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SOLUTION OF BASIC EQUATIONS OF ELECTRODYNAMICS 167

(, )
However, in general, it is known and defined by the given sources of electric current

only. Finally, the vector equation (4.6) can be written as
2 (,)
1 (,)
2
∇ (, ) − − = (, ) (4.11)
0
2 2
In a similar manner, we obtain applying the ( x) operation to both parts of the second equation
in (4.1)
2
1 (,) (,)
2 (, ) (4.12)
∇ (, ) − − = − ×
0
2 2
The vector equations (4.11) and (4.12) are customarily called the wave equations. Note that
each of them is equivalent to three (not six) scalar equations.
4.1.2 Wave Equation in Space-Frequency Domain. Wavenumber.
In Chapter 1 we showed that the time-frequency conversion comes to the replacement of the
differentiation operator with the factor . Therefore, the second derivative in time domain
must be equivalent to the factor ∙ = − . Therefore, the wave equations (4.11) and
2
(4.12) can be written as

2
2
∇ (, ) + ( − )(, ) = (, ) � (4.13)
0
2
2
∇ (, ) + ( − )(, ) = − × (, )
0

Here = [(rad/s) ∙ (s/m) = rad/m] is called the wavenumber or propagation number and
⁄
one of the fundamental physical parameters characterizing the energy of EM waves. Indeed, let
us recall the EM wave duality discussed shortly in Chapter 1 and 3. According to this principle,
EM wave can be considered as a flow of N photons each carrying the discrete portion of
⁄
energy = = ℎ. In the vacuum = = 2/ or = /2. Then the single
ℎ
photon energy is = � � = 3.1615∙ 10 −26 [Joule]. Thereby, the energy carried by

2
whole EM wave is equal to = 3.1615 ∙ 10 −26 () and proportional to its wavenumber
k. If so, the EM energy can be directly measured in wavelengths at any frequency as it is
customary in the optical band.
A quick look at (4.13) reveals that the complex number called the complex propagation
constant and denoted by symbol
= � − = − (4.14)
�
2
1
0
2
can be interpreted as the complex wavenumber in the lossy medium ( ≠ 0). Since is a
�
complex number, EM waves should experience loss when they propagate. The loss level is
defined by the attenuation coefficient . Evidently, in (4.14)
2
2
2
4
√2 = �� + ( ) + ⎫
1
0
⎪
(4.15)
⎬
4
2
√2 = �� + ( ) − 2⎪
0
2
⎭