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long as all fields are expressed in the frame of the moving surface. We can also derive
boundary conditions for a deforming surface moving with arbitrary velocity by using
equations (2.177)–(2.178). In this case all fields are expressed in the laboratory frame.
Proceeding through the same set of steps that gave us (2.194)–(2.197), we find
ˆ n 12 × (H 1 − H 2 ) + (ˆ n 12 · v)(D 1 − D 2 ) = J s , (2.220)
ˆ n 12 × (E 1 − E 2 ) − (ˆ n 12 · v)(B 1 − B 2 ) =−J ms , (2.221)
ˆ n 12 · (D 1 − D 2 ) = ρ s , (2.222)
ˆ n 12 · (B 1 − B 2 ) = ρ ms . (2.223)
Note that when ˆ n 12 · v = 0 these boundary conditions reduce to those for a stationary
surface. This occurs not only when v = 0 but also when the velocity is parallel to the
surface.
The reader must be wary when employing (2.220)–(2.223). Since the fields are mea-
sured in the laboratory frame, if the constitutive relations are substituted into the bound-
ary conditions they must also be represented in the laboratory frame. It is probable that
the material parameters would be known in the rest frame of the material, in which case
a conversion to the laboratory frame would be necessary.
2.9 Fundamental theorems
In this section we shall consider some of the important theorems of electromagnetics
that pertain directly to Maxwell’s equations. They may be derived without reference to
the solutions of Maxwell’s equations, and are not connected with any specialization of
the equations or any specific application or geometrical configuration. In this sense these
theorems are fundamental to the study of electromagnetics.
2.9.1 Linearity
Recall that a mathematical operator L is linear if
L(α 1 f 1 + α 2 f 2 ) = α 1 L( f 1 ) + α 2 L( f 2 )
holds for any two functions f 1,2 in the domain of L and any two scalar constants α 1,2 .A
standard observation regarding the equation
L( f ) = s, (2.224)
where L is a linear operator and s is a given forcing function, is that if f 1 and f 2 are
solutions to
L( f 1 ) = s 1 , L( f 2 ) = s 2 , (2.225)
respectively, and
s = s 1 + s 2 , (2.226)
then
f = f 1 + f 2 (2.227)
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