Page:SahaElectrodynamics.djvu/17

The law of parallelogram of velocities hold up to the first order of approximation. We can put

$U^{2}=\left({\frac {\partial x}{\partial t}}\right)^{2}+\left({\frac {\partial y}{\partial t}}\right)^{2},\ w^{2}=w_{\xi }^{2}+w_{\eta }^{2}$,

and

$\alpha =\tan ^{-1}{\frac {w}{w_{\xi }}}$

i.e., $\alpha$ is put equal to the angle between the velocities v, and w. Then we have—

$U={\frac {\left[(v^{2}+w^{2}+2vw\ \cos \ \alpha )-\left({\frac {vw\ \sin \ \alpha }{c}}\right)^{2}\right]^{\frac {1}{2}}}{1+{\frac {vw\ \cos \ \alpha }{c^{2}}}}}$

It should be noticed that v and w enter into the expression for velocity symmetrically. If w has the direction of the ξ-axis of the moving system,

$U={\frac {v+w}{1+{\frac {vw}{c^{2}}}}}$

From this equation, we see that by combining two velocities, each of which is smaller than c, we obtain a velocity which is always smaller than c. If we put $v=c-\chi$, and $w=c-\lambda$ where $\chi$ and $\lambda$ are each smaller than c,

$U=c{\frac {2c-\chi -\lambda }{2c-\chi -\lambda +{\frac {\chi \lambda }{c^{2}}}}}<c$.

It is also clear that the velocity of light c cannot be altered by adding to it a velocity smaller than c. For this case,

$U={\frac {c+v}{1+{\frac {cv}{c^{2}}}}}=c$