Measuring the speed of light and the speed of the medium.

Let us consider a medium and an observer that is moving through that medium. Say that the magnitude of the velocity of light with respect to the medium is $c$ and say that the velocity of the observer with respect to the medium is $\vec{v}$. Can we construct a device that allows us to measure the velocity of the observer with respect to the medium?

We consider a simple device constructed by a measuring rod $L$, a pulsing light source and a sensor at one end of the measuring rod $L$ and a mirror on the other end of the measuring rod $L$. If this measuring rod $L$ is moving with respect to the medium such that $\phi$ is the angle between the measuring rod and the velocity of the measuring rod – then we find

$\displaystyle \frac{2L}{\Delta T} = c \frac{1 - \beta^2}{\sqrt{ 1 - \beta^2 \sin^2(\phi) }}$

This can be written as

$\displaystyle \left( \frac{\Delta T}{2L} \right)^2_\phi = \frac{1}{c^2} \frac{1 - \frac{1}{2}\beta^2}{1 - \beta^2} + \frac{1}{c^2} \frac{ \frac{1}{2} \beta^2 \cos(2\phi)}{1 - \beta^2}$

Say the we have three measuring rods – the main measuring rod and other measuring rods at angle $\pm 45^\circ$ then we have

$\displaystyle \left( \frac{\Delta T}{2L} \right)^2_{\phi \pm \frac{1}{4} \pi} = \frac{1}{c^2} \frac{1 - \frac{1}{2}\beta^2}{1 - \beta^2} \mp \frac{1}{c^2} \frac{ \frac{1}{2} \beta^2 \sin(2\phi)}{1 - \beta^2}$

Now it is clear that

$\displaystyle 2 \left( \frac{\Delta T}{2L} \right)^2_{\phi} - \left( \frac{\Delta T}{2L} \right)^2_{\phi + \frac{1}{4} \pi} - \left( \frac{\Delta T}{2L} \right)^2_{\phi - \frac{1}{4} \pi} = \frac{1}{c^2} \frac{\beta^2}{1-\beta^2} \cos(2\phi)$

and

$\displaystyle \left( \frac{\Delta T}{2L} \right)^2_{\phi - \frac{1}{4} \pi} - \left( \frac{\Delta T}{2L} \right)^2_{\phi + \frac{1}{4} \pi} = \frac{1}{c^2} \frac{\beta^2}{1-\beta^2} \sin(2\phi)$

Which gives

$\displaystyle \tan(2\phi) = \frac{\displaystyle \left( \frac{\Delta T}{2L} \right)^2_{\phi - \frac{1}{4} \pi} - \left( \frac{\Delta T}{2L} \right)^2_{\phi + \frac{1}{4} \pi}}{\displaystyle 2 \left( \frac{\Delta T}{2L} \right)^2_{\phi} - \left( \frac{\Delta T}{2L} \right)^2_{\phi + \frac{1}{4} \pi} - \left( \frac{\Delta T}{2L} \right)^2_{\phi - \frac{1}{4} \pi}}$

and

$\displaystyle \left(\frac{1}{c^2} \frac{\beta^2}{1-\beta^2}\right)^2 = \left\{ 2 \left( \frac{\Delta T}{2L} \right)^2_{\phi} - \left( \frac{\Delta T}{2L} \right)^2_{\phi + \frac{1}{4} \pi} - \left( \frac{\Delta T}{2L} \right)^2_{\phi - \frac{1}{4} \pi} \right\}^2 + \left\{ \left( \frac{\Delta T}{2L} \right)^2_{\phi - \frac{1}{4} \pi} - \left( \frac{\Delta T}{2L} \right)^2_{\phi + \frac{1}{4} \pi} \right\}^2$

but also

$\displaystyle \left(\frac{1}{c^2} \frac{2-\beta^2}{1-\beta^2}\right)^2 = \left\{ \left( \frac{\Delta T}{2L} \right)^2_{\phi - \frac{1}{4} \pi} + \left( \frac{\Delta T}{2L} \right)^2_{\phi + \frac{1}{4} \pi} \right\}^2$

thus

$\displaystyle \left( \frac{\beta^2}{2-\beta^2} \right)^2 = \left\{1 - \frac{\displaystyle 2 \left( \frac{\Delta T}{2L} \right)^2_{\phi}}{\displaystyle \left( \frac{\Delta T}{2L} \right)^2_{\phi - \frac{1}{4} \pi} + \left( \frac{\Delta T}{2L} \right)^2_{\phi + \frac{1}{4} \pi}} \right\}^2 + \left\{ \frac{\displaystyle \left( \frac{\Delta T}{2L} \right)^2_{\phi - \frac{1}{4} \pi} - \left( \frac{\Delta T}{2L} \right)^2_{\phi + \frac{1}{4} \pi}}{\displaystyle \left( \frac{\Delta T}{2L} \right)^2_{\phi - \frac{1}{4} \pi} + \left( \frac{\Delta T}{2L} \right)^2_{\phi + \frac{1}{4} \pi}} \right\}^2$

which allows to solve $\beta$, and

$\displaystyle \begin{array}{rcl} \displaystyle \frac{2}{c} &=& \displaystyle \sqrt{ \left\{ \left( \frac{\Delta T}{2L} \right)^2_{\phi - \frac{1}{4} \pi} + \left( \frac{\Delta T}{2L} \right)^2_{\phi + \frac{1}{4} \pi} \right\}^2 }\\\\&&\displaystyle -\sqrt{ \left\{ 2 \left( \frac{\Delta T}{2L} \right)^2_{\phi} - \left( \frac{\Delta T}{2L} \right)^2_{\phi + \frac{1}{4} \pi} - \left( \frac{\Delta T}{2L} \right)^2_{\phi - \frac{1}{4} \pi} \right\}^2 + \left\{ \left( \frac{\Delta T}{2L} \right)^2_{\phi - \frac{1}{4} \pi} - \left( \frac{\Delta T}{2L} \right)^2_{\phi + \frac{1}{4} \pi} \right\}^2 }\end{array}$

which allows to solve $c$. So we can find the speed of light with only one clock – assuming that there is some medium for light propagation. The question is if such an experiment can be performed as $\Delta T$ is very small. Once we can find the speed of light by using only one clock then we are able to synchronize two remote clocks. We may wonder what that does with the principles of Relativity?

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