Revisiting the Flaws of the Light Clock in Special and General Relativity

[Mentz114]

OK How do you distinguish between two inertial frames where the worldlines are totally reciprocal mirror images of each other?

Each worldline taken between two events, has a unique proper time associated with it.

WHere the passage of proper time is equivalent in each diagram as depicted by the length of the worldline how do you decide which one is accurate?

I don't follow you. Any segment of any worldline has a proper time.

Given kev's three inertial frames; if in fact you derive a difference in elapsed time between frames how do you explain this without an implication of actual motion on the part of some frame??

How do you explain the length of a piece of string ?

Out of curiosity what are you talking about " my persisting in ignoring proper time"
ANd what do you mean time dilation has nothing to do with elapsed time on clocks??
Do you think time dilation has nothing to do with the slope of worldlines and vice versa?

You're making a puzzle where there isn't one. Observing a moving clock may give you the impression that it is running slower than yours, but that's an instantaneous velocity dependent measurement. To get the proper time you have to integrate ( sum the infinitesimals) of this factor over the whole journey. It's a postulate of SR that this gives the correct elapsed time on the local clock.

The Wiki page is not bad, I recommend a quick look.

http://en.wikipedia.org/wiki/Proper_time

 

[CPL.Luke]

I think the issue here is that there is no basis for deciding who actually went on the trip and who did not.

according to the observer on Earth the twin on the spaceship went on the trip, according to the observer on the spaceship the Earth went on a trip away from him.http://www.scientificamerican.com/article.cfm?id=how-does-relativity-theor&page=2

the above article goes through the full paradox including a nice space time diagram,

I disagree with parts of the resolution, namely the distinction between which object left the reference frame, Ialso need to do the math to see if it works out from the spaceships perspective

I still believe that a key point is that the spaceship is not a valid reference frame for the entire journey, because it does undergo acceleration at 3 points, meaning that without very careful general relativity work, you would not be able to calculate the proper time of the planet earth.

also illuminating is the Hafele-Keating experiment, which experimentally tested the resolution of the twin paradox.

http://en.wikipedia.org/wiki/Hafele-Keating_experiment

 

[yuiop]

In your example all the movement is inertial and there are no inertial accelerations thus how can you make definitive statements about one clock going slower than another?

In my example you are right that I can not make definitive statements about the relative clock rates of spatially separated, beyond the trivial observation that any two ideal clocks that are at rest wrt each other tick at the same rate even when they are they spatially separated and moving relative to the observer. However, I can make a definitive statement about the elapsed proper time between any two timelike events.

Let us say A remains at rest in frame A. All references to coordinate measurements will mean measurements made by observers at rest in frame A. B passes A at coordinate time zero at a coordinate velocity of +0.8c. After a coordinate time of 10 years, B passes C who is going in the opposite direction with a coordinate velocity of -0.8c. The coordinate distance between event (B passing A) = event(B,A) and event(B,C) is 8 lightyears. Other observers in different reference frames will disagree with the coordinate times, distance and velocities measured by A and will also disagree on what clock A reads at event(B,C), but all observers will agree that 6 years of proper time elapses on clock B between events (A,B) and (B,C). (Definitive statement 1.) Eventually clock C passes A at event(C,A). All observers agree that 6 years of proper time passes on clock C between events (B,C) and (C,A). (Definitive statement 2). All observers will agree that 20 years of proper time elapses on clock A between events (B,A) and (C,A). (Definitive statement 3). All observers agree that the combined elapsed proper time between the 3 events (B,A), (B,C) and (C,A) is 12 years (Definitive statement 4) and that this proper time interval is less that the proper time interval between events (B,A) and (C,A). (Definitive statement 5).

Here I am defining the proper time between any two timelike events (the invariant time interval) as being the time measured by an inertially moving ideal clock that is coincident at both events. I am also using the term "all observers" to mean inertial observers that are not necessarily at rest in frame A.

In relation to your example with A, B, C, and D I follow all that you say but I do not see how that contradicts that acceleration changes the rate that a clock ticks with respect to one that does not accelerate. For C, when A and B accelerate to increase the relative velocity with respect to C, the relative clock rates diverge. For D however A and B accelerate (decelerate) to decrease the relative velocity with respect to D, thus the clock rates converge.

You make a nice observation here, but we are left with the problem that even when we know the proper acceleration of a given clock, there is still ambiguity about its clock rate relative to other clocks with relative motion, depending upon which reference frame the comparisons are made from. As you quite rightly pointed out, what looks like a acceleration of a clock with resultant slowing down down of the clock in one frame looks like deceleration of the clock and resultant speeding up of the clock in another frame.

How so? Please explain how you come to that conclusion.

I was talking about an old fashioned alternative method of clock synchronisation or comparison of elapsed clock times by "slow transport" of clocks. This is probably an unnecessary distraction in this thread and I probably shouldn't have mentioned it. If you are still curious about slow clock transport synchronization see http://en.wikipedia.org/wiki/One-way_speed_of_light#Slow_transport or google the terms.

This paper might be of interest to you: http://arxiv.org/ftp/physics/papers/0610/0610226.pdf

Yes, thanks, it is of interest. The authors are trying yet another way to make it clear to students that acceleration is not the resolving factor in the twins paradox, but unfortunately their approach seems even less intuitive (to me anyway) than the well known demonstrations.

 

[Mentz114]

I still believe that a key point is that the spaceship is not a valid reference frame for the entire journey, because it does undergo acceleration at 3 points, meaning that without very careful general relativity work, you would not be able to calculate the proper time of the planet earth.

No. General relativity has nothing to do with it. Nor is the question whether the spaceship is a 'valid frame' (?).

It's the proper time !

 

[gnomechompsky]

Yes, I am assuming the time dilation equations of SR are correct. What is important in the twins paradox is the path length through spacetime. On the outward trip of one of the the twins, there is some ambiguity of the time dilation of the twins relative to each other. When the paths are plotted on a spacetime diagram, it is true that the path of the Earth twin looks longer from the traveling twins frame and vice versa. There is always some ambiguity in clock rates when the clocks are not at rest with one another. However, after the traveling twin returns home, and the two clocks are alongside each other again, there is no ambiguity. The path of the traveling twin through spacetime is ALWAYS longer than the Earth twins path, from the point of view of ANY inertial observer. If we define relative time dilation in terms of spacetime path lengths, then the ambiguities can be made to disappear when the two clocks come to rest wrt to each other even if they are spatially separated.

I don't understand this. An inertial observer moving at half the velocity of the traveller could legitimately see the traveller's and the Earth twin's path lengths as being the same.

Whilst this thread has gotten quite big, the point I am actually trying to bring out (as you are probably aware), is how do we know the time dilation SR equations are valid if the light clock thought experiment can be applied equally to the "traveller" and the earth-bound twin, because there are no special frames of reference?

Happy to ignore acceleration completely.

 

[yuiop]

I don't understand this. An inertial observer moving at half the velocity of the traveller could legitimately see the traveller's and the Earth twin's path lengths as being the same.

This is true on the travellers outward leg, when the twins are spatially separated. The third observer (C) with intermediate velocity says equal proper times have elapsed for the Home twin (A) and the traveling twin (B), when B arrives at the turnaround point, while from B's point of view, A has the longer path and the shortest elapsed proper time and A says B has the longest path and shortest elapsed proper time. Everyone has a different opinion and ambiguity reigns. Once B turns around and arrives back home, observers A, B and C will all agree that B has traveled a longer path through spacetime and than A and all will agree that B has less elapsed proper time than A. All ambiguity is removed once the clocks return to a common location and the differences are unambiguously accounted for in all frames in terms of differences in path lengths.

Whilst this thread has gotten quite big, the point I am actually trying to bring out (as you are probably aware), is how do we know the time dilation SR equations are valid if the light clock thought experiment can be applied equally to the "traveller" and the earth-bound twin, because there are no special frames of reference?

Happy to ignore acceleration completely.

As mentioned in other posts by Mentz, it is better to think in terms of proper time between events, which can be unambiguously defined, rather than in terms of relative time dilation of spatially separated clocks with relative motion, which is ambiguous. When two clocks have non-zero relative motion, there is by definition no single reference frame in which they are both at rest and no way to unambiguously define their relative instantaneous clock rates.

Even when B has turned around and is on his way home, it is still possible to find reference frames in which more time has elapsed for B than for A. It is not the turn around (or the acceleration involved) that removes the ambiguity. It is the arrival of the two clocks at common location that removes the ambiguity about the elapsed proper times, but if the twins do not come to rest wrt each other, they will still disagree about their instantaneous relative clock rates at the final passing event.

 

[gnomechompsky]

This is true on the travellers outward leg, when the twins are spatially separated. The third observer (C) with intermediate velocity says equal proper times have elapsed for the Home twin (A) and the traveling twin (B), when B arrives at the turnaround point, while from B's point of view, A has the longer path and the shortest elapsed proper time and A says B has the longest path and shortest elapsed proper time. Everyone has a different opinion and ambiguity reigns. Once B turns around and arrives back home, observers A, B and C will all agree that B has traveled a longer path through spacetime and than A and all will agree that B has less elapsed proper time than A.

I still don't think you are justifying this. If all movement is relative, and A,B and C were all in inertial frames of reference, why has B gone through the longer path when he returns to A? Is it not equally valid to say that A (Earth) moved away from B then returned to B?

 

[yuiop]

I still don't think you are justifying this. If all movement is relative, and A,B and C were all in inertial frames of reference, why has B gone through the longer path when he returns to A? Is it not equally valid to say that A (Earth) moved away from B then returned to B?

Your third observer C would certainly never see A (Earth) moving away from B and then returning to B. Observer B would feel the acceleration as he changed direction, while observer A never feels a change in direction, so the two situations are certainly not equivalent or symmetrical from that point of view. The observer that feels or meausres proper acceleration is the one that has really changed direction and acceleration is absolute in that sense. I can only suggest you get some graph paper and try drawing the the space versus time diagrams yourself. That is one of the best ways to get an intuitive understanding of what is happening.

Note: You said "If all movement is relative, and A,B and C were all in inertial frames of reference, why has B gone through the longer path when he returns to A? the important word is "were". If B has returned to A, B has not remained in an inertial frame.

 

[Passionflower]

I was talking about an old fashioned alternative method of clock synchronisation or comparison of elapsed clock times by "slow transport" of clocks. This is probably an unnecessary distraction in this thread and I probably shouldn't have mentioned it. If you are still curious about slow clock transport synchronization see http://en.wikipedia.org/wiki/One-way_speed_of_light#Slow_transport or google the terms.

I am familiar with the term but I still like to understand what you mean, do you have the calculations? Note that if two clocks run with a different rate and they want to meet halfway they will disagree about where halfway actually is, slow transport or not.

 

[CPL.Luke]

Mentz,

I think this is the time where you have to demonstrate how the observer in the spaceship would go about calculating the proper time of the observer on the earth, from their reference frame.

you are correct that the proper times will differ, however with just plane SR, you can't calculate the proper time of the Earth correctly (from the space ships perspective). If you attempt to do this in SR you will find that the earthbound twin has aged less than the spacebound twin.

EDIT: by valid frame I meant inertial frame, the space ships frame is non inertial at various points, thus in order to properly calculate the proper time you would have to sum over these areas, while you could do this for certain special examples (such as rindler space) the solution is not so simple as to look at the proper time.

Ignoring the non-inertial parts of the spacetime diagram causes the calculation to fail from the spaceships perspective.

 

[Austin0]

Let us say A remains at rest in frame A. All references to coordinate measurements will mean measurements made by observers at rest in frame A. B passes A at coordinate time zero at a coordinate velocity of +0.8c. After a coordinate time of 10 years, B passes C who is going in the opposite direction with a coordinate velocity of -0.8c. The coordinate distance between event (B passing A) = event(B,A) and event(B,C) is 8 lightyears. Other observers in different reference frames will disagree with the coordinate times, distance and velocities measured by A and will also disagree on what clock A reads at event(B,C), but all observers will agree that 6 years of proper time elapses on clock B between events (A,B) and (B,C). (Definitive statement 1.) Eventually clock C passes A at event(C,A). All observers agree that 6 years of proper time passes on clock C between events (B,C) and (C,A). (Definitive statement 2). All observers will agree that 20 years of proper time elapses on clock A between events (B,A) and (C,A). (Definitive statement 3). All observers agree that the combined elapsed proper time between the 3 events (B,A), (B,C) and (C,A) is 12 years (Definitive statement 4) and that this proper time interval is less that the proper time interval between events (B,A) and (C,A). (Definitive statement 5).

Here I am defining the proper time between any two timelike events (the invariant time interval) as being the time measured by an inertially moving ideal clock that is coincident at both events. I am also using the term "all observers" to mean inertial observers that are not necessarily at rest in frame A.

It would seem that you have demonstrated that frame C was actually moving relative to frame A as the inertial clock located in C and moving toward A had less elapsed proper time than inertial clock A moving toward clock C at the same relative v.

Explanation?

 

[gnomechompsky]

Your third observer C would certainly never see A (Earth) moving away from B and then returning to B.

Disagree. If C was at 1/2 the velocity of B (and also began his return to C at the same time) he would see both moving away from him (and then towards him) at the same speed.

Observer B would feel the acceleration as he changed direction, while observer A never feels a change in direction, so the two situations are certainly not equivalent or symmetrical from that point of view.

Ok, I can agree on this, but as I said in my first post, the only asymmetry in the Twins Paradox comes from acceleration.

However, what is it about the light clock thought experiment that makes it valid on B from A's frame of reference, but not valid on A from B's frame of reference? The acceleration doesn't matter, because B will still see A's light clock as "dilating", on both the outward and return journey. Is there another way to derive the time dilation equation without this paradox?

I can only suggest you get some graph paper and try drawing the the space versus time diagrams yourself. That is one of the best ways to get an intuitive understanding of what is happening.

I can't do this yet, because it would seem to be easily possible to reverse the space/time graph from the frame of reference for B and get the same result.

 

[Austin0]

Originally Posted by kev
Let us say A remains at rest in frame A. All references to coordinate measurements will mean measurements made by observers at rest in frame A. B passes A at coordinate time zero at a coordinate velocity of +0.8c. After a coordinate time of 10 years, B passes C who is going in the opposite direction with a coordinate velocity of -0.8c. The coordinate distance between event (B passing A) = event(B,A) and event(B,C) is 8 lightyears. Other observers in different reference frames will disagree with the coordinate times, distance and velocities measured by A and will also disagree on what clock A reads at event(B,C), but all observers will agree that 6 years of proper time elapses on clock B between events (A,B) and (B,C). (Definitive statement 1.) Eventually clock C passes A at event(C,A). All observers agree that 6 years of proper time passes on clock C between events (B,C) and (C,A). (Definitive statement 2). All observers will agree that 20 years of proper time elapses on clock A between events (B,A) and (C,A). (Definitive statement 3). All observers agree that the combined elapsed proper time between the 3 events (B,A), (B,C) and (C,A) is 12 years (Definitive statement 4) and that this proper time interval is less that the proper time interval between events (B,A) and (C,A). (Definitive statement 5).

Here I am defining the proper time between any two timelike events (the invariant time interval) as being the time measured by an inertially moving ideal clock that is coincident at both events. I am also using the term "all observers" to mean inertial observers that are not necessarily at rest in frame A.

Suppose there is a frame D such that A and C are traveling relative to it, at 0.5c and -0.5c
respectively. At event (B,C) D.. t'''=0 and the spatial interval between A and C is dx''' with observed clock times in A and C of t₀ and t''₀

As observed in D ...A and C meet at dx'''/2 with observed clock times in A and C of T and T''
In both frames dx'''/ 0.5 = dt , dt'' from this it would seem to follow that dt = T-t₀ =T''-t''₀= dt'' Wouldn't this be equal elapsed proper time observed in both frames between events (B,C) and (C,A) ?

Do you see some reason why this would not apply??

 

[Mentz114]

Mentz,
I think this is the time where you have to demonstrate how the observer in the spaceship would go about calculating the proper time of the observer on the earth, from their reference frame.

Given any worldline, one can calculate the elapsed time between two events on it. Furthermore, all observers will agree on this. So the elapsed time as defined above will be the same from all IFRs.

..., however with just plain SR, you can't calculate the proper time of the Earth correctly (from the space ships perspective). If you attempt to do this in SR you will find that the earthbound twin has aged less than the spacebound twin.

I don't agree with this. As I've said, the proper time is the same in all frames.

EDIT: by valid frame I meant inertial frame, the space ships frame is non inertial at various points, thus in order to properly calculate the proper time you would have to sum over these areas, while you could do this for certain special examples (such as rindler space) the solution is not so simple as to look at the proper time.

In the classic twins 'paradox', we don't need Rindler coords, and the resolution is to ignore the instantaneous velocity dependent time-dilation and use the proper time to calculate what the elapsed times are.

Ignoring the non-inertial parts of the spacetime diagram causes the calculation to fail from the spaceships perspective.

The proper time interval is found by integrating along the whole worldline. It doesn't matter if parts are non-inertial.

I don't understand why the differential ageing of the twins is seen as some sort of paradox. Usually the pro-paradox arguments are

1. symmetry - both twins see the other as moving.
That's irrelevant, it's the proper times that give the elapsed times
2. each sees the others clock running slowly
also irrelevant.

 

[gnomechompsky]

1. symmetry - both twins see the other as moving.
That's irrelevant, it's the proper times that give the elapsed times.

How do we derive the formula for proper time? Is it from our geometric understanding of the light clock thought experiment? We can't use proper time to resolve the Twin Paradox if we have not yet justified the premise on which it is based.

Is it possible to derive proper time without using the light clock?

 

[Mentz114]

How do we derive the formula for proper time? Is it from our geometric understanding of the light clock thought experiment? We can't use proper time to resolve the Twin Paradox if we have not yet justified the premise on which it is based.

Is it possible to derive proper time without using the light clock?

In Minkowski spacetime, the infinitesimal line element is given by this formula

<br /> c^2d\tau^2=c^2dt^2-dx^2-dy^2-dz^2<br />

and this forms the basis of the calculation. It is the geometry of Minkowski spacetime, and a postulate of SR is that \tau is the time measured by a clock on the worldline.

I don't see that the light clock comes into at all.

 

[gnomechompsky]

Well Minkowski must have derived the formula somehow. What experiment did he do to get this?

Also, it would seem that the formula above doesn't address the paradox because either side of the equation can equally be applied to either observer. Why is the twin on Earth not on a world line moving away from the Traveller?

 

[Mentz114]

Well Minkowski must have derived the formula somehow. What experiment did he do to get this?

Good question. There is experimental support for SR, but I think Minkowski just intuited the line element.

Also, it would seem that the formula above doesn't address the paradox because either side of the equation can equally be applied to either observer.

The formula applies to whatever segment of worldline you want it to, and it will give a number.

Why is the twin on Earth not on a world line moving away from the Traveller?

From the travellers point of view the Earth is moving away.

I don't know how you are with spacetime diagrams but here's two that show a traveller from the Earth frame, and the Earth from the travellers frame on his outbound trip.

Reading the propertimes off the graphs we find

earth frame : Earth elapsed time ~13
first leg frame : Earth elapsed time sqrt[(14.5*14.5)-(5.5*5.5)] ~ 13

The point is that applying the proper time formula, you get same elapsed time on a segment from any frame.

 

[gnomechompsky]

On those graphs why is Earth always on the straight line? Why can't we make a graph where the traveller is on the straight line and Earth reverses direction?

 

[Mentz114]

Elapsed time on a non-inertial worldline

This graph shows a traveller doing a smooth trip out and back. The worldline is given by

x =0.01 (t² - 1 )⁴

using numerical integration the elapsed time on the traveller's clock is 1.9998564 compared with 2.0000 on the Earth clock.

 

[gnomechompsky]

This graph shows a traveller doing a smooth trip out and back. The worldline is given by

x =0.01 (t² - 1 )⁴

using numerical integration the elapsed time on the traveller's clock is 1.9998564 compared with 2.0000 on the Earth clock.

I think you misunderstood my question. Referring back to the original two graphs, why can we not just switch the reference frames so it is Earth that is moving away and then doing a return journey?

Same with the graph above, the blue curved line could equally be Earth from the frame of reference of the Traveller.

 

[CPL.Luke]

Mentz you are assuming that it is minkowski space for the traveler, if that was the case then the equations you show would have both parties disagreeing on who was younger.

The problem is solved from the travelers frame when you realize that the traveler passes through rindler space each time he accelerates, the Earth is always in minkowski space.

To illustrate this, in post 50 you performed the calculation from the Earth's perspective, can you perform it from the travelers perspective? your solution is correct, but for the wrong reasons.

 

[Mentz114]

I think you misunderstood my question. Referring back to the original two graphs, why can we not just switch the reference frames so it is Earth that is moving away and then doing a return journey?

Because finding rocket motors powerful enough to accelerate the Earth is a problem. But if you could do it, then the Earth's worldline would be curved and its proper time shorter.

Mentz you are assuming that it is minkowski space for the traveler, if that was the case then the equations you show would have both parties disagreeing on who was younger.

Proper time is invariant under Lorentz transformation, and any instant on the curved world line is connected to the Earth worldline by a Lorentz transformation.

The problem is solved from the travelers frame when you realize that the traveler passes through rindler space each time he accelerates, the Earth is always in minkowski space.

It is always Minkowski space. There's no 'Rindler' spacetime, only Rindler coords in Minkowski spacetime.

To illustrate this, in post 50 you performed the calculation from the Earth's perspective, can you perform it from the travelers perspective? your solution is correct, but for the wrong reasons.

Please explain why the calculation is wrong. Seriously, if you think that the elapsed times are frame dependent for the trip under question then demonstrate it.

 

[gnomechompsky]

Because finding rocket motors powerful enough to accelerate the Earth is a problem. But if you could do it, then the Earth's worldline would be curved and its proper time shorter.

Ah, so we are back to the point in my original post, it is the acceleration "felt" by the object that generates the asymmetry in the Twins Paradox.

However, even if we know that it is the Space traveller accelerating, according to the formulas for proper time and SR time dilation, acceleration is not a factor.

 

[Mentz114]

Mentz you are assuming that it is minkowski space for the traveler, if that was the case then the equations you show would have both parties disagreeing on who was younger.

To return to this, if you applied the rules, then any point on the curved worldline has a tangent at which we can assume a comoving inertial frame with velocity equal to the tangent. This is still connected to the Earth frame by Lorentz transformation so the calculation of the Earth proper time would give the same answer. Using this approach there is no disagreement.

Ah, so we are back to the point in my original post, it is the acceleration "felt" by the object that generates the asymmetry in the Twins Paradox.

However, even if we know that it is the Space traveller accelerating, according to the formulas for proper time and SR time dilation, acceleration is not a factor.

I've forgotten a long time ago what this thread was about

 

[Austin0]

Because finding rocket motors powerful enough to accelerate the Earth is a problem. But if you could do it, then the Earth's worldline would be curved and its proper time shorter.
Proper time is invariant under Lorentz transformation, and any instant on the curved world line is connected to the Earth worldline by a Lorentz transformation.
Please explain why the calculation is wrong. Seriously, if you think that the elapsed times are frame dependent for the trip under question then demonstrate it.

You and kev seem to have both ignored post #43 in which I posed a possible counter for kev's demonstration of the problem (with only inertial frames). And a different conclusion wrt elapsed proper time.
If there is some flaw in what I presented I would be glad to learn of it.

You have consistently failed to address the OP's real question.
Disregarding acceleration as a factor (which you have in fact included above) but simply operating with the essential kinematic assumption of SR, what would prevent a perfectly symmetrical Minkowski diagram of the Earth from the perspective of the accelerated frame??
A mirror image??
A plot of the spacetime locations of the Earth relative to the accelerated frame.
I.e. The Earth having the same curved segments, inertial segments etc.
If this was done you would then have totally reciprocal and mutually exclusive elapsed proper times as derived from integrated worldlines between events , no??
As far as I have seen the OP has a relevant question and having read a great many twin threads it appears that there is no real consensus as to the proper resolution and whether acceleration is a crucial criteria or not. Everyone agrees there is no real paradox and agrees on the outcome but there still seems to be questions worth pursuing regarding both acceleration and simultaneity. IMHO

 

[matheinste]

You and kev seem to have both ignored post #43 in which I posed a possible counter for kev's demonstration of the problem (with only inertial frames). And a different conclusion wrt elapsed proper time.
If there is some flaw in what I presented I would be glad to learn of it.

You have consistently failed to address the OP's real question.
Disregarding acceleration as a factor (which you have in fact included above) but simply operating with the essential kinematic assumption of SR, what would prevent a perfectly symmetrical Minkowski diagram of the Earth from the perspective of the accelerated frame??
A mirror image??
A plot of the spacetime locations of the Earth relative to the accelerated frame.
I.e. The Earth having the same curved segments, inertial segments etc.
If this was done you would then have totally reciprocal and mutually exclusive elapsed proper times as derived from integrated worldlines between events , no??
As far as I have seen the OP has a relevant question and having read a great many twin threads it appears that there is no real consensus as to the proper resolution and whether acceleration is a crucial criteria or not. Everyone agrees there is no real paradox and agrees on the outcome but there still seems to be questions worth pursuing regarding both acceleration and simultaneity. IMHO

No matter how you dress it up, or what scenario you propose, in SR, the proper time along a worldline, that is the time measuired by a clock that is present at all points along the worldline, is a measure of the spacetime distance along that worldline. There s no ambiguity and no lack of consensus on this point. Whether the Earth accelerates or the rocket accelerates or both accelerate, whether symmetrically or not, the proper time along the wordlines can measured by a clock traveling along the wordline or calculated by any observer, and compared. There will no disagreement about the reuslts.

As for the effect of acceleration on clocks, the clock hypothesis assumes ther is no effect and this is borne out by laboratory performed experiments to a very high degree. Of course for diiferential time intervals, which of couse require curved wordlines to describe them geometrically, we need acceleration to produce them, but it this curvature of the spactime path which leads to the different proper times, or not, along two wordlines. The acceleration is the cause of this curvature but has no direct effects on clocks. The compared readings on the clocks are in effect, loosely speaking, a measure of the relative curvature of the wordlines.

Matheinste

 

[Mentz114]

I go along with Matheinste's post above which summarises the situation very well. I wish the 'twin paradox' would go away ( although it's fun sometimes talking about it).

 

[yossell]

I go along with Matheinste's post above which summarises the situation very well. I wish the 'twin paradox' would go away ( although it's fun sometimes talking about it).

Oh - I found understanding why the twins paradox is not a paradox an essential part of my coming to understand relativity. The barn paradox too. I think they should be as much a part of a relativity course as the Lorentz equations.

 

[Mentz114]

Oh - I found understanding why the twins paradox is not a paradox an essential part of my coming to understand relativity. The barn paradox too. I think they should be as much a part of a relativity course as the Lorentz equations.

Yes, you're right. The barn and pole came up as an exam question once for me. It seemed to me that I understood SR much better when I fathomed space-time geometry, which answers most questions. When in doubt, draw a diagram.