Is Faster-Than-Light Travel Possible Relative to Other Celestial Bodies?

[matheinste]

Saying the spaceship is in motion is meaningless.

Matheinste.

 

[eekf]

Isn't this reasonong back to front. Clocks in a common inertial reference frame are synchronized to make them show the same time,not to make light speed isotropic. The standard synchronization process depends on the average of the two way directional speed of light being constant which it assumed to be. If clocks are not first synchronized how can you measure speed.
Matheinste

Actually we are trying to measure elapsed time between events at different points in the same reference frame, not speed. To do that we have to synchronise our clocks in the reference frame at the speed it is moving.

Please look at the following quoted from Einstein 1905:
"We have so far defined only an ``A time'' and a ``B time.'' We have not defined a common ``time'' for A and B, for the latter cannot be defined at all unless we establish by definition that the ``time'' required by light to travel from A to B equals the ``time'' it requires to travel from B to A. Let a ray of light start at the ``A time'' from A towards B, let it at the ``B time'' be reflected at B in the direction of A, and arrive again at A at the ``A time'' ."

From your last posting I understand that you actually agree that you have to resynchronise the clocks at v=0.9c before you can measure or calculate anything sensical in SRT, especially if you are comparing the times of different events at different positions.

Do I understand you correctly?

 

[matheinste]

------From your last posting I understand that you actually agree that you have to resynchronise the clocks at v=0.9c before you can measure or calculate anything sensical in SRT, especially if you are comparing the times of different events at different positions.------

If the clocks,during acceleration remained at rest relative to each other then they remain in synch. But I believe whether they remain at rest relative to each other depends upon the acceleration being of a certain type.

Yes clocks do need to be synchronized, or resynchronized IF required, to make time measurements. To calculate speed you need to know both time and distance.

All events take place in every reference frame. We usually measure the time between two events from the perspective of a single inertial reference frame at our convenience.

Matheinste.

 

[vin300]

Eh, now visualise. Light does not, like a person in the spaceship move at the velocity of the ship.
The standard velocity addition formula is
vector(R) = vector(u)+vector(v) [Sorry I don't know using tex]
where R is the resultant velocity, u is the velocity of the object in the f of the ship and v is the velocity of the ship w.r.t. a stationary frame

Relativistic vel addition is
vector(R)=[vector(u)+vector(v)]/1+ dot(uv)/c^2

If you use this formula, you get the velocity of light as c in all frames

Whatever be the velocity of the ship, velocity of light in the ship remains c.

In the frame of reference of the observer on the earth, the length of the ship contracts, he sees the length of the ship to be L/(gamma)
The time for the light from the back to reach the front is
t =L/c+v(gamma)
The time to reach the back is L/c-v(gamma)

I had not used the idea that the length of the ship is contracted for the stationary observer that's a discrepancy in the calculation

 

[Ich]

I would like to know what you believe the whole purpose of clock (re-)synchronisation in each reference frame is, if it is not to get c isotropic?

Hmm... it's about synchronizing clocks?
Anyway, that doesn't matter. Either you synchronize clocks to make c isotropic, or you synchronize clocks assuming that c is isotropic.
If you agree that, following SR, the clocks are out of synch (matheinste is wrong on that point), why do you state "I believe most modern interpretations are that the clocks will stay synchronised because they are in the same reference frame."?
How could this be a question of interpretation, as long as we agree on the terminology (e.g. what "synchronization" means)?

Please tell me what ATM theory is about.

You will tell us. Sooner or later.

 

[matheinste]

If you use this formula, you get the velocity of light as c in all frames

Whatever be the velocity of the ship, velocity of light in the ship remains c.

In the frame of reference of the observer on the earth, the length of the ship contracts, he sees the length of the ship to be L/(gamma)
The time for the light from the back to reach the front is
t =L/c+v(gamma)
The time to reach the back is L/c-v(gamma)

I apologise. I was in the ship's frame.

Matheinste.

 

[matheinste]

Hello Ich,

Rergarding accelerated clocks,perhaps this answer I received a long time ago in another thread confused me. I cannot remember who it was from but it was certainly one of the regular, knowledgeable responders. In answer to the question "will the previously synchronized clocks be in synch,as viewed by acomoving observer, after the acceleration the reply was-----
--------Assuming that the world lines of the two clocks and the world line of this observer guy all look the same in the original rest frame (except for their starting position), the answer is definitely yes. This is an immediate consequence of the definition of proper time and the postulate that clocks measure proper time.

You may have read about the scenario where the clocks are attached to opposite ends of a rigid rod. (It's been discussed in this forum a few times). The rod will be getting shorter (Lorentz contracted) when its speed increases (in the original rest frame). So the rear must be accelerating faster than the front, and later it must be decelerating faster than the front. In this case, the world lines will not be identical.

They might still have the same proper time though, e.g. if the deceleration profile is the "opposite" of the acceleration profile, so that the first half of the world line of the rear is a mirror image of the second half of the world line of the front and vice versa. But they may not have the same proper time in general. --------

This is the reason I mentioned a certain kind of acceleration was needed for clocks to stay in synch. I am however not sure of my ground and would welcome any clarification.

Matheinste.

 

[Ich]

Hello matheinste,

--------Assuming that the world lines of the two clocks and the world line of this observer guy all look the same in the original rest frame (except for their starting position), the answer is definitely yes. This is an immediate consequence of the definition of proper time and the postulate that clocks measure proper time.

I think I remember this discussion.
The clocks stay in synch wrt their initial rest frame.
One can see that "immediately" (as is claimed) because, starting from two simultaneous events in the initial frame, after "running through" identical world line segments, the clocks will show the same time. And the respective events will still have an equal time coordinate in the initial frame.
That means that they are out of synch in their new common rest frame.

 

[matheinste]

Thanks Ich,

That seems to make sense. So the clocks are in synch when viewed from their initial frame and so MUST be out of synch when viewed by someone in their new inertial frame as they cannot be in synch in two frames in relative inertial motion with respect to each other.

Matheinste.

 

[A.T.]

There will be no Doppler shift between two objects moving at the same velocity whether you are using sound in a medium propagating sound or a medium (or non-medium) propagating light.

(...)

Apart from the distance increase of the path in the moving frame, you will therefore also have a Doppler shift in the 90 degree arm which will increase in relation to your velocity

You are contradicting yourself.

The MM-setup doesn't rely on clock synchronization or exact distance measurement, and would not give a null result in water with sound. I cannot follow your argument about the Doppler shift in the 90°-arm, that miraculously cancels out the expected phase shift. If I move trough a sound medium parallel to a wall and make a loud sound, I receive the echo without any Doppler shift. It doesn't even matter if the wall also moves with me or rests in the medium. This sound travels exactly the same way, it would in the 90°-arm of a sound-MM-setup.

I have not checked these calculations, but have been lead to believe that you get a Null result for all velocities.

(...)

I am not trying to start a new belief or theory.

 

[eekf]

You are contradicting yourself.

The MM-setup doesn't rely on clock synchronization or exact distance measurement, and would not give a null result in water with sound. I cannot follow your argument about the Doppler shift in the 90°-arm, that miraculously cancels out the expected phase shift. If I move trough a sound medium parallel to a wall and make a loud sound, I receive the echo without any Doppler shift. It doesn't even matter if the wall also moves with me or rests in the medium. This sound travels exactly the same way, it would in the 90°-arm of a sound-MM-setup.

Let me handle the Doppler shift first. When you are moving through the water, the sound you emit into the water will have a Doppler shift depending on your velocity with respect to the water. Another object moving at the same velocity, will not detect this Doppler shift, as it will have the inverse shift when receiving. If I am wrong on this, Doppler shift has been redefined.

Secondly let me handle the 90°-arm. If I send out a sound wave at exactly 90° when moving through the water, I can not get the reflection back, as it will come back behind me.
The sound reflection I get back can only one that I have aimed slightly in front of me. As it therefore has a forward component, the wave in the water will have a Doppler shift component as explained before. This shift will be a function of v/c. If I am wrong on this, we have to re-evaluate basic physics or geometry, or we have assumed that sound travels balistically and its speed is dependent on its source.

You are right. The MM-setup doesn't rely on clock synchronization or exact distance measurement. It relies on waves interfering, with the presumption in the calculations that the wave traveling at 90° is interfering with the one at 0° with no Doppler effect in the waves. To presume the wave at 90° is interfering you have to assume that light travels ballistically. To presume no Doppler shift in either the arm or straight, you have to assume the "medium" traveling at the same speed as your system (i.e. no absolute medium).
I assume light does not travel ballistically (I think most scientists will agree). I assume there will be a Doppler effect to the vacuum (I think most scientists will disagree). We all know that the results did not tie up with what was expected from the calculations. So which of the assumptions are wrong?

 

[vin300]

Secondly let me handle the 90°-arm. If I send out a sound wave at exactly 90° when moving through the water, I can not get the reflection back, as it will come back behind me.

If you send out a sound wave at exactly 90 deg to the reflecting plane you get it back at exactly 90 deg, that is the law of reflection

 

[Al68]

May I just note that there was a time when people believed it impossible to move faster than sound.

Maybe, and maybe there still are. But scientists didn't believe it. It's been known for centuries that celestial objects had great speeds relative to the speed of sound.

I've also heard the claim that people once thought that nothing heavier than air could fly. Yeah, maybe people that had never seen a bird.

 

[vin300]

Sound does not bend as light to make up transformations.It follows the same path in every frame

First line deleted, second line is wrong, replaced with sound is a pressure wave, it's velocity varies with frames

 

[A.T.]

When you are moving through the water, the sound you emit into the water will have a Doppler shift depending on your velocity with respect to the water. Another object moving at the same velocity, will not detect this Doppler shift, as it will have the inverse shift when receiving.

This is correct and furthermore, if the other object just reflects the wave back, the sender will receive the same frequency he produced. None of them measures a changed frequency, and since the ends of the arm in the MM-setup are also at rest to each other, there will be no frequency shift as well. So I'm still puzzled how you come up with a Doppler shift in the 90° arm.

Secondly let me handle the 90°-arm. If I send out a sound wave at exactly 90° when moving through the water, I can not get the reflection back, as it will come back behind me.

Oh really? But this works fine with light. It always comes back if you shine it at a reflector facing you. So you admit, that light and sound do not behave in the same way, as you originally claimed in this thread.

The sound reflection I get back can only one that I have aimed slightly in front of me.

You don't have to (and cannot easily) aim the sound. Just send it out at a broad angle, and some of it will return reflected by the wall at the end of the arm, with the same frequency you sent out

To presume no Doppler shift in either the arm or straight, you have to assume the "medium" traveling at the same speed as your system

This is again contradicting what you correctly stated in the first quoted part. Namely that regardless their movement in the medium, two observers at rest to each other will not measure any frequency change of their signals. But an observer at rest to the medium of course will observe a different frequency.

And I think this is exactlly the root of your confusion regarding the interfering waves in the sonic-MM-setup:
- For the parallel arm you take the perspective of the moving receiver where no frequency change occurs, as you described in the first quoted part.
- For the perpendicular arm you take the perspective of the medium where a Doppler-shift occurs, because the sender is moving in the medium.
You mix up reference frames and come up with the wrong conclusion that the two interfering waves have different wavelengths. This is not the case. In the medium's frame they both have the same Doppler-shift. At the interference screen both have none (see first quote).

As a side note. Even if two interfering waves had a changing wavelength ratio, it is hard to see how that would cancel out a change in phase shift between the two, so that the interference pattern stays the same. Not that this really matters to the MM-setup, where the interfering waves have the same wavelength.

 

[vin300]

Oh really? But this works fine with light. It always comes back if you shine it at a reflector facing you. So you admit, that light and sound do not behave in the same way, as you originally claimed in this thread.



.

It works fine with sound too, the difference being lesser measured time in case of light and velocity decrease in case of sound(velocity will not decrease as much because of lesser measured time at rel speeds)

 

[eekf]

If you send out a sound wave at exactly 90 deg to the reflecting plane you get it back at exactly 90 deg, that is the law of reflection

I agree, but if I am not there anymore because I am moving with respect to the water, it will come back behind me. This does not have to do with reflection, but with the fact that I am moving with respect to the water which propagates the wave at a fixed speed (not related to my velocity).
Therefore for it to reach me (when I am moving), I have to send it at a different angle.

 

[eekf]

You don't have to (and cannot easily) aim the sound. Just send it out at a broad angle, and some of it will return reflected by the wall at the end of the arm, with the same frequency you sent out.

I think you are missing the point completely. The actual waves interfering and observed have not traveled at 90 deg with respect to each other as assumed by MM. The angle changes depending on which direction you are pointing your equipment to with respect the fixed ether (assumed by MM calculations).

This is again contradicting what you correctly stated in the first quoted part. Namely that regardless their movement in the medium, two observers at rest to each other will not measure any frequency change of their signals. But an observer at rest to the medium of course will observe a different frequency.

I am actually not contradicting myself. The interference between the waves is a function of the waves in the medium (in this case water), not of the observer. The observer is observing the interference pattern.

And I think this is exactlly the root of your confusion regarding the interfering waves in the sonic-MM-setup:
- For the parallel arm you take the perspective of the moving receiver where no frequency change occurs, as you described in the first quoted part.
- For the perpendicular arm you take the perspective of the medium where a Doppler-shift occurs, because the sender is moving in the medium.
You mix up reference frames and come up with the wrong conclusion that the two interfering waves have different wavelengths. This is not the case. In the medium's frame they both have the same Doppler-shift. At the interference screen both have none (see first quote).

I do not quite understand what you are trying to say here. MM does not measure pulses, but interference between continuous waves. In adddition the direction of measurement is constantly changed in a session (recalibrated every session). There will be Doppler shift in both arms, as well as a change in angle between the two light paths.
If you assume an ether drift (as MM did), you have to bring the angular shift between the light paths in the arms dependent on v/c as well as the Doppler shifts into calculation, or are you claiming that to calculate assuming an ether drift you should not bring this into account?

As a side note. Even if two interfering waves had a changing wavelength ratio, it is hard to see how that would cancel out a change in phase shift between the two, so that the interference pattern stays the same. Not that this really matters to the MM-setup, where the interfering waves have the same wavelength.

Sic. Are you claiming that assuming an ether drift (as MM did) you should not bring Doppler shift into account?

 

[A.T.]

If you send out a sound wave at exactly 90 deg to the reflecting plane you get it back at exactly 90 deg, that is the law of reflection

I agree, but if I am not there anymore because I am moving with respect to the water, it will come back behind me.

This is true if you use a sonic-laser (focused sound beam that doesn't disperse at reflection). And this behavior is very different from a laser, which always comes back to you, regardless how you and the mirror (at rest to each other) are moving relative to anything. Do you now understand the difference between light and sound?

 

[A.T.]

Just send it out at a broad angle, and some of it will return reflected by the wall at the end of the arm, with the same frequency you sent out.

The actual waves interfering and observed have not traveled at 90 deg with respect to each other as assumed by MM.

In the reference frame of the apparatus they have traveled at 90 deg with respect to each other. Your apparent contradictions arise from constantly mixing up reference frames.

If you assume an ether drift (as MM did), you have to bring the angular shift between the light paths in the arms dependent on v/c as well as the Doppler shifts into calculation, or are you claiming that to calculate assuming an ether drift you should not bring this into account?

I am claiming that the interference screen will always measure the same frequency for both signals. And you have already explained very well why: The source, the receiver and all reflecting elements are at rest to each other.

The interference between the waves is a function of the waves in the medium (in this case water), not of the observer.

But the Doppler-effect depends on the observer. And you claim that the Doppler-effect affects the interference.

As a side note. Even if two interfering waves had a changing wavelength ratio, it is hard to see how that would cancel out a change in phase shift between the two, so that the interference pattern stays the same. Not that this really matters to the MM-setup, where the interfering waves have the same wavelength.

Are you claiming that assuming an ether drift (as MM did) you should not bring Doppler shift into account?

In the quoted text I'm just curious how you could cancel out a change in the interference pattern due to a phase shift, by adjusting the frequencies of the two waves. Because that is your explanation for the null-result of MM. But to answer you question: Yes I think there is no frequency change between the screen and source, for neither of the signals.

 

[eekf]

In the reference frame of the apparatus they have traveled at 90 deg with respect to each other. Your apparent contradictions arise from constantly mixing up reference frames.

Sorry if this confuses you. I am trying to use sound to evaluate the calculations made by MM (provided the assumptions they claim to have made). I am trying my best to keep reference frames apart. As I said before: I am not trying anything new. I am not trying to start a new theory.

I am claiming that the interference screen will always measure the same frequency for both signals. And you have already explained very well why: The source, the receiver and all reflecting elements are at rest to each other.

etc..

I agree they will measure the same frequency, as they get combined to the same direction again (using the mirrors). However, in the MM experiment, you are effectively using the wavelengths of the light to measure the difference in path lengths the light is travelling. As such the wavelength in the "absolute" reference frame in the various arms are crucial.

 

[eekf]

I am claiming that the interference screen will always measure the same frequency for both signals. And you have already explained very well why: The source, the receiver and all reflecting elements are at rest to each other.

Even if the reflecting elements are stationary to the source and receiver (which I did not imply), unless sound is traveling ballistically, it will not hit the reflecting element (using the MM assumptions). Unless the sound wave is directed to have the same v component (wrt medium) as the source reflector and receiver (i.e. not 90 degrees), it will not hit the reflector, unless the reflector is behind me.

 

[A.T.]

In the reference frame of the apparatus they have traveled at 90 deg with respect to each other. Your apparent contradictions arise from constantly mixing up reference frames.

Sorry if this confuses you.

You are rather confusing yourself here.

However, in the MM experiment, you are effectively using the wavelengths of the light to measure the difference in path lengths the light is travelling.

The actual path lengths are not the point here. The arms don't have exactly the same length anyway. What you measure by the phase shift is the time delay of arrival between the signals. If that changes, the waves would propagate at different speed depending on their direction in the frame of the instrument. Light waves never do this. Sound waves do, if the instrument moves in the medium.

As such the wavelengths in the "absolute" reference frame in the various arms are crucial.

No they are not. The wavelengths in the arms don't affect the time the signal needs to travel both ways in an arm. And only this time shift determines the phase shift in the interference wave, after reunion.

Even if the reflecting elements are stationary to the source and receiver (which I did not imply), unless sound is traveling ballistically, it will not hit the reflecting element (using the MM assumptions). Unless the sound wave is directed to have the same v component (wrt medium) as the source reflector and receiver (i.e. not 90 degrees), it will not hit the reflector, unless the reflector is behind me.

Well I have already addressed this:

This is true if you use a sonic-laser (focused sound beam). And this behavior is very different from a light-laser, which always hits the mirror and comes back to you, regardless how you and the mirror (at rest to each other) are moving relative to anything. Do you now understand the difference between light and sound?

 

[eekf]

The actual path lengths are not the point here. The arms don't have exactly the same length anyway. What you measure by the phase shift is the time delay of arrival between the signals.
etc..

As far as I understand the experiment tried to measure the changes in the path length differences as the table is rotated using a constant light signal (no instantaneous pulses). Nobody ever claimed the arms have to be the same length.

However their differences give an interference pattern. As the table is rotated, it was postulated that the interference pattern has to change given an ether velocity relative to the system as the light path lengths (in the absolute frame) change, due to the movement of the system through it. We all know they did not change.

The calculations did not take into consideration the angle between the light paths or the Doppler effect.

If the difference in number of waves in the different arms (in an absolute reference frame) stays the same regardless of the velocity v for any mathematical model, it would predict a null result, i.e. no change in interference.

If I am mistaken in the original experiment and its setup, please point me to alternative original documents.

If I am mistaken in my understanding that no fringe shifts in interference pattern indicates the number of waves in the arms (and the phases between the two waves when "recombined") stay constant, please correct me.

As far as I understand, if they do not stay the same (in all reference frames, regardless of mathematical model you use) you will have fringe shifts, which does NOT correspond with the actual measured result. A valid model therefore per definition MUST predict the difference between the number of waves in the arms to be constant for all v. If I am wrong on this, please point me to some documentation on interference patterns, because I obviously do not understand it.

 

[vin300]

I agree, but if I am not there anymore because I am moving with respect to the water, it will come back behind me. This does not have to do with reflection, but with the fact that I am moving with respect to the water which propagates the wave at a fixed speed (not related to my velocity).
Therefore for it to reach me (when I am moving), I have to send it at a different angle.

http://en.wikipedia.org/wiki/File:Time-dilation-002.svg
In the image, light is sent out at 90 deg, but from the perspective of an observer through the normal to the plane, the wave goes diagonal, because the velocity of the moving observer adds vectoriallyto perpendicular velocity as seen by the traveling observer.
The latter still sees every point on the diagonal lines move perpendicular to his motion.
The same happens with sound, that is why it doesn't come behind you.
There is no distinction between reflection of light and that of sound, objecting to A.T.'s cloudy point

 

[vin300]

Wy isn't the image visible?
http://en.wikipedia.org/wiki/File:Time-dilation-002.svg"

 

[vin300]

Anyway, there are practical problems with doing MM with sound .A sound splitter which allows vibration in two directions only is to be used which becomes a thoght. Doing it underwater while in motion is something this forum doesn't allow to write(a foul word)
If it is somehow possible, then you hear beats(or not, depending on the difference in frequencies, it gives beats or diff tones)

 

[A.T.]

There is no distinction between reflection of light and that of sound, objecting to A.T.'s cloudy point

It is not about the difference in reflection of light and sound. It is about how to aim a theoretical sonic-laser to hit a reflector at rest in your frame and at 90° to your movement trough the sound's medium. Unlike with light you cannot just aim it at the reflector, because the beam will stay behind with the medium and miss the reflector. This was correctly stated by eekf but he still fails to see the difference between light and sound here.

 

[GoodPR]

Eekf, to say that anything your letting off at 90 would not come back to you does not apply to light as it does in water. In water you can travel (theoritcally) faster than the sound through the water, however you can not move faster than light, so therefore it could reflect back to you from 90, eventually (if your talking about a stationary reflection point)

As for this huge argument your having over the time it will take light to get from between the two mirrors. Relative velocity will not change any measurements you take inside the ship.
If it changed it would mean that when you looked to the back of your ship it would seem long. Since the whole ship is traveling the same speed, there would be no apparent effect on either the time it takes the light to travel, or doppler effect, similar to the boats traveling the same speed. Instead being inside the ship you see the rest of the universe as being stretched and doppler shifted.

Although, in regards to how Eekf originally talked about light seeming to travel + or - your velocity +c I understand what he is getting at, despite that we always know we observe light to be c regardless of velocity.


Also note in your example of the two boats, that in order for them to not notice a doppler shift between each other, they have to be traveling with the same velocity and traveling parallel to each other.

 

[eekf]

http://en.wikipedia.org/wiki/File:Time-dilation-002.svg
In the image, light is sent out at 90 deg, but from the perspective of an observer through the normal to the plane, the wave goes diagonal, because the velocity of the moving observer adds vectoriallyto perpendicular velocity as seen by the traveling observer.
The latter still sees every point on the diagonal lines move perpendicular to his motion.
The same happens with sound, that is why it doesn't come behind you.
There is no distinction between reflection of light and that of sound, objecting to A.T.'s cloudy point

My apologies vin300. I read up on it again last night, and you are right. I think Bradley showed this in the 18th century allready. The direction of the wavefront (in the water) will be at an angle related to v/c because of multiple consecutive waves forming a wavefront. I therefore do not have to aim in front of me.

I think the wave will still have a Doppler shift though.

 

[eekf]

Also note in your example of the two boats, that in order for them to not notice a doppler shift between each other, they have to be traveling with the same velocity and traveling parallel to each other.

Same velocity. Direction of travel is inherent in velocity, as it is a vector. Speed is something different though.

 

[eekf]

As for this huge argument your having over the time it will take light to get from between the two mirrors. Relative velocity will not change any measurements you take inside the ship.

Actually I think we have allready covered that earlier. If you change your velocity, unless you re-synchronise your clocks, you will have incorrect (improper) time measurements. Not elapsed time measurements at any point, but the synchronisation of your clocks at different positions in the ship will be out. You have to accept this as it is part and parcel of SRT.

It will therefore be impossible inside the ship to tell what happened first by reading the clocks, unless you re-synchronise.

Typically people will not pitch up for meetings at the right time :}

 

[eekf]

Anyway, there are practical problems with doing MM with sound .A sound splitter which allows vibration in two directions only is to be used which becomes a thoght. Doing it underwater while in motion is something this forum doesn't allow to write(a foul word)
If it is somehow possible, then you hear beats(or not, depending on the difference in frequencies, it gives beats or diff tones)

I agree it would be very difficult or impossible. I do not expect a frequency shift. If a non-null result can be found (which please I have only now actually started doing the calculations taking Doppler into account) you will only get a beat whilst you rotate the table.

I have actually stated that I want use calculations for a MM sound experiment to evaluate the MM predictions as for the effects taken into account in their predictions provided the assumptions they made.

Currently I have found that just looking at the speed of light and the movement of the platform through a stationary environment is not enough, as the experiment actually measures the light path lengths according to wavelengths. You therefore have to take Doppler effects into account.

It may turn out that taking Doppler into account actually predicts a bigger result.

 

[A.T.]

as the experiment actually measures the light path lengths according to wavelengths.

No it doesn't. How would you do that? Multiply the number of waves with wavelength? You have no idea how many waves there are in the arms, because you don't know the exact arm lengths anyway. And the number of waves doesn't matter. All that matters is if the signal-run-time difference between the arms (and therefore the phase shift) changes, when you change the orientation of the arms.

You therefore have to take Doppler effects into account.

No

 

[GoodPR]

Same velocity. Direction of travel is inherent in velocity, as it is a vector. Speed is something different though.

I disagree, If you are traveling towards a boat that is also traveling towards you, light would blue shift, your boats would hit the waves faster and thus increasing their frequency in much the same way. If you were traveling directly away from each other waves would take longer to travel to you and would have a reduced frequency. Any relative direction between those and it will be some mix of the two, with 90 being unchanged, you have to travel parallel, but not necessarily beside each other.

Actually I think we have allready covered that earlier. If you change your velocity, unless you re-synchronise your clocks, you will have incorrect (improper) time measurements. Not elapsed time measurements at any point, but the synchronisation of your clocks at different positions in the ship will be out. You have to accept this as it is part and parcel of SRT.

There must be something I'm missing here, why would your clocks ever become un synchronized? Did your entire ship not accelerate at the same speed?

 

[vin300]

To do theMM with sound, use two sources of same freq in the two dir with tubes