Page:DeSitterGravitation.djvu/23

The orbit thus remains fixed and plane after the transformation. Since

${\begin{aligned}&a=\sin \ i\ \sin \ \Omega ,\\&b=-\sin \ i\ \cos \ \Omega ,\\&c=\cos \ i,\end{aligned}}$

we find easily

${\begin{aligned}\sin \ &id\ \Omega =q_{1}\mathrm {C} \left(\alpha \ \cos \ \Omega +\beta \ \sin \ \Omega \right)\\&di=q_{1}\mathrm {C} \left[\left(\alpha \ \sin \ \Omega -\beta \ \cos \ \Omega \right)\cos \ i-\gamma \ \sin \ i\right]\end{aligned}}$

(40)

If $\mathrm {L}$ and $\mathrm {B}$ are the longitude and latitude of the positive half of the axis of the transformation we have—

$\alpha =\cos \ \mathrm {L} \ \cos \ \mathrm {B} ,\qquad \beta =\sin \ \mathrm {L} \ \cos \ \mathrm {B} ,\qquad \gamma =\sin \ \mathrm {B} ,$ $\mathrm {C} =\sin \ i\ \cos \ \mathrm {B} \ \sin(\Omega -\mathrm {L} )+\cos \ i\ \sin \ \mathrm {B} ,$ $\alpha \ \cos \ \Omega +\beta \ \sin \Omega =\cos \ \mathrm {B} \ \cos(\Omega -\mathrm {L} ),$ $\alpha \ \sin \ \Omega -\beta \ \cos \Omega =\cos \ \mathrm {B} \ \sin(\Omega -\mathrm {L} ).$

If the plane to be transformed is the plane of ($x,y$) itself, we have in the system ($x,y,z,t$) $i_{0}=0$. The transformed position of the plane is then defined by

$\tan \ i'_{0}={\frac {q_{1}\gamma {\sqrt {\alpha ^{2}+\beta ^{2}}}}{1-q_{1}\left(1-\gamma ^{2}\right)}},\qquad \tan \ \Omega '_{0}={\frac {\alpha }{-\beta }},$

or

$\tan \ i'_{0}={\tfrac {1}{2}}q_{1}{\frac {\sin \ 2\mathrm {B} }{1-q_{1}\cos ^{2}\mathrm {B} }},\qquad \Omega '_{0}=\mathrm {L} +90^{\circ }.$

(41)

Let the plane of ($x,y$) be the ecliptic, and consider another plane of which the inclination and node in the system ($x,y,z,t$) are $i$ and $\Omega$. In the system ($x',y',z',t'$) its inclination and node on the plane of ($x',y'$) are $i+di$ and $\Omega +d\Omega$, as given by the formulæ (40). Let its inclination on the transformed ecliptic be $i+\Delta i,\Omega +\Delta \Omega$. Taking unity for the denominator of $\tan i'_{0}$ in (41) we find easily—

${\begin{alignedat}{2}\sin \ i\Delta \Omega &=\sin \ id\ \Omega &&-q_{1}\gamma \cos \ i(\alpha \ \cos \ \Omega +\beta \ \sin \Omega ),\\\Delta i&=di&&+q_{1}\gamma \cos \ i(\beta \ \cos \ \Omega -\alpha \ \sin \Omega ),\end{alignedat}}$

or

$\Delta \Omega =q_{1}\cos ^{2}\mathrm {B} \ \sin(\Omega -\mathrm {L} )\cos(\Omega -\mathrm {L} )$ ${\begin{aligned}\Delta i=q_{1}\sin \ i[\cos ^{2}\mathrm {B} \ \sin ^{2}(\Omega -\mathrm {L} )-\ &\sin ^{2}B\ \cos \ i],\\&-q_{1}\ \sin ^{2}i\ \sin \ \mathrm {B} \ \cos \ \mathrm {B} \ \sin \ (\Omega -\mathrm {L} ).\end{aligned}}$

If the transformed plane be the equator, we have—

$\Omega =\Upsilon =pt,\qquad i=\epsilon ,$