Speed of the light and dilation of time

[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.

 

[Austin0]

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.

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.

I think he is assuming that the signals are only being sent back along the system of mirrors comparable to the fiberoptic situation. Not considering signals sent ahead during the second half of the circumnavigation.
Given this restriction he would be correct about mutual redshift throuhout iMO

 

[GAsahi]

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.

I think that you are splitting hairs, the only reason for the presence of the cable or the mirrors is for the two observers to exchange signals. The point that you keep missing is that the two observers measure the same effect, a redshift for separation motion.

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

Not if the communication is made over the same arc of circle, extending in the same direction. You get alternating patterms only if you switch the signal direction midway. The cable avoids this direction switching. Once again, the point is that BOTH observers measure the same effect (both measure redshift OR both measure blueshift), contrary to gjwelsjr's claims.

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

BOTH see EITHER redshift OR blueshift. Contrary to gjwellsjr claim that started all this, you cannot have one of them measure redshift and the other measure blueshift. Thank you for making my point, can you explain this to gjwellsjr?

 

[GAsahi]

I think he is assuming that the signals are only being sent back along the system of mirrors comparable to the fiberoptic situation. Not considering signals sent ahead during the second half of the circumnavigation.
Given this restriction he would be correct about mutual redshift throuhout iMO

Thank you for making my point. Please explain this to gjwellsjr.

 

[PAllen]

BOTH see EITHER redshift OR blueshift. Contrary to gjwellsjr claim that started all this, you cannot have one of them measure redshift and the other measure blueshift. Thank you for making my point, can you explain this to gjwellsjr?

As normally defined (measure frequency of light received over vacuum in the most direct path from the emitter to the receiver), there can be, and is asymmetry.

The simplest case is the one Gwellsjr has proposed: an observer stationary in an inertial frame, and an observer moving in a circle around said 'stationary' observer. Do you disagree that:

- the stationary observer will always see redshift for light emitted by the circular moving observer

- the circular moving observer will always see blue shift for light emitted by the stationary observer.

 

[GAsahi]

As normally defined (measure frequency of light received over vacuum in the most direct path from the emitter to the receiver), there can be, and is asymmetry.

The simplest case is the one Gwellsjr has proposed: an observer stationary in an inertial frame, and an observer moving in a circle around said 'stationary' observer. Do you disagree that:

- the stationary observer will always see redshift for light emitted by the circular moving observer

- the circular moving observer will always see blue shift for light emitted by the stationary observer.

Neither of the above is the claim that gjwellsjr made that got all this started. Read the claim he made, please.

 

[Austin0]

Thank you for making my point. Please explain this to gjwellsjr.

I guess you missed this one

austin0

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.`

Your original erroneous assumption was that the effects in a linear situation would also apply to the situation in his scenario.I.e. Circular path and direct communication of signals.

Originally Posted by ghwellsjr

But in this thread, the stationary clock is not at the pole but at the equator which complicates things. And if they could see through the earth, they would each continue to see the other ones clock fluctuating in its tick rate, but on average, the zipping clock would see the station clock as going faster than its own and the station clock would see the zipping clock as going slower than its own. That's why I said in post #20 "on average"..

Failing in that you have simply rewritten his scenario , turned it into a linear situation to artificially support your original incorrect claims.
This is particularly ironic as

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.

here you are faulting ghwelljr for the very thing you were doing. Extrapolating from the Doppler effect in inertial frames. Having committed yourself you seem unable to simply admit that within his stated context he was completely right.

 

[GAsahi]

I guess you missed this one

I didn't miss anything, I am simply ignoring your posts.

 

[PAllen]

Neither of the above is the claim that gjwellsjr made that got all this started. Read the claim he made, please.

That claim is obviously correct. What do you disagree with? [Edit: and please, let's not fuss that the formula quoted is only accurate to first order in v^2/c^2]

Some specific points:

-When they two observers get back together and compare clocks (having synchronized them the last time they were together), the circular moving clock is behind.

-Directly observed doppler is exactly as Gwellsjr has described: each sees a periodic pattern of red and blue shift, but the circular track observer's pattern has more blueshift.

- There is no direct way to measure 'time flow' of a distant object. Any such statement is an interpretation. You have many choices:

-- You can report what you actually see on the hands of a distant clock, in which case the description will match the Doppler: each sees the other clock moving slow, then fast, but the proportion is such that the circular observer sees the stationary observer's clock moving fast more of the time.

-- You can try to remove doppler according to some model to get some proposed 'underlying time rate'. One way to do this is using 'radar simultaneity'. This would have the feature that when they are near each other (whether separating or coming together), each interprets the other as slow, but at other points, the circular observer interprets the stationary clock as going much faster. Meanwhile, the stationary clock interprets the circular clock as slow always.

-- Any rational method of imputing a time rate to the other clock must match the fact that on meeting, total time is asymmetric. Thus, there must be asymmetry in any rational way of imputing mutual time rate.

-- It is (IMO) rather silly to argue much about mutual time rate since it is unmeasurable. The only measurables are Doppler (described above), and time comparisons at meeting points.

 

[GAsahi]

That claim is obviously correct. What do you disagree with? [Edit: and please, let's not fuss that the formula quoted is only accurate to first order in v^2/c^2]
.

You keep missing the obvious, in his post, gjwellsjr claims that the two observers start together and are reunited, meaning that they are BOTH on the circumference, and NOT, as you claim, one in the center and the other one on the circumference.

that two clocks starting together and one of them taking a circular path arriving back at the other clock will accumulate less time

 

[PAllen]

You keep missing the obvious, in his post, gjwellsjr claims that the two observers start together and are reunited, meaning that they are BOTH on the circumference, and NOT, as you claim, one in the center and the other one on the circumference.

They are two separate scenarios. I have tried to be clear in each case which I refer to. The post you are replying two is strictly about one observer stationary in some inertial frame, and another following a circular path that meets the stationary at one point. I have described two completely different Doppler patterns for the two scenarios. I am not missing anything at all.

 

[GAsahi]

They are two separate scenarios. I have tried to be clear in each case which I refer to.

We are talking about gjwellsjr scenario, not the scenarios that you keep introducing. It is this scenario that gjwellsjr introduced his mistake. In fact the scenario was clearly specified by the OP from post 1:

If I am 35 years old and I am in a train sitting in a special tube that is built around the entire circumference of the planet. Putting aside all of the questions of practibility and g forces and the like, if the train I am in accelerates to just a hair under the speed of light and I travel on this train until my 5 year old son is 35 years old, at which time my train wil come to a complete stop and I will disembark my "light speed" train and meet my son at the "train station" when I walk out, will we both be the same age?

The post you are replying two is strictly about one observer stationary in some inertial frame, and another following a circular path that meets the stationary at one point. I have described two completely different Doppler patterns for the two scenarios. I am not missing anything at all.

Could you please stop moving the goalposts with your scenarios. Stay on topic. Thank you.

 

[PAllen]

We are talking about gjwellsjr scenario, not the scenarios that you keep introducing. In fact the scenario was introduced by the OP from post 1:

Please stop moving the goalposts with your scenarios. Stay on topic. Thank you.

In what way do you think post #113 is not the same scenario as the OP? Because it dispenses with a tube? I see only two distinct scenarios:

The OP, which is what I describe in #113 (sans tube and earth, which only complicate the issue for SR); and the additional case introduced by Gwellsjr in #34 of a central stationary observer (so there is no meeting of the observers). I have tried to make clear in each post, which of these I refer to.

 

[GAsahi]

-Directly observed doppler is exactly as Gwellsjr has described: each sees a periodic pattern of red and blue shift,

It is good that now you are now sticking with the correct scenario, not with the two scenarios that have nothing to do with this thread.

The issue is that BOTH observers measure redshift OR blueshift AT the SAME TIME , contrary to the claim by Gwellsjr that one of the observers measures redshift while the other measures blueshift (and vice-versa). Actually, this error has first been introduced by gjwellsjr at post 20. I pointed out to you this several times. This is the root of my disagreement with gjwellsjr. Do we really need 100+ posts to get this?

 

[PAllen]

Since there seems to be enormous miscommunication about scenarios, let me give simple equations for what I think are the two distinct cases in this thread. Let us propose standard coordinates (t,x,y,z) in flat spacetime. r is an arbitrary radius, v an arbitrary speed.

OP scenarios:

stationary observer has world line (t,r,0,0).

circular moving observer has world line (t,x,y,z) = (t, r*cos(vt/r), r*sin(vt/r),0)

In any posts where I don't talk about a central observer, this is what I mean.

Central observer scenario:

central observer has world line (t,0,0,0)

circular observer has same world line as above.

I believe in most, if not all, posts about the central observer case, I have used the word 'central' or some other clear language. In posts not using 'central' (obviously including any post that mentions the observers meeting) I am talking about the OP case. #113 was purely about the OP case.

 

[jcsd]

It is good that now you are now sticking with the correct scenario, not with the two scenarios that have nothing to do with this thread.

The issue is that BOTH observers measure redshift OR blueshift AT the SAME TIME , contrary to the claim by Gwellsjr that one of the observers measures redshift while the other measures blueshift (and vice-versa). Actually, this error has first been introduced by gjwellsjr at post 20. I pointed out to you this several times. This is the root of my disagreement with gjwellsjr. Do we really need 100+ posts to get this?

He said 'on average'. To be honest you don't even need to write a single equation to see why he must be correct - if it wasn't the case the two observers wouldn't measure a difference in the time interval.