Time Dilation in Twin Paradox: Exploring the Puzzling Reality

[JesseM]

it is exactly what he 'sees' as long as he takes light travel time into account. where are you getting the idea that it isnt. this is simple relativity.

Light travel time depends on knowing the distance that the signal was when it was emitted, and this depends on your choice of coordinate system as well. I object to your use of the word "perceive", which makes it sound like it's just a straightforward observation that doesn't depend on choosing a particular coordinate system in which to define measurements. For example, if the twin that turns around uses an inertial coordinate system where he is at rest during the outbound phase, and then continues to use that same coordinate system during the inbound phase (rather than switching to a new coordinate system where he is at rest during the inbound phase), then he will find that the inertial twin's clock is ticking faster than his own during the inbound phase, not slower. This will be true even if the only thing he uses the coordinate system for is to calculate the distance of the inertial twin from himself when the light from each clock tick was emitted, in order to figure out how long ago each tick "really" happened by subtracting the light travel time.

 

[granpa]

the phrase 'what he sees' implies that he is using his current frame in which he is at rest.

 

[JesseM]

the phrase 'what he sees' implies that he is using his current frame in which he is at rest.

Most physicists seem to use the word "observes" for what is happens in a given observer's rest frame, while "sees" refers to actual visual appearances.

 

[granpa]

lol.

 

[JesseM]

Well, whatever you think of the terminology, my original point stands: "perceives" makes it sound too physical, when in fact your statement depends on the non-inertial observer first picking one coordinate system during the outbound leg and another during the inbound leg, any time you have a non-inertial observer the choice of what coordinate system(s) represent his "perceptions" is pretty arbitrary. I could equally well invent a coordinate system where the non-inertial observer is at rest and in which the inertial twin's clock alternates between ticking faster and slower, and then say based on this that the non-inertial twin "percieves" the inertial twin's alternates between fast and slow ticking throughout the journey--this statement would be no more or less physical than your own.

 

[neopolitan]

...(therefore - ed.) you accept that it is the fact that he accelerates during the turn around that causes him to actually age less ...

This, along with the discontinuity, is a sticking point.

Having confirmed that Fredrik does not believe there is a real discontinuity, I thought this reduced ageing due to acceleration was dead too.

Is it possible to garner informed opinion on this: which is true?

A. Acceleration has no effect on relative ageing, per se, only the resultant trajectory through spacetime is relevant.

B. Acceleration does affect relative ageing directly but for some reason only at the middle of a two way journey which ends up with the traveller back at the origin. (Note this is acceleration, not resisted force such as gravity. I am trying to stick to SR as much as possible.)

C. Acceleration does affect relative ageing directly, during all accelerations during the journey. (Note this is acceleration, not resisted force such as gravity. I am trying to stick to SR as much as possible.)

In my understanding, the answer is A. I also interpret DaleSpam's response as A.

cheers,

neopolitan

 

[granpa]

This, along with the discontinuity, is a sticking point.

Having confirmed that Fredrik does not believe there is a real discontinuity, I thought this reduced ageing due to acceleration was dead too.

Is it possible to garner informed opinion on this: which is true?

A. Acceleration has no effect on relative ageing, per se, only the resultant trajectory through spacetime is relevant.

B. Acceleration does affect relative ageing directly but for some reason only at the middle of a two way journey which ends up with the traveller back at the origin. (Note this is acceleration, not resisted force such as gravity. I am trying to stick to SR as much as possible.)

C. Acceleration does affect relative ageing directly, during all accelerations during the journey. (Note this is acceleration, not resisted force such as gravity. I am trying to stick to SR as much as possible.)

In my understanding, the answer is A. I also interpret DaleSpam's response as A.

cheers,

neopolitan

now please don't take my comments out of context. I was speaking to one particular person with whom I was looking for semething we could agree on. and I was speaking in the broadest possible terms. acceleration doesn't effect aging directly. time dilation affects aging.

but we all agree that it is acceleration that is the key to understainding the twin paradox.

 

[neopolitan]

now please don't take my comments out of context. I was speaking to one particular person with whom I was looking for semething we could agree on. and I was speaking in the broadest possible terms. acceleration doesn't effect aging directly. time dilation affects aging.

but we all agree that it is acceleration that is the key to understainding the twin paradox.

I wasn't ascribing a particular point of view, granpa, I was trying to clarify exactly what your point of view is, and whether there are informed people who actually think that acceleration does affect the ageing.

I agree that, in so much as the twin which undergoes acceleration and returns to the unaccelerated twin and that means that one twin experiences two inertial frames while the other experiences one, then, yes, acceleration is key. But it is not the acceleration, per se, but the consequence of the acceleration.

And acceleration causes no time dilation directly. Different inertial frames cause time dilation.

I assume that you do not think that acceleration causes time dilation, but I can't put my hand on my heart and say that I know that you don't. (I could I guess, but I would not be being genuine :) )

cheers,

neopolitan

 

[Dale]

Any reference frame where the stationary twin is aging more slowly on both legs is a non-inertial reference frame.

that is exactly what we are saying. why do you say that as though it disagrees with what we are saying.

I haven't been following what you have been saying, but this disagrees with what Frederik has been saying. He has been claiming that you can obtain the twin paradox without reference to a non-inertial frame. Therefore I have been pointing out how every time he thinks he is avoiding a non-inertial frame he is in fact not avoiding it.

If this is an accurate summary of DaleSpam's view I think I'd agree with you and disagree with DaleSpam--after all, if we want to integrate along a curve to find its length from point A to C in Euclidean space, we're free to pick some point B along the curve between A and B, and use one cartesian coordinate system to do an integral which gives us the length from A to B, and a different cartesian coordinate system to do an integral which gives us the length from B to C, and then add these two lengths. Here we are not using a single non-cartesian coordinate system, but rather adding results from two different cartesian coordinate systems, which is perfectly valid in this situation.

But that is exactly what you are not doing. If you are to correctly perform such an integration you must do a change of variables. If you do not do a change of variables then you are implicitly using a single non-cartesian coordinate system.

$$\int_a^c f(x) \, dx=\int_a^b f(x) \, dx+\int_b^c f(x) \, dx=\int_a^b f(x) \, dx+\int_{b'}^{c'} f\left(x'\right) \, dx'\neq \int_a^b f(x) \, dx+\int_b^c f\left(x'\right) \, dx'$$
where the primed variables are the same points but in a different cartesian coordinate system.

Since in the analysis of the twins paradox you do not properly transform between the two inertial coordinate systems you are not doing two separate integrals in cartesian/inertial coordinates, but rather one integral in a non-cartesian/non-inertial coordinate system.

You cannot possibly obtain the twin paradox without using a non-inertial coordinate system and treating it as an inertial coordinate system. Therefore all that needs to be taught is how to identify and avoid non-inertial coordinate systems.

 

[JesseM]

If this is an accurate summary of DaleSpam's view I think I'd agree with you and disagree with DaleSpam--after all, if we want to integrate along a curve to find its length from point A to C in Euclidean space, we're free to pick some point B along the curve between A and B, and use one cartesian coordinate system to do an integral which gives us the length from A to B, and a different cartesian coordinate system to do an integral which gives us the length from B to C, and then add these two lengths. Here we are not using a single non-cartesian coordinate system, but rather adding results from two different cartesian coordinate systems, which is perfectly valid in this situation.

But that is exactly what you are not doing. If you are to correctly perform such an integration you must do a change of variables. If you do not do a change of variables then you are implicitly using a single non-cartesian coordinate system.

Of course I'm doing a change of variables--for the first integral from A to B you'd use the position coordinates of the first coordinate system in order to get the coordinate-independent length of the curve from A to B, and for the second integral from B to C you'd use the position coordinates of the second coordinate system in order to get the coordinate-independent length from B to C, then you'd add up the two coordinate-independent lengths to get the total length from A to C. I don't understand why you think this change of variables conflicts with what I said above in the quoted section, though.

Since in the analysis of the twins paradox you do not properly transform between the two inertial coordinate systems you are not doing two separate integrals in cartesian/inertial coordinates, but rather one integral in a non-cartesian/non-inertial coordinate system.

I don't understand what you mean by "transform between the two inertial coordinate systems" in this context. If different coordinate systems didn't disagree about simultaneity, you could certainly use one coordinate system to find the coordinate-independent proper time elapsed on the inertial twin's worldline between the moment the other twin departed and the moment the other twin turned around, and then use a different coordinate system to find the coordinate-independent proper time elapsed on the inertial twin's worldline between the moment the other twin turned around and the moment they reunited. Since proper times are coordinate-independent, you could then just add these to find the total time elapsed on the inertial twin's clock between the moment they separated and the moment they reunited, no need for any transformation of these times. Of course, since the two coordinate systems don't define simultaneity the same way this procedure is wrong because it leaves a section of the inertial twin's worldline unaccounted for, but that difference in simultaneity between the two frames is really the only reason to reject this procedure.

 

[phyti]

The twin paradox appeared after considering the effects of SR. To avoid
accelerations, the travel times can be extended so as to make the reversal
neglible. The gamma expression for time dilation specifically states that speed
relative to c is the determining factor. When you eliminate the reversal, you trade
off the gradual catching up of signals from the Earth twin for a discontinuous
'jump' in Earth time, and must insert this (reset the Earth clock).
It can be shown geometrically that the twin on the two leg trip, ages slower on at
least one of the legs, and less in total than the twin on the one leg trip.

 

[Dale]

Of course I'm doing a change of variables--for the first integral from A to B you'd use the position coordinates of the first coordinate system in order to get the coordinate-independent length of the curve from A to B, and for the second integral from B to C you'd use the position coordinates of the second coordinate system in order to get the coordinate-independent length from B to C, then you'd add up the two coordinate-independent lengths to get the total length from A to C.

If you do such a change of variables then the twin paradox does not occur. The twin paradox only occurs if you fail to transform coordinates at the turnaround (thereby implicitly using a non-inertial reference frame).

Of course, since the two coordinate systems don't define simultaneity the same way this procedure is wrong because it leaves a section of the inertial twin's worldline unaccounted for, but that difference in simultaneity between the two frames is really the only reason to reject this procedure.

That and the fact that it is mathematically incorrect. Just because there might be some pairs of coordinate systems where the transformation of one or more coordinates is identity doesn't mean that you can skip the transform.

 

[JesseM]

If you do such a change of variables then the twin paradox does not occur. The twin paradox only occurs if you fail to transform coordinates at the turnaround (thereby implicitly using a non-inertial reference frame).

What specifically do you mean by "transform coordinates at the turnaround" in this example? Let's stick to the example of calculating the length of a curve in ordinary 2D Euclidean space, where I say that using two different coordinate systems to calculate two parts of the length is perfectly valid. Do you agree that length along a curve is coordinate-invariant, so if I calculate the length from point A to point B in one cartesian coordinate system, and I calculate the length from point B to C in another, I can just add them together to get the length from point A to C? Do you think I have to "transform coordinates at point B" here, and am therefore "implicitly using a non-cartesian coordinate system"?

That and the fact that it is mathematically incorrect. Just because there might be some pairs of coordinate systems where the transformation of one or more coordinates is identity doesn't mean that you can skip the transform.

I don't understand what you mean by "the transformation of one or more coordinates is identity"--what specific coordinates do you think I am transforming in this example?

 

[Dale]

Do you think I have to "transform coordinates at point B" here, and am therefore "implicitly using a non-cartesian coordinate system"?

Most definitely you must transform the coordinates. Let's say that in some Cartesian coordinate system A = (ax,ay) and the coordinates of that same point in some other Cartesian coordinate system is A' = (ax',ay'). Similarly for B and C. Because the norm is invariant we have |A-B| + |B-C| = |A-B| + |B'-C'| != |A-B| + |B-C'| != |A-B| + |(bx,by')-C'|.

Any coordinate system in which either of the last two equations would be true would necessarily be a non-Cartesian coordinate system and would therefore use a different form of the norm that the standard Cartesian norm.

 

[JesseM]

Most definitely you must transform the coordinates. Let's say that in some Cartesian coordinate system A = (ax,ay) and the coordinates of that same point in some other Cartesian coordinate system is A' = (ax',ay'). Similarly for B and C. Because the norm is invariant we have |A-B| + |B-C| = |A-B| + |B'-C'| != |A-B| + |B-C'| != |A-B| + |(bx,by')-C'|.

What does notation like |A-B| represent in this context? Is it $$\sqrt{(ax - bx)^2 + (ay - by)^2 }$$? And when you say "the norm is invariant", the norm of what? I thought we were talking about curves, not vectors. But let's pick a very simple example of a curve--a straight line segment with endpoints A and C, and a point along the line between them labeled B. Say we have two cartesian coordinate systems, in the first the coordinates of A are (ax,ay), the coordinates of B are (bx,by), the coordinates of C are (cx,cy), and in the second coordinate system the coordinates are (ax',ay'), etc. So can't we say that, using the pythagorean theorem in the first coordinate system, the distance from A to B is $$\sqrt{(bx - ax)^2 + (by - ay)^2 }$$, and using the pythagorean theorem in the second coordinate system, the distance from B to C is $$\sqrt{(cx' - bx')^2 + (cy' - by')^2 }$$? And once we have actually evaluated those expressions, since each length is coordinate-invariant, can't we just add the two to get the total length from A to C? I don't see where in this a coordinate transformation has been done.

 

[Dale]

But let's pick a very simple example of a curve--a straight line segment with endpoints A and C, and a point along the line between them labeled B.

Sorry I wasn't clear. That is exactly what I was thinking. We should probably also be explicit that we are considering only rotations and translations.

So can't we say that, using the pythagorean theorem in the first coordinate system, the distance from A to B is $$\sqrt{(bx - ax)^2 + (by - ay)^2 }$$, and using the pythagorean theorem in the second coordinate system, the distance from B to C is $$\sqrt{(cx' - bx')^2 + (cy' - by')^2 }$$? And once we have actually evaluated those expressions, since each length is coordinate-invariant, can't we just add the two to get the total length from A to C? I don't see where in this a coordinate transformation has been done.

The coordinates (bx',by') and (cx',cy') are the result of the coordinate transformation of B and C into the primed frame.

 

[JesseM]

The coordinates (bx',by') and (cx',cy') are the result of the coordinate transformation of B and C into the primed frame.

Sure, but any time you use multiple coordinate systems to analyze the same physical scenario, naturally you must know the coordinate transformation that relates them. Are you arguing that anytime we juggle different cartesian coordinate systems (or different inertial coordinate systems in SR) when making calculations about coordinate-invariant quantities, we are implicitly using a single non-cartesian/non-inertial coordinate system? Suppose instead of breaking the path from A to C into two segments, we instead calculated the entire length from A to C in the first coordinate system, then again calculated the entire length from A to C in the second coordinate system. Aren't we just doing calculations in two separate cartesian coordinate systems here, rather than implicitly using a single non-cartesian coordinate system?

 

[Dale]

Are you arguing that anytime we juggle different cartesian coordinate systems (or different inertial coordinate systems in SR) when making calculations about coordinate-invariant quantities, we are implicitly using a single non-cartesian/non-inertial coordinate system?

No that is not my argument. I'm sorry JesseM, but I don't know how much clearer I can be than posts 189 and 184, and it really frustrates me that you are arguing against such an obvious point as the fact that the twin paradox does not occur unless you try to treat the traveling twin's non-inertial frame as an inertial one. Geometrically it is stunningly obvious that any frame that straightens the travelers worldline is non-inertial.

Suppose instead of breaking the path from A to C into two segments, we instead calculated the entire length from A to C in the first coordinate system, then again calculated the entire length from A to C in the second coordinate system. Aren't we just doing calculations in two separate cartesian coordinate systems here, rather than implicitly using a single non-cartesian coordinate system?

Yes, and when you do that the twin paradox does not arise since all inertial frames agree on the length of each path.

 

[JesseM]

No that is not my argument. I'm sorry JesseM, but I don't know how much clearer I can be than posts 189 and 184

If you would address my questions it might clear it up--I really don't see how the notation you gave in 189 and 184 is relevant to what I'm doing, which is computing the length of two segments of a curve in two separate coordinate systems and then adding the lengths of these segments to find the total length.

and it really frustrates me that you are arguing against such an obvious point as the fact that the twin paradox does not occur unless you try to treat the traveling twin's non-inertial frame as an inertial one.

The most "naive" version of the twin paradox arises because one imagines that the non-inertial twin has a single frame which is used throughout the problem, but I have gotten into arguments with people who put forward a more "sophisticated" version, where they acknowledge that the non-inertial twin has different inertial rest frames at different times and suggest there's no reason why you can't just use different frames for each segment of the journey. My point is that if it weren't for simultaneity issues, this more sophisticated argument would be totally valid, just as it's totally valid to use two different cartesian coordinate systems to calculate the length of different segments of a curve, and then add up the length of each segment to get the total length.

Suppose instead of breaking the path from A to C into two segments, we instead calculated the entire length from A to C in the first coordinate system, then again calculated the entire length from A to C in the second coordinate system. Aren't we just doing calculations in two separate cartesian coordinate systems here, rather than implicitly using a single non-cartesian coordinate system?

Yes, and when you do that the twin paradox does not arise since all inertial frames agree on the length of each path.

Only if the segments you are using actually join with one another correctly at the ends. The point is that if you erroneously say "Ok, I'll use frame #1 to calculate the proper time on the inertial twin's worldline between the moment the twins separate and the moment the non-inertial twin turns around, and I'll use frame #2 to calculate the proper time on the inertial twin's worldline between the moment the non-inertial twin turns around and the moment they reunite", then these two segments do not in fact cover the entire worldline of the inertial twin, because the point on the inertial twin's worldline that's simultaneous with the turnaround in frame #1 is not the same as the point on the inertial twin's worldline that's simultaneous with the turnaround in frame #2. If different frames agreed on simultaneity, the procedure I describe above would be fine, but they don't so it's not.

 

[Al68]

And acceleration causes no time dilation directly. Different inertial frames cause time dilation.

Hi neoploitan,

Sounds like a matter of semantics to me, since the change of reference frames is caused by the acceleration. And gravitational time dilation is even derived from SR, considering accelerated reference frames.

Might I suggest looking at a little known and very different (and not very popular) Twins paradox resolution: http://en.wikisource.org/wiki/Dialog_about_objections_against_the_theory_of_relativity

This is Einstein's resolution, and actually considers the ship to be stationary during the turnaround, and Earth's clock to run faster than the ship's in the ship's accelerated frame, resulting in the Earth twin aging more overall.

Al

 

[neopolitan]

Hi neoploitan,

Sounds like a matter of semantics to me, since the change of reference frames is caused by the acceleration. And gravitational time dilation is even derived from SR, considering accelerated reference frames.

Might I suggest looking at a little known and very different (and not very popular) Twins paradox resolution: http://en.wikisource.org/wiki/Dialog_about_objections_against_the_theory_of_relativity

This is Einstein's resolution, and actually considers the ship to be stationary during the turnaround, and Earth's clock to run faster than the ship's in the ship's accelerated frame, resulting in the Earth twin aging more overall.

Al

Hi Al,

I don't mean it to sound like semantics. What I am trying to get at is an idea whether anyone actually believes that something funny happens during the turnaround due to the acceleration.

It seems to me that Fredrik and Jesse are saying that during the turnaround there is a change of trajectory which results in a simultaneity view change, sort of like "B" looking at "A"'s 7.2 year clock before turn around and looking past 25.6 years worth of clocks during the change of trajectory to end up looking at "A"'s 32.8 year clock. But this is a simultaneity perspective change, not any sudden ageing of "A".

It also seems to me that some people want to say that during the turnaround, either "A" ages suddenly, or "B" somehow jumps forward or moves quickly forward from a moment simultaneous with "A" having aged 7.2 years to a moment simultaneous with "A" having aged 32.8 years.

I don't think that is semantics.

If everyone pretty much agrees with the first then there is no problem, if a significant number of people agree with the second, I think there is a problem. It sounds terrible to say it, but I wonder if Einstein was on the ball with that text referenced.

I thought that it is "resisting" gravity, as we do by virtue of standing on the surface of a planet, that is instrumental in creating the time dilation effects. A body which is allowed to fall with gravity takes the shortest path through spacetime and will age more slowly, won't it? A body which is allowed to fall takes on a velocity in the same way as rocket ship alters trajectory when accelerated. For that reason, I would have thought that acceleration would not cause any "speeding ahead".

My understanding is also that acceleration has been shown, experimentally, not to cause any time dilation effect.

I would appreciate informed comment, and a polite request for clarification if anyone doesn't understand what the hell I am talking about.

cheers,

neopolitan

 

[Jonathan Scott]

The thing that happens suddenly is no more mysterious than the fact that if you turn your head, things which were ahead of you are SUDDENLY beside you and so on. It isn't the turning of your head which moves things; it is merely that once you have turned it, you see things differently.

Changing velocity instantly changes your point of view. Our normal view of time, space and simultaneity is a simplified illusion which works only because we do not often encounter relativistic speeds, so we have to use mathematics to model how it would really look. The set of events considered to be simultaneous with one's own clock depends on velocity in a very similar way to the way that "beside me" depends on the direction in which you are facing.

 

[granpa]

The thing that happens suddenly is no more mysterious than the fact that if you turn your head, things which were ahead of you are SUDDENLY beside you and so on. It isn't the turning of your head which moves things; it is merely that once you have turned it, you see things differently.

Changing velocity instantly changes your point of view. Our normal view of time, space and simultaneity is a simplified illusion which works only because we do not often encounter relativistic speeds, so we have to use mathematics to model how it would really look. The set of events considered to be simultaneous with one's own clock depends on velocity in a very similar way to the way that "beside me" depends on the direction in which you are facing.

first post to make good sense in a long time.

 

[Al68]

It sounds terrible to say it, but I wonder if Einstein was on the ball with that text referenced.

Hi neoploitan,

Well, I've asked about it on this forum and was very surprised that no one seems to have an opinion, or be familiar with it. The result is the same as the standard version, which makes sense if gravitational time dilation is just SR time dilation in an accelerated frame.

A body which is allowed to fall takes on a velocity in the same way as rocket ship alters trajectory when accelerated.

I think it would be more accurate to say that a body which is allowed to fall takes on a velocity relative to Earth's surface in the same way as Earth takes on a velocity relative to an accelerated rocket. If I throw a ball up in the air, it will turnaround due to my (proper) acceleration relative to the ball, not the other way around.

My understanding is also that acceleration has been shown, experimentally, not to cause any time dilation effect.

Some would object to the word "cause" semantically, but I would say that gravitational time dilation is proven, and it's "caused" by acceleration (relative to inertial frames). And time dilation is a result of relative velocity which is "caused" by acceleration.

I would appreciate informed comment, and a polite request for clarification if anyone doesn't understand what the hell I am talking about.

Well, I would agree that the standard resolution is not a satisfactory explanation. It really only explains how the SR math works. And the jump ahead in Earth's clock is a result of looking at a single event from two frames. Notice that right before the ship reaches earth, in the ship's return frame, the time the ship first left Earth is 66.66 yrs ago, and only 40 yrs has elapsed on earth. In the ship's return frame, the Earth clock reading zero is simultaneous with the ship's clock reading -42.66 yrs, if I did the math right. It's just a convention used to assign times to events in different frames, "simultaneity issues" don't "cause" anything to happen to anyone or their clocks.

Unless one is of the school of thought that the laws of physics are caused by mathematics. (A school with way too many members, IMO.)

Al

 

[Al68]

I thought that it is "resisting" gravity, as we do by virtue of standing on the surface of a planet, that is instrumental in creating the time dilation effects.

There's also the example of differential clock rates between clocks at the "top" and "bottom" of a long rocket. The result is the same for 1G acceleration whether the rocket is firing its thrusters in deep space or sitting on a launchpad on earth. The rocket is accelerating the same relative to an inertial frame either way.

Al

 

[Dale]

Hi JesseM,

OK, here is one last attempt.

My objection to simultaneity-based resolutions is not a technical one, it is a pedagogical one. I understand the simultaneity explanations and their validity. However, it seems that you do not accept the validity or completeness of the spacetime geometric explanation. If the spacetime geometric explanation is valid and complete then, IMO, it is preferable pedagogically for several reasons:

1) it avoids explicit simultaneity issues which is the most difficult concept for students to learn
2) it allows an opportunity to teach students how to identify and avoid non-inertial coordinate systems
3) it is generally applicable including non-inertial coordinate systems, curved spacetimes, and arbitrary worldlines
4) it reinforces Minkowski geometry and the modern way to think about relativity

OK, having explained my motivation, here is how I would present the spacetime geometric approach to the twin paradox to students.

First, I would draw the spacetime diagram in the stay-at-home twin's rest frame and calculate the spacetime intervals for both twin's paths. I would spend some time talking about Minkowski geometry and how (in contrast to Euclidean geometry) the longest timelike interval between two events is a striaght line.

Second, I would talk about the traveler's point of view and draw the "mirror" spacetime diagram where the traveling twin has a vertical worldline. I would calculate the spacetime intervals and obtain the twin paradox. I would point out that the reference frame that I drew was a non inertial reference frame and mention that the usual laws, including the formula for the spacetime interval, only apply for inertial frames. At that point I would expect a rather lengthy discussion about non-inertial frames including physical features and geometrical features.

Third, I would Lorentz transform the original spacetime diagram into the inertial frame where the traveler was at rest during the first leg, and the inertial frame where the traveler was at rest during the second leg. I would show that the conclusion is the same in each inertial frame. I would then emphasize the point that the special theory of relativity says that the laws of physics are the same in all inertial frames, but not non-inertial frames. I would further mention that there is no inertial frame where the traveler is at rest the whole time.

If a student wanted to do the "two inertial frames and add them up" approach I would ask them to work the problem. If they correctly transform from the "first leg" frame to the "last leg" frame then they will get the correct answer, no paradox. If they do not correctly do the transform then I would point out how they are accidentally using the non-inertial frame described above.

 

[neopolitan]

Hi JesseM,

OK, here is one last attempt.

My objection to simultaneity-based resolutions is not a technical one, it is a pedagogical one. I understand the simultaneity explanations and their validity. However, it seems that you do not accept the validity or completeness of the spacetime geometric explanation. If the spacetime geometric explanation is valid and complete then, IMO, it is preferable pedagogically for several reasons:

1) it avoids explicit simultaneity issues which is the most difficult concept for students to learn
2) it allows an opportunity to teach students how to identify and avoid non-inertial coordinate systems
3) it is generally applicable including non-inertial coordinate systems, curved spacetimes, and arbitrary worldlines
4) it reinforces Minkowski geometry and the modern way to think about relativity

OK, having explained my motivation, here is how I would present the spacetime geometric approach to the twin paradox to students.

First, I would draw the spacetime diagram in the stay-at-home twin's rest frame and calculate the spacetime intervals for both twin's paths. I would spend some time talking about Minkowski geometry and how (in contrast to Euclidean geometry) the longest timelike interval between two events is a striaght line.

Second, I would talk about the traveler's point of view and draw the "mirror" spacetime diagram where the traveling twin has a vertical worldline. I would calculate the spacetime intervals and obtain the twin paradox. I would point out that the reference frame that I drew was a non inertial reference frame and mention that the usual laws, including the formula for the spacetime interval, only apply for inertial frames. At that point I would expect a rather lengthy discussion about non-inertial frames including physical features and geometrical features.

Third, I would Lorentz transform the original spacetime diagram into the inertial frame where the traveler was at rest during the first leg, and the inertial frame where the traveler was at rest during the second leg. I would show that the conclusion is the same in each inertial frame. I would then emphasize the point that the special theory of relativity says that the laws of physics are the same in all inertial frames, but not non-inertial frames. I would further mention that there is no inertial frame where the traveler is at rest the whole time.

If a student wanted to do the "two inertial frames and add them up" approach I would ask them to work the problem. If they correctly transform from the "first leg" frame to the "last leg" frame then they will get the correct answer, no paradox. If they do not correctly do the transform then I would point out how they are accidentally using the non-inertial frame described above.

Very good post, Dale.

The only thing I would add is that it might be worth discussing the fact that:

(total time elapsed for stayathome)² = (total time elapsed for traveller)² plus (total distance traveled by traveller, according to stayathome)²

or, alternatively ...

minus (total time elapsed for stayathome)² plus (total distance traveled by traveller, according to stayathome)² = minus (total time elapsed for traveller)²

There are two reasons for doing this, one, because it is true and, two, because some numerically minded people like myself will work this out anyway, and they should be taught the correct interpretation of this.

That may be S² = -t² + x² ... etc, I am not totally sure what the "correct interpretation" is.

cheers,

neopolitan

PS I just want to clarify that:
"total distance traveled by traveller, according to stayathome" = velocity of separation x time elapsed according to stayathome

In Fredrik's original diagram that is: v = 0.8c, t = 40 years and distance traveled = 32 lightyears.

I think that if you do it the wrong way, then you end up with an incorrect equation (which would work only if the stayathome twin ages 14.4 years). Due to treating a non-inertial frame as an inertial frame. I need to take some time to jot down figures, but not right now.
PPS Jotted down the figures, worked it out and, lo and behold, you get the missing years. That is (in Fredrik's example):

minus (total time elapsed for traveller)² plus (total distance traveled by stayathome, according to traveller)²
= - (24²) + (0.8 x 24)² = -655.63 = - (25.6²)

Add the 14.4 years, then you have 40 years.

How can that be interpreted? or is it a huge coincidence?

 

[phyti]

Referring to the drawing, A moves vertically while B moves along t1 with a spatial
axis of simultaneous events s1. As A reverses for the return leg, his s-axis
rotates to the position s2. In the real world this rotation would occur for a
finite time interval, and B would perceive A's rate of activity speed up. For the
purpose of resolving the twin aging issue, it is instantaneous, and thus referred
to as a 'jump', misleading but still accounting for A's time between the two
positions of B's s-axis. The reason for this approach being, the aging issue can be
explained within the framework of SR in terms of time dilation. Even though
acceleration is required to change time lines, it can be arbitrarily small in
comparison to the total distance, and not significantly alter the results.

As an alternate approach, B could synchronize his clock with A while moving past A,
a third person C could synchronize his clock with B at the reversal point, while
moving on the return leg, i.e., electronic communication of times and uniform
constant motion for all involved.

The purpose of the solution of the problem is to demonstrate the relationship of
time rate with path taken. If the path between two events is longer, the speed must
be greater, therefore the time is shorter because of time dilation (a function of v/c).

 

[matheinste]

Hello phyti.

I agree with all that you say and do not have any problem with the resolution of the twin paradox in its various forms. The problem that some people will have to your third person addition to the scenario is that at the event where B and C meet or cross, what B and c will observe will be different, as they are in different inertial frames. C will of course observe what B would have observed immediately after the turnaround event. This is in no way a criticism of what you have said but in the light of objections earlier in this thread i think that this will be an objection from those who do not like the idea of the consequences of a very fast turnaround ( a rapid ageing of A ), albeit not discontinuous, in the first place.

Matheinste

 

[boysherpa]

General Question re Twin Paradox

I read through this thread - interesting. Nice discussions.

Please note that I am not confused by this discussion, nor am I attempting to disprove it. I am fully aware of its implications, as in how the GPS satellite system corrects for relativistic effects to provide clock data. I simply am trying to understand what is meant by time in the context of the discussion. Specifically, is a separate time "dimension" (for lack of a better term) necessary to the discussion, or is time merely a convenience, and is that what creates the seeming "paradox".

Let's use Einstein's ruler example. Suppose we wish to measure space, but the ruler, being in space, is coupled to the thing being measured. Can its measurement be taken as able to reveal a change in that thing? What is the objective, incorruptible measurement device, and what is the thing actually being measured?

Well, to me, time is what clocks measure. So, I don't know what it means to say time flows differently in the two reference frames. Which observer determines this, or is it only inferred by the twins when they meet up? The latter seems to be the case.

I think the clock with the traveling twin has no choice but to tic differently, not because it measures a different time, but because its operation is dilated. The human body might be considered a clock, with chemical and telomeric tics, and thus travel-twin experience dilated "age". But does this mean there is a time which is slowed? I don't think this is proven on spacetime diagrams, i think it is weakly inferred but not substantiated.

But then, I could be full of beans.

 

[Fredrik]

I simply am trying to understand what is meant by time in the context of the discussion. Specifically, is a separate time "dimension" (for lack of a better term) necessary to the discussion, or is time merely a convenience, and is that what creates the seeming "paradox".

You could draw spacetime diagrams for things that happen in non-relativistic mechanics too. If we do, the time "dimension" is just a convenience. Everyone agrees which "slices" of this spacetime should be thought of as space. This is however not the case in SR.

Suppose we wish to measure space, but the ruler, being in space...

Your question is already ill-defined. In SR, what "slice" of spacetime you would consider space depends on your velocity. What's "space" to you is a mixture of space and time to me.

Let's be a bit more precise: If we both assign spatial and temporal coordinates to events using a grid of rulers and synchronized clocks (your grid being at rest relative to you, and mine being at rest relative to me), and your grid assigns the same time coordinate to events X and Y, my grid would assign different time coordinates to them, unless we're moving at the same velocity.

Well, to me, time is what clocks measure. So, I don't know what it means to say time flows differently in the two reference frames. Which observer determines this, or is it only inferred by the twins when they meet up? The latter seems to be the case.

What clocks measure is called "proper time". That's a coordinate-independent concept, but it's only well-defined along some curves in spacetime. No one says that time "flows" differently in two frames, or that time "flows" at all. What we're saying is that if X and Y are two specific events, with time coordinates t_X and t_Y in one inertial frame, and time coordinates t'_X and t'_Y in another, t_X-t_Y is usually not equal to t'_X-t'_Y. This is where SR differs from pre-relativistic physics, where we would always have t_X-t_Y=t'_X-t'_Y.

I think the clock with the traveling twin has no choice but to tic differently, not because it measures a different time, but because its operation is dilated. The human body might be considered a clock, with chemical and telomeric tics, and thus travel-twin experience dilated "age". But does this mean there is a time which is slowed? I don't think this is proven on spacetime diagrams, i think it is weakly inferred but not substantiated.

The traveling twin will be younger when they meet again because what a clock measures is the integral of $\sqrt{dt^2-dx^2}$ along the curve that represents its motion. (dx is =0 everywhere along A's world line, but not along B's).

 

[Dale]

The only thing I would add is that it might be worth discussing the fact that:

(total time elapsed for stayathome)² = (total time elapsed for traveller)² plus (total distance traveled by traveller, according to stayathome)²

Hi neopolitan,

Sorry about the delay, finally had some time to consider this, and it seems to me that your formula can be derived from the spacetime intervals in the stay-at-home frame. I will use S for spacetime interval (or proper time), T for coordinate time, and X for coordinate distance. A following h is for the stay-at-home twin and a following t is for the traveling twin. And units where c=1.

Sh² = Th² - Xh²
St² = Tt² - Xt²

Since Xh = 0 and Th = Tt we can subtract the two equations and get

Sh² - St² = Xt²
or
Sh² = St² + Xt²

Which is what you have written above.

 

[neopolitan]

Hi neopolitan,

Sorry about the delay, finally had some time to consider this, and it seems to me that your formula can be derived from the spacetime intervals in the stay-at-home frame. I will use S for spacetime interval (or proper time), T for coordinate time, and X for coordinate distance. A following h is for the stay-at-home twin and a following t is for the traveling twin. And units where c=1.

Sh² = Th² - Xh²
St² = Tt² - Xt²

Since Xh = 0 and Th = Tt we can subtract the two equations and get

Sh² - St² = Xt²
or
Sh² = St² + Xt²

Which is what you have written above.

I certainly get "Xh = 0", but I am a bit lost with "Th = Tt" since I am trying to think about a time interval about which "A" and "B" will agree, and I can't come up with one. According to wikipedia ...

In special relativity, the coordinate time (relative to an inertial observer) at an event is the proper time measured by a clock that is at the same location as the event, that is stationary relative to the observer and that has been synchronised to the observer's clock using the Einstein synchronisation convention.

http://en.wikipedia.org/wiki/Coordinate_time"

Wrt to spacetime intervals, which are the basis of the S² equation, wikipedia's entry doesn't really lend itself to pasting here, but http://en.wikipedia.org/wiki/Spacetime_interval#Space-time_intervals" is the link.

I am not saying you are wrong, I follow all the logic and agree with it, but I can't quite get my head about one of your antecedants.

Then we get to the difficult part of interpreting the result of subtracting one equation from another to end up with the result. Is the method you present actually useful, pedagogically?

I was sort of hoping to have some useful interpretation of the numbers.

My shot from the hip, a while back, was a sort of economic approach. I'll rehash it for new readers.

You could think of it as if "A" and "B" have a certain amount of spacetime currency. This currency is enough to "pay" for the travel from one event in spacetime to another event. Say further that if both "A" and "B" stay at rest relative to each other, then these events share the same spatial location, that is they only differ in the time axis.

"A" decides to do just that and exchanges spacetime currency for the maximum amount of time currency, no space currency. "B", on the other hand, decides to do some traveling and exchanges spacetime currency for a mix of time currency and space currency.

The total amounts, however, add up to the same - using the equation we are discussing (using Dale's version):

Sh^² = St^² + Xt^²

Note that it doesn't matter a jot what sort of accelerations or manoeuvres "B" carries out - so long as we accept that Xt here is a simplification and "B" is not restricted to a single axis.

cheers,

neopolitan

PS I need to clarify something - the idea that "A" gets all time currency should be understood to be exchanging to what is, in "A"'s frame, purely time currency. Another observer may see that "A" was never really stationary at all, and had some space currency (ie an inertial velocity). This would be the same as the third observer being away that "A" has a little nest egg which is never counted by "A" in everyday financial considerations. The third observer will note that "B" would also have the same little nest egg, and the calculations would end up like this:

Sh^² + (little nest egg to a third observer)^² = St^² + Xt^² + (little nest egg according to a third observer)^²

Sorry about being pedantic, but it helps to cover yourself sometimes

 

[Dale]

I certainly get "Xh = 0", but I am a bit lost with "Th = Tt" since I am trying to think about a time interval about which "A" and "B" will agree, and I can't come up with one.

They don't agree on any coordinate time interval (T), they only agree on proper times (S).

Th and Tt are the time between the "departure" and "reunion" events in the stay-at-home twin's frame. None of the stuff in your formula or in my derivation of it has any reference to any measurement made in any frame other than the stay-at-home twin's frame.

 

[neopolitan]

They don't agree on any coordinate time interval (T), they only agree on proper times (S).

Th and Tt are the time between the "departure" and "reunion" events in the stay-at-home twin's frame. None of the stuff in your formula or in my derivation of it has any reference to any measurement made in any frame other than the stay-at-home twin's frame.

Ok.

Given that all measurements are made in the stayathome twin's frame, where is the pedagogical advantage in ignoring that, under the circumstances, Sh = Tt = Th and so then substituting we can arrive at:

St² = Sh² - Xt²

and so

Sh² = St² + Xt²

cheers,

neopolitan