employ the 1 × 4 series matrix, formed of differential symbols,—
| [part]/[part]x, [part]/[part]y, [part]/[part]z, [part]/i[part]t,| or (63) | [part]/[part]x_{1} [part]/[part]x_{2} [part]/[part]x_{3} [part]/[part]x_{4} |
For this matrix I shall use the shortened from "lor."[1]
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
| K_{1} K_{2} K_{3} K_{4} |
where K_{k} = [part]S_{1k}/[part]x_{1} + [part]S_{2k}/[part]x_{2} + [part]S_{3k}/[part]x_{3} + [part]S_{4h}/[part]x_{4},
When by a Lorentz transformation A, a new reference system (x´_{1} x´_{2} x´_{3} x_{4}) is introduced, we can use the operator
lor´ = | [part]/[part]x_{1}´ [part]/[part]x_{2}´ [part]/[part]x_{3}´ [part]/[part]x_{4}´ |
Then S is transformed to S´= Ā S A = | S´_{hk} |, so by lor 'S´ is meant the 1 × 4 series matrix, whose element are
K'_{k} = [part]S´_{1k}/[part]x_{1}´ + [part]S´_{2k}/[part]x_{2}´ + [part]S´_{3k}/[part]x_{3}´ + [part]S´_{4k}/[part]x_{4}´.
Now for the differentiation of any function of (x y z t) we have the rule [part]/[part]x_{k}´ = [part]/[part]x_{1} [part]x_{1}/[part]x_{k}´ + [part]/[part]x_{2} [part]x_{2}/[part]x_{k}´ + [part]/[part]x_{3} [part]x_{3}/[part]x_{k}´ + [part]/[part]x_{4} [part]x_{4}/[part]x_{k}´
= [part]/[part]x_{1} a_{1k} + [part]/[part]x_{2} a_{2k} + [part]/[part]x_{3} a_{3k} + [part]/[part]x_{4} a_{4k}.?]
so that, we have symbolically lor´ = lor A.
- ↑ Vide note 17.
