Page:AbrahamMinkowski2.djvu/3

three-dimensional space $x,y,z$, the three first components of vector $V_{I}^{4}$ constitute a three-dimensional vector $V^{3}$, ${\mathfrak {r}}$, the fourth ($u$) a three-dimensional scalar ($S^{3}$).

A four-dimensional vector of the second kind $\left(V_{II}^{4}\right)$ denotes a system of six magnitudes, which are transformed like the following expressions, formed by the components $x_{1},y_{1},z_{1},u_{1}$ and $x_{2},y_{2},z_{2},u_{2}$ of two $V_{I}^{4}$:

(1)

$\left\{{\begin{array}{ccccc}{\mathfrak {a}}_{x}=\left|{\begin{array}{cc}y_{1}&z_{1}\\y_{2}&z_{2}\end{array}}\right|,&&{\mathfrak {a}}_{y}=\left|{\begin{array}{cc}z_{1}&x_{1}\\z_{2}&x_{2}\end{array}}\right|,&&{\mathfrak {a}}_{z}=\left|{\begin{array}{cc}x_{1}&y_{1}\\x_{2}&y_{2}\end{array}}\right|;\\\\{\mathfrak {b}}_{x}=\left|{\begin{array}{cc}x_{1}&u_{1}\\x_{2}&u_{2}\end{array}}\right|,&&{\mathfrak {b}}_{y}=\left|{\begin{array}{cc}y_{1}&u_{1}\\y_{2}&u_{2}\end{array}}\right|,&&{\mathfrak {b}}_{z}=\left|{\begin{array}{cc}z_{1}&u_{1}\\z_{2}&u_{2}\end{array}}\right|.\end{array}}\right.$

Obviously, when projecting into three-dimensional space, $V_{II}^{4}$ is composed of two $V^{3}$, which, in the symbolism of ordinary vector analysis, we can write:

(1a)	${\mathfrak {a}}=\left[{\mathfrak {r}}_{1}{\mathfrak {r}}_{2}\right],\ {\mathfrak {b}}={\mathfrak {r_{1}}}u_{2}-{\mathfrak {r}}_{2}u_{1}$

From two $V_{I}^{4}$:

${\mathfrak {r}},\ u$ and ${\mathfrak {r}}_{1},\ u_{1}$

we can compose a four-dimensional scalar ($S^{4}$) as follows:

(2)	$S=xx_{1}+yy_{1}+zz_{1}+uu_{1}={\mathfrak {rr}}_{1}+uu_{1}$

Conversely, from any four-dimensional scalar $\varphi (x,y,z,u)$, we obtain (derived with respect to their coordinates) a $V_{I}^{4}$:

(3)	$X={\frac {\partial \varphi }{\partial x}},\ Y={\frac {\partial \varphi }{\partial y}},\ Z={\frac {\partial \varphi }{\partial z}},\ U={\frac {\partial \varphi }{\partial u}}.$

So the operators

${\frac {\partial }{\partial x}},\ {\frac {\partial }{\partial y}},\ {\frac {\partial }{\partial z}},\ {\frac {\partial }{\partial u}}$

transform as the components of a $V_{I}^{4}$, and these operators were denoted by Minkowski as the components of the operator "lor".

We can compose a $S^{4}$ from four $V_{I}^{4}$, which determines the space of the parallelepiped of the four vectors:

(4)	$\varphi =\left|{\begin{array}{ccccccc}x&&y&&z&&u\\x_{1}&&y_{1}&&z_{1}&&u_{1}\\x_{2}&&y_{2}&&z_{2}&&u_{2}\\x_{3}&&y_{3}&&z_{3}&&u_{3}\end{array}}\right|$

If we apply scheme (3) to $S^{4}$, we obtain a $V_{I}^{4}$, which is composed of three other $V_{I}^{4},\ {\mathfrak {r}}_{1}u_{1},\ {\mathfrak {r}}_{2}u_{2},\ {\mathfrak {r}}_{3}u_{3}$, whose components are:

(5)

$\left\{{\begin{array}{ccc}X={\frac {\partial \varphi }{\partial x}}=\left|{\begin{array}{ccccc}y_{1}&&z_{1}&&u_{1}\\y_{2}&&z_{2}&&u_{2}\\y_{3}&&z_{3}&&u_{3}\end{array}}\right|,&&Y={\frac {\partial \varphi }{\partial y}}=\left|{\begin{array}{ccccc}z_{1}&&x_{1}&&u_{1}\\z_{2}&&x_{2}&&u_{2}\\z_{3}&&x_{3}&&u_{3}\end{array}}\right|,\\\\Z={\frac {\partial \varphi }{\partial z}}=\left|{\begin{array}{ccccc}x_{1}&&y_{1}&&u_{1}\\x_{2}&&y_{2}&&u_{2}\\x_{3}&&y_{3}&&u_{3}\end{array}}\right|,&&U={\frac {\partial \varphi }{\partial u}}=-\left|{\begin{array}{ccccc}x_{1}&&y_{1}&&z_{1}\\x_{2}&&y_{2}&&z_{2}\\x_{3}&&y_{3}&&z_{3}\end{array}}\right|;\end{array}}\right.$