Past exam of the mathematics course of the University of Cambridge / 2007 / iii / Paper 80 / 2

Use ${\hat{\psi }}_{\mathrm{sc}}(\nu )=\int {\psi }_{\mathrm{sc}}(x,0)e^{-i\nu x}dx$, inverse factor $1/(2\pi )$, and $k=\omega /c$. Fourier transform the homogeneous Helmholtz equation givesAn upward radiating field has no incoming component from $z=+\infty$. For $q(\nu )=\sqrt{k^2-{\nu }^2}$, take the nonnegative real root when $|\nu |<k$ and the positive imaginary root when $|\nu |>k$. Consequently the outgoing angular spectrum isThe propagating terms go upward; the evanescent terms decay with increasing $z$. Grazing values are understood by a radiation limit.

There is a reference-plane qualification in the question. For an arbitrary rough surface the full line $z=0$ need not lie in the medium: already $f(x)=h>0$ puts it in vacuum. In that case the stated physical trace is undefined without continuation. The exact representation is obtained instead on a plane $z=z_{\ast }>\sup f$,One can formally set ${\hat{\psi }}_{\mathrm{sc}}={\hat{\psi }}_{\ast }{e}^{-iq{z}_{\ast }}$, but convergence of that continued spectrum down to the rough boundary is an additional assumption. This is the reference-plane validity of a rough-surface angular spectrum. The first boxed formula is exact in an appropriate homogeneous upper half-space; it is not automatically an exact representation at every point of a completely arbitrary rough domain. Small-height perturbation theory in part (c) uses the mean plane order by order.

For a complex harmonic field use complex conjugation in its wave-field correlation: $m=⟨{\psi }_{\mathrm{sc}}(x_1,z){\psi }_{\mathrm{sc}}(x_2,z{)}^{\ast }⟩$. Put $A(\nu )={\hat{\psi }}_{\mathrm{sc}}(\nu )$ and let $C(\nu ,{\nu }^{\prime })=⟨A(\nu )A({\nu }^{\prime }{)}^{\ast }⟩$. Direct substitution of part (a), before imposing any homogeneity, givesThis is the general answer. It need not depend only on $\xi =x_1-x_2$: even a deterministic sum of two distinct outgoing plane waves has cross terms depending on the midpoint. The requested separation-only wave-field correlation therefore presumes translation-homogeneous two-point statistics.

Under that assumption define the uncentered spectrum by $C(\nu ,{\nu }^{\prime })=2\pi S(\nu )\delta (\nu -{\nu }^{\prime })$. Then the height propagation of an angular-spectrum correlation isEquivalently, for the spectral measure of a stationary random field $F(d\nu )=S(\nu )d\nu /(2\pi )$, write $m=\int e^{i\nu \xi -2\mathrm{Im}qz}F(d\nu )$. This version includes coherent spectral atoms and avoids squaring Dirac delta functions. The propagating spectrum $|\nu |\le k$ is independent of height, while an evanescent contribution is multiplied by $e^{-2\sqrt{{\nu }^2-k^2}z}$. At $\xi =0$ the formula gives the mean wave intensity.

If the brackets instead mean the spatial autocorrelation of a finite-energy deterministic field, define $m_L(\xi ,z)=\int {\psi }_{\mathrm{sc}}(x+\xi ,z){\psi }_{\mathrm{sc}}(x,z{)}^{\ast }dx$. Parseval's identity gives the same integral with $S(\nu )=|{\hat{\psi }}_{\mathrm{sc}}(\nu ){|}^2$. That finite-energy formula must not be applied unmodified to an infinite stationary field or an infinite plane wave, whose transforms require a spectral-measure convention.

Write the unit-amplitude incident plane wave as ${\psi }_i=e^{i{p}_{i}x-i{q}_{i}z}$, with $q_i=\sqrt{k^2-p_i^2}>0$. The flat Dirichlet reflected field is ${\psi }_s^{[0]}=-e^{i{p}_{i}x+i{q}_{i}z}$, so the flat total field vanishes at $z=0$ and has derivative ${\partial }_z{\psi }^{[0]}(x,0)=-2i{q}_i{e}^{i{p}_{i}x}$.

For a controlled height expansion take $f=\epsilon h$ with fixed regular bounded shape $h$. Taylor expansion at the moving boundary yieldsThus the first-order scattered trace is ${\psi }_s^{[1]}(x,0)=2i{q}_{i}f(x)e^{i{p}_{i}x}$, and its transform follows by the Translation property of the Fourier transform:The first formula is the roughness correction; the second includes the zeroth-order reflection. The small-height assumption controls the Taylor expansion for a fixed profile; arbitrarily fine surface scales can require additional derivative control, not merely a small height at fixed $k$.

If the mean roughness height is a constant $\bar{f}$, as for stationary roughness, then $⟨\hat{f}(\nu -p_i)⟩=2\pi \bar{f}\delta (\nu -p_i)$. Consequently the specular first-order mean reflection isIts tangential wavenumber is unchanged and its normal wavenumber is reversed relative to incidence, which is precisely specular reflection. For zero mean height the first-order mean correction is zero, but the full mean reflected field is not zero: it still contains the flat reflection.

Small height alone does not imply the statistical conclusion. For example, deterministic mean height $\bar{f}(x)=a\cos Kx$ with $|ka|\ll 1$ produces first-order mean sidebands at $p_i\pm K$ and is not purely specular. Constant mean is sufficient at first order; stationarity supplies it and, with suitable translation-invariant statistics, constrains higher-order coherent reflection as well.