Page 40 - Electromagnetics

Subtracting again, we have

∂B 0 ∂D 0
E 0 · (∇× H 0 ) − H 0 · (∇× E 0 ) = H 0 · + E 0 · ,
∂t ∂t
hence

∂D 0 ∂B 0
−∇ · (E 0 × H 0 ) = E 0 · + H 0 ·
∂t ∂t
by (B.44). Integrating both sides throughout V and using the divergence theorem on the
left-hand side, we get

∂D 0 ∂B 0
− (E 0 × H 0 ) · dS = E 0 · + H 0 · dV.
S V ∂t ∂t
Breaking S into two arbitrary portions and using (B.6), we obtain

∂D 0 ∂B 0
E 0 · (ˆ n × H 0 ) dS − H 0 · (ˆ n × E 0 ) dS = E 0 · + H 0 · dV.
∂t ∂t
S 1 S 2 V
Now if ˆ n × E 0 = 0 or ˆ n × H 0 = 0 over all of S, or some combination of these conditions
holds over all of S, then

∂D 0 ∂B 0
E 0 · + H 0 · dV = 0. (2.13)
V ∂t ∂t
This expression implies a relationship between E 0 , D 0 , B 0 , and H 0 . Since V is arbitrary,
we see that one possibility is simply to have D 0 and B 0 constant with time. However,
since the fields are identically zero for t < 0, if they are constant for all time then those
constant values must be zero. Another possibility is to have one of each pair (E 0 , D 0 )
and (H 0 , B 0 ) equal to zero. Then, by (2.9) and (2.10), E 0 = 0 implies B 0 = 0, and
D 0 = 0 implies H 0 = 0.Thus E 1 = E 2 , B 1 = B 2 , and so on, and the solution is unique
throughout V . However, we cannot in general rule out more complicated relationships.
The number of possibilities depends on the additional constraints on the relationship
between E 0 , D 0 , B 0 , and H 0 that we must supply to describe the material supporting
the field — i.e., the constitutive relationships. For a simple medium described by the
time-constant permittivity
and permeability µ, (13) becomes

∂E 0 ∂H 0
E 0 ·
+ H 0 · µ dV = 0,
V ∂t ∂t
or
1 ∂
(
E 0 · E 0 + µH 0 · H 0 ) dV = 0.
2 ∂t V
Since the integrand is always positive or zero (and not constant with time, as mentioned
above), the only possible conclusion is that E 0 and H 0 must both be zero, and thus the
fields are unique.
When establishing more complicated constitutive relations, we must be careful to en-
sure that they lead to a unique solution, and that the condition for uniqueness is un-

= 0 implies that the tangential
derstood. In the case above, the assumption ˆ n × E 0
S
components of E 1 and E 2 are identical over S — that is, we must give specific values of
these quantities on S to ensure uniqueness. A similar statement holds for the condition

ˆ n × H 0 = 0. Requiring that constitutive relations lead to a unique solution is known
S
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