How do we know the interferometer is squared to the ether wind (one arm is parallel to it)?
We do not know that the interferometer has one arm moving along the direction of the ether wind. In fact, in general we would assume that the ether wind is in some random direction. We just made this assumption because the parallel case is easy to calculate, and then showed how there’s a difference when the apparatus is rotated by . That’s the main idea: that the fringe shift depends on angle with respect to the wind, so it will change as you rotate the apparatus. No matter which direction the wind is in, an ether wind will cause fringes to shift as you rotate the apparatus.
Related Questions
- March 26, 2004: On other weather sites I have seen "streamline" depictions where lines are drawn parallel to the wind. When the lines come together, is that the same as "convergence"?
- How do we know the interferometer is squared to the ether wind (one arm is parallel to it)?
- What if the ether wind happens to be coming directly from above?
