For this matrix I shall use the shortened from lor.
Then if S is, as in (62), a space-time matrix of the II. kind, by lor S' will be understood the 1✕4 series matrix

where
(64)
|
|
When by a Lorentz transformation
, a new reference system
is introduced, we can use the operator

Then S is transformed to
, so by lor' Sis meant the 1✕4 series matrix, whose element are

Now for the differentiation of any function of (x y z t) we have the rule
|
,
|
so that, we have symbolically

Therefore it follows that
(65)
|
,
|
i.e., lor S behaves like a space-time vector of the first kind.
If L is a multiple of the unit matrix, then by lor L will be denoted the matrix with the elements