Page:IgnatowskiBemerkung.djvu/2

 $$D$$ shall denote the determinant

then it follows from (9) and (10)

{{MathForm2|(12)|$$\left.\begin{align} p'= & \frac{s_{1}}{D}; & & & s'= & -\frac{s}{D}\\ p_{1}'= & -\frac{p_{1}}{D}; & & & s_{1}'= & \frac{p}{D}. \end{align}\right\}$$}}

Now we take in $$K$$ and $$K'$$ two elements $$dx$$ and $$dx'$$ of such length, so that they are equal when brought to mutual rest. If we now synchronously measure $$dx'$$ from $$K$$ (thus $$dt=0$$), then we obtain

If we synchronously measure $$dx$$ from $$K'$$ (thus $$dt'=0$$), then it follows accordingly

Both systems $$K$$ and $$K'$$ are now equally valid, and $$dx$$ and $$dx'$$ have the same length when brought to mutual rest. Consequently, the lengths measured from both systems must be equal. Thus

From that and (12) it follows

Let us now follow the motion of any substantial point or any phenomenon in space, and denote the corresponding velocity by $$\mathfrak{v}$$ or $$\mathfrak{v}'$$. Then it can be simply demonstrated due to (6), that

where

Since $$\mathfrak{v}$$ is totally arbitrary, it is clear that $$p, s$$ etc. cannot depend on $$\mathfrak{v}$$. Let us assume that the movable point rests with respect to $$K'$$. Then $$\mathfrak{v}'=0$$ and $$\mathfrak{v}=q\mathfrak{c}_{0}$$. From that and from (17) we obtain

Because of the preceding things we obtain by similar considerations

so that we can write

{{MathForm2|(21)|$$\left.\begin{align} dx' &=pdx-pqdt\\ dt' &=p_{1}dx+pdt. \end{align}\right\}$$.}}

It only remains to determine $$p_{1}$$ and $$p$$, because the unprimed quantities can be obtained from (12).

For that purpose we introduce a third coordinate system $$K''$$, which moves in the same direction $$\mathfrak{c}_{0}$$ with velocity $$q_{2}$$ measured in $$K$$. The velocity of $$K'$$ as measured in $$K''$$ is $$q_{1}$$. We denote by $$\overline{p_{1}},p,q_{1}$$ the quantities analogous to $$p_{1},p,q$$ for couple $$KK'$$, and by $$p,p_{1},q_{2}$$ the ones for couple $$KK$$. Then it can easily be demonstrated that the following relation exists:

Since every fraction contains mutually independent quantities here, we can see that it can only be a constant, which we denote by $$-n$$. Thus we eventually obtain

{{MathForm2|(23)|$$\left.\begin{align} dx' &=pdx-pqdt\\ dt' &=-pqndx+pdt. \end{align}\right\}$$}}

Furthermore, it follows from (15) and (12)

$p^{2}=\frac{1}{1-q^{2}n}$

or

From (24) it follows, that $$n$$ (which we can denote as a universal space-time constant) is the reciprocal square of a velocity, thus an absolute-positive quantity.

We see that we obtained transformation equations similar to those of, except that $$n$$ is used instead of $$\tfrac{1}{c^{2}}$$. However, the sign is still undetermined, because we could have set the positive sign under the square root in (24) as well.

Now, in order to determine the numerical value and the sign of $$n$$, we have to look at the experiment. Since we haven't used in the previous derivation any special physical phenomenon, it follows that we can determine $$n$$ by using an arbitrary phenomenon, and we always must obtain the same value for $$n$$, since $$n$$ is indeed a universal constant.

For instance, we can measure the length of a moving meter synchronously. If the measurement shows that it has been contracted, then the negative sign is to be chosen, and $$n$$ can be calculated from the contraction. However, it's known that the contraction will be so small that we cannot measure it directly.

We now turn to the electrodynamic equations and especially to the case of a uniformly moving point-charge. We know, besides the relativity principle, that the level-surface of the convection potential of the previous point-charge will be a -Ellipsoid for the resting observer, whose axis ratio is equal to $$\sqrt{1-q^{2}/c^{2}}$$. Now we must conclude due to the relativity principle, that the level-surface of the potential is spherical for an observer co-moving with the point-charge.