Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 346 / 4 / a / Solution

For a statistically homogeneous and isotropic density contrast,

ξ (r) = ⟨ δ (x) δ (x + r)⟩, r = ∣ r ∣ .

Using the Fourier transform convention

δ (x) = \int \frac{d ^{3} k}{( 2 π ) ^{3}} δ_{k} e^{i k \cdot x}

and translational invariance gives

⟨ δ_{k} δ_{k^{'}}^{*} ⟩ = (2 π)^{3} δ_{D} (k - k^{'}) P (k),

with

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

At one point,

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

For a spherical top-hat volume

V = 4 π R^{3} /3

,

σ^{2} (R) = \frac{1}{V ^{2}} \int_{V} d^{3} x_{1} \int_{V} d^{3} x_{2} ξ (∣ x_{1} - x_{2} ∣) .

The overlap of two radius-

R

balls separated by

r \leq 2 R

is

V_{ov} (r) = V (1 - \frac{3 r}{4 R} + \frac{r ^{3}}{16 R ^{3}}) .

Therefore

σ^{2} (R) = \frac{3}{R ^{3}} \int_{0}^{2 R} r^{2} ξ (r) (1 - \frac{3 r}{4 R} + \frac{r ^{3}}{16 R ^{3}}) d r .

For

ξ (r) = (r / r_{0})^{- 1.8}

,

σ^{2} (R) = J_{2} (\frac{r _{0}}{R})^{1.8}, J_{2} = \frac{72}{2 ^{1.8} ( 3 - 1.8 ) ( 4 - 1.8 ) ( 6 - 1.8 )} ≃ 1.865.

Thus

σ_{8} = J_{2} (\frac{5}{8})^{0.9} ≃ 0.895 .

This calculation assumes that the observed galaxies trace the matter field with unit, scale-independent galaxy bias, and that the fitted power law remains valid throughout the top-hat separations.

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