# One-way speed of light

Some authors (referenced in this Wikipedia article ) have alleged that The “one-way” speed of light, from a source to a detector, cannot be measured independently of a convention as to how to synchronize the clocks at the source and the detector – and that this somehow has important implications with regard to the interpretation and application of Special Relativity.

But I beg to differ.

Consider the scenario of two remote systems, B and C, that are stationary with respect to us who, let us say, are at position A. Now let a light signal from B be emitted towards both C and A, and let C, on receiving that signal immediately signal towards us at A. The time delay between our receipt of the two signals will be the light travel time from B to C plus the difference between the light travel times to us from B and C.

So, if we let $t_{XY}$ be the light travel time from X to Y,  then the time difference between the direct and indirect signals starting at B is given by $\Delta t_B = (t_{BC} +t_{CA})-t_{BA}$.

And if we do the process in reverse for signals starting at C we get a time difference of $\Delta t_C = (t_{CB} +t_{BA})-t_{CA}$.

So, if we get the same time difference both ways, then we’ll know that  $(t_{BC} +t_{CA})-t_{BA} = (t_{CB} +t_{BA})-t_{CA}$,

which gives $t_{BC} -t_{CB} =2 (t_{BA}-t_{CA})$.

So in that case the difference between light travel times from B to C and from C to B is exactly twice the difference between the travel times from B to A and C to A.

Now let’s restrict to the case where B and C are separated by a distance $d$ and are both at the same distance $D$ from A, and let $v_{XY}$ be the speed of light in the direction from X to Y and $v_0$ be the speed in the perpendicular direction towards A from the midpoint of the line BC.

Then for the case of equal time differences both ways our previous result becomes $\frac{d}{v_{BC}} -\frac{d}{v_{CB}} =2 (\frac{D}{v_{BA}}-\frac{D}{v_{CA}})$.

In the case that A is midway between B and C (which is what the equidistance gives us in the case of one space dimension) we have  $d=2D$ with $v_{BA}=v_{BC}$ and $v_{CA}=v_{CB}$, so the equal time differences condition is just that $\frac{d}{v_{BC}} -\frac{d}{v_{CB}} =2 (\frac{d/2}{v_{BC}}-\frac{d/2}{v_{CB}})$ – which is clearly always true even if the two one-way speeds are different.

For the case that A is not on the line BC, if we let  $\omega=2\theta$ be the angle at A between AB and AC then $d=2D\sin{\theta}$ and the condition of equal time differences becomes $2D\sin{\theta}(\frac{1}{v_{BC}} -\frac{1}{v_{CB}}) =2D (\frac{1}{v_{BA}}-\frac{1}{v_{CA}})$.

If we let $v_{\theta}$ denote the speed of light in direction at angle $\theta$ from the right bisector of BC, and $T_{\theta}=\frac{1}{v_{\theta}}$, and if the equal time condition is observed in all directions, then we need to find a function $T_{\theta}$ satisfying the conditions $(T_{\theta+\phi}-T_{-\theta+\phi})=\sin{\theta}(T_{\pi/2+\phi}-T_{-\pi/2+\phi})$