The decomposition products of the third rank contravariant tensor in 3+1 space

The third rank tensor contains a scalar, three vectors, two quadrupoles, and an octupole. The tensor of each rank is irreducible, meaning that it is not expressible by any linear combination of tensors of lower rank.

The following calculations are based on the tensor decomposition equation [1]. The third rank tensor is the gradient of the second rank tensor, which was derived in Section 2.

The scalar is -div E, where E is the electrostatic field:

div(grad(Ψ) + 1/c ∂A/∂t).

The first vector is the time derivative of the magnetic field,

∂/∂t (curl A).

The second vector is the gradient of the 3+1 space expansion factor,

grad(div(A) + 3/c ∂Ψ/∂t).

The 4-vector equations of Section 2 do not contain the factor of 3.

The third vector is

div(grad A) - 1/c grad(∂Ψ/∂t).

div(grad A) is the same as ∇2A, which, by a vector identity, is the same as grad(div A) - curl(curl A).

The three vectors are not uniquely specified by the decomposition algorithm [1]. Linear combinations of the three original solutions have been selected for a better orthogonality. The coefficients of the vectors have been dropped.

If the additional constraint is imposed that the whole tensor should be equal to the sum of its decomposition products then the three vectors are uniquely defined. The uniqueness is shown in Section 3.2 below. A particular linear combination of the above vectors must be used. The calculation is mine, but it does not appear to be controversial. The two quadrupoles and the octupole are also shown there.

The fourth rank tensor contains 3 scalars, 6 vectors, 6 quadrupoles, 3 octupoles, and 1 hexadecapole. I will decompose the tensor in 3+1 space one day, but not soon. Sometimes it seems that the fourth rank tensor should play a special role in the four dimensional space.