Page 189 - Maxwell House

SOLUTION OF BASIC EQUATIONS OF ELECTRODYNAMICS 169

In general, the answer is yes. Typically, both functions (in contrast to the source functions) are
continuous with continuous second derivatives. Besides, the wave equations they are satisfied
do not include the problematic source derivatives. To demonstrate it let us substitute (4.19) into
the second equation in (4.1) and group the terms as

2
1 1
2 + � (4.20)
∇ − − = − + ∙ � ∘ + 0
0
0
2
2
2
It looks quite alarming and more complicated than (4.10) and (4.11). But remember that we
have not yet specified how to choose the scalar function . Eventually, we can drastically
simplify the right-hand side of (4.20) if we impose Lorentz’s gauge assuming that
1
∘ + + = 0 (4.21)
0
2

Then our task is to solve the wave equation where only the source current is presented and

not differentiated
2
1
2
∇ − − = − (4.22)
0
0
2
2
Following in much the same way, we can define the scalar potential as the solution of
2
1
2
∇ − − = − (4.23)
0
2
2
It is worth to note that all the time derivatives in case of the time-independent EM fields (i.e.
electrostatic and magnetostatic) are equal to zero and
2
∇ = −
0
� (4.24)
2
∇ = −

The equations in (4.24) can be solved independently since the electric and magnetic fields get
decoupled (check (4.19)).
It may seem surprising at first, but it is unnecessary to define at all the scalar potential u solving
(4.23) except the case of time-independent EM fields. Indeed, once the vector potential has
been obtained as the solution of (4.22), the vector can be collected from the second equity in
(4.19) and as it will be done, the vector follows from the first equation of (4.1). It means we
won!
4.1.4 Are Vector and Scalar Potentials Real?
We can get some hint about the physical entity of these potentials looking at their unit
dimensions under macroscopic electrodynamics. According to (4.18), the scalar potential is
measured in [V] and as we have shown in Chapter 2 proportional to the energy per unit charge
that can be taken from or given to electric fields. The unit dimension of the vector potential
2
(see (4.19)) is [m∙kg∙ s −2 ∙ A −1 = Tesla ∙ m = W/(A/m )]. If so, according to (1.46) this
vector is proportional to the energy per unit element of volume current that can be taken from
or given to magnetic fields. The general circumstances are more complicated because the time-
varying magnetic potential in (4.19) is straight connected to the electric field and, thus, to its