If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

## Physics library

### Course: Physics library>Unit 16

Lesson 1: Michelson and Morley's luminiferous ether experiment

# Michelson–Morley Experiment introduction

Simple idea behind the Michelson‒Morley experiment to detect ether wind.

## Want to join the conversation?

• If you have a fiber-optics cable running the circumference of the Earth at the equator, would a signal traveling with the rotational velocity of the earth complete one circuit in the same amount of time as a signal traveling against the rotational velocity of the earth? Is the same true at the scale of the solar system, galaxy, galaxy cluster, etc. ? • Depends on who's observing. Suppose you're on the equator and sends a signal travelling both directions. Since you sent the signal simultaneously and both rays have the same angular speed relative to you , then they'll perform a lap (360 degrees) simultaneously.
However, if you observe this from outside the earth, simultaneity is lost. Both rays have the same speed, but this time you can see that the equator is moving, so they'll perform a lap in different times.
Keep in mind that simultaneity is not absolute.
• Did the Michelson-Morley experiment fail because it was unable to detect the effect of a luminiferous ether or because they failed to understand that light travels at a constant speed regardless of the reference point? • Failed for both those reasons.
1. If there was a universal frame of reference (ie. ether), then their experiment would have detected a change in speed of light relative to Earth's motion.
2. Since special relativity was unknown to them, then there was no reason to believe ahead of time that spacetime changes relative to motion to keep the speed of light constant in all reference frames.
• If the speed of light speed up when travel in the same directions of the ether and slow down when it does the opposite travel, the blue photon will reached the the center, the middle mirror at same time that the orange photon does. If this is truth, both waves will reached the detecter at same time. So no way to prove the existence of the luminiferous ether. What is wrong in this my interpretation? • Let's say that in ether light would move at c. Also, consider Earth moving at a velocity v relative to the ether, and that, at some day, the ray that was drawn vertically is parallel to v (this last consideration is just for simplicity). Furthermore, the distance travelled by this ray from the middle to the other mirror is s1, while the ray that did not reflected at first travelled s2 until it reached the mirror.
time it takes for the reflected ray to reach the middle again, with ether: s1/(c+v) + s1/(c-v)
time it takes for the reflected ray to reach the middle again, without ether: s1/c + s1/c
s1/(c+v) + s1/(c-v) =/= 2.s1/c
Thus, it does make a difference.
However, that's not the important thing about the experiment, testing this only once would prove nothing, since you don't know the speed of Earth relative to the ether, nor the speed of light (with enough precision).
What matters is the fact that you can spin the apparatus around and observe no change at all. Were there a luminiferous ether, the difference in time taken for light to travel through s1 and s2 would change according to the orientation of the apparatus. To explain further, let's say the reflected ray takes a time Tr to hit the middle mirror and come back, while the other ray takes a time Tp to do the same. For each value of Tr - Tp, you would obtain a unique wave pattern. So, let's say in a given day, you measure a wave pattern corresponding Tr - Tp = 2 nanoseconds (random number). If you were to realize the experiment with the same equipment 3 months later, due to Earth's movement, you should measure a different value of Tr - Tp, because the rays' orientation relative to the ether would be totally different, thus giving you a different wave pattern. Or, more pragmatically, you could just spin the apparatus around, and you would still have no change at all.
• How is it possible to build it so precise that the outward mirrors have exactly the same distance from the inside (half-silver) mirror? • I still don't quite understand how the ether-affected ray travels longer/shorter time compared to the vertical one.. Not sure if we can use average velocity here, but if we do, then V(av)=((speed of light+speed of ether)+(speed of light - speed of ether))/2 = speed of light. For simplicity, I'll use random numbers - ((1000+1)+(1000-1))/2=1000, which is the same as initial velocity. Although when I use WolframAlpha, it gives different time for the two cases. I feel like I'm missing some little thing.. I would be very thankful if someone could explain it to me =) • Let's use an absurd example. You're in a zeppelin, travelling with an airspeed of 50 kph with a 49 kph wind at your back. So your groundspeed is 99 kph, and you go 100 kilometres in a shade over an hour. Then you turn around to go home. Your ground speed is now 50-49=1 kph, and it will take you 100 hrs to complete your trip. So you're comparing 200/50 = 4 hrs for the round trip with no wind, to 101+ hrs with the strong wind.
• As we know light would travel at different speeds at vacuum than let's say in glass. So, is this entire experiment performed in vacuum, to ensure that the experiment is not affected by the refractive index of any material ? • I think the test was done in country-side room with mirrors in air. You are right. M-M exp assumed either could be detected in air. That could have been wrong, and I expect M-M knew that. if the light signal travels at a speed relative to the air in the room and not relative to the either that would hide the either. The effect of room air slows light by 1.0003 or 3 parts in 10000. Speed through either would have to be 0.0003C or 90 km/s to have similar effect. I expect M-M understood that air might dominate the either effect but moved ahead with what they could accomplish at the time.
(1 vote)
• When the light ray goes up, it speeds up due to the "ether"... and when it returns, it slows down. So, Maybe there is a luminous ether but the experiment was just a fail as the same light ray speeds up and slows down by the same rate... • Do the math - let vy be your speed, and ve be the ether speed, both with respect to the ground. If distance s=vt, then t=s/v. So going out, tg = s/(vy+ve) and coming back,
tc = s/(vy-ve). Therefore, tg+tc = s/(vy+ve) + s/(vy-ve). Finding the common denominator,
tg+tc = (svy-sve + svy+sve)/((vy+ve)(vy-ve)) = 2svy/(vy^2-ve^2). With no ether, the total time would be tg+tb=2s/vy. The only way 2svy/(vy^2-ve^2)=2s/vy is if ve=0 - in other words there's no ether effect.   