Notes 13 and 14.
We have denoted the four-vector ω by the matrix | ω_{1} ω_{2} ω_{3} ω_{4} |. It is then at once seen that ω̄ denotes the reciprocal matrix
| ω_{1} |
| ω_{2} |
| ω_{3} |
| ω_{4} |
It is now evident that while ω^1 = ωA, ω̄^1 = A^{-1}ω̄
[ω, s] The vector-product of the four-vector ω and s may be represented by the combination
[ωs] = ω̄s - [=s]ω
It is now easy to verify the formula [function]^1 = A^{-1}[function]A. Supposing for the sake of simplicity that [function] represents the vector-product of two four-vectors ω, s, we have
[function]^1 = [ω^1s^1] = [ω̄^1s^1 - [=s]^1ω^1]
= [A^{-1} ω̄sA - A^{-1}sω̄A]
= A^{-1}[ω̄s - sω̄]A = A^{-1}[function]A.
Now remembering that generally
[function] = ρφ + ρ*φ*.
Where ρ, ρ* are scalar quantities, φ, φ* are two mutually perpendicular unit planes, there is no difficulty in seeming that
[function]^1 = A^{-1}[function]A.
Note 15. The vector product (w[function]). (P. 36).
This represents the vector product of a four-vector and a six-vector. Now as combinations of this type are of
