Use the sign convention in the question,

V (r) = k_{0}^{2} [n^{2} (r) - 1] .

(1)

Then the total field satisfies

(\nabla^{2} + k_{0}^{2}) ψ = - V ψ .

(2)

The outgoing free-space Green function is

G_{0} (r - r^{'}) = \frac{e ^{i k_{0} ∣ r - r^{'} ∣}}{4 π ∣ r - r ^{'} ∣},

(3)

with

(\nabla^{2} + k_{0}^{2}) G_{0} = - δ

. Hence the Lippmann-Schwinger equation is

ψ (r) = ψ_{i} (r) + \int_{D} G_{0} (r - r^{'}) V (r^{'}) ψ (r^{'}) d^{3} r^{'} .

(4)

Replacing the unknown interior total field by the incident field gives the first Born approximation

ψ_{B} (r) = ψ_{i} (r) + \int_{D} G_{0} (r - r^{'}) V (r^{'}) ψ_{i} (r^{'}) d^{3} r^{'} .

(5)

For the Rytov approximation, put

ψ = ψ_{i} e^{ϕ}

. After division by

ψ

, the wave equation gives

\nabla^{2} ϕ + (\nabla ϕ)^{2} + 2\nabla lo g ψ_{i} \cdot \nabla ϕ = - V .

(6)

Neglecting the quadratic term

(\nabla ϕ)^{2}

makes

ψ_{i} ϕ

obey the same inhomogeneous equation as the first Born scattered field. Thus

ϕ_{1} (r) = \frac{1}{ψ _{i} ( r )} \int_{D} G_{0} (r - r^{'}) V (r^{'}) ψ_{i} (r^{'}) d^{3} r^{'},

(7)

and

ψ_{R} (r) = ψ_{i} (r) e^{ϕ_{1} (r)} .

(8)

Its power series begins

ψ_{R} = ψ_{i} (1 + ϕ_{1} + O (ϕ_{1}^{2})) = ψ_{B} + O (V^{2}),

(9)

so the two approximations agree through first order in the scattering potential.

ii

 0  0

Solution

 0  0

Let the incident plane wave be

ψ_{i} (r) = e^{i k_{0} r_{0} \cdot r},

(1)

and define the scattering vector

q = k_{0} (r - r_{0}) .

(2)

For

r

much larger than the diameter of

D

, the far-field pattern follows from

G_{0} (r - r^{'}) \sim \frac{e ^{i k_{0} r}}{4 π r} e^{- i k_{0} r \cdot r^{'}} .

(3)

Writing

V (q) = \int_{D} V (r^{'}) e^{- i q \cdot r^{'}} d^{3} r^{'},

(4)

the Born result is

ψ_{B} (r) \sim e^{i k_{0} r_{0} \cdot r} + \frac{e ^{i k_{0} r}}{4 π r} V (q) .

(5)

The corresponding Rytov logarithmic perturbation is

ϕ_{1} (r) \sim a (r) V (q), a (r) = \frac{e ^{i k_{0} (r - r_{0} \cdot r)}}{4 π r} .

(6)

Therefore

ψ_{R} (r) \sim e^{i k_{0} r_{0} \cdot r} exp [a (r) V (q)] .

(7)

iii

 0  0

Solution

 0  0

Since

n = 1 + W

with small variance,

V = k_{0}^{2} (2 W + W^{2}) = 2 k_{0}^{2} W + O (W^{2}) .

(1)

To leading order,

V

is therefore a centered stationary Gaussian random field. Let its autocorrelation function of a random field be

C_{V} (s) = ⟨ V (r^{'}) V (r^{'} + s) ⟩ .

(2)

The

W^{2}

contribution gives higher-order mean and non-Gaussian corrections and is consistently omitted at this order.

The far-field Rytov approximation from part ii has unit incident intensity and

I (r) = ⟨ e^{ϕ_{1} + ϕ_{1}^{*}} ⟩, ϕ_{1} = a V (q) .

(3)

The real random variable

X = ϕ_{1} + ϕ_{1}^{*}

is centered Gaussian. Its moment-generating function gives

I = exp (\frac{1}{2} ⟨ X^{2} ⟩) .

(4)

Define

J_{-} (q) = \int_{D} \int_{D} C_{V} (r_{1} - r_{2}) e^{- i q \cdot (r_{1} - r_{2})} d^{3} r_{1} d^{3} r_{2},

(5)

J_{+} (q) = \int_{D} \int_{D} C_{V} (r_{1} - r_{2}) e^{- i q \cdot (r_{1} + r_{2})} d^{3} r_{1} d^{3} r_{2} .

(6)

Then

⟨ ∣ ϕ_{1} ∣^{2} ⟩ = ∣ a ∣^{2} J_{-}, ⟨ ϕ_{1}^{2} ⟩ = a^{2} J_{+},

(7)

and hence

I (r) = exp {∣ a (r) ∣^{2} J_{-} (q) + Re [a (r)^{2} J_{+} (q)]} .

(8)

This expression depends only on the two-point autocorrelation of the scattering potential. In the weak-fluctuation expansion it becomes

I = 1 + ∣ a ∣^{2} J_{-} + Re (a^{2} J_{+}) + O (C_{V}^{2}) .

(9)

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Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 335 / 2

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