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NEOCLASSICAL THEORY OF INTERACTION 61

In case of perfect conductor or real conductor in superconductivity, electrons move without
collisions, = ∞ and according to (2.32) = 0 in any moment of time. Nevertheless, the

experiments show that the current density stays finite. If so, the infinite conductivity in

(2.28) makes sense if and only if || → 0 as soon as → ∞. More comprehensive consideration
requires quantum mechanical treatment.
2.2.8 Ohm’s Law

Let demonstrate how the primary engineering tool for circuit
analysis, Ohm’s law, follows from (2.30). First of all, consider
the resistor model as a small rectangular parallelepiped inside
the solid conductive material and assume that the electric field
vector is oriented as shown in Figure 2.2.11. The potential
difference (voltage drop) or work produced by electric

field can be calculated using (1.21) as
Figure 2.2.11 Resistor
model = ∫ ∘ [V] (2.33)

Here one of the possible contour L shown by blue dotted line connects two points O1 and O2 on
end-walls. The net electrical current can be calculated (see (1.4)) as a flux of through the

parallelepiped lateral surface area A = w d

= ∯ ∘ = ∯ ∘ [A] (2.34)

The relationship between the potential and current can be mathematically described as
1 ∫ ∘
= = [Ω] (2.35)
∯ ∘

This ratio is called the resistance and measured in Ohms. In other words, this Ohm’s law tells
us that the current through a conductor between two points is directly proportional to the

potential difference across the two points or

. = (2.36)
1

Before making any estimation of R, it is time now to look more carefully at the definition of
potential in (2.33). Eventually, the linear integral in the numerator (2.33) generally depends

not only on the length but on the configuration of
chosen path L. If so, we cannot define the resistance
in (2.35) as one and only one value. An infinite
number of the different path can be drawn between
infinite combination of two distinct points, and for
the same element infinite number of various
Figure 2.2.12 Different integration resistances can be calculated. Let us look at this
paths problem more carefully.
Suppose the linear integral in (2.33) was calculated along different paths between two
points and as shown in Figure 2.2.12. Taking into account our desire to get the unique
2
1
amount of potential we must request that ∫ ∘ = ∫ ∘ . However, ∫ ∘ =

1 2 2