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

[GoodPR]

Did you read what you quoted and stop? Because I immediately said the same thing as you did after the quote.

 

[eekf]

I am going to comment on quite a few submissions made before.

1. 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. There will be a Doppler shift on the transmit side, and the inverse at the receive side.

2. You would be able to set up an experiment such as the Michelson-Morley experiment for both environments. When you do this (remembering that we actually know that the speed of sound can be exceeded) you can now check to see if you can measure a speed relative to an "absolute rest" environment, which in the case of sound would be the medium.

However, what becomes suddenly clear if you use the sound environment, is that the sound going 90 degrees to your velocity, will not reach you when reflected. In order for the sound you send sideways to reach you, you have to send it slightly ahead of you, depending on your speed. If you travel close to the speed of sound in the medium, it will only reach you if you send it out close to straight ahead.

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 (wrt the medium), corresponding (either totally or to a large degree) with the shift in phase you are expecting because of the longer path length the light has to travel in an "absolute frame" when your velocity relative to this frame increases.

Interference will be in the "absolute rest" frame, before you measure your intervals. You may also have to consider the phase shifts and possibly Doppler effects at the "mirrors".

I have not checked these calculations, but have been lead to believe that you get a Null result for all velocities. Please do not quote me on this. I have the document where this is described by an 80 year old scientist, but have not worked through it. Do the calculations.

3. Assuming the constancy of the speed of light through the vacuum (or sound in a sound medium) is a fact (and we have very little if anything to doubt this), the question becomes if the speed of light in each reference frame is aconstant in both directions. The Lorentz transformation assumes this to be true and then just works with the speed of light c.

Now it is important to consider this: I can either assume the speed of light in both directions is c constant in any moving frame, and use the average of the time to an object and back to determine the distance, or I can assume that they are different, but that by assuming it constant I can have a model which describes mathematics for any frame without knowledge of my velocity relative to my "absolute rest" frame, bearing in mind that I might get some discrepancies.

The part I struggle with most when assuming that the actual relative speed of light is the same in both directions, is the Sagnac effect. The Sagnac effect is used in (amongst others) laser Gyro's and is also needed in keeping the GPS system up to date. In gyro's it is used to send light in two directions to come back to the starting point, where the difference in travel time between the two pulses are measured, when the gyro is rotating.

I have read that SRT explains this effect just as well as other models because the "extra" length causes the time difference. If I evaluate a small straight section of the gyro, I can however not find how to explain this if the assumption is that the constant c is the same in both directions. The velocity of the environment is constant. There should therefore not be any path length difference between the directions the light is traveling in according to SRT.

4. I want you to understand that I am not claiming that it is possible to move faster than light. If atoms are actually held together through EM fields (as we believe currently), and there is a "preferred absolute reference frame" (which most mainstream modern scientists do not believe), the atoms may actually fly apart when approaching or exceeding light speed.

I am saying that the Lorentz transformation prevents traveling faster than the speed of light, but that the mathematical models we use to represent the world has restrictions and assumptions (believes) built into them. We have to try and understand those restrictions.

We humans have a tendency to restrict ourselves with our own knowledge or beliefs. When we "know" it is impossible to go faster than light or for an object to be heated by a colder object or for energy to be created, we tend to overlook opportunities.

The more knowledge we accumulate, the more we tend to restrict ourselves by what we know. Most of the times we do not even realize that we have made a subtle change to what we assume (believe), and that that change is restricting us.

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

 

[jreelawg]

My take is that relativity only makes sense mathematically not intuitively.

 

[Razzor7]

My take is that relativity only makes sense mathematically not intuitively.

It makes as much intuitive sense as any theory. It's just wha' happens.

 

[GoodPR]

EEkf
I don't even see what your trying to say, everything you said appeared to be in agreeance with all modern theories, the only thing you seemed to make a statement on was that the more knowledge we seem to gain, the more we close our minds to other knowledge. On that I'd have to agree, I think any GUT out there is going to be a simple equation, at face value.

 

[eekf]

My take is that relativity only makes sense mathematically not intuitively.

The only way I could understand it was to plot a spacecraft traveling at 0.8c and emmitting a "spherical" light pulse every second. After 4 seconds (as seen from a stationary frame), I drew the position of the light pulses with compasses. I ended up with a drawing looking exactly like the drawings in high school physics books representing the sound waves of an aircraft approaching the speed of sound.

The problem comes in if you consider the positions of the light pulses from the spacecraft 's point of view. If you use the same clock synchronisation (not rate) in the spacecraft 's reference frame, you end up with the first light pulse being much further behind the spacecraft than in front of it. In addition, the light pulse behind would seem to have traveled much faster than the speed of light (actually 1.8c).

If you use this mathematical model, you end up with calculations which have the speed of light dependent on the direction you are traveling in and not isotropic. By adjusting your clocks, you can however rescale to have the light in both directions move at c. You end up that you have to change the scaling of your x-axis as well as your clocks if you want the transformation to be consistent through all reference frames.

If you consider the same setup for sound in air, you realize that the ony reason you can (or have to) do this is because the speed of sound in the medium is a constant, independent of your relative movement. If the air was traveling with you (i.e. in a closed van), you can use the same transformations, but the speed of sound (as viewed from outside the van) would not be isotropic.

 

[eekf]

EEkf
I don't even see what your trying to say, everything you said appeared to be in agreeance with all modern theories, the only thing you seemed to make a statement on was that the more knowledge we seem to gain, the more we close our minds to other knowledge. On that I'd have to agree, I think any GUT out there is going to be a simple equation, at face value.

Actually what I am saying is not in complete agreeance with modern theories, or should I say better: the interpretations given to modern theories.

Consider a spacecraft 1 lightsecond long with two independent atomic clocks (one in the front and one in the back) which is synchronised when it is in "a stationary Earth" reference frame. It then accelerates to a speed of 0.9c. Once stabilised at this speed, a light pulse is sent from the front to the rear at t=0 and reflected back to the front from the rear.

I am saying that the time difference measured between transmitting from the front till reflection at the rear will be much lower than the time difference measured from the reflection to the reception at the front.

As such I am saying that clocks have to be re-synchronised after reaching 0.9c for c to be measured constant in both directions.

I believe most modern interpretations are that the clocks will stay synchronised because they are in the same reference frame.

 

[matheinste]

eekf,

But your interpretation would allow you to detect absolute motion.

Matheinste

 

[eekf]

eekf,

But your interpretation would allow you to detect absolute motion.

Matheinste

Possibly. I believe experiments like this have been done. See http://xxx.lanl.gov/abs/astro-ph/0608223.

 

[eekf]

eekf,

But your interpretation would allow you to detect absolute motion.

Matheinste

Another recent paper: http://blog.hasslberger.com/docs/Cahill_Experiment.pdf.

The conclusions of this paper (amongst others):
1. Speed of light is anisotropic (eight different experiments).
2. Fitzgerald-Lorentz contraction is real effect in interferometers.

It might be that if the Doppler shift because of the fact that the 90 degree light in the arm in interferometer equipment can not come back to interfere (interference can only be with light that is not at 90 degrees and angle is dependent on v) might even indicate no contraction needed.

 

[Ich]

I believe most modern interpretations are that the clocks will stay synchronised because they are in the same reference frame.

I believe that
1. Interpretations must not disagree on facts, therefore, if your "interpretation" does not agree with mainstream results, it is a disguised ATM theory.
2. If ATM proponents would spend only a tenth of their effort disproving the mainstream on actually learning what it says, there'd be no more ATM proponents.

Please calculate your example in SR, and present the result here. If you can't, you're surely not in a position to challenge anything. If you can, your "belief" is moot.

 

[vin300]

Here's the values:If 0.9c is its velocity, the timr taken from front to back in the frame of reference of the ship is 1/1.9s which is 0.52s and the time taken for the reverse is
1/0.1s,i.e. 10s

 

[eekf]

I believe that
1. Interpretations must not disagree on facts, therefore, if your "interpretation" does not agree with mainstream results, it is a disguised ATM theory.
2. If ATM proponents would spend only a tenth of their effort disproving the mainstream on actually learning what it says, there'd be no more ATM proponents.

Please calculate your example in SR, and present the result here. If you can't, you're surely not in a position to challenge anything. If you can, your "belief" is moot.

I agree with vin300's calculation.

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?

Please tell me what ATM theory is about. I do not know it. I am not trying to disguise any theory. I am trying to make sense of what I see - especially in the mathematics.

 

[matheinste]

I agree with vin300's calculation.

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?

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.

vin300's calculation assumes that the light inside the spaceship travels different distances. It does not. Light emitted from the centre of the ship travels the same distance from emitter to front and back to emitter as it does from emitter to rear and back to emitter.

Matheinste

 

[vin300]

vin300's calculation assumes that the light inside the spaceship travels different distances. It does not. Light emitted from the centre of the ship travels the same distance from emitter to front and back to emitter as it does from emitter to rear and back to emitter.

Matheinste

Er a little misconception Light inside the spaceship does go different distances in the f of stationary observer because the spaceship itself is in motion.If you'd like to know how derive it's simple t(back to front) is [L/c-v ]t(f to b) is [L/c+v]

 

[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.