The Paradox of Aging in Relativity: Resolving the Twin Paradox with a Twist

[Ibix]

Now we combine A and B into C, the "patched together" coordinate system. We note that point P has two different time coordinates. This is the problem, points are supposed to have unique coordinates.

If we draw the set of points at some time T , which we can call the "time axis", and which are the horizontal or near-horizontal lines in the diagrams above, then a point must have only one time, it's not allowed for it to be assigned two different times. But we can see that point P does lie on two different time-axis, it has two different times associated with it.

What's confusing me is that different coordinate values for the same point aren't always a problem. For example, you can't cover S² with one coordinate chart. The textbook solution is something like stereographic projection, where you define two charts each excluding one point and overlapping everywhere else. Obviously the co-ordinates aren't equal in the overlap region - so why is the fact that the coordinates aren't equal in the overlap bad in this twin paradox case? I can see that it is bad (there exist timelike paths that cannot be parameterised by their proper time, as Fredrik noted in #44), but is there a way to tell from the definition of the charts that it's bad?

Perhaps I should start a new thread, as I'm not sure this is entirely on topic.

 

[Fredrik]

(there exist timelike paths that cannot be parameterised by their proper time, as Fredrik noted in #44),

That's not what I noted. Every timelike curve can be parametrized by proper time. I was just saying that different coordinate systems assign different coordinates to events (I didn't mean to suggest that this is a problem), and that we would have a real problem (the theory would be nonsense) if there had been two valid ways to calculate a coordinate-independent quantity like the proper time of a timelike curve, and those ways yield different results.

 

[PhoebeLasa]

If, for a given observer, the current age of a distant object is arbitrary, then it is a meaningless concept. If it is a meaningless concept, then we shouldn't even be talking about the basic time dilation result at all (that a distant clock moving wrt us is ticking slower than our own clocks).

 

[ghwellsjr]

If, for a given observer, the current age of a distant object is arbitrary, then it is a meaningless concept. If it is a meaningless concept, then we shouldn't even be talking about the basic time dilation result at all (that a distant clock moving wrt us is ticking slower than our own clocks).

No, meanings come from definitions. Different definitions produce different meanings. Each different reference frame is a different definition for Coordinate Time and that's why they produce different a "current age" for each different reference frame. What's the problem?

 

[PeterDonis]

There is a physical reason why acceleration can be felt and measured as an invariant whereas the derivative of velocity is not invariant. For that, we are saying that acceleration is measured in the space and metric tensor defined by the universe.

You may be saying something true here, but I'm not sure. The true statement is that the metric determines what states of motion are freely falling and what states of motion are not freely falling, i.e., accelerated. But we don't measure acceleration by measuring the metric; we measure it with an accelerometer. That is, we don't have to determine acceleration indirectly (by measuring the metric and then calculating something about our state of motion); we can determine it directly. The acceleration we directly measure must be consistent with other measurements we make that tell us about the metric, of course.

 

[Evanish]

I was thinking about this scenario a while back.

I never came up with a solution I really liked so I kept thinking about it.

Now I'm thinking that maybe the distance between the two observers changes in such a way that it makes up for the time difference.
https://lh5.googleusercontent.com/1TJ6R_4zbJIYyHSrPRL3SyEgfOGRDSHjGWkADjKAqIvYYyKgpWdZHzsA6-ZhwtDVDuV2Qg98mFNlOcilqEjld211clA1oTTzkiVydwNA4x9d898GmUw
For simplicity's sake let's assume there is another person at the start who is observing all this. We can call him Mike. When Fred is moving away from Mike at a constant velocity the distance Mike observes him at is increasing in such a way that events between the two reference frames match up. Basically space is lengthening.

Picture that the three people have a device that produces a radio pulse, senses the radio pulse of the others and measures the distance they are away from each other. Mike would observe that Fred's time is slower, and if he used the speed of light plus measured distance between himself and Fred to calculate when the two pulses happened in relationship to each other he would find that the two pulses happened simultaneously. This is because the distance between them would have lengthened by the amount necessary to make it happen.

I wish I was better at math so I could describe this more clearly. At any rate interesting stuff.

 

[PeterDonis]

Now I'm thinking that maybe the distance between the two observers changes in such a way that it makes up for the time difference.

"Distance" is frame-dependent, so in a sense it does change depending on which observer's frame you use. But you seem to be leaving out relativity of simultaneity; you can't do a proper analysis without including that.

Mike would observe that Fred's time is slower, and if he used the speed of light plus measured distance between himself and Fred to calculate when the two pulses happened in relationship to each other he would find that the two pulses happened simultaneously.

If the two pulses were emitted simultaneously according to Mike, then they can't have been emitted simultaneously according to Fred, because simultaneity is relative, i.e., frame-dependent.

My advice: draw a spacetime diagram. First draw it in the frame of the observer who remains at rest at the starting point. Then transform to Mike's and Fred's frames (note that there are two for each of them, one for the constant velocity segment outbound and one for the constant velocity segment returning) to see how things look there.

 

[ghwellsjr]

My advice: draw a spacetime diagram. First draw it in the frame of the observer who remains at rest at the starting point. Then transform to Mike's and Fred's frames (note that there are two for each of them, one for the constant velocity segment outbound and one for the constant velocity segment returning) to see how things look there.

I already drew these diagrams in post #6 except that I made the accelerations instantaneous. There's no point in making things more complicated by spreading the accelerations out over time.

Note that there are actually just three diagrams but it is Mike that remains at rest (he gets one frame) and there are two more frames, since Bob's starting frame is the same as Fred's ending frame and vice versa.

You can copy my diagrams and draw in the paths of the radio signals in all three diagrams and see that they start and end at the same Proper Times for all observers no matter which frame you use.

If you want to see how a particular observer establishes the distance away of the other observers, just use a radio signal emitted by one observer that echoes off another observer and have the first observer use the time interval between sending and receiving as the two significant points of measurement. By assuming that the radio signals took the same amount of time to get to the other observer as it took for the return signal to get back and by assuming that the signals travel at c, the first observer can establish a time and distance to the other observer. Repeating these measurements over and over again will allow each observer to create a chart for each of their reference frames, even for the non-inertial observers. The concept is extraordinarily simple but it is tedious. Try it, you'll like it.

 

[PeterDonis]

I already drew these diagrams in post #6

Yes, I see you did. I should have expected that.

 

[Evanish]

Thanks ghwellsjr. I'll try to do what you suggested.

 

[Dale]

Mathematically, the derivative of a relative function is relative.

That is simply not correct. Whether or not something is relative is a question of the laws of physics, not merely whether it is a derivative of a relative function.

In fact, the derivative of a relative function can be invariant. This is easiest to see in Newtonian physics where the usual formulation of Newton's 2nd law is clearly invariant under translations and Galilean boosts, but not under accelerations. The velocity is relative but the derivative of velocity is not.

 

[Dale]

I think this is a restatement of the idea that coordinate charts can only be combined if any overlap is "smooth". Presumably this isn't a smooth overlap, but I'm not grasping what makes it non-smooth.

Think about the surface of the Earth (just the surface, so it is a 2D curved space). Now, even without any coordinates defined you can still do a lot of things. You can talk about the length of a road, or the angle of intersection between two roads. You can name points and give directions by reference to named points (on Main street, half a mile past Bob's Hardware).

Now, if you want, you can map physical points on the surface of the Earth to mathematical points in R2. This mapping between points on the Earth and points in R2 is called a chart, and the result can be put on a piece of paper (a cartographer's map).

There are many different ways to do that mathematical mapping (cartographers call them projections) and all are equally valid. Some preserve distances, some preserve angles, some are just easy to draw or read, and so forth.

The charts all have three important features
1. They map from open subsets of the surface of the Earth to open subsets of R2
2. They are smooth, meaning that nearby points on the Earth are also nearby in R2
3. They are invertible, meaning that for every point in the open subset of R2 there is one and only one point in the open subset of the surface

So the actual problem is not non smoothness (feature 2), it is non invertibility (feature 3).

 

[Ibix]

The charts all have three important features
1. They map from open subsets of the surface of the Earth to open subsets of R2
2. They are smooth, meaning that nearby points on the Earth are also nearby in R2
3. They are invertible, meaning that for every point in the open subset of R2 there is one and only one point in the open subset of the surface

So the actual problem is not non smoothness (feature 2), it is non invertibility (feature 3).

Got it. Somehow I'd missed the invertibility requirement, even though (on a re-read) that's what George was getting at back in #42. And it's kind of an obvious requirement for a chart, really. Now you've pointed it out, anyway.

Thank you.

 

[FactChecker]

FactChecker said: ↑
Mathematically, the derivative of a relative function is relative.

That is simply not correct. Whether or not something is relative is a question of the laws of physics, not merely whether it is a derivative of a relative function.

Please explain that. For instance: If the velocities from Twins 1 and 2's perspective always numerically satisfy v₁ = -v₂ , then why wouldn't we always have the numerical identity d/dt ( v₁ ) = d/dt ( -v₂ ) = -d/dt ( v₂ )?

 

[Dale]

Please explain that. For instance: If the velocities from Twins 1 and 2's perspective always numerically satisfy v₁ = -v₂ , then why wouldn't we always have the numerical identity d/dt ( v₁ ) = d/dt ( -v₂ ) = -d/dt ( v₂ )?

Yes, clearly. But if the velocities always satisfy d/dt ( v₁ ) = -d/dt ( v₂ ) then that does not imply that v₁ = -v₂. The question is, which is a valid law of physics. In relativistic physics neither would be a valid law of physics, but in Newtonian physics you could have a scenario where d/dt ( v₁ ) = -d/dt ( v₂ ) derived from a valid law of physics, e.g. Hooke's law or Newton's law of gravitation.

If you built a scenario using Hooke's law and Newton's laws such that d/dt ( v₁ ) = -d/dt ( v₂ ) and you transformed it to a different frame where Hooke's law and Newton's laws were still satisfied then you would find that d/dt ( v₁ )' = -d/dt ( v₂ )' . If you built a scenario using Hooke's law and Newton's laws such that v₁ = -v₂ and you transformed it to a different frame where Hooke's law and Newton's laws were still satisfied then you would find that v₁' ≠ -v₂'. Veocities are relative, accelerations are invariant.

 

[Ibix]

Got it. Somehow I'd missed the invertibility requirement

The naive solution to a reference frame for the traveling twin doesn't work because the map is not invertible - there is a patch of space-time that is covered by both coordinate systems and a patch covered by neither. It seems like a naive solution to that would be to split the space-time diagram along some line through the turnaround event, such as the stay-at-home twin's "now" plane through that event, and use one set of coordinates before the line and one set after.

I think that's invertible, but fails as a coordinate chart due to not being open - you cannot have an open ball anywhere along the boundary.

Is that right?

You need to go with something more sophisticated, such as the Dolby and Gull paper linked upthread.

 

[Dale]

The naive solution to a reference frame for the traveling twin doesn't work because the map is not invertible - there is a patch of space-time that is covered by both coordinate systems and a patch covered by neither.

Exactly.

It seems like a naive solution to that would be to split the space-time diagram along some line through the turnaround event, such as the stay-at-home twin's "now" plane through that event, and use one set of coordinates before the line and one set after.

Yes, you can make that kind of a split as long as you are careful and clear about it.

I think that's invertible, but fails as a coordinate chart due to not being open - you cannot have an open ball anywhere along the boundary.

Is that right?

Yes, that is true, but usually you can get around that kind of problem by careful specifications of the boundaries (i.e. use only strict inequalities)

 

[FactChecker]

Yes, clearly. But if the velocities always satisfy d/dt ( v₁ ) = -d/dt ( v₂ ) then that does not imply that v₁ = -v₂. The question is, which is a valid law of physics. In relativistic physics neither would be a valid law of physics, but in Newtonian physics you could have a scenario where d/dt ( v₁ ) = -d/dt ( v₂ ) derived from a valid law of physics, e.g. Hooke's law or Newton's law of gravitation.

If you built a scenario using Hooke's law and Newton's laws such that d/dt ( v₁ ) = -d/dt ( v₂ ) and you transformed it to a different frame where Hooke's law and Newton's laws were still satisfied then you would find that d/dt ( v₁ )' = -d/dt ( v₂ )' . If you built a scenario using Hooke's law and Newton's laws such that v₁ = -v₂ and you transformed it to a different frame where Hooke's law and Newton's laws were still satisfied then you would find that v₁' ≠ -v₂'. Veocities are relative, accelerations are invariant.

I am just saying this:

Velocity of Twin 2 in Twin 1 reference frame == - Velocity of Twin 1 in Twin 2 reference frame
d/dt ( Velocity of Twin 2 in Twin 1 reference frame ) = d/dt ( - Velocity of Twin 1 in Twin 2 reference frame ) = - d/dt( Velocity of Twin 1 in Twin 2 reference frame )
The derivative of Twin 2 velocity in Twin 1 reference frame == - The derivative of Twin 1 velocity in Twin 2 reference frame.
So the derivative of velocity is just as relative as the velocity itself is.

I think this is the basic reason that people see the Twins' situations as symmetric and thus the Twins "Paradox". To address the difference between the derivative and acceleration, one should point out that the Universe provides the space that acceleration is defined wrt. The preference for inertial reference frames to define acceleration bothered Einstein greatly and motivated his work on GR. He stated that in no uncertain terms.

 

[Dale]

So the derivative of velocity is just as relative as the velocity itself is.

No, it isn't, as explained above.

Your justification is completely irrelevant because it does not address what makes something "relative". It is like saying that an apple is just as sweet as a cherry because they are the same color.

How do you know that velocity is relative? You cannot use the justification that velocity is the 0th derivative of velocity. You also cannot use the justification that if v=-v then v is just as relative as -v. So what is it about velocity that makes it relative?

 

[pervect]

What's confusing me is that different coordinate values for the same point aren't always a problem. For example, you can't cover S² with one coordinate chart. The textbook solution is something like stereographic projection, where you define two charts each excluding one point and overlapping everywhere else. Obviously the co-ordinates aren't equal in the overlap region - so why is the fact that the coordinates aren't equal in the overlap bad in this twin paradox case? I can see that it is bad (there exist timelike paths that cannot be parameterised by their proper time, as Fredrik noted in #44), but is there a way to tell from the definition of the charts that it's bad?

Perhaps I should start a new thread, as I'm not sure this is entirely on topic.

It really depends on how carefully you think as to what sort of trouble you get yourself into. I don't think you'll find much guidance from textbooks on how to deal with situations where you assign multiple coordinate labels to the same points. I'm not aware of any textbooks or papers that cover this issue, which means you are sort of on your own if you go this route. (Or find some papers to talk about it, perhaps0. The fact that you have multiple charts in a manifold (which I think is what you're referring to) isn't really the same thing as giving a point multiple coordinates.

Let's go back to high school geometry for a second. There are two general approaches to it. One is the coordinate-free approach, based on Euclid''s axioms, such as "a straight line can be drawn between any two points". You don't need to use coordinates in this approach. The contrasting approach is one of analytic geometry, where you assign coordinates to every points, and use algebra to solve your geometry problems.

In the later case, you often implicitly assume that there is a 1:1 correspondence between your numbers, and the geometry. If you remove this requirement, and aren't careful, your analytic approach won't give the same answer to the geometrical questions you ask that the coordinate-free approach gives.

But it should - mathematically, the coordinate free approach and the coordinate based approach are supposed to represent the same underlying concept.

So, if you are able to do the coordinate-free approach, and the coordinate-based approach, you have at least a chance of spotting any errors you might make by the non-standard approach of assuming you can assign points multiple coordinates.

If you are totally relying on the analytic approach to geometry (I've seen PF posters do this, they seem unable to grasp the idea that one can do geometry without coordinates), you can easily confuse yourself into incorrect conclusions when you relax the rule that every point must have unique coordinates. You can't rely on the uniqueness and existence theorems of your algebraic problems to say anything about the uniqueness and existence of the geometrical problems. You can also (and I've seen this happen on PF) confuse yourself when you have a situation where you don't assign coordinates to every point, because you have some points that you don't assign coordinates to.

I think the cleanest thing to say about accelerated observers is this: "The" coordinate system of an accelerated observer exists locally, but it doesn't cover all of space-time, only a partial region of it.

 

[Ibix]

It really depends on how carefully you think as to what sort of trouble you get yourself into. I don't think you'll find much guidance from textbooks on how to deal with situations where you assign multiple coordinate labels to the same points. I'm not aware of any textbooks or papers that cover this issue, which means you are sort of on your own if you go this route. (Or find some papers to talk about it, perhaps0. The fact that you have multiple charts in a manifold (which I think is what you're referring to) isn't really the same thing as giving a point multiple coordinates.

I think I had things backwards. I thought the problem with the naive solution was the overlap region, and was trying to contrast the supposed problem with multiple coordinate charts, yes. From DaleSpam's #62 I gather that the fundamental problem is the uncovered region, which makes it non-invertible. From you, I gather that the overlap region is just a trap for the unwary - you can end up "double counting" events in that region.

To paraphrase what (I think) you are saying, I'll use an S² example. Africa appears on both north- and south-polar stereographic projections of Earth. That doesn't mean there are two Africas, obviously, but with more complex (or just less familiar) geometries it's possible to fail to realize that the maps overlap and conclude that there are two Africas. So when someone asks about events "happening and then happening again" in the twin paradox (or Andromeda paradox or whatever), they're asking why there are two Africas, instead of realising that one Africa appears on both maps.

So, if you are able to do the coordinate-free approach, and the coordinate-based approach, you have at least a chance of spotting any errors you might make by the non-standard approach of assuming you can assign points multiple coordinates.

If you are totally relying on the analytic approach to geometry (I've seen PF posters do this, they seem unable to grasp the idea that one can do geometry without coordinates), you can easily confuse yourself into incorrect conclusions when you relax the rule that every point must have unique coordinates.

Presumably the point about coordinate free thinking is (or is related to) why @bcrowell was recommending abstract index notation on a recent thread(?)

 

[Dale]

From DaleSpam's #62 I gather that the fundamental problem is the uncovered region, which makes it non-invertible. From you, I gather that the overlap region is just a trap for the unwary - you can end up "double counting" events in that region

Sorry that I was unclear. Perhaps I should just link to what I think is the best introduction to this material (ch 2 here http://preposterousuniverse.com/grnotes/ )

Remember that a coordinate chart is a mapping between an open subset of spacetime (the manifold) to an open subset of R4 (the coordinates). It is the mapping from spacetime to R4 that must be one to one in order to have a valid coordinate chart. The open subset of spacetime need not cover all of spacetime, and the open subset of R4 need not be all of R4. So the uncovered regions are not a real problem, just something to be aware of.

The problem is when one event in spacetime maps to multiple points in R4 in a single chart. When this happens the mapping between spacetime and the coordinates is not one to one and so you cannot go back and forth between the coordinates and the manifold. That is the problem with the naive approach. It can cause "time to go backwards", and in those regions of spacetime it is not one-to-one and therefore not valid. The "fix" is to exclude those regions from the open subset so that what is left is a valid chart that covers only part of the manifold.

I apologize if I am confusing you. Hopefully, Carroll's notes are more clear than I am.

 

[PhoebeLasa]

I think the cleanest thing to say about accelerated observers is this: "The" coordinate system of an accelerated observer exists locally, but it doesn't cover all of space-time, only a partial region of it.

I don't understand why it doesn't cover all of spacetime, at least in the case of the assumed flat universe of special relativity. If I draw a Minkowski diagram using some (arbitrary) inertial reference frame, the world line of the accelerating traveler can be plotted. At any point on that world line, "now" for the traveler is an infinitely-long straight line with a slope between -45 degrees and +45 degrees (exclusive). That straight "now" line extends infinitely far in both directions ... it covers all of space. And each point on the world line corresponds to a different time, and has a potentially different "now" line. So what part of spacetime is being missed, by the collection of those "now" lines? Have I misunderstood you?

 

[PeterDonis]

So what part of spacetime is being missed, by the collection of those "now" lines?

The problem isn't what points the now lines reach; it's the fact that they cross at the Rindler horizon of the accelerating traveler. So you can't use those now lines as coordinate lines at or beyond the Rindler horizon, since coordinate lines can't cross in a valid chart.

 

[DrGreg]

So what part of spacetime is being missed, by the collection of those "now" lines?

In the case of Rindler coordinates, often chosen to be used for an observer undergoing constant proper acceleration, the answer is the white area in this diagram.(The observer is located at x=1.)

Image attribution: Dr Greg, Wikimedia Commons, CC-BY-SA-3.0.

 

[Ibix]

I'm conflating two issues, I think, and confusing myself.

I have a map-book covering the town where I live. Each page is a different chart with coordinates related to those on other pages by translation. Each page only covers a finite region, and there's a small overlap between pages.

This is perfectly fine, both the overlap and the fact that there is no map of the next town. It's analogous to George's usual three space-time diagrams which overlap and do not cover what's happening around a black hole 100ly away.

What would not be fine would be to photocopy two pages, tape them side by side (overlap and all) and to call that a good map. That's what the naive solution of hacking George's diagrams in two and taping them together is. It can be fixed by removing the overlap, as I suggested, although you need to be careful at the join.

I was conflating the existence of the overlap of the maps (fine) with the act of casually taping them together (not fine). And attempting to resolve the resulting confusion by adding in the fact that the maps don't cover the next town.

Have I got it this time?

I am actually studying Carroll's notes, with mixed results it would seem.

 

[Dale]

I'm conflating two issues, I think, and confusing myself.

I have a map-book covering the town where I live. Each page is a different chart with coordinates related to those on other pages by translation. Each page only covers a finite region, and there's a small overlap between pages.

This is perfectly fine, both the overlap and the fact that there is no map of the next town. It's analogous to George's usual three space-time diagrams which overlap and do not cover what's happening around a black hole 100ly away.

What would not be fine would be to photocopy two pages, tape them side by side (overlap and all) and to call that a good map. That's what the naive solution of hacking George's diagrams in two and taping them together is. It can be fixed by removing the overlap, as I suggested, although you need to be careful at the join.

I was conflating the existence of the overlap of the maps (fine) with the act of casually taping them together (not fine). And attempting to resolve the resulting confusion by adding in the fact that the maps don't cover the next town.

Have I got it this time?

I am actually studying Carroll's notes, with mixed results it would seem.

This is one of the best explanations I have ever seen. I plan on referencing it or repeating it in the future when this question comes up.

 

[PhoebeLasa]

(In reply to Pervect): I don't understand why it doesn't cover all of spacetime, at least in the case of the assumed flat universe of special relativity. If I draw a Minkowski diagram using some (arbitrary) inertial reference frame, the world line of the accelerating traveler can be plotted. At any point on that world line, "now" for the traveler is an infinitely-long straight line with a slope between -45 degrees and +45 degrees (exclusive). That straight "now" line extends infinitely far in both directions ... it covers all of space. And each point on the world line corresponds to a different time, and has a potentially different "now" line. So what part of spacetime is being missed, by the collection of those "now" lines? Have I misunderstood you?

OK, I think I might understand now what Pervect meant, and what I was missing in my understanding. The Minkowski diagram I described can be imagined to have an infinite planar area, and the collection of points on that diagram corresponds to all of spacetime (in this simple example with only one spatial dimension). If the traveler were unaccelerated, then the slope of all of his "now" lines would be the same, and the collection of all those "now" lines would obviously sweep through all of the points of the Minkowski diagram. But if the slopes are changing as the accelerating observer moves along his world line, I can see that there might be some regions of the diagram that don't get "swept". It is true that for each instant of the traveler's life, his "now" line for that instant does cover all of space, and it does give an age for the object at each of those spatial points. But I can see now that the collection of all the spacetime points on the diagram that are swept by the collection of all those "now" lines doesn't necessarily include all the points on the diagram. I've seen a specific example of this somewhere: if the traveler starts accelerating at 1g (for example) at some instant, and never stops that acceleration, then if some other person is a particular, specific distance away from the traveler at the start of the acceleration, that other person's age (according to the traveler) will never change at all. So, according to the traveler, there is no "future" for that other person, and that is an example of a portion of spacetime not being covered. I think this is probably equivalent to the "Rindler Horizon".

 

[PeterDonis]

the collection of all the spacetime points on the diagram that are swept by the collection of all those "now" lines doesn't necessarily include all the points on the diagram

This is true (and it is indeed one way of describing the Rindler Horizon), and it is a problem, but it's not the only problem. The other problem is the one I mentioned, that the "now" lines cross (at the origin of the diagram DrGreg drew). Beyond the crossing point, there is a region of spacetime that the "now" lines do "sweep" through, but they still can't be used as coordinate lines there by the traveler, because "now" lines can't cross in a valid coordinate chart.

 

[Ibix]

This is one of the best explanations I have ever seen. I plan on referencing it or repeating it in the future when this question comes up.

You're very welcome to do so. I'm like the stay-at-home twin - I take the longest possible time to get there, but I get there in the end. :D

 

[Dale]

I take the longest possible time to get there, but I get there in the end. :D

I don't know if you have grappled with this stuff before, but it may make you feel better to know that it took me 7 years of sporadic study to understand SR.

 

[Ibix]

I studied both SR and GR nearly 20 years ago as an undergrad, and didn't really get them. Ghwellsjr's Minkowski diagrams here were a revelation. I suddenly realized a year or so ago that between my Kindle, Carroll's notes, mobile internet (especially PF), a 90 minute commute and a lot more self-confidence than I had as a young(er) man, I had all the tools to be able to get them now. We'll see.

Thanks (to all) for your help.

 

[PhoebeLasa]

It really depends on how carefully you think as to what sort of trouble you get yourself into. I don't think you'll find much guidance from textbooks on how to deal with situations where you assign multiple coordinate labels to the same points. I'm not aware of any textbooks or papers that cover this issue, which means you are sort of on your own if you go this route. (Or find some papers to talk about it, perhaps. The fact that you have multiple charts in a manifold (which I think is what you're referring to) isn't really the same thing as giving a point multiple coordinates.
[...]

I understand that there is a requirement (invertibility of charts) that is necessary in general relativity for "knitting" charts together. But I don't see why any "knitting" is necessary in special relativity, where spacetime is assumed to be infinite and flat (Minkowskian) everywhere.

 

[Ibix]

It isn't. But if you want to construct a chart in which the traveling twin is always at rest, you can't use one of the standard inertial frame charts because the traveller isn't always inertial. You've either got to (carefully) patch together inertial charts or develop a more sophisticated chart.

 

[PeterDonis]

I don't see why any "knitting" is necessary in special relativity, where spacetime is assumed to be infinite and flat (Minkowskian) everywhere.

It isn't.

More precisely, it isn't if you use a single inertial coordinate chart to describe the entire scenario. But a traveller who moves non-inertially for any part of his trajectory will not be always at rest in such a chart. So, as you say, if you want a chart in which the traveller is always at rest, you're going to confront the issue we've been discussing one way or another.

 

[Ibix]

More precisely...

I was coming back to expand on my "it isn't", which was a bit too pithy, but I see I don't have to.

 

[ghwellsjr]

It isn't. But if you want to construct a chart in which the traveling twin is always at rest, you can't use one of the standard inertial frame charts because the traveller isn't always inertial. You've either got to (carefully) patch together inertial charts or develop a more sophisticated chart.

I'm going to show you how to patch together inertial charts to construct a chart in which one of the traveling twins is always at rest. I am using the OP's scenario in which both twins travel away from their starting point at 0.5c in opposite directions and then after some time (I picked seven years according to their own clocks) they each turn around and come back at 0.5c taking another seven years to get back together. I'm going to redraw the diagrams I made in post #6 except that I'm leaving out the blue observer who remains at rest and I'm also relocating the origin of the diagrams to be at the red twin's turnaround point and then I'm drawing in light signals at yearly intervals to show what each twin sees of the other twin.

Ok, here's the first diagram showing the twins' initial and final rest frame with the light signals going each year from the black twin to the red twin:

​

Some things to note: during the first part of red's trip, he sees black's time progressing at one third the rate of his own because it takes him three years to see one year on black's clock. After red turns around, he sees black's clock progressing the same as his own for about four years and finally at the end of the trip, red sees black's clock progressing at three times the rate of his own.

One other thing to note is that when both twins accelerate away from each other at their Proper Time of -7 years, they start to experience Time Dilation because now their speed is 0.5c in this IRF which makes gamma or the Time Dilation factor be 1.1547. You can see that the dots marking off one-year increments of time for them is stretched out by that amount compared to the Coordinate Time and after 7 years of Proper Time it is 8.0829 years of Coordinate Time. You can see that their Proper Time of -7 years occurs at the Coordinate Time of slightly negative of -8 years and their Proper Time of 7 years occurs at the Coordinate Time of slightly more than 8 years at which point they both come to rest in this IRF and their Time Dilation becomes 1 (or goes away).

Now let's transform to the frame in which the red twin is at rest during the first part of the trip. This frame also includes the time when the black twin is at rest during the last part of his trip:

Please note that all the observations that the red twin made in the first diagram are the same in this diagram, in other words during the first leg, red sees black's clock ticking one third the rate of his own. Then after turn round, he sees their clocks ticking at the same rate and near the end, he sees black's clock ticking at three times the rate of his own.

Also note that both twins were traveling at -0.5c prior to their separation which gave them a Time Dilation of 1.1547 but when they separated, red came to rest and had no Time Dilation and black accelerated to -0.8c which gave him a Time Dilation of 1.667. When red turns around at the Coordinate Time of 0, he is traveling at -0.8c which now gives him a Time Dilation of 1.667. You can see that after 3 years of his Proper Time, the Coordinate Time is 5 years (1.667 times 3).

Next we want to transform to the IRF in which red is at rest after he starts his trip back to black:

I'm not going to go into all the details of this diagram since they are similar to the previous one but you should verify that all the observations that the red twin makes of the black twin are the same as in the two previous diagrams.

Now we want to repeat the above three IRF's but this time I will draw in the light signals that arrive each year at the black twin from the red twin. I won't repeat all the details and notes but you should verify that they are what is expected.

Here' the IRF in which both twins start out and end up at rest:

Next we have the IRF in which the red twin is at rest during the first leg of his trip:

And finally, the IRF in which the red twin is at rest for the last leg of his trip:

Now because there is a limit of six images per post, I have to continue this in the next post.

 

[ghwellsjr]

Remember the goal here is to construct a chart in which the red twin is always at rest by patching together inertial charts from the previous post. I didn't bother to show the patches for the initial and final periods of rest but simply show the red twin's path continually at rest. The black twin is also present during those two periods of time but not explicitly shown.

We start with the naive approach which is to take portions of pairs of charts in which the red twin is at rest and simply patch them together. First we take the bottom half of the second chart in the previous post and the top half of the third chart (the green line shows the patch between the two portions):

​

We see a couple obvious problems with this approach: first, it does not include all of the black observer's path including all his time from -4 years to 4 years and second, the light signals that go from the black twin to the red twin are discontinuous.

We can do a similar thing with the last two charts on the previous post:

Same problems.

The next post will show a better way.

 

[ghwellsjr]

Instead of cutting the inertial charts along a horizontal line of simultaneity, we can solve one of our problems by cutting along a diagonal line of the path of a light signal. The first case again involves charts two and three above:

​

Now there's no discontinuities in the light signals going from the black twin to the red twin but there is a discontinuity in the path of the black twin. We'll fix that later but first we want to patch together the last two charts from the previous post along a different light signal path:

Now the question is how to get rid of the discontinuities in the black twin's path. To do this, we will superimpose the above two charts on to one chart:

Now we note that what we really want is for each signal that starts with the red twin to be reflected off the black twin and to traverse back to the red twin. If only the black twin were positioned at the intersection of those two signals, we would achieve our goal. Let's put dots at all of those positions:

Now all we have to do is connect the black twin's path through those dots and get rid of the extraneous paths, dots and signals and we will have achieved our goal:

Note that this chart is invertible. We can use it to construct the black twin's non-inertial frame showing the path of the red twin. Here is a chart showing the signals that the black twin would send to the red twin and the reflections coming back to the black twin:

You can verify that the timings of the signals in this non-inertial frame match the timings of the signals in the Inertial Reference Frames.

If we had repeated this exercise showing how the red twin creates the path of the blue observer and included both in the same chart, then we could also create a chart for the rest frame of the blue observer.

Note that the conventional Time Dilation doesn't work for a non-inertial chart. Instead we see some time compression.

Now after having done all this, I would like to say that it doesn't add any meaning to any IRF. We do it just for fun. There is no added benefit since all properly drawn charts include the same information. They just present the observations and measurements of the observers in a different way, that is, according to different coordinates.

Any questions?

 

[Ibix]

Very nice - thanks George.

Am I right in thinking that you've constructed the same coordinate chart that Dolby & Gull use in the paper linked from #20?

 

[ghwellsjr]

Very nice - thanks George.

Am I right in thinking that you've constructed the same coordinate chart that Dolby & Gull use in the paper linked from #20?

Yes, except that the first time I did this, I didn't realize that it was the same but I was glad to learn that I was not the first to discover this method of making a non-inertial chart.

 

[PeterDonis]

George, this is the best easy visual presentation of the issues in constructing non-inertial charts that I have seen. Great job!

 

[Ibix]

Yes, except that the first time I did this, I didn't realize that it was the same but I was glad to learn that I was not the first to discover this method of making a non-inertial chart.

I actually remember that thread, and not understanding what was going on. Progress!

 

[PhoebeLasa]

[...]
if you want a chart in which the traveler is always at rest, you're going to confront the issue we've been discussing one way or another.

I've just realized that there is a particular similarity between (1) the special relativity scenario where the traveling twin (motionless wrt his home twin at some sufficiently large distance from her) begins and then continues a constant acceleration away from her forever, and (2) the general relativity scenario where someone falls through the event horizon of a spherically-symmetrical non-rotating black hole.

In the case of the black hole (if I understand it correctly), an observer "at infinity" (effectively infinitely far away from the black hole) will say that the age of the person falling toward the black hole will approach a finite limit, and that the person will never quite reach the event horizon, nor will he ever quite reach that limiting finite age. So, for the "infinitely-far-away" observer, the future ages of that falling person never happen. But the falling person himself will say that he passes through the event horizon, and that his age continues to smoothly increase as he continues his falling inside the black hole.

In the case of the special relativity scenario, the traveler [when his "rest" frame (in which he is always at the spatial origin) is determined using the "montage" of the "momentarily co-moving inertial reference frames" (his "MCMIRF" frame)] will say that the home twin's age will continuously increase, but will approach a finite limit. So he will say that she never gets older than that finite, limiting age ... her future ages (which she certainly will say she does reach) never happen at all, according to him.

So, in both the GR black hole scenario, and in the SR infinitely-long-lasting acceleration scenario, the observers (the "infinitely-removed" person in the GR case, and the accelerating traveler in the SR case) are each using a reference frame that doesn't "see" the missing future ages of the other person in each scenario). What I don't understand, though, is why this situation seems to be considered to be acceptable and "valid" in the GR scenario, but is considered (at least commonly on this forum) as unacceptable and "invalid" in the SR scenario.

 

[PeterDonis]

for the "infinitely-far-away" observer, the future ages of that falling person never happen.

No, that's not correct. What is correct is that the far-away observer will never see the future ages of the falling person. But he has no justification for saying that, since he will never see them, they will never happen. Making that unwarranted inference is a common error, but it's still an error.

What I don't understand, though, is why this situation seems to be considered to be acceptable and "valid" in the GR scenario, but is considered (at least commonly on this forum) as unacceptable and "invalid" in the SR scenario.

You are incorrectly describing the situation. The GR and SR scenarios are both the same in this respect: one observer never sees a certain portion of spacetime, which includes the "future ages" of the other observer. That's all there is to it.

 

[Dale]

What I don't understand, though, is why this situation seems to be considered to be acceptable and "valid" in the GR scenario, but is considered (at least commonly on this forum) as unacceptable and "invalid" in the SR scenario.

How can you possibly not understand this point. It has been explained to you over and over and over again. As near as I can tell it has been explained to you every single time that you have posted by multiple people in multiple ways during multiple threads.

A valid coordinate chart does not need to cover all of spacetime. Both charts you describe are alike in this feature

A valid coordinate chart does need to be invertible. The two charts differ in this required feature.

Please read chapter 2 here
http://preposterousuniverse.com/grnotes/

 

[PhoebeLasa]

After reading Carroll's 2nd chapter, I think I've learned two important things:

1) First, I think I've learned that, when spacetime is curved, there is generally no single coordinate system that can cover the entirety of spacetime. When that is the case, it is necessary to smoothly knit together more than one coordinate system, so that all of spacetime can be covered. And in order to guarantee that these multiple coordinate systems can be smoothly knitted together, they must be invertible.

2) Second, I think I've learned that, when spacetime is flat (Minkowskian) everywhere, there is no requirement that all coordinate systems must be invertible, because there is at least one coordinate system which can cover the entirety of spacetime. No knitting-together of multiple coordinate systems is required to get complete coverage, so invertibility of all coordinate systems is not required. In particular, I see nothing in Carroll's 2nd chapter that prohibits the use of the "MMCMIRF" coordinate system (the coordinate system consisting of the "montage" of "momentarily-co-moving-inertial-reference-frames") as the "rest" coordinate system of an accelerating traveler, PROVIDED that the spacetime is flat (Minkowskian) everywhere. The fact that the MMCMIRF isn't invertible, and the fact that it doesn't always cover all of spacetime, does not appear to be disqualifying, according to my reading of Carroll.

I am basing my above thinking on the following quotes from Carroll's 2nd chapter:

Quotes from Carroll in support of my thinking in item #1 above:

"So a chart is what we normally think of as a coordinate system on some open set, and an atlas is a system of charts which are smoothly related on their overlaps."

"We therefore see the necessity of charts and atlases: many manifolds cannot be covered with a single coordinate system."

"The entire manifold is constructed by smoothly sewing together these local regions."Quotes from Carroll in support of my thinking in item #2 above:

"Why was it necessary to be so finicky about charts and their overlaps, rather than just covering every manifold with a single chart? Because most manifolds cannot be covered with just one chart."

"Nevertheless, it is very often most convenient to work with a single chart, and just keep track of the set of points which aren’t included."

 

[Ibix]

Maps must be invertible by definition. It isn't stated in the clearest terms in Carroll's notes, but it is stated. He defines a map as a one-to-one function, then notes that any map is onto its image. A function that is both one-to-one and onto is invertible. Therefore maps must be invertible by their definition. Their is no exception for flat space.

Think about what you are saying for a minute (as I should have done). A non-invertible map means that you can draw the chart but can't use it to navigate, or that you can use it to navigate but cannot draw it. Neither is useful.

 

[pervect]

It might be helpful to look at MTW's "Gravitation section $6.3". You can probably get it through google if you search for "Constraints on size of an accelerated frame" and look at the google book results. "Constraints on size of an accelerated frame" is the section title for section $6.3 of this textbook.

My $.02 is that it appears to be standard practice to insist that coordinate systems assign unique labels to every point - and that this approach seems to me to be less confusing than the alternative of attempting to deal with multiple labels.

If there was some strong benefit to having coordinate systems with non-unique labels, it might be worth investigating the issue in more detail - I'm not aware of any advantages to such a practice.

 

[PeterDonis]

I think I've learned that, when spacetime is curved, there is generally no single coordinate system that can cover the entirety of spacetime.

Yes.

When that is the case, it is necessary to smoothly knit together more than one coordinate system, so that all of spacetime can be covered.

Yes.

And in order to guarantee that these multiple coordinate systems can be smoothly knitted together, they must be invertible.

What you mean here is that the mapping between the two coordinate systems must be invertible. This is quite true: in order to "knit" two charts together, the mapping between them must be invertible in any part of spacetime that both charts cover.

However, that is not the same as having a single coordinate chart, considered as a mapping between points in spacetime and 4-tuples of real numbers, being invertible. The latter sense of "invertible" is the one people have been discussing here.

I think I've learned that, when spacetime is flat (Minkowskian) everywhere, there is no requirement that all coordinate systems must be invertible, because there is at least one coordinate system which can cover the entirety of spacetime.

First of all, even if we apply the word "invertible" to maps between charts, instead of to a single chart (see above), this does not follow. Yes, in Minkowski spacetime any inertial coordinate chart covers all of spacetime (and there are an infinite number of such charts). However, that in no way removes the requirement that if you choose to use some non-inertial chart on a portion of Minkowski spacetime, the mapping between that chart and any other chart that you use to cover the rest of spacetime (which could be an inertial chart or another non-inertial chart) must be invertible, because if you are using a non-inertial chart at all, as Ibix pointed out, you are imposing on yourself the requirement of "knitting" that chart together with other charts to cover all of spacetime, and doing that works the same whether spacetime is flat or curved.

However, none of this has anything to do with the fact that even a single inertial chart covering all of Minkowski spacetime must still be an invertible map between points in Minkowski spacetime and 4-tuples of real numbers. Any map that does not have that property is simply not a valid coordinate chart, period.