Page:PlummerAberration.djvu/13

264 13. The process of eliminating $$\lambda_{1},\ \mu_{1},\ \nu_{1}$$ and θ is perfectly simple and straightforward, if rather complicated. We find

$1-\left(\lambda_{1}\cos\theta+\mu_{1}\sin\theta\right)\sin\beta_{10}=\frac{\left(1+\lambda\ \sin\beta_{0}\right)\cos^{2}\beta_{1}}{\left\{ 1+\left(\lambda\ \cos\alpha+\mu\ \sin\alpha\right)\sin\beta_{1}\right\} \left(1-\cos\alpha\ \sin\beta_{0}\ \sin\beta_{1}\right)}$|undefined

whence

$\nu_{10}=\nu\ \cos\beta_{0}/\left(1+\lambda\sin\beta_{0}\right).$

Also

$\begin{array}{r} \sin\beta_{10}\left(\mu_{1}\cos\theta-\lambda_{1}\sin\theta\right)\left\{ 1+\left(\lambda\ \cos\alpha+\mu\ \sin\alpha\right)\sin\beta_{1}\right\} \left(1-\cos\alpha\ \sin\beta_{0}\ \sin\beta_{1}\right)\\ \\=\cos\beta_{1}\left\{ \mu\left(\cos\alpha\ \sin\beta_{1}-\sin\beta_{0}\right)-\left(\lambda+\sin\beta_{0}\right)\sin\alpha\ \sin\beta_{1}\right\} \end{array}$

whence

$\mu_{10}\sin\beta_{10}=\frac{\cos\beta_{0}\left\{ \mu\left(\cos\alpha\ \sin\beta_{1}-\sin\beta_{0}\right)-\left(\lambda+\sin\beta_{0}\right)\sin\alpha\ \sin\beta_{1}\right\} }{\left(1+\lambda\ \sin\beta_{0}\right)\left(1-\cos\alpha\ \sin\beta_{0}\ \sin\beta_{1}\right)}.$

Similarly a little reduction gives

$\lambda_{10}\sin\beta_{10}=\frac{\left(\lambda+\sin\beta_{0}\right)\left(\cos\alpha\ \sin\beta_{1}-\sin\beta_{0}\right)+\mu\ \sin\alpha\cos^{2}\beta_{0}\sin\beta_{1}}{\left(1+\lambda\ \sin\beta_{0}\right)\left(1-\cos\alpha\ \sin\beta_{0}\ \sin\beta_{1}\right)}.$|undefined

We come now to the interpretation of these expressions. The components of V1 along and perpendicular to V0 are V1 cos α, V1 sin α. Hence the apparent components of the relative velocity of E as observed by S are

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

or

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

Hence if χ be the angle between V0 and the resultant $$V_{01}=U\ \sin\beta_{01}$$, we have

$\sin\beta_{01}\cos\chi=\left(\cos\alpha\ \sin\beta_{1}-\sin\beta_{0}\right)/\left(1-\cos\alpha\ \sin\beta_{0}\ \sin\beta_{1}\right)$

$\sin\beta_{01}\sin\chi=\sin\alpha\ \cos\beta_{0}\sin\beta_{1}/\left(1-\cos\alpha\ \sin\beta_{0}\ \sin\beta_{1}\right)$

and we deduce

$\cos\beta_{01}=\cos\beta_{0}\cos\beta_{1}/\left(1-\cos\alpha\ \sin\beta_{0}\ \sin\beta_{1}\right).$

We see that $$\beta_{01}=\beta_{10}$$, and when we introduce $$\lambda_{0},\ \mu_{0},\ \nu_{0}$$ and χ into the last expressions for $$\lambda_{10},\ \mu_{10},\ \nu_{10}$$ we obtain simply

$\lambda_{10}=\lambda_{0}\cos\chi+\mu_{0}\sin\chi,\ \mu_{10}=\mu_{0}\cos\chi-\lambda_{0}\sin\chi,\ \nu_{10}=\nu_{0}.$

The interpretation is that the sky as seen by the observer E and referred to the direction in which he observes himself to be moving relative to S, is identical with the sky as seen by the observer S, and referred to the direction in which he observes E to be moving relative to himself. Thus the observations of E,