Page 178 - Electromagnetics

By virtue of (B.44)we have

[(∇× Q) · (∇× P) − P · (∇ × {∇ × Q})] dV = [P × (∇× Q)] · dS.
V S
We now interchange P and Q and subtract the result from the above, obtaining

[Q · (∇ × {∇ × P}) − P · (∇ × {∇ × Q})] dV =
V

[P × (∇× Q) − Q × (∇× P)] · dS. (3.162)
S
Note that ˆ n points outward from V . This is Green’s second identity for vector fields.
Now assume that V contains a magnetic material of uniform permeability µ and set
c

P = A(r ), Q = ,
R
in (3.162)written in terms of primed coordinates. Here c is a constant vector, nonzero
but otherwise arbitrary. We first examine the volume integral terms. Note that
c c c
!
2
∇ × (∇ × Q) =∇ × ∇ × =−∇ +∇ ∇ · .
R R R
By (B.162)and (3.58)we have
1
c
1
1 1
2 2 2 2
∇ = ∇ c + c∇ + 2 ∇ ·∇ c = c∇ =−c4πδ(r − r ),
R R R R R
hence
c
!

P · [∇ × (∇ × Q)] = 4πc · Aδ(r − r ) + A ·∇ ∇ · .
R
Since ∇· A = 0 the second term on the right-hand side can be rewritten using (B.42):

∇ · (ψA) = A · (∇ ψ) + ψ∇ · A = A · (∇ ψ).
Thus
1
" #

P · [∇ × (∇ × Q)] = 4πc · Aδ(r − r ) +∇ · A c ·∇ ,
R
where we have again used (B.42). The other volume integral term can be found by
substituting from (3.129):
1

Q · [∇ × (∇ × P)] = µ c · J(r ).
R
Next we investigate the surface integral terms. Consider
c
$ !%

ˆ n · P × (∇ × Q) = ˆ n · A × ∇ ×
R
1 1
" #

= ˆ n · A × ∇ × c − c ×∇
R R
1
" #

=−ˆ n · A × c ×∇ .
R
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