What is the effect of the twin paradoxon?

[Ibix]

One can simplify this example of the different travel times of twins considerably and reduce it to the core of what is happening.

Unfortunately, your simpler example misses the point, which was that you can have identical accelerations and different aging. This strongly suggests that acceleration is not really relevant to differential aging. In fact, it's just one way to make the teins meet twice so that their clocks can be directly compared.

After the meeting of A and B, the display of A's clock lags behind that of B's passing clocks.

Maybe. That would depend on facts you haven't specfied about where A and B were when their clocks were zeroed.

without acceleration (without this jump to the passing clock of B) no "twin paradox" is possible.

This statement is not correct. It is only true in flat spacetime with a trivial topology, and the fact that it's not true in general tells you that acceleration is not key to the differential aging. It's only the means you use to allow the twins to meet twice.

 

[Dale]

That is, in a Minkowski spacetime with trivial topology, one cannot have the former without the latter. What physically distinguishes causation from correlation in this context?

The real universe is not well described by Minkowski spacetime. In very ordinary curved spacetimes you can easily get different ages without any proper acceleration. You can also get proper acceleration without different ages. So proper acceleration is neither necessary nor sufficient for different ages.

 

[Sagittarius A-Star]

If you meant that a traveler shouts out the display of his watch to an oncoming traveler as he passes, that would be relativistic bookkeeping, but not physics.

A usual clock contains an oscillator, a counter and a display. What the counter does, is bookkeeping. There is no problem, to divide the work between two clocks.

 

[Dale]

A usual clock contains an oscillator, a counter and a display. What the counter does, is bookkeeping. There is no problem, to divide the work between two clocks.

Also, it is entirely physically possible for the two clocks to simply happen to read the same as they pass. Regardless how those clocks come to read the same, the fact remains that the time at the final comparison is smaller.

There is asymmetry, but no acceleration in this scenario. All the acceleration ever did was make the twins asymmetric. Other asymmetries suffice, not just acceleration.

 

[Peter Strohmayer]

Unfortunately, your simpler example misses the point, which was that you can have identical accelerations and different aging.

This is not correct: in #53 I described two identical accelerations (jumps to the passing encounter clock).

In fact, it's just one way to make the teins meet twice so that their clocks can be directly compared.

I have simplified the example: it is enough for one twin to switch to the other's frame of reference once to make him age less.

you haven't specfied about where A and B were when their clocks were zeroed

I find this objection too sophistical. In all these examples, the clocks are set to zero when the twins meet. I described this in #20.

It [i.e. "without acceleration (without this jump to the passing clock of B) no "twin paradox" is possible"] is only true in flat spacetime with a trivial topology, and the fact that it's not true in general tells you that acceleration is not key to the differential aging. It's only the means you use to allow the twins to meet twice.

See above: It does not matter if the twins meet twice.

My statement applies to the acceleration due to recoil. I never claimed that this acceleration would play a role in slower aging in a gravitational field. It does not follow from a denial of this claim that recoil acceleration would not play a role even in a gravitationally-free field.

Rather, the question shifts to whether recoil acceleration and gravitational acceleration affect an observer's world line in a similar way. Can a heavy twin be equated with a recoil-accelerated twin? I do not feel called upon to answer this question. In any case, the answer does not change the dependence of the twin paradox on the recoil acceleration in a gravitationally free region.

 

[Dale]

I have simplified the example: it is enough for one twin to switch to the other's frame of reference once to make him age less

You are simply using the acceleration to have them start agreeing on simultaneity, which is always a matter of convention anyway. Even under that approach, it is during the inertial motion that the disagreement with the other’s clocks begins and that disagreement accumulates the longer the inertial motion continues.

However, it is just as easy for an observer to use a frame in which they are not at rest. Both observers could choose to use any inertial frame they agree on.

They could both use the frame of the other observer, there the one twin’s slow aging occurs on the first leg and the acceleration stops the one twin’s slow aging, with the one twin being younger.

They could both use the inertial frame where the one twin is initially at rest. In that frame the other twin is aging more slowly. The acceleration of the one twin makes the one twin age more slowly too. But in this frame the other twin is the younger.

They could also use an inertial frame where both twins are moving, with equal and opposite speed. Then both twins are at all times the same age, both before and after the acceleration.

It does not matter if the twins meet twice

It really does matter. As described above, if they do not meet twice then the final state is frame dependent. Any answer becomes possible.

 

[Dale]

The symmetry can only be broken by acceleration.

I think this may be your key lack of understanding. It is good to think in terms of symmetry, but it is incorrect that acceleration is the only thing that can break the symmetry.

In the three-observer scenario there is no acceleration. The symmetry is broken by the use of two inertial clocks instead of one. In curved spacetime you can have a twin scenario with no acceleration and the symmetry is broken by the curvature of the spacetime.

Furthermore, not only can you break the symmetry without acceleration, you can also accelerate without breaking the symmetry. This was described for the counter-rotating uniform circular motion. But it would apply for many other acceleration profiles too.

I think that there are two concepts you will need to correct before continuing:

1) the twins must meet at two events (at least) in order for their clock comparison to be frame invariant

2) acceleration breaks the symmetry in the usual scenario, but in general acceleration is neither necessary nor sufficient for broken symmetry

Please think on those ideas before continuing.

 

[Peter Strohmayer]

Let me ask you a closely related question. Given a triangle with sides A, B, and C each side opposite angle a, b, and c respectively. We know that A<B+C. Would you say that the angle a caused the difference in length B+C−A?

No, I wouldn't.

 

[Peter Strohmayer]

As described above, if they do not meet twice then the final state is frame dependent.

In order to have a basis for my thinking, I ask for the following clarifications:

We agree that in the example I gave, neither twin accelerates at the beginning of the trip (the twins are supposed to set their clocks to zero as they pass each other).

You use the terms "twins" and "observers" synonymously. (You speak of "two observers"). Then you speak of the two "observers" (=twins) agreeing on a frame of reference. So they obviously represent one and the same (imaginary) observer. - Could we then agree to call a twin an observer, if we take as a basis a reference frame in which the respective twin rests (at most up to the time of the first and at the same time only acceleration considered here)? It would be possible to choose an observer other than one of the twins (e.g. one moving relative to each of the twins), but we do not choose such an observer because it would obviously complicate the analysis.

Instead of speaking of "one twin" and "the other twin", I would suggest referring to the twins as twin A (which rests respectively moves to the left) and twin B (which moves to the right respectively rests), so that there are no misunderstandings (e.g. about which twin is accelerated when and to what speed, so that slower aging would be stopped; or about the claim that both twins would have aged the same since they set their clocks to zero
before and after "the acceleration", which would be impossible with a single acceleration of one of the twins I spoke of).

 

[Dale]

You use the terms "twins" and "observers" synonymously.

Yes

Then you speak of the two "observers" (=twins) agreeing on a frame of reference. So they obviously represent one and the same (imaginary) observer.

This is exactly the issue I am addressing. You have in your mind that “observer” = “frame”, such that if two observers agree to use the same reference frame they are the same observer. This is a result of the usual didactic approach where scenarios are populated by observers that each use their rest frame.

The principle of relativity states that all inertial frames are equivalent. So an observer may use any inertial frame. They may use one where they are at rest or they may use one where they are moving. I, as an observer, may use the frame where the ground is at rest and say “I went to the store”. I am not required to use the frame where I am at rest and say “the store came to me”.

Could we then agree to call a twin an observer, if we take as a basis a reference frame in which the respective twin rests (at most up to the time of the first and at the same time only acceleration considered here)?

This is precisely what I wish not to do. I am deliberately dissociating observers/twins from reference frames, as permitted by the principle of relativity.

Instead of speaking of "one twin" and "the other twin", I would suggest referring to the twins as twin A (which rests respectively moves to the left) and twin B

That is fine. The “one twin” and the “other twin” was your terminology which I simply continued.

 

[Dale]

@Peter Strohmayer I decided to go ahead and illustrate the issue so that maybe it is super-clear. This is the scenario you described where the blue line is the worldline of the inertial twin and the orange line is the worldline of the accelerating twin. On each worldline the first dot is $\tau=0 \ years$, the second dot is $\tau=4 \ years$, and the third dot is $\tau=8 \ years$ on each twin's clock. The speed is $v=3/5 \ c$.

They could both use the frame of the other observer, there the one twin’s slow aging occurs on the first leg and the acceleration stops the one twin’s slow aging, with the one twin being younger.

This frame is moving at $v=0$ with respect to the inertial twin.

They could both use the inertial frame where the one twin is initially at rest. In that frame the other twin is aging more slowly. The acceleration of the one twin makes the one twin age more slowly too. But in this frame the other twin is the younger.

This frame is moving at $v=3/5$ with respect to the inertial twin.

They could also use an inertial frame where both twins are moving, with equal and opposite speed. Then both twins are at all times the same age, both before and after the acceleration.

This frame is moving at $v=1/3$ with respect to the inertial twin.

By not having them co-located the relativity of simultaneity means that different frames disagree on which times are simultaneous. The result is that at the end you are free to choose an inertial frame where either twin is older or where they are the same age. It is entirely a matter of convention, and each is completely valid.

 

[Vanadium 50]

Tangent Thread

Huh. Usually in relativity you get hyperbolic tangents.

 

[Peter Strohmayer]

This is precisely what I wish not to do.

An "observer" is an ideal, massless, imaginary measuring station at the center of a chosen frame of reference.
It leads to misunderstandings, to call a mass point an "observer", which cannot observe anything, but only produce a world line.
Only if a twin fulfills the function of the mentioned measuring station in a chosen reference system, it should be called "observer".

That is fine. The “one twin” and the “other twin” was your terminology which I simply continued.

This is what I said:
-----
„#1: Each of the twins is at the center of a long synchronized clock line. As the twins move past each other, they set the clocks directly in front of them - and thus their entire row of clocks - to "zero". The twin's clock continues to move past the other twin's row of clocks. Comparing the display of his clock with the display of the meeting clock, the twin notices that the display of his clock is lagging more and more behind the display of the just passing meeting clock. This is a symmetric process. Neither twin ages faster or slower as a result."

#34: I assume two clock lines (reference systems) moving towards each other. After the meeting of A and B, the display of A's clock lags behind that of B's passing clocks. This time difference (which occurs symmetrically at B with respect to the passing clocks of A) increases proportionally to the length of the journey.

If A decides to jump over to one of the passing clocks of B (this is an acceleration in infinitesimal steps over the integral of all infinitesimal speed differences), then he takes the time of his clock A unchanged with him, but now rests in the frame of reference of B in the middle of the clocks of B, which now show the same time for A as for B. The display of A's clock (his lifetime) now remains constantly behind the displays of all clocks of B. This means that he has actually remained younger compared to B.

The difference in age increases the later A decides to jump to one of B's passing clocks (i.e. to accelerate and equalize the relative speed with respect to B). This does not change the fact that without acceleration (without this jump to the passing clock of B) no "twin paradox" is possible.“

A = accelerating twin, B= inertial twin.
-----
To compare the age (time elapsed since meeting) of the twins (by the "observer" described above), it is not necessary for them to be at the same place. The twins do not have to pull each other's beards. It is sufficient for them to rest in the same inertial frame, even if they are far apart in space.

I decided to go ahead and illustrate the issue so that maybe it is super-clear.

Your diagrams do not show what I have described.

The point is to determine which time the clock of the clock line of B just passing A shows when A decides to jump to this clock of B.

This cannot be read from the Minkowski diagrams you draw. You do not include the world line of the encounter clock. With proper analysis, it turns out that the age difference resulting does not depend on the choice of reference frame.

Here are the two graphs showing how a recoil acceleration leads to an age difference (strictly speaking, this schematic simplification is only valid for an infinitesimal acceleration step):
Red=Inertial twin B:

Green = accelerating twin:

different frames disagree on which times are simultaneous

Yes.

you are free to choose an inertial frame where either twin is older or where they are the same age

No.

 

[PeterDonis]

No.

You are incorrect here, and @Dale is correct. He showed you explicitly how. I strongly advise you to be careful, as you are getting close to a warning at this point.

 

[PeterDonis]

To compare the age (time elapsed since meeting) of the twins (by the "observer" described above), it is not necessary for them to be at the same place.

If you want the comparison to be frame invariant, it is.

Again, I strongly advise you to be careful. You are getting close to a warning.

 

[PeterDonis]

As the twins move past each other, they set the clocks directly in front of them - and thus their entire row of clocks - to "zero".

This is impossible as you describe it. You cannot magically set an entire row of clocks instantaneously. Each twin can set the clock directly in front of them that is moving with them; but they can only propagate that setting to other clocks moving with them at the speed of light.

 

[Nugatory]

To compare the age (time elapsed since meeting) of the twins (by the "observer" described above), it is not necessary for them to be at the same place. The twins do not have to pull each other's beards. It is sufficient for them to rest in the same inertial frame, even if they are far apart in space.

This is simply wrong. Even when they are at rest in the same inertial frame (a frame-independent way of expressing this constraint would be to say that their relative velocity is zero) if they are not colocated then different frames will calculate and observe A to be older than B, A to be younger than B, or both to be the same age.

 

[Dale]

An "observer" is an ideal, massless, imaginary measuring station at the center of a chosen frame of reference.

No, again, this is exactly the issue I am addressing. An observer is anything that can make an observation, whether that is a person or a measuring device or a particle or whatever. Describing them as “ideal” is fine (“massless” is not).

But the frame of reference is separate from the observer. For an inertial observer there exists an inertial reference frame, called the observer’s frame, in which it is at rest and at the origin. But the reference frame is not part of the observer any more than your shoes are part of you.

An observer may use any reference frame, even more easily than you may change your shoes. That is the principle of relativity.

In this analysis I am deliberately dissociating the observers and the reference frames. I am deliberately using reference frames where the observers are not at rest. This is the principle of relativity.

Your diagrams do not show what I have described.

They describe exactly this:

I have simplified the example: it is enough for one twin to switch to the other's frame of reference once to make him age less

As shown in the diagrams this specific claim of yours is false.

This is what I said

You said a lot. That is not what I was reacting to. I was reacting to the specific statement I quoted. That is why the quote feature exists.

To compare the age (time elapsed since meeting) of the twins (by the "observer" described above), it is not necessary for them to be at the same place. The twins do not have to pull each other's beards. It is sufficient for them to rest in the same inertial frame, even if they are far apart in space

Not if you want the comparison to be frame invariant, as I showed.

to determine which time the clock of the clock line of B

The issue that you are still neglecting is that neither B nor A must use a line of clocks at rest with respect to them.

you are free to choose an inertial frame where either twin is older or where they are the same age

No

So you reject the principle of relativity. That is definitely “personal theory” territory.

 

[Peter Strohmayer]

He showed you explicitly how. I strongly advise you to be careful, as you are getting close to a warning at this point.

So you reject the principle of relativity. That is definitely “personal theory” territory.

Did the "age of twin A" mean the proper time of twin A elapsed since the meeting?
Was the proper time of a twin (his age) invariant?

 

[Peter Strohmayer]

This is impossible as you describe it.

Is it possible that each twin synchronizes its line of clocks according to Einstein's method? Is it possible that each of the twins considers the time, which their clocks indicate, when they pass each other later, as "zero point" of their time counts?

 

[Dale]

Did the "age of twin A" mean the proper time of twin A elapsed since the meeting?
Was the proper time of a twin (his age) invariant?

The proper time along any timelike worldline between two fixed events is indeed invariant.

The problem is that by having them meet only once you are only fixing one event. The other events are not fixed but depend on the frame-variant simultaneity convention.

In other words although $d\tau$ is unequivocally invariant, $\tau =\int_a^b d\tau$ depends on both of the end points $a$ and $b$. If both are invariant then $\tau$ is also invariant. But if either depend on the reference frame then $\tau$ also does.

They must meet twice in order to fix the two events $a$ and $b$.

Is it possible that each twin synchronizes its line of clocks according to Einstein's method?

Yes, and it is also possible that each twin uses an Einstein synchronized line of clocks that they are moving relative to. That is part of the principle of relativity that you seem to be missing.

 

[vanhees71]

An "observer" is an ideal, massless, imaginary measuring station at the center of a chosen frame of reference.
It leads to misunderstandings, to call a mass point an "observer", which cannot observe anything, but only produce a world line.

An "observer" is an ideal MASSIVE point, moving along a TIMELIKE worldline. You can define a local (in general non-inertial) non-rotating frame of reference by Fermi-Walker transport of a tetrad along this worldline. Note that "non-roting" is meant wrt. the observer. In general relative to a fixed inertial frame of reference this reference frame will be rotating. That's the famous Wigner rotation giving rise to the Thomas precession of spins.

 

[Peter Strohmayer]

you are only fixing one event. The other events are not fixed but depend on the frame-variant simultaneity convention

Again, I strongly advise you to be careful. You are getting close to a warning.

Isn't what happens above (#46 and #48)(ct0=0, x0=0; ct'=0, x'0=0) an event a?

Isn't what happens above (ct1=5, x1=3; ct'1=4, x'1=0) an event b?

Doesn't it follow from event B (because of ct'1=4) that twin A (green) has aged 4 years from event a to event b (proper time)?

Doesn't it follow from event B (because of ct1=5) that twin B (red) has aged 5 years from event a to event b (proper time)?

Is this true, even though twin B (red) (as a mass point at x=0) is not personally present at event b (although the world line of twin B does not intersect the world line of twin A)?

If yes, does this have something to do with the fact that the clock of twin B and the clock of his line of clocks which meets twin A at event b when he jumps (and equalize the relative speed of the meeting clock by recoil acceleration) run synchronously?

Wasn't the age difference caused by the one-time recoil acceleration?

Do the twins age at the same rate after event b?

Would this mean that without acceleration the age difference does not change after event b?

Is the elapsed proper time of A (green) of 4 years and the elapsed proper time of B (red) of 5 years between the events a and b the same from the point of view of all reference frames?

 

[Nugatory]

Isn't what happens above (#46 and #48)(ct0=0, x0=0; ct'=0, x'0=0) an event a?

Yes.

Isn't what happens above (ct1=5, x1=3; ct'1=4, x'1=0) an event b?

Yes.

Doesn't it follow from event B (because of ct'1=4) that twin A (green) has aged 4 years from event a to event b (proper time)?

Yes, because events A and B are both on twin A's worldline (and assuming that twin A's path between the two events is an inertial straight line).

Doesn't it follow from event B (because of ct1=5) that twin B (red) has aged 5 years from event a to event b (proper time)?

No. Event B is not on twin B's worldline so twin B's aging between events A and B is not defined (The spacetime interval between events A and B is of course defined, but it has nothing to do with twin B's age or proper time). If we choose an event C somewhere on twin B's worldline then the proper time between events A and C would be how much twin B aged between those two events.

Is this true, even though twin B (red) (as a mass point at x=0) is not personally present at event b (although the world line of twin B does not intersect the world line of twin A)?

No, because of the "no" answer above.

If yes, .....

We don't need to go here because the "If yes" condition is not satisfied.

Do the twins age at the same rate after event b?

"Same rate" is not a meaningful concept here. Post #11 of this thread explains why.

Would this mean that without acceleration the age difference does not change after event b?

It's not defined, and again see post #11.

Is the elapsed proper time of A (green) of 4 years and the elapsed proper time of B (red) of 5 years between the events a and b the same from the point of view of all reference frames?

All frames agree about the elapsed proper time between events A and B, and that this proper time is the amount by which twin A aged (because both events are on A's worldline and A followed a straight path between them). All frames also agree that this proper time has nothing to do with twin B's age.

 

[Peter Strohmayer]

Event b is not on twin B's worldline so twin B's aging between events a and b is not defined

Is it not permissible, then, to conclude that two children born at the same time in different places will always be the same age at any given time, from the point of view of the frame of reference in which they rest, because their world lines do not cross?

 

[Ibix]

Is it not permissible, then, to conclude that two children born at the same time in different places will always be the same age at any given time, from the point of view of the frame of reference in which they rest, because their world lines do not cross?

Using that frame only, yes. Using other frames, no. This is the point of the twin paradox: the twins meet, separate, and meet again, and everyone agrees on their accumulated age difference between the meetings. In your example, almost everyone disagrees.

 

[Peter Strohmayer]

Using that frame only, yes. Using other frames, no.

Is not the age of the children (their proper time) an invariant?

 

[Ibix]

Is not the age of the children (their proper time) an invariant?

There age when? It's the "when" that is not invariant unless they meet.

 

[Sagittarius A-Star]

Is not the age of the children (their proper time) an invariant?

two children born at the same time in different places

And because they were born at different places, it is not invariant, if they were born at the "same time".

 

[Nugatory]

Is it not permissible, then, to conclude that two children born at the same time in different places will always be the same age at any given time, from the point of view of the frame of reference in which they rest, because their world lines do not cross?

It's neither more nor less permissible than concluding that one of them will always be older than the other when we calculate "ages" using coordinates in which they are both moving at a constant non-zero velocity - it all depends on our essentially arbitrary choice of coordinates.

When you inserted that additional "from the point of view" qualifier you redefined "same age" so that it is frame-dependent and no longer has any physical significance, and furthermore chose a redefinition that makes it true by definition: define "same age" to mean "has the same time coordinate" and then choose time coordinates that advance at the rate as proper time and of course they will age at the same rate.

 

[robphy]

Is not the age of the children (their proper time) an invariant?

The proper-time age of a child is the elapsed time on its wristwatch between two events $A_1$ and $A_2$ on that child's worldline (akin to an arc length along a specific curve between two points). For two specified events along that worldline, that elapsed proper-time is invariant.

To compare the ages of two children,
you have to specify events $A_1$ and $A_2$ on child-A's worldline
and events $B_1$ and $B_2$ on child-B's worldline.

For distinct spacelike-separated events $A_1$ and $B_1$ ,
not all observers will regard $A_1$ and $B_1$ as simultaneous (i.e. as "being at the same time")
.... and similarly for $A_2$ and $B_2$.

 

[PeroK]

When I first studied SR, I assumed that acceleration must be the ultimate cause of the differential ageing in twin paradox. Then I learned about Minkowski spacetime. Then someone on this forum posted the idea of replacing the turnaround with the communication of a clock reading to an inbound spacecraft. And it was clear that acceleration was only a physical constraint and a geometric red herring.

This process of continually updating one's knowledge of a subject is called learning. It's the opposite of religiously adhering to an established view in the face of evidence to the contrary.

 

[Dale]

born at the same time in different places

This is a frame variant concept.

from the point of view of the frame of reference in which they rest

In that frame it is true. Not in other frames. They do not need to use the frame in which they are at rest.

Do you accept the principle of relativity?

 

[Dale]

And it was clear that acceleration was only a physical constraint and a geometric red herring.

Yes. The acceleration breaks the symmetry, but it is not the only way to break the symmetry. Acceleration is neither necessary nor sufficient for different aging, so it cannot be the cause.

 

[Dale]

Is not the age of the children (their proper time) an invariant?

As I mentioned previously $d\tau$ is an invariant. $\tau=\int_a^b d\tau$ is invariant only if $a$ and $b$ are.