Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 335 / 2

Let $V=k_0^2(n^2-1)$ be the scattering potential supported in $D$. The background Helmholtz equation and its outgoing kernel give the scalar Lippmann-Schwinger equationReplacing the unknown field in the integral by the incident field gives the Born approximation for scalar wave scattering:For the Rytov approximation, work where ${\psi }_i\ne 0$ and set $\psi ={\psi }_i{e}^{\varphi }$. Dividing the wave equation by the total field and subtracting the incident equation givesNeglect the quadratic gradient term. If ${\varphi }_1={\psi }_{s,B}/{\psi }_i$, direct differentiation shows $\Delta {\varphi }_1+2\nabla \log {\psi }_i\cdot \nabla {\varphi }_1=-V$, soExpanding the exponential in the potential gives ${\psi }_R={\psi }_i+{\psi }_{s,B}+O(V^2)={\psi }_B+O(V^2)$. Thus both approximations have the same first-order field, although exponentiation retains a particular family of higher powers.

For unit plane wave incidence ${\psi }_i=e^{i{k}_0{\hat{\mathbf{r}}}_0\cdot \mathbf{r}}$, define the scattering wavevector difference $\mathbf{q}=k_0({\hat{\mathbf{r}}}_0-\hat{\mathbf{r}})$ andThe outgoing Green function for the three-dimensional Helmholtz equation givesThe Born far-field pattern is $A$, a Fourier transform sample of the potential. Expanding the far-field Rytov exponential in $1/r$ gives the same leading outgoing term, while the logarithmic representation preserves its accumulated phase and amplitude correction.

Use the leading far-field kernel from the previous part. DefineThen ${\varphi }_1={\int }_{D}KV$ and, because the incident intensity is one,To first order in weak contrast, $V=2\mu k_0^{2}W+O({\mu }^2)$. Its linearized part is a centered stationary Gaussian random field, with autocorrelation function of a random field $C_V(\mathbf{s})=4{\mu }^2{k}_0^4{C}_W(\mathbf{s})$. The exponential moment of a Gaussian linear functional therefore gives the usual Gaussian intensity in the first Rytov approximation:Equivalently the exponent is $\mathbb{E}|{\varphi }_1{|}^2+\mathrm{Re}\mathbb{E}{\varphi }_1^2$. The second term generally matters because a finite real random medium need not generate a circular complex Gaussian amplitude.

If the full stated potential $V=k_0^2(2\mu W+{\mu }^2{W}^2)$ is retained, it is not Gaussian and has mean $m_V={\mu }^2{k}_0^2$. To retain every term through order ${\mu }^2$ in the intensity of this first Rytov field, the cumulant expansion instead givesHere $C_V^{\mathrm{cov}}=\mathbb{E}[(V-m_V)(V^{\prime }-m_V)]$; the raw autocorrelation differs by $m_V^2=O({\mu }^4)$. The stated remainder uses Gaussian finite-dimensional statistics of $W$, for which odd total moments vanish. This mean term is absent if one first truncates the potential to $2\mu k_0^{2}W$. Both formulas concern the first Rytov approximation, and exponentiating it does not supply the omitted higher-order scattering corrections.