Файл: Doicu A., Wriedt T., Eremin Y.A. Light scattering by systems of particles (OS 124, Springer, 2006.pdf

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B.4 Translations

287

Using the integral representation for the translation coe cients, making the transformation ϕ0 → ϕ0 + π, changing the variable of integration from β to π − β and using the identities (A.24) and (A.25), yields

Amn,m n (−kr0) = (1)n+n Amn,m n (kr0) ,

Bmn,m n (−kr0) = (1)n+n +1Bmn,m n (kr0) .

Further, since

Amn,m n (kr0) = (1)n+n A−m n ,−mn (kr0) ,

Bmn,m n (kr0) = (1)n+n +1B−m n ,−mn (kr0) ,

we obtain

Amn,m n (−kr0) = A−m n ,−mn (kr0

Bmn,m n (−kr0) = B−m n ,−mn (kr0

) ,

(B.74)

) .

(B.75)

Recurrence relations for the scalar and vector addition theorem has also been given by Chew [32, 33], Chew and Wang [35] and Kim [117]. The relationship between the coe cients of the vector addition theorem and those of the scalar addition theorem has been discussed by Bruning and Lo [29], and Chew [32].


C

Computational Aspects

in E ective Medium Theory

In this appendix we compute the basic integrals appearing in the analysis of electromagnetic scattering from a half-space of randomly distributed particles. Our derivation follows the procedures described by Varadan et al. [236], Tsang and Kong [223, 226], and Tsang et al. [228].

C.1 Computation of the Integral Imm1 n

The integral Imm1 n is

Imm1 n = ejKe·r0p u3m −mn (ksrlp) dV (r0p) ,

Dp

where Ke = Ksez , and the integration domain Dp is the half-space z0p 0, excluding a spherical volume of radius 2R centered at r0l. The volume integral can be transformed into a surface integral by making use of the following result. Let u and v be two scalar fields satisfying the Helmholtz equation in the bounded domain D, with the wave numbers Ks and ks, i.e.,

u + Ks2u = 0, v + ks2v = 0.

Then, from Green’s theorem we have

 

1

 

 

 

 

 

 

 

D uvdV =

 

D (vu − uv) dV

 

ks2 − Ks2

 

1

 

 

∂u

 

∂v

 

 

=

 

 

v

 

− u

 

dS,

 

ks2 − Ks2

S

∂n

∂n

where S is the boundary surface of the domain D, and n is the outward unit normal vector to S. Using this result we transform the integral Imm1 n as follows:


290 C Computational Aspects in E ective Medium Theory

z

Op

r0p rlp

r0l

O

S

θlp 2R

Ol SR RSz

Fig. C.1. Integration surfaces SR , Sand Sz

1

 

1

 

 

3

 

 

 

 

e

jK

r

 

Imm n =

 

 

 

 

 

 

um −mn (ksrlp)

 

 

e·

0p

k2

K2

 

∂n

 

 

s

s SR SSz

 

 

 

 

 

 

 

 

 

 

 

ejKe·r0p

∂um3 −mn

(k r

 

) dS (r

 

) .

 

 

 

 

 

 

 

∂n

 

s

lp

 

0p

 

 

 

 

 

 

The integral can be decomposed into three integrals

Imm1 n = Imm1,R n + Imm1,∞ n + Imm1,z n ,

where Imm1,R n is the integral over the spherical surface SR of radius 2R cen-

1,∞

tered at r0l, Imm n is the integral over the surface of a half-sphere Swith radius Rin the limit R→ ∞, and Imm1,z n is the integral over the xy- plane Sz (the plane z = 0). The choice of the integration surfaces is shown in Fig. C.1.

Using the identity r0p = r0l + rlp and replacing, for convenience, the variables rlp, θlp and ϕlp by r, θ and ϕ, respectively, we obtain

Imm1,R

n =

1

 

ejKsz0l

 

 

 

 

 

 

 

 

 

 

ks2 − Ks2

 

 

 

 

 

 

 

 

 

 

 

 

 

!h(1)

(ksr) P |m −m|

(cos θ) ej(m −m)ϕ

ejKe·r

 

 

 

 

 

 

× SR

 

n

 

 

n

 

 

 

∂r

 

 

ejKe·r

 

h(1) (ksr) P |m −m| (cos θ) ej(m −m)ϕ" dS(r),

 

∂r

 

 

 

 

n

 

 

 

n

 

 

 

 

 

 

whence, taking into account the series representation

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

2n

+ 1

 

 

 

 

 

ejKe·r =

2jn

 

 

 

 

 

 

jn (Ksr) Pn (cos θ)

(C.1)

 

 

2

 

 

 

 

 

 

n =0

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 



C.1 Computation of the Integral Imm1 n

291

for Ke · r = Ksr cos θ, the orthogonality of the associated Legendre functions and the relation Ke · r0l = Ksz0l, we end up with

1,R

16πR3

 

2n + 1

 

 

 

jn

 

 

 

 

ejKsz0l δmm Fn (Ks, ks, R) ,

Imm n =

 

 

 

 

 

 

2

2

 

2

 

where

(ksR) (KsR)

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

(1)

 

Fn (Ks, ks, R) = (ksR) hn

(2ksR) jn (2KsR)

(KsR) h(1)n (2ksR) (jn (2KsR)) .

1,∞

To compute Imm n we use the stationary point method. Using the inequality Im{Ks} > 0 and the asymptotic expressions of the spherical Hankel functions of the first kind

h(1)n (ksr) =

and

d

(1)

 

 

 

hn

(ksr)

dr

(j)n +1ejksr

ksr

,

= j(j)n +1ejksr , r

1,∞

we see that Imm n vanishes as R→ ∞.

To evaluate Imm1,z n we pass to cylindrical coordinates, integrate over ϕ, and obtain

1,z

 

 

 

 

 

 

2π

 

(1)

 

 

 

 

 

 

 

 

Imm n

=

 

 

 

 

 

 

 

δmm

 

jKshn

(ksr) Pn (cos θ)

k2

K2

 

 

 

 

 

 

 

s

 

s

 

0

 

 

 

 

 

 

"

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

(1)

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

hn (ksr) Pn (cos θ)

 

 

 

 

ρ dρ.

 

 

 

∂z

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

z=−z0l

 

Further, using Kasterin’s representation [228, 258]

 

 

 

 

 

 

 

 

(1)

 

 

 

 

 

 

 

 

 

 

 

 

 

n

 

1

 

(1)

 

 

 

hn (ksr) Pn (cos θ) = (j)

Pn

 

 

 

 

 

 

h0 (ksr)

jks

∂z

with

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

(1)

(ksr) =

 

ejksr

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

h0

 

 

 

 

 

,

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

jksr

 

 

 

 

 

 

 

 

 

 

we find that

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

1,z

2π

 

 

 

 

 

 

n

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Imm n =

 

 

 

 

(j) δmm

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

k2

K

2

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

s

 

s

 

 

 

 

 

1 ejks

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

ρ2+z2

× 0

 

 

 

jKs

 

 

Pn

 

 

 

 

 

 

 

 

 

ρ dρ.

 

 

 

∂z

 

jks

∂z

 

jks

 

 

 

 

 

 

 

 

 

ρ2 + z2

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

z=−z0l