[Hal (OP)]

 It is a basic requirement of special relativity theory (SRT)  that all relatively moving inertial observers agree on an observable (interference fringe shift, for example). It is shown that SRT trivially leads to a disagreement on the observables ( interference fringe shift) in two relatively moving inertial reference frames.

Suppose that the interference fringe shift of a light speed experiment predicted in frame S is N, and the fringe shift predicted in frame S’ for the same experiment is N’. It is a requirement of relativity theory that there should be an agreement on the observable (the fringe shift) in both frames, i.e.  N = N’. Let us see if this is actually the case.

We know that,

                       N  = c ∆t / λ      and      N'  =  c ∆t' / λ'

But,

                    ∆t'  =  γ  ( ∆t  -  v ∆x/c2 )          and    λ'   =   γ λ

 Therefore,

                N'  =  c ∆t' / λ'   =   c ( γ  ( ∆t  -  v ∆x/c2 ) ) / (γ λ) =   (c ∆t / λ  ) -  ( v ∆x/(cλ)  =  N -  ( v ∆x/(cλ) ≠ N

 

Galilean relativity, however, does not lead to such disagreement, as shown below.

                                    ∆x'  =  ∆x  -  v ∆t

                                   ∆t'  =  ∆t

                                        c'  =  c ± V  ⟹  c' / c = 1± v/c

                                         f' = f   ⟹ c' / λ'  =  c / λ  ⟹  λ'  = λ (1± v/c)

 Therefore,

                                 N'  = ( c' ∆t' ) / λ'  =  ( c ± V ) (∆t) / λ (1± v/c)  = c ∆t / λ = N

 

 

[Eternal Student]

Hi.

It is a basic requirement of special relativity theory (SRT)  that all relatively moving inertial observers agree on an observable

     That doesn't seem to be true.   The length of a pole or of a barn is an observable and it very much depends on the frame of reference you are using.    Hence, the usual SR examples of length contraction and getting a long pole into a short barn etc.
     It's agreed that the laws of physics should be the same in all inertial frames,  or that events in one frame are mapped to events in another frame on a 1-to-1 basis  -  but there's nothing to demand that an arbitrary observable or measurable quantity will be the same for two observers moving relative to each other.  Indeed, it often cannot be the same for the two frames of reference (or for the observers at rest in those frames, if you prefer to phrase it that way).

Best Wishes.

[Hal (OP)]

Hi.

It is a basic requirement of special relativity theory (SRT)  that all relatively moving inertial observers agree on an observable

     That doesn't seem to be true.   The length of a pole or of a barn is an observable and it very much depends on the frame of reference you are using.    Hence, the usual SR examples of length contraction and getting a long pole into a short barn etc.
     It's agreed that the laws of physics should be the same in all inertial frames,  or that events in one frame are mapped to events in another frame on a 1-to-1 basis  -  but there's nothing to demand that an arbitrary observable or measurable quantity will be the same for two observers moving relative to each other.  Indeed, it often cannot be the same for the two frames of reference (or for the observers at rest in those frames, if you prefer to phrase it that way).

Best Wishes.

See Kennedy-Thorndike experiment, Wikipedia.

" For constant ΔN, i.e. for the fringe shift to be independent of velocity or orientation of the apparatus, it is necessary that the frequency and thus the wavelength λ be modified by the Lorentz factor.  .  .  .   "

[Eternal Student]

Hi.

    That Wikipedia entry and the experiment described there concerns movement relative to the "Luminiferous Aether".   The equipment (and the person observing the interference patterns) was moving relative to the Aether.   The only expectation is that movement relative to the Aether wouldn't make any difference.

Suppose that the interference fringe shift of a light speed experiment predicted in frame S is N, and the fringe shift predicted in frame S’ for the same experiment is N’.

    What you seem to have done in your calculations is to have two different observers in relative motion to each other.   One of those can be stationary with respect to the equipment (they move with the equipment).  The other observer would be moving past the equipment and looking at the interference pattern as they race past it.    The observer who was moving relative to the equipment doesn't need to see the same interference pattern.   All sorts of observable distances between two things (like distances between fringes) can be different for them.

Best Wishes.

[Hal (OP)]

Hi.

    That Wikipedia entry and the experiment described there concerns movement relative to the "Luminiferous Aether".   The equipment (and the person observing the interference patterns) was moving relative to the Aether.   The only expectation is that movement relative to the Aether wouldn't make any difference.

Suppose that the interference fringe shift of a light speed experiment predicted in frame S is N, and the fringe shift predicted in frame S’ for the same experiment is N’.

    What you seem to have done in your calculations is to have two different observers in relative motion to each other.   One of those can be stationary with respect to the equipment (they move with the equipment).  The other observer would be moving past the equipment and looking at the interference pattern as they race past it.    The observer who was moving relative to the equipment doesn't need to see the same interference pattern.   All sorts of observable distances between two things (like distances between fringes) can be different for them.

Best Wishes.

The hypothetical experiment consists of mirrors fixed in the lab and a moving observer/detector (see attached pdf). This observer/detector, say observer A, is the one who sees the fringes (fringe shifts). The problem is to predict the fringe shift observed by the observer A in two relatively moving inertial frames (S and S' ) , one of which is the lab frame (S). It is found that the fringe shift predicted in S disagrees from that predicted in S'.

[Eternal Student]

Hi.

It is found that the fringe shift predicted in S disagrees from that predicted in S'.

OK.

I haven't studied the pdf for various reasons (I'm wary of downloading anything on this computer).   However, on the face of what you've said... that is OK isn't it?   You are reporting the fringe shift as measured in different frames of reference, so you have reported ΔN and ΔN'  - but ΔN' is of very little interest in the lab frame.

    Distances can be 1 unit along the x-axis of frame S but they can be 2 units along the x' axis in frame S'.    There's no reason why they have to be same in different frames of reference.    You can put another person at rest in frame S' and they can shout out to the person in the lab frame  "I have calculated ΔN'.  That fringe shift should be 2 units in my primed ' co-ordinate system"....
The person in the lab frame can reply...."interesting... but I don't have an x' axis.... what is it supposed to be in my co-ordinates?"..... 
"hold on a moment,  I'll calculate it.....turns out it will be 1 unit of distance in your co-ordinate system".....
 "oh...good because that is what I have".

Best Wishes.

[Hal (OP)]

Hi.

It is found that the fringe shift predicted in S disagrees from that predicted in S'.

OK.

I haven't studied the pdf for various reasons (I'm wary of downloading anything on this computer).   However, on the face of what you've said... that is OK isn't it?   You are reporting the fringe shift as measured in different frames of reference, so you have reported ΔN and ΔN'  - but ΔN' is of very little interest in the lab frame.

    Distances can be 1 unit along the x-axis of frame S but they can be 2 units along the x' axis in frame S'.    There's no reason why they have to be same in different frames of reference.    You can put another person at rest in frame S' and they can shout out to the person in the lab frame  "I have calculated ΔN'.  That fringe shift should be 2 units in my primed ' co-ordinate system"....
The person in the lab frame can reply...."interesting... but I don't have an x' axis.... what is it supposed to be in my co-ordinates?"..... 
"hold on a moment,  I'll calculate it.....turns out it will be 1 unit of distance in your co-ordinate system".....
 "oh...good because that is what I have".

Best Wishes.

But there is only one observer/detector of the fringes and he/she can observe only one value of fringe shift. The moving observer/detector could be a camera, for example. Say, the intelligent camera records the fringe shift. The problem is that the fringe shift predicted in S and S' disagree, which is a contradiction because the camera cannot record two different values of fringe shift at the same time.

[Halc]

The problem is that the fringe shift predicted in S and S' disagree, which is a contradiction because the camera cannot record two different values of fringe shift at the same time.

What a particular observer measures is an objective (frame independent) fact, and thus that observation would be the same in both S and S'. Relativity theory does not predict otherwise.

[Hal (OP)]

The problem is that the fringe shift predicted in S and S' disagree, which is a contradiction because the camera cannot record two different values of fringe shift at the same time.

What a particular observer measures is an objective (frame independent) fact, and thus that observation would be the same in both S and S'. Relativity theory does not predict otherwise.

Strict, rigorous application of special relativity and Lorentz transformations always leads to a contraciction, except in the case of null results. That is, if a null fringe shift is predicted in S, then a null result is predicted in S'. SRT leads to a contradiction in the case of non-null results, such as in the case of Sagnac effect, and the hypothetical experiment shown in the attached pdf.

[Eternal Student]

Hi again.

The moving observer/detector could be a camera, for example. Say, the intelligent camera records the fringe shift. The problem is that the fringe shift predicted in S and S' disagree, which is a contradiction because the camera cannot record two different values of fringe shift at the same time.

    If I've understood what you said,  the camera doesn't record or show two different values.   It's moving and it records (takes a picture of) what the fringe shifts look like to the camera.  For that the fringe shifts in the frame S' are what will be observed and what is shown on that picture.
    The camera cannot take a picture of what it looks like to the observer in the lab frame, S, the camera is not at rest in frame S and so it doesn't see that. 

Best Wishes.

[Hal (OP)]

Hi again.

The moving observer/detector could be a camera, for example. Say, the intelligent camera records the fringe shift. The problem is that the fringe shift predicted in S and S' disagree, which is a contradiction because the camera cannot record two different values of fringe shift at the same time.

    If I've understood what you said,  the camera doesn't record or show two different values.   It's moving and it records (takes a picture of) what the fringe shifts look like to the camera.  For that the fringe shifts in the frame S' are what will be observed and what is shown on that picture.
    The camera cannot take a picture of what it looks like to the observer in the lab frame, S, the camera is not at rest in frame S and so it doesn't see that. 

Best Wishes.

S is the lab frame, and S' is moving with velocity V relative to S in the +x direction. The camera is moving with velocity V0
( ≠ V ) in the +x direction relative to S. So the camera can be moving both in S and S'.

The camera is just moving with velocity V0 and taking pictures of the fringes and possibly compute the fringe shift also. The problem is to predict the fringe shift both in S and in S'.

In SRT "what it looks like to S (or to S' ) " means what is predicted in S ( or in S' ). Therefore, what it (the fringe shift)  looks like to S should agree with what it looks like to S', otherwise it would be a contradiction.

If I have understood what you said, what it looks like to S (S' ) in this case does not mean what fringe shift is observed by some other observer/detector in S (S') because that is another (unrelated) problem. We have only one interferometer and a co-moving camera.

[Origin]

S is the lab frame, and S' is moving with velocity V relative to S in the +x direction. The camera is moving with velocity V0
( ≠ V ) in the +x direction relative to S. So the camera can be moving both in S and S'.

No the camera can't be moving in both S and S'!
You said there are 2 frames S and S'.  You said the camera is moving relative S.  So that means the camera is not moving in frame S'.  Unless there is a frame S'' that your not telling us about.

[Halc]

S is the lab frame, and S' is moving with velocity V relative to S in the +x direction. The camera is moving with velocity V0
( ≠ V ) in the +x direction relative to S. So the camera can be moving both in S and S'.

Fine.

The camera is just moving with velocity V0 and taking pictures of the fringes and possibly compute the fringe shift also.

A camera (preferably a video camera if it is measuring something dynamic like a fringe shift) is an objective point of view. It doesn't compute anything. You do (or not as the case may be). What the camera records has nothing to do with any particular reference frame. It also has no direct correspondence to what is actually going on in frame S'' in which it is stationary. Don't confuse a frame with a point of view. Two relatively stationary observers looking at the same thing will measure completely different things such as the frequency of a moving light emitter.

The problem is to predict the fringe shift both in S and in S'.

Nope. The problem is to predict what the camera will see, which is utterly independent of any selection of frame.

In SRT "what it looks like to S (or to S' ) " means what is predicted in S ( or in S' ).

This assertion is dead wrong twice. 1) A frame doesn't define an observer, so what it 'looks like to some frame' is meaningless. 2) What it looks like to some observer has nothing to do with any selection of frame.

You keep making this same mistake over and over. I see little reason to continue to point it out.

No the camera can't be moving in both S and S'!

Why not? There are thousands of possible frames and it can only be stationary in one of those. These two are both apparently not it. Point is, choice of frame (just an abstract method to assign coordinates to events) is irrelevant. A camera is physical (not abstract) and sees what it sees regardless of how one chooses to assign abstract numbers to the events involved.

[Origin]

Why not? There are thousands of possible frames and it can only be stationary in one of those. These two are both apparently not it.

The point I was making is that the OP is so sloppy about frames that it is hard to decipher what he is trying to say.

[Hal (OP)]

S is the lab frame, and S' is moving with velocity V relative to S in the +x direction. The camera is moving with velocity V0
( ≠ V ) in the +x direction relative to S. So the camera can be moving both in S and S'.

Fine.

The camera is just moving with velocity V0 and taking pictures of the fringes and possibly compute the fringe shift also.

A camera (preferably a video camera if it is measuring something dynamic like a fringe shift) is an objective point of view. It doesn't compute anything. You do (or not as the case may be). What the camera records has nothing to do with any particular reference frame. It also has no direct correspondence to what is actually going on in frame S'' in which it is stationary. Don't confuse a frame with a point of view. Two relatively stationary observers looking at the same thing will measure completely different things such as the frequency of a moving light emitter.

The problem is to predict the fringe shift both in S and in S'.

Nope. The problem is to predict what the camera will see, which is utterly independent of any selection of frame.

In SRT "what it looks like to S (or to S' ) " means what is predicted in S ( or in S' ).

This assertion is dead wrong twice. 1) A frame doesn't define an observer, so what it 'looks like to some frame' is meaningless. 2) What it looks like to some observer has nothing to do with any selection of frame.

You keep making this same mistake over and over. I see little reason to continue to point it out.

No the camera can't be moving in both S and S'!

Why not? There are thousands of possible frames and it can only be stationary in one of those. These two are both apparently not it. Point is, choice of frame (just an abstract method to assign coordinates to events) is irrelevant. A camera is physical (not abstract) and sees what it sees regardless of how one chooses to assign abstract numbers to the events involved.

Yes, my hypothetical camera not only takes pictures of the fringes but also can compute the fringe shift by image processing.

Actually it was Eternal Student who first said " What it looks like to S' "  , so your comment should be directed  to him.

My claim is crystal clear : rigorous application of SRT leads to a disagreement between the fringe shifts predicted in S and in S' .

[Halc]

The point I was making is that the OP is so sloppy about frames that it is hard to decipher what he is trying to say.

I see what you mean. The goal posts seem to have moved again.
Now the 'camera' is making up its own fiction instead of recording what it actually sees/measures.

[Eternal Student]

Hi.

Actually it was Eternal Student who first said " What it looks like to S' "  , so your comment should be directed  to him.

    I can't actually find where I said that.

It's difficult to keep all the right words in, all of the time and what I actually said was this:

The camera cannot take a picture of what it looks like to the observer in the lab frame, S, the camera is not at rest in frame S and so it doesn't see that.

   Which is better but there's still a minor omission in that.    The observer wasn't just IN the lab frame, S     they were stationary in the lab frame, S.      I think it's quite common to say   "an observer in frame S"  and everyone assumes or understands that you mean an observer at rest in that frame and naturally using the co-ordinates associated with that frame of reference.    There are just too many words and some shortening has to happen sometimes.

Best Wishes.