12—Tensors 304I am seeking a wave solution for the field, so assume a solutionE~
(
~r,t
)
=E~ 0 ei~k.~r−ωt. Each∇brings
down a factor ofi~kand each time derivative a factor−iω, so the equation is
−~k
(~
k.E~ 0
)
+k^2 E~ 0 =μ 0 ω^2 α
(~
E 0
)
+μ 0 0 ω^2 E~ 0 (12.26)
This is a linear equation for the vectorE~ 0.
A special case first: A vacuum, so there is no medium andα≡ 0. The equation is
(
k^2 −μ 0 0 ω^2
)~
E 0 −~k
(~
k.E~ 0
)
= 0 (12.27)
Pick a basis so thatzˆis along~k, then
(
k^2 −μ 0 0 ω^2
)~
E 0 −z kˆ^2 zˆ.E~ 0 = 0 or in matrix notation,
(
k^2 −μ 0 0 ω^2
)
1 0 0
0 1 0
0 0 1
−k^2
0 0 0
0 0 0
0 0 1
E 0 x
E 0 y
E 0 z
This is a set of linear homogeneous equations for the components of the electric field. One solution
forE~is identically zero, and this solution is uniqueunlessthe determinant of the coefficients vanishes.
That is (
k^2 −μ 0 0 ω^2
) 2 (
−μ 0 0 ω^2
)
= 0
This is a cubic equation forω^2. One root is zero, the other two areω^2 =k^2 /μ 0 0. The eigenvector
corresponding to the zero root has components the column matrix( 0 0 1 ), orE 0 x= 0andE 0 y= 0
with thez-component arbitrary. The field for this solution is
E~=E 0 zeikzz,ˆ then ∇.E~=ρ/ 0 =ikE 0 zeikz
This is a static charge density, not a wave, so look to the other solutions. They are
E 0 z= 0, withE 0 x, E 0 yarbitrary and~k.E~ 0 =∇.E~= 0
E~=(E 0 xxˆ+E 0 yyˆ)eikz−iωt and ω^2 /k^2 = 1/μ 0 0 =c^2
This is a plane, transversely polarized, electromagnetic wave moving in thez-direction at velocity
ω/k= 1/
√
μ 0 0.
Now return to the case in whichα 6 = 0. If the polarizability is a multiple of the identity, so that
the dipole moment density is always along the direction ofE~, all this does is to add a constant to 0 in
Eq. (12.26). 0 → 0 +α=, and the speed of the wave changes fromc= 1/
√
μ 0 0 tov= 1/
√
μ 0
The more complicated case occurs whenαis more than just multiplication by a scalar. If the
medium is a crystal in which the charges can be polarized more easily in some directions than in others,
αis a tensor. It is a symmetric tensor, though I won’t prove that here. The proof involves looking at
energy dissipation (rather, lack of it) as the electric field is varied. To compute I will pick a basis, and