Michelson–Morley experiment: Did it disprove the existence of ether?

[harrylin]

If I understand correctly then the device is designed such that in the frame where the device is stationary the light arrives simultaneously and the clocks are synchronized so the light is detected as simultaneous.

If that is correct then in the frame where it is moving the light arrives at different times (not simultaneous) but the clocks are also not synchronized. The de-synchronization of the clocks exactly offsets the time difference between the arrival of the light so the light is detected as simultaneous even though it is not.

Note that simultaneity [Edit: distant simultaneity] doesn't play a role in MMX; Roi's discussion would be more fitting in the new thread on relativity of simultaneity.

 

[Dale]

Note that simultaneity doesn't play a role in MMX

I don't know that I would go that far. True, there are no synchronized clocks in the experiment, but if the light from the two arms does not return simultaneously then that causes a phase shift between the arms which is reflected in the interference pattern. I think it is safe to say that simultaneity has little to do with the MMX, but not nothing.

 

[harrylin]

I don't know that I would go that far. True, there are no synchronized clocks in the experiment, but if the light from the two arms does not return simultaneously then that causes a phase shift between the arms which is reflected in the interference pattern. I think it is safe to say that simultaneity has little to do with the MMX, but not nothing.

Sorry, I meant the distant simultaneity that Roi now brought up.
That really belongs in the relativity of simultaneity thread.

 

[Dale]

Agreed, on both points.

 

[Buckleymanor]

Light speed is not a field, just as sound speed is not grass!

I am totally confused now are you sure that lightspeed is not a traveling field.

 

[Russell E]

Note that simultaneity [Edit: distant simultaneity] doesn't play a role in MMX...

Not true. The extended mechanical arms of the apparatus always maintain exactly the same equilibrium configuration in terms of the co-moving system of inertial coordinates, with the relative simultaneity implicit in those coordinate systems. So distant simultaneity plays a crucial role in the relativistic explanation of the Michelson-Morley experiment. Of course, like any single experiment considered in isolation, this experiment could be explained in many different ways (e.g., by a ballistic theory), but it's only meaningful to consider an experiment in the context of what has been established by other experiments. In that context, recognizing the physical significance of Lorentz's "local time" (i.e., the relativity of simultaneity) is crucial for any rational account of Michelson's result.

 

[harrylin]

I am totally confused now are you sure that lightspeed is not a traveling field.

They are not the same kind of concepts - just as boiling temperature is not water!
Water does boil at the boiling temperature, and fields do expand at the speed of light, as far as we know.

 

[harrylin]

Not true. The extended mechanical arms of the apparatus always maintain exactly the same equilibrium configuration in terms of the co-moving system of inertial coordinates, with the relative simultaneity implicit in those coordinate systems. So distant simultaneity plays a crucial role in the relativistic explanation of the Michelson-Morley experiment. Of course, like any single experiment considered in isolation, this experiment could be explained in many different ways (e.g., by a ballistic theory), but it's only meaningful to consider an experiment in the context of what has been established by other experiments. In that context, recognizing the physical significance of Lorentz's "local time" (i.e., the relativity of simultaneity) is crucial for any rational account of Michelson's result.

And how does Michelson measure that, do you think?

 

[Russell E]

And how does Michelson measure that, do you think?

Like any single experiment, considered in isolation, Michelson's observation of a lack of fringe shifts doesn't really signify anything, because it could be explained in many different ways (e.g., by a ballistic theory). But we know that it's only meaningful to consider an experiment in the context of what has been established by other experiments. I suggest you read about the numerous experiments and observations that led Lorentz to develop his theorem of corresponding states. The "length contraction" which he inferred from Michelson's result (the lack of fringe shifts) is due to the fact that the extended mechanical arms of the apparatus always maintain exactly the same equilibrium configuration in terms of the co-moving system of inertial coordinates, with the relative simultaneity implicit in those coordinate systems. This is why distant simultaneity plays a crucial role in the relativistic explanation of the Michelson-Morley experiment. The experiment (in context) compells us to recognize the physical significance of Lorentz's "local time" (i.e., the relativity of simultaneity) for the equilibrium configurations of mechanical entities.

 

[harrylin]

Like any single experiment, considered in isolation, Michelson's observation of a lack of fringe shifts doesn't really signify anything, because it could be explained in many different ways (e.g., by a ballistic theory). But we know that it's only meaningful to consider an experiment in the context of what has been established by other experiments. I suggest you read about the numerous experiments and observations that led Lorentz to develop his theorem of corresponding states. The "length contraction" which he inferred from Michelson's result (the lack of fringe shifts) is due to the fact that the extended mechanical arms of the apparatus always maintain exactly the same equilibrium configuration in terms of the co-moving system of inertial coordinates, with the relative simultaneity implicit in those coordinate systems. This is why distant simultaneity plays a crucial role in the relativistic explanation of the Michelson-Morley experiment. The experiment (in context) compells us to recognize the physical significance of Lorentz's "local time" (i.e., the relativity of simultaneity) for the equilibrium configurations of mechanical entities.

Yes, I know about those. Roi started to discuss the synchronization of distant clocks, thus I referred to the simultaneity convention of special relativity and the related thread. Sorry if I wasn't clear! MMX is not concerned with the convention to set distant clocks to "local time".

 

[roineust]

At this point what I understand is that, in words, the reason for this device that I drew, to act exactly the same, no matter what velocity is, would be:

True, there is time dilation, but the speed of light in one way, is not the same between two frames, it is even not the same (the one way speed of light) within one frame, under certain conditions.

Correct?

 

[Dale]

At this point what I understand is that, in words, the reason for this device that I drew, to act exactly the same, no matter what velocity is, would be:

True, there is time dilation, but the speed of light in one way, is not the same between two frames, it is even not the same (the one way speed of light) within one frame, under certain conditions.

If you are using inertial reference frames then the one way speed of light (in vacuum) is always c under all conditions.

Did you not read my explanation above?

 

[harrylin]

At this point what I understand is that, in words, the reason for this device that I drew, to act exactly the same, no matter what velocity is, would be:

True, there is time dilation, but the speed of light in one way, is not the same between two frames, it is even not the same (the one way speed of light) within one frame, under certain conditions.

Correct?

I don't know what you mean with "not the same within one frame, under certain conditions". But I answered those questions more or less in #112 and #128.

Usually people mean with "frame", a single frame of measurement. According to special relativity (which is only valid at constant gravitation), the speed of an undisturbed light ray is everywhere the same.
Moreover, the laws of special relativity are made for a special kind of frame: clocks must be set in such a way that the measured one-way light speed is the same everywhere. This is being discussed in the thread on relativity of simultaneity. Do you understand the train example? If not, please ask your questions there!

 

[roineust]

Dalespam,

Of course I read your explanation.

And then I thought, that you were talking about the one way vs. two way speed of light.

Did you refer by de-synchronization to something else, e.g. a result of moving the light detector clocks apart, so that 'moving them apart' movement, no matter even if slow, creates de-synchronization, which amounts exactly to the time dilation created by the frame velocity itself?

 

[Dale]

And then I thought, that you were talking about the one way vs. two way speed of light.

I understand your diagram to have light going in only one direction, so the two-way speed of light is not relevant.

Did you refer by de-synchronization to something else, e.g. a result of moving the light detector clocks apart, so that 'moving them apart' movement, no matter even if slow, creates de-synchronization, which amounts exactly to the time dilation created by the frame velocity itself?

Exactly. If you have two clocks which are not at the same location and they are synchronized in one frame then, by the relativity of simultaneity, they will not be synchronized in any other frame.

In your drawing you show that the detector clocks are separated by a distance of ~1/3 the total length of the lower path, so if the detector clocks are synchronized in the rest frame they will not be synchronized in the moving frame. This desynchronization will exactly compensate for the time dilation on the upper path.

 

[harrylin]

[..]so that 'moving them apart' movement, no matter even if slow, creates de-synchronization, which amounts exactly to the time dilation created by the frame velocity itself?

Only slow clock transport allows to keep the synchronization (in good approximation).
Consequently, if clocks are synchronized in a moving frame and then slowly moved apart along the direction of motion, then in the stationary frame the clocks will look de-synchronized. And that effect is due to time dilation.

Is that what you meant?

 

[Dale]

What you are saying is, that it is actually not the moving apart of the clocks, but the acceleration (or is it also, or only, the constant speed) engaged on clocks, while and because they are not measuring at the same place, even inside the same frame, although they are both experiencing in that frame the exact same acceleration (or constand speed), right?

There is no acceleration involved. I was assuming that your device was purely inertial and that the reference frames were also purely inertial.

The desynchronization is inherent in the definition of the inertial frames themselves (Einstein synchronization convention) and is not related to mechanical acceleration.

And also you are saying that there is no 'mechanical' or 'by words' explanation, from this point on, and at that phase of understanding further, but only a mathematical description available further, right?

And that this barrier, of 'the end of spoken words ability to describe', is maybe, the exact historical and major point of departure, of modern science from classical science?

I can make lots of spoken words on this topic. The words are merely a translation from math to English, but that is true of Newtonian physics too.

 

[harrylin]

Dalespam, harrylin, ghwellsjr, anyone else:

For me, at least for a while, it is the end of my investigation on the subject, although I don't know if for the right reasons…

So can you please take one last look at this rephrased description and tell me if you agree:

What you are saying is, that it is actually not the moving apart of the clocks, that is actually, absolutely possible to synchronize in the stationary frame, but the difference of constant speed between the frames, engaged on the clocks, while and because they are not measuring time at the same place, although both inside the same frame, although they are both experiencing in that frame the exact same constant speed , right?

And also, you are saying that there is no 'mechanical' or 'by words' explanation, from this point on, and at that phase of understanding further, but only a mathematical description available further, right?

Otherwise, why would the exact same velocity on the two clocks would de-synchronize them? 'Just' because while at the same frame, they are some distance apart? It is because it is because, 'because' means: math says so and verified by experiments, and there is no middle way between the math and the experiments...?

Correct?

Thanks,

Roi.

Hmm.. I think that you now have it wrong, because you jumped to (wrong) conclusions.

The exact same velocity on two clocks can not de-synchronize them; but two clocks can only move apart if they do not have the exact same velocity.
The difference in time dilation causes the de-synchronization.

And note that such a de-synchronization does not happen on distant clocks during acceleration: then you must change the clocks yourself (or keep them together during acceleration, and move them slowly apart afterwards).

 

[roineust]

harrylin,

I have made a few more small changes and added a diagram, please take a look if this is what you meant.

 

[Dale]

What you are saying is, that it is actually not the moving apart of the clocks, that is actually, absolutely possible to synchronize in the stationary frame, but the difference of constant speed between the frames, engaged on the clocks, while and because they are not measuring time at the same place within the moving frame, although both are inside the same moving frame, although they are both experiencing in that moving frame the exact same relative constant speed , right?

I can't make heads or tails of this paragraph.

And also, you are saying that there is no 'mechanical' or 'by words' explanation, from this point on, and at that phase of understanding further, but only a mathematical description available further, right?

Otherwise, why would the exact same velocity on the two clocks would de-synchronize them? 'Just' because while at the same frame, they are some distance apart? It is because it is because, 'because' means: math says so and verified by experiments, and there is no middle way between the math and the experiments...?

All of physics is mathematical and experimental. For instance, what is force absent Newton's law f=ma?

 

[phyti]

roineust,
Using SR synch method, A sets clocks 1 & 2 to match his central clock, without knowing his speed through space. The round trip transit times are equal.
U and anyone else who does not have a velocity equal to A, sees c1 & c2 out of synch.
If the clocks were moved from the center in opposite directions,each would move at a different speed.

clocksynch.gif

 

[ghwellsjr]

Roi, I have no idea what your diagram is depicting but I don't really care. You should not be trying to question the ideas of Special Relativity or trying to find experiments that you can understand that will prove Special Relativity to you. You should be learning Special Relativity. After you understand it and the reason why it is a useful theory to understanding reality, then you probably won't care about your diagram either.

Let me just emphasize some points that you have consistently questioned:

*There is no way to detect an absolute rest frame.

*There is no way to build an absolute clock.

*Time dilation is a real phenomenon.

*Only clocks that are at rest with respect to each other can be synchronized.

*Length contraction is a real phenomenon.

OK?

I have tried to help you learn what relativity is all about without using any mathematics, just animations, but your last comment to me on your thread was:

ghwellsjr,

Ok,
It seems you added quite detailed clarifications here,
So, it will take me some time now, to get familiar with the new post.

And then you went off on this long side track asking for experimental proof of length contraction and time dilation, and finally ending in more of your diagrams that nobody can understand.

This is not the way to make progress. I still believe I can help you understand Special Relativity without getting into math through the use of animations. I urge you to forget asking questions about your own ideas and focus on what I have presented to you and ask questions only about what you don't understand in my animations and my descriptions of them. I would suggest that you go back and read all my posts on the above linked thread and make sure you understand those animations and ask any question you may have about them. Are you willing to do this?

 

[roineust]

Dalespam, harrylin, ghwellsjr, anyone else:

Can you please take one last look at this attached diagram, and rephrased description, and tell me if you agree: (please note that synchronization is always done before the clocks are apart, and that, in the diagram example, the clocks [A,B] are synchronized and moved apart, only when they are in the stationary frame).

What you are saying is, it is absolutely possible to keep them synchronize (A,B) in relation to one another, after they are moved apart, within the stationary frame. What it is, that makes the device, with the exact same configuration, work the same, in both the moving and stationary frames (in contrast to the diagram) - has to do with what happens in the moving frame, which is: Time dilation in the middle clock (C), within the moving frame, coupled with the fact, that both clocks (A,B) are measuring time, at different positions, within the moving frame. The same different positions (A,B) as in the stationary frame, now in the moving frame, makes the difference.

So the difference that compensates for C clock time dilation, and makes the device work always the same in both frames, can occur in two ways: 1. When acceleration occurs - if the clocks are first synchronized and moved apart, in the stationary frame (as in the above example and diagram), Or: 2. While the action of moving the clocks apart occurs (relatively to the stationary frame, and not to each other), if synchronization is done in the moving frame. Is that correct?

Thanks,

Roi.

 

[Dale]

What you are saying is, it is absolutely possible to keep them synchronize (A,B) in relation to one another, after they are moved apart, within the stationary frame. What it is, that makes the device, with the exact same configuration, work the same, in both the moving and stationary frames (in contrast to the diagram) - has to do with what happens in the moving frame, which is: Time dilation in the middle clock (C), within the moving frame, coupled with the fact, that both clocks (A,B) are measuring time, at different positions, within the moving frame. The same different positions (A,B) as in the stationary frame, now in the moving frame, makes the difference.

Yes. This is known as the relativity of simultaneity.

So the difference that compensates for C clock time dilation and makes the device work always the same in both frames, can occur in two ways: 1. When acceleration occurs - if the clocks are synchronized, and moved apart, in the stationary frame (as in the above example and diagram), Or: 2. While the action of moving the clocks apart occurs (in relation to the stationary frame), if synchronization is done in the moving frame.

It is more basic than that. Neither acceleration nor moving the clocks apart is essential since you can construct scenarios without acceleration or moving the clocks apart. The point of the relativity of simultaneity is that different reference frames disagree about whether or not two spatially separated events are simultaneous. It has more to do with the definition of simultaneity (aka the Einstein synchronization convention) than with the movement of the clocks.

And also, you are saying that there is no 'more mechanical' or 'only one ideal spoken words' explanation, from this point on, but at that phase of understanding further, only is a mathematical description available.

This is rather irritating. You have asked this exact same question twice already and I have responded twice already. Please don't repeat your same question a third time, instead respond to my answers. If there is something about them that you don't understand then ask for clarification. Repeating yourself a third time will get the same answer a third time (or an irritated response like this). This form of conversation seems to be a bad habit with you as I have often had you repeat questions that I already answered. Frankly, it is a bit rude.

 

[roineust]

No problem,
Consider the last section just a leftover.
I took it out.

Thanks a lot.
Roi.

 

[harrylin]

Dalespam, harrylin, ghwellsjr, anyone else:

Can you please take one last look at this attached diagram, and rephrased description, and tell me if you agree: (please note that synchronization is always done when the clocks are not apart, and that, in the diagram example, the clocks [A,B] are synchronized and moved apart, only when they are in the stationary frame).

OK then, this is for me the last time that I reply to this in this MMX thread.

Your clocks A and B are synchronized and moved apart when the system is in rest, and next the whole system with clocks is brought to a certain speed - I will assume a speed parallel to the drawing.

Then the one-way speed of light will not appear isotropic in that moving system because the clocks are not synchronized according to the standard SR convention.

What you are saying is, it is absolutely possible to keep them synchronize (A,B) in relation to one another, after they are moved apart, within the stationary frame.

[edit:] If you mean, if they are moved apart, they will remain in sync according to the frame in which they are in rest? Yes. That is however relative, not absolute.

What it is, that makes the device, with the exact same configuration, work the same, in both a moving and stationary frame (in contrast to the diagram) - has to do with what happens in the moving frame, which is: Time dilation of the time delay of the middle clock (C), within the moving frame, coupled with the fact, that both clocks (A,B) are measuring time, at different positions, within the moving frame. The same different positions (A,B) as in the stationary frame, now in the moving frame, makes the difference.
So the difference that compensates for C clock time dilation and makes the device work always the same in both frames, can occur in two ways: 1. When acceleration occurs - if the clocks are synchronized, and moved apart, in the stationary frame (as in the above example and diagram), Or: 2. While the action of moving the clocks apart occurs (in relation to the stationary frame), if synchronization is done in the moving frame.

Huh?! No, the main point of relativity is that you can choose which frame you call "moving" or "stationary"! Correcting your errors, we get:

The difference that compensates for C clock time dilation and makes the device work always the same in both states of motion, can occur in several ways, for example:

1. If the clocks A and B were synchronized, and moved apart before acceleration (as in the above example and diagram), and then re-synchronized by means of light signals.
Or:
2. If those clocks are synchronized and equally moved apart relative tot the system after the acceleration, they will remain synchronized according to the synchronization convention.

And also, you are saying that there is no 'more mechanical' or 'only one ideal spoken words' explanation, from this point on, but at that phase of understanding further, only is a mathematical description available.

Is that correct?

Thanks,

Roi.

No, of course not- just read what we wrote, none of us were saying that!

Success,
Harald

 

[phyti]

1. If the clocks A and B were synchronized, and moved apart before acceleration (as in the above example and diagram), and then re-synchronized by means of light signals.
Or:
2. If those clocks are synchronized and equally moved apart relative tot the system after the acceleration, they will remain synchronized according to the synchronization convention.

Success,
Harald

1. agree

2. If they move within the rest frame, they are no longer a part of that frame.
The frame has an unknown velocity which will not be zero, therefore each clock will move to its destination at different speeds and be out of synch.

 

[harrylin]

1. agree

2. If they move within the rest frame, they are no longer a part of that frame.
The frame has an unknown velocity which will not be zero, therefore each clock will move to its destination at different speeds and be out of synch.

Clocks that are equally moved apart will remain in sync as measured in the frame in which they were (and are again) in rest. Obviously that means that they must then be out of sync as measured in a frame in which they are moving.

 

[Dale]

If they move within the rest frame, they are no longer a part of that frame.

I don't like this phrasing. They are material objects and as such they cannot be "a part of that frame" to begin with. Frames are essentially just coordinate systems, so material objects do not go in or out of frames and they are never a part of a frame. I would have said it:

"If they move within the rest frame, they are no longer stationary in that frame."

 

[phyti]

I don't like this phrasing. They are material objects and as such they cannot be "a part of that frame" to begin with. Frames are essentially just coordinate systems, so material objects do not go in or out of frames and they are never a part of a frame. I would have said it:

"If they move within the rest frame, they are no longer stationary in that frame."

I didn't know we had to submit our posts for review or editing.
On examining your statement, there seems to be redundancy (red or blue).

"If they move within the rest frame, they are no longer stationary in that frame."
Sorry you were not appointed a role model for society, but keep trying!

Coordinate systems have at least one reference object (the origin), right? A frame has a system of clocks, right? Aren't clocks objects?
In the frame with its origin at Earth center, is there a clock at Alpha Centauri, no. Do we need one there, no.
A single observer with a single clock and laser is sufficient to make measurements where feasible.

The point being made was physics, not grammar.
If the object is moving relative to the frame or any object that is at rest in that frame, its clock rate will be different, until it arrives at its destination in that frame. It will run at its previous rate but will be out of synch with the master clock which did not move. Synchronization is used here as equal simultaneous readings, which is the purpose of the method used in SR.