Page:Ueber die Ablenkung eines Lichtstrals von seiner geradlinigen Bewegung.djvu/3

and the center of the attracting body a straight line, then this line will be the major axis of the curved one for the trajectory of light; by describing over and under this line two fully congruent sides of the curved line. —

Fig. 3

C (Fig. 3) shall now be the center of the attracting body, A is the location at its surface. From A, a light ray goes into the direction AD or in the horizontal direction, by a velocity with which it traverses the way v in a second. Yet the light ray, instead of traveling at the straight line AD, will be forced by the celestial body to describe a curved line AMQ, whose nature we will investigate. Upon this curved line after the time $\epsilon$ (calculated from the instant of emanation from A), the light ray is located in M, at the distance CM = r from the center of the attracting body. g be the gravitational acceleration at the surface of the body. Furthermore CP = x, MP = y and the angle MCP = $\phi$. The force, by which the light in M will be attracted by the body into the direction MC, will be $2gr^{-2}$. This force can be decomposed into two other forces,

${\frac {2g}{r^{2}}}\cos \phi$ und ${\frac {2g}{r^{2}}}\sin \phi$,

into the directions x and y; and for that we obtain the following two equations (s. Traité de mécanique céleste par Laplace, Tome I, pag. 21)

${\frac {ddx}{dt^{2}}}=-{\frac {2g}{r^{2}}}\cos \phi$

(I)

${\frac {ddy}{dt^{2}}}=-{\frac {2g}{r^{2}}}\sin \phi$

(II)

If we multiply the first of these equations by $-\sin \phi$, the second one by $\cos \phi$, and sum them up, then we obtain:

${\frac {ddy\ \cos \phi -ddx\ \sin \phi }{dt^{2}}}=0$

(III)

Now we multiply the first one by $\cos \phi$, the second one by $\sin \phi$ and and sum them together, then we obtain: