Ibix said:
No, the Michelson-Morley experiment confirms that the two-way speed of light is isotropic. It doesn't (and cannot) say anything about the one-way speed of light, which was the question in this thread.
Let me work this through...
Let's call the 2-way (or all way) length of one arm L1 and the corresponding length of the other (perpendicular) arm L2. Then we have the speed of light in the North, South, East, and West direction will be cN, cS, cE, and cW respectively.
And we will apply this additional constraint: L1 and L2 will not be precisely equal. They will be off by at least a few wavelengths - enough to be noticeable as the experiment is conducted.
If the measurement is first made with L1 oriented North/South, the interference pattern will reflect the difference L1(1/cN+1/cS)−L2(1/cE+1/cW). Then as the device is rotated clockwise, the interference pattern reflects a range of other cX vs. c(X+π/2) 2-way combinations, and since the interference pattern is not shifting, all of these also yield the same value.
That is: T=L1(1/cX+1/(cX+π))−L2(1/(cX+π/2)+1/(cX−π/2)) is a constant.
So after 90 degrees, you will have:
T=L1(1/cN+1/cS)−L2(1/cE+1/cW)=L2(1/cN+1/cS)−L1(1/cE+1/cW).
0=(L1−L2)((1/cN+1/cS)−(1/cE+1/cW))
(1/cN+1/cS)=(1/cE+1/cW)
More generally:
(1/cX+1/(cX+π))=(1/(cX+π/2)+1/(cX−π/2))
At this point, we cannot even claim that the 2-way path is isotropic - there could be some pattern (like a square) that repeats every 90 degrees. But if we repeat the experiment with (for example) the arms at 1.5 radians to each other instead of 90 degrees, we demonstrate that the two-way is isotropic.
This only says that the sum of the two time periods remains a constant as the direction changes.
So it seems you (
@Ibix ) are right.
Letting C=T/(L1−L2)=1/cX+1/(cX+π) (a constant).
The the average time for each opposing direction would then always be proportional to C/2.
We can define a periodic function f(X) that indicates how much the transit time changes from the average:
f(X)=1/cX−(C/2)
So C=1/cX+1/(cX+π)=(C/2+f(X))+(C/2+f(X+π/2))
Thus f(x)=−f(X+π/2).
So if we add one more constraint, that f(X) is a continuous function, we can demonstrate that there is at least one direction X0 such that f(X0)=−f(X0+π/2)=0. For direction X0, the speed of light is the same in both directions.
If there really is no way to determine f(X) for an inertial reference frame, then by Occam's Razor, we should take f(X)=0.