B Relativity: Twin Paradox - Is Age Determinable?

[PeroK]

Thank you all very much for your in depth analyses of this and the discussions. I have a much clearer picture now. And the thing to remember (that I keep forgetting) is "velocity relative to the frame...velocity relative to the frame". The traveler had a velocity relative to the Earth Mars frame and this was the determining factor as to why he aged (when seen from the Earth Mars frame). Any other frame would give a different result as to the age difference because of relativity of simultaneity and synchronization issues, but again, I was really just interested in the age as measured in the Earth Mars frame. My head feels much clearer.

Yes, in a frame where the Earth-Mars system was moving (in the direction towards the Earth) then the traveller would be seen to decelerate to a slower speed, before accelerating back to the same speed as the Earth-Mars system. In this frame, more time would have elapsed on the traveling clock than a clock on Earth or Mars.

In that reference frame, the twin that traveled to Mars would be older.

 

[FactChecker]

Not really. The frame you are talking about is the one where the outbound twin and stay-at-home have equal and opposite velocities. On the return leg, the traveling twin has to travel very fast to catch up with the Earth, so a large age difference is completely predictable.

I see your point and stand corrected.

 

[FactChecker]

There are a number of problems with your analysis in this thread. You have a problem here:

Consider a third traveller, who accelerates very rapidly, then decelerates very rapidly without having traveled very far from Earth. The acceleration and deceleration phases could be identical to those of a space traveller, who continued at their cruising relativistic speed for some time (before decelerating).

This third traveller will have experienced minimal differential ageing despite having experienced the same acceleration and deceleration as the space traveller.

This shows that nothing special happens during an acceleration. There is no rapid ageing.

The differential ageing is entirely a function of the time spent traveling at relativistic speeds (*).

(*) PS More generally, it depends on the entire path through spacetime; and not on the periods of acceleration.

My two cents:
The situation where both frames are inertial and no acceleration occurs is well known. There is no preferred inertial reference frame and symmetry holds. Both observers see the other as aging slower. That is true for the entire time that there is no acceleration. So it can not account for a mutually recognized differential aging without considering a change in velocity. The GR answer when there is acceleration is that the acceleration is equivalent to a gravitational field. Another observer far away is farther in the gravitational field and is effected more. Therefore, the acceleration causes a person far away to age (in the perspective of the accelerating observer) more than a nearby person. This same effect should be shown in SR when the path of an observer changes velocity.

 

[PeroK]

My two cents:
The situation where both frames are inertial and no acceleration occurs is well known. There is no preferred inertial reference frame and symmetry holds. Both observers see the other as aging slower. That is true for the entire time that there is no acceleration. So it can not account for a mutually recognized differential aging without considering a change in velocity. The GR answer when there is acceleration is that the acceleration is equivalent to a gravitational field. Another observer far away is farther in the gravitational field and is effected more. Therefore, the acceleration causes a person far away to age (in the perspective of the accelerating observer) more than a nearby person. This same effect should be shown in SR when the path of an observer changes velocity.

We work quite hard on PF to dispel the myths that:

a) Acceleration is the key to the twin paradox
b) To study acceleration we need GR, not just SR.

Quite explicitly, the twin paradox takes place in flat spacetime, which is the realm of SR. There is no gravity for which you would need GR.

Moreover, the twin paradox is about the lengths of different paths through the flat spacetime of SR. It's a geometric property that can be demonstrated with no acceleration. The "change in velocity" can equally well be achieved through an instantaneous change in IRF.

Therefore, the acceleration causes a person far away to age (in the perspective of the accelerating observer) more than a nearby person. This same effect should be shown in SR when the path of an observer changes velocity.

Here you are confusing an "accelerating reference frame", with the acceleration of an object in an IRF.

 

[FactChecker]

We work quite hard on PF to dispel the myths that:

a) Acceleration is the key to the twin paradox
b) To study acceleration we need GR, not just SR.

Quite explicitly, the twin paradox takes place in flat spacetime, which is the realm of SR. There is no gravity for which you would need GR.

To say that you have another way to solve it is not the same as saying that GR is false. If you want to argue against the GR equivalence principle, or how it is used then I will need to leave that argument for people who know more than I do.

Moreover, the twin paradox is about the lengths of different paths through the flat spacetime of SR. It's a geometric property that can be demonstrated with no acceleration. The "change in velocity" can equally well be achieved through an instantaneous change in IRF.

This reliance on an instantaneous change of IRF to say that acceleration is irrelevant seems ok to me. I can accept that the results are the same, but I do not agree that either necessarily invalidates the other. It seems like a leap of faith to tie an instantaneous IRF change to a physical end result like twins having different ages when the traveling one gets back to earth. That is switching from one IRF to another when the two do not agree with each other. But I can accept it. I have as easy a time accepting the equivalence principle in this simple application.

 

[PeroK]

To say that you have another way to solve it is not the same as saying that GR is false. If you want to argue against the GR equivalence principle, or how it is used then I will need to leave that argument for people who know more than I do.This reliance on an instantaneous change of IRF to say that acceleration is irrelevant seems ok to me. I can accept that the results are the same, but I do not agree that either necessarily invalidates the other. It seems like a leap of faith to tie an instantaneous IRF change to a physical end result like twins having different ages when the traveling one gets back to earth. That is switching from one IRF to another when the two do not agree with each other. But I can accept it. I have as easy a time accepting the equivalence principle in this simple application.

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

The equivalence principle does not say that acceleration is equivalent to gravity. In particular, it definitely does not say that an accelerating object can be considered subject to a gravitational potential and subject to gravitational time dilation.

 

[FactChecker]

How does one even define an inertial reference frame without mentioning acceleration directly or in a disguised form (as a change in velocity wrt other reference objects)? How does one decide when to switch "instantly" from one IRF to another in determining the solution to the twins paradox? Without reference to some external knowledge or influence (eg. acceleration, reference to a third body, etc.), one can not distinguish the "stationary" twin from the "traveling" twin. IMHO, the attempts to completely ignore acceleration is flawed in the most fundamental ways.

 

[DaveC426913]

IMHO, the attempts to completely ignore acceleration is flawed in the most fundamental ways.

I didn't get the sense anyone was "completely ignoring" acceleration, but PeroK made a valid point, showing how it is not the acceleration itself that results in time dilation (I know you read this and responded already; just posting for clarity):

Consider a third traveller, who accelerates very rapidly, then decelerates very rapidly without having traveled very far from Earth. The acceleration and deceleration phases could be identical to those of a space traveller, who continued at their cruising relativistic speed for some time (before decelerating). This third traveller will have experienced minimal differential ageing despite having experienced the same acceleration and deceleration as the space traveller.

This shows that nothing special happens during an acceleration. There is no rapid ageing.

The differential ageing is entirely a function of the time spent traveling at relativistic speeds (*).

 

[FactChecker]

I didn't get the sense anyone was "completely ignoring" acceleration, but PeroK made a valid point, showing how it is not the acceleration itself that results in time dilation (I know you read this and responded already; just posting for clarity):

The change in the IRF is acceleration, either gradual or instantaneous. And that is the exact time when the aging of far away objects happens. At all other times, both observers see the other as aging slower.

I feel that this is going in circles and will leave the discussion to others. I just accept both ways of looking at it and am not convinced that there is a real conflict in the two views.

 

[DaveC426913]

The change in the IRF is acceleration, either gradual or instantaneous.

Yes. But (at the risk of being repetitive), the point is the acceleration itself does not result in the time dilation.

As witnessed in PeroK's example where an identical acceleration/deceleration curve can result in virtually no discrepancy in aging. It is the time spent moving at relativistic velocity that causes the discrepancy.

(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.)

And that is the exact time when the aging of far away objects happens.

No. The aging occurs during any time spent at relativistic velocities - whether for 1 second or for a month.
And the time spent getting to that speed can be arbitrarily short.

 

[Nugatory]

How does one decide when to switch "instantly" from one IRF to another in determining the solution to the twins paradox?

Everything is always in all frames and you can switch which frame you use to analyze the problem at any time and you will get the correct answer, or you can analyze the problem without ever switching frames.

Frame-changing only appears in the discussion of the twin paradox because there is no inertial frame in which the traveling twin is at rest; therefore any attempt to use such a frame to calculate the time elapsed on either clock must yield bogus results (and indeed the “paradox” is the result of taking the bogus result at face value).

 

[FactChecker]

No. The aging occurs during any time spent at relativistic velocities - whether for 1 second or for a month.
And the time spent getting to that speed can be arbitrarily short.

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?

 

[DaveC426913]

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?

But we're not talking about what they observe in-transit - we're talking about what they measure once planetside, and checking their clocks.

@Ibix offered to try to draw some Minkowski diagrams. That will illustrate what they see in-transit, and how the observation of each other's slowing is resolved.

 

[phinds]

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?

That's not what he said. He is pointing out that the age DIFFERENCE, that you don't see until they get back together, is greater the longer the time that the traveler travels.

EDIT: I see Dave beat me to it.

 

[FactChecker]

In SR, consider two IRFs whose clocks are Einstein synchronized. In SR, when two observers are moving wrt each other, the other frame's Einstein-synchronized clocks are always drifting off so that the trailing clocks indicate ahead of what you think they should and the farther back, the worse the error. The leading clocks indicate behind of what you think they should. If the traveling twin instantly turns around, the Earth IRF clock suddenly switches from a trailing position to a leading position. So it appears to the traveling twin that the Earth IRF clock has jumped from indicating behind to indicating ahead. That is, the Earth IRF clock suddenly ages a great deal. The farther away the Earth is, the greater its jump in age is. This is the SR mathematical treatment of an instantaneous change of direction of the traveling twin. It coincides exactly with the turn around of the traveling twin. So the aging of the Earth twin occurs at the instant of the turnaround. The amount of aging is determined by the distance of the traveling twin from Earth.

 

[FactChecker]

That's not what he said. He is pointing out that the age DIFFERENCE, that you don't see until they get back together, is greater the longer the time that the traveler travels.

EDIT: I see Dave beat me to it.

No. He didn't talk about the amount of aging. He talked about when the aging occurs. The amount of aging is determined by the distance between the twins. The aging occurs when the twin turns around. In Einstein-synchronized IRFs we can assume that observers all along the path can observe and reliably report back what they see. In such IRFs, they will always see the other frame's clocks running slow and aging slow. It is only at the moment of turn around that the traveling twin can record that the Earth twin ages too fast.

 

[DaveC426913]

In SR, consider two IRFs whose clocks are Einstein synchronized. In SR, when two observers are moving wrt each other, the other frame's Einstein-synchronized clocks are always drifting off so that the trailing clocks indicate ahead of what you think they should and the farther back, the worse the error. The leading clocks indicate behind of what you think they should. If the traveling twin instantly turns around, the Earth IRF clock suddenly switches from a trailing position to a leading position. So it appears to the traveling twin that the Earth IRF clock has jumped from indicating behind to indicating ahead. That is, the Earth IRF clock suddenly ages a great deal. The farther away the Earth is, the greater its jump in age is. This is the SR mathematical treatment of an instantaneous change of direction of the traveling twin. It coincides exactly with the turn around of the traveling twin. So the aging of the Earth twin occurs at the instant of the turnaround. The amount of aging is determined by the distance of the traveling twin from Earth.

Well, there's nothing "instant" about the change from a distant observer moving slowly to moving quickly. You will not see the clock "jump". The transition is smooth, even if distorted and asymmetrical.

Again, let's wait for the Minkowski diagrams. That will make it all easy to discuss.

 

[FactChecker]

Well, there's nothing "instant" about the change from a distant observer moving slowly to moving quickly. You will not see the clock "jump".
As the traveling twin decelerates, he will see his counterpart speed up to normal, so that, when he reaches rest wrt to Earth (even if only instantaneously), the twin on Earth will now be aging at a normal rate.

Again, let's wait for the Minkowksi diagrams. That will make it all easy to discuss.

This is all true and it shows that the Earth twin aging process can only occur when velocity changes. I have been discussing an instantaneous turnaround, but the same thing applies here. This is the SR way of mathematically handling changes in velocity. When there is no change in velocity, there can be no observed fast aging of the Earth twin. The length of the inertial flight only determines what amount of aging there will be when the traveling twin turns around. It does not determine when that aging occurs. The aging happens when the traveling twin turns around.

 

[DaveC426913]

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.

 

[DaveC426913]

Ah. This one is more realistic:

 

[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)