Page:Ueber die Ablenkung eines Lichtstrals von seiner geradlinigen Bewegung.djvu/4



To reduce in these equations the number of variable quantities, we want to express x and y by r and $$\phi$$. We easily see that

$$x=r\ \cos \phi$$ and $$y= r\ \sin \phi$$.

If we differentiate, then we will obtain:

$$dx=\cos \phi\ dr- r \sin \phi\ d\phi$$, und $$dy =\sin  \phi\ dr+ r \cos  \phi\ d\phi$$.

And if we differentiate again,

$$ddx=\cos \phi\ ddr- 2 \sin \phi\ d\phi\ dr- r\ \sin \phi\ dd\phi- r\ \cos \phi\ d\phi^2$$,

and

$$ddy=\sin \phi\ ddr+ 2 \cos \phi\ d\phi\ dr+ r\ \cos \phi\ dd\phi- r\ \sin \phi\ d\phi^2$$,

If we substitute these values for ddx and ddy in the previous equations, the we obtain from (III):

$$\frac {ddy\ \cos \phi- ddx\ \sin \phi} {dt^2} = \frac {2d\phi\ dr+ r\ dd\phi} {dt^2}$$.

Thus we have

And furthermore by (IV),

To make equation (V) a true differential quantity, we multiply it by rdt, thus:

$$\frac {2r\ d\phi\ dr+ r^2\ dd\phi} {dt} =0$$,

and if we again integrate, we will obtain:

$$r^2d\phi=C\ dt$$,

where C is an arbitrary constant magnitude. To specify C, we note that $$r^2 d\phi (=r\ rd\phi)$$ is equal to: the double area of the small triangle which described the radius vector r in the time dt. The double area of the triangle that is described in the first second of time, is however: = AC · v; thus we have C = AC · v. And if we assume the radius AC of the attracting body as unity, what we will always do in the following,