Absorption Cross Section

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The power absorbed by a small particle is

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(1.2.16)

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The absorption cross section is

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...!...1E-1 2 211

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(1.2.17)

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For the case of a sphere, putting (1.2.6) and (1.2.16) in (1.2.17) gives

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= k E~ 41l" 3 E

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2 3E

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(1.2.18)

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If we take the ratio of scattering cross section to absorption cross section, we find

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eT - -_ , - , - - - - 1 - 2(ka)3\ E - E s

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(1.2.19)

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Thus the relative magnitude of (ka) and E~/E.

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eT s

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2.3 Rayleigh Scattering by an Ellipsoid

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Let a, b, and c be the half axes length of the ellipsoid respectively in Xb, fib, and Zb directions where a, b, and c are all much less than >.... Then the internal electric field Eint induced by the incident electric field E i is

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( Xb' E-) Eint = Xb 1 + Vd~a

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+ Yb

fib' 1 + Vd~b

E.) + Zb (Zb .Vd~C ) E1+

(1.2.20)

where (1.2.21)

2.3 Rayleigh Scattering by an Ellipsoid

= {Ooo

io io

(s + a2 ) [(s

+ a2 )(s + b2 )(s + c2 )J2

(1.2.22) (1.2.23) (1.2.24)

A b = {Ooo A c = {Ooo

+ b2 ) [(s + a2 )(s + b2 )(s + c2 )J2 + a2 )(s + b2 )(s + c2 )J2

(s + c2 ) [(s

The integral of the sum of A a, A b , and A c can be performed analytically to give

2 A a + A b + A c = -b (1.2.25) a c For spheroids, a = b and the integrals in A a , A b , and A c can be integrated analytically. The axis Zb is the axis of symmetry, and

Aa = Ab =

~ (a;e - Ac)

(1.2.26)

For prolate spheroids, c > a = band

1 1l = - 3"3 ( 2e + In - - ee

(1.2.27)

where (1.2.28) is the eccentricity. For oblate spheroids, e < a and

Ac =

where

c3 f2

( 1 - 1 tan

-1) f

(1.2.29)

(1.2.30) For a thin disk, we can treat it as a special case of the oblate spheroid by letting a c in (1.2.29)-(1.2.30). Then,

f~- l

(1.2.31)

2 2 Ac = - - = 2 3

e j2

(1.2.32) (1.2.33)

= Ab = 0

1 ELECTROMAGNETIC SCATTERING BY SINGLE PARTICLE

Substituting (1.2.32)-(1.2.33) in (1.2.20) gives

E int = Xb(Xb . Ed + Yb(Yb . E i ) + Zb(Zb . Ei)~

(1.2.34)

The interpretation of (1.2.34) is that the tangential component of the incident electric field penetrates into the thin disk while the normal component of the incident electric field is changed by a factor of E/ Ep when it penetrates into the thin disk.

(1.2.36) Hence

(1.2.37)

Since

(1.2.38)

it follows from (1.2.37) and (1.2.38) that

fvv =

Vs . F . Vi

(1.2.39a)

(1.2.39b)

fvh = Vs . F . hi

fhv =

fhh =

hs . F . Vi hs . F . hi

= 47rI fvv I

(1.2.39c) (1.2.39d)

The bistatic scattering cross sections are defined by

(J'vv(k s , k i )

(1.2.40a)

2.4 Scattering Dyads

tJvh(k s , k i ) =

47rI fvh I

(1.2.40b) (1.2.40c) (1.2.40d)

tJhv(k s , k i ) = 4 7r lfhvl tJhh(k s , k i ) = 47r lfhhl

In the backscattering direction (monostatic radar)

ks = -ki

so that

(1.2.41) (1.2.42)

and (1.2.43) Note that there is no negative sign in (1.2.43) while there are negative signs in (1.2.41) and (1.2.42).

Example 1:

For spheres,

a; (E

= Ab = Ac =

1) where k(,;3) so that 1

Ep/E

is the relative permittivity.

(1.2.44)

This gives

= fo(:EbXb + YbYb + ZbZb) = f ol

(1.2.45)

In backscattering direction

fhv fhh

= Vs . f ol. Vi = fa = Vs . fa! . hi = 0 = hs . fa! Vi = 0 = hs . f 1. hi = 0

(1.2.46a) (1.2.46b) (1.2.46c)

(1.2.46d)

Thus

Evs) 0 ( E = (fa

0) (EVi)

E hi

(1.2.47) (1.2.48)

tJvh

= (Thv =

(1.2.49)

1 ELECTROMAGNETIC SCATTERING BY SINGLE PARTICLE

Thus a characteristic of backscattering from a sphere is that (Tvv = (Thh and there is no cross-polarization. If the incident wave is right-hand circularly polarized, then E i = Vi +ihi so that E vi = 1 and Ehi = i. Substituting into (1.2.47) gives E vs = fa and Ehs = -ifa so that E s = fo(v s - ih s ). The scattered wave is left-hand circularly polarized.

Example 2:

For the case of thin disks

Ac =

then

2 and A a a c

(1.2.50)

where (1.2.51) Let ki = :h and ks = -Xb for backscattering. Also let Vi = Zb = vs, hi = fib, and hs = -fib' Then fvv = fD.!:... and Ihh = - fD; fhv = fvh = O. If lOp 10, tp then (Tvv (Thh. Thus the polarization dependence of the backscattering cross section contains information on the shape of the object.

Integral Representations of Scattering and Born Approximation

3.1 Integral Expression for Scattering Amplitude

In this section, we derive the integral expression for the exact scattering amplitude. Recall from equation (1.2.4) that the far-field scattered field from a small particle is

_ k 2 eikr E s = --4-Vo(Ep 'TrIOr

A A _

k s x (k s X E int )

(1.3.1)

Consider a scatterer of arbitrary size, shape, and inhomogeneous permittivity Ep(1'), as shown in Fig. 1.3.1. Let dv' be a differential volume around point 1", which has permittivity Ep(1"). Then the differential contribution to

3.1 Integral Expression for Scattering Amplitude

Figure 1.3.1 Scatterer of finite size and arbitrary shape. The permittivity of the scatterer is Ep(r'), and;pl is a point within the particle. The observation direction is k8 and the distance between the point 1" and the observation point Pis R = IF - 1"1.

the far-field scattered field dEs from dv' is

' -I] dEs = - k e 47fER k s x [' x (-I - E) ks Ep(r) Eint(r) dv ,

where R = 11' - 1"1 is the distance between 1" and observation point observation point is far away, then

2 ikR

(1.3.2)

r. If the

(1.3.3)

R r:::: r - l' . ks

We approximate the phase term in (1.3.2) by

(1.3.4)

and the amplitude by

"-J -

2 ikr

(1.3.5)

Making the far-field approximation on the scattered field and integrating over the volume of the scatterer gives

- = - k47fEr ksx [, X e , Es ks

fff -'] JJJ dx'dy'dz'(Ep(r') - E)Eint(r')e- 'kk ., r

(1.3.6)

The expression in (1.3.6) is the exact far-field scattering amplitude for particles of arbitrary size and shape and inhomogeneous permittivity. However, the internal field Eint(r') is an unknown quantity. To calculate rigorously Eint(r'), one has to solve Maxwell's equations. For example, Eint(r') is dependent on the coherent wave interaction among different parts of the particle.