Page:VaricakRel1910a.djvu/2

If the velocities $v_{1}$ and $v_{2}$ enclose the angle α, and

${\frac {v_{1}}{c}}=\operatorname {th} \,\ u_{1},\ {\frac {v_{2}}{c}}=\operatorname {th} \,u_{2},$

then lay off the line $OA=u_{1}$ from point O into the direction of $v_{1}$, and apply the line $AC=u_{2}$ under the angle α. The resultant corresponds to the line $OC=u$. In the Lobachevskian triangle OAC the relation is given

$\operatorname {ch} \,u=\operatorname {ch} \,u_{1}\ \operatorname {ch} \,u_{2}+\operatorname {sh} \,u_{1}\operatorname {sh} \,u_{2}\cos \ \alpha$

If we denote herein

$\operatorname {ch} \,u_{i}={\frac {1}{\sqrt {1-\left({\frac {v_{i}}{c}}\right)^{2}}}},\ \operatorname {sh} \,u_{i}={\frac {\frac {v_{i}}{c}}{\sqrt {1-\left({\frac {v_{i}}{c}}\right)^{2}}}},$

then we obtain after some simple transformations the general Einsteinian addition theorem for velocities. In the case $\alpha ={\tfrac {\pi }{2}}$, we have

$\operatorname {ch} \,u=\operatorname {ch} \,u_{1}\operatorname {ch} \,u_{2}$

or

$\operatorname {th} ^{2}u=\operatorname {th} ^{2}u_{1}+\operatorname {th} ^{2}u_{2}-\operatorname {th} ^{2}u_{1}\operatorname {th} ^{2}u_{2}$

respectively

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

That this addition is not commutative can be easily seen from the first figure of Sommerfeld that, however, we now have to interpret as a figure in the Lobachevskian plane. Additionally we have to put

$OA=BD=u_{1},\ AC=OB=u_{2}$ $OC=OD=u.$

In hyperbolic geometry the angle sum in any triangle is smaller than two right angles. Hence

$\alpha _{1}+\alpha _{2}<{\frac {\pi }{2}},$

thus OD does not coincide with the direction of OC. For the direction difference $\delta =\sphericalangle COD$ we find

$\operatorname {cotg} \,\delta =\operatorname {tg} \,\left(\alpha _{1}+\alpha _{2}\right)={\frac {\operatorname {th} \,u_{1}\operatorname {sh} \,u_{1}+\operatorname {th} \,u_{2}\operatorname {sh} \,u_{2}}{\operatorname {sh} \,u_{1}\operatorname {sh} \,u_{2}-\operatorname {th} \,u_{1}\operatorname {th} \,u_{2}}}$ $={\frac {v_{1}^{2}{\sqrt {1-\left({\frac {v_{2}}{c}}\right)^{2}}}+v_{2}^{2}{\sqrt {1-\left({\frac {v_{1}}{c}}\right)^{2}}}}{v_{1}v_{2}\left(1-{\sqrt {1-\left({\frac {v_{1}}{c}}\right)^{2}}}{\sqrt {1-\left({\frac {v_{2}}{c}}\right)^{2}}}\right)}}$

It is also

$\operatorname {tg} \,{\frac {\delta }{2}}=\operatorname {th} \,{\frac {u_{1}}{2}}\operatorname {th} \,{\frac {u_{2}}{2}}$

If $v_{1}$ and $v_{2}$ are not in the xy-plane, but arbitrarily in space, then we obtain six terminal points, while above we only had points C and D.

In addition, I want to show by some examples, how Einstein's formulas of can be interpreted as real in the Lobachevskian geometry.

Equations (3) in § 5 of the mentioned paper of Einstein define (in respect to the stationary system S) the velocity components $u_{x},u_{y},u_{z}$ of a point uniformly moving in relation to S'. If $u_{z'}=0$, then

${\frac {u_{y}}{u_{x}}}={\frac {u_{y'}}{u_{x'}+v}}\cdot {\sqrt {1-\left({\frac {v}{c}}\right)^{2}}}$

If we take $c=\infty$, then the straight line upon which that point is moving, encloses the angle λ with the x-axis. However, if c remains finite and equal to the propagation velocity of light in empty space, then we find the direction coefficients of that straight line as:

$\operatorname {tg} \,\lambda '={\frac {\operatorname {tg} \,\lambda }{\operatorname {ch} \,u}}$

If u is the hypotenuse and ${\tfrac {\pi }{2}}-\lambda$ is an acute angle in the right angled Lobachevskian triangle, then $\lambda '$ is the second acute angle. It will be the smaller, the greater the translation velocity of S'. For v = c we have $\lambda '=0$.

We define s as the extension of a stationary electron in the direction of the x-axis. If it is set into motion with velocity v in the same direction, then its contracted extension is

$s'={\frac {s}{\operatorname {ch} \,u}}$

Upon the distance line y = u having the x-axis as its center line and u as its parameter, and beginning from its intersection point U with the ordinate axis, we measure the length UM = s. The abscissa of M is $s'$.

In the same way the dilation of a clock in uniform motion relative to a reference frame can be interpreted.

If we further put

$\varphi =\Pi \left(u_{1}\right),\ \varphi '=\Pi \left(u'_{1}\right),\ {\frac {v}{c}}=\operatorname {th} \,u,$