HansH said:
then something should be wrong in my assumptions.
...as we have been telling you for some time now.
Let's do this properly. See the diagram in
post #1. In the usual Einstein coordinates we describe light leaving S, traveling to A or B, and being reflected to R.
The light leaves S at ##x=0##, ##y=0##, ##t=0##.
It arrives at A/B at ##x=\pm X##, ##y=0##, ##t=X/c##.
It arrives at R at ##x=0##, ##y=Y##, ##t=X/c+(X/\cos\alpha)/c=\frac Xc(1+\sec\alpha)##.
Now let's work in a frame where the speed of light is not isotropic. To do this, we simply apply the coordinate transform ##x'=x##, ##y'=y##, ##t'=t+\kappa x##, where ##\kappa## is a constant with dimensions of inverse velocity and ##|\kappa|<\frac 1c## so that our spatial planes remain spacelike.
The light leaves S at ##x'=0##, ##y'=0##, ##t'=0##.
It arrives at A/B at ##x'=\pm X##, ##y'=0##, ##t'=X/c\pm\kappa X##.
It arrives at R at ##x'=0##, ##y'=Y##, ##t'=\frac Xc(1+\sec\alpha)##.
Note that the only actual difference from the Einstein case is the arrival time of the light at A and B.
From this we can deduce the speed of light along AR and BR. The distance traveled is ##X\sec\alpha## and the time taken is ##\frac Xc(1+\sec\alpha)-(X/c\pm\kappa X)## (i.e., the arrival time at R minus the arrival time at A/B), so the speed is $$c_\pm=\frac{c}{1\mp \kappa c\cos\alpha}$$There are a few interesting cases. First, if ##\kappa=0## this clearly reduces to the isotropic case. Second, if ##\alpha=0## then AR and BR are antiparallel and we have a 1d problem in which we get ##c_\pm=c/(1\mp\kappa c)##. It's easy to see that the average speed, ##c_{av}##, satisfies ##\frac{2X}{c_{av}}=\frac{X}{c_+}+\frac{X}{c_-}##, and hence that ##c_{av}=c## for all ##\kappa## - i.e. that the two-way speed of light is preserved.
If you get different equations from these you are doing something wrong.