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
![{\displaystyle \left|K_{1},\ K_{2},\ K_{3},\ K_{4}\right|}](https://wikimedia.org/api/rest_v1/media/math/render/svg/65fa788e3868585c0ebfa95568e9f2cfec035970)
where
(64)
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When by a Lorentz transformation
, a new reference system
is introduced, we can use the operator
![{\displaystyle lor'=\left|{\frac {\partial }{\partial x'_{1}}},\ {\frac {\partial }{\partial x'_{2}}},\ {\frac {\partial }{\partial x'_{3}}},\ {\frac {\partial }{x'_{4}}}\right|}](https://wikimedia.org/api/rest_v1/media/math/render/svg/feafb4d8fb8dc63a65139b0f6230afb3d75c39d5)
Then S is transformed to
, so by lor' Sis meant the 1✕4 series matrix, whose element are
![{\displaystyle K'_{k}={\frac {\partial S'_{1k}}{\partial x'_{1}}}+{\frac {\partial S'_{2k}}{\partial x'_{2}}}+{\frac {\partial S'_{3k}}{\partial x'_{3}}}+{\frac {\partial S'_{4k}}{\partial x'_{4}}}\qquad (k=1,2,3,4)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/9acdba098b93936afddf0b070017c270debbbd14)
Now for the differentiation of any function of (x y z t) we have the rule
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so that, we have symbolically
![{\displaystyle lor'=lor\ ({\mathsf {A}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/868897335957645316d49fcf7be45b788ef03676)
Therefore it follows that
(65)
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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