Relativity: Twin Paradox - Is Age Determinable?

[FactChecker]

Here's a simple one.

It's overly simplified because it illustrates infinite acceleration. (the traveling twin's path is not curved, as it would be with realistic acceleration).

That's important, because realistic acceleration means that in reality, the red and blue lines will not intersect at the midpoint - so there will be no "jump" from blue to red - it is smooth, if rapid.

View attachment 245235

Notice that the change from red to blue EDIT: blue to red occurs exactly when the traveling twin turns around. That is when he is accelerating. This is the Minkowski diagram representation of acceleration.

 

[DaveC426913]

Notice that the change from red to blue occurs exactly when the traveling twin turns around. That is when he is accelerating. This is the Minkowski diagram representation of acceleration.

See second diagram. Acceleration is not instant.

Note also that deceleration and negative acceleration (back toward Earth) are the same thing.

Mars-bound traveler actually beings accelerating at point 3, not at the midway point 4.5.

 

[FactChecker]

I am trying to say that all these approaches fit together and are not in conflict. When something is true, it can often be looked at in many consistent ways. The Minkowski diagram includes representations of changes in velocity (accelerations). To say that accelerations do not play a role is to say that these Minkowski diagrams are wrong.

 

[DaveC426913]

To say that accelerations do not play a role is to say that these Minkowski diagrams are wrong.

The first diagram is indeed wrong. It illustrates infinite acceleration (for simplicity).
The second diagram is the correct one.

And his acceleration (toward Earth) actually begins at point 3, not the midpoint.

 

[FactChecker]

The first diagram is indeed wrong. It illustrates infinite acceleration.
The second diagram is the correct one.

Or the first one is just on a scale where one can not see the smooth turnaround. The difference is less important than the similarities: when the motion is inertial, nothing unusual happens. It is only during the turnaround acceleration that the Earth twin can indisputably age more. The amount of aging is determined by the distance between the twins. The timing of the aging is determined by the timing of the turnaround acceleration.

 

[Dale]

If you are saying that they can choose to disagree by picking other methods of synchronization, then I will not argue. But I think that Einstein-synchronization has some serious logical advantages in this application.

The point isn’t whether or not to use Einstein synchronization. The point is in which frame to use it. In the earth-mars frame you get one answer, but in other frames using Einstein synchronization gives you a different answer.

It is only during the turnaround that the Earth twin can indisputably age more.

Well, that depends on the details of the specific non-inertial reference frame used. In my favorite coordinates the aging of the Earth twin is accelerated in the traveling twin over a longer period of time. Specifically, the time from when a light signal from the Earth twin will reach the traveler during the acceleration until the time when a light signal from the traveler during the acceleration will reach the Earth twin.

 

[DaveC426913]

The net effect is thus:

1. M and E age at the same rate while at Earth.
2. As M accelerates away from Earth, he will observe E aging slower.
3. If he shuts off his engines, E will continue to age at the same slow rate.
4. As he begins his decel (acceleration toward Earth) nearing Mars, E's slow aging will lessen until he is aging at a normal rate.
5. As M continues to accelerate (toward Earth) it reverses his course and E's aging will accelerate, now starting to appear slightly older.
6. As M continues to accel toward Earth, E will continue to age rapidly until M starts his decel.

The upshot is that, when M begins his acceleration, E continues to age slowly, even though the slowness begins to decrease.

 

[FactChecker]

The point isn’t whether or not to use Einstein synchronization. The point is in which frame to use it. In the earth-mars frame you get one answer, but in other frames using Einstein synchronization gives you a different answer.

I agree. I was wrong.

Well, that depends on the details of the specific non-inertial reference frame used. In my favorite coordinates the aging of the Earth twin is accelerated in the traveling twin over a longer period of time. Specifically, the time from when a light signal from the Earth twin will reach the traveler during the acceleration until the time when a light signal from the traveler during the acceleration will reach the Earth twin.

I think this will take me a while to grasp. It this moment, I am happy with anything that ties it to the velocity change for the twin to turn around.

 

[DaveC426913]

I agree. I was wrong.I think this will take me a while to grasp. It this moment, I am happy with anything that ties it to the velocity change for the twin to turn around.

Yes. Velocity change (specifically, sign from + to -). Not acceleration change.

 

[FactChecker]

The net effect is thus:

1. M and E age at the same rate while at Earth.
2. As M accelerates away from Earth, he will observe E aging slower.
3. If he shuts off his engines, E will continue to age at the same slow rate.
4. As he begins his decel (acceleration toward Earth) nearing Mars, E's slow aging will lessen until he is aging at a normal rate.
5. As M continues to accelerate (toward Earth) it reverses his course and E's aging will accelerate, now starting to appear slightly older.
6. As M continues to accel toward Earth, E will continue to age rapidly until M starts his decel.

The upshot is that, when M begins his acceleration, E continues to age slowly, even though the slowness begins to decrease.

I really like that.
Just to make it more complete, I would add a step 7 where the return is inertial and E appears to M to age slower. But that will not make up for the aging of the prior steps 4 and 5.
There should also probably be a step 8, where M decelerates to Earth speed. This will cause E to lose relative age, but not much since the distance between E and M is relatively small.

 

[DaveC426913]

Just to make it more complete, I would add a step 7 where the return is inertial and E appears to M to age slower.

mm. On the return, E will still appear to age rapidly. (See straight segment between 6 and 7).

There should also probably be a step 8, where M decelerates to Earth speed. This will cause E to lose relative age, but not much since the distance between E and M is relatively small.

E will still appear to age rapidly, but the rapidity will decrease until they are both aging at the same rate.

The diagram shows this.

As long as M is closing the gap with Earth, E will appear to age rapidly.

All lines on the return trip are diagonally going NW to SE (i.e. E is aging faster than M):

(I see that there is a "missing feature" in this diagram. It's not wrong, it's just not easy to plot it on a timeline).

I was writing up a description:
"Earth-Mars Return trip, 12 months Earth time, 9 months ship time"
and planned to describe each point of the traveller's journey as if they are months.

But The traveller is not checking his clock at regular intervals! His checks (0,1,2,3,4,5,6,7,8,9) are not evenly spaced. eg. the passage of time between traveller's 4 > 5 and 6 > 7 are of quite different durations.

 

[phinds]

No. He didn't talk about the amount of aging.

Hm ... I can only think that we are interpreting the following VERY differently:

(A ship that accelerates at 5gs to .9c and then immediately decelerates back to rest may have a very small discrepancy, whereas a ship that accelerates at 5gs to .9c and stays there for a month will have a much larger discrepancy.)

I am only able to interpret that as being about the amount of aging.

EDIT: By the way, I feel that my posts in this thread are likely coming across as being overly argumentative. That is not my intent.

 

[Nugatory]

It is only during the turnaround acceleration that the Earth twin can indisputably age more.

Indisputably? We cannot say without ambiguity when the turnaround happens, which makes it rather easy to dispute that proposition (and any other claim that anything not colocated must have happened at the the time of the turnaround).

If the traveling twin is receiving continuous time broadcasts from the Earth twin (say the Earth twin broadcasts the time on their clock once every second) they will find no discontinuity during the turnaround acceleration; instead the faster aging of the Earth twin is spread out across the entire return leg. Surely that is sufficient reason for the traveling twin to dispute the proposition that the Earth twin's excess aging happened during the turnaround?

What's really going on here: Any attempt to assign the age difference to anyone part of the journey is going to be pretty much arbitrary. It's as if you were to drive directly from Paris to Berlin while I took a longer route through Livorno; certainly I covered more kilometers than you, but there's no non-arbitrary way of saying which specific kilometers on my route were the "extra" ones.

 

[FactChecker]

During inertial flight, the clocks and people in other, relatively moving IRFs always appear to have slow clocks and be aging slower. So if there are unaccelerated flight segments, the traveling twin thinks that the Earth twin is aging less rapidly. The only time when the traveling twin can think that the Earth twin is aging faster is when he is accelerating toward Earth.

 

[PeterDonis]

It seems like this thread could benefit from a link to the Usenet Physics FAQ article series on the twin paradox:

http://www.math.ucr.edu/home/baez/physics/Relativity/SR/TwinParadox/twin_paradox.html
Many of the issues raised in this thread are discussed there.

 

[PeterDonis]

relatively moving IRFs always appear to have slow clocks and be aging slower

It depends on what you mean by "appear". If you read the "Doppler Shift Analysis" from the Usenet Physics FAQ article series I linked to, you will see that (as @Nugatory pointed out), if the traveling twin is watching the stay at home twin through a telescope, he will see the stay at home's clock running faster throughout his return leg (i.e., as soon as he turns around). So as far as what actually "appears" in the telescope image, the statement of yours quoted above is simply wrong.

What you really mean by "appear" is that, after adjusting for light travel time, the traveling twin will calculate that the stay at home twin's clock is running slow compared to his own during both legs (outbound and return). But this calculation also involves a simultaneity convention, and that convention changes from the outbound leg to the return leg. (The "Time Gap Objection" page in the FAQ talks about this.) Not to mention that it's a strange use of language to use the word "appear" to refer to something calculated, while what actually appears in the telescope image is the opposite. (Unfortunately this abuse of language is so common in discussions of relativity that it comes naturally to anyone.)

 

[PeterDonis]

The equivalence principle has no relevance to the twin paradox. That is a fundamental misunderstanding.

This is too extreme. The Usenet Physics FAQ I linked to has an "Equivalence Principle Analysis" page that discusses this issue.

 

[FactChecker]

What you really mean by "appear" is that, after adjusting for light travel time, the traveling twin will calculate that the stay at home twin's clock is running slow compared to his own during both legs (outbound and return).

This is close to what I meant. I think it's equivalent. But my thinking is that all IRFs have their set of Einstein-synchronized clocks and recorders everywhere that report what is happening where the other moving IRF observer is. My use of "appear" was careless, but I meant that someone/something in the "stationary" IRF directly beside the moving observer records and reports the moving clock and aging with a time tag of the stationary IRF.

But this calculation also involves a simultaneity convention, and that convention changes from the outbound leg to the return leg. (The "Time Gap Objection" page in the FAQ talks about this.)

Certainly. I always assume that an IRF with no acceleration has a set of clocks everywhere which have been Einstein-synchronized at all times. The clock times would need to be re-Einstein-synchronized immediately after any acceleration.

Not to mention that it's a strange use of language to use the word "appear" to refer to something calculated, while what actually appears in the telescope image is the opposite. (Unfortunately this abuse of language is so common in discussions of relativity that it comes naturally to anyone.)

Yes. Again, I apologize. I'm sure that there is a lot of more precise terminology that I do not know.

 

[FactChecker]

It seems like this thread could benefit from a link to the Usenet Physics FAQ article series on the twin paradox:

http://www.math.ucr.edu/home/baez/physics/Relativity/SR/TwinParadox/twin_paradox.html
Many of the issues raised in this thread are discussed there.

At first glance, this link looks excellent. I will study it.

 

[PeterDonis]

The clock times would need to be re-Einstein-synchronized immediately after any acceleration.

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.

 

[FactChecker]

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.

 

[PAllen]

I agree.

I think my post in another thread is relevant here:

https://www.physicsforums.com/threa...r-the-twin-paradox.973392/page-5#post-6194401

 

[PeroK]

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.

 

[PeroK]

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. "

 

[Ibix]

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:

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.

 

[PeroK]

... on the other hand, a full and comprehensive explanation of the twin paradox is:

$d\tau^2 = dt^2 - dx^2 - dy^2 - dz^2$

 

[Ibix]

... 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.

 

[morrobay]

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?

 

[FactChecker]

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?

@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.

 

[morrobay]

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)

 

[PeroK]

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.

 

[PAllen]

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.

 

[vanhees71]

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!

 

[morrobay]

Isn't the proper time equal to the space-time interval in this situation? So obviously less proper time for traveler.

 

[PAllen]

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.