Page:PlummerAberration.djvu/11

262 suppose that he is moving relatively to S in the direction of some point E' on the great circle SE; and if he is to deduce the position P0 in space from the observed position of the star P1, the point E' must lie also on the great circle P0P1. Since P0, P1, E' are points on the sides of the triangle SPE, and S, E, E' are points on the sides of the triangle PP0P1, we have

$\begin{array}{ccccc} \frac{\sin SE'}{\sin EE'} & \cdot & \frac{\sin EP_{1}}{\sin P_{1}P} & \cdot & \frac{\sin PP_{0}}{\sin P_{0}S}=1\\ \\\frac{\sin P_{0}E'}{\sin P_{1}E'} & \cdot & \frac{\sin P_{1}E}{\sin PE} & \cdot & \frac{\sin PS}{\sin P_{0}S}=1.\end{array}$|undefined

Let

$PS=\phi_{0},\ P_{0}S=\phi'_{0},\ PP_{0}=\phi_{0}\ \phi'_{0}$

$PE=\phi_{1},\ P_{1}E=\phi'_{1},\ PP_{1}=\phi_{1}-\phi'_{1}.$

Hence

$\frac{\sin\ SE'}{\sin\ EE'}=\frac{\sin\left(\phi_{1}-\phi'_{1}\right)}{\sin\phi'_{1}}\cdot\frac{\sin\phi'_{0}}{\sin\left(\phi_{0}-\phi'_{0}\right)}$|undefined

$\frac{\sin P_{0}E'}{\sin P_{1}E'}=\frac{\sin\phi_{1}}{\sin\phi'_{1}}\cdot\frac{\sin\phi'_{0}}{\sin\phi{}_{0}}.$|undefined

The first of these shows that

$\frac{\sin\left(\phi_{1}-\phi'_{1}\right)}{\sin\phi'_{1}}\div\frac{\sin\left(\phi_{0}-\phi'_{0}\right)}{\sin\phi'_{0}}=const.$|undefined

for all stars, which happens to be true according to the ordinary theory of aberration. But it is not true according to the "relativity" law of aberration, which gives

$\frac{\sin\left(\phi-\phi'\right)}{\sin\phi'}=\frac{\sin\beta+(1-\cos\beta)\cos\phi}{\cos\beta}.$

In fact it appears that a law of aberration, which would explain the absence of a visible effect arising from the secular motion by supposing that the corrected observation coincides in space with the standard observation, would require simultaneously

$\sin(\phi-\phi')/\sin\phi'=f_{1}(V),\ \sin(\phi-\phi')/\sin\phi=f_{2}(V),$

and this is not possible. We have then to look in another direction for the explanation of the way in which the "relativity" law effects a compensation, and for the necessary hint I am indebted to Mr. Eddington, who very kindly supplied me with a particular case which was both simple and illuminating.

12. A general proof will now be given. Let the ether-velocity of the observer S be $$V_{0}=U\ \sin\beta_{0}$$, and of the observer E be $$V_{1}=U\ \sin\beta_{1}$$, and let the angle between them be α, being measured from V0 towards V1. The components of V0 along and perpendicular