I Reichenbach Synchronisation: Proving i=r

wnvl2
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I use Reichenbach synchronisation. The one-way speed of light (OWSOL) in the x and y-direction is ##\frac{c}{2\epsilon}## and in the reverse direction it is for both ##\frac{c}{2(1-\epsilon)}## such that the average round trip speed of light is ##c##. For any choice of ##\epsilon## the physical laws should remain the sames as for ##\epsilon = \frac{1}{2}##.

My goal is to prove that the angle of incidence equals the angle of reflection. When using Huyghens it is obvous that I get a different result.

How can I prove ##i = r##?
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I use units where ##c=1## and Anderson coordinates so ##\kappa=2 \epsilon-1##. The transform between Minkowski coordinates (lower case) and Anderson coordinates (upper case) is: $$ t=T+\kappa X$$ $$x=X$$ $$y=Y$$ $$z=Z$$

So, in the Minkowski coordinates we can write the equation of an ingoing null geodesic as $$r^\mu(t,x,y,z) = \left( \lambda, -\frac{\lambda}{\sqrt{2}} , \frac{\lambda}{\sqrt{2}} ,0 \right)$$ for ##\lambda<0## and the equation of an outgoing null geodesic as $$r^\mu(t,x,y,z) = \left( \lambda, \frac{\lambda}{\sqrt{2}} , \frac{\lambda}{\sqrt{2}} ,0 \right)$$ for ##0<\lambda##.

Then transforming from the Minkowski coordinates to the Anderson coordinates we get $$r^\mu(T,X,Y,Z) = \left( \lambda + \frac{\kappa\lambda}{\sqrt{2}} , -\frac{\lambda}{\sqrt{2}} , \frac{\lambda}{\sqrt{2}} ,0 \right)$$ and $$r^\mu(T,X,Y,Z) = \left( \lambda - \frac{\kappa\lambda}{\sqrt{2}} , \frac{\lambda}{\sqrt{2}} , \frac{\lambda}{\sqrt{2}} ,0 \right)$$

So the transformation does not affect the spatial coordinates, just the temporal coordinate. So the angle is the same in both cases.
 
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Do you have a link about "Anderson coordinates"?
 
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