 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2006/iii/paper-69/1/a/solution

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

u = r - r^{'}

. Substituting the Fourier transforms and changing variables from

(r, r^{'})

(u, r^{'})

gives

E [B_{i} (k) B_{j} (k^{'})] = \int d^{3} u d^{3} r^{'} C_{ij} (u) e^{- i k \cdot u} e^{- i (k + k^{'}) \cdot r^{'}} = C_{ij} (k) \int d^{3} r^{'} e^{- i (k + k^{'}) \cdot r^{'}} .

E [B_{i} (k) B_{j} (k^{'})] = (2 π)^{3} δ (k + k^{'}) C_{ij} (k) .

The plus sign occurs because neither Fourier transform has undergone complex conjugation. For a real magnetic field, the version with

B_{j} (k^{'})^{*}

instead has

δ (k - k^{'})

. Homogeneity determines dependence on the displacement vector; dependence only on its norm needs statistical isotropy as well.

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