Page:VaricakRel1910b.djvu/3

 $$\frac{U}{c}=\frac{1}{3\cdot10^{3}}+\frac{1}{3\cdot3^{3}\cdot10^{9}}+\frac{1}{5\cdot3^{5}\cdot10^{15}}+\dots$$

and the representative length U is by the way about 3mm greater then 100km. The velocity of 100000 km/sec corresponds to a length of 103700 km.

We additionally consider two velocities of &beta;-rays calculated in the famous experiments of, and to which the velocity relations 0,7202 and 0,9326 are connected. They amount ca. 216060 and 279780 km/sec and they were represented by the lengths of 272400 and 503400 km.

For v = c we have U = &infin;.

In graphical illustration these relations can be easily summarized.



If we take u as abscissa and $$\tfrac{v}{c}$$ as ordinate, then (1) will be represented by curve T. The straight line G or the first term in the infinite row (2), corresponds to the ordinary definition $$u=\tfrac{v}{c}$$. The straight line is the inflexion tangent of T in O, so it fits well to the curve in the very far surrounding of the coordinate origin.

Fig. 1 can provide us good services for the composition of velocities. In addition to the resultant length U we can immediately take the corresponding velocity v from the figure.

If the velocities $$v_1$$ and $$v_2$$ enclose the angle &alpha; = 0, i.e. if they lie in the same direction, then $$u=u_{1}+u_{2}$$. The resultant velocity follows from the formula

$$\operatorname{th}\, u=\frac{\operatorname{th}\, u_{1}+\operatorname{th}\, u_{2}}{1+\operatorname{th}\, u_{1}\operatorname{th}\, u_{2}}$$

or

$$v=\frac{v_{1}+v_{2}}{1+\frac{v_{1}v_{2}}{c^{2}}}$$

Although the resultant is smaller as the sum of the components, it will be represented (as in ordinary mechanics) by a length equal to the sum of the lengths representing the components. It is namely in this case $$U=U_{1}+U_{2}$$.

The figure in the 6th Göttingen lecture by Poincaré is not in agreement with this definition.

If we compose two equal velocities to which corresponds the length $$U_1$$, then the resultant will be represented by the length $$2U_1$$.

2. Lorentz-Einstein transformation as translation
We take this transformation in the form as it was given by me.

{{MathForm2|(4)|$$\left.\begin{array}{c} l'=-x\ \operatorname{sh}\, u+l\ \operatorname{ch}\,\ u\\ x'=x\ \operatorname{ch}\, u-l\ \operatorname{sh}\,\ u\\ y'=y,\ z'=z.\end{array}\right\} $$}}

The variables x, y, z, l I interpret as homogeneous coordinates in an three-dimensional space. If X, Y, Z are an right angled coordinates, &xi;, &eta; &zeta; the perpendiculars that fall from point M upon the planes of the right angles coordinate system, and r is the distance of that point from the coordinate origin, then we get the following equations from the easily constructed figure

These coordinates satisfy, as it is easily seen, the quadratic equation

This is therefore the invariant of first kind of the transformation defined by equations (4), or by

If we take

then we get to distance areas