Page:BumsteadContraction.djvu/13

We must also observe that the "apparent" acceleration $\scriptstyle {(f_{1}^{\prime },~f_{2}^{\prime })}$ differs from the "true" acceleration not only on account of the different scale of length in the $\scriptstyle {x}$ direction, but also because of the larger unit of time given by a moving clock. Thus

${\begin{array}{c}\scriptstyle {f_{1}^{\prime }={\frac {1}{1-\beta ^{2}}}f_{1}{\frac {x^{\prime }}{x}}}\\\scriptstyle {f_{2}^{\prime }={\frac {1}{1-\beta ^{2}}}f_{2}.}\end{array}}$

(8)

In equations (6) put $\scriptstyle {\frac {x}{r}}$ for $\scriptstyle {\cos \theta ,}$ and $\scriptstyle {\frac {y}{r}}$ for $\scriptstyle {\sin \theta ;}$ put for $\scriptstyle {\mathbf {E} }$ its value from (1) and for $\scriptstyle {r}$ its value from (7); substituting in (8) the values thus obtained for $\scriptstyle {f_{1}}$ and $\scriptstyle {f_{2},}$ we obtain. ${\begin{aligned}&\scriptstyle {f_{1}^{\prime }={\text{V}}^{2}{\frac {e}{r^{\prime 2}}}x^{\prime }}\\&\scriptstyle {f_{2}^{\prime }={\text{V}}^{2}{\frac {e}{r^{\prime 2}}}y}\end{aligned}}$ The resultant "apparent" acceleration will thus be $\scriptstyle {\mathbf {f} ^{\prime }={\text{V}}^{2}{\frac {e}{r^{\prime 2}}}\mathbf {r} _{1}^{\prime }}$ When $\scriptstyle {\mathbf {r} _{1}^{\prime }}$ is an "apparent" unit vector in the direction $\scriptstyle {r^{\prime }.}$

When there is relative motion of the planet with respect to the sun, however, the compensation is not perfect. In fact, deviations from the Newtonian law may be introduced which would not exist if the longitudinal and transverse masses were equal. This may be most easily seen when the attracting body is at rest in the ether with a planet moving about it; in this case the force given by electrical theory is the ordinary electrostatic force; it will be in the direction of the radius vector and will vary according to the inverse square of the distance. But the resultant acceleration will not be along the radius vector if the longitudinal and transverse masses are different. Let $\scriptstyle {\phi }$ be the angle between the radius vector and the tangent to the path; and let the forces and accelerations, tangential and normal to the path, be respectively $\scriptstyle {F_{t},~F_{n},~f_{t},~f_{n}.}$ Then ${\begin{array}{c}\scriptstyle {F_{t}={\text{V}}^{2}{\frac {e}{r^{2}}}\cos \phi ;\ F_{n}={\text{V}}^{2}{\frac {e}{r^{2}}}\sin \phi }\\\scriptstyle {f_{t}={\frac {(1-\beta ^{2})^{\frac {3}{2}}}{m_{0}}}F_{t}={\frac {{\text{V}}^{2}e}{m_{0}r^{2}}}(1-\beta ^{2})^{\frac {3}{2}}\cos \phi }\\\scriptstyle {f_{n}={\frac {(1-\beta ^{2})^{\frac {1}{2}}}{m_{0}}}F_{n}={\frac {{\text{V}}^{2}e}{m_{0}r^{2}}}(1-\beta ^{2})^{\frac {1}{2}}\sin \phi }\end{array}}$