Green s functions for planarly layered media

Transcription

1 Green s functions for planarly layered media Massachusetts Institute of Technology lecture notes Introduction: Green s functions The Green s functions is the solution of the wave equation for a point source (dipole). For scalar problems, the wave equation is written as (k 0 = ω ɛµ): and the solution for an unbounded medium is: ( 2 + k 2 0) g( r, r ) = δ( r r ), () g( r, r ) = eik 0 r r 4π r r. (2) From Maxwell s equations in frequency domain with an e iωt dependency, the wave equation for electric field Ē( r) is: Ē( r) k2 0Ē( r) = iω J( r), Therefore, the free-space dyadic Green s function satisfies = 0 for source free case. (3) which solution is G( r, r ) k 2 0G( r, r ) = I δ( r r ), (4) G( r, r ) = ( I + ) k0 2 g( r, r ), (5) (check using (Ig) = g ( g)i and Eq. ()). For the use of Green s functions in scattering problems, it is useful to express the Green s function in the same coordinates as the problem, which can be rectangular, cylindrical, spherical, etc. Here we shall concentrate on the rectangular representation (Cartesian).

2 2 Section 2. Cartesian coordinates 2 Cartesian coordinates 2. Scalar Green s function The formulae are derived from Eq. () and the Fourier transform of the quantities: g( r, r ) = (2π) 3 δ( r r ) = (2π) 3 d k e i k ( r r ) g( k), d k e i k ( r r ), (6a) (6b) where k = k xˆx + k y ŷ + k z ẑ and d k = dk x dk y dk z. Upon using Eq. (), we write: ( 2 + k0) 2 d k e i k ( r r ) g( k) = d k e i k ( r r ), (7) Introducing the differential operator ( 2 = 2 x y z 2 ) we write: ( 2 + k0) 2 d k e i k ( r r ) g( k) = d k ( 2 + k0)e 2 i k ( r r ) g( k) from which we conclude that = d k ( kx 2 ky 2 kz 2 + k0)e 2 i k ( r r ) g( k) = d k e i k ( r r ), (8) g( k) = kx 2 + ky 2 + kz 2 k0 2. (9) Using Eq. (6a), we therefore need to evaluate the following integral: g( r, r ) = (2π) 3 dk x dk y dk z k 2 x + k 2 y + k 2 z k 2 0 e i k ( r r ). (0) Note that Eq. (0) can be integrated along one of the three axis. In remote sensing application, the vertical axis is usually taken to be the z axis, (xy) being the transverse plane (planar components). We therefore choose to evaluate Eq. (0) along k z, and we split: k = k xˆx + k y ŷ + k z ẑ = k + k z ẑ, r = r + zẑ, r = r + z ẑ. (a) (b) (c)

3 3 We will perform the integral of Eq. (0) in the complex plane, using Cauchy s theorem and the Residue theorem. Before doing this, we have to be careful not to have divergent integrals. Since we integrate in k z, the condition is: If we write k z as k z = k z + ik z (k z R, k z R), we see that lim k eikzz < +. (2) z if z > 0, we have to choose k z > 0, which means that for complex plane integration, we need to deform the contour into the upper plane. if z < 0, we have to choose k z < 0, which corresponds to a deformation into the lower plane. In addition, we see from Eq. (0) that the integrand has a pole at k 2 0 z = k 2 0 k 2 x k 2 y = k 2 0 k 2. (3) We therefore need to evaluate Eq. (0) via the Residue theorem. Calculus: let us just write the integral in dk z for z > 0: + kz 2 k0 2 e i k ( r r ) = 2iπ Res z kz 2 k0 2 e i k ( r r ) z = 2iπ lim k z k 0z k z k 0z (k z k 0z )(k z + k 0z ) ei k ( r r ) e ikz(z z ) = 2iπ 2k 0z e i k ( r r ) e ik 0z (z z ), (4) so that g( r, r ) = i (2π) 2 d k 2k 0z e i k ( r r ) e ik 0z (z z ), for z z > 0. (5) The treatment for z < 0 follows the same reasoning so that we write for all (z z ): (z z ) R, g( r, r ) = i (2π) Dyadic Green s function From Eq. (6) and Eq. (26), we can get the dyadic Green s functions. d k 2k 0z e i k ( r r ) e ik 0z z z. (6) Note that is a dyadic operator (give a dyad when applied to a scalar) and can be exchanged with the integral sign. In addition, it only applies to the exponential terms so that we actually need to evaluate: e i k ( r r ) e ik 0z z z (7)

4 4 2.2 Dyadic Green s function or, by a simple change of variables: e i k r e ik 0z z (8) Calculus: Let us first consider z > 0 and Various derivatives will be: x x... x f(x, y, z) = k2 x f(x, y, z), y f(x, y, z) = k xk y f(x, y, z), and identically for z < 0. At z = 0 however, 2 z 2 eik 0z z = {ik 0z e ik 0z z z } z z (e i k r e ik 0z z ) = f(x, y, z). (9) { ( ) = ik 0z ik 0z e ik 0z 2 z z z + ik 0z e ik 0z z 2 z 2 z = 2ik 0z δ(z) k 2 0 z e ik 0z z. (20) } Using these results, we write: 2 z 2 g( r) = i (2π) 2 = δ(z) (2π) 2 = δ( r) i d k e i k r 2ik 0z δ(z) k 20z e ik 0z z 2k 0z d k e i k r i d k k 0z e i k r e ik 0z z d k k 0z e i k r e ik 0z z. (2) Again, all the other terms of applied to the integrand give k k so that the Green s function becomes: where G( r, r ) = ẑẑ δ( r) k i d k I k k k 0z k0 2 K K d k I k 0z k 2 0 e i k r e i K r for z > 0, for z < 0, (22) k = k xˆx + k y ŷ + k 0z ẑ, K = k xˆx + k y ŷ k 0z ẑ. (23a) (23b) Some notes:

5 5. The Dirac delta function is known as the singularity of the Green s function and is important in calculating the fields in the source region. 2. The different signs ensure that the integral converges for evanescent waves, i.e. when k 2 x + k 2 y > k The square bracket in the expression of the Green s functions can be expressed in terms of superposition of TE and TM waves, as we shall see. 2.3 Superposition of TE and TM waves Based on k, we can form an orthonormal system for TE/TM polarized waves: TE: ê(k 0z ) = k ẑ k ẑ = ˆxk y ŷk x = (ˆxk y ŷk x ), kx 2 + ky 2 k ρ (24a) TM: ĥ(k 0z ) = k 0 ê(k 0z ) k = k 0 z k o k ρ (ˆxk y + ŷk x ) + k ρ k o ẑ. (24b) The three vectors ˆk, ĥ and ê form an orthonormal system, in which: I = ˆkˆk + ê(k 0z )ê(k 0z ) + ĥ(k 0 z )ĥ(k 0 z ). (25) After translating to the origin, we get for the dyadic Green s functions: G( r, r ) = ẑẑ δ( r) k0 2 + i where d k ê(k 0z )ê(k 0z ) + k ĥ(k 0 )ĥ(k z 0 z ) 0z d k ê( k 0z )ê( k 0z ) + k ĥ( k 0 )ĥ( k z 0 z ) 0z e i k ( r r ) e i K ( r r ) for z > z, for z < z, (26) ê( k 0z ) = ê(k 0z ), (27a) ĥ( k 0z ) = ê( k 0 z ) K k 0. (27b) (Note that K, ê( k 0z ) and ĥ( k 0 z ) form another orthonormal set of vectors about K). 2.4 Treatment of layered media Depending upon the medium under study and the location of the source, the kernel of Eq. (26) will have to be modified. To make it more clear, we can gather the terms in the Green s function relative to the source (primed coordinates) and those relative to the source.

6 6 2.4 Treatment of layered media G( r, r )= ẑẑ δ( r) k i d k d k k 0z {ê(k 0z ) e i k r ê(k 0z ) e i k r +ĥ(k 0z ) ei k r ĥ(k 0z ) e i k r } for z > z, {ê( k k 0z ) e i K r ê( k 0z ) e i K r +ĥ( k 0z 0z ) ei K r ĥ( k 0z ) e i K r } for z < z, (28) If we now consider a layered medium problem, with an arbitrary number of layers and a source in the top region (incident wave), we write: with G( r, r ) i0 =. For z < z, i = 0: i { } dk x dk y K e ê( k 0z ) e i K r + k K h ĥ( k 0z ) e i K r 0z (29) K e =ê( k 0z ) e i K r + R T E ê(k 0z ) e k 0z e i k r K h =ĥ( k 0 z ) e i K r + R T M ĥ(k 0z ) e k 0z e i k r (30a) (30b) 2. For region l, i = l: K e =A l ê(k lz ) e i k l r + B l ê( k lz ) e i K l r K h =C l ĥ(k lz ) e i k l r + D l ĥ( k lz ) e i K l r (3a) (3b) (3c) 3. For region t, i = t: where K e =T T E ê( k tz ) e i K t r K h =T T M ĥ( k tz ) e i K t r k lz = kl 2 k2 x ky 2, k l =ˆxk x + ŷk y + ẑk lz, K l =ˆxk x + ŷk y ẑk lz, (32a) (32b) (32c) (33a) (33b) (33c) and the coefficients A l, B l, C l and D l are determined from the boundary conditions. The boundary conditions apply to the tangential electric and magnetic fields. Thus, in terms of Green s functions, we need to satisfy the continuity of ẑ G( r, r ) and ẑ

7 7 G( r, r ). Let us write this at the interface between media (l) and (l + ), by separating the TE and TM components: A l e ik lz z + B l e ik lz z = A l+ e ik lz z + B l+ e ik lz z k lz A l e ik lz z B l e ik lz z = k zl+ A l+ e ik lz z B l+ e ik lz z k lz A l e ik lz z B l e ik lz z = k zl+ A l+ e ik lz z B l+ e ik lz z k l k l+ k lz C l e ik lz z + D l e ik lz z = k zl+ C l+ e ik lz z + D l+ e ik lz z (34a) (34b) (34c) (34d) With the conditions in the first and last layer as: A 0 = R T E, B 0 =, C 0 = R T M, D 0 =, (35a) A t = 0, B t = T T E, C t = 0, D t = T T M. (35b) Before evaluating these coefficients, we can build a recursive scheme to calculate the amplitudes from region l to region l +. For example, it is straightforward to build a propagation matrix for TE modes from Eq. (34a) and (34b): ( ) ( ) A l+ e ikzl+ d l+ = V T E A l e ikzl d l (36) B l+ e ikzl+ d l+ B l e ikzl d l A similar procedure of course applied to the TM modes: ( ) ( ) C l+ e ikzl+ d l+ = V T M C l e ikzl d l D l+ e ikzl+ d l+ D l e ikzl d l In order to end up the recursive method, we have to express the reflection and transmission coefficient in the first and last regions, respectively. We shall only illustrated this point here, as it has been developed in previous classes. Let us consider a plane wave incident from region 0, with its plane of incidence parallel to the (xy) plane. All fields vectors are independent on y, so that y = 0 in Maxwell s equations. Thus, we can decompose the fields into their TE and TM components. We get in region l: TE modes: (37) z E ly, H lx = iωµ l H lz = iωµ l x E ly, ( 2 x 2 + y + ω2 ɛ l µ l ) E ly = 0. (38a) (38b) (38c)

8 8 2.4 Treatment of layered media TM modes: E lx = iωɛ l z H ly, E lz = ( 2 iωɛ l x H ly, x 2 + y + ω2 ɛ l µ l ) H ly = 0. (39a) (39b) (39c) For a TE wave inside the stratified medium: E ly =(A l e ik lz z + B l e ik lz z ) e ikxx, (40a) H lx = k l z ωµ l (A l e ik lz z B l e ik lz z ) e ikxx, (40b) H lz = k x ωµ l (A l e ik lz z + B l e ik lz z ) e ikxx. (40c) By matching the boundary conditions, and upon using the already known notation we get the recursive relation: p l(l+) = µ lk z(l+) µ l+ k lz, (4a) R l(l+) = p l(l+) + p l(l+). (4b) A l = e2iklz dl + /R2 l(l+) e 2ik (l+)z +k lz d l B l R l(l+) /R l(l+) e 2ik (l+) z d l + A, (42) l+ B l+ with the limiting condition: A t B t = 0,, A 0 B 0 = R. (43) Example: for a two-layer medium (t = 2): R = R 0 + R 2 e 2ik z (d d 0 ) + R 0 R 2 e 2ik z (d d 0 ) e2ikzd 0. (44)