Kirchhoff, Fresnel, Fraunhofer, Born approximation and more

Transcription

1 Kirchhoff, Fresnel, Fraunhofer, Born approximation and more Oberseminar, May 2008

2 Maxwell equations Or: X-ray wave fields X-rays are electromagnetic waves with wave length from 10 nm to 1 pm, i.e., 10 8 m to m and frequency from Hz to Hz Maxwell equations (with constitutive relations incorporated) curl E + t µh = 0, curl H t εe = 0 ε = electric permittivity, µ = magnetic permeability James Clerk Maxwell,

3 From Maxwell to Helmholtz curl E + t µh = 0, curl H t εe = 0 µ = µ 0 nonmagnetic material, ε = nε 0 with refractive index n curl curl E + ε 0 µ 0 n 2 E t 2 = 0 Time harmonic waves: E(x, t) = R{E(x) e iωt } k 2 = ε 0 µ 0 ω 2 wave number curl curl E k 2 ne = 0 E + k 2 ne = grad div E Helmholtz equation u + k 2 nu = 0 Hermann von Helmholtz,

4 The main tools 1. The fundamental solution Solution of Φ(x, y) = eik x y 4π x y, x y, x = (x 1, x 2, x 3 ) IR 3, x := Φ + k 2 Φ = δ( y) x x x 2 3 Can be obtained via Fourier transformation or as spherically symmetric solution to the Helmholtz equation. Outgoing spherical wave.

5 The main tools 2. Green s integral theorem bounded domain with sufficiently smooth boundary, u and v sufficiently smooth functions u v dx = u v ν ds From divergence theorem for A = u grad v {u v v u} dx = ν grad u grad v dx { u v ν v u } ds ν

6 Helmholtz representation { } { u[ v+k 2 v] v[ u+k 2 u] dx = u v ν v u } ds ν ν Choose v = Φ(x, ) with x u(x) = { u Φ(x, ) Φ(x, ) u ν ν } ds x { u+k 2 u} Φ(x, ) dy If u + k 2 u = 0 in, then u(x) = { u Φ(x, ) Φ(x, ) u ν ν } ds, x Green s formula, George Green, Helmholtz representation

7 Helmholtz representation If u + k 2 u = 0 in, then u(x) = { u Φ(x, ) Φ(x, ) u ν ν } ds, x Green s formula, Helmholtz representation Can replace by any solution Φ of Φ(x, y) = eik x y 4π x y, x y, Φ + k 2 Φ = δ( y), i.e., can add any solution of the Helmholtz equation to Φ

8 Radiation condition u(x) r iku(x) = o ( ) 1, r = x r Sommerfeld radiation condition, Arnold Sommerfeld Characterizes outgoing waves, i.e., outgoing energy flux Implies Sommerfeld finiteness condition ( ) 1 u(x) = O, r = x r and same order of decay for all derivatives

9 Forward propagation of waves Knowing u on the plane x 3 = 0, we want to compute u on the parallel plane x 3 = z for z > 0. { } u Φ(x, ) Γ u(x) = Φ(x, ) u ds ν Γ R ν ν Γ R R x H R ν { } u Φ(x, ) + Φ(x, ) u ds H R ν ν one can readily show that the second integral vanishes in the limit R However: integrand only O(R 2 ). For details of proof, see yellow book pp. 19.

10 Rayleigh-Sommerfeld integral Γ := {y = (y 1, y 2, 0)}, x = (x 1, x 2, z), x = (x 1, x 2, z) Γ ν { } u Φ(x, ) u(x) = Φ(x, ) u ds Γ ν ν Replace Φ by Φ (x, y) = Φ(x, y) Φ(x, y) x x u(x) = 1 u(y) 2π Γ ν(y) e ik x y x y ds(y) Rayleigh Sommerfeld diffraction integral Lord John William Rayleigh,

11 Rayleigh-Sommerfeld integral Γ := {y = (y 1, y 2, 0)}, x = (x 1, x 2, z), x = (x 1, x 2, z) Γ ν { } u Φ(x, ) u(x) = Φ(x, ) u ds Γ ν ν Replace Φ by Φ N (x, y) = Φ(x, y) + Φ(x, y) x x u(x) = 1 u eik x y (y) 2π Γ ν x y ds(y) Rayleigh Sommerfeld diffraction integral Lord John William Rayleigh,

12 Forward propagation via Fourier transform Apply Fourier transform F with respect to x 1 and x 2. û(ξ 1, ξ 2, z) = 1 e i ξ y u(y 1, y 2, z) dy 2π IR 2 u + k 2 u = 0 2 û z 2 + (k 2 ξ 2 )û = 0 From the two solutions e i k 2 ξ 2 z and e i k 2 ξ 2 z the first is propagating forward, the second backward û(ξ 1, ξ 2, z) = û(ξ 1, ξ 2, 0)e i k 2 ξ 2 z u(, z) = F 1 z Fu(, 0) with the forward propagator z given by ( z v)(ξ) := e i k 2 ξ 2 z v(ξ)

13 Equivalence with Rayleigh Sommerfeld integral x = (x 1, x 2, z), y = (y 1, y 2, 0), ξ = (ξ 1, ξ 2, 0) [F 1 z Fu(, 0)](x) = 1 4π 2 e i x ξ e i k 2 ξ 2 z e i y ξ u(y) dy dξ IR 2 IR 2 = 1 4π 2 u(y) IR2 e i (x y) ξ ei k 2 ξ 2 z IR 2 z i dξ dy k 2 ξ2 = 1 u(y) 2π IR 2 z e ik x y x y dy i IR2 e i (x y) ξ ei k 2 ξ 2 z eik x y dξ =, Weyl expansion 2π k 2 ξ2 x y Hermann Weyl,

14 Kirchhoff approximation u i Γ A ν A { } u Φ(x, ) u(x) = Φ(x, ) u ds Γ A A ν ν { } u i u(x) A ν Φ(x, ) Φ(x, ) ui ds ν Mathematical objections: u = ν u = 0 on Γ A implies u = 0 everywhere No influence of boundary condition on Γ A Better: u(x) 2 A u i ν Φ(x, ) ds, u(x) 2 u i Φ(x, ) ds A ν

15 Fresnel diffraction u i Γ A A u(x) 1 u(y) 2π A ν(y) e ik x y x y dy x = (x 1, x 2, z), y = (y 1, y 2, 0) x y = x 1 x x y + y 2 2 x [x y]2 2 x 3 } {{ } f (x,y) +O ( ) 1 x 2 If x large, k large, x y 1 and A small, then u(x) ik eik x 2π x A u(y) e ikf (x,y) dy Fresnel diffraction, Fresnel region Augustin Jean Fresnel,

16 Fraunhofer diffraction u i Γ A A u(x) 1 u(y) 2π A ν(y) e ik x y x y dy x = (x 1, x 2, z), y = (y 1, y 2, 0) x y = x 1 x x y + y 2 2 x [x y]2 2 x 3 } {{ } f (x,y) If x very large and A very small, then u(x) ik eik x 2π x A +O u(y) e ik 1 x x y dy ( ) 1 x 2 Fraunhofer diffraction, far field pattern Joseph von Fraunhofer,

17 o this via Fourier transform Recall u(, z) = F 1 z Fu(, 0) with the forward propagator z given by ( z v)(ξ) := e i k 2 ξ 2 z v(ξ) Use k 2 ξ 2 k ξ2 2k for small ξ to approximate via «(z Fresnel v)(ξ) := e i k ξ2 z 2k v(ξ)

18 o this via Fourier transform x = (x 1, x }{{ 2, z), } y = (y 1, y 2, 0), ξ = (ξ 1, ξ 2, 0), ex [F 1 Fr z Fu(, 0)](x) = 1 4π 2 IR 2 e i (x y) ξ e i 1 4π 2 eikz IR 2 e i x ξ e i A = ik 2πz eikz u(y) A «k ξ2 z 2k e i y ξ u(y) dy dξ A ξ 2 e i (x y) ξ e i 2k z dξ dy IR 2 (ex y)2 ik u(y)e 2z dy ξ 2 2k z dξ = 2πk (ex y) 2 iz eik 2z, Fresnel integral

19 o this via Fourier transform x = (x 1, x }{{ 2, z), y = (y } 1, y 2, 0) ex u(x) = ik (ex y)2 2πz eikz ik u(y)e 2z dy A ik ex 2 2πz eikz ik e 2z u(y)e ik 1 z x y dy Compared to the Fraunhofer integral u(x) ik ϕ(y) e ik 1 x x y dy 2π x eik x Note: x = z 2 + x 2 z + x 2 2z A A

20 X-ray interaction with matter Or: Helmholtz equation for inhomogeneous medium u + k 2 nu = 0 in IR 3 Recall refractive index n. Assume bounded support of n 1. Scattering of an incident field u i requires for the total field u = u i + u s the inhomogeneous Helmholtz equation and for the scattered field u s the Sommerfeld radiation condition.

21 Forward propagation in matter u + k 2 nu = 0 Consider again forward propagation via u(x) = w(x)e ikz Helmholtz equation equivalent to 2ik w z + 2w + 2 w }{{ z 2 +k 2 (n 1)w = 0 } w Neglecting the second derivative in z direction leads to the paraxial approximation 2ik w z + 2w + k 2 (n 1)w = 0

22 Paraxial approximation u(x) = w(x)e ikz 2ik w z + 2w + k 2 (n 1)w = 0 For free space n = 1 this reduces to 2ik w z + 2w = 0 2ik ŵ z ξ2 ŵ = 0 Rediscover Fresnel approximation of forward propagator u(, z) F 1 Fresnel z Fu(, 0) with «(z Fresnel v)(ξ) := e i k ξ2 z 2k v(ξ)

23 Projection approximation u(x) = w(x)e ikz 2ik w z + 2w + k 2 (n 1)w = 0 Neglecting the Laplacian that couples neighboring rays leads to the projection approximation with solution u(x, z) = u(x, 0) exp 2ik w z + k 2 (n 1)w = 0 ( ikz + k 2i z 0 ) (1 n(x, ζ)) dζ

24 Lippmann Schwinger equation u + k 2 nu = 0 in IR 3 u = u i + u s u s satisfies radiation condition enote the support of n 1 and recall Green s formula { } u Φ(x, ) u(x) = Φ(x, ) u ds { u+k 2 u} Φ(x, ) dy ν ν The scattering problem is equivalent to the integral equation u(x) = u i (x) k 2 Φ(x, y)(1 n(y))u(y) dy, x Bernard A. Lippmann Julian Seymour Schwinger,

25 Lippmann Schwinger equation The scattering problem is equivalent to the integral equation u(x) = u i (x) k 2 Φ(x, y)(1 n(y))u(y) dy, x Can show existence and uniqueness of a solution in any reasonable function space, i.e., C(), L 2 () or Sobolev spaces, see yellow book pp. 214 Fraunhofer approximation, i.e., far field pattern u(x) u i (x) k 2 eik x e ik 1 x x y (1 n(y))u(y) dy, 4π x x large

26 Lippmann Schwinger equation For small contrast k 2 (1 n) can solve Lippmann-Schwinger equation u(x) = u i (x) k 2 Φ(x, y)(1 n(y))u(y) dy, x by successive approximations, i.e., by Neumann series. For small contrast k 2 (1 n) can solve Lippmann-Schwinger equation u m+1 (x) = u i (x) k 2 Φ(x, y)(1 n(y))u m (y) dy, x by successive approximations, i.e., by Neumann series. The Born approximation u B (x) = u i (x) k 2 Φ(x, y)(1 n(y))u i (y) dy, x corresponds to executing only one iteration step with u 0 = u i Max Born,

27 Lippmann Schwinger equation Born approximation u B (x) = u i (x) k 2 has far field u B (x) u i (x) k 2 eik x 4π x Φ(x, y)(1 n(y))u i (y) dy, e ik 1 x x y (1 n(y))u i (y) dy, x x large For plane wave incidence u i (x) = e ik x d with direction d u B (x) u i (x) k 2 eik x e ik 1 x x d y (1 n(y)) dy, x large 4π x Requires Fourier transform of 1 n on the Ewald sphere Ewald := {k(z d) : z S 2 } that is, the sphere of radius k centered at kd. Paul Peter Ewald,