Page 24 - Electrical Properties of Materials

Electromagnetic waves in solids 7

Second, we shall express the current density in terms of the electric field as

J = σE E E . (1.24)

It would now be a little more elegant to perform all the calculations in vector
form, but then you would need to know a few vector identities, and tensors
(quite simple ones, actually) would also appear. If we use coordinates instead,
it will make the treatment a little lengthier, but not too clumsy if we consider
only the one-dimensional case, when

∂ ∂
=0, = 0. (1.25)
∂x ∂y

Assuming that the electric field has only a component in the x-direction (see
the coordinate system in Fig. 1.3), then

e x e y e z

∂ ∂E x

∇ × E E E = 00 = e y , (1.26)

∂z
∂z
E x 00

where e x , e y , e z are the unit vectors. It may be seen from this equation that
the magnetic field can have only a y-component. Thus, eqn (1.23) takes the
simple form
∂E x ∂B y
=– . (1.27)
∂z ∂t
We need further
∗
We have here come face to face with a
e x e y e z
dispute that has raged between physicists

∂ ∂B y and engineers for ages. For some odd
∇ × B = 00 = e x , (1.28)

∂z
reason the physicists (aided and abet-
∂z ted by mathematicians) use the symbol
0 B y 0 √
ifor –1 and the exponent –i(ωt – kz)
to describe a wave travelling in the z-
which, combined with eqn (1.24), brings eqn (1.22) to the scalar form direction. The engineers’ notation is j for
√
–1 and j(ωt – kz) for the exponent. In
this course we have, rather reluctantly,
∂B y ∂E x accepted the physicists’ notations so as
– = μσE x + μ . (1.29)
∂z ∂t not to confuse you further when reading
books on quantum mechanics.
Thus, we have two fairly simple differential equations to solve. We shall
attempt the solution in the form ∗ ω represents frequency, and k is
the wavenumber.
exp {–i(ωt – kz)} (1.30)
E x = E x 0
and
exp {–i(ωt – kz)} . (1.31)
B y = B y 0