it x E 1 = it x E itxH1=itxH in .NET

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The time-averaged power absorbed by the medium 1 is
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Since (3.5.47) it follows that
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1 M"2\7. [Re(E I x HI)
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By using the divergence theorem, (3.5.48) becomes the sum of a surface integral on 5 and a hemispherical surface integral on 5 100 , Since the medium 1 is attenuative, the hemispherical surface integral on Sloo vanishes. Thus Pa =
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The second equality in (3.5.50) is a result of boundary conditions of (3.5.44) and (3.5.45). Now E = E i + E s , H = Hi + H s . Hence
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~Re ~ dB
It is clear that
-* = -AocosOo (3.5.52) Hi 2"7 which is the power impinging upon the half-space medium. In (3.5.52) A o is the area of the half-space medium illuminated by the incident beam. By making use of the divergence theorem and the fact that the upper half-space is a source field, we can convert the last surface integral in (3.5.51) to be that at infinity:
dS E i
dB . E s x H:
-2 Re Js~ dS E s
r --
-* xHs
r = - Jo
r/ d<jJ Jo dO
IEs l2 r 2
= - 47f
fo27r d<jJs fo7r/2 dOs sin Os A
cos 0 2 0
'Y(3Qb (B, Bob)
where 'Y(3Qb (B, Sob) is the bistatic scattering coefficient defined for active remote sensing with incident polarization ab. We note from (3.5.39) that
Hi = !(SOb x ab)eiksob':;:
Thus -
dB (E i x H:
+ E s x H;)
dSe-iksob':;: [ab' n x HsCr ) -
~SOb x ab . n x Es(r)]
Next we make use of the lower part of (2.2.25). We dot that equation with as, which is a vector that lies in the plane perpendicular to k s . Then from (2.2.25), for ks pointing into the lower half-space of region 1 we obtain
dSe- ik...:;: {as
n x Hs(r) - ~ks x as . n x
Since Sob of (3.5.54) and (3.5.55) points into region 1, comparison of (3.5.55) and (3.5.56) gives -!Re dB (E i x H: + E s x H;) 2 Using (3.5.51)-(3.5.57) and (3.5.41), we have
r Js
5.4 Emissivity of Four Stokes Parameters
r iV drwE~(1')IElI2
1 = A 0 cosO 0 { 1 - -4 2 "7 1f (3=v,h
= 81f 2 lim
1 27r d<ps 17r/2 dOs'Y(3aCs, Sob) }
r; iV drwE~(1')IGlO(1', 1'0) . it b 2 r
Thus we obtain the relation
. 11m
161f ; A "7 0 0 cos 0
1 "( ) ( 1=
r 41f L io (3=v,h
drWEl l' GlO 1',1'0 . D:b
= 1-
r/ io
dOs'Y(3a(S, Sob)
The left-hand side of (3.5.59) is in a form similar to the right-hand side of the brightness temperature equation (3.5.38). Let
V(Sob) = v(so) (3.5.60) Then for the case of uniform temperature T, comparing (3.5.59) with (3.5.38) gives TBv(So) = T 1 - - 1 [ 41f (3=v,h
Similarly, by setting it to be
7r 2 / dOs sinOs'Y(3v(S, Sob) ]
we get
/ 1 (3.5.63) TBh(So) = T 1 - 41f d<ps dOs sin Os'Y(3h(S, Sob) ] [ (3=v,h Equations (3.5.61) and (3.5.63) are identical to the results of Kirchhoff's law and the results that were derived by Peake [1959J. Next we let it = P polarization, which is a linear polarization with 45 with respect to vertical polarization and horizontal polarization.
r L io
r io
7r 2
~(V(SOb) + h(Sob)) = ~(V(so) -
Then We let it =
R polarization which is right-hand circular polarized. R = ~(V(SOb) + ih(sob)) = ~(v(so) - ih(so)) (3.5.65)
pi 2 =
"2(V(SO) - h(so)) . G lO . G lO . (v(so) - h(so))
(3.5.66a) (3.5.66b)
IGlO . RI =
~(V(So) -
ih(so)) . ~o . ~o . (V(SO)
+ ih(so))
Using (3.5.38), (3.5.59), (3.5.66a), and (3.5.66b) gives
UB(So) = TBv(So)
- 2T{ 1 =
1 1 ~ I: 1 1
4~ L
+ TBh(So)
7r 2 /
d()s sin ()s'Y(3p(s, SOb)}
7r 2 /
dOs sin Os
[2'Y(3p( S, Sob) - 'Y(3v (s, Sob) - 'Y(3h (s, SOb)] VB(So) = TBv(So)
- 2T{ 1 =
1 1 ~ L 1 1
4~ L
+ TBh(So)
7r 2 /
dOs sin Os'Y;3R(S, SOb)}
dOs sin () s
[2'Y(3R(S, Sob) - 'Y;3v(S, Sob) - 'Y(3h(S, SOb)]
Equations (3.5.67) and (3.5.68) give the relations of the emissivities of third and fourth Stokes parameters in terms of bistatic scattering coefficients. It is interesting to further examine the integrand in (3.5.67) and (3.5.68) of the third and fourth Stokes parameter brightness temperatures. If we take scattering amplitude dyad to be given by