The second rank 3+1 space contravariant tensor is derived below. The second rank tensor contains a scalar, a vector, and a quadrupole .
The symmetric part of the strain tensor is
|∂Ax/∂x + 1/c ∂Ψ/∂t||(∂Ax/∂y + ∂Ay/∂x)/2||(∂Ax/∂z + ∂Az/∂x)/2|
|(∂Ax/∂y + ∂Ay/∂x)/2||∂Ay/∂y + 1/c ∂Ψ/∂t||(∂Ay/∂z + ∂Az/∂y)/2|
|(∂Ax/∂z + ∂Az/∂x)/2||(∂Ay/∂z + ∂Az/∂y)/2||∂Az/∂z + 1/c ∂Ψ/∂t|
Ax, Ay, and Az are the components of the the vector potential A. Ψ is the scalar potential.
The scalar is the trace of the tensor, div(A) + 3/c ∂Ψ/∂t. The scalar is the expansion factor. The quadrupole is obtained by subtracting 1/3 of the expansion factor from the diagonal elements. The quadrupole contains both elongation and shear terms, but these components are not separable.
The antisymmetric part of the strain tensor is
|0||∂Ψ/∂z + 1/c ∂Az/∂t||-∂Ψ/∂y - 1/c ∂Ay/∂t|
|-∂Ψ/∂z - 1/c ∂Az/∂t||0||∂Ψ/∂x + 1/c ∂Ax/∂t|
|(∂Ψ/∂y + 1/c ∂Ay/∂t||-∂Ψ/∂x - 1/c ∂Ax/∂t)||0|
This portion of the tensor is equivalent to the vector
-E = grad(Ψ) + 1/c ∂A/∂t,
where E is the electrostatic field.
A similar tensor can be written in 4-vector form. The 4-gradient is the operator [∂/∂x, ∂/∂y, ∂/∂z, -i/c ∂/∂t], which operates on the 4-potential [Ax, Ay, Az, i Ψ]. The gradient of the vector is the following tensor:
|i ∂Ψ/∂x||i ∂Ψ/∂y||i ∂Ψ/∂z||1/c ∂Ψ/∂t|
The tensor can be made symmetric by adding half of it to half of its transpose, and the Maxwell equations follow directly from the Lagrangian of the symmetric tensor. The trace of the tensor is the Lorentz condition, div(A) + 1/c ∂Ψ/∂t. The factor of 3 in the expansion factor derived above has vanished. Does that mean that the expansion factor of 4-space is different than the expansion factor of 3+1 space? This tensor seems to have the same meaning as the 3+1 space tensor shown above, and the 4-space version is easier to work with.
The symmetric and antisymmetric measuring rods: