I am doing a self-study of A.P. French's book on Special relativity and I am stuck at the Michelson-Morley experiment. I am having a few difficulties in the derivation in the book,

  1. On page 54, it is mentioned that the ether wind and light add up using vector addition and that the resultant velocity relative to the interferometer is given by $\sqrt{c^2 - v^2}$. I do not understand this since if the ether wind is parallel to the horizontal arm, shouldn't the velocity along the vertical arm remain independent of the ether wind speed? Am I missing something here?
  2. On page 64, there is a remark that the MM experiment shows the null result for any random orientation of the arm relative to the horizontal. However, when I try to follow the steps for the usual MM experiment, my equations does not solve the equation $$t_1-t_2 = \frac{2(l_{10} - l_{20})/c}{\sqrt{1 - \frac{v^2}{c^2}}}$$ derived earlier in the literature. Did anybody solve this? If requested, I will post my solution here for better clarity.

1 Answer 1


To your first question, the motion of light through moving aether is thought of identically to movement of a wave on a stream. If one is traveling in a boat against the current, any waves you make perpendicular to the direction of travel will travel downstream—they pick up a horizontal velocity component corresponding to the velocity of the medium, leading in analogy to light to the equation you give for the resultant velocity. The "wave" is an emergent phenomenon of the collective behavior of the medium through which it travels; any collective behavior of the medium as a whole is transmitted to the wave.

As for your second question, I would need to see your solution to guess as to its error and why an equation solution is even needed in the first place. In my mind, it's "obvious" (and easily shown through the triangle inequality) that if a null result is obtained for the "downwind" direction a null result will be obtained for any random orientation, as the effect on the velocity would be expected to be smaller. Perhaps I misunderstand in what sense you mean "horizontal" (The Solar plane? The tangent plane to the Earth's surface? A plan view of the MM apparatus?). In addition, the formula you use to check your result appears to be of relativistic origin. I believe one must use the classical model in order to analyze the power of the results refuting the classical model.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.