How is twin paradox resolved in case of no/zero acceleration?

[San K]

how is twin paradox resolved in case of no/zero acceleration/deceleration?

two twins one on Earth and other on a ship moving at constant velocity (say 80% speed of light).

the other twin does not turn around ..thus no acceleration/deceleration.

how we resolve the paradox now?

twin A will say twin B is moving..thus he/she should age slower
twin B will say twin A is moving...thus he/she should age slower

the paradox is traditionally resolved by the fact that one of them is accelerating/decelerating...because one of the twins has to turn around...etc

but in the case above (and we can build more scenarios) where there is only constant velocity...no acceleration...no turning around...

no meeting of the twins...

 

[tiny-tim]

Hi San K!

no meeting of the twins...

no meeting of the twins, no paradox …

they have to start and finish together (or at least at the same velocity) to be able to compare ages at the same time

(if they have different velocities, two events at the same time for one will not be at the same time for the other)

 

[jtbell]

two twins one on Earth and other on a ship moving at constant velocity (say 80% speed of light).

the other twin does not turn around ..thus no acceleration/deceleration.

how we resolve the paradox now?

For a numerical example of how this works, see the post linked below, and the post that it in turn links to:

https://www.physicsforums.com/showpost.php?p=3394389&postcount=8

 

[Filip Larsen]

You could make a special non-accelerated version of the paradox by using 3 clocks (A, B, C) instead of 2 twins. Clock A stays behind representing the stationary twin. The two other clocks are are moving with a constant speed in opposite direction of each other such that clock B first passes A and then C, and clock C such that it first passes clock B and then A. We then have three separate space-time events, let's call them AB, BC, and CA, where the clocks meet each other in pairs. In event AB we synchronize both A and B so they start with time zero, at BC we set clock C to same time as clock B, and at CA we just write down the value of clock A and C so we can compare them. Note, that we can compare the clock values of two clocks even though they move relative to each other because we allow them to "pass through" each other.

Calculating the relativistic distance between each successive pair of event you will end up with the expected result, namely that the proper time for clock A from event AB to CA will be larger than the proper time for clock B from AB to BC plus proper time for clock C from BC to CA.

 

[PAllen]

You could make a special non-accelerated version of the paradox by using 3 clocks (A, B, C) instead of 2 twins. Clock A stays behind representing the stationary twin. The two other clocks are are moving with a constant speed in opposite direction of each other such that clock B first passes A and then C, and clock C such that it first passes clock B and then A. We then have three separate space-time events, let's call them AB, BC, and CA, where the clocks meet each other in pairs. In event AB we synchronize both A and B so they start with time zero, at BC we set clock C to same time as clock B, and at CA we just write down the value of clock A and C so we can compare them. Note, that we can compare the clock values of two clocks even though they move relative to each other because we allow them to "pass through" each other.

Calculating the relativistic distance between each successive pair of event you will end up with the expected result, namely that the proper time for clock A from event AB to CA will be larger than the proper time for clock B from AB to BC plus proper time for clock C from BC to CA.

Well, you still have a non-inertial path through spacetime being compared to an inertial path, both paths beginning and ending on the same pair of events. In SR, that is the fundamental requirement.

 

[Hurkyl]

You could make a special non-accelerated version of the paradox by using 3 clocks (A, B, C) instead of 2 twins.

Why people fall for the triplet paradox always boggled me.

I can understand why people find "I see your clock going slower than mine, but you see my clock going slower than yours" paradoxical -- they haven't really grasped that the two facts aren't really comparable in the same way as if our clocks are sitting side by side. They haven't internalized that "I see" and "you see" are important.

I can understand why people find the twin paradox paradoxical. The amount of aging is now comparable, but they haven't internalized that "inertial travel" is a prerequisite for the qualitative fact that you see all^* moving clocks running slower than yours.

But people that insist on the triplet paradox being paradoxical baffle me. As far as I can tell, they're so focused on defeating the usual resolution of the twin paradox that they simply don't notice they've also invalidated the (faulty) argument that leads to paradox.



*: Well, I mean properly calibrated clocks

 

[Filip Larsen]

Why people fall for the triplet paradox always boggled me.

I'm not sure what you mean?

I like this formulation of the paradox because it captures the essence of the twin paradox while still allowing only algebraic calculations using relativistic intervals from special relativity. Sure, you can also make an accelerated version but that requires integrals and is likely to have a more limited "audience" when you want to explain what is going in layman's terms.

 

[Hurkyl]

I'm not sure what you mean?

I like this formulation of the paradox because it captures the essence of the twin paradox

I mean the essence of the twin paradox is gone! Vanished! No longer there!

You keep the closed loop, but I can't see anything that remains that someone might find paradoxical.

 

[Filip Larsen]

I mean the essence of the twin paradox is gone! Vanished! No longer there!

You keep the closed loop, but I can't see anything that remains that someone might find paradoxical.

The only thing this version does is to "remove" the accelerating phase. You can imagine that the two clocks B and C representing the traveling twin with the accelerating turn-around phase left out, like if the twin brought a clock with him which he pauses just before he starts to accelerate back towards his brother back home and unpauses just when he stops accelerating; then clock B corresponds to the twin's clock on the outbound coast and clock C corresponds to his clock in the inbound coast.

You could also look at the non-acceleration scenario as a limit when acceleration of the twin goes to infinity. Assuming he can turn around using arbitrarily high acceleration, he can bring the time he spend doing the turn around arbitrarily close to zero. In the limit the twin will use zero proper time and distance to turn around and his clock will read like in the non-accelerated scenario.

To me, the non-accelerated version sound just as paradoxical as the accelerated version, especially since I think the paradox (the apparent contradiction that people feel must be present when they hear about time dilation) stems from the symmetry in stating that you will observe anyone moving relative to you as having time dilation. That two people can observe each others time as running slower do sound paradoxical until you start describing the full scenario that allow you to properly compare clocks.

 

[Hurkyl]

The only thing this version does is to "remove" the accelerating phase.

It does remove something else significant -- the space-bound observer.

Notice that your post continues by considering the ordinary two-clock twin paradox, rather than discussing a three-clock scenario.

accelerating turn-around phase left out,

the non-accelerated version

You didn't leave out the turn-around phase, you just paused the traveling twin's clock. There is still acceleration.

 

[Filip Larsen]

It does remove something else significant -- the space-bound observer.

No, it replaces one observer (the traveling twin) with two observers (clock B and C). And, as I try to argue, you can even think of the two clocks as being "world line segments" of the same clock.

When looking at the time it takes to make the trip back and fourth the only part that is removed in the non-accelerated version is, well, the acceleration.

Notice that your post continues by considering the ordinary two-clock twin paradox, rather than discussing a three-clock scenario. You didn't leave out the turn-around phase, you just paused the traveling twin's clock. There is still acceleration.

In my reply to your earlier comment I outline how you can go from the ordinary twin paradox with an accelerated turn around to a simplified three-clock model with no acceleration that can be analyzed with relativistic intervals only and still get a "resolution" to the paradox. That is why I mention acceleration.

I am not sure what you think I am claiming. I am not claiming that you can make a single twin turn around without acceleration, or that the two scenarios will be identical. I merely say you can make a simplified version without acceleration and end up with a scenario where it is very easy to calculate the resulting difference in proper time between the various clocks and that this scenario is interesting as it can be thought of as an approach the accelerated scenario in the limit where the time used to accelerate becomes insignificant compared to the time used coasting.

You seem a little bend on marking the three-clock scenario as being "invalid" or "bad" in some sense. Does it accurately describe a twin rocketing around in a real rocket? No, it doesn't. Does it offer a resolution to the paradox of having symmetric time dilation? Yes, I believe it does.

 

[Hurkyl]

No, it replaces one observer (the traveling twin) with two observers (clock B and C). And, as I try to argue, you can even think of the two clocks as being "world line segments" of the same clock.

When looking at the time it takes to make the trip back and fourth the only part that is removed in the non-accelerated version is, well, the acceleration.

In the three observer version, there is no round trip. If there is no round trip, the argument of the twin paradox is inapplicable. Auggie, the out-bound observer cannot compare his clock with Terra the terrestrial observer at the start end end of his journey. Neither can Indy, the in-bound observer. Only Stella the space-bound observer who hitches a ride with Auggie then switches over to Indy's ship as it passes by is able to compare clocks with Terra.

If there is a round trip, there is acceleration. (be it smooth and prolonged, or infinite and instantaneous.

 

[Filip Larsen]

In the three observer version, there is no round trip. If there is no round trip, the argument of the twin paradox is inapplicable.

There is the "round trip" of proper time intervals, which (to me) is all that matters. You can make an accelerated scenario, cut out the acceleration to get a non-accelerated version and then still have an explanation through calculations as to why one twin will be older than the other. Makes perfect sense to me.

But I guess you don't consider the resolution of the paradox using Minkowski diagrams [1] valid either then?


[1] http://en.wikipedia.org/wiki/Twin_paradox#Resolution_of_the_paradox_in_special_relativity

 

[lovetruth]

Hi San K!


no meeting of the twins, no paradox …

they have to start and finish together (or at least at the same velocity) to be able to compare ages at the same time

(if they have different velocities, two events at the same time for one will not be at the same time for the other)

I don't see any reason why twins have to start and finish together or why the twins should meet at all? The paradox still remains as both twins observe time dilation in other frame. Both twins are also in inertial frames and thus SR must then hold.

I was going to post the same question in PF but I am glad that someone else thought like me. In the classic explanation of twin paradox, this argument is swept under the carpet by assuming that one twin must accelerate and decelerate. Since we don't know what an accelerating and decelerating observer will observe or God forbid some mind-numbing GR tensor equations are brought into consideration, a simpleton is browbeaten into believing that the twin in the rocket will observe the same as the twin on earth. I think this twin-paradox problem is very important as it raises the question of internal consistency of SR.

 

[pervect]

I think Hurkyl is inviting you to consider the "triangle paradox". The "triangle paradox", usually called the triangle inequality, basically says that if you sum two sides of a triangle, the result will be greater than the third side - and equal only in very special cases. It does with distances what the twin paradox does with times, with one important difference. In the "triangle paradox" aka the triangle inequality, the sum of the distances is greater than the straight-line path, in the "twin paradox", the sum of the times is lesser than the straight line path.

The fact that the sum is lowest is a bit weird, compared to euclidean geometry, but the point is that if you believe that Euclidean geometry is self-consistent, it's not that hard to picture space-time as having a similar, consistent geometry. The only thing that really has to go is the notion of observer-independent simultaneity.

 

[San K]

i quote two that i think i agree with...so i question it further to understand it better...

Hi San K! no meeting of the twins, no paradox …

they have to start and finish together (or at least at the same velocity) to be able to compare ages at the same time

(if they have different velocities, two events at the same time for one will not be at the same time for the other)

hi tiny (but mighty?) tim...:)

you mean one cannot compare because they are not comparable frames of reference?

that have to be at same velocity and place? can put it more precisely?

I can understand why people find "I see your clock going slower than mine, but you see my clock going slower than yours" paradoxical -- they haven't really grasped that the two facts aren't really comparable in the same way as if our clocks are sitting side by side. They haven't internalized that "I see" and "you see" are important.

I can understand why people find the twin paradox paradoxical. The amount of aging is now comparable, but they haven't internalized that "inertial travel" is a prerequisite for the qualitative fact that you see all^* moving clocks running slower than yours.

why are they not comparable?

even if they are not comparable, don't we still have to resolve the issue about...

one twin saying that the other is moving...so he is ageing faster, not me...
and the other twin saying the same thing

or are both the twins right?...till they meet...and it depends who ends up visiting whom at what speed/place/time etc...

i mean..how do we resolve...who is moving? i guess the answer is...we cannot say...who is moving...

unless and until...we get them in the same frame of reference...
is the below statement correct? if not...then modify it to make it correct

you cannot compare any arbitrary frames of references.

..to compare any two events...they have to be in same frame of reference...or "normalized/adjust/accounted-for" to be in the same frame of reference...is that ...why we say...space-time

being at different space/speed...can mean different time?

space and time...are...in a sense...interchangeable?

 

[San K]

The amount of aging is now comparable, but they haven't internalized that "inertial travel" is a prerequisite for the qualitative fact that you see all^* moving clocks running slower than yours.

hi hurkyl, can you write a few more sentences that can help folks internalize "inertial travel"?

 

[Dale]

"Inertial travel" means that you are traveling in a manner such that an accelerometer attached to you reads 0 at all times. If you aren't traveling inertially then "your frame" is non-inertial and so the "usual" physics equations are not correct. Once you have realized that then it is clear that there is no twin paradox.

 

[San K]

case where: one twin moving at constant velocity and the other stationery

the twin that is moving will have velocity and hence momentum...

however you cannot tell between two twin who is moving...relative to the other twin....

when we have only two frames of references...

however if you have a third (or a forth or fifth) reference point, you could tell...

is the above correct?

 

[San K]

"Inertial travel" means that you are traveling in a manner such that an accelerometer attached to you reads 0 at all times. If you aren't traveling inertially then "your frame" is non-inertial and so the "usual" physics equations are not correct. Once you have realized that then it is clear that there is no twin paradox.

hi dale

in the example i initially gave...both the frames would be inertial frames...then

is there no twin paradox because the stationary inertial frame (for twin A) is not comparable with the constant velocity inertial frame (for twin B)?...because they can never meet?

note: acceleration in both cases is zero because one is stationary and other is moving at constant velocity

 

[Hurkyl]

There is the "round trip" of proper time intervals, which (to me) is all that matters. You can make an accelerated scenario, cut out the acceleration to get a non-accelerated version and then still have an explanation through calculations as to why one twin will be older than the other. Makes perfect sense to me.

There are two problems:

• This isn't a resolution of the twin paradox -- you've changed the scenario to eliminate a key component to the (flawed) derivation of paradox.
• Some people find the triplet paradox paradoxical, and you've claimed it's equally paradoxical to the twin paradox. Tell me, what is paradoxical?

You seem to have totally and completely missed the idea of what it means to be a "paradox". The problem is not that people are asking "How do I compute that one twin ages less than the other" -- the problem is that people are asking "I've done the computation two different ways and have contradictory answers!"

 

[Dale]

in the example i initially gave...both the frames would be inertial frames...then

Correct.

is there no twin paradox because the stationary inertial frame (for twin A) is not comparable with the constant velocity inertial frame (for twin B)?

There is no paradox because everything is symmetric. There is nothing to distinguish A from B, and so both A and B make symmetric measurements regarding themselves and each other.

 

[Dale]

case where: one twin moving at constant velocity and the other stationery

the twin that is moving will have velocity and hence momentum...

however you cannot tell between two twin who is moving...relative to the other twin....

when we have only two frames of references...

however if you have a third (or a forth or fifth) reference point, you could tell...

is the above correct?

The number of reference frames is not relevant in any way. Velocity is relative regardless.

 

[Filip Larsen]

Tell me, what is paradoxical?[/list]

You seem to have totally and completely missed the idea of what it means to be a "paradox". The problem is not that people are asking "How do I compute that one twin ages less than the other" -- the problem is that people are asking "I've done the computation two different ways and have contradictory answers!"

My idea of paradox in this context is exactly as it explained in the Wikipedia page [1] that I referred to in my previous post regarding explaining the paradox using a Minkowski diagram (a question you ignored, by the way). Let me quote two passages from it that sums it up for me:

In physics, the twin paradox is a thought experiment in special relativity, in which a twin makes a journey into space in a high-speed rocket and returns home to find he has aged less than his identical twin who stayed on Earth. This result appears puzzling because each twin sees the other twin as traveling, and so, according to a naive application of time dilation, each should paradoxically find the other to have aged more slowly.

and

The paradox centers around the contention that, in relativity, either twin could regard the other as the traveler, in which case each should find the other younger—a logical contradiction. This contention assumes that the twins' situations are symmetrical and interchangeable, an assumption that is not correct.

If you use some other "definition" on what the Twin Paradox is all about, then we have to disagree. In the sense described by the above quotes, I still claim the three-clock scenario brings a resolution to the paradox. The paradox (the apparent symmetry) is resolved by noting that the situation is not symmetric after all, not even if you leave out acceleration, and that you can easily calculate the difference in proper time intervals of the clocks using simple algebraic equations.

There are two problems:

• This isn't a resolution of the twin paradox -- you've changed the scenario to eliminate a key component to the (flawed) derivation of paradox.
• Some people find the triplet paradox paradoxical, and you've claimed it's equally paradoxical to the twin paradox.

I do not and have not claimed the two scenarios to be (physical) equal. What I have been claiming, though, is that is that with regard to the calculation of proper time intervals for the two twins you can consider the three-clock scenario a limiting case of the accelerated scenario where the acceleration is arbitrarily high. Do you disagree with that?

If you still want to discuss this outside your assumption that I have "totally and completely missed the idea" of what this is about, then it would also be nice to know if you consider the explanation using Minkowski diagram a valid resolution of the paradox?


[1] http://en.wikipedia.org/wiki/Twin_paradox

 

[lovetruth]

case where: one twin moving at constant velocity and the other stationery

the twin that is moving will have velocity and hence momentum...

however you cannot tell between two twin who is moving...relative to the other twin....

when we have only two frames of references...

however if you have a third (or a forth or fifth) reference point, you could tell...

is the above correct?

Yes you are correct.
The paradox still remains: Both twins see that they age faster than their counterpart.

 

[Dale]

Yes you are correct.
The paradox still remains: Both twins see that they age faster than their counterpart.

But that is not a paradox. They are symmetric so you would expect them to see each other's clocks symmetrically. Any other result would be paradoxical.

 

[lovetruth]

There is no paradox because everything is symmetric. There is nothing to distinguish A from B, and so both A and B make symmetric measurements regarding themselves and each other.

The symmetry is broken because the twins are in different inertial frame. Each twin should observe time dilation in others frame. I know this result is counter-intuitive and paradoxial but this is what relativity says.

 

[Dale]

The symmetry is broken because the twins are in different inertial frame.

Different inertial frames are indistinguishable, so that is symmetrical also. I think you have a misunderstanding of what makes a situation symmetrical or not.

 

[lovetruth]

But that is not a paradox. They are symmetric so you would expect them to see each other's clocks symmetrically. Any other result would be paradoxical.

What about time dilation due to relative velocity of the observers/twins.

 

[Dale]

What about time dilation due to relative velocity of the observers/twins.

That is completely symmetrical, they each observe the other to be dilated by the same amount.

 

[lovetruth]

Different inertial frames are indistinguishable, so that is symmetrical also

Different inertial frames are distinguishable because they are moving at different speed. If inertial frames were indistinguishable, they must be equivalent and not relative. But we don't have Theory of Equivalency instead of Theory of Relativity.

 

[lovetruth]

That is completely symmetrical, they each observe the other to be dilated by the same amount.

If each twin observes other twin's time to be dilated, how can their time be same?

 

[jtbell]

Follow the link in post #3 of this thread.

 

[tiny-tim]

Hi San K!

no meeting of the twins, no paradox …

they have to start and finish together (or at least at the same velocity) to be able to compare ages at the same time

(if they have different velocities, two events at the same time for one will not be at the same time for the other)

hi tiny (but mighty?) tim...:)

you mean one cannot compare because they are not comparable frames of reference?

that have to be at same velocity and place? can put it more precisely?

people who are further away look smaller

if A is a long way from B, then A says "B looks smaller", and B says "A looks smaller" …

where's the paradox??​

there would only be a paradox if A was standing next to B (or if they were moving past each other, but they make their measurements at the exact moment they are passing) …

then it is paradoxical for each to regard the other as shorter

(of course, that paradox does exist … it's called the Lorentz-Fitzgerald contraction! )

similarly, there's no clock paradox unless the two clocks start at the same velocity, and finish at the same velocity (position doesn't matter)

 

[Hurkyl]

The paradox (the apparent symmetry) is resolved by noting that the situation is not symmetric after all, not even if you leave out acceleration, and that you can easily calculate the difference in proper time intervals of the clocks using simple algebraic equations.

You cannot resolve a paradox by justifying one half of the contradiction. You have to explain why the other half of the contradiction is invalid.

A resolution of the twin paradox has to explain what is wrong with the argument

Stella sees that Terra is moving with respect to her, and thus her clock running slowly for the entire trip. Therefore, when Stella returns home, she finds that Terra's clock reads less time than Stella's clock does​

Showing an asymmetry does not demonstrate a flaw with the above argument. It simply defeats the supplementary argument that Terra and Stella's points of view are indistinguishable.

One can hope that demonstrating the asymmetry, or that providing a correct way to compute things, could prompt the confused person into resolving the the paradox himself. But you also risk the rather dangerous possibility of pushing the person into accepting doublethink -- to accept the resolution despite retaining the belief that there is nothing actually wrong with the argument above.