Past exam of the mathematics course of the University of Cambridge / 2006 / iii / Paper 69 / 1 / a

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- Solution a

Solution

Statistical homogeneity makes the two-point correlation depend on the displacement

u = r - r^{'}

. Substituting the Fourier transforms and changing variables from

(r, r^{'})

to

(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^{'}} .

The last integral is a Dirac delta function, so

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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