# Does Michelson Morley experiment assume that the Earth drags aether?

My knowledge is that the purpose of Michelson Morley experiment was to measure the velocity of Earth with respect to aether. It was assumed that the Earth does not "drag" the aether or interfere with it, as it was shown by measures on stellar aberration by Bradley.

Nevertheless, considering the red light ray in picture, in order to hit the mirror this ray must travel at the same velocity of earth horizontally, otherwise it would surely miss the mirror. So does this mean that in the description of the experiment it is assumed that the Earth drags the aether? Otherwise why should the ray have the horizontal velocity of Earth (which allows it to hit the mirror) instead of just going on a straight line and miss the mirror?

The light ray only need to be aimed towards the future position of the mirror. Let $\Delta t$ be the time it takes for the wave front to move from the first splitter mirror to the second mirror (horizontal on your diagram). The latter has moved by $v\Delta t$ and therefore light has to travel a distance $\sqrt{L^2 + (v\Delta t)^2}$ to reach it, and that distance shall be equal to $c\Delta t$ as light propagates at the speed $c$ in all directions in the Aether frame where your diagram is drawn, which yield

$$\Delta t = \frac{L}{\sqrt{c^2-v^2}},$$

which is half of $T_t$ on your diagram, as it should be.

But the light ray is not purposely "aimed toward the future position of the mirror". It is only the incoming light ray reflecting by the splitter mirror. Naively applying the reflection law "incident angle = reflected angle", we would expect the reflected ray to go vertically on your diagram, so what is going on? In fact, this reflection law is not valid for a moving mirror. So first, let's see what is the reflection law in the case you wondered about. The following diagram is the same as yours with extra information relevant to the question at hand. The light rays are in red. The angle between the mirror plane and the speed of the lab frame with respect to the Aether is $\varphi=\pi/4$; the incident angle is $\alpha=\pi/4$; the reflected angle is $\beta=\pi/4+\theta$. Then the argument I developed in my first paragraph implies that

$$\sin\theta=\frac{v\Delta t}{c\Delta t}=\frac{v}{c}.$$

Since $v/c \ll 1$, the angle $\theta$ is very small and therefore $\sin\theta \approx\theta$, hence putting everything together

$$\beta = \frac{\pi}{4} + \frac{v}{c}.\tag{1}$$

This is the reflection law in this configuration. It is a special case of the general law. Let me update the above diagram of mine to present the general case. The reflection law valid when $v/c \ll 1$ is

$$\beta = \alpha + 2\frac{v}{c}\sin\varphi\sin\alpha.\tag{2}.$$

As you can see, with $\alpha=\varphi=\pi/4$, eqn. (2) gives indeed eqn. (1).

The last step is then to derive (2). There is an excellent paper Gju04 working out the law of reflection for any speed $v$ by using Huygens' principle. Another paper of the same author Gju04bis derives it using Fermat's principle. It should be noted that Einstein did derive the formula in his seminal relativity paper!

[Gju04] Aleksandar Gjurchinovski. Reflection of light from a uniformly moving mirror. American Journal of Physics, 72(10):1316–1324, 2004.

[Gju04bis] Aleksandar Gjurchinovski. Einstein’s mirror and fermat’s principle of least time. American Journal of Physics, 72(10):1325–1327, 2004.

• Thanks a lot for your answer! If I may, I can't understand why "the light ray only need to be aimed towards the future position of the mirror": in a interferometer (an ideal one) the red light ray should have a unique precise direction, perpendicular to the blue ray, which (I think) does not correspond to the future position of the mirror. Could you say something more on this point? Since there is only one ray how can it be surely directed in the exact direction that is the right one to get the mirror? Aug 27 '17 at 7:33
• The law of refraction on a moving mirror is not just "incident angle = reflection angle": it depends on the mirror speed. This dependence is in such a way that the reflected beam will hit the second mirror. I will add to my answer to explain you how that work later.
– user154997
Aug 27 '17 at 10:51
• I have just did so, only introducing the subject and referring you to the literature for the demonstration of the reflection law from a moving mirror. If you don't easily have access to it, I can give you a gist of the derivation, either the Huygens' one or the Fermat's one, whichever you are more comfortable with. But it may again take a few days!
– user154997
Aug 28 '17 at 15:51
• I am pretty sure Michelson and Morley knew about the dual property: in the lab frame, the usual law of reflection (incident angle = reflected angle) hold, even though there is an Ether wind. The tough problem was not reflection actually but refraction, and it is Fresnel who worked out the solution in the early 19th century with the partial drag, to avoid starlight aberration. I am pretty sure that if refraction was worked out, reflection must have been too since some telescopes use mirrors, and this issue had to be addressed too to address starlight aberration.
– user154997
Aug 28 '17 at 21:56
• But whether Michelson knew about applying Fermat or Huyghens to deduce the right reflection law in the Aether frame, I don't know. Most papers I read only go as far back as Einstein. But I am not a professional historian of science! Einstein who derived the formula using Lorentz transforms, obviously, so a different story altogether.
– user154997
Aug 28 '17 at 21:58