Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 335 / 2

For a slowly varying envelope $E=\psi e^{-ikx}$, the paraxial approximation to the Helmholtz equation isBecause $n=1+\mu W$ and ${\mu }^2\ll 1$, passage through a sufficiently thin phase screen producesIts modulus is one at the screen exit. Beyond the screen, $n=1$, so the parabolic wave equation is $E_x=i{E}_{zz}/(2k)$. A Taylor expansion in propagation distance givesSincewe findIt follows thator, equivalently, $|E{|}^2=1-x{\varphi }^{\prime \prime }/k+O(x^2)$. Thus random phase curvature produces local focusing and defocusing: free-space diffraction converts phase fluctuations into amplitude fluctuations immediately after the screen.

To first order in the weak fluctuation, $n^2-1=2\mu W+O({\mu }^2)$, soForintroduce the signs $s=(1,-1,1,-1)$. The product rule gives

Assume that $W$ is a zero-mean stationary Gaussian random field, that its longitudinal correlation length is short compared with the envelope's evolution scale, and that the propagation distance is long compared with that correlation length. The forward Markov approximation then neglects diffraction during one correlation length. Applying the Furutsu–Novikov formula closes the last average at second order in $\mu$. Define the integrated longitudinal autocorrelation function of a random fieldThenwhereWriting $C_{ij}=C(z_i-z_j)$ and using the evenness of the covariance gives the equivalent expressionThe derivation also assumes paraxial propagation, weak scattering, sufficient regularity to interchange differentiation and expectation, and statistical homogeneity in both coordinates. Without the short-correlation approximation, the Gaussian identity produces a nonlocal longitudinal memory integral rather than this local closed equation.

Substitute $m_4=e^{\Psi }$ into the given equation. The chain rule givesThe incident reduced field is the constant one, so $m_4(-\Delta x)=1$ and $\Psi (-\Delta x)=0$. During a thin-screen crossing, $\Psi =O(\Delta x)$. Its mixed-derivative contribution integrates to $O(\Delta x)$. More precisely, its contribution to $\Psi$ is $O((\Delta x{)}^2)$, while the product of its first derivatives contributes only at still higher order. Thereforeand hence

In free space $Q=0$, and the fourth moment obeysFor the two-dimensional Fourier transformthe Fourier transform of a derivative turns this equation into the ordinary differential equationThusUsing the screen-exit value from part (a), the inverse transform givesthrough first order in $\Delta x$. Equivalently, the free-space fourth-moment propagator has kernel