Page 370 - Electromagnetics

leaves the expressions (5.17) and (5.18) unchanged. This is called a gauge transformation,
and the choice of a certain alters the specification of ∇· A e . Thus we may begin with
the Coulomb gauge as our baseline, and allowany alteration of A e according to (5.20)
2
as long as we augment ∇· A e by ∇· ∇ =∇ .
Once ∇· A e is specified, the relationship between the potentials and the current J
can be found by substitution of (5.17) and (5.18) into Ampere’s law. At this point
we assume media that are linear, homogeneous, isotropic, and described by the time-
i
invariant parameters µ, , and σ. Writing J = J + σE we have
2
1 i ∂A e ∂ A e ∂
∇× (∇× A e ) = J − σ − σ∇φ e − − ∇φ e . (5.22)
µ ∂t ∂t 2 ∂t
Taking the divergence of both sides of (5.22) we get
∂ ∂ 2 ∂
i
0 =∇ · J − σ ∇· A − σ∇· ∇φ e − ∇· A e − ∇· ∇φ e . (5.23)
∂t ∂t 2 ∂t
Then, by substitution from the continuity equation and use of (5.19) along with ∇·∇φ e =
2
∇ φ e we obtain
∂ i 2 2
ρ + ∇ φ e =−σ∇ φ e .
∂t
For a lossless medium this reduces to
2
i
∇ φ e =−ρ / (5.24)
and we have
i
ρ (r , t)

φ e (r, t) = dV . (5.25)
V 4π R
We can obtain an equation for A e by expanding the left-hand side of (5.22) to get
2 ∂
i
2
∇ (∇· A e ) −∇ A e = µJ − σµ ∂A e − σµ∇φ e − µ ∂ A e − µ ∇φ e , (5.26)
∂t ∂t 2 ∂t
hence
2 ∂
2
i
∇ A e − µ ∂ A e =−µJ + σµ ∂A e + σµ∇φ e + µ ∇φ e
∂t 2 ∂t ∂t
under the Coulomb gauge. For lossless media this becomes
2
2 ∂ A e i ∂
∇ A e − µ =−µJ + µ ∇φ e . (5.27)
∂t 2 ∂t
Observe that the left-hand side of (5.27) is solenoidal (since the Laplacian term came
from the curl-curl, and ∇· A e = 0), while the right-hand side contains a general vector
i
field J and a lamellar term. We might expect the ∇φ e term to cancel the lamellar
i
portion of J , and this does happen [91]. By (5.12) and the continuity equation we can
write the lamellar component of the current as
∇ · J (r , t) ∂ ρ (r , t) ∂
i i
i

J (r, t) =−∇ dV = ∇ dV = ∇φ e .
l
V 4π R ∂t V 4π R ∂t
Thus (5.27) becomes
2
∂ A e
i
2
∇ A e − µ =−µJ . (5.28)
s
∂t 2
© 2001 by CRC Press LLC