Page 474 - Electromagnetics

Here G(r|r ), called the “static Green’s function,” represents the potential at point r

produced by a unit point source at point r .
In Chapter 3 we find that G(r|r ) = 1/4π|r − r |. In a variety of problems it is also

useful to have G written in terms of an inverse Fourier transform over the variables x
r
and y. Letting G form a three-dimensional Fourier transform pair with G, we can write
1 ∞
r jk x x jk y y jk z z
G(r|r ) = 3 G (k x , k y , k z |r )e e e dk x dk y dk z .
(2π)
−∞
Substitution into (A.51) along with the inverse transformation representation for the
delta function (A.4) gives
1 2 ∞ r jk x x jk y y jk z z

∇ G (k x , k y , k z |r )e e e dk x dk y dk z
(2π) 3
−∞
1 ∞ jk x (x−x ) jk y (y−y ) jk z (z−z )

=− e e e dk x dk y dk z .
(2π) 3 −∞
We then combine the integrands and move the Laplacian operator through the integral
to obtain
1 ∞ 2 r jk·r jk·(r−r ) 3

∇ G (k|r )e + e d k = 0,
(2π) 3
−∞
where k = ˆ xk x + ˆ yk y + ˆ zk z . Carrying out the derivatives,
1 ∞ 2 2 2 r − jk·r jk·r 3

−k − k − k z G (k|r ) + e e d k = 0.
y
x
(2π) 3
−∞
2
2
2
Letting k + k = k and invoking the Fourier integral theorem we get the algebraic
x y ρ
equation
2 2 r − jk·r

−k − k G (k|r ) + e = 0,
ρ z
r
which we can easily solve for G :
e − jk·r
r
G (k|r ) = 2 2 . (A.52)
k + k
ρ z
Equation (A.52) gives us a 3-D transform representation for the Green’s function.
Since we desire the 2-D representation, we shall have to perform the inverse transform
over k z . Writing
1 ∞
xy r jk z z

G (k x , k y , z|r ) = G (k x , k y , k z |r )e dk z
2π
−∞
we have

1 ∞ e − jk x x e − jk y y e jk z (z−z )
xy
G (k x , k y , z|r ) = dk z . (A.53)
2
2π k + k 2
−∞ ρ z
To compute this integral, we let k z be a complex variable and consider a closed contour in
the complex plane, consisting of a semicircle and the real axis. As previously discussed,
we compute the principal value integral as the semicircle radius →∞, and find that
the contribution along the semicircle reduces to zero. Hence we can use Cauchy’s residue
theorem (A.14) to obtain the real-line integral:

1 e − jk x x e − jk y y e jk z (z−z )
xy

G (k x , k y , z|r ) = 2π j res .
2
2π k + k 2
ρ z
© 2001 by CRC Press LLC