H y ¼
ffiffi ffi
2
p
E
μv
cos ωt cos kz þ C,
where v is a light velocity in the dielectric medium and C is an integration constant.
Removing C and putting
H
E
μv
,
we have
H y ¼
ffiffi ffi
2
p
H cos ωt cos kz:
Using a vector expression, we have
H ¼ e 2
ffiffi ffi
2
p
H cos ωt cos kz:
Thus, E (ke 1 ), H (ke 2 ), and n (ke 3 ) form the right-handed system in this order. As
noted in Sect. 8.8, at the interface (or wall) the electric field and magnetic field form
nodes and antinodes, respectively. Namely, the two fields are in-quadrature.
Let us calculate electromagnetic energy of the dielectric medium within a cavity.
In the present case the cavity is meant as the dielectric sandwiched by a couple of
metal layer. We have
W ¼ W e þ W m ,
ð9:142Þ
where W is the total electromagnetic energy; W e and W m are electric and magnetic
energies, respectively. Let the length of the cavity be L. Then the energy per unit
cross-section area is described as
W e ¼
ε
2
Z L
0
E
2 dz and W m ¼
μ
2
Z L
0
H
2 dz:
ð9:143Þ
Performing integration, we get
W e ¼
ε
2
LE
2 sin
2
ωt,
W m ¼
μ
2
LH
2 cos
2
ωt ¼
μ
2
L
E
μv
2
cos
2
ωt ¼
ε
2
LE
2 cos
2
ωt,
ð9:144Þ
where we used 1/v
2
¼ εμ with the last equality. Thus, we have
376
9 Light Quanta: Radiation and Absorption