Page:VaricakRel1910a.djvu/3

 then we obtain from the first formula, the principle of :

$$\nu'=\nu\left(\operatorname{ch}\, u-\operatorname{th}\, u_{1}\operatorname{sh}\, u\right)=\frac{\nu\,\operatorname{ch}\,\left(u-u_{1}\right)}{\operatorname{ch}\, u_{1}}$$

The parallel angle $$\varphi=0$$ corresponds to the length $$u_{1}=\infty$$, and so we have in this case

$$\nu'=\nu e^{-u}\,$$

It is also

$$\nu'=\nu\left(1-u+\frac{u^{2}}{2!}-\frac{u^{3}}{3!}+\dots\right)$$

Higher powers of u can be neglected for small values, and we can replace $$\operatorname{th}\,\tfrac{v}{c}$$ by $$\tfrac{v}{c}$$. The preceding formula goes over into the expression of 's principle of ordinary mechanics:

$$\nu'=\nu\left(1-\frac{v}{c}\right)$$

Note, that in the primed reference frame $$v'=-v$$.

The ratio of the frequencies $$\nu$$ and $$\nu'$$ in the formula $$\nu'=\nu e^{-u}$$ can be represented as the ratio of two limiting circular arcs between two common axes.

The expression of aberration is transformed into

$$\operatorname{th}\, u'_{1}=\operatorname{th}\,\left(u_{1}-u\right)$$

The aberration equation is thus

$$u'_{1}=u_{1}-u\,$$

A light ray T coming from an infinitely distant light source, is striking the x-axis at point M under the acute angle φ. We lay off the line $$MN=u$$ toward the increasing abscissa, and from N apply the an parallel to T. This parallel $$T'$$ encloses the angle $$\varphi'$$ with the x-axis.



If $$\varphi=\tfrac{\pi}{2}$$, also $$u_{1}=0$$, hence $$u'_{1}=-u$$, and the angle $$\varphi'$$ goes over into its supplement $$\varphi'_{1}$$.

The formulas of relativity theory are very simplified in this interpretation. For example, for a moving electron of mass &mu; we have

instead of

$$\frac{\mu}{\left(\sqrt{1-\left(\frac{v}{c}\right)^{2}}\right)^{3}}$$ and $$\frac{\mu}{1-\left(\frac{v}{c}\right)^{2}}$$

If UM (Fig. 1) represents the longitudinal mass, then ON is the transverse mass.

The radius of curvature R of the orbit, when a magnetic force acting normal to the velocity of the electron is present, will be

$$R=\frac{c^{2}\mu}{\epsilon N}\operatorname{sh}\, u$$

instead of

$$R=c^{2}\frac{\mu}{\epsilon}\cdot\frac{\frac{v}{c}}{\sqrt{1-\left(\frac{v}{c}\right)^{2}}}\cdot\frac{1}{N}$$

A body in uniform translational motion in the direction of the increasing x-coordinate, has according to relativity theory the kinetic energy

$$K_{0}=\mu c^{2}\left\{ \frac{1}{\sqrt{1-\left(\frac{v}{c}\right)^{2}}}-1\right\} $$

where &mu; denotes its mass in the ordinary sense. We can write this in a simpler way

$$K_{0}=2\mu c^{2}\operatorname{sh}^{2}\frac{u}{2}$$

Instead of $$\operatorname{sh}^{2}\tfrac{u}{2}$$ we can write $$\operatorname{tg}\, p$$, where p is the area of a isosceles quadrilateral with two right angles. Its three sides enclosing two right angles, have a length of u.

Here it was presupposed that this body is not subjected to external forces. However, if external forces act on that body, which are in equilibrium with each other, i.e. they don't accelerate the body, then according to the investigations of its kinetic energy is (oddly enough) increased by

$$\Delta E=-\frac{\left(\frac{v}{c}\right)^{2}}{\sqrt{1-\left(\frac{v}{c}\right)^{2}}}\sum\left(\xi K_{\xi}\right)$$

According to our definition this expression goes over into

$$\Delta E=-\operatorname{th}\, u\,\operatorname{sh}\, u\sum\left(\xi K_{\xi}\right)$$

We take a an right angled triangle. If the legs a