Why can light traverse unequal lengths in an equal amount of time?

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Offline butchmurray

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Measured from within the inertial frame of the Michelson Morley experiment the lengths of the perpendicular arms are equal.

The condition of no significant fringe confirms that light traverses the lengths of the perpendicular arms in an equal amount of time.

The frame of the Michelson Morley experiment is in relative motion compared to an observer at relative rest.

The length of the arm of the Michelson Morley experiment that is in the direction of motion is contracted and the length of the arm perpendicular to the direction of motion is not contracted judged from relative rest.

The speed of light is constant and the same for all observers.

How can light traverse unequal lengths of the perpendicular arms of the Michelson Morley experiment in an equal amount of time judged from relative rest?
I was not smart enough to know it was impossible to do what I did.

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Offline David Cooper

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I've posted a link into your thread in New Theories which shows how it works through interactive diagrams. If a moderator thinks the content of that page is acceptable for the Physics forum, I'll leave it for them to decide if a link to it from here is okay, but they may consider it to be evangelising as the page describes LET in detail.

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Offline butchmurray

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Any other explanations?

Thanks,
Butch
I was not smart enough to know it was impossible to do what I did.

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Offline old guy

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"...the lengths of the perpendicular arms are equal."

"How can light traverse unequal lengths of the perpendicular arms..."

The actual lengths of the arms are equal. The observed lengths, from a frame moving relative to the arms, are unequal. The image of the arms is distorted by relative motion. The arms stay the same length.
Different observations of images do not create different lengths of the actual arms.

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Offline old guy

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Mods,
Why didn't my reply post. It says "click here to view og's post," but nothing happens. More censorship or just a tech glitch?

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Offline JP

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Mods,
Why didn't my reply post. It says "click here to view og's post," but nothing happens. More censorship or just a tech glitch?

More censorship.  You've been asked to stop evangelizing before, and your posts were shrunk.  Please keep it to New Theories.

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Offline yor_on

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Didn't we have a lengthy discussion about that before?
Or several :)
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Offline yor_on

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What exactly does "The frame of the Michelson Morley experiment is in relative motion compared to an observer at relative rest." mean here? If you're at rest with 'something' it will have no motion relative you, if it has, you can't be at rest with it Butch?
"BOMB DISPOSAL EXPERT. If you see me running, try to keep up."

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Offline mirormimic

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How can light traverse unequal lengths of the perpendicular arms of the Michelson Morley experiment in an equal amount of time judged from relative rest? 
 
If i shine a light toward a reflective surface (mirror) then the light traverses the length (distance) from the light source to the reflective plane surface. If the light is located between two reflective planes ( surfaces ) then the light traverses both distances in an equal amount of time. ( the speed of light) . Yet these two mirror planes will now begin to "stretch" the distance by perpetuating the light seemingly infinitely. As soon as the light is received by both planes then the light is perpetuated ( copied) back into the perspective mirrors. This creates the illusion of the light source being copied, as well stretches the distance. every subsequent reflection and distance that does not represent the surface reflection is ...virtual. Virtual light ..virtual distance. This creates the illusion of the expansion of space and the illusion of a single real source of light taking on infinite virtual subsidiary ( myriad) forms. Thus as soon as the light source ray reaches the plane the only real distance traversed is from the light source to the mirror surface. Yet all corresponding perpetual reflections will appear simultaneous to the surface reflection. This makes it appear that light travels faster or that: " light travels (unequal) lengths ( virtual distances NOT representing the real distance occurring explicitly on the surfaces, perspectively) in an equal amount of time." In other words light is only traveling the real distance from its point to the reflective plane. Because mirror to mirror reflection as reflective of light creates the illusion of the reproduction of light and the lengthening of distance this causes a relative observer to assume that light is traveling faster. This also will create the illusion that light can be in two places ( or myriad places) at the same time.
A picture is worth........!

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Offline mirormimic

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This illustrates my last post. I will explain more comprehensively relative to inquiry.
A picture is worth........!

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Offline butchmurray

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What exactly does "The frame of the Michelson Morley experiment is in relative motion compared to an observer at relative rest." mean here? If you're at rest with 'something' it will have no motion relative you, if it has, you can't be at rest with it Butch?
yor_on,
I don’t understand your comment. The MMX is in relative motion. The observer is at relative rest.

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As soon as the light is received by both planes then the light is perpetuated ( copied) back into the perspective mirrors.
mirormimic,
In the MMX the light is not perpetuated back into the mirrors. The light paths are perpendicular to each other.

I apologize for the delay. So much time passed without further responses, I stopped checking for a while.

Thank you,
Butch
I was not smart enough to know it was impossible to do what I did.

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Offline butchmurray

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Any other explanations?

Thanks,
Butch
I was not smart enough to know it was impossible to do what I did.

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Offline CliffordK

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The average two-way speed of light is always constant.

Most current experiments use some sort of interferometry to measure the difference in distance and time from which the speed is deduced.

So, for example A-->B + B-->A

Adding additional points doesn't really change this.  So, if one has A-->B-->C-->A  Then in Cartesian coordinates, one can assign A (Xa,Ya), B (Xb,Yb), C (Xc,Yc), and your problem reduces to:
Xa-->Xb-->Xc-->Xa in one direction, and Ya-->Yb-->Yc-->Ya in the other direction.  So, one ends up measuring the average velocity along the X axis and the average along the Y axis for any arbitrary X and Y axis.

Unfortunately, there are many synchronization problems limiting the ability to measure the one-way speed of light to an accuracy of 5+ decimal places where one might expect to potentially find a difference in the one way speed of light, A-->B and B-->A if such a difference would actually exist.
« Last Edit: 23/10/2012 02:16:40 by CliffordK »

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Offline butchmurray

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I understand and agree with you on your point.

But, my quandary is that with the MMX in relative motion the arm in the direction of motion is contracted and the arm perpendicular to the direction of motion is not contracted judged from relative rest. The amount of time for light to traverse both arms is equal and verified by the lack of significant fringe.

So, judged from relative rest the arms are different lengths yet the time for light to traverse the arms is equal.

How can light traverse the unequal lengths in the same amount of time judged from relative rest if the speed of light is constant and the same for all observers? v=d/t; v is equal for both arms and t is equal for both arms but d is unequal for the arms judged from relative rest.

Thanks,
Butch
I was not smart enough to know it was impossible to do what I did.

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Offline butchmurray

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Please, does anyone have the answer to this question?

Thanks,
Butch
I was not smart enough to know it was impossible to do what I did.

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Offline JP

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I haven't had the time to work it through (I suspect it's not quite trivial, but I'm not that great at relativity).  You're going to have to consider the relativistic doppler shift as well, since you're in motion with respect to the various mirrors in the interferometer.  I suspect the resolution comes when you consider the frequency shift, since seeing a bright fringe depends on frequency as well as time.

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Offline butchmurray

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Don’t get on the wrong track.

The observer is at relative rest.
The MMX is in relative motion.
In the experiment there is virtually no fringe, so Doppler shift is not a consideration.

Thanks,
Butch
I was not smart enough to know it was impossible to do what I did.

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Offline JP

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The observer is at relative rest.
The MMX is in relative motion.

Those statements aren't useful in relativity, though, since relative motion is between the observer and something else.  In this case, the only thing that matters is what the observer sees, so how is the observer moving relative to the experimental apparatus?  Is the observer moving with respect to the apparatus or not?

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Offline butchmurray

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The apparatus is in the moving frame.
The observer is in the rest frame.
The observer judges the arm of the MMX that is in the direction of motion as contracted and the arm that is perpendicular to the direction of motion as not contracted.
I was not smart enough to know it was impossible to do what I did.

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Offline JP

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The apparatus is in the moving frame.
The observer is in the rest frame.

There are no such things as absolute "moving" or "rest frames" in relativity.  What matters is what is moving with respect to what else.  If you mean that the observer is moving with respect to the experiment, and along one of it's two arms, then what I said about Doppler shifts is important.  I don't know if it fully answers the question, but it's going to need to be taken into account.

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Offline butchmurray

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There are no such things as absolute "moving" or "rest frames" in relativity. What matters is what is moving with respect to what else.
There are 2 inertial frames, K and K’.
Relative to frame K, frame K’ is in motion.
Relative to frame K’, frame K is at rest.

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If you mean that the observer is moving with respect to the experiment, and along one of it's two arms, then what I said about Doppler shifts is important.
The MMX is in frame K’. 
The observer is in frame K.
With special relativity relative motion in respect to only one of the arms is not possible.

The arm of the MMX that is in the direction of motion is contracted and the arm that is perpendicular to the direction of motion is not contracted judged by the observer in frame K.
In the experiment the amount of time is equal for the light to traverse both arms, verified by the lack of significant fringe.
The speed of light is constant and the same for all observers.

How can light, having constant speed, traverse the unequal lengths in an equal amount of time judged from relative rest?

Does this clarify my question?

Thank you,
Butch

I was not smart enough to know it was impossible to do what I did.

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Offline JP

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I think I see what you're getting at, but there are several major problems with the terminology you use (they don't match the terms used in relativity textbooks), which makes understanding precisely what you're getting at hard.
 
There are 2 inertial frames, K and K’.
Relative to frame K, frame K’ is in motion.
Relative to frame K’, frame K is at rest.

If K' is in motion relative to K, then K is in motion relative to K'. 

I think what you're saying is that K contains the experiment.  An observer in K sees the experiment at rest.  An observer in K' sees it in motion.  You're comparing observations by an observer in K and an observer in K'.

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In the experiment the amount of time is equal for the light to traverse both arms, verified by the lack of significant fringe.
The speed of light is constant and the same for all observers.
I take this to mean that as observed in frame K, the light takes the same time to traverse both arms, with no significant fringes: true.

My point still stands.  It is not possible for the light to take the same time.  It is possible, however, to see no significant fringe even if the time taken on each arm is unequal.  Fringes occur when the product of frequency times time taken is not equal (or technically is not a multiple of 2*pi).  Relative motion with respect to the experiment changes the time taken, but it also changes the frequency of the observed light due to the relativistic Doppler shift, both of which will have to be accounted for to predict fringes.  I suspect this is how you resolve the paradox, but I'll leave the math to someone else.  :) 
« Last Edit: 26/10/2012 16:52:58 by JP »

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Offline butchmurray

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If K' is in motion relative to K, then K is in motion relative to K'.
Yes. And, either frame can be considered at rest with respect to the other frame. In this circumstance K’ is considered as in relative motion.

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I think what you're saying is that K contains the experiment.
No. K’ contains the experiment.

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An observer in K sees the experiment at rest.  An observer in K' sees it in motion.
The opposite. An observer in K sees the experiment in motion.

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You're comparing observations by an observer in K and an observer in K'.
Yes, but only the lengths of the arms. The arms are equal judged from within K’ (the only comparison). As for the fringe, none was observed within K’ so none was produced and equal time for light to traverse both arms was verified. The condition of ‘no significant fringe’ is true for all observers so the equal time for light to traverse both arms is also true for all observers.   

What is under consideration is what the observer in K sees when looking at the experiment in K’.

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I take this to mean that as observed in frame K, the light takes the same time to traverse both arms, with no significant fringes: true.
Almost. As observed from frame K with the experiment in frame K’.

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My point still stands.  It is not possible for the light to take the same time.
That is my point!!! The condition of ‘no significant fringe’ for the experiment is true for all observers. So, the equal time for light to traverse both arms is also true for all observers including observers who judge the arms to be of unequal length.

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It is possible, however, to see no significant fringe even if the time taken on each arm is unequal.  Fringes occur when the product of frequency times time taken is not equal (or technically is not a multiple of 2*pi).  Relative motion with respect to the experiment changes the time taken, but it also changes the frequency of the observed light due to the relativistic Doppler shift, both of which will have to be accounted for to predict fringes. I suspect this is how you resolve the paradox,…
Same as above. The condition of ‘no significant fringe’ for the experiment is true for all observers. So, the equal time for light to traverse both arms is also true for all observers even though judged from relative rest (frame K) the lengths of the arms are unequal.

That is the paradox. What is the solution?

Thank you VERY much,
Butch
I was not smart enough to know it was impossible to do what I did.

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Offline JP

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"No significant fringe" does not mean equal time.  It means the product of time x frequency is equal. 

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Offline butchmurray

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"No significant fringe" does not mean equal time.  It means the product of time x frequency is equal.

You’re right. ‘No significant fringe’, however, does mean that the time for light to traverse one of the arms compared to the time for light to traverse the other arm did not change (remained equal) as the frequency was identical for the light in both arms.

The condition of ‘no significant fringe’ for the experiment is true for all observers. So, the time remained equal for light to traverse both arms of unequal length judged from relative rest. True?

BTW
I can’t find any resources that discuss light projected in the direction of motion within a moving frame judged from relative rest. Do you or anyone else know of any? (I’m certain there aren't any)

Thank you,
Butch
I was not smart enough to know it was impossible to do what I did.

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Offline JP

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The condition of ‘no significant fringe’ for the experiment is true for all observers. So, the time remained equal for light to traverse both arms of unequal length judged from relative rest. True?

Nope.  From relative rest, the arms are of equal length, so the time is obviously the same.  In motion, they appear as different lengths, so the time is different.  The frequency will also change, which has to be accounted for.

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Offline butchmurray

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JP,

First and foremost, thank you for your time and expertise.

Secondly, as not to get bogged down by avoidable misunderstandings:

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From relative rest, the arms are of equal length, so the time is obviously the same.
I take this to mean at rest with the apparatus in the relatively moving frame. Correct?

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In motion, they appear as different lengths, so the time is different. The frequency will also change, which has to be accounted for.
I take this to mean judged by an observer in the frame that is at rest relative to the moving frame containing the apparatus. Correct?

Thank you,
Butch
I was not smart enough to know it was impossible to do what I did.

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Offline JP

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There are two frames, K and K', moving with respect to each other. 

The experiment is at rest if viewed from frame K'. 

If an observer in frame K' observes the experiment, both arms appear to be equal, the light traveling down both arms seems to be the same frequency, and he sees no fringes.

If an observer in frame K observes the experiment, one arms will appear to be length contracted, so the time taken for the light to travel down back along the two arms will no longer be equal.  However, the frequency of the light down the two arms will no longer be equal due to the Doppler shift.  Since frequency multiplied by time determines the condition for a bright fringe, this observer will probably also see a bright fringe.  Again, I haven't worked out the math.

I'm not sure how much clearer I can be.  If you don't understand what I'm getting at here, perhaps someone else can explain better.

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Offline butchmurray

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I'm not sure how much clearer I can be.  If you don't understand what I'm getting at here, perhaps someone else can explain better.
You’re doing great! Can you please explain:

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an observer in frame K' observes the experiment … sees no fringes.
If the apparatus did not produce fringe
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an observer in frame K observes the experiment… see a bright fringe.
How could any observer see fringe from the apparatus?

Thanks,
Butch

I was not smart enough to know it was impossible to do what I did.

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Offline JP

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I really don't know how to say it much clearer: frequency multiplied by time taken is what you measure in your Michelson interferometer.  Both time taken and frequency depend on the relative motion of the observer and experiment, so you need to take them into account.  If you don't include frequency, you're going to make wrong predictions.

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Offline butchmurray

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I am not trying to be nor want to appear to be argumentative.

My point is that the experiment did not produce significant fringe.

All observers see the same experiment.

No observer can see significant fringe from the experiment if the experiment didn’t produce significant fringe.

Do you see my point? Or, is there a flaw with my logic?

Thanks,
Butch
I was not smart enough to know it was impossible to do what I did.

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Offline David Cooper

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I can’t find any resources that discuss light projected in the direction of motion within a moving frame judged from relative rest. Do you or anyone else know of any? (I’m certain there aren't any)

That's because the light has to be observed by the observer at what you call relative rest in accordance with the speed limit on light in that observer's own frame. You cannot expect him to be able to see light moving at any other speed through his frame, but that's what you appear to be expecting him to do. As always, though, you can only measure the speed of light on a round trip, so what he observes isn't necessarily giving a true picture.



If an observer in frame K observes the experiment, one arms will appear to be length contracted, so the time taken for the light to travel down back along the two arms will no longer be equal.  However, the frequency of the light down the two arms will no longer be equal due to the Doppler shift.  Since frequency multiplied by time determines the condition for a bright fringe, this observer will probably also see a bright fringe.  Again, I haven't worked out the math.

No observer sees a bright fringe. The easiest way to resolve this thing is to put lots of semi-silvered mirrors into the experiment to reflect some of the light out sideways to the observer observing from another frame. When you do this, that observer will see the progress of the light through the apparatus travel in accordance with the speed limit for light of his own frame, and the frequency of the light will remain constant throughout from his point of view too, though it will be at a lower frequency than would be observed by an observer moving with the apparatus.

If instead of the semi-silvered mirrors we allow our stationary observer to stick his head into the beam as the apparatus goes past him, he will then see the frequency of the light change depending on which direction it's going in relative to the apparatus, but that has no bearing on what happens when the fringe is formed (or not formed) because at the location where that happens (or doesn't happen) we have two light beams merged together again and travelling along the same path with the same frequency.
« Last Edit: 29/10/2012 22:31:36 by David Cooper »

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Offline JP

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No observer sees a bright fringe
Actually, all observers see bright fringes.  The Michelson interferometer produces a fringe pattern.  The central fringe may be bright or dark depending on the relative optical path lengths (distance divided by frequency) in the two arms. 


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The easiest way to resolve this thing is to put lots of semi-silvered mirrors into the experiment to reflect some of the light out sideways to the observer observing from another frame. When you do this, that observer will see the progress of the light through the apparatus travel in accordance with the speed limit for light of his own frame, and the frequency of the light will remain constant throughout from his point of view too, though it will be at a lower frequency than would be observed by an observer moving with the apparatus.
The frequency shifts each time the light bounces off a moving mirror.  This is one of the ways that RADAR works to pick up the speed of targets.  If you're moving WRT the experiment, you're moving longitudinally with respect to one arm and transversely with respect to the other which seems like it should impart 2 different Doppler shifts along those two arms.  Similarly, the length contraction of those arms shouldn't be identical.

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If instead of the semi-silvered mirrors we allow our stationary observer to stick his head into the beam as the apparatus goes past him, he will then see the frequency of the light change depending on which direction it's going in relative to the apparatus, but that has no bearing on what happens when the fringe is formed (or not formed) because at the location where that happens (or doesn't happen) we have two light beams merged together again and travelling along the same path with the same frequency.
I wouldn't be surprised if you added up the total path lengths for both paths and got equal answers for the moving and stationary observers (with respect to the experimental reference frame), but this argument isn't sufficient to prove it.  I completely agree that along the last path segment, both beams pick up the same phase shift, but the question is whether they pick up identical phase shifts along the prior segments for both the moving and stationary observers.

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Offline butchmurray

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Hypothetically, there is an indicator on the apparatus that illuminates for any condition other than ‘insignificant fringe’.

At the apparatus the indicator is never illuminated.

The indicator is never illuminated for any observer.

The condition of ‘insignificant fringe’ is true for all observers.

Is this logic realistic?


Thanks,
Butch
I was not smart enough to know it was impossible to do what I did.

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Offline David Cooper

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You certainly aren't going to get a different answer for different observers when it comes to the business of whether there's a null result or not. There will always be a null result, and at no point will anything show up that can show one frame to be a preferred frame (or closer to it) than any other.

The trouble with this whole business is that you can pick any frame and set the speed of light across the apparatus by it, thereby determining how the frequency must be changed at the mirrors and how far the light has to go through that frame to complete the trips along the arms, but all frames provide the same end result, no matter what happened along the way. By banning all discussion of simultaneous events at a distance, SR allows you to ignore the contradictions in the accounts generated by the analysis from the point of view of observers in different frames, but if you lift that ban you can actually create a version of SR which eliminates these and all the other contradictions normally found in SR simply by allowing there to be something similar to a preferred frame of reference (meaning that not all frames are equal): this still maintains the important difference between SR and LET by continuing to provide the Spacetime aspect of the model which is necessary for it to be used as a base for GR.

I think Butch is essentially trying to disprove SR on the basis of these contradictions, but they are just a side show - they demonstrate that there is some false dogma built into SR, but none of it is critical to the real functionality of SR. He is demanding that a stationary observer should be able to see light move through the moving apparatus at the speed of light relative to the frame of reference of the apparatus rather than to his own frame of reference, and he's demanding this on the basis of Einstein's claim that the speed of light is the same for all observers. At the same time, that stationary observer would obviously also have to be able to see light progressing through the moving apparatus at the speed of light relative to his own frame of reference, so contradiction is impossible to avoid. I think he's misinterpreting Einstein's claim, because although the contradictions are indeed there in SR, they aren't quite as overt as that.

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Offline JP

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I think Butch is essentially trying to disprove SR on the basis of these contradictions, but they are just a side show - they demonstrate that there is some false dogma built into SR, but none of it is critical to the real functionality of SR.

David, the moderators have asked you many times to stop stating your opinions about SR as facts.  You have made many good contributions to the forum, so please don't make us moderate you further on this topic.

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Offline butchmurray

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Since the state of the hypothetical indicator is the same for all observers the frequency x time relationship of the light in the perpendicular arms of the apparatus is also the same for all observers.

Can there be any doubt that the time for light to traverse the perpendicular arms of the apparatus is the same for all observers?

Butch
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Offline JP

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Can there be any doubt that the time for light to traverse the perpendicular arms of the apparatus is the same for all observers?

Oh yes, the measured time for light to traverse both arms changes if you move with respect to the apparatus.  Why?  Because the light in the transverse arm has no apparent longitudinal motion when you're at rest, and your motion gives it longitudinal motion WRT you.  It travels a longer total path from your reference frame at a constant speed, so it takes longer.

I found this today.  It's basically the experiment you're setting up, with all the details of deriving how it works (and why you get a null result).  The frequency shift is important (and they account for it in that site via time dilation):
http://en.wikipedia.org/wiki/Kennedy%E2%80%93Thorndike_experiment

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Offline David Cooper

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I think Butch is essentially trying to disprove SR on the basis of these contradictions, but they are just a side show - they demonstrate that there is some false dogma built into SR, but none of it is critical to the real functionality of SR.

David, the moderators have asked you many times to stop stating your opinions about SR as facts.  You have made many good contributions to the forum, so please don't make us moderate you further on this topic.

Sorry - I try to write everything carefully to avoid doing stating things directly as truths when I have no doubt that they are true, but I don't spot it every time because I'm hypersensitive to contradictions due to my work in A.I. where they're used to invalidate theories (or the parts of theories which generate those contradictions) - when I can see that something is wrong, I can't help seeing that it is wrong. It in an incontrovertible fact that contradictions are generated by SR in its standard form, but I don't see why this is such a big problem for you or for science as SR does not depend on the parts which generate those contradictions for any of its vital functionality. I don't see why it should cause you to switch into suppression mode whenever that comes to the surface. It will inevitably keep coming to the surface because there are millions of people out there who reject SR (and even science as a whole) on the basis of these unnecessary contradictions. In the near future, you're also going to have AGI systems making exactly the same point in posts to this forum (and to every other science forum), and they won't be programmed to make incorrect objections: they'll simply be analysing the theories and commenting on them in accordance with their rational analysis of them. Contradictions aren't good - they need to be eliminated if you are genuinely seeking truth.

[Edit was to changed "falsify" into "invalidate" - don't know why I used the wrong word there and failed to pick up on it while checking text after posting.]
« Last Edit: 01/11/2012 21:00:26 by David Cooper »

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Offline JP

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David, I have nothing against discussions of what constitutes good science and discussing interpretations of the mathematics of theories, but I'm feeling like a broken record on this point: let's keep such discussions to New Theories. 

Future posts going down this path will be moved to New Theories without discussion.

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Offline butchmurray

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I looked at The Kennedy–Thorndike experiment and it became clear my original question was inadvertently misleading. The question should have been:

“Why can light traverse the non-contracted length of a light path in the same amount of time light traverses the contracted length of that light path?”

This is the question that I cannot find an answer for. Sorry for the confusion, JP.


Thank you,
Butch
I was not smart enough to know it was impossible to do what I did.

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Offline JP

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I thought I had stated this clearly earlier, but reviewing my prior posts, I didn't.

“Why can light traverse the non-contracted length of a light path in the same amount of time light traverses the contracted length of that light path?”

It doesn't.  They take different times.  The reason you still get a null result of the experiment is explained in that link.

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Offline butchmurray

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Quote
It doesn't.  They take different times.  The reason you still get a null result of the experiment is explained in that link.
I understand that interferometer arms/light paths of unequal length can be set to lengths that result in the condition of no significant fringe notwithstanding their unequal lengths and light traverses those unequal lengths in different amounts of time.

Me:
Quote
“Why can light traverse the non-contracted length of a light path in the same amount of time light traverses the contracted length of that light path?”
Notice it refers to the non-contracted and contracted lengths for that (singular) light path. I am not referring to an interferometer for which the arms set to unequal lengths. I am referring to a non-contracted length and a length contracted by Lorentz–FitzGerald contraction.

To be clear, my question is in reference to the MMX only for which the arms are set to an equal length within the frame of the apparatus.

I sense somewhat understandable impatience that I think is due to a misunderstanding of the question and its purpose. The facts as I know them lead to an inconsistency in the interpretation of the null result of the MMX. I need to know whether my facts are correct or not and if not where my misunderstanding is. Additionally, I am unable to locate resources that address the properties of light in light paths contracted by Lorentz–FitzGerald contraction, the heart of my quandary. I’ve asked the Forum previously and I am asking again if anyone knows where I can find this information or credibly inform me that no such information exists. That is the purpose and, hopefully, an insight into the question.

That said, following is proof of the legitimacy and further clarification of my question with facts, as I know them. There is one caveat. I’m almost certain there is a credible reference to the following statement but I can’t locate it at the moment. Until I confirm the source the following should be presumed an assumption. “An event observed by any observer in observable by all observers.”

Definitions:
Frame K’ – the relatively moving inertial frame which contains the apparatus of the MMX
Frame K – the inertial frame, which is at relative rest


For the MMX its two arms, which are light paths, were set to an equal length. It was verified at the apparatus that light took an equal amount of time to traverse the perpendicular arms/light paths.

At the apparatus each light wave from the monochromatic source was split by a half-silvered mirror and produced a pair of light waves. One of each pair of the resultant light waves was projected onto each of the two perpendicular arms/light paths of the apparatus. Each pair of light waves was reflected directly back to the half-silvered mirror and recombined. The pair of light waves that resulted from the split at the half-silvered was the same pair of light waves that were recombined at the half-silvered mirror. For the same pair of light waves that was split from a single light wave to arrive back at the half-silvered mirror at the same time and recombine perfectly the light waves must traverse their respective arms/light paths in an equal amount of time. Since the split pairs of light waves arrived back at the half-silvered mirror at the same time at the apparatus and recombined, they arrived back at the half-silvered mirror at the same time for all observers. THE ASSUMPTION: *An event observed by any observer in observable by all observers. The event is the recombining of the same pair of light waves that resulted from the split.*

Therefore, the time for the pair of light waves to traverse the perpendicular arms/light paths was equal for all observers to include an observer in frame K that was at relative rest. But, for relative velocities >0<c for the relatively moving frame K’ which contains the apparatus, the arm/light path of the apparatus which is in the direction of motion is contracted by the Lorentz–FitzGerald contraction factor and the arm/light path perpendicular to the direction of motion is not contracted judged from frame K, which is at relative rest.

So, how can light traverse the contracted and the non-contracted arms/light paths of the apparatus in an equal amount of time judged from frame K, which is at relative rest?

Butch
I was not smart enough to know it was impossible to do what I did.

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Offline butchmurray

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The above distilled to its essence:

A light path can contract due to Lorentz contraction.
The light in that light path does not contract.

Can those two be reconciled?


Butch
I was not smart enough to know it was impossible to do what I did.

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Offline JP

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I don't know how else to say this: 

The light does contract, in a sense, since it's frequency shifts due to time dilation.  This is sufficient to make observers who are in K and K' agree on the results of the experiment.  To agree, the same number of wavelengths have to exist down both paths.  The paths contract, but time dilation (or if you prefer, length contraction) keeps the total number of wavelengths in each path the same.

If that doesn't make sense still, I suggest you read up a bit on how time dilation and length contraction are related, especially on that wiki link I provided previously.  Or perhaps someone else here can explain it more clearly.

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Offline butchmurray

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JP,
Thank you so much for your time and patience.


The dilemma stated more exactly:

Think of this light path as the arm of the MMX that’s in the direction of motion. The proper length of this light path and the length of this light path contracted by Lorentz contraction are not equal.

Light traverses these unequal lengths in an equal amount of time.

Above, the constant speed of light and Lorentz contraction obviously conflict. AT LEAST ONE OF THE TWO IS INVALID!


Albert Einstein (1879–1955).  Relativity: The Special and General Theory.  1920.
XIV.  The Heuristic Value of the Theory of Relativity
http://www.bartleby.com/173/14.html
Quote from Einstein:
“OUR train of thought in the foregoing pages can be epitomised in the following manner. Experience has led to the conviction that, on the one hand, the principle of relativity holds true, and that on the other hand the velocity of transmission of light in vacuo has to be considered equal to a constant c. By uniting these two postulates we obtained the law of transformation for the rectangular co-ordinates x, y, z and the time t of the events which constitute the processes of nature. In this connection we did not obtain the Galilei transformation, but, differing from classical mechanics, the Lorentz transformation.     
The law of transmission of light, the acceptance of which is justified by our actual knowledge, played an important part in this process of thought. Once in possession of the Lorentz transformation, however, we can combine this with the principle of relativity, and sum up the theory thus:     
Every general law of nature must be so constituted that it is transformed into a law of exactly the same form when, instead of the space-time variables x, y, z, t of the original co-ordinate system K, we introduce new space-time variables x', y', z', t' of a co-ordinate system K'. In this connection the relation between the ordinary and the accented magnitudes is given by the Lorentz transformation. Or, in brief:General laws of nature are co-variant with respect to Lorentz transformations.     
This is a definite mathematical condition that the theory of relativity demands of a natural law, and in virtue of this, the theory becomes a valuable heuristic aid in the search for general laws of nature. If a general law of nature were to be found which did not satisfy this condition, then at least one of the two fundamental assumptions of the theory would have been disproved.”

Someone at Cambridge should probably look into this.

The world of physics will change starting NOW.

Butchmurray
Thorntone E. Murray
I was not smart enough to know it was impossible to do what I did.

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Offline JP

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Butch, please don't use the mainstream fora to suggest that special relativity is wrong.  The New Theories forum is the proper place for such suggestions.

Thanks!

On the physics:

Regarding your claims: your assumptions are wrong, and therefore so are the conclusions you draw from them.  Time dilation and length contraction go hand in hand, and when you're trying to analyze the Michelson-Morely experiment from a moving reference frame, you need to account for them both.  It's the number of wavelengths of light that exist in each arm that matters, and that depends on both the frequency of the light and length of the arms.  If you're in a moving frame WRT the experiment, both of those change, not just the length.

Claiming that the length shifts is true.  Claiming that the length is the only thing that matters in the MM result is false.  Therefore, you're reaching incorrect conclusions.

I really can't be more clear than this.  Perhaps someone else can explain it better.  If not, I'd suggest reading up on time dilation and the Doppler shift.  It really does explain this effect fully.
« Last Edit: 04/11/2012 21:03:30 by JP »

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Offline butchmurray

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Thank you,
Butch
I was not smart enough to know it was impossible to do what I did.

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Offline sciconoclast

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This might help:
                          Instead of stating that light appears to travel at the same speed to any observer try stating that light within any frame appears to travel at the same speed to an observer in that frame the same as light in any other frames appears to travel to observers in those frames.

 The frame in most of these experiments is the Earth's gravitational field. The Sagnac experiment in which light is sent through mirrors on a turntable that represents motion independent of that of the Earth's gravitational field results in different arrival times.

The Michelson-Gale experiment, on the other hand, which uses the Earth as a turntable does not produce a phase shift.

Einstein predicted the results for both these experiments incorrectly. Later they were deemed to be in agreement with relativity. It is often said that Einstein was the worst interpreter of his own theories.

An interesting anomaly to relativity occurred when there were two separate rectangular paths with one having a larger portion parallel to the Earth"s rotation and the other a larger portion perpendicular to it. Within each course there is no phase shift. However, when the results of one are compared to the other there is a difference. This enabled the experimenters to calculate the rotational velocity of Earth which agreed with that calculated from the star field.

Getting back to the original question, the distances that appear to be the same to an observer within a frame will also appear to be traveled by light in the same amount of time to the same observer. However, to any observer in space that is not rotating with the Earth the distances will appear different and if he, or she, or it, is sent the arrival times the speed of light will appear to have varied.
 
Just ask the guys working with orbiting communication net works.