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

[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.

 

[Passionflower]

The compared readings on the clocks are in effect, loosely speaking, a measure of the relative curvature of the wordlines.

That is incorrect, curvature or relative curvature has nothing to do with it. If it did then the clock hypothesis would be false.

Think of the twin paradox where the traveling twin follows the path of two edges of a triangle while the at home twin has a straight line. In Minkowski spacetime the total length (which is equal to the elapsed proper time) of the edges of the triangle is smaller than the straight line between them, which of course opposite in Euclidean space, a reversal of the triangle inequality.

 

[matheinste]

That is incorrect, curvature or relative curvature has nothing to do with it. If it did then the clock hypothesis would be false.

Think of the twin paradox where the traveling twin follows the path of two edges of a triangle while the at home twin has a straight line. In Minkowski spacetime the total length (which is equal to the elapsed proper time) of the edges of the triangle is smaller than the straight line between them, which of course opposite in Euclidean space, a reversal of the triangle inequality.

Any depiction of realistic acceleration requires a curved path to represent it in a spacetime diagram. I do however see the point you are trying to make and perhaps my choice of words was not appropriate.

Thanks for the help but I am pretty much at home with the properties of the geometry required for the spacetime metric.

Matheinste.

 

[phyti]

For simplicity the times are referenced as A or B followed by the year.
Twin B leaves twin A moving at .8c, reverses direction at B12, and returns.
Fig. 1 shows A's view of B's trip. The axis of simultaneity (gray) for B is (A7.2, B12) outbound, and (B12, A32.8) inbound. The instantaneous jump from A7.2 to A32.8 is due to excluding any period of acceleration for B to transfer from the outbound to the inbound frame of reference. The ratio of B-time to A-time is 24/40 = .60.

Fig. 2 is B's view using the Einstein simultaneity convention. The discontinuous motion of A at A4 reflects the switching of frames without acceleration. The extreme distortion of times and locations, using this convention, is noted with A4 simultaneous with B-36, 36 years before they parted! The ratio of A-time to B-time is 4/6.7 = 36/60 = .60 for both path segments.

At this point the slow clock rate is reciprocal.

Fig. 3 is B's view using a horizontal axis of simultaneity, i.e. a translation of positions, and A moving at -.8c. Since the initial conditions place A in the 'chosen' static frame, B must be moving, therefore if he had the means to measure the 1-way light speed, it would not be c the absolute speed through space, but the speed relative to him, c-v and c+v (magenta). The ratio of A-time to B-time is 4/2.4 = 36/21.6 = 1.67, in agreement with the result for the closed path. The time dilation is now asymmetrical as calculated by both A and B. The extreme space and time shifts are also removed.

An example of relative speeds.
The question: How much time is required for a car moving at 60 mph, to overtake a car moving at 50 mph with a 1 mile lead?
The answer: distance/(v1-v2) = 1/(60-50) = 1/10 = .1 hr = 6 min.
It's the relative or closing speed that determines the answer. Neither car would expect the other to approach at 60 mph. If the lead car used 60 for the chase car rate, the initial separation would have been 60*.1 hr = 6 miles, not 1 mile. The absolute car speed is relative to the ground. The relative car speed is relative to the other car. They are two different types of relations. If light replaces the chase car, its speed c is relative to space, defined as an invisible but fixed frame of reference, and its relative speed as c-v, with v the speed of the object being chased. The fact that relative light speed is different from c, doesn't contradict its absolute speed, no more than the 10 mph closing speed alters the 60 mph chase car speed.

Fig. 4 shows a more realistic case with a short period of acceleration for B transitioning between frames. B would explain the curved portion of A's motion as resulting from an equivalent g-field during his acceleration. This also provides an asymmetrical view with 40 A events to 24 B events.

The simultaneity definition is the source of the 'paradox' where unequal path lengths are defined as equal, for the purpose of preserving constant light speed.
https://www.physicsforums.com/attachments/27694

 

[Austin0]

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

Although everything you have said here is true it does not actually address anything I said.
There is no disagreement regarding the fundamental interpretation of worldlines and integrated path lengths as proper time. I certainly have no question in this regard and from what I have read from the OP that is not his point or question either.
As far as the clock hypothesis goes, I have myself on more than one occasion cited it as a counter to the acceleration argument with regard to non-reciprocal dilation.
So all of your post is another straw man argument aimed at questions never asked and positions never stated.
In the meantime totally ignoring explicit questions and arguments.
kev made , in his own words "difinitive " statements wrt his own scenario with only inertial frames.
Part of this was the assertion that his conclusions would be agreed upon by all inertial observers.
I offered a simple and clear example of one possible inertial frame that seemed to disagree with this definitive statement.
I could very well be wrong. If that is so then it should be simple for you to show me this and demonstrate your case.

I offered the simple premise that simply based on kinematics and coordinate systems it is easily possible to plot the Earth's time/space positions relative to the accelerated system with the accelerated system at rest. AS per the OP's repeated question. If this is done the application of the same integration of the Earth's worldline , the same interpretation of proper time , would produce the reciprocal but opposite conclusion.
This is not done for other reasons. I.e. an acclerated frame is not valid, acceleration is real etc etc etc.
I am not saying these are not valid reasons only that to simply say proper time is the intergration of the worldline simply begs the question.
I certainly have no answer as to the best resolution to the twin question or non-reciprocal dilation , only the belief that it represents a meaningful question

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).

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''
For both frames dx'''*0.5/ 0.5c = 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??

 

[matheinste]

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

My apologies if my reply was irrelevant. I was commenting on the above part of your post.
I'm afraid when I come accoss the words "Twin paradox" in the same paragraph as "no real consensus as to the proper resolution" I cannot help myself, its an automatic reaction. Most regular posters here have developed a mechanism for overcoming such reactions. I must try harder to do the same

Matheinste.

 

[yuiop]

I offered the simple premise that simply based on kinematics and coordinate systems it is easily possible to plot the Earth's time/space positions relative to the accelerated system with the accelerated system at rest. AS per the OP's repeated question. If this is done the application of the same integration of the Earth's worldline , the same interpretation of proper time , would produce the reciprocal but opposite conclusion.

I will probably end up having to draw some diagrams for this explanation, but this is my initial attempt using only words.

In the space time diagram of the stay on Earth twin (ignoring gravity) the Earth twins wordline is vertical and the traveling twin has the longer path corresponding to the other two sides of a triangle. In this space time diagram, all horizontal lines are lines of simultaneity according to the Earth twin and in this diagram longer paths unambiguously represent paths with the shortest proper time interval when clocks share an initial starting location and share a final end location.

Now if we try and draw the space time diagram of the traveling twin, it is no longer true that all lines of simultaneity according to the traveling twin are horizontal and parallel. Some event E1 on the traveling twins worldline that the traveling twin considers to happen simultaneously with event E1' on the Earth twins worldline, can also be simultaneous with some other later event E2' on the Earth twins worldline. Because lines of simultaneity in the traveling twins space time diagram causes ambiguous interpretation of simultaneous events in his own frame, it is difficult to draw any definitive conclusions from the traveling twins space time diagram and you can certainly cannot conclude that that a longer worldline in the spacetime diagram of a non-inertial observer is the path of least proper time. This is only true for spacetime diagrams of inertial observers and the traveling twin does not qualify as an inertial observer.

 

[yossell]

My apologies if my reply was irrelevant. I was commenting on the above part of your post.
I'm afraid when I come accoss the words "Twin paradox" in the same paragraph as "no real consensus as to the proper resolution" I cannot help myself, its an automatic reaction.

I don't even know what consensus is, so I'm not in a position to comment. There is of course great consensus about the resolution of paradox, but I'm sure Austin0 never meant to question this.

 

[Austin0]

I will probably end up having to draw some diagrams for this explanation, but this is my initial attempt using only words.

In the space time diagram of the stay on Earth twin (ignoring gravity) the Earth twins wordline is vertical and the traveling twin has the longer path corresponding to the other two sides of a triangle. In this space time diagram, all horizontal lines are lines of simultaneity according to the Earth twin and in this diagram longer paths unambiguously represent paths with the shortest proper time interval when clocks share an initial starting location and share a final end location.

Now if we try and draw the space time diagram of the traveling twin, it is no longer true that all lines of simultaneity according to the traveling twin are horizontal and parallel. Some event E1 on the traveling twins worldline that the traveling twin considers to happen simultaneously with event E1' on the Earth twins worldline, can also be simultaneous with some other later event E2' on the Earth twins worldline. Because lines of simultaneity in the traveling twins space time diagram causes ambiguous interpretation of simultaneous events in his own frame, it is difficult to draw any definitive conclusions from the traveling twins space time diagram and you can certainly cannot conclude that that a longer worldline in the spacetime diagram of a non-inertial observer is the path of least proper time. This is only true for spacetime diagrams of inertial observers and the traveling twin does not qualify as an inertial observer.[/QUOTE]

It is completely unneccessary to draw diagrams. Particulary since you would attempt to draw them with the assumptions derived from drawings with the Earth as the rest frame.
You are missing the point. If you remove the physics i.e. the fact that accelerometers tell us the space twin is accelerating. Based on the purely kinematic assumption of SR that motion is relative there is no reason why the lines of simultaneity in a diagram with the space twin at rest would not be parallel. WIth the Earth lines with varying degrees of slope.
With the Earth worldline with curved segments. Etc
I am in no way suggesting this is a productive or valid course to follow only that in the end , this course is essentially eliminated on the basis of your last sentence , which in fact I had made explicit in previous posts in this thread and don't disagree with. ANd this very point is also explicitly the question of the OP.
You have in this thread attempted to eliminate the physics, acceleration, as a determining factor for a preferred frame i.e. the Earth and show it with only inertial frames involved..
You have still ignored my post relevant and counter to your demonstration, why is this??

 

[Austin0]

I don't even know what consensus is, so I'm not in a position to comment. There is of course great consensus about the resolution of paradox, but I'm sure Austin0 never meant to question this.

Nope not me. At least not the consensus that there is in fact no paradox.
But there are resolutions that do not directly involve acceleration as a crucial determinent.
And a certain lack of consistency between various valid and logical resolutions IMHO

 

[yossell]

Whoooossshhhhh!

 

[Austin0]

My apologies if my reply was irrelevant. I was commenting on the above part of your post.
I'm afraid when I come accoss the words "Twin paradox" in the same paragraph as "no real consensus as to the proper resolution" I cannot help myself, its an automatic reaction. Most regular posters here have developed a mechanism for overcoming such reactions. I must try harder to do the same

Matheinste.

No aps neccessary. If you misinterpreted my words as meaning there was any question at all about there being a satisfactory resolution I can understand your reaction.
As I understood the OP he was also not questioning that there was a resolution but only if there was a resolution that did not involve acceleration as a deciding factor.

 

[matheinste]

If the scenario is the usual one, twin leaves earth, SAME twin returns to Earth then acceleration cannot be avoided.

In other proposed scenarios such as the one where we have an observer who remains on earth, a traveller passing Earth outbound, setting his clock to Earth time in passing, meeting a third observer traveling inbound, setting clocks to be the same when they meet at the agreed turnaround point, and this third observer not stopping at Earth on the way back but comparing clocks in passing, then acceleration by none of the observers is involved. But the out and inbound travellers do not share the same inertial frame and do not agree about the age of the earthly observer when they meet at what would have been the turnaround point in the original scenario. The result will of couse be the similar, the inbound observer's clock will have a reading less than the Earth clock.

Whatever scenario is proposed in SR, the solution is availible and follows logically from the axioms.

Matheinste.

 

[Austin0]

If the scenario is the usual one, twin leaves earth, SAME twin returns to earth then acceleration cannot be avoided.

In other proposed scenarios such as the one where we have an observer who remains on earth, a traveller passing Earth outbound, setting his clock to Earth time in passing, meeting a third observer traveling inbound, setting clocks to be the same when they meet at the agreed turnaround point, and this third observer not stopping at Earth on the way back but comparing clocks in passing, then acceleration by none of the observers is involved. But the out and inbound travellers do not share the same inertial frame and do not agree about the age of the earthly observer when they meet at what would have been the turnaround point in the original scenario. The result will of couse be the similar, the inbound observer's clock will have a reading less than the Earth clock.
Whatever scenario is proposed in SR, the solution is availible and follows logically from the axioms.

Matheinste.

Acceleration cannot be avoided to create the actual occurences but it is not neccessarily crucial to the analysis or resolution. It has been done with reciprocal dilation throughout and the final outcome attributed to relative simultaneity.
Here again you have asserted that the conclusion would be the same regarding purely inertial frames but have not examined or countered my explicit exception to this claim.
Why is this? It is vey simple and right there. DO you not understand it , was I not clear enough perhaps?

 

[matheinste]

Acceleration cannot be avoided to create the actual occurences but it is not neccessarily crucial to the analysis or resolution. It has been done with reciprocal dilation throughout and the final outcome attributed to relative simultaneity.
Here again you have asserted that the conclusion would be the same regarding purely inertial frames but have not examined or countered my explicit exception to this claim.
Why is this? It is vey simple and right there. DO you not understand it , was I not clear enough perhaps?

I have lost track a bit here, my fault. If you propose a single question or counter example, or guide me to the particular post, I would be happy to try and answer or at least explain why I cannot answer it.

Matheinste.

 

[CPL.Luke]

I agree with Matheinste,

The point I was trying to make is that if you pretend that the spaceship instantly accelerates in the opposite direction, then the problem is symmetrical and there is a paradox.

however once you acknowledge that the spaceship must undergo acceleration, then you can just use the rindler metric, which is

ds²=-a²x²c²dt²+dx²

the problem is no longer symmetrical, and once you calculate it out both parties should agree on the amount of proper time which has elapsed.

alternatively if your careful with how the observers will observe each others clocks you'll get the right answer. Just as Austin0 described.

 

[Austin0]

I have lost track a bit here, my fault. If you propose a single question or counter example, or guide me to the particular post, I would be happy to try and answer or at least explain why I cannot answer it.

Matheinste.

It is in post #64 but here is the question:

Originally Posted by kev
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).

Originally Posted by Austin0
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) ..in D.. t'''=0 and the spatial interval between A and C is dx''' with observed clock times in A of t₀ and in C of t''₀

As observed in D ...A and C meet at dx'''/2 with observed clock times in A of (T) and in C of (T'') event (C,A)
For both frames dx'''*0.5/ 0.5c = 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??

 

[Austin0]

Hi kev I kept waiting for a responce and then sort of forgot but not completely, ;-)
Nor have I forgotten your telling me that if I could demonstrate non-reciprocal time dilation between inertial frames it would mean demonstrating real motion. So I am hoping for some clarification here.

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).

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?

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

As observed in D ...A and C meet at dx'''/2 with observed clock times in A of (T) and in C of (T'') this is event (C,A) as per kev.
For both frames,( A and C) ---- dt and dt'' = dx'''*0.5/ 0.5c 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??

 

[yuiop]

Hi kev I kept waiting for a responce and then sort of forgot but not completely, ;-)
Nor have I forgotten your telling me that if I could demonstrate non-reciprocal time dilation between inertial frames it would mean demonstrating real motion. So I am hoping for some clarification here.

Sorry about the delay. I have a number of threads I would like to analyse and a number of projects I am working on such as working out the equation for the time dilation of an accelerating light clock and doing a java program to demonstrate 2D relativistic transformations for light clocks, barn-pole paradox, Thomas rotation etc. but the spare time I have for such things seems to be taken up defending unjustified attacks on my work from you-know-who.

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?

The trouble here is that different observers will disagree with where A was located spatially relative to event(B,C) and what time was showing on A's clock at event(B,C), because A was not located at that event.

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

As observed in D ...A and C meet at dx'''/2 with observed clock times in A of (T) and in C of (T'') this is event (C,A) as per kev.
For both frames,( A and C) ---- dt and dt'' = dx'''*0.5/ 0.5c 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??

O.K, you are right that this is how things would look from D's point of view, but that is just D's point of view and D is not a final arbitrator. At event(B,C), D will consider A to be 3.46 ly away to the left and C to be 3.46 ly to the right. From D's POV , C is coming towards him at 0.5c so C arrives at D's location in 3.46/0.5 = 6.9282 years by D's clock and A arrives at the same time. Because B is moving at 0.5c relative to D, D will work out the the proper time elapsed on B's clock is 6.9282*sqrt(1-0.5^2) = 6 years. So yes, for this leg of the journey D considers the elapsed times on A and C's clocks to be the same. However, for the first leg of the journey, when B is going away from A, D sees A coming towards him at 0.5c (same as in the final leg) but B is going away from him at (0.8-0.5)/(1-0.8*0.5)=0.92857c so the time dilation of B's clock on the outward journey is much greater than the time dilation of A's clock, according to D. The first leg of the experiment took 14 years of A's proper time in D's frame. This equates to 14/sqrt(1-0.5^2)=16.1658 years by D's clock. D calculates that the proper time of B's clock on the outward journey "during the same time" is 16.1658*sqrt(1-0.92857^2)=6 years. Overall, D says 20 years of proper time elapsed on A's clock and 12 years of proper time elapsed on the combined times of B and C and all is good.

As you can see from the above calculations, D says A was 3.46*2=6.92 lys away from event(B,C) "at the time" B and C crossed paths, while A says he was 8 lys away from B and C "at the time" B and C crossed paths, so A and D differ on where A was "at the time" B and C crossed and what time was showing on A's clock "at the time" B and C crossed.

Hope that helps.

 

[Austin0]

Hi kev
Well you know I agree completely with you about the problems with the acceleration explanation and have said some of the same things you just posted , so i would be happy if you came up with a totally inertial scenario.

You didn't mean to make a suggestion of absolute motion, but if there is any way of measuring non-reciprocal time dilation, then that certainly implies a way of measuring absolute motion.

So you went directly from saying the above to yhr brlow

All observers will agree that 20 years of proper time elapses on clock A between events (B,A) and (C,A). (Definitive statement 3).While only 12 years elapse on B and C

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.

You ignored this one.If you were successful with this exercise, it would be demonstrating real motion would it not?

O.K, you are right that this is how things would look from D's point of view, but that is just D's point of view and D is not a final arbitrator. So yes, for this leg of the journey D considers the elapsed times on A and C's clocks to be the same.

Yu are right D is not the final arbiter just an exception to agreement , on the other hand what frame is the final arbiter if there is not frame independant agreement?

However, for the first leg of the journey, when B is going away from A, D sees A coming towards him at 0.5c (same as in the final leg) but B is going away from him at (0.8-0.5)/(1-0.8*0.5)= 0.92857c
Well you didn't like my advice ,what can I say, "subtraction of velocities equation"?
so the time dilation of B's clock on the outward journey is much greater than the time dilation of A's clock, according to D. The first leg of the experiment took 14 years of A's proper time in D's frame. This equates to 14/sqrt(1-0.5^2)=16.1658 years by D's clock. D calculates that the proper time of B's clock on the outward journey "during the same time" is 16.1658*sqrt(1-0.92857^2)=6 years. Overall, D says 20 years of proper time elapsed on A's clock and 12 years of proper time elapsed on the combined times of B and C [B and all is good.

Well maybe not quite 100% just yet. .

How exactly does B take 16.1658 years on D's clock to travel approx. 6.9 ly at 0.92867c ?

After a coordinate time of 10 years, B passes C who is going in the opposite direction with a coordinate velocity of -0.8c. .

So this time we have 14 years of proper time for the first leg instead of 10 , but why quibble because more importantly frame E disagrees with this completely.
Frame E Moving 0.5c --> + x relative to A and ( - 0.5c) relative to B and origen colocated with both at ( B,A)
At event (B,C) in E at t'''' and x₂''''' and position of A at t'''', at x₁''''' with dx'''''=x₂''''' - x₁''''' etc etc

I am sure you see the rest of this written on the wall. yes??

As you can see from the above calculations, D says A was 3.46*2=6.92 lys away from event(B,C) "at the time" B and C crossed paths, while A says he was 8 lys away from B and C "at the time" B and C crossed paths, so A and D differ on where A was "at the time" B and C crossed and what time was showing on A's clock "at the time" B and C crossed.

He said , she said :-)

 

[starthaus]

It can also be seen that I have taken the time dilation due to acceleration into account and it is insignificant compared to the velocity time dilation and an unnecessary complication.

This is false, the only reason that there is such a disparity between the values is that you have chosen a very long "cruising" time and a very short period of acceleration. In reality, if you had chosen comparable values to the "cruise" time T_c vs. the acceleration time T_a you would have found that the difference in proper time between the two twins is equally spread between the two effects and, that both depend on acceleration a:

d \tau=2T_c/ \sqrt{1+(aT_a/c)^2}+4\frac{c}{a} arcsinh(aT_a/c)

for the accelerating twin

vs.dt=2T_c+4T_a

for the inertial twin. If you make T_a closer to T_c than you made them in your skewed example, you will find, to your surprise, that the effects are comparable.
One more thing, the above has nothing to do with the clock "hypothesis"

 

[JesseM]

This is false, the only reason that there is such a disparity between the values is that you have chosen a very long "cruising" time and a very short period of acceleration. In reality, if you had chosen comparable values to the "cruise" time T_c vs. the acceleration time T_a you would have found that the difference in proper time between the two twins is equally spread between the two effects and, that both depend on acceleration a:

d \tau=2T_c/ \sqrt{1+(aT_a/c)^2}+4\frac{c}{a} arcsinh(aT_a/c)

for the accelerating twin

It depends what you mean by "depend on acceleration a". It's true that you can express the time elapsed as a function of the proper acceleration as well as the time spent accelerating, but it is also always possible to express it as a function only of the velocity as a function of time v(t) for the traveling twin in some inertial frame. For any brief interval of time dt where the traveling twin has velocity v in that interval, the traveling twin will age by \sqrt{1 - v^2/c^2} \, dt, so if t₀ and t₁ are the times that the traveling twin departs from and reunites with the inertial twin, the total elapsed time for the traveling twin can be calculated with the integral \int_{t_0}^{t_1} \sqrt{1 - v(t)^2/c^2} \, dt. You can see that this integral depends only on the velocity as a function of time, not the acceleration, and this is exactly what is meant by the clock hypothesis.

 

[starthaus]

It depends what you mean by "depend on acceleration a". It's true that you can express the time elapsed as a function of the proper acceleration as well as the time spent accelerating, but it is also always possible to express it as a function only of the velocity as a function of time v(t) for the traveling twin in some inertial frame. For any brief interval of time dt where the traveling twin has velocity v in that interval, the traveling twin will age by \sqrt{1 - v^2/c^2} \, dt, so if t₀ and t₁ are the times that the traveling twin departs from and reunites with the inertial twin, the total elapsed time for the traveling twin can be calculated with the integral \int_{t_0}^{t_1} \sqrt{1 - v(t)^2/c^2} \, dt. You can see that this integral depends only on the velocity as a function of time, not the acceleration, and this is exactly what is meant by the clock hypothesis.

I know perfectly well how the proper time us calculated, so please, explain the term in \frac{c}{a} arcsinh(aT_a/c). Is this a function of v only?

 

[JesseM]

I know perfectly well how the proper time us calculated, so please, explain the term in \frac{c}{a} arcsinh(aT_a/c). Is this a function of v only?

No, but you don't need to include that term in order to calculate the elapsed time on the accelerating clock, instead you can just use the clock's v(t). Again, the clock hypothesis just says it is possible to calculate the elapsed proper time in the way I described, not that it is the only way to calculate it.

 

[starthaus]

No, but you don't need to include that term in order to calculate the elapsed time on the accelerating clock,

This is false. The term you are trying to explain way is the time elapsed during acceleration. So, one more time, please explain the term \frac{c}{a} arcsinh(aT_a/c) using v only.

 

[JesseM]

This is false. The term you are trying to explain way is the time elapsed during acceleration.

My claim was that you don't need to include that term if you want to calculate "the time elapsed during acceleration" (although that's one way of calculating it), instead you can find the function giving velocity as a function of time v(t) for the accelerating clock in some inertial frame, and then calculate the elapsed proper time between the moment of the start of acceleration at t₀ and the moment of the end of acceleration at t₁ using the integral \int_{t_0}^{t_1} \sqrt{1 - v(t)^2/c^2} \, dt. Do you disagree that this is another valid way to calculate the elapsed proper time?

 

[starthaus]

My claim was that you don't need to include that term if you want to calculate "the time elapsed during acceleration" (although that's one way of calculating it), instead you can find the function giving velocity as a function of time v(t) for the accelerating clock in some inertial frame, and then calculate the elapsed proper time between the moment of the start of acceleration at t₀ and the moment of the end of acceleration at t₁ using the integral \int_{t_0}^{t_1} \sqrt{1 - v(t)^2/c^2} \, dt. Do you disagree that this is another valid way to calculate the elapsed proper time?

.

Yes, the derivation is all based on the \int_{t_0}^{t_1} \sqrt{1 - v(t)^2/c^2} \, dt, this is well known but that is not the point, the final answer is a function of acceleration as shown in post #80 . Do you disagree with that?

The other point that I was making to kev (that you also missed) is that he has inadvertenly ignored the difference due to the acceleration by making the acceleration period very small wrt the cruising period. If the two periods are comparable, the acceleration effect cannot be neglected. Do you disagree with that?

 

[JesseM]

.

Yes, the derivation is all based on the \int_{t_0}^{t_1} \sqrt{1 - v(t)^2/c^2} \, dt, this is well known but that is not the point

Well it's my point, you can define it in terms of a function that includes a dx/dt term but no d^2 x / dt^2 term. Perhaps the confusion is that according to the http://math.ucr.edu/home/baez/physics/Relativity/SR/rocket.html a, v(t) does itself include the term a (if the velocity is zero at t=0 then we have v(t) = at / sqrt[1 + (at/c)^2]) so the result when you evaluate the integral will include an a term too, but this a is just a constant which tells you the proper acceleration, it is not the same as the time-derivative of the coordinate velocity v(t) (or the second time derivative of the coordinate position x(t)) i.e. d^2 x / dt^2. The "clock hypothesis" just says that the total elapsed time can be found by an integral where the only derivative of of x(t) in the integrand is the first derivative i.e. dx/dt.

The other point that I was making to kev (that you also missed) is that he has inadvertenly ignored the difference due to the acceleration by making the acceleration period very small wrt the cruising period. If the two periods are comparable, the acceleration effect cannot be neglected. Do you disagree with that?

I would say that talking about the "acceleration effect" is a way of speaking which is apt to confuse people if they have also heard of the clock hypothesis which says that time dilation is only a function of velocity (which is understood to mean that the elapsed time can be calculated using an integral of the form I gave above). It would be less confusing to say that if the two periods are comparable, you have to take into account the way the velocity was changing continuously during the period of acceleration if you want to calculate the elapsed time.

 

[starthaus]

but this a is just a constant which tells you the proper acceleration,

Proper acceleration, far from being "just a constant" is a measurable physical entity. The final result, the difference in the elapsed proper time of the twins is a proper function of proper acceleration.

it is not the same as the time-derivative of the coordinate velocity v(t) (or the second time derivative of the coordinate position x(t)) i.e. d^2 x / dt^2. The "clock hypothesis" just says that the total elapsed time can be found by an integral where the only derivative of of x(t) in the integrand is the first derivative i.e. dx/dt.

True. As true as the fact that , when expressing v as a function of a, the final result is not a useful function of v but a function of a.

I would say that talking about the "acceleration effect" is a way of speaking which is apt to confuse people if they have also heard of the clock hypothesis which says that time dilation is only a function of velocity (which is understood to mean that the elapsed time can be calculated using an integral of the form I gave above).

Only if such people are unaware of basics physics.

It would be less confusing to say that if the two periods are comparable, you have to take into account the way the velocity was changing continuously during the period of acceleration if you want to calculate the elapsed time.

In order to be precise, one always needs to take into account the accelerated period.

 

[Passionflower]

All these "acceleration is not relevant" arguments are rather silly because if we do not take proper (or inertial) acceleration into account we cannot possibly measure which twin ages more!

Watch how ridiculous the "paradox" becomes when we consider acceleration to be irrelevant:

Have two twins co-located with a relative velocity of zero. They separate with an instant velocity of v, when they are a distance x removed from each other, their relative velocity instantly becomes -v. As soon as they meet their instant velocity becomes 0 again. Which twin ages more?

The "paradox" is only a paradox for those who consider acceleration to be irrelevant!

 

[starthaus]

All these "acceleration is not relevant" arguments are rather silly because if we do not take proper (or inertial) acceleration into account we cannot possibly measure which twin ages more!

Watch how ridiculous the "paradox" becomes when we consider acceleration to be irrelevant:

Have two twins co-located with a relative velocity of zero. They separate with an instant velocity of v, when they are a distance x removed from each other, their relative velocity instantly becomes -v. As soon as they meet their instant velocity becomes 0 again. Which twin ages more?

The "paradox" is only a paradox for those who consider acceleration to be irrelevant!

Yes, it is the typical approach for the people that prefer dumbed down versions of reality beacuse the math necessary to describe the realistic situations, is "too hard".

 

[yuiop]

This is false, the only reason that there is such a disparity between the values is that you have chosen a very long "cruising" time and a very short period of acceleration. In reality, if you had chosen comparable values to the "cruise" time T_c vs. the acceleration time T_a you would have found that the difference in proper time between the two twins is equally spread between the two effects and, that both depend on acceleration a:

d \tau=2T_c/ \sqrt{1+(aT_a/c)^2}+4\frac{c}{a} asinh(aT_a/c)

for the accelerating twin

I know perfectly well how the proper time us calculated, so please, explain the term in \frac{c}{a} asinh(aT_a/c). Is this a function of v only?

The correct equation for the proper time d \tau of the traveling twin with cruising time T_c and acceleration time T_a (both as measured in the inertial stay at home twins rest frame) can be stated as a function of v only like this:

d \tau=\frac{T_c}{\gamma}+\frac{c T_a}{v \gamma} \, \, asinh(v \gamma/c)

Where v is the cruising velocity and \gamma is 1/\sqrt{1-v^2/c^2}.

For a numerical example let us say that the total cruising time T_c is 10 years and the total acceleration time T_a is also 10 years, (both measured in the rest frame of the stay at home twin) so that the inertial twin ages by 20 years during the round trip of the travelling/accelerating twin. The cruising velocity of the traveling twin is 0.8c relative to the stay at home twin and the gamma factor is 1/0.6. Just to be absolutely clear, the traveling twin goes away from home at 0.8c for 5 yers, then spends 5 years slowing down to a stop, another 5 years accelerating to 0.8c and finally the last 5 years cruising at 0.8c for a round trip time of 20 years, all as measured by the inertial twin.

The proper time that elapses on the traveling twins clock in years is:

d \tau=10*0.6+10*0.6/0.8* \,asinh(0.8/0.6) = 6+8.23959 = 14.2396

Note that most of the time dilation happens during the cruise phase even though equal times are spent cruising and accelerating according to the inertial twin. This is because on average the instantaneous relative velocity during the acceleration phase is less than the cruising relative velocity and the acceleration has not increased the time dilation.

Your equation is wrong and predicts the proper time of the traveling twin to be longer than the stay at home twins' proper time. Basically you screwed up the factors when hacking your derivation. You didn't get it from any textbook did you?

 

[JesseM]

Proper acceleration, far from being "just a constant" is a measurable physical entity.

Of course it's measurable, but it's a constant in the equation for elapsed proper time of an observer with constant proper acceleration, which is where the equation you wrote down came from (that equation wouldn't be used for calculating the elapsed proper time for an observer whose proper acceleration was varying). And if the proper acceleration is varying, you can still calculate v(t) and then derive the elapsed proper time from that.

True. As true as the fact that , when expressing v as a function of a, the final result is not a useful function of v but a function of a.

OK, but a is not a variable in the equation v(t) = at / sqrt[1 + (at/c)^2]) (or in the equation for proper time as a function of coordinate time which you wrote down), the only variable that v is a function of is the coordinate time t. Regardless of the rocket's motion (even if it has a varying proper acceleration), you can always express v as function where the only variable is t, though the function may of course include some numerical constants. And I understand the "clock hypothesis" to mean, first of all, that in the limit as we look at arbitrarily short time intervals dt, the amount of time elapsed on a clock during that interval will be equal to \sqrt{1 - v^2/c^2} \, dt, so even if two different clocks are at different positions or have different instantaneous accelerations, as long as they have the same instantaneous velocity they will have the same instantaneous rate of ticking in a given inertial frame. And second, as a corollary of the above, I understand the clock hypothesis to mean that if you know v(t) for a given clock between t₀ and t₁, the integral \int_{t_0}^{t_1} \sqrt{1 - v(t)^2/c^2} \, dt will give the elapsed proper time for this clock between t₀ and t₁. Both of these are nontrivial statements about how the laws of physics work in our universe, for example we could write down an alternate set of hypothetical laws where the instantaneous time dilation did depend on the instantaneous acceleration, so two clocks with the same instantaneous velocity could have different instantaneous rates of ticking. Hopefully you don't disagree with any of the above? Assuming you don't, my point was just that talking about the time dilation as a function of acceleration could potentially confuse someone who doesn't fully understand the clock hypothesis, so from a pedagogical point of view it's better not to talk in that way, even though nothing you said was incorrect on a technical level.

Only if such people are unaware of basics physics.

Well, one of the points of the forum is to teach about basic physics to people who aren't necessarily aware of it. You brought up the clock hypothesis (or were you responding to someone else having brought it up earlier), so I just wanted to clarify for people reading what it means to say in the context of the clock hypothesis that time dilation is not a function of acceleration, which might confuse someone who just saw you saying that the elapsed time was a function of a.

In order to be precise, one always needs to take into account the accelerated period.

True, but before you said:

The other point that I was making to kev (that you also missed) is that he has inadvertenly ignored the difference due to the acceleration by making the acceleration period very small wrt the cruising period. If the two periods are comparable, the acceleration effect cannot be neglected. Do you disagree with that?

I took this to mean you were speaking about what was needed to get an approximately correct figure for the elapsed time, since even if the two periods are not comparable and the acceleration period is very small, any finite period of acceleration needs to be taken into account if you want a perfectly precise figure for the elapsed time (say, in a theoretical word-problem where the length of the acceleration period is known with total precision).

 

[Passionflower]

Well, one of the points of the forum is to teach about basic physics to people who aren't necessarily aware of it.

Then the first thing to teach should be that the \gamma factor between two observers is determined by their acceleration both proper and inertial. If we know that we can calculate the time dilation as \gamma is a factor in this. And so of course the total dilation depends on acceleration, as \gamma is determined by acceleration. Or that if we know the time dilation we can calculate the acceleration. These are useful things.

However if one starts with "oh, they are moving but we don't know how this started" then all bets are off since one cannot calculate time dilation from that. All one can do is dwell on completely inconsequential observed time dilation and length contraction using completely unphysical planes of simultaneity. "They keep going forever with a relative speed x, both observers measure the other observers clock to go slower and their length to be shorter So what? You think that should be the basics?

 

[yuiop]

All these "acceleration is not relevant" arguments are rather silly because if we do not take proper (or inertial) acceleration into account we cannot possibly measure which twin ages more!

Watch how ridiculous the "paradox" becomes when we consider acceleration to be irrelevant:

Have two twins co-located with a relative velocity of zero. They separate with an instant velocity of v, when they are a distance x removed from each other, their relative velocity instantly becomes -v. As soon as they meet their instant velocity becomes 0 again. Which twin ages more?

The "paradox" is only a paradox for those who consider acceleration to be irrelevant!

Hi Passionflower. You have a sort of point and it is just as confusing to say that relative proper times is only a function of relative velocities. It is better to say the twins paradox is resolved by the asymmetry of path lengths in spacetime as analysed from any inertial reference frame. Mostly the arguments here are semantic and philosophical as others have mentioned. The paradox can be analysed either in terms of velocity and spacetime path lengths or in terms of acceleration. Both methods are equally valid mathematically, but the acceleration explanation can be confusing to newcomers. You can calculate the time dilation of an accelerating particle in terms of the sum of its instantaneous velocities, but then you might ask how much additional time dilation should we factor in for the acceleration and the answer is none! On the other hand you can calculate the the time dilation of the same particle in terms of its acceleration and distance away at any instant and arrive at the same answer and in this case no additional time dilation is attributed to instantaneous velocity. They are two sides of the same coin. Basically v\gamma=aT_a and you can either use v\gamma or aT_a in the calculations (but not both).

I do agree that any differential ageing involves acceleration, but in some situations where both twins experience equal proper acceleration, you also have to factor in where and when in spacetime the acceleration occurred and this basically boils down to comparing path lengths in space time.

Starthaus gives an equation where the proper time of the constant velocity leg of the journey is a function of the acceleration at the turn around (which is mathematically correct when he gets his equation right), but philosophically it is not ideal, because it implies that the proper time that elapses during the constant velocity leg is applied retrospectively once we find out what the acceleration is at the turn around. Let us say that the twin launches away from Earth with constant velocity and passes Alpha Centauri at constant velocity. What is the proper time that elapses on the traveling twins clock between leaving Earth and passing Alpha Centauri? The acceleration explanation can not give an answer, because it does not know where,when or how quickly the twin is going to turn around, unless the clock has some sort of predictive ability. Note that there is a definitive proper time elapsed on the traveling clock between the events "passing Earth" and "passing Alpha Centauri" even if the traveling clock was always moving at constant velocity and no acceleration was involved.

 

[Passionflower]

The paradox can be analysed either in terms of velocity and spacetime path lengths or in terms of acceleration. Both methods are equally valid mathematically, but the acceleration explanation can be confusing to newcomers.

Really? So then please explain to me mathematically how without any acceleration (proper or inertial) you can say which twin ages more. Use any example you like.

Mostly the arguments here are semantic and philosophical as others have mentioned.

They become philosophical, and frankly irrelevant, as soon as one completely ignores acceleration as there is absolutely nothing to calculate in that case.

 

[JesseM]

The correct equation for the proper time d \tau of the traveling twin with cruising time T_c and acceleration time T_a (both as measured in the inertial stay at home twins rest frame) can be stated as a function of v only like this:

d \tau=\frac{T_c}{\gamma}+\frac{c T_a}{v \gamma} \, \, asinh(v \gamma/c)

Where v is the cruising velocity and \gamma is 1/\sqrt{1-v^2/c^2}.

It depends on how you choose to express things, your method avoids having to know the value of the proper acceleration while starthaus' equation assumed we did know it. If we do know the value of the proper acceleration a, we also know the rocket will be decelerating from v to 0 for T_a/2 and accelerating from 0 to v for T_a/2, and according to the http://math.ucr.edu/home/baez/physics/Relativity/SR/rocket.html , the proper time for a rocket that accelerates with constant proper acceleration a starting from a speed of 0 would be given by tau = (c/a)*arcsinh(aT/c), so for T=T_a/2 we get tau = (c/a)*arcsinh(aT_a/2c), and by the symmetry we can double that to get the total proper time elapsed from the beginning of the acceleration to the end, giving tau = (2c/a)*arcsinh(aT_a/2c). So, I get a total elapsed time of:

d \tau=\frac{T_c}{\gamma}+\frac{2c}{a} \, \, arcsinh(a T_a / 2c)

If we want to express it without reference to a, another equation on the relativistic rocket page says that velocity v as a function of coordinate time T would be v(T) = aT / sqrt[1 + (aT/c)^2], so if we know it takes a time of T_a/2 to go from 0 to the cruising speed v, that means that v = aT_a / 2 * sqrt[1 + (aT_a/2c)^2], which means:

v^2 = a^2 T_a^2 / 4*[1 + (a^2*T_a^2 / 4c^2)]
4v^2 + 4*v^2*a^2*T_a^2 / 4c^2 = a^2 * T_a^2
4v^2 + (v^2/c^2)*a^2*T_a^2 = a^2*T_a^2
4v^2 = a^2*T_a^2*(1 - v^2/c^2)
a^2 = 4v^2 / (T_a^2 * (1 - v^2/c^2))
a = 2v*gamma/T_a

Substituting that into my equation above:

d \tau=\frac{T_c}{\gamma}+\frac{c T_a}{v \gamma} \, \, arcsinh(v\gamma /c)

...which is the same as the equation you wrote down.

edit: whoops, I read your post hastily and thought you were objecting to the part of starthaus' post where he wrote the equation \frac{c}{a} arcsinh(aT_a/c) for the proper time during the acceleration phase, but now I think you probably objecting to this other equation starthaus posted:

d \tau=2T_c/ \sqrt{1+(aT_a/c)^2}+4\frac{c}{a} asinh(aT_a/c)

Maybe in this equation starthaus is actually letting T_a represent half the acceleration period, which would explain why he uses asinh(aT_a/c) rather than asinh(aT_a/2c) as I did, but why is there a 4 out in front? I don't know where the first term comes from either...

 

[starthaus]

The correct equation for the proper time d \tau of the traveling twin with cruising time T_c and acceleration time T_a (both as measured in the inertial stay at home twins rest frame) can be stated as a function of v only like this:

d \tau=\frac{T_c}{\gamma}+\frac{c T_a}{v \gamma} \, \, asinh(v \gamma/c)

Where v is the cruising velocity and \gamma is 1/\sqrt{1-v^2/c^2}.

If you do the proper calculations, you'll find out that v\gamma=aT_a

Your equation is wrong

Err, no.

and predicts the proper time of the traveling twin to be longer than the stay at home twins' proper time.

Err, also no.

Basically you screwed up the factors when hacking your derivation. You didn't get it from any textbook did you?

No, I didn't hack it and I didn't get it from a textbook.

The proper time that elapses on the traveling twins clock in years is:

d \tau=10*0.6+10*0.6/0.8* \,asinh(0.8/0.6) = 6+8.23959 = 14.2396

...which clearly contradicts your earlier claim that the contribution of the acceleration period is negligible.

 

[starthaus]

d \tau=2T_c/ \sqrt{1+(aT_a/c)^2}+4\frac{c}{a} asinh(aT_a/c)

I guess in this equation starthaus is actually letting T_a represent half the acceleration period, so the second term is correct. Not sure where he got the first term though...

...from knowing math and physics

 

[JesseM]

Then the first thing to teach should be that the \gamma factor between two observers is determined by their acceleration both proper and inertial.

Not necessarily, what about two observers who pass one another moving inertially in space and note their ages as they pass? And even if you assume they started at rest relative to one another, if you assume acceleration was quasi-instantaneous (as is commonly assumed in these kinds of problems to make the math simpler), then the question of which one accelerated initially is irrelevant to figuring out their elapsed time, all that matters is which one accelerated to turn around once they were a significant distance apart. Then all you have to do is pick a frame, find the velocity of the one that turned around both before and after the turnaround (as well as the time between departure and turnaround in the frame you chose), as well as the velocity of the one that moved inertially between departure and reunion, and from that you can figure out how much elapsed time both have experienced between departure and reunion.

 

[yuiop]

Substituting that into my equation above:

d \tau=\frac{T_c}{\gamma}+\frac{c T_a}{v \gamma} \, \, arcsinh(v\gamma /c)

...which is the same as the equation you wrote down.

edit: whoops, I read your post hastily and thought you were objecting to the part of starthaus' post where he wrote the equation \frac{c}{a} arcsinh(aT_a/c) for the proper time during the acceleration phase, but now I think you probably objecting to this other equation starthaus posted:

d \tau=2T_c/ \sqrt{1+(aT_a/c)^2}+4\frac{c}{a} asinh(aT_a/c)

I guess in this equation starthaus is actually letting T_a represent half the acceleration period, so the second term is correct. Not sure where he got the first term though...

If you start with my equation:

d \tau=\frac{T_c}{\gamma}+\frac{c T_a}{v \gamma} \, \, arcsinh(v\gamma /c)

and substitute aT_a for v\gamma and \sqrt{1+(aT_a/c)^2} for \gamma you end up with the equation Starhaus should have ended up with if hadn't blundered:

d \tau=T_c/ \sqrt{1+(aT_a/c)^2}+\frac{c}{a} arcsinh(aT_a/c)

Even if you allow for the fact that Starhaus might of meant T_c and T_a were the the cruising time and accelerating time of the the outwrad leg only, rather than the round trip time, the equation he should have ended up with is:

d \tau= 2T_c/ \sqrt{1+(aT_a/c)^2}+2\frac{c}{a} arcsinh(aT_a/c)

Either way, Starthaus messed up.

However, he is right that it does produce the mathematical oddity that the proper time of the outward leg is apparently determined by the acceleration at the end of the leg, which just shows you have to be careful in how you physically interpret the equations.