Page:PlummerAberration.djvu/12

to V₁ are $V_{0}\cos \alpha ,\ -V_{0}\sin \alpha$. Hence for E the components of the apparent velocity of S are (by § 6)

${\frac {V_{0}\cos \alpha -U\ \sin \beta _{1}}{1-V_{0}\cos \alpha \sin \beta _{1}/U}},\ {\frac {-V_{0}\sin \alpha \ \cos \beta _{1}}{1-V_{0}\cos \alpha \sin \beta _{1}/U}}$

or

${\frac {U\left(\cos \alpha \ \sin \beta _{0}-\sin \beta _{1}\right)}{1-\cos \alpha \sin \beta _{0}\sin \beta _{1}}},\ {\frac {-U\ \sin \alpha \ \sin \beta _{0}\sin \beta _{1}}{1-\cos \alpha \sin \beta _{0}\sin \beta _{1}}}.$

Now E must infer that his own velocity relative to S, $V_{10}=U\ \sin \beta _{10}$, has these components reversed in sign; and if θ be the angle between V₁ and the resultant we have

$\sin \beta _{10}\cos \theta =\left(\sin \beta _{1}-\cos \alpha \ \sin \beta _{0}\right)/\left(1-\cos \alpha \sin \beta _{0}\sin \beta _{1}\right)$

$\sin \beta _{10}\cos \theta =\sin \alpha -\sin \beta _{0}\ \cos \beta _{1}/\left(1-\cos \alpha \sin \beta _{0}\sin \beta _{1}\right)$

and we deduce

$\cos \beta _{10}=\cos \beta _{0}\ \cos \beta _{1}/\left(1-\cos \alpha \sin \beta _{0}\sin \beta _{1}\right).$

We take the z-axis perpendicular to V₀ and V₁ throughout, and

(i) The x-axis parallel to V₀. If (λ, μ, ν) are the direction cosines of a star in its true position, S observes this star in the direction (by § 7)

$\lambda _{0}={\frac {\lambda +\sin \beta _{0}}{1+\lambda \sin \beta _{0}}},\ \mu _{0}={\frac {\mu \ \cos \beta _{0}}{1+\lambda \sin \beta _{0}}},\ \nu _{0}={\frac {\nu \ \cos \beta _{0}}{1+\lambda \sin \beta _{0}}}.$

(ii) We turn the x-axis through the angle a to bring it into the direction of V₁. The direction cosines of the star in its true position become

$\lambda \ \cos \alpha +\mu \ \sin \alpha ,\ \mu \ \cos \alpha -\lambda \ \sin \alpha ,\ \nu ,$

and E will observe it in the direction

${\begin{array}{lr}\lambda _{1}=&\left(\lambda \ \cos \alpha +\mu \ \sin \alpha +\sin \beta _{1}\right)/\left\{1+\left(\lambda \ \cos \alpha +\mu \ \sin \alpha \right)\sin \beta _{1}\right\}\\\\\mu _{1}=&(\mu \ \cos \alpha -\lambda \ \sin \alpha )\cos \beta _{1}/\left\{1+\left(\lambda \ \cos \alpha +\mu \ \sin \alpha \right)\sin \beta _{1}\right\}\\\\\nu _{1}=&\nu \ \cos \beta _{1}/\left\{1+\left(\lambda \ \cos \alpha +\mu \ \sin \alpha \right)\sin \beta _{1}\right\}\end{array}}$

(iii) We turn the x-axis through the further angle θ to bring it into the direction of V₁₀. The direction cosines of the star in its apparent position become

$\lambda _{1}\cos \theta +\mu _{1}\sin \theta ,\ \mu _{1}\cos \theta -\lambda _{1}\sin \theta ,\ \nu _{1},$

and when E has corrected this position for his observed relative velocity V₁₀ he will infer that the star lies in the direction

${\begin{array}{lr}\lambda _{10}=&\left(\lambda _{1}\ \cos \theta +\mu _{1}\ \sin \theta -\sin \beta _{10}\right)/\left\{1-\left(\lambda _{1}\ \cos \theta +\mu _{1}\ \sin \theta \right)\sin \beta _{10}\right\}\\\\\mu _{10}=&(\mu _{1}\ \cos \theta -\lambda _{1}\ \sin \theta )\cos \beta _{10}/\left\{1-\left(\lambda _{1}\ \cos \theta +\mu _{1}\ \sin \theta \right)\sin \beta _{10}\right\}\\\\\nu _{10}=&\nu _{1}\ \cos \beta _{10}/\left\{1-\left(\lambda _{1}\ \cos \theta +\mu _{1}\ \sin \theta \right)\sin \beta _{10}\right\}.\end{array}}$

These have to be compared with $\lambda _{0},\ \mu _{0},\ \nu _{0}$, remembering that the axes have been turned through an angle α + θ.