Speed of the light and dilation of time

[DrGreg]

...the Doppler effect is not frame invariant...

To show that the Doppler effect is frame invariant you don't have to do any maths at all -- just look at the definition $\nu_r / \nu_s$.

$\nu_s$ is measured by the source using the source's own proper-time clock.

$\nu_r$ is measured by the receiver using the receiver's own proper-time clock.

All observers, whatever coordinates they are using, agree on what the source and receiver's proper times are (associated with a $2\pi$ phase change), therefore agree on the two frequencies, therefore agree on the Doppler factor.

 

[GAsahi]

The doppler effect is completely determined by the relative velocity of source and target. Both of these velocities are frame dependent, but the relative velocity between emitter at event of emission and receiver at event of detection is frame invariant. Thus all observers agree on the doppler measured by a given detector from a given source.

Even if the relative velocity between source and receiver were frame invariant, the Doppler effect is not frame invariant by virtue of the fact that:

$$\frac{\nu_o}{\nu_r}=\frac{1-\frac{v}{c} cos (\theta)}{\sqrt{1-(v/c)^2}}$$

$\theta$, the angle between the relative velocity and the wave vector is known not to be frame invariant (see the aberration formula).

Mathematically, relative speed is defined by parallel transport of 4-velocity from one event to another (in SR, this is path independent, thus unique), then dot product of transported source 4-velocity with unit 4-vector 4-orthogonal to target 4-velocity. Dot products are invariant - period. (In a standard inertial frame in SR, parallel transport leaves a vector unchanged).

Dot product of 4-vectors is indeed frame-invariant. Unfortunately, the vectors involved in the Doppler effect are THREE - vectors, not 4-vectors:

$$\frac{\nu_o}{\nu_r}=\frac{1-\frac{\vec{v}\vec{e}}{c}}{\sqrt{1-(v/c)^2}}$$

whre $\vec{e}$ is the wave vector. The dot product of two 3-vectors is not frame invariant.

 

[jartsa]

Let us consider observer A circling around observer B, and one instantaneous moment of this situation. It looks like A is just passing B at high speed, so we can guess A and B see a mutual time dilation.

 

[PAllen]

Even if the relative velocity between source and receiver were frame invariant, the Doppler effect is not frame invariant by virtue of the fact that:

$$\frac{\nu_o}{\nu_r}=\frac{1-\frac{v}{c} cos (\theta)}{\sqrt{1-(v/c)^2}}$$

$\theta$, the angle between the relative velocity and the wave vector is known not to be frame invariant (see the aberration formula).

This is wrong too. The angle here is the angle as measured in the comoving inertial frame of the detector. This is frame invariant.

Let me ask this: Do you really think the measurement on a given instrument depends on who looks at the instrument?

E
Dot product of 4-vectors is indeed frame-invariant. Unfortunately, the vectors involved in the Doppler effect are THREE - vectors, not 4-vectors:

$$\frac{\nu_o}{\nu_r}=\frac{1-\frac{\vec{v}\vec{e}}{c}}{\sqrt{1-(v/c)^2}}$$

whre $\vec{e}$ is the wave vector. The dot product of two 3-vectors is not frame invariant.

This is incorrect. The doppler is determined by the source 4-velocity (at emission event) represented in the basis of the target frame at event of detection. This is determined by dot products of the source 4-velocity with the target basis unit vectors, as I previously explained. This result is manifestly frame invariant.

 

[DeepSpace9]

I totally understand how moving at the speed of light can make you travel into the future, but how does it make you go back in time?

 

[GAsahi]

This is wrong too. The angle here is the angle as measured in the comoving inertial frame of the detector. This is frame invariant.

...measured from an arbitrary frame in motion wrt the frame of the detector this angle changes (via the aberration formula).

Let me ask this: Do you really think the measurement on a given instrument depends on who looks at the instrument?

Wrongly formulated question, not "who looks at the instrument" but "from what frame is the measurement made". This is the essence of judging frame invariance.

This is incorrect.

Look up p.62, formula (90), "The theory of relativity", C. Moller.
Look up p.114, formula (311a) , Theory of relativity", W. Pauli,
Look up...

 

[PAllen]

...measured from an arbitrary frame in motion wrt the frame of the detector this angle changes (via the aberration formula).

The only measurement that matters for Doppler is the angle measured from the detector's frame.

.

Wrongly formulated question, not "who looks at the instrument" but "from what frame is the measurement made". This is the essence of judging frame invariance.

The measurement is made by a detector. Any measurement by any detector is frame invariant.

.Look up p.62, formula (90), "The theory of relativity", C. Moller.
Look up p.114, formula (311a) , Theory of relativity", W. Pauli,
Look up...

I read Pauli 40 years ago. I think I understand what he wrote perfectly well.

 

[GAsahi]

The only measurement that matters for Doppler is the angle measured from the detector's frame.

From the perspective of another frame, moving wrt. the detector, the angle changes.

I read Pauli 40 years ago. I think I understand what he wrote perfectly well.

This is not a valid answer, the formula I cited can be found in both Moller and Pauli, and in any standard textbook, perhaps it is time for you to read Pauli again.

 

[PAllen]

From the perspective of another frame, moving wrt. the detector, the angle changes.

That is not relevant. You misinterpret the physics of the Doppler formula. The angle in the formula is the angle as measured from the detector frame.

This is not a valid answer, the formula I cited can be found in both Moller and Pauli, and in any standard textbook, perhaps it is time for you to read Pauli again.

The formula for Doppler is not the issue. The issue is your misinterpretation of the variables in the formula. I certainly don't need to re-read Pauli. I have a copy and enjoy it regularly. The issue, again, is physical understanding of the formulas in it.

 

[GAsahi]

That is not relevant. You misinterpret the physics of the Doppler formula. The angle in the formula is the angle as measured from the detector frame.

No one denies that. How does the formula transform from the perspective of a moving frame , this is the issue being discussed.

This is not a valid answer, the formula I cited can be found in both Moller and Pauli, and in any standard textbook, perhaps it is time for you to read Pauli again.

The formula for Doppler is not the issue. The issue is your misinterpretation of the variables in the formula. I certainly don't need to re-read Pauli. I have a copy and enjoy it regularly. The issue, again, is physical understanding of the formulas in it.

The "variable" in the formula is , contrary to your claim, a 3-vector, not a 4-vector. While the dot product of two 4-vectors is invariant, the dot product of 3-vectors is not. I suggest that you re-read Pauli. While you are at it, read (or get) Moller. His book is better than Pauli's.

 

[PAllen]

No one denies that. How does the formula transform from the perspective of a moving frame , this is the issue being discussed.






The "variable" in the formula is , contrary to your claim, a 3-vector, not a 4-vector. I suggest that you re-read Pauli. While you are at it, read (or get) Moller. His book is better than Pauli's.

The 3-vector in the formula is constructed by taking a 4-vector's dot products with the basis vectors of the detector frame. These dot products are frame invariant.

Let's focus on physics. If you say two detectors in different states of motion measure different Doppler, of course, this is true. It is also true that a given detector's measurement of Doppler is invariant. This is about all there is to the physics of the situation. Do you dispute any of this? If so, you are simply wrong.

 

[GAsahi]

The 3-vector in the formula is constructed by taking a 4-vector's dot products with the basis vectors of the detector frame. These dot products are frame invariant.

There is no 4-vectors in any of the two derivations I cited. You can check your copy of the Pauli book. Even better, get the Moller book, it is a better book.

Let's focus on physics. If you say two detectors in different states of motion measure different Doppler, of course, this is true.

So, are we done here?

 

[PAllen]

There is no 4-vectors in any of the two derivations I cited. You can check your copy of the Pauli book. Even better, get the Moller book, it is a better book.
So, are we done here?

I don't know. Gwellsjr was simply stating that the relative velocity and doppler between a given source and detector was frame invariant, and calculated a specific example showing this. Everything about his calculation and interpretation is correct. You disputed this. So I don't know if we are done. Gwellsjr said nothing about two different detectors in different states of motion measuring the same thing.

[as for 3-vectors versus 4-vectors: an event at a world line has an associated tangent 4-vector. If you express this in a particular basis, you get a 3-vector (3-velocity). The way to get this expression is to form dot products of the 4-velocity with a frame's basis vectors. The basis vector's of a detector's frame are unit 4-vectors, and the dot products are invariant. I would readily state that Pauli does not use this more modern language, but (unlike you) he makes no mistakes at all about the physics of the situation under discussion. He understands that the variables in the Doppler formula relate to measurements in the detector frame, and thus are determined purely by the motion of detector and source, and not by the motion of some other observer.]

 

[ghwellsjr]

Post #24:

Yes, on average, your son would see you in slow motion and you would see all the people in fast motion, but I wouldn't say "thousands of times actual Earth speed", just thousands of times your speed.

This is false, time dilation is symmetrical, each observer sees the other in slow motion.

Post #54:

One thing about Doppler, since it is a local effect, you can calculate it using any frame and it will be the same in all other frames. I realize that I have not calculated it in a non-inertial frame in which the equator clock is at rest but I already know the answer from the frame in which the pole clock is at rest so I don't have to do all that extra work. But if you don't believe it, then I invite you to do the calculation in whatever frame would convince you rather than to just continually claim that the equator clock won't see the pole clock running faster than itself.

And since I have complied with your request, can you please comply with mine:

I invite you to do the calculation in whatever frame would convince you rather than to just continually claim that the equator clock won't see the pole clock running faster than itself.

And please answer my request:

I invite you to do the calculation in whatever frame would convince you rather than to just continually claim that the equator clock won't see the pole clock running faster than itself.

Now I've done everything you've asked, will you please do what I requested:

I invite you to do the calculation in whatever frame would convince you rather than to just continually claim that the equator clock won't see the pole clock running faster than itself.

We're not done until GAsahi performs my simple request to show that Einstein's claim that the equator clock is going more slowly than the pole clock also applies to the pole clock that it is going more slowly than the equator clock and that

time dilation is symmetrical, each observer sees the other in slow motion.

.

 

[GAsahi]

[as for 3-vectors versus 4-vectors: an event at a world line has an associated tangent 4-vector. If you express this in a particular basis, you get a 3-vector (3-velocity). The way to get this expression is to form dot products of the 4-velocity with a frame's basis vectors. The basis vector's of a detector's frame are unit 4-vectors, and the dot products are invariant. I would readily state that Pauli does not use this more modern language, but (unlike you) he makes no mistakes at all about the physics of the situation under discussion.

You are trying to move the goalposts, I posted a formula for the Doppler effect in terms of 3-vectors (post 72), you claimed that it was wrong (post 74). I gave you two references that show the formula to be correct (post 76).

He understands that the variables in the Doppler formula relate to measurements in the detector frame, and thus are determined purely by the motion of detector and source, and not by the motion of some other observer.]

There is no reference to the invariance of the Doppler formula in the Pauli text. Nor is it any reference in the Moller text.

 

[PAllen]

You are trying to move the goalposts, I posted a formula for the Doppler effect in terms of 3-vectors (post 72), you claimed that it was wrong (post 74). I gave you two references that show the formula to be correct (post 76).

No, I said your interpretation of the variables in it was wrong. In a different post, I showed where you made an error computing relative velocity between source and target in different frames, and that after correcting the error, the result is that it is the same.

 

[GAsahi]

We're not done until GAsahi performs my simple request to show that Einstein's claim that the equator clock is going more slowly than the pole clock also applies to the pole clock that it is going more slowly than the equator clock and that.

It is very simple really: Have the observer traveling around the Equator unroll a fiberoptic cable connecting him with the stationary observer. From time to time, each observer sends a pulse towards the other observer (not at the same time). Each observer will measure the other's frequency redshifted by : $\sqrt{\frac{1-\omega R/c}{1+\omega R/c}}$. The Doppler effect is mutual, the roles of the source and the receiver are interchangeable. You are mixing the Doppler effect with Einstein's calculation of total elapsed time. In that experiment, the roles of the inertial and the non-inertial observer are not interchangeable.

 

[GAsahi]

No, I said your interpretation of the variables in it was wrong..

Here is exactly what you said:

This is incorrect. The doppler is determined by the source 4-velocity (at emission event) represented in the basis of the target frame at event of detection. This is determined by dot products of the source 4-velocity with the target basis unit vectors, as I previously explained. This result is manifestly frame invariant.

 

[PAllen]

Here is exactly what you said:

My explanation is referring to how the 3 vectors in the formula are actually frame invariant because they are referred to the basis determined by the 4-velocity of the detector. I may not have been clear as I should have been, but the thrust of my statement was this: the definition of the variables in the formula make the result frame invariant.

 

[PAllen]

It is very simple really: Have the observer traveling around the Equator unroll a fiberoptic cable connecting him with the stationary observer. From time to time, each observer sends a pulse towards the other observer (not at the same time). Each observer will measure the other's frequency redshifted by : $\sqrt{\frac{1-\omega R/c}{1+\omega R/c}}$. The Doppler effect is mutual, the roles of the source and the receiver are interchangeable. You are mixing the Doppler effect with Einstein's calculation of total elapsed time. In that experiment, the roles of the inertial and the non-inertial observer are not interchangeable.

The Doppler effect is mutual between two inertial world lines (each acting as source and target relative to the other). Between an inertial world line and a non-inertial world line it is not mutual. A light source in forced circular motion around an inertial target is always red shifted. A target in forced circular motion about an inertial light source will see its light always blue shifted. The simplest way to derive this latter fact is to compute, in the inertial frame, the proper time interval for the circular target between arrival of successive circular wave fronts emitted by the central source.

 

[GAsahi]

The Doppler effect is mutual between two inertial world lines (each acting as source and target relative to the other). Between an inertial world line and a non-inertial world line it is not mutual. A light source in forced circular motion around an inertial target is always red shifted.

Sure, it is a pure form of transverse Doppler effect. This is NOT what my post is about, it is about the Doppler effect between an observer stationary on the circumference and one traveling along the circumference. There is no central "target". Both source and receiver are on the circumference (one is stationary wrt the circumference, the other one isn't).

The simplest way to derive this latter fact is to compute, in the inertial frame, the proper time interval for the circular target between arrival of successive circular wave fronts emitted by the central source.

There is no "central source", please pay attention to the scenario.

 

[PAllen]

Sure, it is a pure form of transverse Doppler effect. This is NOT what my post is about, it is about the Doppler effect between an observer stationary on the circumference and one traveling along the circumference. There is no central "target".



There is no "central source", please pay attention to the scenario.

Sure there is. Under discussion (at least the discussion I was responding to) was polar clock or light source, and an equatorial clock or light source, under the assumption of removing the Earth (so as to remove gravitational effects not known to Einstein or anyone else in 1905), while keeping the motions the same and noting that the polar clock/source is the inertial one (they can't both be). In this scenario, light from a polar source will be always seen blue shifted by an equatorial detector. Light from an equatorial source will be seen always red shifted by a polar detector.

 

[GAsahi]

Sure there is. Under discussion (at least the discussion I was responding to) .

Once again, you are responding to the wrong scenario, the scenario in discussion has both observers on the circumference.

 

[PAllen]

Once again, you are responding to the wrong scenario, the scenario in discussion has both observers on the circumference.

I was responding to the discussion of the scenario introduced in post #34, not the earlier, more complex scenario. However, on that one, I agree with Gwellsjr that while both observers would see a periodic pattern of red and blue shift, the inertial observer would see, on average, more red shift, and the non-inertial obaserver would see, on average, more blue shift.

 

[ghwellsjr]

Once again, you are responding to the wrong scenario, the scenario in discussion has both observers on the circumference.

That's not what you responded to earlier:

We're not done until GAsahi performs my simple request to show that Einstein's claim that the equator clock is going more slowly than the pole clock also applies to the pole clock that it is going more slowly than the equator clock...

It is very simple really: Have the observer traveling around the Equator unroll a fiberoptic cable connecting him with the stationary observer. From time to time, each observer sends a pulse towards the other observer (not at the same time). Each observer will measure the other's frequency redshifted by : $\sqrt{\frac{1-\omega R/c}{1+\omega R/c}}$. The Doppler effect is mutual, the roles of the source and the receiver are interchangeable. You are mixing the Doppler effect with Einstein's calculation of total elapsed time. In that experiment, the roles of the inertial and the non-inertial observer are not interchangeable.

I thought you were proposing that the "observer traveling around the Equator" was actually at a fixed location on the Equator and "traveling" due to the rotation of the Earth which is what Einstein was talking about. And I thought the stationary observer was at one of the poles which is what Einstein was talking about. And I thought you proposed a fiberoptic cable connecting the two to take care of the problem due to the curvature of the earth.

Did you actually have something else in mind?

 

[GAsahi]

That's not what you responded to earlier:

I thought you were proposing that the "observer traveling around the Equator" was actually at a fixed location on the Equator and "traveling" due to the rotation of the Earth which is what Einstein was talking about.

There is no rotation, the "traveller" moves along the Equator and unrolls a fiberoptic cable as he travels. The "stay at home" observer is just stationary on a point on the Equator.

And I thought the stationary observer was at one of the poles which is what Einstein was talking about. And I thought you proposed a fiberoptic cable connecting the two to take care of the problem due to the curvature of the earth.

Did you actually have something else in mind?

Yes, see above.

 

[ghwellsjr]

There is no rotation, the "traveller" moves along the Equator and unrolls a fiberoptic cable as he travels. The "stay at home" observer is just stationary on a point on the Equator.

So the Earth is not rotating?

Does the "traveller" finally get back to the "stay at home" observer and if so, are they communicating via a fiberoptic cable that goes around the earth?

 

[GAsahi]

So the Earth is not rotating?

Does the "traveller" finally get back to the "stay at home" observer and if so, are they communicating via a fiberoptic cable that goes around the earth?

The "earth" is a non-rotating flavor.
The "traveller" gets back to the "stay at home".
The "twins" exchange signals via the fiberoptic cable unrolled by the traveller.
The "traveller" transmits the data only in the direction contrary to his direction of motion.
The "stay at home" sends data "chasing" the traveller".

 

[Austin0]

Easy, in frame F, the realtive speed between source and receiver is $v$. In frame F', moving with speed $u$ wrt F, the relative speed is $w=\frac{u \pm v}{1 \pm uv/c^2}$. So, your "proof" falls appart right from the start. You missed in your calculations that $\beta=v/c$ is frame - VARIANT. Now, that I showed your mistake, redo your post 4 correctly, please.

What do you think w represents??
Taken at face value from F' if u = velocity of F wrt F' then v would represent the velocity of the source wrt F' in the additions equation.
In which case w would represent the velocity of the source as measured in F =v

On the other hand if you are running the formula from F then w would be the velocity of the source as measured in F'
Which is it?

Please stop butting in. If you do not know that speed is a frame variant quantity (this answers your question about the meaning of w), you should not interfere in this thread. Thank you.

Yes I understand the meaning of w and the additive velocities equation. I was unsure of which frame you were applying it from and what significance you were attaching to the result.

It is also now clear that I was unsure of what you thought you were doing with the Additions equation because you were trying to apply it incorrectly. As demonstrated by ghwellsjr and his correct application.

I came by an understanding that relative velocities between two frames was invariant simply through the meaning and use of the additions formula. This of itself necessarily implies frame invariance of calculated velocities ,which coupled with the understanding that Doppler was a simple function of relative velocity necessarily implies that measured Doppler is invariant.

The velocity of anyone frame is obviously a relative quantity.
But as far as I know the internally measured relative velocity between any two frames S and S' is frame invariant.
Do you think this is incorrect?

Yes. Please stop.

Well here you are not only rude but simply factually wrong

Sure, it is a pure form of transverse Doppler effect. This is NOT what my post is about, it is about the Doppler effect between an observer stationary on the circumference and one traveling along the circumference. There is no central "target". Both source and receiver are on the circumference (one is stationary wrt the circumference, the other one isn't).
please pay attention to the scenario.

Once again, you are responding to the wrong scenario, the scenario in discussion has both observers on the circumference.

SO now we have come full circle back to my post #44

This is further compounding the confusion, the Doppler effect on frequency follows a different set of rules in accelerated frames. The revolving observer is continuously accelerating , so you cannot extrapolate from the Doppler effect in inertial frames. The only thing that you got right is the fact that, quantitatively, the observers notice a mutual blueshift when they approach each other. When they are separating from each other, they are experiencing a mutual redshift.
In both cases the effect is mutual i.e. you cannot have:.

#44

I think in this instance ghwellsjr is correct.
The mutual red and blue shift you are referencing here is regarding the relationship of received signal frequency to emitted (proper) frequency. Obviously valid but it does not necessarily imply that the received signals are going to be shifted in that manner relative to their own proper rates.
To put it in context:
If we assume a gamma of 31.4 and a circular path of 314 ls then when the moving clock reaches the opposite side, the inertial clock will have emitted 157 1 second signals.
100 of those signals will still be in transit meaning that 57 signals would have been received by the accelerated system. The accelerated system at this point would have emitted 5 signals and have an elapsed time of 5 sec.
Obviously this is an approximation based on the assumption of angular velocity close to c..

Here I provided a clear appraisal of your position and the basis of your erroneous assumption. I.e. That the reciprocal aspect of linear Doppler shift would apply in this context.
I also gave a simple demonstration of specific conditions which clearly showed the error of your assumption.
Rationally you had two options:
1) Show that my calculations were wrong or 2) recognize that your position was in error.
That you did not effect option 1) , I view as tacit admission that you could not.

Well of course it is your picture as per the conditions of the original scenario. I did not provide more explanation as I assumed the extrapolations were self evident.

GAsahi was claiming that your claim was incorrect because Doppler shift was reciprocal so I provided an example with your boundary conditions where this was clearly controverted.
The first half of the trip was the interval where the signals would be red shifted.
If in that interval the accelerating system received more signals than proper time it is clear that those signals were NOT red shifted relative to that systems proper time.

It is also clear that although the instantaneous relative velocity/gamma would vary with angle of motion wrt the inertial clock the same basic relationship would necessarily pertain. I.e. The accelerating system would receive more signals than sent. If those signals were video footage then as you say the motion of the inertial actors would appear sped up..

Apparently my assumption of logical facility and self evident extrapolation was misplaced in your case because
you also did not choose option 2) but instead proceeded to ignore this evidence and continue with a course of obfuscation and false premises. Not to mention your rude arrogance towards me without refuting or even addressing anything specific i offered.
Do you seriously think the needless complication of your fiberoptic cable could possibly alter the fundamental relationship of clock rates in this scenario? Or that the results of any alternative conditions could negate the conclusions derived from my or ghwellsjr's scenarios where you are clearly in error?

 

[PAllen]

The "earth" is a non-rotating flavor.
The "traveller" gets back to the "stay at home".
The "twins" exchange signals via the fiberoptic cable unrolled by the traveller.
The "traveller" transmits the data only in the direction contrary to his direction of motion.
The "stay at home" sends data "chasing" the traveller".

A fiber optic cable will add significant complications:

- speed of light is not c and not invariant in a fiber optic
- refraction is frame dependent in rather complex ways (see, e.g. http://www.mathpages.com/rr/s2-08/2-08.htm)

The scenario without a cable is not hard to analyze in the inertial frame. Then, you can rely on the frame invariance of measured Doppler. The clearest statement of why this is true was given by Dr. Greg way back in post #71.

 

[ghwellsjr]

The "earth" is a non-rotating flavor.
The "traveller" gets back to the "stay at home".
The "twins" exchange signals via the fiberoptic cable unrolled by the traveller.
The "traveller" transmits the data only in the direction contrary to his direction of motion.
The "stay at home" sends data "chasing" the traveller".

Why would you dream up this kind of a scenario? It has nothing to do with anything else in this thread. It has nothing to do with Einstein's scenarios. A non-rotating "earth" has no equator and no poles. Can you please forget about this scenario and focus on Einstein's scenario?

We have one observer at a fixed location on the equator with a precise clock that emits a flash of light once per second. He is non-inertial because he is rotating with the surface of the earth. We have a second observer at the South pole with an identical clock that also emits a flash of light once per second. He is inertial. There is a fiberoptic cable running between the two observers that communicates the light flashes between the two observers. Do you still claim that the two observers will detect flashes from the other observer at the same rate they send them?

 

[GAsahi]

A fiber optic cable will add significant complications:

- speed of light is not c and not invariant in a fiber optic

Nevertheless, fiberoptics gyroscopes work just fine. If you are really concened about the second order effects, you can replace the fiberoptics cable with a sequence of mirrors strategically placed. The point of the setup is to show that both observers measure the same effect, i.e. a redshift.

The scenario without a cable is not hard to analyze in the inertial frame. Then, you can rely on the frame invariance of measured Doppler. The clearest statement of why this is true was given by Dr. Greg way back in post #71.

I know how to analyze it in BOTH the inertial and the non-inertial frame. There is a very good formalism that covers all the imaginable cases (including the cases when either the receiver or the light source are in the center). Once again, for the scenario in discussion the point is that both observers measure the same effect, i.e. a redhift.

 

[GAsahi]

Why would you dream up this kind of a scenario? It has nothing to do with anything else in this thread. It has nothing to do with Einstein's scenarios. A non-rotating "earth" has no equator and no poles. Can you please forget about this scenario and focus on Einstein's scenario?

A rotating Earth makes BOTH observers non-inertial. If you know how to solve the problem, you will find out that nothing changes, the observers agree that they are measuring the incoming frequency is redshifted.

We have one observer at a fixed location on the equator with a precise clock that emits a flash of light once per second. He is non-inertial because he is rotating with the surface of the earth.

Yes.

We have a second observer at the South pole

Nope, the second observer is located on the Equator as well. How else would "the observers start TOGETHER" as you claim here? You have flip-flopped between scenarios so many times that it is difficult to keep track but this is the scenario in discussion.

 

[Austin0]

Nevertheless, fiberoptics gyroscopes work just fine. If you are really concened about the second order effects, you can replace the fiberoptics cable with a sequence of mirrors strategically placed. The point of the setup is to show that both observers measure the same effect, i.e. a redshift.






I know how to analyze it in BOTH the inertial and the non-inertial frame. There is a very good formalism that covers all the imaginable cases (including the cases when either the receiver or the light source are in the center). Once again, for the scenario in discussion the point is that both observers measure the same effect, i.e. a redhift.

In this case there would be mutual redshift but so what?
You have simply completely restructured the parameters to create a whole new scenario.
Effectively turning it into a linear case. This has no bearing whatsoever to ghwellsjr's scenario or conclusions within his given parameters. It is simply a red herring.

 

[PAllen]

Nevertheless, fiberoptics gyroscopes work just fine. If you are really concened about the second order effects, you can replace the fiberoptics cable with a sequence of mirrors strategically placed. The point of the setup is to show that both observers measure the same effect, i.e. a redshift.

You can't gloss over this. The fiber optic medium cannot have the same relative motion to both observers. If it is unrolled so as to be stationary with respect to the inertial observer, ti will be moving rapidly relative to the the other observer. For the non-inertial observer, the speed of light through the cable will then depend on direction.

Replacing with mirrors won't make the situation symmetric either. If the mirrors are stationary with respect to the inertial observer, they are moving rapidly on different trajectories relative to the non-inertial observer. [Edit: However, under reasonable assumptions, the motion of the mirrors won't matter, and there would be symmetric red shift for the light received from this specific path]

Finally, no one defines Doppler in such baroque way. Defined as normally:each observer measuring frequency of light the other emits - the emission being by the same process for each, as received through the vaccuum, then, as stated many times:

Each will observer a periodic pattern of red and blue shift.

The pattern will not be identical for both. The inertial observer will have more red shift, the non-inertial observer more blue shift.