Past exam of the mathematics course of the University of Cambridge / 2012 / iii / Paper 78 / 3

Use the scattering potential convention $V=k_0^2(n^2-1)$. With time dependence $e^{-i\omega t}$ the outgoing Green function is $G_0(r,r^{\prime })=e^{i{k}_0|r-r^{\prime }|}/(4\pi |r-r^{\prime }|)$ and satisfies $(\Delta +k_0^2)G_0=-\delta$. The total-field Lippmann-Schwinger equation isReplacing the field in the integral by the incident field gives the first Born approximation for scalar wave scattering:For the first Rytov approximation, write $\psi ={\psi }_0{e}^{\chi }$ on a region where the incident field is nonzero and a continuous complex logarithm can be chosen. The logarithmic wave perturbation equation isThe last dot product is bilinear, not a squared modulus. Dropping it leaves a linear equation for ${\chi }_1$. Multiplication by ${\psi }_0$ shows that ${\psi }_0{\chi }_1$ satisfies the same outgoing inhomogeneous Helmholtz equation as ${\psi }_1$. HenceIntroduce a small multiplier $\lambda$ on $V$. Then ${\psi }_1=O(\lambda )$ andThe two total-field approximations agree to first order. Exponentiating the first logarithmic wave perturbation does not supply the generally missing second-order Born series contribution.

Write the observation point as $r\hat{r}$, put $q=k_0(\hat{r}-{\hat{r}}_0)$, and take ${\psi }_0(r\hat{r})=e^{i{k}_{0}r{\hat{r}}_0\cdot \hat{r}}$. For a bounded medium, the far-field approximation for an outgoing source usesFor finite-distance accuracy require $r\gg \ell$ and $k_0{\ell }^2/r\ll 1$ as well as being in the radiation region. Define the Fourier transform sample $S(q)={\int }_{D}V(r^{\prime })e^{-iq\cdot r^{\prime }}{d}^3{r}^{\prime }$ and the known factorThen the required total Born approximation for scalar wave scattering and total Rytov approximation areThe first Born approximation far-field pattern is $S(q)/(4\pi )$. In both formulas the incident field is retained; an outgoing scattered term alone would not be a total-field answer.

The zero-mean fluctuation is $W$, not the refractive index $n$: $⟨n⟩=1$. Also the full scattering potential has a nonzero mean and is not Gaussian:The usual Gaussian closure requires $W$ to be a jointly Gaussian random field, not merely to have a Gaussian distribution at each point. We first give the intended leading weak-fluctuation result with the linearized potential $V_1=2\mu k_0^{2}W$, then account for the quadratic term.

Keep the definitions $q,B$ from the preceding solution and let ${\chi }_1=B{\int }_D{V}_1(y)e^{-iq\cdot y}dy$. Because the incident field has unit modulus,The exponent is a real Gaussian linear functional $Z={\int }_{D}h{V}_1$. If $C_W(y-z)=⟨W(y)W(z)⟩$, the linearized potential has autocorrelation function of a random field $C_1(s)=4{\mu }^2{k}_0^4{C}_W(s)$. Complete the square in the one-dimensional Gaussian density of $Z$: $⟨e^Z⟩=\exp (⟨Z⟩+\mathrm{Var}Z/2)$. Here its mean is zero, soThis is Gaussian intensity in the first Rytov approximation. In a more explicit complex notation, putThenThe second term is the pseudo-covariance of a complex random variable contribution. It does not vanish for an arbitrary finite real medium. The real-kernel expression shows that the combined exponent is nonnegative, even though its two complex-form contributions need not be separately nonnegative.

For the full potential, Gaussian moment pairing theorem gives its centered covariance functionIts uncentered autocorrelation is $R_V(s)=C_V(s)+{\mu }^4{k}_0^4$. The first Rytov approximation is linear in $V$, but its intensity is an exponential average; applying a Gaussian formula exactly to this quadratic potential would be incorrect. A consistent expansion through order ${\mu }^2$ isThe first term is the coherent mean-potential correction. Odd powers vanish by the symmetry of the joint Gaussian field. The variance term can equally use $C_V=R_V-{\overline{V}}^2$ through this order. Thus the mean shift must not be discarded while retaining the order-${\mu }^2$ fluctuation intensity.

An exact expression within the far-field, first-Rytov approximation model is also available when the full $n^2$ is retained. Assume the restriction of $W$ to $D$ is a Gaussian element of the real Hilbert space $L^2(D)$, with covariance operator $C$ of kernel $C_W(y-z)$. Let $H$ denote multiplication by $h$, ${\ell }_h=2\mu k_0^{2}h$, $Q={\mu }^2{k}_0^2{C}^{1/2}H{C}^{1/2}$ and $b_h=C^{1/2}{\ell }_h$. The Gaussian quadratic exponential moment isHere the determinant is the Fredholm determinant; require $I-2Q$ strictly positive for finiteness. Diagonalize the self-adjoint trace-class operator $Q$ and complete the square in each Gaussian coordinate to derive the formula. On a bounded $D$, unit pointwise variance makes $\mathrm{tr}C=|D|$, so the covariance is trace class under the stated measurable $L^2$ hypothesis. Small enough $\mu$ satisfies the positivity condition. Expanding the determinant and inverse recovers the preceding mean and covariance terms. For small $\mu$ this is also determined by the potential autocorrelation: since $C_W\in [-1,1]$,using the branch near zero. This makes explicit the extra joint-Gaussian assumption behind an exact correlation-based answer.

Finally, Gaussian marginals alone are insufficient. Take $X\sim N(0,1)$ and independent random signs $S_j$. The stationary sequence $W_j=S_{j}X$ has Gaussian marginals and covariance $⟨W_i{W}_j⟩={\delta }_{ij}$, just like independent standard Gaussian variables. However $⟨e^{t(W_1+W_2)}⟩=(1+e^{2t^2})/2$, whereas for the independent Gaussian pair it is $e^{t^2}$. The different intensities show why the joint hypothesis is needed. The mean-zero hint is interpreted for $W$, and the linearized Gaussian formula is not claimed to be exact for $V=k_0^2(n^2-1)$.