Gaussian intensity in the first Rytov approximation Created 2026-10-05 Updated 2026-10-06
Let with unit-modulus incidence and . For a centered real Gaussian random field , the exponential moment of a Gaussian linear functional givesHere is the autocorrelation function of a random field. Both the covariance and the generally nonzero term enter the exponent; dropping the latter is not justified for an arbitrary finite real random medium.
Past exam of the mathematics course of the University of Cambridge 2012 iii Paper 57 3 b Solution Created 2026-10-03 Updated 2026-10-07
For a statistically homogeneous isotropic dimensionless field, write its two-point autocorrelation function of a random field as andThe formal real-space scale invariance condition is for all . Changing variable in the first expression and equating the nonzero-mode Fourier transforms gives , equivalently . Taking yieldsThus a scale-invariant inflationary power spectrum has equal variance per logarithmic wavenumber interval, rather than equal power per Fourier volume.
There is an important mathematical qualification to the literal covariance condition. A nonzero power spectrum over all scales has , divergent at both endpoints; it is not the covariance of a finite-variance ordinary field. Moreover, exact dilation invariance of a continuous isotropic makes it constant for and, by continuity, at zero: its spectrum can then consist only of a zero-mode delta measure. The nontrivial cosmological statement consequently concerns nonzero modes, with cutoffs or subtraction of an unobservable constant. For example the finite subtracted covarianceis invariant under simultaneous rescaling of . This is the precise infrared qualification of a scale-invariant covariance; a regulated unsubtracted generally shifts by an additive constant under dilation. The formal derivation above gives the intended nonzero-mode scaling, with this qualification rather than an impossible finite-variance premise.
Past exam of the mathematics course of the University of Cambridge 2012 iii Paper 78 3 c Solution Created 2026-10-03 Updated 2026-10-14
The zero-mean fluctuation is , not the refractive index : . Also the full scattering potential has a nonzero mean and is not Gaussian:The usual Gaussian closure requires 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 , then account for the quadratic term.
Keep the definitions from the preceding solution and let . Because the incident field has unit modulus,The exponent is a real Gaussian linear functional . If , the linearized potential has autocorrelation function of a random field . Complete the square in the one-dimensional Gaussian density of : . 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 . The first Rytov approximation is linear in , but its intensity is an exponential average; applying a Gaussian formula exactly to this quadratic potential would be incorrect. A consistent expansion through order 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 through this order. Thus the mean shift must not be discarded while retaining the order- fluctuation intensity.
An exact expression within the far-field, first-Rytov approximation model is also available when the full is retained. Assume the restriction of to is a Gaussian element of the real Hilbert space , with covariance operator of kernel . Let denote multiplication by , , and . The Gaussian quadratic exponential moment isHere the determinant is the Fredholm determinant; require strictly positive for finiteness. Diagonalize the self-adjoint trace-class operator and complete the square in each Gaussian coordinate to derive the formula. On a bounded , unit pointwise variance makes , so the covariance is trace class under the stated measurable hypothesis. Small enough satisfies the positivity condition. Expanding the determinant and inverse recovers the preceding mean and covariance terms. For small this is also determined by the potential autocorrelation: since ,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 and independent random signs . The stationary sequence has Gaussian marginals and covariance , just like independent standard Gaussian variables. However , whereas for the independent Gaussian pair it is . The different intensities show why the joint hypothesis is needed. The mean-zero hint is interpreted for , and the linearized Gaussian formula is not claimed to be exact for .
Past exam of the mathematics course of the University of Cambridge 2014 iii Paper 72 1 c ii Solution Created 2026-10-03 Updated 2026-10-06
Let the stationary scattering potential fluctuation have autocorrelation function of a random fieldBecause is centered, this is also its covariance function. Multiplying the two real wave phase integrals and taking expectations givesThe two minus signs cancel. This is the wave phase variance, since the wave phase mean is zero. More generally the phase covariance in the first Rytov approximation replaces the first kernel by and the second by . Finite observation/scattering windows, or suitable weighted-integrability hypotheses, make these double integrals well-defined in a stationary infinite-medium model.
The correlation needed here is that of the scattering potential fluctuation. With the printed , the covariance of a squared random field isIt involves a fourth moment of , so its value is not generally determined by the ordinary two-point correlation alone. If one additionally assumes a zero-mean Gaussian random field, Isserlis theorem yields . That assumption is not printed and must not be inserted silently. Alternatively, for a physical weak fluctuation , gives . The general answer uses ; either reduction to a refractive index two-point correlation requires an extra assumption.
Past exam of the mathematics course of the University of Cambridge 2015 iii Paper 76 1 d Solution Created 2026-10-03 Updated 2026-10-06
Write and . Substituting into part (a), without prematurely replacing by its linear part, givesTaking the expectation yields the exact mean equation within the paraxial approximation:The correlation term cannot be discarded just because : the field depends on the same random medium as . Nor can be factored exactly from the given variance alone. The printed statistics do not specify the autocorrelation function of a random field, so they do not determine a unique local closed equation for .
A useful covariance-dependent equation follows to second order in . Defineand letbe the Fresnel propagator kernel for one transverse coordinate. The Duhamel principle applied to the linear fluctuation term giveswhere . Multiply by and average. Also . Since , replacing by in terms already multiplied by preserves second-order accuracy. We obtain the weak-fluctuation memory equation for the coherent fieldThis perturbation statement is for fixed propagation intervals with sufficient covariance regularity and moment bounds. The kernel is understood as an oscillatory integral. If the field is jointly Gaussian, odd perturbative contributions vanish and the remainder can be improved under the corresponding expansion hypotheses. Joint Gaussianity itself still leaves the covariance unspecified. The term is the quadratic refractive-index shift in the coherent field; it comes from the mean of .
For the usual local answer, impose the additional Markov approximation for a random medium. Its longitudinal correlation length must be much shorter than the distance on which the envelope varies, and diffraction across that correlation distance must not resolve substantial transverse covariance variation. Away from the entry layer, replace the slowly varying factors in the memory integral by their local values and extend the longitudinal integral to infinity. Define the integrated covarianceFor an integrable real stationary covariance, is even, and is its zero-longitudinal-frequency spectral density. The conventional leading weak-index model keeps only the linear random potential ; in that model the local first-moment equation isThe factor can be checked independently in the white-noise model. If is a Brownian field with , writeConverting the Stratonovich integral to an Itô integral adds . The remaining stochastic integral has zero mean, giving the boxed equation andIf the finite-correlation model retains consistently through order , the local second-order expansion also has , adding the phase factor . This phase is computed before the white-noise idealization: squaring ideal Gaussian white noise is not the finite-variance operation .
The need for a correlation assumption can be seen without any closure argument. The stationary Gaussian random field , with a single standard normal variable , has all the printed one-point statistics. In the linear weak-index model it givesIts attenuation is quadratic in propagation distance, whereas the Markov model gives linear-distance exponential attenuation. The local attenuation equation requires the additional correlation/Markov assumption; it does not follow from statistical homogeneity and unit variance alone.
Past exam of the mathematics course of the University of Cambridge 2015 iii Paper 76 2 d Solution Created 2026-10-03 Updated 2026-10-06
Let the medium's autocorrelation function of a random field and power spectrum use the explicit conventionSince , only the linear term in is needed to obtain the cross-section through second order:Squaring and taking the expectation givesThis is the finite-volume formula in the Born approximation for scalar wave scattering; the zero mean of does not make its squared scattered amplitude vanish. For a jointly centered Gaussian random field, the third-order contribution is zero, so the next perturbative contribution is fourth order when that expansion is valid.
Introduce the overlap volume . Changing variables rewrites the double integral asWhen the linear dimensions of a regular scattering volume are large compared with the correlation length, and is integrable, over the significant covariance range. This gives the finite-volume random scattering spectrum resultThere are two distinct approximations here: the weak-scattering expansion in and the negligible-boundary-overlap approximation in the volume-to-correlation-scale ratio. If instead denotes the power spectrum of the index fluctuation , then and the external factor is absorbed into .
Statistical homogeneity alone allows an anisotropic spectrum, so its argument is generally the vector . If statistical isotropy is additionally assumed, depends only onThe scalar notation is valid in that isotropic case. For the alternative convention , the bulk formula is . Changing the sign of the scattering vector does not change the spectrum of a real stationary scalar field; changing the Fourier transform normalization does change the displayed factors. Both conventions have therefore been stated explicitly.
Past exam of the mathematics course of the University of Cambridge 2019 iii Paper 335 2 iii Solution Created 2026-10-03 Updated 2026-10-05
Use the leading far-field kernel from the previous part. DefineThen and, because the incident intensity is one,To first order in weak contrast, . Its linearized part is a centered stationary Gaussian random field, with autocorrelation function of a random field . The exponential moment of a Gaussian linear functional therefore gives the usual Gaussian intensity in the first Rytov approximation:Equivalently the exponent is . The second term generally matters because a finite real random medium need not generate a circular complex Gaussian amplitude.
If the full stated potential is retained, it is not Gaussian and has mean . To retain every term through order in the intensity of this first Rytov field, the cumulant expansion instead givesHere ; the raw autocorrelation differs by . The stated remainder uses Gaussian finite-dimensional statistics of , for which odd total moments vanish. This mean term is absent if one first truncates the potential to . Both formulas concern the first Rytov approximation, and exponentiating it does not supply the omitted higher-order scattering corrections.
Past exam of the mathematics course of the University of Cambridge 2023 iii Paper 335 2 ii Solution 2026-09-28
Assume that 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 . Define the integrated longitudinal autocorrelation function of a random fieldThenwhereWriting 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.
Past exam of the mathematics course of the University of Cambridge 2024 iii Paper 335 2 iii Solution Created 2026-09-28 Updated 2026-10-03
Since with small variance,To leading order, is therefore a centered stationary Gaussian random field. Let its autocorrelation function of a random field beThe contribution gives higher-order mean and non-Gaussian corrections and is consistently omitted at this order.
The far-field Rytov approximation from part ii has unit incident intensity andThe real random variable is centered Gaussian. Its moment-generating function givesDefineThenand henceThis expression depends only on the two-point autocorrelation of the scattering potential. In the weak-fluctuation expansion it becomes
For a parabolic wave equation with , a zero-mean stationary real field of unit variance, and , finite-correlation perturbation theory givesHere is the coherent field, is the autocorrelation function of a random field and is the one-coordinate Fresnel propagator. Expand the random solution once using the Duhamel principle, multiply by and average to obtain the memory term. The local phase term comes from the term in . The expansion is for fixed propagation distances with suitable covariance regularity and moment bounds. A Markov approximation is an additional scale assumption that can turn this integral equation into a local attenuation equation.