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

[harrylin]

harrylin,

I don't understand exactly what you mean by cross trip, but if you mean from one bank of the river, to the other bank, and the current speed is 2kph, and the boat speed is 5kph, and the river width is, say 1km, then I think the time to cross to the other bank would be:

1/(root((2^2)+(5^2)))=1/(root(29))= ~0.185 hours= ~11 minutes.

Would that be correct?

Roi.

I don't think that I wrote "cross trip" - see below!

But you already answered another question that is relevant for MMX.

And yes your answer is almost right - it's just trigonometry.
Funny enough, it's a question that Michelson also had wrong the first time that he calculated it:

The boat has a constant speed of 5 km/h, let's call that velocity c.
Now replace the boat by light.
Also, replace the river bank by the MMX apparatus (moving at velocity v).
Then, in 1881 Michelson calculated 0.20 hours for your example! But when he did his famous experiment in 1887, he calculated it correctly and he found more than 0.20 hours. Do you understand why it will take longer when there is a strong river current?

But I wrote "How do you calculate the time that the motorboat needs to go from one place to another? and back?" and "Yes that's right - for a boat sailing along with the river. And how long does the return trip take?".
The end question is: how long will it take to go from A to B downstream and immediately back upstream to A?

If you have that, and you solved the little error in your calculation here above, then you can do Michelson's calculations. And after that, you may have the basic knowledge to follow the things we discussed with you.

 

[Buckleymanor]

I don't think that I wrote "cross trip" - see below!

But you already answered another question that is relevant for MMX.

And yes your answer is almost right - it's just trigonometry.
Funny enough, it's a question that Michelson also had wrong the first time that he calculated it:

The boat has a constant speed of 5 km/h, let's call that velocity c.
Now replace the boat by light.
Also, replace the river bank by the MMX apparatus (moving at velocity v).
Then, in 1881 Michelson calculated 0.20 hours for your example! But when he did his famous experiment in 1887, he calculated it correctly and he found more than 0.20 hours. Do you understand why it will take longer when there is a strong river current?

But I wrote "How do you calculate the time that the motorboat needs to go from one place to another? and back?" and "Yes that's right - for a boat sailing along with the river. And how long does the return trip take?".
The end question is: how long will it take to go from A to B downstream and immediately back upstream to A?

If you have that, and you solved the little error in your calculation here above, then you can do Michelson's calculations. And after that, you may have the basic knowledge to follow the things we discussed with you.

Now replace the boat with light.
Is this not a supposition that is unsupported.
How can you .

 

[roineust]

harrylin,

I think that the answer to your original question, which now I probably understand, is 5kph-2kph=3kph, and 21km/3kph=7hours. So the round trip time would be: 3hours+7hours=10hours.

Roi.

 

[harrylin]

harrylin,

I think that the answer to your original question, which now I probably understand, is 5kph-2kph=3kph, and 21km/3kph=7hours. So the round trip time would be: 3hours+7hours=10hours.

Roi.

Yes indeed!

Now the same question, which you already started, if one wants to cross a 1 km wide river straight over and then return back home.
If you will just aim your boat straight to the other side, then you will arrive on the other side after 0.2 hours, but the stream will have drifted you from course, like this (you want to get to from A B but you end up somewhere else):

Sketch by a swimmer floating in the water:

A ->v
----------
¦
¦c
¦ river
----------
B ->v

Sketch by an observer on the wall:

A
----------
¦
.¦ river ->
..¦
----------
B

So, you should aim your boat slightly upstream in order to get straight to the other side. How long will it take to get straight to the other river bank, like this:

A
----------
/
/ river ->
/
----------
B Harald

 

[harrylin]

Now replace the boat with light.
Is this not a supposition that is unsupported.
How can you .

The calculation is the same! The boat in this example has a constant speed in the water, just as light is supposed to have a constant speed in space.

 

[Buckleymanor]

The calculation is the same! The boat in this example has a constant speed in the water, just as light is supposed to have a constant speed in space.

That is the point.The boat might have a constant speed in this example in water but it does not consistently behave this way.However light is supposed to have a constant speed in space allways which in part is derived from the boats behavioure in water.
Is it not comparing different mediums with the same calculations and arriveing at the same conclusion because the numbers fit.

 

[harrylin]

Oops I put some arrows in the wrong direction.
Roi, here it is corrected with a little elaboration:

Yes indeed, you got the first calculation right!

Now the same question, which you already started, if one wants to cross a 1 km wide river straight over and then return back home.
If you will just aim your boat straight to the other side, then you will arrive on the other side after 0.2 hours, but the stream will have drifted you from course, like this (you want to get to from A B but you end up somewhere else):

Sketch by a swimmer floating in the water (the river banks moving to the left):

A <-- v
----------
¦
¦c
¦ river
----------
B <--v

Sketch by an observer on the wall (the river moving to the right):

A
----------
¦
.¦ river -->
..¦
----------
B

So, you should aim your boat slightly upstream in order to get straight to the other side. How long will it take to get straight to the other river bank, like this:

A
----------
/
/ river -->
/
----------
B Edit: I'll sketch how to calculate it with trigonometry.
The boat goes diagonally to the other side at speed c in the water:

<--- speed of A
|.../
|../c
|./
<--- speed of BHarald

 

[harrylin]

That is the point.The boat might have a constant speed in this example in water but it does not consistently behave this way.However light is supposed to have a constant speed in space allways which in part is derived from the boats behavioure in water.
Is it not comparing different mediums with the same calculations and arriveing at the same conclusion because the numbers fit.

Dear Buckleymanor,

An illustration is rarely perfect, and obviously a boat is not a wave. It's just an easy to understand illustration of how to calculate the propagation of a light ray between moving points.
If Roi understands how a boat is delayed due to relative motion, he can also calculate and understand how light propagates in a moving MMX interferometer; then he may be able to follow the discussions in this thread including the answers to his questions.

 

[harrylin]

It does not show that there is no ether, it shows that the Earth is not rotating through an ether. There are other possibilities such as an ether that is dragged along with Earth that would show the same result, but these are mostly considered superfluous.

Actually, that second part of that answer is completely wrong!
MMX was by design insensitive to rotation. For detecting rotation the Sagnac device was used and Michelson made a larger version to detect the rotation of the earth.
- http://en.wikipedia.org/wiki/Michelson–Gale–Pearson_experiment

 

[roineust]

I think that would be:

root((5kph^2)-(2kph^2))=root(21)kph=~4.6kph
1km/4.6kph=~13minutes, one way straight line.
13minutes*2=~26minutes, round trip from bank to bank straight lines.

Is that correct?

 

[harrylin]

I think that would be:

root((5kph^2)-(2kph^2))=root(21)kph=~4.6kph
1km/4.6kph=~13minutes, one way straight line.
13minutes*2=~26minutes, round trip from bank to bank straight lines.

Is that correct?

Yes that is correct!
And if the river were 21 km wide, the round trip would take about 2x4.6= 9.2 hours.

Now compare that with your earlier calculation for the round trip over the same distance downstream and upstream; you will see that that differs. For that trip you calculated 10 hours.

That is the basis of Michelson's experiment.
He assumed that the Earth is flowing through a light ether, and that it would be possible to detect that motion by comparing the return trip of light in two different directions. In his time it was impossible to directly measure the round trip time over such a short distance, but a change in the roundtrip time could be made visible with his interferometer.
However, he did not find a significant difference. So, he had made a mistake, but where?

Lorentz and Fitzgerald proposed that perhaps matter contracts because it is held together by electromagnetic fields. Heaviside had calculated for charges in motion that those fields contract in the direction of motion, by what is now know as the Lorentz factor.

For the calculation above, the Lorentz factor is (I write SQRT where you write root):
1 / SQRT(1 - 4/25) = 1/0.92.

So (here the boat example falls flat on its face but just consider the calculation!), if the distance of the embankment would shrink like that due to the motion of the river, then the return time as you first calculated would not be 10 hours but 9.2 hours. That is exactly the same as what you get for a trip to the other side of the river and back!
As a matter of fact, it would be the same in all directions.

If you now look again at my post #131, you may be able to understand it this time.
There I only discuss the one-dimensional problem. You may recognize c+v as the light going in the opposite direction as the apparatus, and c-v as the apparatus running ahead of the light, so that the light must catch up - and that takes longer.

Here follow the same formula's with the numbers of the boat example plugged in.

Michelson thought:
t1 = L/(c+v) t1 = 21/(5+2)
t2 = L/(c-v) t2 = 21/(5-2)
t1+t2 = T = [L(c-v)+L(c+v)] / [(c+v)(c-v)]
T = 2L * c /(c² - v²)
T = 2L/c * 1/(1 -v²/c²) T = 2x(21/5) x 1/(1-4/25) = 10

Cheers,
Harald

 

[Buckleymanor]

Lorentz and Fitzgerald proposed that perhaps matter contracts because it is held together by electromagnetic fields. Heaviside had calculated for charges in motion that those fields contract in the direction of motion, by what is now know as the Lorentz factor.

I would like Roi to understand light propagation and clear up some of my own misunderstandings on the subject as well.
What's unclear is that the Lorenze Fitzgerald contraction appears to be a propasition.
Light propagation is a form of electromagnetism which does not contract because it's speed is constant, where as electromagnetic fields do.
So why do some fields contract but not others.

 

[harrylin]

I would like Roi to understand light propagation and clear up some of my own misunderstandings on the subject as well.
What's unclear is that the Lorenze Fitzgerald contraction appears to be a propasition.
Light propagation is a form of electromagnetism which does not contract because it's speed is constant, where as electromagnetic fields do.
So why do some fields contract but not others.

How could a propagation speed "contract"? Instead, fields are maintained at the speed of light.

This proposition necessarily followed from special relativity, which was the outcome of the above considerations: in order to explain such null effects as of MMX, all material objects and fields have to contract in the direction of motion if we assume that the speed of light is constant in all directions.

 

[roineust]

Please see attached diagram,

I still don't understand, why there is no way to explain, in very general terms, or words, and without having light going both ways, but only one way, why this arrangement will not work as I think it will, while in the moving frame.

I know you are trying to explain to me many times, why both beams will always arrive together, no matter what the velocity is...But I still don't understand why.

Is there no middle way between understanding in words, and understanding in math?

 

[Dale]

I have tried to make some calculations, and got to the conclusion, that such an arrangement, will become possible, only when atomic clocks will reach the 10^-21 seconds accuracy level

That sounds like a numerical precision error. Would you like to post your work, and I can run it on Mathematica using arbitrary-precision numbers.

 

[roineust]

Thanks DaleSpam,

The idea is that the light in this arrangement travels a certain way, and that the arrangement is accelerated to a certain velocity. The presumable difference in indication of clock light detectors (at the end), arises supposedly from the time dilation caused by velocity, within the middle way clock (the rhombus shape), so there is a need to be able to detect such a time dilation difference between the two clock light detectors in the moving apparatus.

Therefore, the size of such a device, to be accelerate to a certain velocity, is dependent on the way that is chosen in order to accelerate it. I think there could be two different ways. Either using a Jet, Rocket, Satellite etc...which is in the speed range of 0.0001C-0.000001C, or, if it is possible to make this device really really small, say micro to nano size (I saw that there already exists an atomic clock on a chip), then it might be possible to accelerate the device to say, 0.05C?

For the type of device on a jet, the size could be no more than, say 10 meters long?
It takes light to cross 10 meters ~0.00000003 seconds? at 0.00001C, time dilation factor is ~1.00000000004. Multiplication of these numbers gives ~10^-9.

For the type of device in an accelerator, the size would be, say 0.0001 meter? It takes light to cross this length, ~0.0000000000003 second, and at 0.05C the time dilation factor is ~1.0018. Multiplication of these numbers gives ~10^-13.

Now I see that I might be wrong with the estimation of 10^-21.

It seems to be 10^-9 to 10^-13.

Is this calculation correct?

If it is, then it seems that technologically wise, it is very feasible, and even not astronomically costly to build such an experiment (the jet version).

So maybe such an experiment has already been done in the past, since such accuracy was already achievable some decades ago?

If such an experiment was not done, then of course it all comes back, to my effort to understand ,why is it so clear, that there is no need to build such an experiment?

Why is it so clear, that this device will always have the light clock detectors synchronized as they were at the beginning, no matter what the velocity is?

 

[Dale]

For the type of device on a jet, the size could be no more than, say 10 meters long?
It takes light to cross 10 meters ~0.00000003 seconds? at 0.00001C, time dilation factor is ~1.00000000004. Multiplication of these numbers gives ~10^-9.

Well, a jet flies at more like 10^-6 c where the time dilation factor is 1 + 5 10^-13 or 2 parts in 10^12. This is orders of magnitude less than the current precision of atomic clocks (~1 parts in 10^17).

If such an experiment was not done, then of course it all comes back, to my effort to understand ,why is it so clear, that there is no need to build such an experiment?

Why is it so clear, that this device will always have the light clock detectors synchronized as they were at the beginning, no matter what the velocity is?

It is very clear because this device operates purely on EM principles, and the time dilation of EM phenomena is well studied and well established.

When you mentioned that you had done calculations I was thinking something more along the lines of a thourough analysis of the device itself. Specifically, in the rest frame you have the following:
A) button push, electrical transmission to light, optical transmission to crystal, transmission through crystal, optical transmission to detector, electrical transmission to comparator
B) button push, electrical transmission to light, optical transmission to mirror, reflection, optical transmission to detector, electrical transmission to comparator

Based on assumptions about the geometry of the device and the speed of the signals in the electrical wires and the crystal you should be able to determine if the comparator registers a synchronous activation or not. Then by using standard relativity (time dilation, length contraction, velocity addition, etc.) you should be able to boost the experiment to a different reference frame and determine if the comparator registers a synchronous activation or not. From first principles you are guaranteed that the answers will be the same, so if you get a different answer then it is probably due to numerical precision errors.

 

[harrylin]

Please see attached diagram,

I still don't understand, why there is no way to explain, in very general terms, or words, and without having light going both ways, but only one way, why this arrangement will not work as I think it will, while in the moving frame.

I know you are trying to explain to me many times, why both beams will always arrive together, no matter what the velocity is...But I still don't understand why.

Is there no middle way between understanding in words, and understanding in math?

Understanding in only words or in only math is half-baked, and often mistaken!
A good understanding needs all the help possible, from words, drawings and math.

But I already answered your question with math and words in #131! It is impossible in your drawing that both signals go in only one way (have you ever heard of Escher?). PS I hope that you did notice that the signal from the button to the bottom light is going in the other direction...

Please make sure that you now understand the time dilation of a signal that is going in two directions, as I explained in #131.

Or at least, that you now understand why it takes longer to return in the moving system.

 

[Buckleymanor]

How could a propagation speed "contract"? Instead, fields are maintained at the speed of light.

This proposition necessarily followed from special relativity, which was the outcome of the above considerations: in order to explain such null effects as of MMX, all material objects and fields have to contract in the direction of motion if we assume that the speed of light is constant in all directions.

Note that you have mentioned that fields have to contract.So why can you rule out light speed, once it has propagated.
I can't see a clear distinction between the two types of fields or how experimentialy one is drawn.

 

[harrylin]

Note that you have mentioned that fields have to contract.So why can you rule out light speed, once it has propagated.
I can't see a clear distinction between the two types of fields or how experimentialy one is drawn.

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

 

[Buckleymanor]

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

So light propagation is not a field even though its part of the electromagnetic spectrum and can't be said to contract.
Allthough electromagnetic fields maintained at lightspeed which hold matter together can?
Sorry I am just too dim to understand why one state can and one can't.

 

[harrylin]

So light propagation is not a field even though its part of the electromagnetic spectrum and can't be said to contract.
Allthough electromagnetic fields maintained at lightspeed which hold matter together can?
Sorry I am just too dim to understand why one state can and one can't.

A state of motion is not a state of being... For example, do you think that a contracted loudspeaker can affect the speed of sound? Inversely, if a loudspeaker were made up of sound waves (don't think about how it could be done), would a moving loudspeaker have the same shape as one in rest?

 

[roineust]

harrylin, Dalespam, Buckleymanor.

Thanks for trying to put things into words.
Please Look at the new diagram here.

I hope that I am not bringing the question into places, that can not be explained by words, no matter what...

Roi.

 

[harrylin]

harrylin, Dalespam, Buckleymanor.

Thanks for trying to put things into words.
Please Look at the new diagram here.

I hope that I am not bringing the question into places, that can not be explained by words, no matter what...

Roi.

In the top picture, do you mean that the clock adds a delay time?

This situation is very different from before, as the two detectors are not at the same place along the direction of motion.

In relativity, the simultaneity of distant clocks is "relative": when you bring a system in motion the clocks that appeared synchronous now do not anymore appear in sync - just as you found.

What people do for a standard independent measurement system is to adjust the clocks in the moving system so that the light appears to take the same time both ways. Then your signals will appear to arrive at the same time again for observers in the moving system.

And if you try to calculate it, don't forget that also the moving system is contracted. It's a complicated calculation, but in words what happens is that you make the one-way-speed equal to the average two-way speed. And because the average two-way speed works out, then also the one-way speed works out.

 

[Dale]

Thanks for trying to put things into words.
Please Look at the new diagram here.

I hope that I am not bringing the question into places, that can not be explained by words, no matter what...

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.

 

[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".