Past exam of the mathematics course of the University of Cambridge / 2010 / iii / Paper 70 / 2

Write $\mathbf{r}=(y,z)$ and prescribe a deterministic entrance field $E_0(\mathbf{r})$. Substitution of the carrier into the Helmholtz equation givesThe usual weak-index paraxial model drops $E_{xx}$ and linearizes $n^2-1=2\mu W+O({\mu }^2)$, givingAll the mean-field formulas below use this model. If the ${\mu }^2{W}^2$ term in $n^2$ is retained, its extra mean and fluctuations must also be included; Gaussian averaging of the linear random potential alone is then insufficient.

For a colored medium, taking expectations gives the exact propagation identityStationarity and Gaussianity alone do not close the last term. To make its dependence precise, let $C(s,\mathbf{r})=⟨W(x+s,{\mathbf{r}}_0+\mathbf{r})W(x,{\mathbf{r}}_0)⟩$ and let $U_W(x,s;\mathbf{r},{\mathbf{r}}^{\prime })$ be the random paraxial propagator. Assuming a jointly Gaussian random field, rather than only Gaussian one-point marginals, the functional derivative of the solution isThe Furutsu–Novikov formula therefore yieldsThis is a propagation equation for the mean, with the remaining correlation displayed rather than silently replacing it by a product of means.

The general solution can be written explicitly as a Gaussian-averaged phase-screen product. Put $\Delta =x/N$, $x_j=j\Delta$, ${\mathbf{r}}_N=\mathbf{r}$ and letbe the integral kernel of the Fresnel propagator. Successive free-propagation and random-phase steps, followed by Gaussian averaging, give the colored Gaussian paraxial mean propagatorFor smooth colored media this is the time-slicing representation, with oscillatory integrals understood by the usual Fresnel regularization. At finite $N$ the exponential follows directly from the characteristic function of the jointly Gaussian phase sum. It exhibits why a general transverse covariance cannot be replaced by its value at zero separation along every diffraction path.

For completeness, the unlinearized paraxial model has $E_x=DE+iaWE+i(k{\mu }^2/2)W^{2}E$ and the corresponding mean equation includes $i(k{\mu }^2/2)⟨W^{2}E⟩$. Its colored-medium solution uses the same time-slicing integral, replacing the Gaussian factor byThis follows by completing the square in the finite-dimensional Gaussian integral, with the determinant branch continuous from $\mu =0$. It supplies the generic colored-medium mean when the $W^2$ term is retained. A square of ideal white noise is not defined by the covariance in part (b); that limit uses the linear random-potential model and its stated drift.

If transverse diffraction of the fluctuations is neglected, the simpler straight-ray approximation givesIndeed $E=E_0\exp [ia{\int }_0^{x}W(s,\mathbf{r})ds]$ and the variance of its phase is $2a^2{\int }_0^x(x-s)C(s,0)ds$. If $W$ is constant transversely, this formula is exact with $E_0$ replaced by the free diffracted field $e^{xD}{E}_0$. For example $W=Z$, a single standard normal variable constant throughout space, satisfies stationarity and unit variance but gives $m=e^{-a^2{x}^2/2}{e}^{xD}{E}_0$. Thus a constant-rate exponential attenuation is not a consequence of part (a)'s assumptions alone. It becomes the closed paraxial result under the longitudinal white-noise assumption in the next part.

Interpret the longitudinal delta covariance as a white-noise limit. It is a generalized random field, not a pointwise unit-variance function. Let $B_x(\mathbf{r})$ have independent longitudinal increments withThe smooth-medium limit gives the Stratonovich equation $dE=DEdx+iaE\circ dB$. Conversion to Itô form contributes the local variance drift:The Itô stochastic integral has zero mean, so coherent attenuation in a white-noise random medium satisfiesHere $U_x=e^{ix{\Delta }_{\perp }/(2k)}$ is the Fresnel propagator. Equivalently,For a constant entrance envelope this is simply its constant value times $e^{-\gamma x}$.

Put $B_n(r)={\mu }^{2}A(r)$, the transverse covariance coefficient of the refractive-index fluctuations. The spectrum required here is the transverse spectrum of that longitudinal covariance coefficient, or equivalently the longitudinal-zero-frequency slice of a prelimit stationary medium. Radial symmetry refers to the transverse plane; exact white noise in one direction is not full three-dimensional isotropy. With the Cartesian Fourier convention,angular integration gives $S_F(\nu )=2\pi {\int }_0^{\infty }{B}_n(r)J_0(\nu r)rdr$. Its inverse givesThe printed transform hint mixes two different normalizations. Its radial transform and inverse constitute the self-reciprocal Hankel pair, whose spectrum is $S_H=S_F/(2\pi )$, rather than the unnormalized Cartesian transform also displayed there. Using that Hankel convention consistently yields insteadThese are the same physical solution after converting spectra. The distinction is Fourier-Hankel normalization for an isotropic spectrum; $J_0(0)=1$ fixes the zero-separation covariance in either convention.

For the given Gaussian radial spectrum, the only needed integral isIf $S$ is interpreted with the self-reciprocal Hankel pair in the supplied hint, the coherent field isIn this convention $B_n(r)=2{\mu }^{2}L{e}^{-r^2/L^2}$ and hence $B_n(0)=2{\mu }^{2}L$, directly verifying the attenuation coefficient. If instead the stated spectrum is the unnormalized Cartesian Fourier transform, use its inverse normalization and obtainIts transverse covariance coefficient is then $B_n(r)={\mu }^{2}L{e}^{-r^2/L^2}/\pi$. The apparent difference is precisely the inconsistent Fourier/Hankel normalization in the printed hint, not a difference in stochastic propagation. For a unit constant entrance field the Fresnel factor is one; coherent intensity is the square of the displayed mean amplitude.