I Reichenbach Synchronisation: Proving i=r

[wnvl2]

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 $i = r$?

 

[Dale]

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.

 

[wnvl2]

Do you have a link about "Anderson coordinates"?

 

[Dale]

Do you have a link about "Anderson coordinates"?

Yes, here it is https://www.sciencedirect.com/science/article/pii/S0370157397000513