ΔΣ = 0
Now if for ΔΣ we choose the four-dimensional parallelopiped with sides (dx₁, dx₂, dx₃, dx₄), we have then
Div S = ∂s₁/∂x₁ + ∂s₂/∂x₂ + ∂s₃/∂x₃ + ∂s₄/∂x₄ = lor S.
If f denotes a space-time vector of the second kind, lor f is equivalent to a space-time vector of the first kind. The geometrical significance can be thus brought out. We have seen that the operator ‘lor’ behaves in every respect like a four-vector. The vector-product of a four-vector and a six-vector is again a four-vector. Therefore it is easy to see that lor S will be a four-vector. Let us find the component of this four-vector in any direction s. Let S denote the three-space which passes through the point Q (x₁, x₂, x₃, x₄) and is perpendicular to s, ΔS a very small part of it in the region of Q, dσ is an element of its two-dimensional surface. Let the perpendicular to this surface lying in the space be denoted by n, and let fs n denote the component of f in the plane of (sn) which is evidently conjugate to the plane dσ. Then the s-component of the vector divergence of f because the operator lor multiplies f vectorially.
= Div fs = Lim (∫ fs ndσ)/ΔS.
Δs = 0
Where the integration in dσ is to be extended over the whole surface.
If now s is selected as the x-direction, Δs is then a three-dimensional parallelopiped with the sides dy, dz, dl, then we have
and generally