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(1.5.8)
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Balancing Eq. (1.5.8) to first order gives
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+ VPI = 0
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(1.5.9)
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From (1.5.3), we have
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V . [(Po + PI) V d + %t
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Balancing Eq. (1.5.10) to first order yields
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+ PI)
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(1.5.10)
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V 'PoV I + -
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(1.5.11)
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Use of the constitutive equation (1.5.7) in (1.5.11) gives
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(1.5.12)
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1 ELECTROMAGNETIC SCATTERING BY SINGLE PARTICLE
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From (1.5.9) and (1.5.12) we can form the wave equation for PI, with wave velocity c: - V PI
+ c2 at 2
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1 8 2pI
(1.5.13)
The boundary conditions at an interface between two media are: (i) continuity of pressure PI and (ii) continuity of the normal component of velocity VI.
Scattering by Spheres, Cylinders, and Disks
Mie Scattering
Mie scattering is scattering of electromagnetic waves by a sphere of radius
a and permittivity Ep [van de Hulst, 1957; Bohren and Huffman, 1983]. Let kp = WVI-"E p and the sphere be centered at the origin (Fig. 1.6.1).
Because of the symmetry of a sphere, it is convenient to use the scattering plane orthonormal system to express the scattering amplitudes. To get
we let the incident wave be along the direction, so that ki = Z. Also let the observation direction ks be in the y-z plane, with <P = 90 ; then (1.6.1) 8=0
L = is = x =-4>
2i = y
_ E t. - eteikz - _ ~ L..J
(1.6.2) (1.6.3)
is =
(1.6.4)
The incident field is given by, from (1.4.68),
'n (2n + 1) L..J z ( 1) n n+ n=I==-I,I
[e i ' C-=n(O,O)R9 M =n (k r, e"')
'Y=n
''f'
ei . B_=n(O, 0) RgN=n(kr, 0, <P)]
(1.6.5)
'Y=n To solve the boundary value problem, we let the scattered field be
E s - _ ~ '" L..J L...J
n=I==-I,I
n (2n + 1) [_ ei . C-=n(O, 0) b M (k 0 "') ( 1) n =n r, ,'f' n n+ 'Y=n
6.1 Mie Scattering
Figure 1.6.1 Geometry of a sphere with radius a and permittivity propagates in the +z direction.
The incident wave
"(mn and let the internal field be
E mt
.ei "')] + z . B-mn(O, 0) an N mn (k r, 0,'I'
(1.6.6)
' " .n(2n+1) [ei.C-mn(O,O)dRM (k 0"') L.J z n(n + 1) mn n 9 mn pr, ,'I'
n=l m=-l,l
"(mn Note that the internal field satisfies the vector wave equation with wavenumber kp The boundary conditions are the continuity of n x E and n x H which is proportional to it x V x E. At r = a Balancing the
_ z.ei ILmn (O, 0) en R9 N mn (kpr, () ,'I' "')]
(1.6.7)
r x (E i + E s ) = r x E int and r x N components gives,
(1.6.8)
respectively,
(1.6.9) (1.6.10)
jn(ka) - bnhn(ka) = dnjn(kpa) [kajn(ka)]' [kahn(ka)]' [kpajn(kpa)] , -'------'- - an = en ka ka kpa Next we match at r = a r x (V x E i + V x E s ) = r x V X E int
(1.6.11)
1 ELECTROMAGNETIC SCATTERING BY SINGLE PARTICLE
We have V x M = kN and V x N = kM with k replaced by k p for the internal field. Thus we have, using (1.6.11) and (1.6.5)-(1.6.7),
[kajn(ka)l' - bn [kahn(ka)l' = dn [kpajn(kpa)l' kjn(ka) - ankhn(ka) = enkpjn(kpa)
(1.6.12) (1.6.13)
Solving (1.6.9) and (1.6.12) gives bn and dn . Solving (1.6.10) and (1.6.13) gives an and Cn. We find
(1.6.14a)
(1.6.15a)
where
= _ jn(kpa) [kajn(ka)]' - jn(ka) [kpajn(kpa)]'
jn(kpa)[kahn(ka) ], - hn ( ka)[kpajn(kpa)]'
() 1.6.15b
We can make use of the Wronskian [Abramowitz and Stegun, 1965]
jn(z)h~(z) - hn(z)j~(z) = i[jn(Z)Yn-l(Z) - Yn(Z)jn-l(Z)] = 2"
(1.6.16)
where Yn(z) is the spherical Neumann function. Then
n C =
k~a2jn(kpa)[kahn(ka)]' _
ikpa k2a2hn(ka)[kpajn(kpa)]'
( 1.6.17a
i 1 dn = ka jn (kpa)[kah n (ka)]' _ hn(ka)[kpajn(kpa)]'
(1.6.17b)
From (1.4.56)-(1.4.57) and (1.6.6), in the far field
_ ie ikr E s = -,;;:-
(2n + 1) { n(n+ 1) bn [ei . C-mn(O, 0)] Crnn(O,</J)
A _ _
n=lm=-l,l
+ an [ei . B-mn(O, 0)] Bmn(O, </J)}
Substituting (1.4.72)-(1.4.75) in (1.6.18), we have, summing over m in (1.6.18),
(1.6.18)
= 1,-1
Es =
-~ L
. ikr
t+ ) +
1) {
an [l\ (x - iy)] [ihn(cosO)
+ ,i7Tn(cosO)]eic/>
6.1 Mie Scattering
- an [ei' (5: + iy)]
+ ibn rei . (x -
[0 Tn (COS 0) - i1Tn(cosO)]e-iq, iy)] [0 i1Tn (cos 0) - Tn (cos 0)] eiq,
(1.6.19)
+ ibn [e; . (x + iy)) [Ii i~n( COS 9) + ~ Tn(cos 9)] e -i<I }
To get 111(8) and 121(8), we follow equations (1.6.1)-(1.6.4) and set ei = x, <P = 90 0 , so that ii = is = x = - , and 2s = O. Substituting these in (1.6.19), we find
_ ie ikr ~ E s = ---y;;:-<P
(2n + 1) n(n + 1) [an1Tn(COS 0)
+ bnTn(COS 0)]
(1.6.20)
Thus, since 0 = 8, we have from (1.6.20)
(1.6.21) (1.6.22)
where
(1.6.23)
To get fI2(8) and 122(8), we set ei = and 2s = O. This time (1.6.19) yields
y, <P = 90
so that
2i = y, is = - ,
(1.6.24)
_ .eikr ~ 00 (2n + 1) E s = z-k 0 [anTn(COSO) r n=l n n + 1
L ( )
+ bn1Tn (cosO)]
Thus
fI2(8) = 0
(1.6.25) (1.6.26)
122(8)