Page:VaricakRel1912.djvu/6

 is valid. One only has to graphically illustrate the components as it was explained above, and to use an trigonometry for the calculation of the resultant.

$$v_1$$ and $$v_2$$ are two velocities, that enclose the angle &alpha; with each other. The lengths $$U_{1}, U_{2}$$ with measure units $$u_{1}, u_{2}$$ correspond to them according to the relations

Then lay off the line $$OA=U_{1}$$ from point O into the direction of $$v_1$$, and apply the line $$AB=U_{2}$$ under the angle &alpha;. The resultant corresponds to the line $$OB=U$$. In the an triangle OAB the relation is given

If we denote herein

then we obtain

$$\sqrt{1-\left(\frac{v}{c}\right)^{2}}=\frac{\sqrt{1-\left(\frac{v_{1}}{c}\right)^{2}}\sqrt{1-\left(\frac{v_{2}}{c}\right)^{2}}}{1+\frac{v_{1}v_{2}\cos\alpha}{c^{2}}}$$

and after some transformations

$$\begin{array}{c} \left(\frac{v}{c}\right)^{2}=\frac{\left(\frac{v_{1}}{c}\right)^{2}+\left(\frac{v_{2}}{c}\right)^{2}-\left(\frac{v_{1}v_{2}}{c^{2}}\right)^{2}-1}{\left(1+\frac{v_{1}v_{2}\cos\alpha}{c^{2}}\right)^{2}}+1\\ \\=\frac{\left(\frac{v_{1}}{c}\right)^{2}+\left(\frac{v_{2}}{c}\right)^{2}-\left(\frac{v_{1}v_{2}}{c^{2}}\right)^{2}\cdot\left(1-\cos^{2}\alpha\right)+\frac{v_{1}v_{2}\cos\alpha}{c^{2}}}{\left(1+\frac{v_{1}v_{2}\cos\alpha}{c^{2}}\right)^{2}}\end{array}$$

Eventually, from this it follows

and that is 's addition law of velocities. In non-euclidean vector notation we can write it in the form