_f_ = ρφ + ρ*φ*.
Where ρ, ρ* are scalar quantities, φ, φ* are two mutually perpendicular
unit planes, there is no difficulty in seeming that
_f_^1 = A⁻¹_f_A.
Note 15.
The vector product (_w__f_). (P. 36).
This represents the vector product of a four-vector and a six-vector.
Now as combinations of this type are of frequent occurrence in this
paper, it will be better to form an idea of their geometrical meaning.
The following is taken from the above mentioned paper of Sommerfeld.
“We can also form a vectorial combination of a four-vector and a
six-vector, giving us a vector of the third type. If the six-vector be
of a special type, _i.e._, a piece of plane, then this vector of the
third type denotes the parallelopiped formed of this four-vector and
the complement of this piece of plane. In the general case, the
product will be the geometric sum of two parallelopipeds, but it can
always be represented by a four-vector of the 1st type. For two pieces
of 3-space volumes can always be added together by the vectorial
addition of their components. So by the addition of two 3-space
volumes, we do not obtain a vector of a more general type, but one
which can always be represented by a four-vector (loc. cit. p. 759).
The state of affairs here is the same as in the ordinary vector
calculus, where by the vector-multiplication of a vector of the first,
and a vector of the second type (_i.e._, a polar vector), we obtain a
vector of the first type (axial vector). The formal scheme of this
multiplication is taken from the three-dimensional case.
Let A = (A_{_x_}, A_{_y_}, A_{_z_}) denote a vector of the first type, B
= (B_{_y z_}, B_{_z x_}, B_{_x y_}) denote a vector of the second type.
From this last, let us form three special vectors of the first kind,
namely—
B_{_x_} = (B_{_x x_}, B_{_x y_}, B_{_x z_}) }
B_{_y_} = (B_{_y x_}, B_{_y y_}, B_{_y z_}) } (B_{_i k_} = - B_{_k
i_}, B_{_i i_} = 0).
B_{_z_} = (B_{_z x_}, B_{_z y_}, B_{_z z_}) }
Since B_{_j j_} is zero, B_{_j_} is perpendicular to the _j_-axis. The
_j_-component of the vector-product of A and B is equivalent to the
scalar product of A and B_{_j_}, _i.e._,
(A B_{_j_},) = A_{_x_} B_{_j x_} + A_{_y_} B_{_j y_} + A_{_z_} B_{_j
z_}.
We see easily that this coincides with the usual rule for the
vector-product; _e. g._, for _j_ = _x_.
(AB_{_x_}) = A_{_y_} B_{_x_ _y_} - A_{_z_} B_{_z_ _x_}.
Correspondingly let us define in the four-dimensional case the product
(P_f_) of any four-vector P and the six-vector _f_. The _j_-component
(_j_ = _x_, _y_, _z_, or _l_) is given by
(P_f__{_j_}) = P_{_x_}_f__{_j_ _x_} + P_{_y_}_f__{_j_ _y_} +
P_{_w_}_f__{_j_ _z_} + P_{_z_}_f__{_j_ _l_}
Each one of these components is obtained as the scalar product of P, and
the vector _f__{_j_} which is perpendicular to j-axis, and is obtained
from _f_ by the rule _f__{_j_} = [(_f__{_j_ _x_}, _f__{_j_ _y_},
_f__{_j_ _z_}, _f__{_j_ _l_}) _f__{_j_ _j_} = 0.]
We can also find out here the geometrical significance of vectors of the
third type, when _f_ = φ, _i.e._, _f_ represents only one plane.
We replace φ by the parallelogram defined by the two four-vectors U, V,
and let us pass over to the conjugate plane φ^*, which is formed by the
perpendicular four-vectors U^*, V^*. The components of (Pφ) are then
equal to the 4 three-rowed under-determinants D_{_x_} D_{_y_} D_{_z_}
D_{_l_} of the matrix
| P_{_x_} P_{_y_} P_{_z_} P_{_l_} |
| |
| U_{_x_}^* U_{_y_}^* U_{_z_}^* U_{_l_}^* |
| |
| V_{_x_}^* V_{_y_}^* V_{_z_}^* V_{_l_}^* |