Cu - (| u | ρ′)/√(1 - u²)
= (cu - | u |ρ)/√(1 - u²)
= Ju/(1 - u²)
and the component in a perpendicular direction is Cu = Jū.
This space-vector is connected with the space-vector J = C - ρu, which we denoted in [§8] as the conduction-current.
Now by comparing with Φ = -ωF, the relation (E) can be brought into the form
{E} s + (ωṡ)ω = - σωF,
This formula contains four equations, of which the fourth follows from the first three, since this is a space-time vector which is perpendicular to ω.
Lastly, we shall transform the differential equations (A) and (B) into a typical form.