / (k r)

. Hence

ξ (r) = \frac{1}{2 π ^{2}} \int_{0}^{\infty} k^{2} P (k) \frac{sin ( k r )}{k r} d k .

Conversely, when the transform is defined,

P (k) = 4 π \int_{0}^{\infty} r^{2} ξ (r) sin (k r) / (k r) d r

. The stated convention fixes every

2 π

factor.

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