In the MM experiment, the speed of one light beam was supposed to slow down as it moved against the flow of aether, but when it bounced off the mirror, it would be moving in the opposite direction--with the flow of aether and therefore should be moving faster. The average speed of light on that arm of the experiment should have been exactly the same as the light flowing across the current of aether as the aether flow terms cancel each other.

Why isn't this the same as a high school physics problem involving boats in a current or people on a moving sidewalk where the current term just drops out because you add speed in one direction and subtract it in the reverse direction?

It seems like a properly adjusted MM apparatus, would always measure the speed of light on the 2 axes as being the same REGARDLESS of whether there is aether or not.

What am I missing?

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    $\begingroup$ "What am I missing?" Sitting down and actually doing the math involved, perhaps. The speed in both arms is affected by the "flow" and for the up-down arm it is affected differently in the two directions, but the net result is not the same for the two arms. You can't wave your hands at it, you have to calculate it. $\endgroup$ – dmckee Apr 14 at 2:04

The point of the experiment is that in the aether, light along one arm of the interferometer will return at a different time than light along the other arm.

Forget about light for now and think of one of those moving walkways that you find in an airport (https://en.wikipedia.org/wiki/Moving_walkway). If I walk to one end of the walkway and back, that meaning I get a constant increase in speed going one way, and a constant decrease in speed coming back, will I have taken the same amount of time as if I didn't use the walkway at all? It turns out that the answer is no.

To see this, suppose the length of the walkway is $d$ and I walk with a speed $v$. If the walkway moves with a speed $v_w$, then I will be moving at a speed $v+v_w$ when I walk with the walkway, and a speed $v-v_w$ when I walk against it. The time it takes me to walk there and back is therefore $$\Delta t = \frac{d}{v+v_w} +\frac{d}{v-v_w}$$ as long as $v_w<v$ (which is has to be or I would never get back). This is in contrast to the time it would take with no walkway at all: $$\Delta t' = \frac{d}{v}+\frac{d}{v}$$ which is always less than the time it takes on the walkway whenever $v_w \neq 0$. So in fact, the average velocities are not the same, as it takes me a longer time to walk the same distance on the walkway.

It is the same idea with the Michelson-Morley experiment. If the aether existed, the light ray that is parallel to the flow of the aether would experience a speed-up in one direction, and a slow-down in the other, which means it would take longer to return than the beam perpendicular to it. This time delay would cause the two beams to interfere when recombined, and this interference pattern can be viewed on a screen. Notice that the only solution to the equation $\Delta t = \Delta t'$ is when $v_w=0$, meaning that no interference pattern implies that there is no walkway, which in this case means no aether that changes the relative speed of light.

  • $\begingroup$ So the speed increase and decrease only cancel each other for constant time trips, not constant distance trips. Thank you very much! $\endgroup$ – mike663 Apr 14 at 13:11

At the time, the experiment did not disprove the existence of the "aether". What it did was demonstrate by experiment that if the aether did indeed exist, it would have to possess some extremely peculiar properties: on the one hand, it would have to allow material objects to pass through it without resistance (so as to not interfere with the motion of planets), while at the same time it also had to be somehow dragged along with an object that was moving through it (so as to yield a null result in the M-M experiment).

The "aether drag" hypothesis was seriously considered within the physics community as a way to preserve the aether and escape the consequences of the M-M test, but it did not get very far.

  • $\begingroup$ The MM experiment was explained by Lorentz in his Theory of Electrons, for which he received the Nobel Prize in 1902. The explanation involved applying the Lorentz transformation to the aether thus effectively making the aether relative. $\endgroup$ – safesphere Apr 14 at 8:21
  • $\begingroup$ Safesphere, do you have any idea why Lorentz didn't take the next step and beat Einstein to special relativity? I know the idea was "in the air" at that time, but it seems he got really close- or no? $\endgroup$ – niels nielsen Apr 14 at 17:46
  • $\begingroup$ In physics, formulas represent a theory; explanations represent an interpretation. According to Wikipedia, "Because the same mathematical formalism occurs in both, it is not possible to distinguish between LET [Lirentz Ether Theory] and SR [Einstein Special Relativity] by experiment". Therefore SR is an interpretation of LET (a part of the Theory of Electrons). So Lorentz already received the Novel Prize for relativity (among other things). The relative aether is undetectable, so it's existence becomes inconsequential. SR does away with the aether, but this does not change any formulas. $\endgroup$ – safesphere Apr 14 at 18:08
  • $\begingroup$ The existence of the relative aether is inconsequential in a flat Minkowski soacetime, but is significant in a curved soacetime. So Einstein eventually had to go back to the existence of aether by 1920: "according to the general theory of relativity space is endowed with physical qualities; in this sense, therefore, there exists an ether. According to the general theory of relativity space without ether is unthinkable". In a modern language this is called the intrunsic curvature and most people don't even understand how it relates to the aether, but Einstein did cause he was a lot smarter. $\endgroup$ – safesphere Apr 14 at 18:19
  • $\begingroup$ Is space itself playing the role of aether? Does Einstein's description say what would happen or would not happen if you shot a beam of light outward from the "edge" of space into whatever space is expanding into? $\endgroup$ – mike663 Apr 14 at 19:06

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