Polarimetry of homogeneous half-spaces 1

Transcription

1 Polarimetry of homogeneous half-spaces 1 M. Gilman, E. Smith, S. Tsynkov special thanks to: H. Hong Department of Mathematics North Carolina State University, Raleigh, NC Workshop on Symbolic-Numeric Methods for Differential Equations and Applications Courant Institute of Mathematical Sciences, New York, July 20, Work supported by AFOSR, grants # FA , FA , and FA M. Gilman (Mathematics, NCSU) Polarimetry of homogeneous half-spaces NYU, July 20, / 34

2 Overview Preliminaries Overview Scattering by homogeneous half-spaces: extension of a textbook case: Fresnel reflection coefficients; inverse problem: material properties from scattered waves; quantifier elimination for polarimetry with uniaxial dielectric tensor. Inhomogeneous half-spaces: vertical stratification, Leontovich boundary conditions, rough surface scattering. M. Gilman (Mathematics, NCSU) Polarimetry of homogeneous half-spaces NYU, July 20, / 34

3 Preliminaries Motivation Motivation: radar imaging Synthetic aperture radar (SAR) is an established signal processing technology that retrieves the averaged reflectivity image-detail?img=17281#.WyuwpKknbUI TerraSAR-X (spaceborne) SAR scattering model assumes independent point scatterers: acceptable for detection and tracking; not helpful for reconstruction of scatterer properties (e.g., aboveground biomass, significant wave height). M. Gilman (Mathematics, NCSU) Polarimetry of homogeneous half-spaces NYU, July 20, / 34

4 Preliminaries Physical preliminaries Electromagnetic waves Electric E and magnetic H fields obey the Maxwell s equations in material: 1 H + curl E = 0, div H = 0, c t 1 D curl H = 4π c t c j, div D = 0. The material is characterized by dielectric ε and conductivity σ tensors: D = ε E, j = σ E. In isotropic dielectric with losses: ε = εi, σ = σi D = εe, j = σe. In vacuum: ε = 1, σ = 0 D = E, j = 0. M. Gilman (Mathematics, NCSU) Polarimetry of homogeneous half-spaces NYU, July 20, / 34

5 Transverse waves Preliminaries Electromagnetic waves Homogeneous isotropic perfect dielectric is described by ε = εi, where ε is piecewise constant and σ = 0 ; hence 1 H c t + curl E = 0, yield, away from discontinuities of ε(r): 1 (εe) curl H = 0, div(εe) = 0 c t E = ε 2 c 2 t 2 E Plane waves: solutions to the Maxwell s equations of the form E(r, t) e i(k,r) iωt where εω = k c, H = c ω k E (for linear problems, take either real or imaginary part) Transversality: div E = div H = 0 (k, E) = (k, H ) = 0. M. Gilman (Mathematics, NCSU) Polarimetry of homogeneous half-spaces NYU, July 20, / 34

6 Linear polarizations Preliminaries Refraction-reflection problem Plane waves in vacuum: solutions to the Maxwell s equations of the form (for linear problems, take either real or imaginary part): E(r, t) e i(k,r) iωt, ω = kc, k = k, H = k E/k with E k, H k, E H, E = H. Planar interface: is characterized by the normal n. If we have a planar interface and k n, we can define the incidence plane and two linear polarizations: Horizontal polarization (or E-polarization) E(r, t) = E A e i(k,r) iωt, Vertical polarization (or H-polarization) H (r, t) = H A e i(k,r) iωt, E A = E 0 n (k/k), H (r, t) = k E(r, t)/k. vector amplitude incidence plane H A = H 0 n (k/k), E(r, t) = k H (r, t)/k. vector amplitude incidence plane M. Gilman (Mathematics, NCSU) Polarimetry of homogeneous half-spaces NYU, July 20, / 34

7 Polarimetry Preliminaries Refraction-reflection problem Any plane wave can be represented as a linear combination of two waves in basic linear polarizations (with a possible phase shift). We can do so for the incident and reflected waves. The reflecting properties can be fully described by a 2 2 matrix relating amplitudes of scattered and incident waves: S HV S = [ S HH ], such that Er A = S E A S VH S VV If we control E A i and measure E A r then we can determine S. Entries of S may be complex max. 8 measurements (only 7 if the absolute phase is unavailable). This is how many scatterer parameters we potentially can reconstruct. i. M. Gilman (Mathematics, NCSU) Polarimetry of homogeneous half-spaces NYU, July 20, / 34

8 Homogeneous half-spaces Isotropic material. Direct problem Perfect isotropic dielectric: D = εe, j = 0 The simplest setup to test the methodology: the field in each half-space is described by a scalar equation (ε(z) 2 t 2 )U = 0, ε(z) = 1, z > 0, ε, z < 0, (*) where U is any field component. Born approximation (i.e., perturbations, or weak scatterer): δ 1, U = U (0) + U (1), ε = 1 + (ε 1), U (1) δ U (0), ε 1 δ. The incident field is U (0) ; it satisfies (*) everywhere with ε = 1, so (ε(z) 2 t 2 )U(1) = (ε(z) 1) 2 t 2 U(0). Linearization: throw away (ε 1)U (1) δ 2. M. Gilman (Mathematics, NCSU) Polarimetry of homogeneous half-spaces NYU, July 20, / 34

9 Homogeneous half-spaces Solution in two domains Isotropic material. Direct problem Linearized equation: ( 2 t 2 )U(1) = (ε(z) 1) 2 t 2 U(0). For r = (x, y, z) we take U (0) = u (0)A e i(kir ωt), where k i = (K i, 0, q i ). The interface is (z = 0). Fourier representation: U(x, y, z, t) = û(z)e ikix iωt, where U is any field component; so the incident field is û (0) (z) = u (0)A e iqiz ; scattered field: ( d2 dz 2 + q2 i )û(1) (z) = θ( z)(ε 1)k 2 û (0) (z) = θ( z)(ε 1)k 2 u (0)A e iq iz, where θ( ) is a step function. For z > 0, û (1) (z) is also a plane wave; for z < 0, û (1) (z) is a forced oscillation; note the resonance. General solution without RHS + particular solution with RHS : û (1) u (0)A Be iqiz, z > 0 (B is the scattering coefficient), = u (0)A (Az + C)e iqiz, z < 0, A = ik2 (ε 1) q. i M. Gilman (Mathematics, NCSU) Polarimetry of homogeneous half-spaces NYU, July 20, / 34

10 Homogeneous half-spaces Isotropic material. Direct problem Origin of the linearly growing term The linearly growing term Aze iqiz is unphysical. This term is an artifact of the Born approximation, in particular ignoring the change of the vertical wavenumber: k = ε ω c, q(exact) t = ε ω2 c 2 K2 i, ω q(born) t = q i = 2 c 2 K2 i. For z 0, the difference between the two periodic solutions is e iq(exact) t z e iq(born) z = e iq(born) z (e i(q(exact) t t e iq(born) t z ( i(q (exact) t t q (Born) t )z 1) q (Born) t )z) e iq(born) t z ( i) 1 (ω/c) 2 (ε 1) z. 2 q i A Hence, the transmitted part of the Born solution is valid in the vicinity of the interface (important for interface conditions). M. Gilman (Mathematics, NCSU) Polarimetry of homogeneous half-spaces NYU, July 20, / 34

11 Homogeneous half-spaces Interface conditions Isotropic material. Direct problem Needed where div D = div(ε E) does not exist. Relate the free-space (F) and material (M) sides of the interface E x,y (F) = E x,y (M), H x,y (F) = H x,y (M). For plane waves U = Û(z)e ik ix iωt, the interface conditions reduce to two conditions for a single variable, typically the component that is normal to the incidence plane (xz): Horizontal polarization, E = (0, E y, 0), H = (H x, 0, H z ): Ê y (F) = Ê y (M), dê y dz = dê y dz. (F) (M) Vertical polarization, H = (0, H y, 0), E = (E x, 0, E z ): dĥ y dz = (ε 1 dĥ y dz ), Ĥ y (F) = Ĥ y (M). (F) (M) M. Gilman (Mathematics, NCSU) Polarimetry of homogeneous half-spaces NYU, July 20, / 34

12 Homogeneous half-spaces Reflection coefficients Isotropic material. Direct problem We linearize the Interface conditions and use universal notation û for Ê y and Ĥ y according to the polarization: both polarizations: û (1) (F) = û (1) (M), horizontal: vertical, linearized: dû (1) dz (F) dû (1) dz (F) = dû(1) dz, (M) = dû(1) dz (ε 1) dû(0) (M) dz z=0 From û (1) u (0)A Be iqiz, z > 0, = u (0)A ( ik2 (ε 1) q z + C)e iqiz, z < 0, i we obtain: horizontal: B S HH = (ε 1) k2, 4q 2 i vertical: B S VV = (ε 1)( 1 2 k2 ) = S 4q 2 HH + ε 1 2. i Both expressions coincide with the Fresnel formulas as ε 1 0. M. Gilman (Mathematics, NCSU) Polarimetry of homogeneous half-spaces NYU, July 20, / 34

13 Homogeneous half-spaces Isotropic material Summary for isotropic dielectric Test case of the methodology. Direct problem: Real-valued reflection coefficients. Obtained correct asymptotics for reflection coefficients. Single scatterer parameter: (ε 1). No cross-polarized scattering: can satisfy the differential equations and interface conditions within a single linear polarization. Inverse problem: Inversion is straightforward: (ε 1) = 4q2 i k 2 S HH. 2 out of 4 entries of the matrix S. The data has only 1 out of 8 degrees of freedom because S VV = Q = K2 i q 2 i S HH k 2 does not depend on the properties of the scatterer. M. Gilman (Mathematics, NCSU) Polarimetry of homogeneous half-spaces NYU, July 20, / 34

14 Homogeneous half-spaces Isotropic lossy dielectric Isotropic material Finite isotropic conductivity: j = σe. For E = Ẽe iωt, and similarly for H and D: curl H = ik D + 4π c σẽ = ik(ε + i4π ω σ)ẽ = ikε Ẽ. Use ε instead of ε for the case of perfect dielectric. Need σ/ω 1 to use in Born approximation. Direct problem: two parameters of the scatterer: (ε 1) and σ/ω; reflection coefficients are complex; otherwise similar to the perfect dielectric. Inversion: (ε 1) = 4q2 i k 2 S HH (similar to the case of perfect dielectric). Note that S VV /S HH is independent of the material properties need the absolute phase to detect σ. M. Gilman (Mathematics, NCSU) Polarimetry of homogeneous half-spaces NYU, July 20, / 34

15 Dielectric tensor Homogeneous half-spaces Anisotropic material Anisotropic material: D i = ε ij E j, ε εi. From energy considerations, ε is symmetric. For a real symmetric matrix ε, choose (x, y, z ) such that D i = diag(ε x, ε y, ε z ) ij E j. Uniaxial model: two (rather than three) independent parameters (i.e., matrix eigenvalues): ε x = ε y = ε ; ε z = ε ; e z is the optical axis. We still require ε 1 1 and ε 1 1. Let e z = (α, β, γ), α 2 + β 2 + γ 2 = 1, and ζ = ε ε, then ε xx = ε + α 2 ζ, ε yy = ε + β 2 ζ, ε zz = ε + γ 2 ζ, ε xy = ε yx = αβζ, ε xz = ε zx = αγζ, ε yz = ε zy = βγζ. For η = ε 1, we have, up to the first order: η xx = 1/ε xx, η yy = 1/ε yy, η zz = 1/ε zz, η xy = η yx = ε xy, η xz = η zx = ε xz, η yz = η zy = ε yz. M. Gilman (Mathematics, NCSU) Polarimetry of homogeneous half-spaces NYU, July 20, / 34

16 Homogeneous half-spaces Governing equations Anisotropic material. Direct problem We can reduce the Maxwell s equations dê y dz = ikĥ x, dĥ y dz = ik ˆD x, dê x dz ik iêz = ikĥ y, dĥ x dz ik iĥz = ik ˆD y, ik i Ê y = ikĥ z, ik i Ĥ y = ik ˆD z, to the following system for Ê y and Ĥ y : (η yy d 2 dz 2 + k2 K 2 i η yy)ê y = k(iη xy d dz + K iη yz )Ĥ y, (η xx d 2 dz 2 + k2 K 2 i η zz)ĥ y 2iη xz K i dĥ y dz = i k (iη yzk i η xy d dz )(d2 Ê y dz 2 K2 i Ê y ), and substitute Ê x and Ĥ x expressed via Ê y and Ĥ y into Ê x,y (F) = Ê x,y (M), Ĥ x,y (F) = Ĥ x,y (M). M. Gilman (Mathematics, NCSU) Polarimetry of homogeneous half-spaces NYU, July 20, / 34

17 Homogeneous half-spaces Anisotropic material. Direct problem Calculating scattering coefficients In Born approximation, the system of two equations decouples in a different way for different polarizations of the incident field. For example, when Êi = (0, E (0)A e iq iz, 0), then Ĥ y Ĥ (1) E (0)A, and in (η yy d 2 dz 2 + k2 K 2 i η yy)ê y = k(iη xy d dz + K iη yz )Ĥ y, we drop RHS δ 2 and then linearize LHS to obtain dz 2 + q2 i )Ê(1) y (z) = θ( z)(ε 1)k 2 E (0)A e iq iz + ICs S HH, ( d2 whereas for Ĥi = (0, H (0)A e iq iz, 0) and Ê y Ê (1) H (0)A we linearize both sides: ( d2 dz 2 + d q2 i )Ê(1) y = k(iη xy dz + K iη yz )Ĥ y (0) + ICs S HV. M. Gilman (Mathematics, NCSU) Polarimetry of homogeneous half-spaces NYU, July 20, / 34

18 Homogeneous half-spaces Direct and inverse problems Anisotropic material Direct (α, β, γ, ξ, ζ) S and inverse S (α, β, γ, ξ, ζ) problems: S HH = 1 4 k 2 q 2 i (ξ + β 2 ζ), S HV = 1 k (α + K i γ)ζβ, 4 q i q i S VV = 1 4 ((ξ + α2 ζ) K2 i q 2 (ξ + γ 2 ζ)), S VH = 1 4 i 1 = α 2 + β 2 + γ 2 (components of e z ), k (α K i γ)ζβ, q i q i where k q i and K i q i parameters, k 2 = K 2 i + q 2 i ; ξ = ε 1, ζ = ε ε. Inverse problem: 4 polynomials of 3rd degree and 1 polynomial of 2nd degree. Degrees of freedom vs. orientation of optical axis (S ij R): 2 if β = 0 because S HV = S VH = 0; 3 if (α = 0 and βγ 0) OR (γ = 0 and αβ 0) because S HV ± S VH = 0; 4 if αβγ 0. M. Gilman (Mathematics, NCSU) Polarimetry of homogeneous half-spaces NYU, July 20, / 34

19 Homogeneous half-spaces Anisotropic material Solvability of the inverse problem Proposition: The system of equations can be solved with respect to ε, ε, α, β, and γ for the given S HH, S VV, S HV, S VH, and θ inc if and only if (S VV + VS HH ) 2 4WS HV S VH, (**) where W = q2 i K 2 i k 2 = cos 2 θ inc sin 2 θ inc = cos 2θ inc. If α = 0 then S HV = S VH, and for θ inc > π/4 we have W < 0 RHS < 0 in (**) the problem always has a solution. So, (**) puts an additional constraint on the values of the reflection coefficients and thus implies a limitation of solvability of the linearized inverse problem. Derivation of (**) has taken several weeks... M. Gilman (Mathematics, NCSU) Polarimetry of homogeneous half-spaces NYU, July 20, / 34

20 Homogeneous half-spaces Symbolic computations Quantifier elimination method rules! Hoon Hong to me Dear Mikhail, I applied a general method and obtained the following result (in about 3 seconds):... The following system has a real solution for a,b,g,z,x 0 = -s1-1/4*k^2*(x+b^2*z), 0 = -s2 + 1/4*((x+a^2*z)-K^2*(x+g^2*z)), 0 = -s3-1/4*k*(a-k*g)*z*b, 0 = -s4 + 1/4*k*(a+K*g)*z*b, 0 = -1 + a^2 + b^2 + g^2, 0 = - k^2 + K^2 + 1 iff K^4*s1^2-2*K^4*s1*s2+K^4*s2^2+4*K^4*s3*s4-2*K^2*s1^2 +2*K^2*s2^2+s1^2+2*s1*s2+s2^2-4*s3*s4 > 0 M. Gilman (Mathematics, NCSU) Polarimetry of homogeneous half-spaces NYU, July 20, / 34

21 Homogeneous half-spaces Anisotropic material Summary for anisotropic dielectric Direct problem: Real-valued reflection coefficients. Multiple scatterer parameters. Can produce cross-polarized scattering. Inverse problem: All 4 entries of the matrix S. The data has 2 to 4 (out of 8) degrees of freedom. Special condition of solvability (because the material parameters must be real). If we allow uniaxial conductivity, we can get all 8 degrees of freedom. Separate solvability problems for Re(S) and for Im(S). Biaxial dielectric may serve as an example of underdetermined system of equations (not done yet). M. Gilman (Mathematics, NCSU) Polarimetry of homogeneous half-spaces NYU, July 20, / 34

22 Inhomogeneous half-spaces Overview Inhomogeneous half-spaces Main deficiency of homogeneous half-space models: only specular reflection, no backscattering. Backscattering is the primary configuration for remote sensing. Options to obtain backscattered signal: horizontally inhomogeneous material; horizontally inhomogeneous boundary conditions; horizontally inhomogeneous shape of the interface. M. Gilman (Mathematics, NCSU) Polarimetry of homogeneous half-spaces NYU, July 20, / 34

23 Inhomogeneous half-spaces Horizontally inhomogeneous material Horizontally inhomogeneous material For ε = ε(x, y), Maxwell s equations reduce to E grad div E = ε c 2 2 E, div(εe) = 0. t2 Plane wave E e i(k,r) iωt is not a solution (try ε = 1 + a sin(2k i x)). Linearization is needed to move forward. We consider an acoustic problem U = ε 2 c 2 U for U(r, t) where t2 U = U (0) + U (1) and ε = ε (0) + ε (1) (x, y), U (1) U (0), ε (1) ε (0), whereas ε (0) 1 is not necessarily small. Linearization yields U (0) as the Fresnel solution for the horizontal polarization: U (0) U (0) = i + U (0) r, z > 0, U (0) t, z < 0. M. Gilman (Mathematics, NCSU) Polarimetry of homogeneous half-spaces NYU, July 20, / 34

24 Linearization Inhomogeneous half-spaces Horizontally inhomogeneous material In Fourier domain: we obtain ε (1) (x, y) = 1 (2π) 2 ˆε (1) (K x, K y )e i(kxx+kyy) dk x dk y, ( d2 dz 2 + q2 )û (1) = 0, z > 0, ( d2 dz 2 + q 2 )û (1) = k 2 (1) u (0)Aˆε rz B e iq, z < 0, where û (1) = û (1) (K x, K y, z), ˆε (1) B = ˆε(1) (K x K i, K y ), q 2 = k 2 K 2 x K 2 y, q 2 = k 2 ε (0) K 2 x K 2 y, q 2 r = k 2 ε (0) K 2 i. Backscattering for z > 0 means K x = K i, K y = 0, i.e., ˆε (1) B ˆε(1) ( 2K i, 0). This is the Bragg harmonic of ε(x, y). M. Gilman (Mathematics, NCSU) Polarimetry of homogeneous half-spaces NYU, July 20, / 34

25 Inhomogeneous half-spaces Reflection coefficients Horizontally inhomogeneous material General solution: Be iqz, z > 0, û (1) = u (0)A Ce iq z + A 1 e iq rz, z < 0 and q /= q r, Ce iq z + A 2 ze iq rz, z < 0 and q = q r. Interface conditions: continuity for û (1) and dû (1) /dz at z = 0. Reflection coefficients: 2ˆε (1) k 2 q i B B = (q + q)(q r + q )(q r + q i ), q q r, 2ˆε (1) k 2 q i B (q r + q)2q r(q r + q i ), q = q r. B is insensitive to the resonance in the material (i.e., two expressions for B coincide if q = q r). M. Gilman (Mathematics, NCSU) Polarimetry of homogeneous half-spaces NYU, July 20, / 34

26 Inhomogeneous half-spaces Horizontally inhomogeneous material Summary for inhomogeneous material Scalar problem for inhomogeneous isotropic dielectric. Linearization of ε about ε (0). The scattering coefficient is proportional to ˆε (1) B. Vector problem (i.e., Maxwell s equations rather than wave equation) is huge (check [GST, 2017]). It demonstrates depolarization if K y 0. No depolarization for backscattering. Not done yet: inhomogeneous anisotropic material. We expect to obtain depolarization in backscattering. M. Gilman (Mathematics, NCSU) Polarimetry of homogeneous half-spaces NYU, July 20, / 34

27 Inhomogeneous half-spaces Horizontally inhomogeneous material Polarization: additional information about the target Polarimetry is useful for segmentation/classification of terrains: RAMSES (Airborne SAR) /Urban_Classification_3Drendering.pdf Polarimetry reveals, e.g.: forested land by high HV; bare ground and grass by high HH-VV; although the models are still semi-empirical. M. Gilman (Mathematics, NCSU) Polarimetry of homogeneous half-spaces NYU, July 20, / 34

28 Inhomogeneous half-spaces Reflection-only setups Alternatives for refraction-reflection problem We can consider reflection-only problem because we are not interested in the field below the interface. Advantage: simpler setup. Disadvantage: boundary properties should represent bulk properties. Possible options: Variable boundary conditions on a plane surface (example: Leontovich boundary condition). Homogeneous boundary conditions on a non-plane surface (i.e., rough surface scattering). Combination of the two inhomogeneities. M. Gilman (Mathematics, NCSU) Polarimetry of homogeneous half-spaces NYU, July 20, / 34

29 Inhomogeneous half-spaces Reflection-only setups Leontovich boundary condition Let ε = ε (0) = const. In Fourier domain: û(k x, K y, z) = e iωt U(x, y, z, t)e i(kxx+kyy) dx dy, consider the regular reflection-refraction problem û A i (K x, K y )e iqz + û A r (K x, K y )e iqz, z > 0, û(k x, K y, z) = û A t (K x, K y )e iq z, z < 0, where q 2 = k 2 Kx 2 Ky 2, q 2 = k 2 ε (0) Kx 2 Ky 2. If û and û/ z are continuous at z = 0, then we can show that û(k x, K y, z) = i ε (0) k 2 (Kx 2 + Ky 2 ) û(k x, K y, z). z z=+0 z=+0 Hence, in the coordinate domain, the relation between U and U/ z will be non-local. M. Gilman (Mathematics, NCSU) Polarimetry of homogeneous half-spaces NYU, July 20, / 34

30 Inhomogeneous half-spaces Reflection-only setups Leontovich boundary condition, cont d For ε (0) 1, we can simplify ε (0) k 2 (K 2 x + K 2 y ) ε (0) k, so û(k x, K y, z) = i ε (0) k û(k x, K y, z) z z=+0 z=+0 U(x, y, z, t) = i z z=+0 ε (0) k U(x, y, z, t). z=+0 These are local boundary conditions, called Leontovich conditions. Generalization: U(x, y, z, t) = i ε(x, y)k U(x, y, z, t), ε(x, y) 1. z z=+0 z=+0 M. Gilman (Mathematics, NCSU) Polarimetry of homogeneous half-spaces NYU, July 20, / 34

31 Inhomogeneous half-spaces Reflection-only setups Leontovich boundary condition, cont d We don t need a solution for z < 0 because we have boundary conditions at z = 0. With ε(x, y) = ε (0) + ε (1) (x, y), ε (1) ε (0), and U = U (0) + U (1), U (0) = û A i e i(k ix q i z) iωt, U (1) U (0), we can obtain û A r (K x, K y ) û A i = ˆε(1) B q i (ε (0) ) 3/2 k. this is the reflection coefficient ˆε (1) B = ˆε(1) (K x K i, K y ). The reflection coefficient coincides with that obtained in the reflection-refraction problem for ε (0) 1. Need more validation. Possible next step: vector problem and anisotropic ε (1). M. Gilman (Mathematics, NCSU) Polarimetry of homogeneous half-spaces NYU, July 20, / 34

32 Inhomogeneous half-spaces Rough surface scattering Reflection-only setups Intensely studied topic, important for remote sensing of sea surface (see, e.g., [Beckmann & Spizzichino, 1963], [Bass & Fuks, 1979], [Voronovich, 1998], [Bruno et al., 2002]). The simplest setting: Dirichlet problem at z = h(x, y) for the scalar (Helmholtz) equation (strip e iωt factor): ( + k 2 )(u (0) i + u (0) r + u (1) ) = 0, z > h(x, y), (u (0) i + u (0) r + u (1) ) = 0, z = h(x, y). In Fourier domain: û (1) (K x, K y, z) = u (1) (x, y, z)e i(kxx+kyy) dx dy, ĥ(k x, K y ) = h(x, y)e i(kxx+kyy) dx dy, we have for the first order field: ( d2 dz 2 + q2 )û (1) = 0, where q 2 = k 2 K 2 x K 2 y. M. Gilman (Mathematics, NCSU) Polarimetry of homogeneous half-spaces NYU, July 20, / 34

33 Inhomogeneous half-spaces Reflection-only setups Rough surface scattering, cont d Linearization of the domain and boundary conditions for q i h 1: û (0) i u (1) (x, y, 0) = (u (0) i + u (0) r ) z=h(x,y), (***) (K x, K y ) z=h(x,y) = u (0)A e i(k ix q i z) e i(kxx+kyy) dx dy z=h(x,y) i u (0)A i e ikix (1 iq i h(x, y))e i(kxx+kyy) dx dy = u (0)A i ((2π) 2 δ(k x K i )δ(k y ) iq i ĥ B ), where ĥ B = ĥ(k x K i, K y ), and similarly for û (0) r, so (***) yields û (1) (K x, K y, 0) = u (0)A i 2iq i ĥ B. Using û (1) = u (0)A i Be iqz we find reflection coefficient to be 2iq i ĥ B. Vector problem no depolarization [Voronovich, 1998]. M. Gilman (Mathematics, NCSU) Polarimetry of homogeneous half-spaces NYU, July 20, / 34

34 Summary This talk dealt with the following topics: Need for physics-based scattering models for inverse problems in radar imaging. Polarimetry and degrees of freedom of the scattering matrix. Half-space models of scatterer: horizontally homogeneous and horizontally inhomogeneous. Single-domain models: Leontovich boundary condition and rough surface scattering. Not mentioned: pulsed/modulated signal, interaction between surface harmonics, speckle, coherence, non-linearities in the scatterer, etc. THANK YOU! M. Gilman (Mathematics, NCSU) Polarimetry of homogeneous half-spaces NYU, July 20, / 34

35 Backup slides Degrees of freedom: special cases 2 d.o.f. if β = 0; 3 d.o.f. if α = 0 or γ = 0. z y K i plane of incidence k = (K i, 0, q i ) q i x interface γ c = (α, β, γ) β α free space material M. Gilman (Mathematics, NCSU) Polarimetry of homogeneous half-spaces NYU, July 20, / 34

36 Backup slides Anisotropic material. Monte-Carlo simulations Limited solvability: θinc < π/4; α = 0 Ô SHV = SVH M. Gilman (Mathematics, NCSU) Polarimetry of homogeneous half-spaces NYU, July 20, / 34

37 Backup slides Limited solvability: blow-up Anisotropic material. Monte-Carlo simulations M. Gilman (Mathematics, NCSU) Polarimetry of homogeneous half-spaces NYU, July 20, / 34

38 Backup slides Anisotropic material. Monte-Carlo simulations Unlimited solvability: θ inc > π/4; α = 0 S HV = S VH M. Gilman (Mathematics, NCSU) Polarimetry of homogeneous half-spaces NYU, July 20, / 34