Page 188 - Maxwell House

168 Chapter 4

4.1.3 Electrodynamic Potentials in Space-Time Domain

In principle, the wave equations (4.13) we have obtained straight from Maxwell’s equations
can be solved analytically or numerically. The main inconvenience is that the vector functions
defining the sources distribution on the right-hand side of these equations not only are relatively
complicated but according to (4.9), (4.10) and (4.12) they should be differentiated in space- and
time-domain. Taking into account that these sources are typically concentrated in small
1
volumes such differentiation often leads to the functions with singularities like -function . If
so, the possibility of direct numerical solution becomes questionable. That is why the equations
(4.13) are mostly used to define EM fields in free-of-sources areas.

To get around this problem let us take a quick look at Maxwell’s equations (4.1). According to
the fourth equation ∘ = 0 meaning that the magnetic field is divergenceless, i.e. source
free, across any closed area and purely curled. If so, this vector can be represented as = ×
where ≠ 0 is some arbitrary constant and, evidently, ∘ = ∘ ( × ) ≡ 0 (see
Appendix). Assuming for convenience of the following transformations that = 1/ we
0
can obtain using (4.1)

(×)
× = − = − = − × (4.16)
0

Grouping all terms on the left-hand side we have

× � + � = 0 (4.17)

According to this expression, the combination of the vector electric field and the time
differential is purely conservative vector field similar to static electric field. That means that

in such combination of fields no work can be done, for example, by moving a charge sensor
along any closed path. To build such conservative field we must accept that

+ = − ∙ = −grad() (4.18)

since × ( ∙ ) ≡ 0 (see Appendix) for any differentiable scalar function . In Chapter 2 we
established that the positive potential means the reduction in electric field strength. That
explains the minus sign of the right-hand term in (4.18).

Assume for a moment that we have found the functions and customary called the vector
and scalar potential, respectively. Then we can calculate the electric and magnetic field as

= − ∙ −
� (4.19)
1
= ×
0
Following the brief moment of delight that we have succeeded in the field estimation, we
recognized that the question is prompted: ‘is it really worth to look for two new field functions
and instead of and if the additional affords (see (4.19)) is required to find the fields?’

1 Recall that () is equal to 0 elsewhere except the point = 0 where it reiterates the infinity value.