Page:AbrahamMinkowski2.djvu/4

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

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:

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):

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

In the electrodynamics of, 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"

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

($$\mathfrak{q}$$ 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 $$V_{I}^{4}$$

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