Solving the Twin Paradox with Lorentz Transformation

[Fredrik]

Are you saying you think the geometric approach does resolve the question without recourse to the phenomenon of acceleration?

It's very clear that he isn't saying that. He said that acceleration is involved.

Wouldn't you agree it was true that the worldline of the Earth , as plotted in the accl. frame, purely on the basis of spacetime coordinates (time and position measurements) would be a mirror identical shape.

I haven't checked the math, but I'd be surprised if it turned out to have the same shape. In general, "the" accelerating frame might not even cover the region of spacetime that Earth is in, except at the beginning and end of the trip. So in general, Earth's world line doesn't even make it into the diagram.

With exactly the same path length and equivalent time?

If you're talking about proper time, that has nothing to do with coordinates.

((2)) Here you seem to be referring to actual gravitational time dilation due to the Earth's mass.

He isn't. This is SR, so mass has no effect on spacetime geometry. Consider the diagram here and suppose that one of the curves is the world line of the astronaut twins. The straight lines with different slopes are the simultaneity lines. Imagine Earth's world line as a vertical line drawn some distance to the right. The greater the distance, the fewer simultaneity lines will intersect Earth's world line. So if the distance is greater, the accelerating frame will assign a smaller coordinate time difference to two specific events on Earth (e.g. the event 5 years after the departure and the event 5 years before the return).

Gravitational dilation as applied in a static mass does not involve velocity . Conceptually it involves acceleration which everybody has concluded is "real" and is distinctly different from inertial velocity.

No, it's not that simple. As I mentioned in one of my previous posts, two clocks attached to the front and rear of an accelerating rocket will not tick at the same rate. This is because the rigidity of the rocket will give the clock at the rear a higher velocity in the frame where they both started out at rest. You may not think of this as "gravitational" time dilation, but this is exactly the same phenomenon that (in GR) causes two clocks on different floors of the same building to tick at different rates. The rigidity of the building makes the lower clock deviate more from geodesic motion, ensuring that its world line has a smaller proper time, or expressed differently, ensuring that it has a higher velocity in the local inertial frame of an observer that's doing geodesic motion.

Dilation may be calculated for an accelerating system on the basis of ICMF's but that does not make Gravitational time dilation the same phenomenon as velocity time dilation.

What I just said explains in what sense they're the same. I objected to Al68's claim that they're the same at first, but I think it's fair to say that they are. I guess I just don't like the term "gravitational time dilation", even when gravity is involved.

Not unless you are going to say that acceleration is the same phenomenon as velocity.

Which I would have no problem with actually.

I would.

By the way, your posts would be easier to read if you used quote tags consistently. Use the "multi quote" buttons if you're going to quote several posts, and edit the result to make the quotes look the way they do in this post.

 

[Austin0]

Originally Posted by MikeLizzi
What I object to is anyone who says the Twins Paradox can be resolved without recourse to the phenomenon of acceleration.

Since acceleration is a bend in a worldline that is exactly what the geometric approach does.

Originally Posted by Austin0
Are you saying you think the geometric approach does resolve the question without recourse to the phenomenon of acceleration?

It's very clear that he isn't saying that. He said that acceleration is involved..

Self evidently acceleration is involved but that was not the point as I see MikeLizzis's
statement.
The resolution as seen in the diagram is based on an interpretaion of the bend in the world line. This interpretaion is founded on the concept that acceleration is real as opposed to inertial motion. That it has a real effect that makes all the inertial motion of the accelerated system between accelerations non-reciprocal.
It is not the small period where the bend takes place that is relevant, it effects the whole path.

BTW I see your point about the quotes and also the need to make my posts shorter and more focused. It really is a pain have to deal with the problem of laying out quotes and retaining context.

 

[Dale]

What I object to is anyone who says the Twins Paradox can be resolved without recourse to the phenomenon of acceleration.

Since acceleration is a bend in a worldline that is exactly what the geometric approach does.

Are you saying you think the geometric approach does resolve the question without recourse to the phenomenon of acceleration?

Sorry about the confusion. I am saying that the geometric approach resolves it with recourse to acceleration, so it should be an acceptable resolution to MikeLizzi.

Wouldn't you agree it was true that the worldline of the Earth , as plotted in the accl. frame, purely on the basis of spacetime coordinates (time and position measurements) would be a mirror identical shape. With exactly the same path length and equivalent time?

No. The components of the metric are different in the accelerating frame. The algebraic form of the worldline is the same, but the metric is what determines geometric things like intervals and angles and curvature. Acceleration is a frame invariant geometric quantity (the radius of curvature of a worldline) that does not depend on the coordinate system used to represent it. You need to separate in your mind the geometry from the coordinates. The worldline is bent geometrically, regardless of whether or not you use similarly bent coordinates to describe it.

That the reason this is not reciprocal and is invalid is because the bend in the worldline in the Earth based spacetime diagram indicates acceleration. And it is that interpretaion of this bend and the assumptions regarding the phenomenon of acceleration that is the actual rationale for invalidating the the observations and calculations from the other perspective.

They are perfectly valid, they just use a different coordinate system to describe the same geometry. Along with the different coordinate system comes different metric components.

EDIT: corrections to refer to components of metric, see below

 

[yossell]

The metric is different in the accelerating frame.

This threw me a bit - I thought the metric is independent of frame and coordinate system. I thought the metric measured the intrinsic geometry of space-time. It's true, the mathematical form of the metric changes in different coordinate systems, so his equation for the line element looks different - is this equation also sometimes called the metric?

 

[Fredrik]

It's the same metric. It has different components in different coordinate systems.

 

[yossell]

Great - thanks Fredrik.

 

[Dale]

It's true, the mathematical form of the metric changes in different coordinate systems, so his equation for the line element looks different

You are right, that would have been a better way for me to say it. The metric is a tensor so it is a single geometric object which has different components in different coordinate systems.

Sorry about the sloppy language trying to explain my earlier sloppy language

EDIT: Fredrik beat me to it.

 

[atyy]

I'm not sure what to make of it in those defined coordinates, since I only glanced at it, and "radar time" and "radar distance" aren't familiar enough to me. I did notice that the rate of acceleration doesn't appear to be part of the metric in the applicable regions, which might? make sense, if it's a shortcut for a real acceleration profile.

Can one achieve a non-instantaneous turn around "slow enough" that it would not kill Stella?

The gradual turn around case is treated on p5-7 of http://arxiv.org/abs/gr-qc/0104077 . You can choose the acceleration to be as small as you like, it will just take longer for the twin to get back. On p7, just below fig. 8, they give the metric in coordinates in which the accelerated twin is stationary. In this case, (acceleration.coordinate distance) appears in the right place for a gravitational potential.

 

[matheinste]

Originally posted in Sept 09

Taken from Seminar Poincare, Einstein 1905-2005 page 106.

The section from which it is taken is called "A comfortable trip for the Langevin traveler", comfortable because the acelerations for the inward and outward journey are just plus and minus g, equivalent to Earth's gravity, and the forces involved at the (instantaneous) turnaround are said to be easily handled by human beings. The article points out that considerable, prohibitive amounts of fuel, comparable to planetary masses, would be required for the longer journeys and the gravitational effects of such large fuel masses are ignored in the calculation. It also assumes that the journey takes place in flat Minkowski spacetime.

<br /> \begin{tabular}{| c | c | c |}<br /> \hline Traveller&#039;s proper time &amp; Earth&#039; proper time\\<br /> \hline 1 year &amp; 1 year 4 days\\<br /> \hline 2years &amp; 2years and 1month \\<br /> \hline 4 years &amp; 4.7 years \\<br /> \hline 8 years &amp; 14.5 years \\<br /> \hline 16 years &amp; 104 years \\<br /> \hline 20 years &amp; 297 years \\<br /> \hline 28 years &amp; 2,200 years\\<br /> \hline 32 years &amp; 5,960 years \\<br /> \hline 40 years &amp; 44,000 years \\<br /> \hline 48 years &amp; 326,000 years \\<br /> \hline 60 years &amp; 5.54x$10^6$ years \\<br /> \hline 72 years &amp; 131x$10^6$ years \\<br /> \hline 84 years &amp; 2.64x$10^9$ years \\<br /> \hline 86 years &amp; 5 billion years \\<br /> \hline<br /> <br /> \end{tabular}<br /> <br /> [/itex]<br /> <br /> Matheinste.

 

[matheinste]

A more detailed explanation and clarification of the method proposed in an almost word for word transcription of the same source:-

------The standard presentation of the “twin paradox” (or “Langevin traveller), which amounts to a direct trip with return between a point of the Earth and some far distant space station, with large uniform velocity v, in both directions, is remarkable by its beautiful pedagogical simplicity. In fact, we can see that it illustrates the Minkowskian triangular inequality. However, since it appeared in the literature, various objections have been eaised whose point was generally to conclude that this was a school example which was probably physically incorrect or at best unrealistic. This type of opinion has also been often endorsed by vulgarizers of special relativity, as a reassuring thought with respect to what looks like a scandal for the common sense.

The main objection was about the instantaneous passage from velocity v to –v when reaching the term of the travel. Such passage had to be produced by a shock, or even if smoothened out by some decelerating device, it seemed to involve so large accelerations that certainly the biological organisms and maybe clocks themselves could not stand such constraints. Now in view of Minkowski’s study of uniformly accelerated motions, one can actually show the possibility of organizing a more comfortable trip in which the traveller would be submitted to a constant acceleration, or deceleration. We even impose, for making the accelerations biologically normal, that its value be precisely equal to the value of the gravity acceleration g on the earth. Of course, we admit that the whole travel will take place in the vacuum, far from any gravitational source, in such a way that the flat Minkowski sapcetime remains a reasonably good approximation to the real spacetime.

Choice of the motion.

The trajectory is along a straight line joining the Earth denoted by A and a space station B considered as at rest with respect to the earth. The travel which is proposed is composed of

A uniformly accelerated motion with acceleration g from A to the middle M of AB.

A uniformly accelerated motion with acceleration –g from M to B (namely a phase of deceleration).

The acceleration –g is maintained and produces the first half of the returning trip from B to M.

The acceleration is shifted from –g to g for producing a uniformly decelerated motion from M to A.

It is clear that the discontinuity of the acceleration from g to –g produced at M is bearable by the physical and biological systems in the spaceship: of the direction of the normal gravity g on earth.it is just felt as a sudden inversion.-----

Matheinste.

 

[matheinste]

Originally posted sept09

This extract is from page 45 of Wolfgang Rindler's book Essential Relativity--Special, General and Cosmological--.

"Like length contraction, so also time dilation can lead to an apparent paradox when viewed by two different observers. In fact, this paradox, the so-called twin or clock paradox (or paradox of Langevin), is the oldest of all the relativistic paradoxes. It is quite easily resolved, but its extraordinary emotional appeal keeps debate alive as generation after generation goes through the cycle of first being perplexed, then elated at understanding (sometimes mistakenly), and then immediately rushing into print as though no one had understood before. The articles that have been published on this one topic are practically uncountable, while their useful common denominator would fill a few pages at most. But while no one can get very excited about pushing long poles into short garages and the like, the prospect of going on a fast trip through space and coming back a few years later to find the Earth aged by a few thousand years--this modern elixir vitae--keeps stirring the imagination."

Matheinste

 

[starthaus]

Originally posted in Sept 09

Taken from Seminar Poincare, Einstein 1905-2005 page 106.

The section from which it is taken is called "A comfortable trip for the Langevin traveler", comfortable because the acelerations for the inward and outward journey are just plus and minus g, equivalent to Earth's gravity, and the forces involved at the (instantaneous) turnaround are said to be easily handled by human beings. The article points out that considerable, prohibitive amounts of fuel, comparable to planetary masses, would be required for the longer journeys and the gravitational effects of such large fuel masses are ignored in the calculation. It also assumes that the journey takes place in flat Minkowski spacetime.

<br /> \begin{tabular}{| c | c | c |}<br /> \hline Traveller&#039;s proper time &amp; Earth&#039; proper time\\<br /> \hline 1 year &amp; 1 year 4 days\\<br /> \hline 2years &amp; 2years and 1month \\<br /> \hline 4 years &amp; 4.7 years \\<br /> \hline 8 years &amp; 14.5 years \\<br /> \hline 16 years &amp; 104 years \\<br /> \hline 20 years &amp; 297 years \\<br /> \hline 28 years &amp; 2,200 years\\<br /> \hline 32 years &amp; 5,960 years \\<br /> \hline 40 years &amp; 44,000 years \\<br /> \hline 48 years &amp; 326,000 years \\<br /> \hline 60 years &amp; 5.54x$10^6$ years \\<br /> \hline 72 years &amp; 131x$10^6$ years \\<br /> \hline 84 years &amp; 2.64x$10^9$ years \\<br /> \hline 86 years &amp; 5 billion years \\<br /> \hline<br /> <br /> \end{tabular}<br /> <br /> [/itex]<br /> <br /> Matheinste.

<br /> <br /> So, if we go away for 16 years (at what fraction of c?) , when we come back, everybody we used to know is dead :-(

 

[atyy]

So, if we go away for 16 years (at what fraction of c?) , when we come back, everybody we used to know is dead :-(

Except for the twin paradox

 

[starthaus]

Except for the twin paradox

So, they aren't dead ? Yiipee!

 

[atyy]

So, they aren't dead ? Yiipee!

No, I meant you can come back to PF or whatever its successor is and still find the twin paradox vigourously discussed by new folks.

 

[starthaus]

No, I meant you can come back to PF or whatever its successor is and still find the twin paradox vigourously discussed by new folks.

Ah, I see, the ones left behind 16 years ago are dead :-(

 

[stevmg]

Originally posted in Sept 09

Taken from Seminar Poincare, Einstein 1905-2005 page 106.

The section from which it is taken is called "A comfortable trip for the Langevin traveler", comfortable because the acelerations for the inward and outward journey are just plus and minus g, equivalent to Earth's gravity, and the forces involved at the (instantaneous) turnaround are said to be easily handled by human beings. The article points out that considerable, prohibitive amounts of fuel, comparable to planetary masses, would be required for the longer journeys and the gravitational effects of such large fuel masses are ignored in the calculation. It also assumes that the journey takes place in flat Minkowski spacetime.

<br /> \begin{tabular}{| c | c | c |}<br /> \hline Traveller&#039;s proper time &amp; Earth&#039; proper time\\<br /> \hline 1 year &amp; 1 year 4 days\\<br /> \hline 2years &amp; 2years and 1month \\<br /> \hline 4 years &amp; 4.7 years \\<br /> \hline 8 years &amp; 14.5 years \\<br /> \hline 16 years &amp; 104 years \\<br /> \hline 20 years &amp; 297 years \\<br /> \hline 28 years &amp; 2,200 years\\<br /> \hline 32 years &amp; 5,960 years \\<br /> \hline 40 years &amp; 44,000 years \\<br /> \hline 48 years &amp; 326,000 years \\<br /> \hline 60 years &amp; 5.54x$10^6$ years \\<br /> \hline 72 years &amp; 131x$10^6$ years \\<br /> \hline 84 years &amp; 2.64x$10^9$ years \\<br /> \hline 86 years &amp; 5 billion years \\<br /> \hline<br /> <br /> \end{tabular}<br /> <br /> [/itex]<br /> <br /> Matheinste.

<br /> <br /> <br /> <blockquote data-attributes="" data-quote="starthaus" data-source="post: 2797150" class="bbCodeBlock bbCodeBlock--expandable bbCodeBlock--quote js-expandWatch"> <div class="bbCodeBlock-title"> starthaus said: </div> <div class="bbCodeBlock-content"> <div class="bbCodeBlock-expandContent js-expandContent "> So, if we go away for 16 years (at what fraction of c?) , when we come back, everybody we used to know is dead :-( </div> </div> </blockquote><br /> v = 0.988094838c

 

[Al68]

Wouldn't you agree it was true that the worldline of the Earth , as plotted in the accl. frame, purely on the basis of spacetime coordinates (time and position measurements) would be a mirror identical shape. With exactly the same path length and equivalent time?

No, it wouldn't. Remember that the coordinate distance of Earth in the ship's accelerated frame changes drastically during turnaround due to the effect of velocity on length contraction. In addition, the worldline of Earth in the ship's frame includes the coordinate acceleration of Earth absent any applied force, which must be accounted for.

((2)) Here you seem to be referring to actual gravitational time dilation due to the Earth's mass. An inverse function of the difstance I assume?

Absolutely not. I was referring to the gravitational time dilation in the accelerated frame of the ship.

I am talking about the psuedo-gravitational time dilation occurring soley within the accelerating frame with zero relevance to Earth's clocks.

Earth's clock is the one being compared to the ship clock in the ship's accelerated frame. Gravitational time dilation factors in the distance between the clocks, because the lorentz transformations factor in the distance between the clocks.

Dilation may be calculated for an accelerating system on the basis of ICMF's but that does not make Gravitational time dilation the same phenomenon as velocity time dilation.

Sure it does, because the time dilation calculated on the basis of ICMF's (using lorentz transformations) is gravitational time dilation wrt the accelerated frame. That's how gravitational time dilation was derived originally by Einstein in 1907.

 

[Ken Natton]

Very aware that this is a thread that has largely run its course and that there may be no-one watching it any more, I nevertheless wanted to come in late with a slightly different perspective from someone who freely confesses to never having studied physics at higher level and for whom most of his knowledge comes from the kind of popular science books many contributors would consider beneath them.

From my viewpoint, what is often not explicitly mentioned in explanations of length contraction and time dilation is the way in which they operate together to maintain distance in spacetime. Two observers in references frames that are in significant motion relative to each other viewing two consequent events may not agree on the distance in space between the events, or on the duration of time between them, but they will agree upon the distance in spacetime between them. If do their calculations using the Minkowski version of Pythagoras, they will get the same answer. With the first event on the origin of a space time diagram, the second event lies on hyperbola for all observers. The only thing they disagree on is what proportion of the travel through spacetime is travel through space and what proportion is travel through time. Perhaps that is an obvious point to many of you, but I’m not sure it is obvious to everyone struggling to understand length contraction and time dilation and it seems to me that it is a critical point.

Applying that, then, to the twin paradox, because the twins start at the same time in the same location and finish at the same time in the same location, clearly their net journeys through spacetime are identical. From the perspective of planet earth, which is where they both start and finish, earthbound twin’s journey through spacetime has negligible amounts of travel through space (not zero, perhaps she went on holiday to Australia a few times while her twin was gallivanting among the stars) and thus is mainly travel through time. Space traveling twin, by some accounts on this thread, traveled to a point 15 or 16 light years away and back. A significant proportion of her journey through spacetime is taken up with travel through space and thus inevitably, a much smaller proportion is taken up with travel through time. Clearly, traveling twin is the one who will be the younger.

Some of the diagrams that lie further up this thread are excellent visualisations of the dynamic situation as traveling twin’s journey unfolds, and I accept these are the best demonstration of the point. But the above reasoning, it seems to me, is very easy to follow, and should remove any doubt that it is the traveling twin who will be the younger and prove that there is no paradox. Or, semantic pedantry considered, at least there is no contradiction of relativity theory.

 

[atyy]

Very aware that this is a thread that has largely run its course and that there may be no-one watching it any more, I nevertheless wanted to come in late with a slightly different perspective from someone who freely confesses to never having studied physics at higher level and for whom most of his knowledge comes from the kind of popular science books many contributors would consider beneath them.

From my viewpoint, what is often not explicitly mentioned in explanations of length contraction and time dilation is the way in which they operate together to maintain distance in spacetime. Two observers in references frames that are in significant motion relative to each other viewing two consequent events may not agree on the distance in space between the events, or on the duration of time between them, but they will agree upon the distance in spacetime between them. If do their calculations using the Minkowski version of Pythagoras, they will get the same answer. With the first event on the origin of a space time diagram, the second event lies on hyperbola for all observers. The only thing they disagree on is what proportion of the travel through spacetime is travel through space and what proportion is travel through time. Perhaps that is an obvious point to many of you, but I’m not sure it is obvious to everyone struggling to understand length contraction and time dilation and it seems to me that it is a critical point.

Applying that, then, to the twin paradox, because the twins start at the same time in the same location and finish at the same time in the same location, clearly their net journeys through spacetime are identical. From the perspective of planet earth, which is where they both start and finish, earthbound twin’s journey through spacetime has negligible amounts of travel through space (not zero, perhaps she went on holiday to Australia a few times while her twin was gallivanting among the stars) and thus is mainly travel through time. Space traveling twin, by some accounts on this thread, traveled to a point 15 or 16 light years away and back. A significant proportion of her journey through spacetime is taken up with travel through space and thus inevitably, a much smaller proportion is taken up with travel through time. Clearly, traveling twin is the one who will be the younger.

Some of the diagrams that lie further up this thread are excellent visualisations of the dynamic situation as traveling twin’s journey unfolds, and I accept these are the best demonstration of the point. But the above reasoning, it seems to me, is very easy to follow, and should remove any doubt that it is the traveling twin who will be the younger and prove that there is no paradox. Or, semantic pedantry considered, at least there is no contradiction of relativity theory.

So the paradox is - in the first case, there is a disagreement but in the second case there is no disagreement, as both agree the non-accelerating twin is older. Why did the disagreement disappear?

 

[Ken Natton]

Ah ha! Engagement! I was beginning to think that this was a party that I had gatecrashed.

I must confess that I am not sure that I follow your reasoning, atyy. I see no disappearing argument. The observers may disagree about time durations and spatial distances, but if they are diligent physicists and do their Minkowski calculations, they do not disagree at all about distance through spacetime, and that applies throughout traveling twin’s journey. When traveling twin arrives home, neither is in any disagreement or confusion about why traveling twin is the younger. Both twins’ spacetime vectors have identical lengths, but traveling twin’s vector leans further along the space axis, while earthbound twin’s vector lies nearly directly up the time axis. Where’s the disappearing disagreement?

 

[Ich]

Hi Ken,

you're definitely on the right track, but there seems to be some confusion left.

Applying that, then, to the twin paradox, because the twins start at the same time in the same location and finish at the same time in the same location, clearly their net journeys through spacetime are identical.

"Net journey" makes sense only if it means spacetime distance. Spacetime distance along a worldline is proper time. The spacetime distance both twins have traveled are most obviously not identical, as their proper times increased differently. It's like one twin took a detour, but with the inverse Phytagoras that means less spacetime distance covered.

Both twins’ spacetime vectors have identical lengths, but traveling twin’s vector leans further along the space axis, while earthbound twin’s vector lies nearly directly up the time axis.

Again, "length of spacetime vector" can only mean spacetime distance. Then you have one vector for the Earth twin, and two for the traveller, the summed length of which is shorter than the length of the single Earth vector.

 

[Austin0]

So the paradox is - in the first case, there is a disagreement but in the second case there is no disagreement, as both agree the non-accelerating twin is older. Why did the disagreement disappear?

Ah ha! Engagement! I was beginning to think that this was a party that I had gatecrashed.

I must confess that I am not sure that I follow your reasoning, atyy. I see no disappearing argument. The observers may disagree about time durations and spatial distances, but if they are diligent physicists and do their Minkowski calculations, they do not disagree at all about distance through spacetime, and that applies throughout traveling twin’s journey. When traveling twin arrives home, neither is in any disagreement or confusion about why traveling twin is the younger. Both twins’ spacetime vectors have identical lengths, but traveling twin’s vector leans further along the space axis, while earthbound twin’s vector lies nearly directly up the time axis. Where’s the disappearing disagreement?

I think atyy was referring to your first interpretation i.e the distance would be the same
in which casae there would be disagreement as both twins would be the same age as opposed to your final interpretion in which there is no disagreement.
I would agree with atyy that is largely a part of the popular "paradox"

 

[Ken Natton]

Huh. I had to know that I wouldn’t have got it quite right. I’m sure you understand, Ich, that in the following, I am not disagreeing with you, I am merely trying to trace the exact source of my error.

The vectors I referred to I had meant to be the spacetime vectors, which are, as I understand it, the world lines. And by net journey, I meant the line that joins the start of the first component vector on the space time diagram with the end of the last component vector, which is, I think, the sum of all the component vectors. So traveling twin’s history involves a complicated sequence of component vectors, but the understanding of the age difference with her twin depends only on this summary vector.

Since they both start at the same location and time and finish at the same location and time, it seemed to me that the two summary vectors would have the same length, but as I visualised it, with both twins’ summary vectors anchored on the origin of my space time diagram, earthbound twin’s vector would lean only very slightly from the time axis (which I understand is calibrated as ct not just t), whereas traveling twin’s vector rotates much further and thus finishes lower down the time axis.

I suppose that if you are telling me that the summary vector lengths are not the same, then I do not quite understand why. I am fairly sure of my assertion that length contraction and time dilation operate to maintain the same spacetime distance for two observers who remain in different inertial reference frames. I understand that is not quite the case for our twins.

 

[matheinste]

Huh. I had to know that I wouldn’t have got it quite right. I’m sure you understand, Ich, that in the following, I am not disagreeing with you, I am merely trying to trace the exact source of my error.

The vectors I referred to I had meant to be the spacetime vectors, which are, as I understand it, the world lines. And by net journey, I meant the line that joins the start of the first component vector on the space time diagram with the end of the last component vector, which is, I think, the sum of all the component vectors. So traveling twin’s history involves a complicated sequence of component vectors, but the understanding of the age difference with her twin depends only on this summary vector.

Since they both start at the same location and time and finish at the same location and time, it seemed to me that the two summary vectors would have the same length, but as I visualised it, with both twins’ summary vectors anchored on the origin of my space time diagram, earthbound twin’s vector would lean only very slightly from the time axis (which I understand is calibrated as ct not just t), whereas traveling twin’s vector rotates much further and thus finishes lower down the time axis.

I suppose that if you are telling me that the summary vector lengths are not the same, then I do not quite understand why. I am fairly sure of my assertion that length contraction and time dilation operate to maintain the same spacetime distance for two observers who remain in different inertial reference frames. I understand that is not quite the case for our twins.

A clock traveling with an observer measures the proper time along the worldline of the observer. It is a measure of the spacetime path length and is frame independent. The proper time measured by a clock with the stay at home twin is not the same as the proper time measured by the traveler and so the spacetime path length is not the same.

Matheinste.

 

[Ken Natton]

Well, all I can say, Austin0 is that I do not understand where my interpretation changes. It seems perfectly consistent to me. Again, I suppose you could be pedantic and say that they would only ever disagree about spatial distances and time durations if they were both taking a Newtonian Mechanical view. If they had both studied relativity, they would understand that distances and times would not appear the same for each other and if they were comfortable with the mathematics, they could even work out exactly what the distances and times would appear to be for each other. But the point is that they don’t really need to. If they do their Minkowski calculations with their own distance and time measurements and arrive at a consistent spacetime distance, then there is no disagreement.

As I have said, I accept that the diagrams that appear further up this thread show best how the two viewpoints change over the duration of traveling twin’s journey. It just seemed to me that there was a much simpler way of understanding why traveling twin is younger when the journey is over, and why such a notion is consistent with relativity theory.

 

[Ich]

And by net journey, I meant the line that joins the start of the first component vector on the space time diagram with the end of the last component vector, which is, I think, the sum of all the component vectors.

Ok, so that's distance "bee-line". Yes, that's the (length of the) sum of the vectors.

So traveling twin’s history involves a complicated sequence of component vectors, but the understanding of the age difference with her twin depends only on this summary vector.

I don't agree. The age differnce comes from the different path lengths. The traveling twin's path is shorter than the Earth path. The reason is the http://en.wikipedia.org/wiki/Triangle_inequality#Reversal_in_Minkowski_space" in Minkowski space. The sum of the lenghts is shorter than the length of the sum.
Conceptually, it's nothing else than the explanation why (in euclidean space) two sides of a triangle are longer than the third. I think you would not explain it with "motion along x adds to motion along y", its easier if you don't refer to the (anyway arbitrary) coordinates and stick to the geometry.

 

[Ken Natton]

Well there you go, I asked for engagement, and now it comes so thick and fast that I can’t keep up with it. Be careful what you wish for, I suppose.

I have read enough of this and other threads to know that you are someone whose comments I must attend to, mathienste, but your comment does not align with the understanding I had taken, that whatever the variance in viewpoints of time and spatial distance, spacetime distances are consistent for all observers. All I can do is return to the text from which I thought I had taken this understanding, and see if I can find where I misunderstood it, or return to you if I conclude that I had not misunderstood it.

 

[yossell]

edit: beaten by ich

The vectors I referred to I had meant to be the spacetime vectors, which are, as I understand it, the world lines.

The spacetime vectors are not the world lines. Assuming that by vector you mean a straight directed segment joining two points, we can identify vectors with straight line journeys. World lines can be curved, and can be segmented - but a vector can't.

And by net journey, I meant the line that joins the start of the first component vector on the space time diagram with the end of the last component vector

I don't know what this means. What's the first component vector? What's the last component vector? Are you breaking the journeys up into a series of smaller journeys and treating the overall journey as a sum of these smaller journeys?

So traveling twin’s history involves a complicated sequence of component vectors, but the understanding of the age difference with her twin depends only on this summary vector.

You need to know the journey to compute the age - the starting point and ending point aren't enough. The path is essential. It's true that minkowski separation between two points is independent of coordinate system - so you're right about that - but you can only read the time off this figure for a clock that travels inertially - i.e. in a straight spatio-temporal line - between the two points.

To work out the time for the twin who travels, if he does it in two distinct journeys, you can just sum the times of the two journey. But, as it's a different path, this sum won't be the same, any more than the sum of two sides of a triangle equals the third.

Hope this helps

 

[Austin0]

Well, all I can say, Austin0 is that I do not understand where my interpretation changes. It seems perfectly consistent to me. Again, I suppose you could be pedantic and say that they would only ever disagree about spatial distances and time durations if they were both taking a Newtonian Mechanical view. If they had both studied relativity, they would understand that distances and times would not appear the same for each other and if they were comfortable with the mathematics, they could even work out exactly what the distances and times would appear to be for each other. But the point is that they don’t really need to. If they do their Minkowski calculations with their own distance and time measurements and arrive at a consistent spacetime distance, then there is no disagreement.

As I have said, I accept that the diagrams that appear further up this thread show best how the two viewpoints change over the duration of traveling twin’s journey. It just seemed to me that there was a much simpler way of understanding why traveling twin is younger when the journey is over, and why such a notion is consistent with relativity theory.

Hi Hen Natton
I think there is a miscommunication here. I doubt that atyy disagreed with your final relativistic interpretation. I know I don;t in any way.

What I was referring to was the contrast between : 1)the Newtonian view and also the view based on the assumption of reciprocal dilation with2)the full relativistic analysis and conclusion.

That the disagreement between the expected results in 1)[i.e same age] and the reality [age differenence]
compared to 2) where the expections agree with the end result,, is a large part of the cause for the popularity and perreneial reoccurrence of the "paradox"

The Newtonian view is the problem for people with no real understanding of SR
The assumprion of reciprocal dilation is the problem for people with some limited understanding of SR. For some people with a broader understanding of the principles involved it is not a problem with regard to the Twin;s "P" itself which is a non=issue.
But it raises questions regarding the various valid and consistent methods of resolution and the role of recirpocal dilation in those methods.

I hope this may clarify things.

 

[Ken Natton]

I’m not really looking to relight the blue touch paper on this. I did want to acknowledge the contributions from yossell and Austin0. I think, for the most part I have sorted out what you guys are telling me – I’m not claiming to fully understand it all, but I can see some of where I went wrong. I suppose it was always a little naïve to think that there was any simpler way of putting it than had already been presented on the thread. I suppose that I need to keep reading and keep thinking.

But of all the responses I got yesterday, the one that really bothers me is matheinste’s post. Pretty central to the understanding that I thought I had taken is the idea that a clock traveling with an observer measures the time in his or her reference frame only. I confess that I have not previously encountered this term ‘proper time’. This seems to carry connotations of some absolute measurement of time, which matheinste’s assertion that

A clock traveling with an observer measures the proper time along the worldline of the observer. It is a measure of the spacetime path length and is frame independent.

seems to confirm. But this is fundamentally against what I thought special relativity asserts.

The ... time measured by a clock with the stay at home twin is not the same as the ... time measured by the traveler ...

is more in line with my understanding, but I don't see why it follows that

...and so the spacetime path length is not the same.

Would someone care to expand on this for me?

 

[yossell]

Hi Ken,

remember that, while times and (spatial)distances are all frame-relative in relativity, the Minkowski separation between two points is *not* frame-relative, and so the Minkowski `length' of a path in space-time is not frame-relative either. That's what Matheinste is referring to by his clock traveling with the observer - it's measuring the Minkowski length of the path and so measures something frame invariant.

Now, for an inertial clock traveling from event A to event B, these two events have the same spatial coordinate in that clock's frame - since the formula for a Minkowski separation between two events calculated in a frame F is: -temporal distance^2 + spatial distance^2 (quantities calculated from within frame F), and since for this frame spatial distance^2 is zero, the proper time alone effectively measures the Minkowski separation.

It's still true that, in other frames, other clocks will disagree about the temporal separation between these two events.

Hopefully that resolves the tension you felt between the first two quoted statements.

As for `so the spacetime length is not the same', I think there is nothing more to this than the idea that two paths between two events do not have the same (Minkowski) length - it is effectively the triangle inequality - the length of a straight path from A to B is not the equal to the length of the straight path from A to C plus the length of the straight path from C to B, even though the VECTOR (AB) equals the VECTOR SUM of (AC) with (CB)

 

[Austin0]

IMO
Your initial feeling that a clock meaqsures the time in its own frame is correct.
This in fact is proper time

Matheniste is referring to the calculation of that time from other frames which is derived from spacetime path length. And will agree with other frames calculations regarding the same frame

Given that the proper elapsed time of the stay at home twin is greater infers that the spacetime path length must be shorter.

I am gald if I have helped.

 

[matheinste]

Hello Ken.

EDIT. Yossell beat ne to it.

The interval is central to the geometry of spacetime and is frame invariant, every observer agrees upon its value. It is analogous to the distance between two points in Euclidean space which can be calculated from the coordinate values of the points, but because of the minus sign in the metric it can take positive, zero or minus values unlike the Euclidean distance which is always positive. Any path in spacetime can be approximated as closely as we wish by a series of small intervals along the path. So the "length" of a spacetime path is invariant because it is the sum of invariant intervals. For a timelike curve, that is one in which all intervals making up the curve are timelike and which is the only type of path along which a body possessing mass can move, the proper time, that is the time interval measured by a clock traveling with the object, is a measure of the path length and is invariant.

Do not confuse the spacetime path length with the spacetme distance (interval) between two events. There are an infinite number of spacetime paths between two events but only one interval. But all the path lengths (proper times) are invariant.The stay at home twin's path length is the same as the spacetime interval but the traveling twin's is not.

An object traveling inertially experiences the greatest possible proper time between two events at which it is present and so the stay at home twin, being inertial, ages more than than his twin whatever path his twin takes between the two events of separation and reuniting.

Matheinste.

 

[Ken Natton]

My opinion, for what it is worth, is who beat whom to reply is not important. Differing perspectives are always useful and those last three replies add up to something very helpful for me. I hope yossell and Austin0 won’t mind if I say that matheinste’s post #184 delivers the most solid boot in the backside in the direction of a better understanding for me. I’m sure you understand that I need to go and chew the cud for a while. I hope to cross paths with you all again very soon.

 

[yossell]

I hope yossell and Austin0 won’t mind ...

On the contrary, I am now a broken man. Expect a stiff letter from my lawyer.

 

[phyti]

Would someone care to expand on this for me?

First you should understand this is an idealized, isolated, thought experiment.
The Earth is not in the 'chosen' fixed frame of reference, and the outbound and inbound legs will not be equal. When a clock moves,it runs slower, a result of the speed of
light in space being constant and independent of its source. The clock used by the Earth twin is moving (relative to the sun, at the least). The clock used by the a-naut is moving
(relative to the earth). Both clocks are thus affected by time dilation, but we can
only measure the difference in speed of both clocks, which is not enough information to decide if the clock rates are different.
SR is designed to produce symmetrical results when comparing one linear motion to another, and the method used to establish simultaneity for a frame of reference produces
distorted times and distances. The simple observations of diverging or converging paths will not determine if there is a difference in the clocks (aging of the twins).

When one twin rejoins the other, the symmetry and uncertainty is removed. For the returning twin, either the outbound or inbound leg, or both had to be at a speed faster
than the speed of the other twin, therefore that clock would have accumulated less time than the other clock, and the loss cannot be totally compensated by any gain in the other
leg of the trip.
The 2-part trip will lose more time than the 1-part trip.