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Author Topic: Why can light traverse unequal lengths in an equal amount of time?  (Read 10326 times)

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?


 

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.
 

Offline butchmurray

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Any other explanations?

Thanks,
Butch
 

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Offline old guy

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Shrunk
"...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.
 

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?
 

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.
 

Offline yor_on

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

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?
 

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.
 

Offline mirormimic

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This illustrates my last post. I will explain more comprehensively relative to inquiry.
 

Offline butchmurray

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Quote
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.

Quote
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
 

Offline butchmurray

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Any other explanations?

Thanks,
Butch
 

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 »
 

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
 

Offline butchmurray

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Please, does anyone have the answer to this question?

Thanks,
Butch
 

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.
 

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
 

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?
 

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.
 

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.
 

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.

Quote
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

 

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 »
 

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.

Quote
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
 

Offline JP

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

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
 

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