Page:AbrahamMinkowski2.djvu/4

if we write this in a vectorial way, we have:

(5a)	${\begin{cases}{\mathfrak {R}}=u_{1}\left[{\mathfrak {r}}_{2}{\mathfrak {r}}_{3}\right]+u_{2}\left[{\mathfrak {r}}_{3}{\mathfrak {r}}_{1}\right]+u_{3}\left[{\mathfrak {r}}_{1}{\mathfrak {r}}_{2}\right],\\U=-{\mathfrak {r}}_{1}\left[{\mathfrak {r}}_{2}{\mathfrak {r}}_{3}\right].\end{cases}}$

Canceling index 3, we write $V_{1}^{4}$ which we obtained:

${\begin{array}{l}{\mathfrak {R}}=u\left[{\mathfrak {r}}_{1}{\mathfrak {r}}_{2}\right]+\left[{\mathfrak {r}},\ {\mathfrak {r}}_{1}u_{2}-{\mathfrak {r}}_{2}u_{1}\right],\\U=-{\mathfrak {r}}\left[{\mathfrak {r}}_{1}{\mathfrak {r}}_{2}\right].\end{array}}$

and introduce $V_{II}^{4}\left\{{\mathfrak {a,b}}\right\}$ instead of $V_{I}^{4}\left\{{\mathfrak {r}}_{1},\ u_{1}\right\}$ and $\left\{{\mathfrak {r}}_{2},\ u_{2}\right\}$, composed from them by rule (1a). Then we have:

(6)	${\begin{cases}{\mathfrak {R}}=ua+[{\mathfrak {rb}}],\\U=-{\mathfrak {ra}},\end{cases}}$

that is, a $V_{I}^{4}$ composed of a $V_{I}^{4}$ and a $V_{II}^{4}$.

We obtain another $V_{I}^{4}$, by permutation of $V^{3}{\mathfrak {a}}$ by ${\mathfrak {b}}$ in expressions (6). To demonstrate this, we form the two $S^{4}$:

${\mathfrak {rr}}_{1}+uu_{1}$ and ${\mathfrak {rr}}_{2}+uu_{2}$

Multiplying them respectively with $V_{I}^{4}$:

$-{\mathfrak {r}}_{2},\ -u_{2}$ and $+{\mathfrak {r}}_{1},\ -u_{1}$

and summing, we construct from $V_{I}^{4}$:

${\begin{array}{l}{\mathfrak {R}}'={\mathfrak {r}}_{1}\left({\mathfrak {rr}}_{2}+uu_{2}\right)-{\mathfrak {r}}_{2}\left({\mathfrak {rr}}_{1}+uu_{1}\right),\\U'=u_{1}\left({\mathfrak {rr}}_{2}+uu_{2}\right)-u_{2}\left({\mathfrak {rr}}_{1}+uu_{1}\right)\end{array}}$

that can be written:

${\begin{array}{l}{\mathfrak {R}}'=u\left({\mathfrak {r}}_{1}u_{2}-{\mathfrak {r}}_{2}u_{1}\right)+\left[{\mathfrak {r}}\left[{\mathfrak {r}}_{1}{\mathfrak {r}}_{2}\right]\right],\\U'=-\left({\mathfrak {r,\ }}{\mathfrak {r}}_{1}u_{2}-{\mathfrak {r}}_{2}u_{1}\right)\end{array}}$

When we introduce $V_{II}^{4}\{a,b\}$ by means of $I_{a}$, then this is resulting in formulas analogous to (6):

(6a)	${\begin{cases}{\mathfrak {R}}'=u{\mathfrak {b}}+[{\mathfrak {ra}}],\\U'=-{\mathfrak {rb}},\end{cases}}$

where ${\mathfrak {a}}$ and ${\mathfrak {b}}$ have changed their place.

In the electrodynamics of Minkowski, four $V^{3}$ take part, i.e, the electric and magnetic excitations ${\mathfrak {D}}$ and ${\mathfrak {B}}$, and two auxiliary vectors ${\mathfrak {E}}$ and ${\mathfrak {H}}$, which form two $V_{II}^{4}$:

${\mathfrak {B}},\ -i{\mathfrak {E}}$ and ${\mathfrak {H}},\ -i{\mathfrak {D}}$

Then we have the $V_{I}^{4}$-"velocity"

${\frac {\mathfrak {q}}{\sqrt {1-{\mathfrak {q}}^{2}}}},\ {\frac {i}{\sqrt {1-{\mathfrak {q}}^{2}}}}$

(${\mathfrak {q}}$ designates the three-dimensional velocity vector related to the speed of light).

If we combine this $V_{I}^{4}$ with $V_{II}^{4}\left\{{\mathfrak {B}},\ -i{\mathfrak {E}}\right\}$ according to scheme (6a), then we obtain the $V_{I}^{4}$

(7)	${\mathfrak {R}}^{e}={\frac {{\mathfrak {E}}+[{\mathfrak {qB}}]}{\sqrt {1-{\mathfrak {q}}^{2}}}},\ U^{e}={\frac {i({\mathfrak {qE}})}{\sqrt {1-{\mathfrak {q}}^{2}}}},$

which were denoted by Minkowski as the "electric rest force". Instead, from the $V_{I}^{4}$-"velocity"