Understanding Michelson Morley Experiment

[Albertgauss]

TL;DR Summary

I have some questions on the details of the relative velocities of the light beams in the ether.

Hi all,

I’m having a little trouble understanding the Michelson Morley experiment. Questions below.

I understand that the light beams in the interferometer were expected to somehow pick up differences in the speed of light when one beam traveled parallel to the direction of motion and the other beam traveled perpendicular to the direction of motion. I also understand how different light speeds would produce the different fringes.

In a popular analogy, there is the land, water moving at the speed “u”, and a raft on the water. The swimmers parallel and perpendicular to the raft represent the light beams. The raft represents the earth--- the vessel in which the lab is----moving through the ether. The water represents the ether current or “ether wind”. I do not know what the land represents, though. If the ether is everywhere, would it not simply be the same as the land?

First, just to make sure the Earth is the “raft” here, the vessel, through which we move through the ether. Michelson’s lab, being on the earth, is the vessel that moves through the ether, is this correct? If there is any change in speed of light to be detected, that change will come about because of the Earth velocity in the ether, is this correct? That is, when we write terms like , the “v” is the velocity of the earth.

In Michelson’s frame, the lab, he’s going to shine both light beams at the speed of light relative to him. I feel like if I am on the raft and I shine a flashlight in either direction, perpendicular or parallel, I would say the speed of that light is the usual “c” in my raft frame. If the speed of light changed due to the presence of the ether, I feel like I would have to be on the ground frame to find it, the land, where I might say that the parallel beam would have a lightspeed of in the direction of motion and on the return trip of the parallel beam back to the lab. Of course, Michelson can’t leave the Earth for some absolute rest frame in space. Yet, somehow, Michelson would have been able to measure the relative velocities of the speed of light on the raft, had they existed. I don’t understand this.

Also, once you’ve explained how the Michelson experiment failed to find the ether, next line of logic is Einstein’s two postulates about the laws of physics the same in all frames and the constancy of the speed of light in all frames. I understand how Einstein reinterpreted the speed of light as a physics law--- no different than electricity, optics, etc.--- that the speed of light should not change depending on the frame you’re in.

I attached a jpeg of how I am trying to imagine the whole picture of this experiment.

https://www.physicsforums.com/attachments/268819But why was it necessary to say that “the laws of physics are the same in all inertial frames”? What physics law were people saying in the ether experiment that wasn’t the same in all frames? It seems perfectly obvious to me that the laws of physics should not change depending if you move in an airplane, car, or go for a walk, so I am missing the profoundness of this statement after the ether experiment.

 

[Ibix]

I think you are overcomplicating the analogy. You only need water, which models the ether and we can say isn't moving, and the raft, which models the lab and may or may not be moving with respect to the water. In the ether model, light always does $c$ with respect to the water because it's a mechanical disturbance of the water, and it doesn't care about the raft. If the raft is moving with velocity $v$ with respect to the water then the velocity of light with respect to the raft is $c-v$ (you can make those vectors if you like).

What physics law were people saying in the ether experiment that wasn’t the same in all frames?

Electromagnetism. Because it isn't invariant under Galilean transform, it looked very much like there was a special frame, the ether frame that would have to be worked into the laws of physics. Einstein, of course, showed that our mathematical issues lay in Newtonian dynamics and Galilean relativity, not electromagnetism, and that there is no special frame for electromagnetic phenomena or anything else we are aware of.

 

[FactChecker]

I don't understand your question. It seems like you should be measuring all velocities in the hypothesized ether reference frame to keep things simple. But you have "ground", "raft", "Earth", etc. and your diagram has the Earth velocity and the ether wind in the same direction. That doesn't make sense to me.

 

[Albertgauss]

@Ibix, that's actually a lot more clear. Yes, that does make sense now. The light waves are propagating in the ether. If I move by them on Earth at the speed of the Earth, then I would expect to see the velocities of the light waves already in the ether to have their velocities adjusted.

In the Micheslon/Earth frame I am moving at the speed of the Earth relative to the ether. I am shooting the light waves at speed c in my frame. Wouldn't I clock them at light speed as they come out of my flashlight? If I threw a baseball (light wave) in the direction of the Earth's velocity, in my frame the baseball (light wave) would move at speed v relative to me. Someone in the ether, watching the baseball go by would say the baseball (light wave) has speed v plus speed the speed of the earth. Or, does the light wave, once it comes out of the flashlight, automatically propagate at c because it is now in the ether?

@Ibix, thanks for letting me know about the Electromagnetism with the preferred frame. I went back and looked in my books and seems the books don't bring this issue up. This was good to know.

@FactChecker, yes, you are right, I am confused. I made the image as best as I could understand the experiment. The trouble I have with measuring everything in the ether frame is that Michelson was on the moving Earth. What would you do to my picture that would make it correct? That's actually what I would be looking for. I admit I was confused about when people talk about the "ether wind" so I got that all tangled up, somehow Michelson was going against the "ether wind" or "with the ether wind". This could be another analogy that probably isn't working well.

 

[Nugatory]

Wouldn't I clock them at light speed as they come out of my flashlight

At light speed relative to you and your flashlight, or at light speed relative to the ether?
That's the question that the MM experiment is asking (although as a matter of history Michelson, Morley, and their contemporaries did not describe it in those terms).

There's a perfectly sensible and internally consistent theory in which the speed of light is $c$ relative to the ether (that's Newtonian mechanics and Galilean relativity) and there's a perfectly sensible and internally consistent theory in which the speed of light is $c$ relative to you, your flashlight, and every inertial observer (that's the theory that Einstein proposed with his second postulate). Michelson-Morley experiments are one way of determining which of these theories properly describes the universe we live in.

 

[Dale]

Or, does the light wave, once it comes out of the flashlight, automatically propagate at c because it is now in the ether?

When a vehicle produces a sound wave that wave moves at the speed of sound relative to the air, not relative to the vehicle. That is the way all waves were understood to work at the time.

 

[Albertgauss]

@Dale, yes, you're right good way of putting that. Sound waves, once emitted, go at the speed in air, absolutely. Thus, for the ether, light waves, once emitted from the flashlight, would go at the speed in ether.
I think I've almost got it.

Point 1: Michelson shines his light, fully expecting, in his frame, for the light to propagate at the speed of light in the ether, according to him.
Point 2: The Earth is moving with respect to the ether, that will be the relative speed "v".

Believe it or not, I was not solid on the first two points before this post. But I am now.

Okay, so now, once Michelson let's the light waves out of his flashlight and should proclaim that these light waves travel at "c" according to him upon release. The light waves travel up and back along their respective interferometer arms. The ether is taking control of the light waves now; how does it modify both light beams to produce changes in perceived light speed? The motion of the emitter does not matter for a wave set to propagation so there shouldn't be any c+v or c-v pieces, etc. Yet, Michelson expects such pieces.

Is this maybe the idea of the "ether wind" I was confused about, a substance of light wave medium that exists in space analogous to a car moving through air? When the car is still, you don't feel the air. When the car goes in motion, the "still air" is now a wind that feels real. If the Earth was still, the light waves would propagate at "c", but the Earth moves through the universe at speed v, and therefore the "ether" pushes or drags on the light beams. Is the analogy closer to how this experiment should be thought?

I do appreciate your help in this. I don't think this experiment is trivial.

 

[A.T.]

Someone in the ether, watching the baseball go by would say the baseball (light wave) has speed v plus speed the speed of the earth.

Then you don't need the ether, because it doesn't do anything. What you describe is the ballistic theory of light, which is consistent with the Michelson Morley experiment, but disproved by other observations:

https://en.wikipedia.org/wiki/Emission_theory

The Michelson Morley experiment disproves the ether theory (where the light propagation speed relative to the ether is constant). But it doesn't imply Special Relativity as the only possible explanation. You need other observations to rule out alternative explanations.

 

[Ibix]

Is this maybe the idea of the "ether wind" I was confused about, a substance of light wave medium that exists in space analogous to a car moving through air? When the car is still, you don't feel the air. When the car goes in motion, the "still air" is now a wind that feels real. If the Earth was still, the light waves would propagate at "c", but the Earth moves through the universe at speed v, and therefore the "ether" pushes or drags on the light beams. Is the analogy closer to how this experiment should be thought?

Broadly yes. However, in ether theory, light is a wave in the ether. Ether doesn't push or drag light rays - the mechanics of the ether mean that waves move at a certain speed with respect to the ether, not the source. That is, the ether was assumed to operate like water. If I throw a stone in a pool the ripples spread at the same rate as the wake of a boat, not caring that the source was a single point-like impulse or a moving object.

So imagine throwing a stone over the side of a moving boat. If I'm floating in the water I'll see the ripples spreading out at the same speed in all directions. You, on the boat, could be (for example) moving at the same speed as the waves, so one side of the ripple pattern would be stationary next to you while the other side of the pattern moves away at double speed.

 

[Dale]

once Michelson let's the light waves out of his flashlight and should proclaim that these light waves travel at "c" according to him upon release.

No, they travel at c relative to the aether, not relative to him. See your point 1 in post 7.

The lab is traveling at v relative to the aether. See your point 2 in post 7.

Therefore the light is traveling at c-v relative to the lab. By vector addition.

So once Michelson let's the light waves out of his flashlight he proclaims (incorrectly) that these light waves travel at c-v according to him upon release.

how does it modify both light beams to produce changes in perceived light speed?

It doesn’t modify them. The light beams were at all times moving at c-v in his frame. When a sound is produced by a moving (relative to the air) vehicle it never travels at the speed of sound relative to the vehicle. It always travels at the speed of sound relative to the air. There is no modification to produce changes in its speed. That is always the speed of the wave.

 

[Janus]

Okay, so now, once Michelson let's the light waves out of his flashlight and should proclaim that these light waves travel at "c" according to him upon release. The light waves travel up and back along their respective interferometer arms. The ether is taking control of the light waves now; how does it modify both light beams to produce changes in perceived light speed? The motion of the emitter does not matter for a wave set to propagation so there shouldn't be any c+v or c-v pieces, etc. Yet, Michelson expects such pieces.

The light is emitted towards a mirror on the end of an arm a distance L away in the direction of the lab's motion.
If there is an ether, the light will travel at c relative to it. The end of the arm however, is moving in the same direction as the light at v relative to the ether. Ergo, the light moving at c has to "catch up" to the mirror moving at v, which had a head start of L. The amount of time it takes for the light to reach the mirror is t1 = L/(c-v).
The light reflects off the mirror, again moving at c relative to the ether but in the opposite direction. Now it is heading back to the emission source. But the emission source is still moving in its original direction, so now the two are rushing towards each other and will meet in t2= L/(c+v).
Total time between emission and return = L/(c-v) + L(c+v) = 2L(c^2-v^2)
Meanwhile, light is sent along a arm of the same length but perpendicular to the motion with respect to the ether. The light travels at c with respect to the ether. But now the mirror is neither receding or approaching relative to the ether. Thus the trip time is L/c both ways, and the total trip time is 2L/c.
So, with an ether, the round trip time for light traveling along the line of motion is different than the round trip time for the light traveling perpendicular to it. ( For example, if v = 0.1c, the time difference works out to be ~ 1%)
This is what Michelson was looking for. ( which would have shown up as the two beams being slightly out of phase with each other upon their return.)

 

[Albertgauss]

Oh, I think I it see now: when Michelson releases the light waves out of his flashlight, they are already traveling at the relative velocities with the c-v, c+v terms. The light waves of the medium ether are already propagating with c-v, c+v terms. I did not understand that until this most recent post. I thought when Michelson released the light waves, he would proclaim them to propagate at their natural speed of “c”. This was an important clarification to make.

This would mean that Michelson would be the “ground observer” in all the analogies. That’s fair enough, instead of saying that Michelson moves with respect to the ether, I can reinterpret the situation to be that Michelson is at rest and the ether moves with respect to him. With that thinking, Michelson is the “ground observer” and the river of the ether flows by him. Or, in another analogy, Michelson is on the ground watching the moving train (the ether) go by him. Let’s say that an observer is embedded in the ether frame lines up with Michelson’s flashlight at the exact moment Michelson releases his light beams. If I am on the train, or in the river, of the ether, I will proclaim the light rays travel at speeds “c”. Michelson is watching the river, train, the ether go by him in a relative velocity, the relative velocity of the earth. With that in mind, it is easy to see how the relative velocities add and subtract according to the mathematical terms Michelson using experiment.

You can see that up until now, I thought (incorrectly), that Michelson was the moving frame. Michelson is the ground frame.

All of these analogies usually precede the discussion of the Michelson experiment, depending on the book you read. The train and the river as the ether are the two most common analogies I have encountered. I think what confuses many people is that when they read that “the Earth moves through the ether”, the mental link (though incorrect) is that Michelson is on the moving train or on a moving raft in the river, and from inside the train or on the raft in the river, Michelson shoots out his light beams. Moving earth, moving vessel. Actually, Michelson is on the ground watching the train, the river, the ether go by him.

By the way, my original jpeg is now completely obsolete in light of this new understanding.

How is this doing?

 

[jbriggs444]

Meanwhile, light is sent along a arm of the same length but perpendicular to the motion with respect to the ether. The light travels at c with respect to the ether. But now the mirror is neither receding or approaching relative to the ether. Thus the trip time is L/c both ways, and the total trip time is 2L/c.

Not so fast. If the light is moving on a path along the arm, its motion relative to the ether is on a diagonal. It has to be aimed "upstream" so that its velocity made good is perpendicular to the flow of the ether. It will be traveling at a speed so that $L^2 + tv^2 = tc^2$ for both one-way trips. The round trip time is then $2t$ and is given by $\frac{2L}{\sqrt{c^2-v^2}}$

 

[Albertgauss]

@Janus and @jbriggs, I understand the math you did. The confusion I've been having is how Michelson ends up as the ground frame and the ether the rest frame of the light waves. I think I have it straightened out now, but just want to confirm. I think I got it sorted out, finally, in my most recent post.

 

[jbriggs444]

@Janus and @jbriggs, I understand the math you did. The confusion I've been having is how Michelson ends up as the ground frame and the ether the rest frame of the light waves. I think I have it straightened out now, but just want to confirm. I think I got it sorted out, finally, in my most recent post.

You can choose to calculate in the ground frame and use a "speed of light" that is adjusted so that the speed in the ether frame comes out as c.

Or you can choose to calculate in the ether frame, use a constant speed of light and account for the fact that the arms are moving.

Both approaches should yield an identical result for the round trip time on a given arm.