[thebrain13]

now I apologize if I started skimming through and not paying enough attention to your arguments, but this is how I believe this problem is resolved.

What you need to pay attention to is where the measurement of time takes place. In this case it is the earth. The earth views both spaceships as younger then it is. All three observers view a different order of time throughout the whole trip. Since they are separated in distance, and they are traveling away from each other, all three would view each other as younger than themselves. see relativity of simultaneity. They would only agree once they are local(no distance between them) and have no relative velocity.

Now it seems like people always get caught up in arguing that both people view the same thing, but If I'm not mistaken a non inertial reference frame doesn't necessarily mean that it is accelerating, it can mean that it was accelerated. The ships are not in the same reference frame as the earth. The earth can lay claim that it was stationary the whole time because thats where the experiment was measured. The earth had no relative motion in relation to it, the pilots did.

 If we decide to measure the experiment at celestial north, a different conclusion would be met. Lets say that the beginning of the experiment is the same two ships travel away at equal speeds in opposite directions. Except for one ship stops at celestial north, then the earth accelerates to meet it, and the other ship has to travel even further, since he traveled south before heading up north. (They have to be united to view the affect otherwise they will all just disagree with each other, and thats not a paradox, its Einsteins prediction) Since the celestial north is where the experiment is being measured, it would view the south traveling pilot as the youngest, the earth and the north moving pilot had spent the same amount of time traveling the same speeds relative to it, therefore they would be the same age as each other. And both would be the same amount older than the other pilot who spent more time traveling in relationship to the observer, celestial north.

[jolly]

This is about time slowing down when you travel at the speed of light- so if time is slower for you when traveling at light speed but time stays the same for some who is not. Then when you meet the other person he has aged more than you.

The time of the journey is not important- 5mins you'll have a slight age increase on the brother that stays at home. A long journey and a big age increase on the brother that stays at home. 

[lightarrow]

As far as each twin is concerned, in my very simple model, there is no 'fixed ' earth or distant star. In fact we have no fixed grid in space for our measurements, at all.
They are both in a deserted part of deep space. I am not allowing you to have any more in your experiment than this.
They could both be going at any speed you like at the start of the experiment. The only things they can see or measure are each other and their ships and their relative motion.
Apart from the fact that one has used his engines (which must play a part in resolving the paradox), when they start to move apart, their relative motions are equal and opposite.
Now take up my previous argument. Neither twin knows which of them has actually 'moved', only that they have changed position, relative to each other.
The actual distances they each observe the other one to have traveled is not really relevant. What is relevant is the fact that, when they finally meet (when their space time vectors coincide again) they will have different lengths of beard!
SR tells us that when the traveling one  (the one who uses his rockets)  returns to the original spot will have aged less because of time dilation and  he will appear younger.
But, from his point of view, it was the other one who went off at speed and then returned. So the other one will appear younger to him.

Yes, this is the "Twin Paradox", more or less as Einstein put it ≈ 100 years ago (it's incredible we are still discussing this, isnt'it?).
You say: <<So the other one will appear younger to him>>.
What I'm trying to convince you is that, actually , SR doesn't say that. It would seem, with a first logical reasoning, that the twin who has used engines "should" see the other twin as younger too. But this is only a "kind-of-logical-intuitive reasoning", and not the real mathematical evaluation, made in a little more time.

How can there not be a paradox there? if A>B, you can't say B>A.
But, if you allow for the effect of acceleration on one of them (which one of them could feel as a 'weight  force' on him), to alter the rate of progress of time on his ship.  That is a GR effect. As has been said previously,  effects on the passage of time  have  been observed, due both to high speeds ( the muon observations - SR) and due to gravitational fields (the Mossbauer effect - GR).

Sophie, it's called "gravitational time-dilation":
http://en.wikipedia.org/wiki/Gravitational_time_dilation
but, as I tried to show you, it's not enough to explain the overall time difference between the twins.
The asymmetry between the twins does come from the fact that one of them is accelerating; the difference in time does come from that fact, but the amount of time difference does not come from it (is this the real paradox?!)
If you could find the very simple (but rigorous) book: "Spacetime Physics, W.H. Freeman & Co., New York 1992- Edwin Taylor and John Archibald Wheeler", talking about how to solve the Twin Paradox, in paragraph 4.10 it says, as a title of a sub-paragraph:
"Do we need General Relativity? NO!"
It's an interesting book (even if with too little formulas for me).

[lyner]

a non inertial reference frame doesn't necessarily mean that it is accelerating, it can mean that it was accelerated.

Inertial means it's not accelerating. Its velocity is not changing whilst you are doing the experiment.
If anything is at a velocity, it must have accelerated at some time, no?
An inertial frame is the only condition under which you can do simple SR calculations.
I don't want to do this experiment on Earth. I don't have to.
My two twins are the only people in my experiment.
I didn't say that they see the same thing. I just said that they part at a time (noted by each of them by setting their identical clocks) and they meet some time later (at the same place in space time - or they wouldn't be meeting).
Their clocks will just not agree if  one of them has been accelerating - and he will have had to accelerate if he is to go away and then come back. There would have  been fewer clicks of the traveller's clock than on the clock of the one who remained (in his inertial frame, all the time).
It doesn't matter what speed the two were doing, relative to some observer on a 'stationary' Earth. That is not relevant.

What you need to pay attention to is where the measurement of time takes place. In this case it is the earth.

The place that the time measurement is taking place is on board each ship - the way time behaves on each ship is how fast the twins are each aging.
The whole point about relativity is that it avoids having to have a 'grid' in space. In fact, it effectively forbids one.
There is no need to involve the Earth or the direction of travel. That just adds confusion.
My experiment (in my last post) takes place miles from anyone else. Each twin is an observer . there are no  other observers. They each see different times on their clocks (and some other things about their distances, masses etc., but I'm not concerned with that).
Nowhere is stationary, in any case - didn't Michelson  and Morley imply that in  their classic experiment?
The earth is spinning and in orbit and the Sun is zapping round the galaxy, which, in turn, is zapping through deep space. (Unless you are an ancient Greek, in which case the earth is at the centre etc. etc.)

[lyner]

Sophie, it's called "gravitational time-dilation":
http://en.wikipedia.org/wiki/Gravitational_time_dilation

Uncle Albert equates gravity and acceleration, doesn't he? They are precisely the same thing; Newton2.
A little acceleration for a long time or a lot for a little time will both have the effect of  changing the RATE of the clocks on the traveller's ship for ever, once the engines are turned off.
btw,  no one can rely on Wikipedia any more than we can really rely on what is posted here.

It would seem, with a first logical reasoning, that the twin who has used engines "should" see the other twin as younger too. But this is only a "kind-of-logical-intuitive reasoning",

Precisely - we are both saying that you can't apply SR in cases where there is not a inertial frame. When one tries- by virtue of saying "it's all relative" even though we're non-inertial, one is stepping into dodgy territory and you get so-called paradoxes.

In an earlier post this evening I quoted two ships passing each other at great speed. In that case they would observe each other's ships clock as being slow. That's ok - there is no paradox in that case. They are both in inertial frames, throughout  the experiment.

Until I have looked into it further, I will just have to disagree with  lightarrow about the  amount of the effect of acceleration. If it's not due to that, what can it be due to?
btw, have you noticed the amount of interest this has generated? []

[Batroost]

Nowhere is stationary,

Indeed, but ONLY the 'travelling' twin would be able to measure an acceleration using, say, an inertial accelerometer. The twin on the earth would not 'experience' a measureable acceleration, and this is why you can't equate the two frames of reference. This is where the asymmetry creeps into the paradox and why the travelling twin really does return younger than the stay-at-home. During any steady-velocity part of the journey both twins would observe the clocks of the other twin to be running slow - this is a symmetrical effect and applies regardless of the direction of travel

There is another affect that we need to remember i.e. In a strong gravitational field, general relativity says that clocks run slower than in a weak gravitational field. So if you really want to make this impressive, send teh travelling twin to orbit close to a black hole for a couple of months!

[lyner]

Yes, batroost - there is no real paradox.

http://www.astrophysicsspectator.com/topics/specialrelativity/Coordinate.html

For what it's worth, there is a good link here which deals with the Twins paradox.
I am quoting it because 1. I found it easily and 2. It seems to agree with what I was saying - well. I would, wouldn't I?
This link:
http://www.astrophysicsspectator.com/topics/specialrelativity/TravelDilation.html
 gives some numbers for would-be starship travelers.

[lightarrow]

Sophie, it's called "gravitational time-dilation":
http://en.wikipedia.org/wiki/Gravitational_time_dilation

Uncle Albert equates gravity and acceleration, doesn't he? They are precisely the same thing; Newton2.

My name is Albert too. []

A little acceleration for a long time or a lot for a little time will both have the effect of  changing the RATE of the clocks on the traveller's ship for ever, once the engines are turned off.

I don't have the possibility to post a simple drawing of space-time diagrams, however you can find many on the web. If I find one I'll post it. You would notice that the amount of acceleration/deceleration and the interval of time it lasts, is not so important, in the evaluation of the age difference of the twins, if they move apart for a long distance and at high relative speed. Infact, what really counts is exactly that: distance travelled and relative speeds.

Imagine an orthogonal cartesian ref. frame; you put time on the vertical axis and distance x on the horizontal.

In the ref. frame of A:
T is the time needed by B to reach Proxima Centaury (5 years, as you remember).
L is earth-Proxima Centaury distance (4 light years).
A vertical segment from (0,0) to (0,2T) represent the world-line of A;
a couple of bent segments, the first from (0,0) to (L,T), the second from (L,T) to (0,2T) represent the world-line of B.
The proper-time of a twin is Δs/c where Δs is the space-time interval between two events in a world line.
So, A's proper time is Δs/c evaluated in A's world-line, from (0,0) to (0,2T).
Since Δs = Sqrt[(cΔt)² - (Δx)²] we have:
Δs(A) = Sqrt[(c*2T)² - 0] = 2cT.
Δs(B) = 2Sqrt[(c*T)² - (L)²]
Let's evaluate the difference of proper times:
Δs(A)/c - Δs(B)/c = 2T - 2Sqrt[(T)² - (L/c)²] =
= 2*5 - 2Sqrt[(5)² - (4)²] =
= 10 - 2Sqrt(9) = 10 - 6 = 4 yars.
As you see, B accelerates around point (L,T) but how much it accelerates doesn't show in the evaluation.

[lightarrow]

http://math.ucr.edu/home/baez/physics/Relativity/SR/TwinParadox/twin_spacetime.html

Minkowski recast Einstein's version of Special Relativity (SR) on a new stage, Minkowski spacetime.  The Twin Paradox has a very simple resolution in this framework.  The crucial concept is the proper time of a moving body.

[lyner]

Code: [Select]

As you see, B accelerates around point (L,T) but how much it accelerates doesn't show in the evaluation.The fact is that he DID accelerate, and did not stay in an inertial frame, - which is what got him to the speed with which you are doing your calculations. It is only the fact that he accelerated that makes him different from the 'stationary' twin. You cannot argue with that, I am sure. Without an accelerometer, neither of the twins would, initially, know  whether it was he or his brother or both of them who was 'moving away' (trains moving out of stations etc.). That was the start of the original idea of SR.
As far as I can see, your calculations are, of course, fine but the link in my last post evaluates how the acceleration has produced precisely the sort effect that your calculations show. This is not surprising;  there are often many ways of showing a result / killing a cat. However - the link  takes more into account and doesn't rely on any talk of a fixed reference frame or a journey to any particular star.   It just describes my simple two twin situation - out in the middle of nowhere  - and determines which of them ages slower.

[lightarrow]

Code: [Select]

As you see, B accelerates around point (L,T) but how much it accelerates doesn't show in the evaluation.The fact is that he DID accelerate, and did not stay in an inertial frame, - which is what got him to the speed with which you are doing your calculations. It is only the fact that he accelerated that makes him different from the 'stationary' twin. You cannot argue with that, I am sure.

Yes, even because I've already said it several times in my posts!

Without an accelerometer, neither of the twins would, initially, know  whether it was he or his brother or both of them who was 'moving away' (trains moving out of stations etc.). That was the start of the original idea of SR.

Of course.

As far as I can see, your calculations are, of course, fine but the link in my last post evaluates how the acceleration has produced precisely the sort effect that your calculations show. This is not surprising;  there are often many ways of showing a result / killing a cat. However - the link  takes more into account and doesn't rely on any talk of a fixed reference frame or a journey to any particular star.   It just describes my simple two twin situation - out in the middle of nowhere  - and determines which of them ages slower.

If the twin A is on earth or in space and there are no stars, space-time diagrams are exactly the same. I talked about earth and Proxima Centaury just to make things easier to understand, but there is no need at all.
See also:
http://mentock.home.mindspring.com/twins.htm
the diagrams relative to this document's explanation are in here:
http://sheol.org/throopw/sr-twin-01.html

A recent discussion on Twin Paradox on Physics Forums:
http://www.physicsforums.com/showthread.php?t=165370&page=2

If you consider the space-time diagram of a twins paradox type experiment, you can see that the twins form a triangle.

Acceleration means that the triangle isn't perfect, that the points of the triangle are actually slightly rounded by the acceleration.

However, this can be made to be a very small effect, by taking the limit of high accelerations.

What the twin paradox says is very similar to the geometrical theorem that the shortest distance between two points is a straight line, or eqivalently, the "triangle inequality" that says that the sum of the length of two sides of a triangle is always longer than the remaining third side.

In the case of the twins paradox, though, the statement becomes that the observer following a geodesic path (analogous to the straight line, especially in the flat geometry of SR) has the longest proper time. The time is the longest, but the distance in the Euclidean geometrical analogy is the shortest. This has to do with the difference between Euclidean geometry with it's ++++ metric signature, and the Lorentzian geomery of SR with it's -+++ signature.

Acceleration (the curvature at the tips of the triangle) really has very little to do with this geometrical result.

That is why people say that the twin paradox isn't about the rounding of the corners due to the acceleration - it's due to the angle between the observers. This angle on a space-time diagram is simply a change of velocity. What is important is not the rate at which the velocity changes (the accleration) , but what is often called "delta-V", the change in velocity.

The geometrical analogy of the "twins pardox" would be the "triangle paradox". The triangle paradox would say:

If I go directly from A to B, I travel the shortest distance. And if I go from A to C to B, I travel a longer distance. It's "paradoxical" (?) that when I go from A to C to B, that this distance is longer than going straight from A to B. Furthermore, if I look at the distance from A to B, that's shorter than going from A to C to B. Now AB is the shortest distance!

Of course people gemerally don't get confused by the "triangle paradox" - it's really not that much more difficult not to get confused by the "twin paradox" either.

So, I repeat that:
1. The Twin Paradox Does arises from the fact that one of them have to accelerate.
2. The age difference has Nothing to do with the amount of acceleration.
3. There is no need of GR to explain it and to make all the necessary computations.

By the way, there isn't even the need of accelerometers to understand who was the one who has accelerated!

If they send each other light pulses at costant rate (in their own ref frame), for example, one light pulse every second, after the complete journey we can understand who was the one who has accelerated just from the distribution of light pulses every twin has received from the other one.

Look at Fig 2 in this document:
http://math.ucr.edu/home/baez/physics/Relativity/SR/TwinParadox/twin_vase.html#curvi-coords

One of them (1) sees a low frequency of pulses for longer time and a high frequency for a shorter time.
The other (2) sees a low frequency of pulses for a shorter time and a high frequency for a longer time (or the two intervals of time are the same, in the limit of infinite deceleration/acceleration at the return point).

The one who has accelerated is (2).

Twin Paradox has always fascinated me!
Alberto.

[lyner]

I can see that we are agreeing about all of this apart from the fact that you say the acceleration is irrelevant. If it weren't for the fact that one was accelerating, there would be no difference in velocity so no effect.

Acceleration (the curvature at the tips of the triangle) really has very little to do with this geometrical result.

The velocity difference gives you the ANGLES in your triangle. which is what gives you something to calculate with. If you had instantaneous speed change then you would have sharp corners, which would look pretty but is not practical. Velocity is merely the integral over time of acceleration -  in practice, giving curves, not sharp corners.  This detail  (curved or sharp) is not really relevant.
Your 'delta v' is a factor from outside SR and accounts for the angle.
The only paradox is there when one ignores this.
The total change in velocity, of course, depends on the acceleration and how long it's accelerating for (the same as impulse  (force times time) can be used in conventional dynamics problems.  I can't see why this presents you with a difficulty. During the time our twin is changing velocity, it is not in an inertial  frame. The reason behind all your calculations  is what happens whilst it is  in a non- inertial frame.
The only situation in which you  can ignore GR is when they are coasting past each other, without any velocity change. If you have acceleration , you can't ignore GR. But why should you want to?
 []

[lightarrow]

I can see that we are agreeing about all of this apart from the fact that you say the acceleration is irrelevant.

Sophie, "irrelevant" for what? Did you understand what I mean when I say this?
I've tried to explain this several times. This is another time. I said that:
The Amount of Acceleration can be made Irrelevant for computing The amount of The age difference.
Not, that Acceleration is irrelevant!

If it weren't for the fact that one was accelerating, there would be no difference in velocity so no effect.

Of Course!

Acceleration (the curvature at the tips of the triangle) really has very little to do with this geometrical result.

The velocity difference gives you the ANGLES in your triangle. which is what gives you something to calculate with. If you had instantaneous speed change then you would have sharp corners, which would look pretty but is not practical. Velocity is merely the integral over time of acceleration -  in practice, giving curves, not sharp corners.  This detail  (curved or sharp) is not really relevant.
Your 'delta v' is a factor from outside SR and accounts for the angle.
The only paradox is there when one ignores this.
The total change in velocity, of course, depends on the acceleration and how long it's accelerating for (the same as impulse  (force times time) can be used in conventional dynamics problems.  I can't see why this presents you with a difficulty. During the time our twin is changing velocity, it is not in an inertial  frame. The reason behind all your calculations  is what happens whilst it is  in a non- inertial frame.

See up.

The only situation in which you  can ignore GR is when they are coasting past each other, without any velocity change. If you have acceleration , you can't ignore GR. But why should you want to?
 []

Sorry, Sophie, but *where exactly* did I use GR to make my calculations?
ds² = (cdt)² - (dx)²
is the Minkowsky metric, valid only in a "Flat" space-time, that is, in SR and NOT in GR.
I repeat, the paradox DOES arise from the fact one of them is accelerating.
But GR IS NOT NECESSARY (= we don't need to use or know that theory) to make all the considerations and calculations to solve the paradox..

[neilep (OP)]

THANK YOU all for an amazing thread !!

[lyner]

Sorry, Sophie, but *where exactly* did I use GR to make my calculations?
ds2 = (cdt)2 - (dx)2
is the Minkowsky metric, valid only in a "Flat" space-time, that is, in SR and NOT in GR.

Your geometry is impeccable and it gives the right result from the point of view of the twin who stays behind because HE is in an inertial frame.  There is nothing wrong with your construction to show what happens on the straight bits, but you are  including changes in velocity at the corners, which, as I keep saying, involves  the traveler not being in an inertial frame all the time.   This is where the GR bit comes in.  It is this effect on him that makes him observe the one who stays behind as being older than him when he returns. It is the traveler who experiences the time-rate  changes at the corners. If you don't allow for this then what is the difference between the observations of the two? There is no reference grid, so why, apart from the acceleration factor, will the experiences not be symmetrical and paradoxical?
Go through the same procedure for the travelling twin and obey the rules; he is not in flat space time all the 'time' so he must see things differently; you can't  use your Minowski metric in the same way for him.
But this is just going round in circles and that, btw, is not an inertial frame, either!

[lightarrow]

Sorry, Sophie, but *where exactly* did I use GR to make my calculations?
ds2 = (cdt)2 - (dx)2
is the Minkowsky metric, valid only in a "Flat" space-time, that is, in SR and NOT in GR.

Your geometry is impeccable and it gives the right result from the point of view of the twin who stays behind because HE is in an inertial frame.  There is nothing wrong with your construction to show what happens on the straight bits, but you are  including changes in velocity at the corners, which, as I keep saying, involves  the traveler not being in an inertial frame all the time.   This is where the GR bit comes in.  It is this effect on him that makes him observe the one who stays behind as being older than him when he returns. It is the traveler who experiences the time-rate  changes at the corners.

What I have colored in blue is not true, and I showed you why.

If you don't allow for this then what is the difference between the observations of the two? There is no reference grid, so why, apart from the acceleration factor, will the experiences not be symmetrical and paradoxical?
Go through the same procedure for the travelling twin and obey the rules; he is not in flat space time all the 'time' so he must see things differently; you can't  use your Minowski metric in the same way for him.
But this is just going round in circles and that, btw, is not an inertial frame, either!

Let's put it in this way:
The Twin Paradox CAN be explained using General Relativity.
Have I ever claimed the contrary?
No. I have said that this "Is not necessary".

Let's say that we have n objects, each of them moving with different velocities and accelerations in a region of space far from any massive object.
Can I find their relative aging using General Relativity?
YES.
Can I find it using Special Relativity?
YES.
How is that possible?
I choose an inertial reference frame and I draw space-time diagrams in that frame; then I make all the necessary computations. Does this inertial ref. frame coincide with one of the object's ref frame, as in the twin prdx? Yes or not, it doesn't matter.
Why should I prefer this reference frame and not another one?
Because computations are simpler.

Let's say you are in the accelerating rocket, and you want to make all the necessary computations to understand if it's you or your twin who will be younger when you meet again.
The fact it's you who is accelerating, you can understand:
1. with accelerometers, or
2. studying the distribution of light signals arriving to you from your twin.

Then how can you make the computations? You can choose:
General Relativity
OR
Special Relativity.

If you draw diagrams using your ref. frame, you have to use GR; if you choose an inertial ref frame (any!) to draw your diagrams, you can use SR. You are not forced to use GR only because you are not in an inertial ref. frame, do you know what I mean? Because you don't have the necessity to use your ref. frame in your notebook's diagrams!

[lyner]

You are sounding quite convincing now.
I'm all in favour of using the simplest method for getting an answer  and I have some chance of following SR in my brain.
BUT (and there's got to be a but) it seems that, using your argument, no one need ever use GR to explain anything. You imply that one can always explain the effects of general relativity by referring your measurements to some inertial frame, elsewhere. This can't be true for all cases, or they wouldn't have bothered to invent GR.
Can we, possibly, reconcile  our differences with the following scenario?
Your traveling twin goes near a huge dead star, which he cannot see, because it is a black dwarf. The gravitational field of this object introduces an acceleration (we, surely, must use GR in this case, mustn't we?),  which  he measures and which he takes into account when he analyses his observations of his non-traveling brother.  He attributes  his  relativistic  effects to acceleration due to his engines  and  does your calculations , making the appropriate corrections etc.  to decide how much younger he will be at the end of the experiment.
Can he possibly still get the correct answer about his relative age after this experience?
If he can, then I can't see why GR should ever be necessary. If he can't, then what is the difference between the acceleration due to his engines and the acceleration due to the gravitational field that he enters?
I am not being argumentative for the sake of it now; I am genuinely mystified .
Is the traveller actually capable of making all the corrections without taking GR into account? Unless he makes allowances for his time dilation (due to acceleration) can he, in fact, determine where he is and how fast he's going?  I guess you will say "yes".
Have you an example of a similar  type of experiment  where GR would HAVE to be  used, so I can get the picture properly?

I notice that not many other people are getting involved with this.
Is there anyone else who can resolve this?

[lightarrow]

You are sounding quite convincing now.
I'm all in favour of using the simplest method for getting an answer  and I have some chance of following SR in my brain.
BUT (and there's got to be a but) it seems that, using your argument, no one need ever use GR to explain anything. You imply that one can always explain the effects of general relativity by referring your measurements to some inertial frame, elsewhere. This can't be true for all cases, or they wouldn't have bothered to invent GR.
Can we, possibly, reconcile  our differences with the following scenario?
Your traveling twin goes near a huge dead star, which he cannot see, because it is a black dwarf. The gravitational field of this object introduces an acceleration (we, surely, must use GR in this case, mustn't we?),  which  he measures and which he takes into account when he analyses his observations of his non-traveling brother.  He attributes  his  relativistic  effects to acceleration due to his engines  and  does your calculations , making the appropriate corrections etc.  to decide how much younger he will be at the end of the experiment.
Can he possibly still get the correct answer about his relative age after this experience?

Good consideration. My compliments to you! I really don't know the answer. I imagine that, if the dead star is far, non rotating, and the travelling twin doesn't approach it much and so he experiences only uniform accelerations inside the rocket, then it's possible to find at least one inertial ref. frame and to draw space-time diagrams, and make all the computations necessary, so that GR is not necessary even in this case, but I'm not sure.

If he can, then I can't see why GR should ever be necessary. If he can't, then what is the difference between the acceleration due to his engines and the acceleration due to the gravitational field that he enters?

In general, in a curved region of space, is not (always?) possible to find an inertial ref. frame, so all that could be impossible of course. I don't know much about it however.

I am not being argumentative for the sake of it now; I am genuinely mystified.

Me too!

Is the traveller actually capable of making all the corrections without taking GR into account? Unless he makes allowances for his time dilation (due to acceleration) can he, in fact, determine where he is and how fast he's going?  I guess you will say "yes".

As I said, maybe it should be possible, in some very simple cases; certainly not in all of them.

Have you an example of a similar  type of experiment  where GR would HAVE to be  used, so I can get the picture properly?

In a curved region of space, not asymptotically minkowskian, with rotating massive objects, I'm pretty sure GR is necessary!  []

I notice that not many other people are getting involved with this.
Is there anyone else who can resolve this?

I hope so!
And I hope we two will make other interesting discussions like this in other threads, in future.
Bye!

[lyner]

What a nice outcome!
We will meet again, I'm sure. Soon.