B Relativity: Twin Paradox - Is Age Determinable?

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The discussion centers on the Twin Paradox in relativity, questioning whether the twin who travels to Mars ages less than the twin who stays on Earth. It is concluded that the aging difference is frame variant and depends on the synchronization convention used. While both twins may perceive the other's clock as running slower, the traveling twin ultimately ages less due to the path taken through spacetime, which is shorter for the traveler. Acceleration plays a role in breaking the symmetry of their experiences, but the key factor is their relative motion and the lack of co-location when comparing clocks. Thus, the determination of age difference is not absolute but varies based on the chosen frame of reference.
  • #91
PeterDonis said:
Yes, but then your claim that the stay at home twin ages during the turnaround turns into the claim that re-synchronizing clocks can cause the stay at home twin to age. Which seems unusual, to say the least.
I agree.
 
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  • #93
FactChecker said:
In SR, when two observers are moving inertially with respect to each other, each observer thinks that the other is aging slower. In that situation, when does the stationary twin get to age faster in the eyes of the traveling twin, as you propose?

It would be a good exercise for you to figure this out in the following variation. You have A, B and C:

A remains on Earth
B travels away from Earth at relativistic speed (*)
C travels towards Earth at the same relativsitic speed in the opposite direction, starting from far away

(*) To take acceleration out of the experiment, we have B accelerate in an orbit and then fly past the Earth, at which point A and B synchronise their clocks. This is the start of the experiment. There is no acceleration for A, B or C during the experiment.

In B's reference frame, the clock at A runs slow for the entire journey. This is simple time-dilation.

When B has traveled for some time, it passes C on the way to Earth. B reports to C that "A's clock has been running slow during the journey". And C then synchronises his clock with B's,

During C's continued journey to Earth, in his frame A's clock is running slow the whole way.

Yet, when C gets to A it is C's clock that reads less time. It reads the B's proper time for the outbound journey + C's proper time for the inbound journey (from the point C and B crossed).

And, yet, in those two inertial reference frames, during the whole experiment it was A's clock that was running slow.

Answers on a postcard, as they say.
 
  • #94
PeterDonis said:
This is too extreme. The Usenet Physics FAQ I linked to has an "Equivalence Principle Analysis" page that discusses this issue.

That is interesting. To quote from the link you posted, what I really meant was:

"The essence of Einstein's first insight into General Relativity was this: (a) you can derive time dilation for uniform pseudo-gravitational fields, and (b) the Principle of Equivalence then implies time dilation for gravitational fields. A stunning achievement, but irrelevant to the twin paradox. "
 
  • #95
Apologies - didn't get to the Minkowski diagrams last night and am a bit pressed for time this morning. In particular, I haven't had a chance to catch up on the thread in detail, so apologies if I'm restating stuff.

The problem with "when is the extra time" is, as always, the relativity of simultaneity. If you use the Einstein synchronisation convention, what does "at the same time as the traveller turn around" mean? It means different things to the inbound and outbound frames:
245249

The fine gray lines are "now" for the inbound and outbound frames, the one with the positive gradient associated with the outbound frame and the one with the negative gradient associated with the inbound frame. Looking at the red line, then, it is tempting to say that the "extra" time happens "during the turnaround". But before you do that, look to the right of the turnaround event. The wedge between the two gray lines is both "before the turnaround" and "after the turnaround" by this logic, so happens twice.

There isn't a solution to this in terms of Einstein synchronisation. If you patch together two inertial frames in this naive way, you inevitably end up with a wedge of spacetime that you haven't accounted for (the bit "during" the acceleration) and a wedge that you double count (the bit that's both before and after the acceleration). Non-instantaneous acceleration doesn't fix this, because those two fine grey simultaneity lines are non-parallel. They will cross somewhere even if you spread them out a bit by curving the corner, and it's that crossing that's the problem.

The solution is to use a non-inertial frame. There are many approaches. One I like is "radar time", which simply asserts that any observer can be equipped with a radar set. If they emit a pulse at time ##t## and receive the echo at ##t+\Delta t## then the echo happened at time ##t+\Delta t/2## at a distance from the observer of ##c\Delta t/2##. Full stop. For an instantaneous acceleration this turns out to mean that (in this scheme) the traveller would use the outbound Einstein frame to describe events in the past lightcone of the turnover, the inbound frame for events in the future lightcone of the turnover, and a betwixt-and-between frame for everything else. The simultaneity planes under this scheme for various scenarios are given in Dolby and Gull's paper.

It's worth noting that whatever scheme you use to define simultaneity, what the twins actually see depends on the Doppler effect. You never see the clocks jump.

The traveller will see the stay-at-home's clock tick slowly until turnover, whereupon he will see it tick quickly until his return (the tick rate jumps in an instantaneous turnaround, but the time shown is continuous). The stay-at-home will see the traveller's clock tick slowly until the light from the turnover reaches home, whereupon the tick rate jumps. Note that the observed behaviour of the clocks is different - the traveller sees the clock rate jump at the midpoint of the journey; the stay-at-home does not see the jump until almost at the end. This is another way of seeing that nobody is surprised by the traveller being younger.
 
  • #96
... on the other hand, a full and comprehensive explanation of the twin paradox is:

##d\tau^2 = dt^2 - dx^2 - dy^2 - dz^2##
 
  • #97
PeroK said:
... on the other hand, a full and comprehensive explanation of the twin paradox is:

##d\tau^2 = dt^2 - dx^2 - dy^2 - dz^2##
The thing this doesn't do is provide an explanation for why naive application of the time dilation formula leads to a paradox. It is indeed a complete and very general solution to the scenario and any other similar one. But it doesn't explain the mistake, which is essentially the same one from Round the World in Eighty Days - forgetting to change your calendar when you change timing convention.
 
  • #98
phinds said:
It sounds like you are continuing to promote the falacious point of view that things slow down for the traveler IN HIS FRAME. That is not true. Neither his clock nor his biological processes slow down in his frame, he's just taking a different path through space-time so the NUMBER of ticks of his clock is different but not the rate at which they occur.

I think we need to be careful not to promote this very misleading point of view.
Yes and that shorter path the traveler takes is the space-time interval :delta s^2 = sqrt (c delta t)^ 2 - (delta x) ^2 unfortunately the symbols. no longer seem available?
 
  • #99
phinds said:
It sounds like you are continuing to promote the falacious point of view that things slow down for the traveler IN HIS FRAME. That is not true. Neither his clock nor his biological processes slow down in his frame, he's just taking a different path through space-time so the NUMBER of ticks of his clock is different but not the rate at which they occur.

I think we need to be careful not to promote this very misleading point of view.

morrobay said:
Yes and that shorter path the traveler takes is the space-time interval :delta s^2 = sqrt (c delta t)^ 2 - (delta x) ^2 unfortunately the symbols. no longer seem available?
@phinds Sorry. I never meant to imply that. If something I said or terms I used implied that, then I'm sorry. I have been reading the link that @PeterDonis gave, and it is very good. It clarifies a lot that I only had a vague understanding of.
 
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  • #100
morrobay said:
Yes and that shorter path the traveler takes is the space-time interval :delta s^2 = sqrt (c delta t)^ 2 - (delta x) ^2 unfortunately the symbols. no longer seem available?
Edit, delta s = (same above)
 
  • #101
Ibix said:
The thing this doesn't do is provide an explanation for why naive application of the time dilation formula leads to a paradox. It is indeed a complete and very general solution to the scenario and any other similar one. But it doesn't explain the mistake, which is essentially the same one from Round the World in Eighty Days - forgetting to change your calendar when you change timing convention.

To summarise. The flat spacetime metric tells you everything you need to know about any path or paths through spacetime. The simplest explanation for the twin paradox, therefore, involves analysis in any IRF. It doesn't need to be the Earth's rest frame. Any IRF will do and the metric does the rest.

This, however, does not explain things in terms of a physical explanation for an accelerating observer. When we try to find an accelerating coordinate system to cover the entire experiment we immediately find ambiguities. The same event may be mapped to different coordinates, for example. This undermines the attemept to give a clear and unambiguous record of what happened where and when in an accelerating reference frame.

This takes us into the background material presented in this thread: attempts to give a consistent coordinate system for the accelerating observer etc. In particular, if a simultaneity convention is seen as something non-physical, then a physical explanation for the accelerating observer is elusive.
 
  • #102
I assume this has been stated earlier, so apologize for the repetition. But there is a simple invariant statement that can be made (whose description will differ by coordinate choice, but not the result):

If you Einstein synchronize the Earth and Mars clock, and synchronize an Earth rocket clock with the Earth clock, then have the rocket travel to mars, then the rocket clock will be behind the Mars clock. This is true no matter what travel path or acceleration is used, but these latter choices determine whether the amount is miniscule or enormous (e.g. a looping path near c per earth, that takes 1k years in earth-mars frame to reach mars).

Obvious simplifications: gravity ignored, Earth and Mars considered mutually stationary.
 
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  • #103
I think one saves oneself a lot of trouble, if one simply uses covariant quantities. If this is not possible, something usually is at least problematic, if not simply ill-defined or wrong, with the way a problem is stated or thought about.

In the case of the "one-way twin paradox" one can simply refer to the proper time of each twin. That's a local concept too, because it just is the time each twin reads off from his or her wristwatch. Using these proper times for the "aging" of the twins thus you get an unanimous answer to who aged more or less compared to the other. Then there's no frame dependence nor is there any paradox to be thought about left!
 
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  • #104
Isn't the proper time equal to the space-time interval in this situation? So obviously less proper time for traveler.
 
  • #105
vanhees71 said:
I think one saves oneself a lot of trouble, if one simply uses covariant quantities. If this is not possible, something usually is at least problematic, if not simply ill-defined or wrong, with the way a problem is stated or thought about.

In the case of the "one-way twin paradox" one can simply refer to the proper time of each twin. That's a local concept too, because it just is the time each twin reads off from his or her wristwatch. Using these proper times for the "aging" of the twins thus you get an unanimous answer to who aged more or less compared to the other. Then there's no frame dependence nor is there any paradox to be thought about left!
But for the one way example here, there is no way to avoid a synchronization assumption, because that is the sole determinant of what the start event is for the Mars clock. There is only one incident of colocation. The interval beginnings are determined solely by a synchronization decision, which can be a physical procedure, thus invariant, but it is still a choice, and effectively defines a frame.
 
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  • #106
PeroK said:
"The essence of Einstein's first insight into General Relativity was this: (a) you can derive time dilation for uniform pseudo-gravitational fields, and (b) the Principle of Equivalence then implies time dilation for gravitational fields. A stunning achievement, but irrelevant to the twin paradox. "

The use of the EP to predict time dilation in real gravitational fields is irrelevant to the twin paradox, yes. But the interpretation that a pseudo-gravitational field (which has time dilation) is present while Stella fires her thrusters, and that this pseudo-gravitational field explains how Terence ages much faster than Stella during the turnaround, is certainly not irrelevant.
 
  • #107
PeterDonis said:
The use of the EP to predict time dilation in real gravitational fields is irrelevant to the twin paradox, yes. But the interpretation that a pseudo-gravitational field (which has time dilation) is present while Stella fires her thrusters, and that this pseudo-gravitational field explains how Terence ages much faster than Stella during the turnaround, is certainly not irrelevant.

I was just thinking though ... what happens if Stella executes the turnaround more than once? Terence must age faster every time. Which means that he must get physically younger during the first phase of the second turnaround. Which seems very unphysical.

To go through the steps:

Stella is heading away, Terence has age ##T## years in Stella's frame.
Stella executes a turnaround of a few hours, say. Terence has age ##T + 1## years, say, in Stella's frame.
Stella brakes and heads away again. Terence is back to age ##T## years (approx) in Stella's frame.
Stella executes the turnaround again and Terence is back to ##T + 1## years.

Or, alternatively, if Stella orbits a distant star at relativistic speeds, then Terence's age is going backwards and forwards during each orbit.
 
  • #108
PeroK said:
he must get physically younger during the first phase of the second turnaround

No, he just ages slower, because he's deeper in the gravity well when Stella is accelerating away from him.

PeroK said:
Stella brakes and heads away again. Terence is back to age ##T## years (approx) in Stella's frame.

I don't see why. Terence would simply be at age ##T + 1## years plus much, much less than a few hours (the time Stella takes to brake and head away again by her own clock).

PeroK said:
Stella executes the turnaround again and Terence is back to ##T + 1## years.

No, he is at ##T + 2## years (plus the small increment of time he aged during the brake and head away again phase).
 
  • #109
PeroK said:
if Stella orbits a distant star at relativistic speeds, then Terence's age is going backwards and forwards during each orbit

Stella has zero proper acceleration in this case (she's in a free fall orbit), so there is no pseudo-gravitational field, so the EP analysis does not apply.

If Terence is indeed far enough away for his "age" to be fluctuating this way if Stella adopts a "naive" extrapolation of her local comoving inertial frame, that just means that this method of extrapolating her local comoving inertial frame does not produce a valid coordinate chart that far away, because the mapping of coordinate time to proper time along Terence's timelike worldline is not one-to-one.
 
  • #110
PeterDonis said:
No, he is at ##T + 2## years (plus the small increment of time he aged during the brake and head away again phase).

That can't be right. The differential ageing relative to Terence can't depend on the number of changes of direction.
 
  • #111
PeterDonis said:
Stella has zero proper acceleration in this case (she's in a free fall orbit), so there is no pseudo-gravitational field, so the EP analysis does not apply.

Yes, forget the star orbit. I meant a powered orbit in any case: fly round, take a look at the solar system and head home!
 
  • #112
PeroK said:
Or, alternatively, if Stella orbits a distant star at relativistic speeds, then Terence's age is going backwards and forwards during each orbit.
No. Terence is always getting older. The rate will alternate between aging slowly and aging rapidly, but it will always be positive.

PeroK said:
That can't be right. The differential ageing relative to Terence can't depend on the number of changes of direction.
When Stella is closing the distance he will age rapidly, when Stella is receding he will age slowly.
 
  • #113
DaveC426913 said:
No. Terence is always getting older. The rate will alternate between aging slowly and aging rapidly, but it will always be positive.When Stella is closing the distance he will age rapidly, when Stella is receding he will age slowly.

The problem is that ageing rapidly and ageing slowly don't cancel out. Let's assume Stella's turnaround is a day, say. Half a day to slow down and half a day to speed back up (or a day to go in a loop).

If Terence ages a year in half a day (Stella's time) and next to nothing in the next half day, then that's still a year older, give or take.

If, therefore, Stella keeps repeating the turnaround, then the years pile up for poor Terence! Unless, of course, there is a part of the repeated turnaround in which Terence actually gets younger.

Fundamentally, the age differential when Stella returns is determined by Stella's velocity profile over the duration. The turnaround, physically, does not do anything special to the age difference (in a short period of Stella's time). It only changes the simultaneity convention.

This is the fundamental problem with ascribing the age differential to something physical during the acceleration phase.
 
  • #114
PeroK said:
The problem is that ageing rapidly and ageing slowly don't cancel out.
Agree. Though not a problem, as you put it..

PeroK said:
If, therefore, Stella keeps repeating the turnaround, then the years pile up for poor Terence!
They sure do. If Stella spends a great deal of time moving at relativistic velocities - in any direction - she's going to come back to a very old Terence.

PeroK said:
Unless, of course, there is a part of the repeated turnaround in which Terence actually gets younger.
Nope.

PeroK said:
Fundamentally, the age differential when Stella returns is determined by Stella's velocity profile over the duration. The turnaround, physically, does not do anything special to the age difference (in a short period of Stella's time). It only changes the simultaneity convention.
Agree! Post 79:
DaveC426913 said:
Yes. Velocity change (specifically, sign from + to -). Not acceleration change. :smile:
Although, note that Terence's very slow aging immediately begins to speed up (to normal) as soon as Stella begins her deceleration (negative acceleration) on approach to the turn around, while she still has positive velocity - and not when her velocity actually reverses.

Though only when her velocity actually flips from positive to negative does Terence start to age faster than her.

So positive acceleration decreases Terence's rate of aging,
while negative acceleration increases his rate of aging
(independent of what that rate was - slower or faster than Stella's - at the time).
 
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  • #115
DaveC426913 said:
No. Terence is always getting older. The rate will alternate between aging slowly and aging rapidly, but it will always be positive.
It will alternate between forward and backward. It will not always be positive. This is a feature of the naive construction of an accelerated frame. For any given [powered] orbital acceleration, the hyperplane of simultaneity in the tangent inertial frame of the traveler will swing back and forth on the worldline of a sufficiently distant stay at home twin.

Analogously, draw a straight line and a curved line side by side vertically up a piece of paper. Draw lines of "simultaneity" at regular intervals perpendicular to the curved line. As one moves up the curved line, the "simultaneous" point on the vertical line will move both up and down.
 
  • #116
PeroK said:
The differential ageing relative to Terence can't depend on the number of changes of direction.

Why not? You're changing Stella's path through spacetime. That changes the differential aging.
 
  • #117
jbriggs444 said:
It will alternate between forward and backward. It will not always be positive
We may be talking past each other here.
Terence will always be getting older (his aging will always be positive) according to Stella.
His rate of aging will decrease and increase - it may be negative or positive compared to Stella's - but it will always be > 0.
 
  • #118
jbriggs444 said:
It will alternate between forward and backward. It will not always be positive. This is a feature of the naive construction of an accelerated frame.

If you have an invalid frame, any statements about relative aging made using that frame are also invalid. So I wouldn't say "it will alternate between forward and backward", since that would be treating statements made using an invalid frame as valid.
 
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  • #119
DaveC426913 said:
Terence will always be getting older (his aging will always be positive).

This will be true if Stella is using a valid frame (more precisely, coordinate chart) that covers both her and Terence during the entire trip.

However, the "frame" that is used to make claims about Terence's aging going backwards is not valid.
 
  • #120
FactChecker said:
During inertial flight, the clocks and people in other, relatively moving IRFs always appear to have slow clocks and be aging slower.

Two additional clocks and a synchronization convention are required to reach this conclusion.

That is, each twin needs two clocks, separated along the line of motion, and synchronized.
 
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