Page:PlummerAberration.djvu/5

and thus the complete transformation is determined:

${\begin{array}{l}\tau =t\ \sec \beta -x\ \tan \beta /U\\\\\xi =x\ \sec \beta -Ut\ \tan \ \beta \\\\\eta =y,\ \zeta =z,\ \sin \beta =v/U.\end{array}}$

5. The physical interpretation of this transformation is simple. The coordinates y and z are unaltered, but we have at any given time t

$\xi _{2}-\xi _{1}=\left(x_{2}-x_{1}\right)\sec \beta .$

The measured distance between two points in the direction of motion is therefore greater than the actual distance in the ratio of sec β : 1. This accords with the idea already entertained that the corresponding dimension of the material system, including any scale which may be used for making measures, is actually diminished in consequence of the motion in the ratio cos β : 1.

At a given position in the ether

${\frac {\partial \tau }{\partial t}}=\sec \beta ,$

which means that a stationary clock made to synchronise with passing clocks keeping "local" time must be accelerated in the ratio sec β : 1 when compared with the standard time of space t. But on the other hand,

${\frac {d\tau }{dt}}=\sec \beta -v\ \tan \beta /U=\cos \beta ,$

which means that a "local" clock moving through the ether with the velocity v has a rate retarded in the ratio cos β : 1 when compared with the "standard" clock. The result of the transformation is that we have established a consistent system of apparent time which is such that if we imagine luminous clock-faces at all points of the moving system indicating local time, those which are at equal measured distances from a given station will appear to show the same time, and this a time differing from the local time of the station by an amount equal to the apparent constant distance divided by U.

6. We now see that if the laws of optics relative to the moving system, expressed in terms of ξ, η, ζ and τ, are formally the same as the laws for a stationary system, expressed in terms of x, y, z and t, there will be no possibility of detecting the fact of motion by any optical experiment made with apparatus which is carried with the system.

Now we have from the transformation of § 4,

${\begin{array}{l}{\frac {d\xi }{d\tau }}=\left({\frac {dx}{dt}}-U\ \sin \beta \right)/\left(1-{\frac {dx}{dt}}\sin \beta /U\right)\\\\{\frac {d\eta }{d\tau }}={\frac {dy}{dt}}\cos \beta /\left(1-{\frac {dx}{dt}}\sin \beta /U\right)\\\\{\frac {d\zeta }{d\tau }}={\frac {dz}{dt}}\cos \beta /\left(1-{\frac {dx}{dt}}\sin \beta /U\right)\end{array}}$