Then by applying rule (45), we have
(58) [Φ.Ψ] = i[ωΩ]*
i.e. Φ_{1}Ψ_{2} - Φ_{2}Ψ_{1} = i(ω_{3}Ω_{4} - ω_{4}Ω_{3}) etc.
The vector Ω fulfils the relation
(ωΩ) = ω_{1}Ω_{1} + ω_{2}Ω_{2} + ω_{3}Ω_{3} + ω_{4}Ω_{4} = 0,
(which we can write as Ω_{4} = i(ω_{x}Ω_{1} + ω_{y}Ω_{2} + ω_{2}Ω_{3}) and Ω is also normal to ω. In case ω = 0, we have Φ_{4} = 0, Ψ_{4} = 0, Ω_{4} = 0, and
[Ω_{1}, Ω_{2}, Ω_{3}?] = | Φ_{1} Φ_{2} Φ_{3} |
|Ψ_{1} Ψ_{2} Ψ_{3} |.
I shall call Ω, which is a space-time vector 1st kind the Rest-Ray.
As for the relation E), which introduces the conductivity σ
we have -ωS = -(ω_{1}s_{1} + ω_{2}s_{2} + ω_{3}s_{3} + ω_{4}s_{4})
= (- | u | C_{u} + ρ)/[sqrt](1 - u^2) = rho´.?]
This expression gives us the rest-density of electricity (see §8 and §4).
Then 61) = s + (ω[=s])ω represents a space-time vector of the 1st kind, which since ωω = -1, is normal to ω, and which I may call the rest-*current. Let us now conceive of the first three component of this vector as the (x-y-z) co-ordinates of the space-vector, then the component in the direction of u is
C_{u} - (| u | ρ´)/[sqrt](1 - u^2) = (c_{u} - | u |ρ)/[sqrt](1 - u^2) = J_{u}/(1 - u^2)(1 - u^2) ?].
and the component in a perpendicular direction is C_{u} = J_{[=u]}.
This space-vector is connected with the space-vector J = C - ρu, which we denoted in §8 as the conduction-*current.
