Page:BumsteadContraction.djvu/15

${\begin{aligned}&\scriptstyle {\mathbf {F} _{t}=\mathbf {E} \cos \psi }\\&\scriptstyle {\mathbf {F} _{n}=\mathbf {E} \sin \psi +|(\mathbf {v} +\mathbf {u} )\times \mathbf {H} |}\end{aligned}}$

in which the term enclosed by vertical lines represents the magnitude only of the vector. $\scriptstyle {\mathbf {H} }$ is perpendicular to the plane containing $\scriptstyle {\mathbf {r} }$ and $\scriptstyle {\mathbf {v} ;}$ let $\scriptstyle {u_{1}}$ be the component of $\scriptstyle {\mathbf {u} }$ in this plane and let $\scriptstyle {w}$ be the resultant of $\scriptstyle {u_{1}}$ and $\scriptstyle {v}$. Then

$\scriptstyle {|(\mathbf {v} +\mathbf {u} )\times \mathbf {H} |=wH={\frac {wv}{{\text{V}}^{2}}}E\sin \theta }$

and

$\scriptstyle {F_{n}=E(\sin \psi +{\frac {wv}{{\text{V}}^{2}}}\sin \theta ).}$

Dividing $\scriptstyle {F_{t}}$ and $\scriptstyle {F_{n}}$ by the longitudinal and transverse masses respectively, we obtain for the accelerations,

${\begin{array}{c}\scriptstyle {f_{t}={\frac {(1-\beta ^{2})^{\frac {3}{2}}}{m_{0}}}\mathbf {E} \cos \psi }\\\scriptstyle {f_{n}={\frac {(1-\beta ^{2})^{\frac {1}{2}}}{m_{0}}}\mathbf {E} (\sin \psi +{\frac {wv}{{\text{V}}^{2}}}\sin \theta )}\end{array}}$

The acceleration perpendicular to the radius vector is

$\scriptstyle {f_{n}=\cos \psi -f_{t}\sin \psi ={\frac {(1-\beta ^{2})^{\frac {1}{2}}}{m_{0}}}\mathbf {E} (\beta ^{2}\sin \psi \cos \psi +{\frac {wv}{{\text{V}}^{2}}}\sin \theta \cos \psi ).}$

Recent estimates make the sun's velocity about 20 kilometers per second, so that $\scriptstyle {\beta ^{2}=0\cdot 45\times 10^{-8};}$ its direction makes an angle with the plane of the earth's orbit of about 55°. When $\scriptstyle {\mathbf {r} }$ is perpendicular to the plane containing $\scriptstyle {\mathbf {v} }$ and the normal to the plane of the orbit, $\scriptstyle {\cos \psi }$ is nearly zero; it must in fact be less than $\scriptstyle {\epsilon }$ (the eccentricity of the orbit) even in the favorable case when the minor axis falls in this position; with the major axis in this position it will be zero. In this position, therefore, the acceleration perpendicular to the radius vector cannot be as much as twice that which was found for the sun at rest. When $\scriptstyle {\mathbf {r} }$ is in the plane containing $\scriptstyle {\mathbf {v} }$ and the normal to the plane of the orbit, $\scriptstyle {\theta =55}$°, $\scriptstyle {\psi <55}$° and $\scriptstyle {w=v.}$ So that the acceleration perpendicular to the radius vector will be less than

$\scriptstyle {{\frac {(1-\beta ^{2})^{\frac {1}{2}}}{m_{0}}}\mathbf {E} \beta ^{2}\sin ~110}$°.

that is its ratio to the acceleration in the direction of the radius will be less than $\scriptstyle {1\cdot 4\times 10^{-9}.}$

In order to be quite certain that astronomical facts are not in conflict with the principle of relativity, it will doubtless be necessary to make detailed comparisons between observation