Speed of the light and dilation of time

[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.

 

[GAsahi]

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)

This is the case being debated, please stop adding your own case. Would you please stay on topic?

 

[PAllen]

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?

I disagree with you. As normally used (Doppler measured in vacuo for light actually received by direct path), the pattern of doppler for circular and stationary (NOT CENTRAL) observer are different. Further, the very phrase 'at the same time' is seriously problematic. For who? The stationary observer? The circular moving observer? They obviously differ as to 'same time', and for the circular traveler, there is no unique way to even define global simultaneity.

I agree with what Gwellsjr said about doppler and about temporal relationships. #113 summarizes what is true for OP scenario.

 

[GAsahi]

He said 'on average'.

"Average" has nothing to do with it, when A measures redshift, B measures redshift. When A measures blueshift, B measures blueshift. gjwellsjr has claimed (at least twice) that A measures redshift while B measures blueshift. I can't believe that it is already past 100 posts and people don't get it.

 

[GAsahi]

I disagree with you. As normally used (Doppler measured in vacuo for light actually received by direct path), the pattern of doppler for circular and stationary (NOT CENTRAL) observer are different.

Yes, you have tried this hair splitting earlier. The point is that they both measure the same type (red OR blue) shift, NOT opposite shifts as gjwellsjr incorrectly claimed.

 

[jcsd]

"Average" has nothing to do with it, when A measures redshift, B measures redshift.

Yes "average" does have everything to do with it as that is what he said, I'm not sure redshift or blueshift had even been mentioned at this point.

When A measures blueshift, B measures blueshift. gjwellsjr has claimed (at least twice) that A measures redshift while B measures blueshift.

The two observers have clocks that run at different speed and no objective "when".

 

[PAllen]

"Average" has nothing to do with it, when A measures redshift, B measures redshift. When A measures blueshift, B measures blueshift. gjwellsjr has claimed (at least twice) that A measures redshift while B measures blueshift. I can't believe that it is already past 100 posts and people don't get it.

People don't get it because it is wrong (except for your scenario of carefully inserted sequence of mirrors, and measuring only on that path). In any normal discussion, one refers to doppler measured through empty space by direct light path. In this case the doppler pattern is not symmetric between the two observers. As to 'same time', I noted in another response that this is not even meaningful.

 

[GAsahi]

Yes "average" does have everything to do with it

It is a red herring.

as that is what he said, I'm not sure redshift or blueshift had even been mentioned at this point.

It has been mentioned, please go back and read the thread. Start reading at post 20.

 

[GAsahi]

People don't get it because it is wrong (except for your scenario of carefully inserted sequence of mirrors, and measuring only on that path). In any normal discussion, one refers to doppler measured through empty space by direct light path. In this case the doppler pattern is not symmetric between the two observers.

Irrelevant, do you really think that the observers measure opposite shifts (one red, the other one blue)? Why is it so difficult for you to admit that you are defending an error?

As to 'same time', I noted in another response that this is not even meaningful.

Hair splitting: throughout the revolution, both observers only see one type of shift if they maintain the direction of emitting signals.

 

[jcsd]

It is a red herring.

It's not a red herring, because that's what was said in the post that you objected to!

It has been mentioned, please go back and read the thread. Start reading at post 20.

So I should start reading the thread at post 20, to see if redshift or blueshift was mentioned before post 20?

 

[GAsahi]

It's not a red herring, because that's what was said in the post that you objected to!

I objected to his incorrect claim that the observers measure opposite type of shifts.

So I should start reading the thread at post 20, to see if redshift or blueshift was mentioned before post 20?

Just start reading at post 20, ok?

 

[PAllen]

Irrelevant, do you really think that the observers measure opposite shifts (one red, the other one blue)? Why is it so difficult for you to admit that you are defending an error?

Because I am not in error. If we place on, each world line I describe for OP scenario, a monochromatic light source and a spectrograph recording light from the other, the recordings on the spectrographs would not have the same shape. What could be clearer than that?

 

[GAsahi]

Because I am not in error. If we place on, each world line I describe for OP scenario, a monochromatic light source and a spectrograph recording light from the other, the recordings on the spectrographs would not have the same shape. What could be clearer than that?

You mean one would be blueshifted and the other one would be redshifted? Yes or No?

 

[PAllen]

You mean one would be blueshifted and the other one would be redshifted? Yes or No?

Both choices are wrong. Each graph would be a smooth, periodic function, but the shape would be different. If you try to compare the graphs at a meeting point, besides the difference in shape, the length of the graphs would be different (assuming identically constructed devices), because of the difference in accumulated proper time. Trying to say which point on one graph corresponds to which point on the other would get into exactly the issues I raised in #113, which is a summary of what I think the important features are.

 

[GAsahi]

Both choices are wrong. Each graph would be a smooth, periodic function, but the shape would be different.

You are not answering the question, you are just attempting to evade it. Obviously there is a redshift from the emitted frequency since the observers are separating. Is the shift the same (towards red) or not? Yes or No?

 

[PAllen]

You are not answering the question, you are just attempting to evade it. Obviously there is a shift from the emitted frequency since the observers are separating. Is the shift the same or not? Yes or No?

I am not evading it in any way. Your choice itself is wrong. Both graphs show redshift and blue shift portions, as stated many times. But the shape is different (as is the length). The circular observer's graph will have a wider period of blue shift compared to the stationary observer's graph.

 

[GAsahi]

I am not evading it in any way.

Sure you are, you have been playing this game for quite a few posts now.

Both graphs show redshift and blue shift portions[/color], as stated many times.

No, they do not. Let me tighten the condition such that you can't try to slip through: the observers emit beams of light in ONE direction only, so there is NO switching from redshift to blueshift, it is redshift throughout. Now, please answer the question.

 

[Austin0]

I think part of the confusion stems from a misunderstanding of the Doppler shift.
There are two components , The intrinsic periodicity of the emitter and the classical shift due to relative motion. In the case of the circular traveler the red shift occurring during the first half of the circut due to classical doppler is steadily decreasing as the instantaneous velocity
relative to the inertial observer decreases with the relative angle of the motion wrt that observer.
The intrinsic relationship of periodicities remains constant as it is based on the angular velocity.

This means that the signals initially received in the moving frame have maximal classical redshift combined with intrinsic blueshift. Reaching the opposite side of the circut there is no classical redshift and it is purely intrinsic blueshift by the gamma factor of the angular velocity.
It is logically quite apparent that at some point in that transit the classical redshift had to have diminished to the point that the observed shift was blue to some degree long before reaching the point where the observed shift was the full undiminished gamma factor.

It is equally obvious that considering an arc segment of travel on the opposite side where the shift is essentially transverse there is no possibility that the stationary observer could observe anything but redshift and equally no possibility the traveling observer could see anything but blueshift..

 

[PAllen]

No, they do not. Let me tighten the condition such that you can't try to slip through: the observers emit beams of light in ONE direction only, so there is NO switching from redshift to blueshift, it is redshift throughout. Now, please answer the question.

I find that completely artificial and counter to the way Doppler is discussed by everyone else. I propose simple physics: a spectrograph and light source at each observer. The observers move through empty space along the given world lines. The spectrographs measure what I claim.

 

[jcsd]

I objected to his incorrect claim that the observers measure opposite type of shifts.

This is what he said is:

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.

And in response you said:

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

Just start reading at post 20, ok?

Because you say so? Well, I have already read most of the thread. I don't particularly want to get bogged down in post facto justififications of what was intially a simple mistake. Clearly there is an average blueshift and an average redshift, also there is no way to talk about "at the same time" in special relativity for observers in different states of motion.

 

[GAsahi]

Clearly there is an average blueshift and an average redshift,

There is no "average". There is no "blueshift". Let me take away this loophole from you: the two observers send signals in ONE direction only, the stationary one in the "chasing" direction, the "traveling" one in the rear direction. Now, do the two observers see the same type of shift? Yes or No?

I don't particularly want to get bogged down in post facto justififications of what was intially a simple mistake.

So, are you admitting that gjwellsjr posts are wrong? Yes or No?

 

[GAsahi]

I propose simple physics: a spectrograph and light source at each observer. The observers move through empty space along the given world lines. The spectrographs measure what I claim.

Please answer the question I asked you.

 

[PAllen]

Please answer the question I asked you.

Sure, though I think it is utterly pointless:

If you have mirrored light tube that can bend light in a perfectly circular path without affecting its speed, then if stationary observer and circular observer send light only into this tube in opposite directions, then when they meet:

- both graphs will show only red shift
- the circular mover's graph will be shorter; his watch will be behind.

So what? All that accomplishes is to remove all the normal relationship between doppler and differential aging, producing an apparent (but not real) paradox.

 

[jcsd]

There is no "average". There is no "blueshift". Let me take away this loophole from you: the two observers send signals in ONE direction only, the stationary one in the "chasing" direction, the "traveling" one in the rear direction. Now, do the two observers see the same type of shift? Yes or No?

Like I said, not getting bogged down.

So, are you admitting that gjwellsjr posts are wrong? Yes or No?

No, I am saying that gjwellsjr said sometthing that was perfectly correct and you mistakenly objected to it and since then we now have pages of you trying to justify your intial mistake.

 

[GAsahi]

Sure, though I think it is utterly pointless:

If you have mirrored light tube that can bend light in a perfectly circular path without affecting its speed, then if stationary observer and circular observer send light only into this tube in opposite directions, then when they meet:

- both graphs will show only red shift

Thank you. Meaning the gjwellsjr claim , that started all this debate is false.

So what? All that accomplishes is to remove all the normal relationship between doppler and differential aging, producing an apparent (but not real) paradox.

No one is discussing any "paradox", we were discussing the error in gjwellsjr claim. Thank you for (finally) acknowledging it.

 

[PAllen]

Thank you. Meaning the gjwellsjr claim , that started all this debate is false.

No, his claim is true, because no one else in the world defines red shift the way you do. Look at any of the thousands of explanations of relation between differential aging and Doppler, and they agree with Gwellsjr and everyone else on this thread except you.

Each time you have referred me to a post of Gwellsjr that is supposedly wrong, I find that it is correct and your critique is wrong (or wrong headed - like defining your own version of Doppler that no one else uses or wants to use).

 

[GAsahi]

No, his claim is true, because no one else in the world defines red shift the way you do.

I don't understand why you have such a hard time admitting that you are wrong.

Look at any of the thousands of explanations of relation between differential aging and Doppler, and they agree with Gwellsjr and everyone else on this thread except you.

The discussion has nothing to do with differential aging and/or its relation to the Doppler effect, please don't start moving the goalposts again, it has to do with the mutual nature of the Doppler effect. Twice , in this thread, you have admitted that both observers are perceiving the same type of shift, i.e. redshift, while gjwellsjr has claimed in two instances (at least) that one observer measures redshift while the other perceives blueshift.

 

[PAllen]

I don't understand why you have such a hard time admitting that you are wrong.

The discussion has nothing to do with differential aging and/or its relation to the Doppler effect, please don't start moving the goalposts again, it has to do with the mutual nature of the Doppler effect. Twice , in this thread, you have admitted that both observers are perceiving the same type of shift, i.e. redshift, while gjwellsjr has claimed in two instances (at least) that one observer measures redshift while the other perceives blueshift.

I am not wrong.

I have admitted what you say only for definition of doppler that is non-standard and (IMO) useless. Meanwhile, what Gwellsjr has claimed about redshift/blue shift is correct for how everyone else in the world measures them. However the circular observer assigns simultaneity between their world line and 'stationary' world line, they will find corresponding events where they measure blue shift while the stationary observer measures redshift (measures using the approach everyone in the world except you uses). [edit: and the same goes for the 'stationary' observer, only they have an obvious convention for simultaneity].

 

[GAsahi]

I am not wrong.

Of course not :-)

However the circular observer assigns simultaneity between their world line and 'stationary' world line, they will find corresponding events where they measure blue shift while the stationary observer measures redshift (measures using the approach everyone in the world except you uses).

Let's make it simpler for you:

1. Observer A accelerates in a straight line away from observer B. The two observers send light signals towards each other. B sends a ray that chases after A and A sends a ray back, along the line connecting them. Do they measure the waves as being:

A1. Mutual Redshift
B1. Mutual Blueshift
C1. One Redshift while the other Blueshift

2. Same problem as above but the trajectory is a half circle. The rays exchanged follow the circular arc.

A2. Mutual Redshift
B2. Mutual Blueshift
C2. One Redshift while the other Blueshift

 

[PAllen]

Of course not :-)



Let's make it simpler for you:

1. Observer A accelerates in a straight line away from observer B. The two observers send light signals towards each other. B sends a ray that chases after A and A sends a ray back, along the line connecting them. Do they measure the waves as being:

A1. Mutual Redshift
B1. Mutual Blueshift
C1. One Redshift while the other Blueshift

2. Same problem as above but the trajectory is a half circle. The rays exchanged follow the circular arc.

A2. Mutual Redshift
B2. Mutual Blueshift
C2. One Redshift while the other Blueshift

(1) Is obvious - both redshift.

(2) At the point where the circular observer's watch is halfway between meet up times, they receive light neither red or blueshifted. At the point where the stationary observer's watch is halfway between meet up times, they receive red shifted light. At points a little beyond halfway as I've defined it, the stationary observer sees redshift and the circular observer sees blueshift.

This is, of course, using the universally recognized defintion of doppler, not your personal definition of measuring through a curved mirror tube.

 

[GAsahi]

(1) Is obvious - both redshift.

Correct.

(2) At the point where the circular observer's watch is halfway between meet up times, they receive light neither red or blueshifted. At the point where the stationary observer's watch is halfway between meet up times, they receive red shifted light.

Correct. If you paid attention, you would have noticed that the path is HALF circle, so you should have stopped here.

This is, of course, using the universally recognized defintion of doppler, not your personal definition of measuring through a curved mirror tube.

Try paying attention to the scenario, gjwellsjr had the two observers "looking at each other through a webcam". How do you think one connects webcams? Hint: through a fiberoptic cable.

At points a little beyond halfway as I've defined it, the stationary observer sees redshift and the circular observer sees blueshift.

You are assuming that the Earth is transparent, aren't you? Last I checked, it wasn't. :-)