How are the twins distinguished?

[Mike_Fontenot]

It is the guy who goes out and comes back even if the accelerations were instaneous that would age more slowly. If they both go out then both and never turn around, as you said, each would perceive the other as aging more slowly.

My posting was not addressing how the total ageing of the two twins compare, when the scenario of the total trip allows that comparison to be mutually agreed upon by each twin. I was addressing only periods when neither twin is accelerating. I've edited my previous posting, to try to make that clearer.

In the case of the standard traveling twin example (with one perpetually inertial ("home") twin, and a traveling twin who is unaccelerated except for an instantaneous direction reversal at the turnaround), each twin correctly concludes that the other twin is ageing more slowly during the two inertial segments of the traveler's trip. Yet they obviously must agree about their two ages when they are reunited. How is that possible?

It is possible, because the traveler will conclude that the home twin's age suddenly increases during the turnaround. When the traveler adds up the three components of the home twin's ageing, the total is exactly the age of the home twin when they are reunited. The two twins agree about the FINAL correspondence between their ages, but they do NOT agree about their corresponding ages during the trip (except for the one instant during the turnaround, when their relative velocity is zero).

Mike Fontenot

 

[stevmg]

My posting was not addressing how the total ageing of the two twins compare, when the scenario of the total trip allows that comparison to be mutually agreed upon by each twin. I was addressing only periods when neither twin is accelerating. I've edited my previous posting, to try to make that clearer.

In the case of the standard traveling twin example (with one perpetually inertial ("home") twin, and a traveling twin who is unaccelerated except for an instantaneous direction reversal at the turnaround), each twin correctly concludes that the other twin is ageing more slowly during the two inertial segments of the traveler's trip. Yet they obviously must agree about their two ages when they are reunited. How is that possible?

It is possible, because the traveler will conclude that the home twin's age suddenly increases during the turnaround. When the traveler adds up the three components of the home twin's ageing, the total is exactly the age of the home twin when they are reunited. The two twins agree about the FINAL correspondence between their ages, but they do NOT agree about their corresponding ages during the trip (except for the one instant during the turnaround, when their relative velocity is zero).

Mike Fontenot

This goes against what JesseM stated and which I summarized so that I could understand it:

This post by JesseM explains the conceptual and simple mathematical approach to this problem

https://www.physicsforums.com/showpost.php?p=2610219&postcount=63

In essence we have twin A and B. If one looks at it from the point of view of twin A's frame of reference (FOR) B moves to the right, turns around and moves to the left. Time in A's FR is proper time as he is moving in time alone. Folks who are moving in time AND space experience less time because of the motion (you know, Lorentz, et al.) Thus B is moving both away and back and experiences less time. JesseM gives a nice quick calculation.

The supposed symmetrical situation is to look from B's FR. In this case A moves left - BUT never stops. Twin B starts moving to catch up with A and eventually does. When all the times are added up the elapsed time for B is the same this way as it was looking at it the first way in the above paragraph.

Guess what! This is NOT a symmetrical situation is it? In the first case one of the twins sits still in space but moves only in time (and gets older) while the other twin moves in space and time and gets older slower. The first twin never moves in space, just in time.

In the second case, BOTH twins move in time and space although one sits still for a while in space before moving in space and time.

These are NOT symmetrical approaches. There's no way to make them symmetrical. JesseM's calculations and working through the problem is self explanatory.

A symmetrical scenario would be to have both twin A and B depart the reference frame in opposite directions at the same speed for the same time and both turn around and come back to meet. In this case, they both would age at the same rate (though not as fast as their triplet brother who remained on Earth) and be the same age when they rejoined. Their triplet brother who remained on Earth would be older than both.

Does this make sense?

 

[Mike_Fontenot]

This goes against what JesseM stated and which I summarized so that I could understand it:

[...]

Does this make sense?

No. In special relativity, space is not "something" on it's own ... it's NOT "stuff" that an object can move (or not move) through. It makes no sense to talk about which object is REALLY moving, and which isn't. "Moving" has meaning only as RELATIVE motion between two (or more) objects.

In the standard traveling twin example, the ONLY thing that distinguishes the twins is the fact that one of them never accelerates, and the other one does.

Mike Fontenot

 

[stevmg]

From the point of view of the reference frame (FR) of twin A, who stays at 0,0 in his FR, twin B ages slower than A both going out and coming in and the sum total of B's jouney in time is less than A's total elapsed time - all in A's FR.

B has two FR's, the one going out and the one coming in. If you select either one, you must stick with that one and again the elapsed time for B in that FR is still less than A. That's what JesseM's calculation shows. In other words you now have three FRs to contend with:
A stationary
B staionary going out
B stationary going in
Any one of them will always add up to elapsed time for A > B-out + B-in.

If you were to stick with FR of a stationary A or a FR of stationary B, B would never rejoin and on their journeys each would see the other as time going slower. But, they never rejoin.

This occurs at constant relative velocities of all three FRs with respect to each other. It does not require acceleration and the time change with under acceleration/ deceleration conditions.

 

[Mike_Fontenot]

B has two FR's, the one going out and the one coming in. If you select either one, you must stick with that one [...]

No. B has ONE reference frame. It is the only reference frame that has direct meaning for him. His reference frame ISN'T an inertial frame. But it IS a perfectly valid and useful reference frame, and it's HIS.

If you want to know what the the world is like, from the traveler's perspective, then you'd better know how to do the calculations in HIS frame. Calculate in other frames if you want to know other people's perspectives. That's the real meaning of the term "frame of reference".

Mike Fontenot

 

[stevmg]

No, Mike, he has two. He changes with respect to himself... He's going then coming.

Wait, let's call for a third party, JesseM.

JesseM - can you help us resolve this one?

Mike - I sent JesseM a message to get on line with this one.

 

[DrGreg]

No. B has ONE reference frame. It is the only reference frame that has direct meaning for him. His reference frame ISN'T an inertial frame. But it IS a perfectly valid and useful reference frame, and it's HIS.

Mike Fontenot

No, Mike, he has two. He changes with respect to himself... He's going then coming.

Well, you're both right, depending on how you look at it.

There are two different inertial frames in which B is at rest for the outbound and inbound journeys respectively. In that sense B has two frames.

But it is also the case that there is a single non-inertial frame in which B is at rest throughout. In that sense B has one frame.

So really it depends what you mean by "frame".

I would also comment that non-inertial frames aren't as easy to define as inertial frames: you have some choice in precisely how to define your non-inertial frame, so there is really more than one non-inertial frame in which B is at rest. For simple linear motion in the absence of gravity, as in this example, however, there is a "natural" choice of frame (where frame simultaneity coincides with co-moving inertial simultaneity). But in other cases, such as rotating frames, problems arise if you want to extend a local frame to be shared by multiple rotating observers. But that's another story...

 

[stevmg]

Well, you're both right, depending on how you look at it.

There are two different inertial frames in which B is at rest for the outbound and inbound journeys respectively. In that sense B has two frames.

But it is also the case that there is a single non-inertial frame in which B is at rest throughout. In that sense B has one frame.

So really it depends what you mean by "frame".

I would also comment that non-inertial frames aren't as easy to define as inertial frames: you have some choice in precisely how to define your non-inertial frame, so there is really more than one non-inertial frame in which B is at rest. For simple linear motion in the absence of gravity, as in this example, however, there is a "natural" choice of frame (where frame simultaneity coincides with co-moving inertial simultaneity). But in other cases, such as rotating frames, problems arise if you want to extend a local frame to be shared by multiple rotating observers. But that's another story...

When you go for the proper times then don't you use the inertial frames? That's an inertial frame.

After all, proper time is the time that elapses in an FR in which a subject does NOT move.

Otherwise, this sounds like an answer we get from our politicians (from time immemorial - even dating back past Charles Dickens.)

 

[JesseM]

When you go for the proper times then don't you use the inertial frames? That's an inertial frame.

After all, proper time is the time that elapses in an FR in which a subject does NOT move.

Otherwise, this sounds like an answer we get from our politicians (from time immemorial - even dating back past Charles Dickens.)

Proper time along a worldline is a frame-invariant quantity--you'll get the same answer for the proper time even if you calculate it in a non-inertial frame (this is more clear in general relativity, where the main point of the 'metric' defining the curvature of spacetime is that it allows you to calculate proper time along any worldline...note that for a large region of curved spacetime, no coordinate system covering this region can really be considered 'inertial', in GR you can only talk about local inertial frames in the limit as the size of the region covered by the coordinate system approaches zero so the effects of curvature approach zero too, see here for more info).

I'll take the "politician's" route and say you're both right in a sense, although the fact that you're arguing about this may suggest you both have too narrow a view of what counts as a good answer to the twin paradox. One way of answering the paradox is to just note that you get the same answer in different inertial frames, and that this has to do with the fact that all inertial frames the non-inertial twin has a higher velocity than the inertial one for at least one half of the journey. Another way is to construct a non-inertial frame for the non-inertial twin, and analyze it from that point of view. But any time you talk about a "non-inertial frame" you have to understand that there are an infinite number of different ways to construct a coordinate system where a non-inertial observer is at rest, unlike with inertial frames where there is only one correct way, so any answers you give about how the other twin's clock behaves in "the frame" of the non-inertial twin depends on the details of how you construct that frame. For example, if you choose to construct the non-inertial frame in such a way that judgments about simultaneity in the non-inertial frame always match up with those of the instantaneous inertial rest frame of the traveling twin, then it's true that the clock of the Earth twin will advance very rapidly during the acceleration phase. But you needn't make that assumption! You could equally well construct a non-inertial rest frame for the traveling twin with an odder definition of simultaneity, such that the Earth clock ticks very slowly for most of the trip, including the acceleration phase, but then it advances very rapidly during the tail end of the trip when the traveling twin is coasting inertially. Since there are an infinite number of ways to construct a non-inertial rest frame for the traveling twin and none are any more "correct" than any other, there's no single answer to how the inertial twin's clock behaves from the "point of view" of the traveling twin (unless you're talking purely about visual appearances rather than frame-dependent statements about time dilation).

 

[Mike_Fontenot]

I would also comment that non-inertial frames aren't as easy to define as inertial frames: you have some choice in precisely how to define your non-inertial frame, [...]

No, you actually don't have any other choice, if you want to avoid contradicting the traveler's own (correctly performed) measurements and calculations.

Mike Fontenot

 

[JesseM]

No, you actually don't have any other choice, if you want to avoid contradicting the traveler's own (correctly performed) measurements and calculations.

There is no single physically correct way for a non-inertial observer to "measure" which events on a distant clock are simultaneous with his own clock-readings, that depends on your choice of coordinate system, and you can construct different non-inertial frames where the non-inertial observer is at rest which have different judgments about simultaneity. Likewise, "calculations" require a choice of coordinate system which involves a choice of simultaneity convention. In general relativity all non-inertial coordinate systems are considered equally valid thanks to the principle of diffeomorphism invariance (see this article), and there's no reason to treat any particular non-inertial frames as "preferred" in special relativity either.

 

[jcsd]

No, you actually don't have any other choice, if you want to avoid contradicting the traveler's own (correctly performed) measurements and calculations.

Mike Fontenot

But all measurments are made by the traveller locally and the calculations he makes are dependent on the choice of frame.

His measurements leave us with an obvious way of defining basis vector fields along his worldline (not that we should always take the most obvious route), however for a spatially extended frames there's no single 'winning' procedure for defining basis vector fields over a larger patch of spacetime by taking an arbitary worldline as our starting point.

 

[Passionflower]

Since there are an infinite number of ways to construct a non-inertial rest frame for the traveling twin and none are any more "correct" than any other, there's no single answer to how the inertial twin's clock behaves from the "point of view" of the traveling twin (unless you're talking purely about visual appearances rather than frame-dependent statements about time dilation).

Some comments here since people may get the wrong impression.

For 'practical' purposes, these "purely visual" e.g. radar and light beam methods are of prime importance. What is 'calculated' to be at the same time has no physical significance whatsoever. It makes good exercises for a test but apart from that it is totally useless. On the other hand we can make interesting and useful exercises using the twin "paradox" for instance by calculating time dilation from radar info, speed or acceleration from radio signals, or acceleration from trip duration etc.

There is also another option that a traveling twin could take and that is to take an inertial clock with him that measures the time back home. By measuring each acceleration (including direction) and duration one could construct such a clock and see 'what time' it is back home.

 

[stevmg]

Twin B (the guy who goes away and comes back) always is younger than twin A though...

Now, I know if you look at twin B as a reference frame and the universe revolves around him, twin A does some weird moving but twin A is still going to be older.

 

[matheinste]

There is also another option that a traveling twin could take and that is to take an inertial clock with him that measures the time back home. By measuring each acceleration (including direction) and duration one could construct such a clock and see 'what time' it is back home.

By an inertial clock do you mean a clock moving at constant velocity, because the traveling clock does not so move at all points of the journey. But I think you mean a clock adjusted continiuosly so as to give the same readings as the home twin. That sounds plausible and Rindler proposes something along those lines. But this is not necessary as the "the time" back home, as reckoned by the traveller can be calculated by the traveller, as it can be for any other observer. This value at any time on the travellers clock is effectively calculated from the traveller's line/hyperplane of simultaneity.

I suspect you already know all this anyway, but it may be of benefit to others.

Matheinste.

 

[Passionflower]

By an inertial clock do you mean a clock moving at constant velocity, because the traveling clock does not so move at all points of the journey. But I think you mean a clock adjusted continiuosly so as to give the same readings as the home twin. That sounds plausible and Rindler proposes something along those lines.

Indeed, a clock that includes an accelerometer and gyroscopes, which adjusts its clock rate each time the spaceship undergoes proper acceleration in any direction.

But this is not necessary as the "the time" back home, as reckoned by the traveller can be calculated by the traveler, as it can be for any other observer.

Well the clock would instantly calculate it. Out of curiosity how would you calculate it if not using the acceleration and gyroscope information?

Alternatively one could use radio beacons, a sort of GPS in space, but for far away trips this is not practical.

Also such a device could be useful for reading space charts as one could calculate chart coordinate speed, coordinate direction and coordinate time. When a spaceship accelerates a few times in a few different directions things get kind of 'twisted' (Thomas precession), think of it as a futuristic compass.

Such a clock would work great in flat spacetime but as we know there is gravitation so we would still have to make adjustments for gravitation.

 

[matheinste]

Indeed, a clock that includes an accelerometer and gyroscopes, which adjusts its clock rate each time the spaceship undergoes proper acceleration in any direction.


Well the clock would instantly calculate it. Out of curiosity how would you calculate it if not using the acceleration and gyroscope information?

Alternatively one could use radio beacons, a sort of GPS in space, but for far away trips this is not practical.

Also such a device could be useful for reading space charts as one could calculate chart coordinate speed, coordinate direction and coordinate time. When a spaceship accelerates a few times in a few different directions things get kind of 'twisted' (Thomas precession), think of it as a futuristic compass.

Such a clock would work great in flat spacetime but as we know there is gravitation so we would still have to make adjustments for gravitation.

To calculte you only need to know your velocity relative to your starting place, earth, but to know this I suppose you need to know your acceleration history. As was pointed out to me in ealier thread, a realizeable physical clock has finite dimension and unless its acceleration profile is of a certain type not all its "parts" have the same velocity during acceleration and so do not share the same line/hypersurface of simultaneity. Gravity, I suppose, would also pose the same problem for a real clock. But I think these things are really just interesting points and not relevant to the present discussion where, I assume, ideal clocks are proposed.

Matheinste.

 

[Passionflower]

But I think these things are really just interesting points and not relevant to the present discussion where, I assume, ideal clocks are proposed.

I think such a clock is very relevant. Can you determine what such a clock, inflight, at any instant of time would measure?

Such a clock would measure the time of the home stayer's clock if he would have traveled to the spaceship's position on a geodesic.

Many of the peculiarities caused by using planes of simultaneity disappear with such an approach.

 

[Janus]

Out of curiosity how would you calculate it if not using the acceleration and gyroscope information?

Doppler shift comes to mind.

 

[Passionflower]

Doppler shift comes to mind.

Yes Doppler shift is an option but certainly not easier than calculating rest time using an accelerometer and 3 gyroscopes in case of consecutive accelerations in arbitrary directions.

 

[JesseM]

I think such a clock is very relevant. Can you determine what such a clock, inflight, at any instant of time would measure?

Such a clock would measure the time of the home stayer's clock if he would have traveled to the spaceship's position on a geodesic.

Many of the peculiarities caused by using planes of simultaneity disappear with such an approach.

I thought the idea of the clock was that it would figure out its velocity relative to the home twin based on its past accelerations, and adjust its rate so that it would tick at the same rate as the home twin's clock, as measured in the home twin's rest frame. In this case the clock wouldn't measure what the home twin's clock would read if he had traveled on a geodesic to meet the traveling twin, instead it would just measure what the home twin's clock does read at any given moment according to the definition of simultaneity used in the home twin's frame. But if that's not the idea you were describing, then what did you mean by "adjusts its clock rate" in the statement "Indeed, a clock that includes an accelerometer and gyroscopes, which adjusts its clock rate each time the spaceship undergoes proper acceleration in any direction". Can you give a formula or descriptive rule for how it adjust its rate based on measurements of G-forces?

 

[Passionflower]

I thought the idea of the clock was that it would figure out its velocity relative to the home twin based on its past accelerations, and adjust its rate so that it would tick at the same rate as the home twin's clock, as measured in the home twin's rest frame. In this case the clock wouldn't measure what the home twin's clock would read if he had traveled on a geodesic to meet the traveling twin, instead it would just measure what the home twin's clock does read at any given moment according to the definition of simultaneity used in the home twin's frame. But if that's not the idea you were describing, then what did you mean by "adjusts its clock rate" in the statement "Indeed, a clock that includes an accelerometer and gyroscopes, which adjusts its clock rate each time the spaceship undergoes proper acceleration in any direction". Can you give a formula or descriptive rule for how it adjust its rate based on measurements of G-forces?

It is the same thing but perhaps I was not very clear in my language. Clock rates can only be compared at the same event, the inertial clock will read exactly the time of the home clock as soon as they meet. In flight however the clock will give the time of the home clock as if it moved on a geodesic to the time reading event.

See for an overview Minguzzi: http://arxiv.org/abs/physics/0411233

 

[starthaus]

It is the same thing but perhaps I was not very clear in my language. Clock rates can only be compared at the same event, the inertial clock will read exactly the time of the home clock as soon as they meet. In flight however the clock will give the time of the home clock as if it moved on a geodesic to the time reading event.

See for an overview Minguzzi: http://arxiv.org/abs/physics/0411233

You like the same paper as I do :-)

 

[Austin0]

Indeed, a clock that includes an accelerometer and gyroscopes, which adjusts its clock rate each time the spaceship undergoes proper acceleration in any direction.


Well the clock would instantly calculate it. Out of curiosity how would you calculate it if not using the acceleration and gyroscope information?

Alternatively one could use radio beacons, a sort of GPS in space, but for far away trips this is not practical.

Also such a device could be useful for reading space charts as one could calculate chart coordinate speed, coordinate direction and coordinate time. When a spaceship accelerates a few times in a few different directions things get kind of 'twisted' (Thomas precession), think of it as a futuristic compass.

Such a clock would work great in flat spacetime but as we know there is gravitation so we would still have to make adjustments for gravitation.

How do you decide which way to recalibrate the clock rate as a function of accelration as determinded by the accelerometer?? I.e. DO you increase the rate with positive acceleration only, so it decreases when thrust is in the opposite direction or do you increase it as long as there is acceleration in any direction?

 

[Passionflower]

You like the same paper as I do :-)

This one http://cdsweb.cern.ch/record/999103/files/0611076.pdf?version=1 is interesting too.

 

[Passionflower]

do you increase it as long as there is acceleration in any direction?

No.

Basically when you accelerate with respect to the home clock the inertial clock rate would have to increase to compensate for the natural time dilation and when you decelerate with respect to the home clock the rate would decrease. As soon as you introduce angles the situation becomes increasingly more complicated (first 2D and finally 3D), just as complicated as general 3D velocity additions but then 'backwards'.

 

[JesseM]

It is the same thing but perhaps I was not very clear in my language. Clock rates can only be compared at the same event, the inertial clock will read exactly the time of the home clock as soon as they meet. In flight however the clock will give the time of the home clock as if it moved on a geodesic to the time reading event.

But that's different from the case I was describing where the moving clock adjusts so that it is always synchronized with the home (inertial) clock in the home clock's inertial rest frame, agreed?

See for an overview Minguzzi: http://arxiv.org/abs/physics/0411233

OK, I'll check it out.

 

[starthaus]

This one http://cdsweb.cern.ch/record/999103/files/0611076.pdf?version=1 is interesting too.

Yes, thank you.

 

[Passionflower]

But that's different from the case I was describing where the moving clock adjusts so that it is always synchronized with the home (inertial) clock in the home clock's inertial rest frame, agreed?

Yes, because then you need planes, or heaven forbid, curved surfaces of simultaneity with all the oddities that are discussed in his topic. You call it 'always synchronized' but as you likely know the results are next to bizarre and I much prefer the above described method.

 

[Austin0]

No.

Basically when you accelerate with respect to the home clock the inertial clock rate would have to increase to compensate for the natural time dilation and when you decelerate with respect to the home clock the rate would decrease. As soon as you introduce angles the situation becomes increasingly more complicated (first 2D and finally 3D), just as complicated as general 3D velocity additions but then 'backwards'.

Well as per what you just described then the "home clock" rate would be decreasing from the time of the initiiation of reverse acceleration and would be decreased though the whole homeward inertial phase as the accelerometer would not distinguish between deccelration and what would be increasing relative velocity after the at rest point. Within the accelerating frame how would observers be able to tell at what point they were at rest with home and had started on the return trip to increase the home clock rate?

 

[Passionflower]

Well as per what you just described then the "home clock" rate would be decreasing from the time of the initiiation of reverse acceleration and would be decreased though the whole homeward inertial phase as the accelerometer would not distinguish between deccelration and what would be increasing relative velocity after the at rest point. Within the accelerating frame how would observers be able to tell at what point they were at rest with home and had started on the return trip to increase the home clock rate?

Sorry but I really cannot follow what you are saying.

 

[JesseM]

Yes, because then you need planes, or heaven forbid, curved surfaces of simultaneity with all the oddities that are discussed in his topic. You call it 'always synchronized' but as you likely know the results are next to bizarre and I much prefer the above described method.

What's bizarre about the results? This is how GPS clocks work after all (they adjust to maintain a constant rate of ticking in an Earth-centered coordinate system, even as their velocities/heights are changing in this coordinate system), and it would be even simpler if you just wanted some clocks to adjust to maintain a constant rate of ticking in an inertial frame in flat spacetime.

One oddity about your method which occurred to me is that if one twin travels away from the home twin, accelerates to turn around, and and then travels back to the home twin inertially, then during the inertial return trip the adjusted clock will actually keep ticking faster and faster relative to a "normal" clock carried by the traveling twin (one which measures proper time), the ratio between the tick rates of the two clocks won't remain constant despite the fact that no acceleration is happening in this leg. For example, if the traveling twin goes out at 0.8c for 10 years in the home twin's frame, accelerates briefly and then returns at 0.8c for another 10 years, then at the time of acceleration at t=10 in the home frame the adjusted clock should read about 6 years, then at the time the twins reunite at t=20 the adjusted clock should read 20 years, but halfway through the return leg at t=15, the traveling twin will be at a distance of 4 light-years from the home twin so if the home twin took a geodesic path to that point the home twin's velocity would have been (4/15)c, so the time on that geodesic path would be 15*sqrt(1 - (4/15)^2) = 14.46 years, more than halfway between 6 and 20.

 

[Passionflower]

What's bizarre about the results? This is how GPS clocks work after all (they adjust to maintain a constant rate of ticking in an Earth-centered coordinate system, even as their velocities/heights are changing in this coordinate system), and it would be even simpler if you just wanted some clocks to adjust to maintain a constant rate of ticking in an inertial frame in flat spacetime.

Are you at all following what is discussed in this topic?

 

[stevmg]

All right - back to basics.

Twin A is the guy who stays at home

Twin B is the guy who takes a trip and comes back

If the twins are girls or one is a boy and the other is a girl or vice versa this same still holds. Forget about the Turner syndrome female and normal male "identical" twins I once referred to.

If twin B moves at 0.6c both away and back, the interval he experiences is 20% less than the interval experienced by twin A.

Am I right?

 

[Mentz114]

All right - back to basics.

Twin A is the guy who stays at home

Twin B is the guy who takes a trip and comes back

If the twins are girls or one is a boy and the other is a girl or vice versa this same still holds. Forget about the Turner syndrome female and normal male "identical" twins I once referred to.

If twin B moves at 0.6c both away and back, the interval he experiences is 20% less than the interval experienced by twin A.

Am I right?

Yes. I get 14 s and 11.2 s. Or, 1 - (0.6)² = 0.64, sqrt(0.64) = 0.80.