Page:PlummerAberration.djvu/12

Jan. 1910. 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)

$\frac{V_{0}\cos\alpha-U\ \sin\beta_{1}}{1-V_{0}\cos\alpha\sin\beta_{1}/U},\ \frac{-V_{0}\sin\alpha\ \cos\beta_{1}}{1-V_{0}\cos\alpha\sin\beta_{1}/U}$|undefined

or

$\frac{U\left(\cos\alpha\ \sin\beta_{0}-\sin\beta_{1}\right)}{1-\cos\alpha\sin\beta_{0}\sin\beta_{1}},\ \frac{-U\ \sin\alpha\ \sin\beta_{0}\sin\beta_{1}}{1-\cos\alpha\sin\beta_{0}\sin\beta_{1}}.$|undefined

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)

$\lambda_{0}=\frac{\lambda+\sin\beta_{0}}{1+\lambda\sin\beta_{0}},\ \mu_{0}=\frac{\mu\ \cos\beta_{0}}{1+\lambda\sin\beta_{0}},\ \nu_{0}=\frac{\nu\ \cos\beta_{0}}{1+\lambda\sin\beta_{0}}.$|undefined

(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.