Spacetime diagram - Twin paradox

[jaumzaum]

I was studying the twin paradox (of Einstein special relativity) and everything was working well until I get to the traveler's spacetime diagram.

First let me introduce the paradox for you to understand the diagram.
Pam is the twin sister of Joe. Pam goes out Earth in 2007 in a spaceship with v = 0.6c (velocity that Jim measures) for a 3 light-years round trip (again, distance that Jim measures). Every year that passes in Pam calendar, she sends a message to Jim ( the message travels with light-speed). So do Jim. At the end, they find out Pam is 2 years younger than Jim.

The spacetime diagram in the Jim's frame of reference is:

But when we try to put this in the Pam's frame of reference it goes like this:

Or, if we draw the message lines too (red lines)

http://img109.imageshack.us/img109/643/33306470.png

I'm confused about the following:
1) What is this discontinuity in the diagram? What does it means?

2) In 2009 (Pam's calendar) Pam sends a New Year message to his brother, and we can see the message intersects the diagram 2 times (in 2011 and 2014, Jim's calendar ). What does it mean? Will Jim receives 2 times the same message sent by Pam?

3) In the discontinuity line, 2 messages sent by Pam arrives. What does it mean? Seen that the turnaround occurs in a very short period of time, will Pam receive this 2 messages at the turnaround, and simultaneously?

 

[PeterDonis]

I was studying the twin paradox (of Einstein special relativity) and everything was working well until I get to the traveler's spacetime diagram.

That's because you are trying to draw a spacetime diagram for a non-inertial frame as if it were an inertial frame. You can't do that; non-inertial frames work differently.

1) What is this discontinuity in the diagram? What does it means?

It means, as above, that you've tried to treat Pam's frame as an inertial frame, when in fact it isn't. There isn't even a single unique way to draw Pam's spacetime diagram; there are multiple ways to construct a non-inertial frame in which Pam is always at rest, and they will give rise to diagrams that look different.

The real question is, why do you feel it necessary to do this to study the twin paradox? You can do everything in the single inertial frame in which the stay-at-home twin is at rest. What's missing?

 

[jaumzaum]

Actually the images and the traveler's diagram are from this site

http://www.einsteins-theory-of-relativity-4engineers.com/twin-paradox-2.html

I just wanted to understand them

 

[PAllen]

Actually the images and the traveler's diagram are from this site

http://www.einsteins-theory-of-relativity-4engineers.com/twin-paradox-2.html

I just wanted to understand them

Basically, they've shown how not to construct coordinates representing the traveling twin at rest. There are perfectly reasonable way to do so, that have no discontinuities if you even slight round the turnaround. They don't look anything like that diagram, which I would call a just strawman to shoot down.

They give, as their main point, that you are much better off sticking to one of the inertial frames.

While I agree with their overall point, their implication that any attempt to represent the experience of the traveling twin in one set of coordinates will be absurd, is, itself absurd.

 

[jaumzaum]

There are perfectly reasonable way to do so, that have no discontinuities if you even slight round the turnaround

How would be the correct way to do that?

 

[Mentz114]

How would be the correct way to do that?

Einstein would probably do it with a pseudo-gravitational field. The method is described ( without maths ) here

http://math.ucr.edu/home/baez/physics/Relativity/SR/TwinParadox/twin_gr.html

This is an article by Einstein on this subject

http://en.wikisource.org/wiki/Dialog_about_objections_against_the_theory_of_relativity

This analysis is not necessary because working out the proper lengths can be done more easily in an inertial frame. But if one insists that there is a valid frame in which the traveling twin remains stationary, this seems to be the only way to get it.

 

[PAllen]

This analysis is not necessary because working out the proper lengths can be done more easily in an inertial frame. But if one insists that there is a valid frame in which the traveling twin remains stationary, this seems to be the only way to get it.

It's a perfectly plausible way for simple trajectories. It isn't the only way, and, as I pointed out in another thread, it will not work for more complex trajectories (e.g. it will not work for a W shaped trajectory for the traveler). That still doesn't mean you can't draw reasonable coordinates with such a traveler stationary. Coordinates based on radar simultaneity will work for such a trajectory. However, the pseudo-gravity field will be very complex in such coordinates.

 

[PAllen]

How would be the correct way to do that?

I don't have software to draw nice pictures. Using the most common simultaneity convention for non-inertial motion, assuming a slight smoothed turnaround, and assuming that x coordinate represents proper distance along a simultaneity surface, then for such coordinates you would get:

- the traveling twin a straight vertical line.

- The home twin would proceed to the left from starting point at less the 45° to vertical until near midway point. Then it would shoot far to the left nearly horizontally, sharply turnaround back, nearly horizontally (with no intersection, loop, or discontinuity); then proceed the rest of the way back in a straight line less the 45° to the vertical.

The wild path for the home twin would still be a geodesic in these coordinates, while the vertical traveling twin path would not be a geodesic. Light rays would have complex paths in these coordinates.

If you marked proper time of the home twin along its path in these coordinates, most of the tick marks would occur on the 'shoot left and back' portion of the trajectory.

 

[PeterDonis]

Basically, they've shown how not to construct coordinates representing the traveling twin at rest.

Looking through the site, this does not appear to be the only entry that's questionable. Their entry on "what is gravity" appears to be even worse:

http://www.einsteins-theory-of-relativity-4engineers.com/what-is-gravity.html

 

[Ben Niehoff]

I'm confused about the following:
1) What is this discontinuity in the diagram? What does it means?

The discontinuity is due to Pam's infinite acceleration during her abrupt change in velocity at the apex of her journey. The instantaneous jump in her velocity causes an instantaneous jump in the direction of her lines of simultaneity.

This discontinuity can be eliminated if we smooth Pam's journey out by rounding off the corner. The simplest way is to replace the sharp corner by a segment of a hyperbola (which represents constant proper acceleration in the Minkowski diagram).

2) In 2009 (Pam's calendar) Pam sends a New Year message to his brother, and we can see the message intersects the diagram 2 times (in 2011 and 2014, Jim's calendar ). What does it mean? Will Jim receives 2 times the same message sent by Pam?

3) In the discontinuity line, 2 messages sent by Pam arrives. What does it mean? Seen that the turnaround occurs in a very short period of time, will Pam receive this 2 messages at the turnaround, and simultaneously?

The apparent double intersections come from straightening out Pam's worldline, which is not straight: it has a kink (or, if you smooth out the corner, a curved region). The last diagram violates the postulates of spacetime diagrams---it represents a non-inertial observer by a straight line. In the process of contorting the spacetime diagram to accomplish that, you end up representing a true, inertial observer as an impossible looped path.

When an observer undergoes acceleration, it is possible that objects in the "spatial future" (that is, outside the lightcone, but ahead of the line of simultaneity) end up in the "spatial past" (outside the lightcone, and behind the line of simultaneity), or vice versa. During the short moments when Pam turns around, large swaths of Jim's worldline quickly migrate from Pam's "spatial future" to her "spatial past", effectively causing Pam to "skip" many of Jim's years. The skipped years are the ones that appear in your triangular loop.

I would not attempt to use your distorted diagram to find out how signals travel between Pam and Jim, because you have essentially torn a rift in spacetime and glued it back together in order to make that diagram. The paths actually taken by light are not going to be the obvious ones as you might try to draw in that picture.

 

[PAllen]

The apparent double intersections come from straightening out Pam's worldline, which is not straight: it has a kink (or, if you smooth out the corner, a curved region). The last diagram violates the postulates of spacetime diagrams---it represents a non-inertial observer by a straight line. In the process of contorting the spacetime diagram to accomplish that, you end up representing a true, inertial observer as an impossible looped path.

I don't think any of this is necessary for this case. Assuming a slightly smoothed turnaround, the whole region to the left of the traveling twin is covered just fine by Fermi-normal coordinates as I outlined in my prior post. The region to the right is where there would be a problem (intersection of simultaneity surfaces).

As I've pointed out in another thread, for a more complex twin trajectory, you couldn't construct valid coordinates using simultaneity based on spacelike geodesics 4-orthogonal to the traveling world line. However, for any 'twin' situation at all, you could still construct coordinates based on radar simulteneity.

 

[Alain2.7183]

Coordinates based on radar simultaneity will work for such a trajectory. However, the pseudo-gravity field will be very complex in such coordinates.

How would you describe that pseudo-gravity field (which would give the same result given by "radar simultaneity")?

 

[PAllen]

How would you describe that pseudo-gravity field (which would give the same result given by "radar simultaneity")?

pseudo-gravity would be defined by the connection components, which are derived from the metric. Given a formula for radar coordinates based on some non-inertial world line (for example), you transform the Minkowski metric and compute the connection components.

If you are looking for a verbal description, or a computation for a specific case, I am not willing to do it. Radar coordinates are simple in concept but lead to very messy metrics.

 

[Mike Holland]

You can visualize Pam's non-moving point of view as follows.

On the first part of the journey, Jim is traveling away from Pam. Then suddenly someone in the opposite direction to Jim switches on a powerful gravitational field. Pam is suspended motionless in this field, but she feels her feet pressing against the floor, which is away from Jim who is now above her head. She sees Jim slow down in this field, until he starts to fall towards her. Then the field switches off and Jim continues to approach her at constant speed.

While this linear gravitational field is switched on, Jim is at a much higher potential in the field than Pam, and as per GR, his clock will advance faster than Pam's. Where she did see it delayed relative to hers, she now sees it advance rapidly until it is well ahead of hers. Then the field switches off and his clock ticks slowly again, but when he gets back she finds his clock is still ahead of hers.

So there we have a description from the point of view of a "stationary" Pam. All we need is a way to switch gravity on and off.

Mike

Just waiting for PAllen and Peter Donis to pull me to pieces!

Edit: OK, so I've ignored the time light takes to travel for Pam to "see" Jim's clock. So let's say she calculates what his clock would be reading in her coordinate system. I know the situation is much more complicated than I have described.

 

[PAllen]

You can visualize Pam's non-moving point of view as follows.

On the first part of the journey, Jim is traveling away from Pam. Then suddenly someone in the opposite direction to Jim switches on a powerful gravitational field. Pam is suspended motionless in this field, but she feels her feet pressing against the floor, which is away from Jim who is now above her head. She sees Jim slow down in this field, until he starts to fall towards her. Then the field switches off and Jim continues to approach her at constant speed.

While this linear gravitational field is switched on, Jim is at a much higher potential in the field than Pam, and as per GR, his clock will advance faster than Pam's. Where she did see it delayed relative to hers, she now sees it advance rapidly until it is well ahead of hers. Then the field switches off and his clock ticks slowly again, but when he gets back she finds his clock is still ahead of hers.

So there we have a description from the point of view of a "stationary" Pam. All we need is a way to switch gravity on and off.

Mike

Just waiting for PAllen and Peter Donis to pull me to pieces!

Edit: OK, so I've ignored the time light takes to travel for Pam to "see" Jim's clock. So let's say she calculates what his clock would be reading in her coordinate system. I know the situation is much more complicated than I have described.

At a glance, this looks fine, and has been explained earlier. But if your explanation helps further, great.

 

[ghwellsjr]

OK, so I've ignored the time light takes to travel for Pam to "see" Jim's clock. So let's say she calculates what his clock would be reading in her coordinate system. I know the situation is much more complicated than I have described.

If you've ignored the paths that light takes to travel, then you haven't addressed jaumzaum's issue, have you? But while we're waiting for your diagram that does that, maybe you could answer Peter's questions:

The real question is, why do you feel it necessary to do this to study the twin paradox? You can do everything in the single inertial frame in which the stay-at-home twin is at rest. What's missing?

 

[A.T.]

The real question is, why do you feel it necessary to do this to study the twin paradox? You can do everything in the single inertial frame in which the stay-at-home twin is at rest. What's missing?

What is missing, is the whole point of the paradox, which is the question:

Why doesn't Pam conclude to age more from her perspective?

Answering this question with:

Who cares about Pam's perspective?

is not very convincing to the inquiring layman.

 

[PAllen]

What is missing, is the whole point of the paradox, which is the question:

Why doesn't Pam conclude to age more from her perspective?

Answering this question with:

Who cares about Pam's perspective?

is not very convincing to the inquiring layman.

Well, there is a question of: "what is meant by Pam's perspective?". When Pam looks out of the rocket portal, takes measurements, etc. coordinates are not involved. Coordinates are an artifact of a mathematical model of reality. Given the philosophy that Einstein called general covariance and is now called diffeomorphism invariance, you have the result that it doesn't matter what coordinates you use to answer any possible physical definition of "Pam's perspective". Given this hard and fast mathematical foundation, why not answer any question about Pam's perspective using the simplest possible coordinates? These would be, any convenient inertial frame. Similarly in GR, nobody worries about having coordinates in which some world line is the time axis; you just pick some computationally convenient coordinates and compute what any observer sees or measures.

 

[A.T.]

why not answer any question about Pam's perspective using the simplest possible coordinates?

What people are seeking is not the mathematically simplest way to get a numerical result. They seek a conceptual understanding of how the situation is not symmetrical, and why both twin's rest frames predict that Pam will age less.

 

[PAllen]

What people are seeking is not the mathematically simplest way to get a numerical result. They seek a conceptual understanding of how the situation is not symmetrical, and why both twin's rest frames predict that Pam will age less.

To which, numerous times, the answer has been given that just as there is no such thing as global frames in GR, there is no such thing as a global non-inertial frame in SR. Thus to the extent you are looking for a rest frame for Pam that is analogous to stay at home rest frame, the only correct answer is that there is no such thing.

(You can construct any number of coordinate systems in which Pam has constant spatial coordinate position, but none of these constitute a global frame for Pam in the sense that there are global inertial frames).

 

[PeterDonis]

They seek a conceptual understanding of how the situation is not symmetrical

It's not symmetrical because Pam feels a force and Jim does not. I've said that a number of times in other twin paradox threads, but I see that I didn't emphasize it in this one; there have been so many threads on this topic lately that I've lost track. So now consider it emphasized here.

and why both twin's rest frames predict that Pam will age less.

This is harder because there isn't a unique way to construct "Pam's rest frame". My point was that you don't have to to derive the result that Pam ages less. If someone absolutely insists on having *some* representation of "Pam's rest frame", then we'll have to pick one. This has been discussed in other recent threads as well.

It looks like I need to get busy on the Twin Paradox FAQ that I have been meaning to draft for this forum.

 

[A.T.]

Personally I like to use space-propertime diagrams to visualize the twins, because you see the age difference directly. In this interactive version there is both types of diagrams, and the three inertial frames of the quick-turnaround version:

http://www.adamtoons.de/physics/twins.swf

As has been mentioned: If you want to have just one rest frame of non-inertial twin, you have to smooth the acceleration to avoid discontinuities. In the simplest case the acceleration is constant, and you have a constant gravity in the rest-frame of the non-inertial twin.

In space-propertime diagrams gravity looks something like this:

http://www.physics.ucla.edu/demoweb/demomanual/modern_physics/principal_of_equivalence_and_general_relativity/curved_time.gif
From: http://www.physics.ucla.edu/demoweb..._and_general_relativity/curved_spacetime.html

So the twin frames would look something like this:

Keep in mind that both worldlines are supposed to have the same length in each diagram, because everything advances at the same rate in space-propertime.

Also note that this is equivalent with throwing up an object (red) from the surface of the Earth (green).

 

[A.T.]

It's not symmetrical because Pam feels a force and Jim does not.

Yes, and that difference in the frame-invariant proper-acceleration means that in both diagrams:
- Jim's worldline is straight
- Pam's worldline is curved

 

[PAllen]

A.T.,We've been over how this is not a valid coordinate chart of Minkowski space because one event in Minkowski space appears in two places on this chart. It could conceivably be treated as a chart of a completely different manifold.

In terms of utility, how do you draw light paths on this? It seems all light paths are on top of each other along the bottom of the diagram. Not very instructive. Also not instructive is intersections of paths on this chart do not represent coincident events in the real world.

I think this highly non-standard tool will only serve to confuse.

 

[PeterDonis]

Yes, and that difference in the frame-invariant proper-acceleration means that in both diagrams:
- Jim's worldline is straight
- Pam's worldline is curved

If you draw the diagrams the way you did, yes. But you can always do a coordinate transformation that makes Pam's worldline look straight and Jim's look curved. It will be harder to read off the physics from such a diagram (which is part of the point of trying to explain to people that non-inertial frames are not quite what they imagine), but you can certainly draw one.

Edit: I should have noted, as PAllen did, that your diagram is not a coordinate chart; the coordinate transformation I was talking about would be done from a standard inertial chart to a non-inertial chart in which Pam's worldline was the "time" axis.

 

[PAllen]

Let me remind to look at my post #8, where I describe carefully how the diagram looks for Pam at rest in a typical coordinate chart. This description is based on Fermi-Normal coordinates for Pam.

 

[A.T.]

In terms of utility, how do you draw light paths on this?

In the interactive version there are check boxes to switch on the light paths
http://www.adamtoons.de/physics/twins.swf

I think this highly non-standard tool will only server to confuse.

I think both diagrams can be used in parallel. It's not like the standard tool (Minkowski diagram) is very successful at providing an intuitive visualization of this (from what I have experienced).

Let me remind to look at my post #8, where I describe carefully how the diagram looks for Pam at rest in a typical coordinate chart. This description is based on Fermi-Normal coordinates for Pam.

I would love to see such a diagram, even if it is only an approximate sketch. But the big problem of the "typical coordinate chart" is of course that you don't see the age difference directly in the diagram.

 

[PAllen]

I think both diagrams can be used in parallel. It's not like the standard tool (Minkowski diagram) is very successful at providing an intuitive visualization of this (from what I have experienced).

.

I don't know about browser issue or whatever, but nothing happens when I check those boxes. I tried refreshing, checking the different 'whose frame' buttons on top, but no light paths appeared.

Can you give a simple answer to how they appear? It seems they can only be as I supposed because they have no proper time.

[Edit: Ok, I see you have to press the animate button. But that just adds more confusion: you draw light paths as horizontal lines from where they originate, so based on proper time of whatever world line emitted them. It's clear as mud where they end; they don't end on the line for the other twin (in general). I don't see the logic of where they end. (Oh, I figured that out).

Additional point: such diagrams are completely un-interpretable except in the presence of the Minkowski diagram, because a twin situation or a situation where two rockets never meet at all could be indistinguishable. ]

 

[A.T.]

I don't think light signals are useful or needed in the space-propertime diagram of the twins. In the Minkowski diagram they are used as an indirect way to show the differential aging. But that is already shown directly in the space-propertime diagram.

 

[Mike Holland]

If you've ignored the paths that light takes to travel, then you haven't addressed jaumzaum's issue, have you? But while we're waiting for your diagram that does that, maybe you could answer Peter's questions:

Jaunzaum's second diagram is incorect. It ignores the length contraction that will make Pam see a much shorter distance between the start and end of her journey while she is traveling. Only as they match speeds at the midpoint of the journey does this distance expand again to match the distance seen by Jim. So Pam sees Jim shoot out further out at turnaround, as has been mentioned before (by Peter, I think).

Also, when she switches on her gravity machine (her rocket motors) she will see Jims clock speed up due to the difference in gravitational potential. So a more correct diagram would be like this -

In answer to Peter's question, his view is that of a mathematicuian. All the correct numbers can be obained from one static reference frame, and one doesn't have to look at the situation from any other point of view. But some of us want a bit more than numbers. We want to be able to visualize the situation from different points of view to enable us to understand it a bit more deeply. Of course, there are limits to what our imagining can achieve - I still like the Bohr atom with its circular and elliptical orbits, and electrons jumping from one orbit to the other!

Mike


NB. I didn't count the year dots on Jims turnaround line.

NB2. The two lines of simultaneity shouldn't meet at 4 on Pams time line - than would assume that turnaround is accomplished in zero time.

 

[PeterDonis]

All the correct numbers can be obained from one static reference frame, and one doesn't have to look at the situation from any other point of view. But some of us want a bit more than numbers. We want to be able to visualize the situation from different points of view to enable us to understand it a bit more deeply.

I should clarify that I am sympathetic to this desire; I just think it's important to understand the limitations of using non-inertial frames. It's particularly important when you start moving on to curved spacetime and GR, where there are *no* global inertial frames, so *all* ways of describing a scenario have limitations.

 

[PAllen]

Jaunzaum's second diagram is incorect. It ignores the length contraction that will make Pam see a much shorter distance between the start and end of her journey while she is traveling. Only as they match speeds at the midpoint of the journey does this distance expand again to match the distance seen by Jim. So Pam sees Jim shoot out further out at turnaround, as has been mentioned before (by Peter, I think).

Also, when she switches on her gravity machine (her rocket motors) she will see Jims clock speed up due to the difference in gravitational potential. So a more correct diagram would be like this -

In answer to Peter's question, his view is that of a mathematicuian. All the correct numbers can be obained from one static reference frame, and one doesn't have to look at the situation from any other point of view. But some of us want a bit more than numbers. We want to be able to visualize the situation from different points of view to enable us to understand it a bit more deeply. Of course, there are limits to what our imagining can achieve - I still like the Bohr atom with its circular and elliptical orbits, and electrons jumping from one orbit to the other!

Mike


NB. I didn't count the year dots on Jims turnaround line.

NB2. The two lines of simultaneity shouldn't meet at 4 on Pams time line - than would assume that turnaround is accomplished in zero time.

Thanks for the drawing. You will see that this is exactly what I described in my post #8. An example of what Peter is referring to about limitations of the approach is try to do this for a sideways W shaped trajectory for Pam (where the center peak of the W does not reach all the way back to the stay at home world line). You find that you cannot construct coordinates of this type at all because the lines of simultaneity for this convention intersect, causing the the home world line to by multiply labeled: for range of events on it, each is given 2 time coordinates, which is inadmissible.

The direct metric method, or picking any inertial reference frame method, or doppler analysis all work fine for a W shaped trajectory, but these lines of simultaneity break down. It simply means there are limitations to that simultaneity convention for more complex non-inertial motion. You can pick a different simultaneity convention, getting a different type of diagram in which Pam is 'at rest', that does work for this case. One example is radar coordinates.

 

[Mike Holland]

Thanks, PAllen, for mentioning the shoot out to the left. I hadn't realized that that happened, but its obvious when you consider Lorentz contraction and decreasing relative velocity.

 

[A.T.]

Additional point: such diagrams are completely un-interpretable except in the presence of the Minkowski diagram, because a twin situation or a situation where two rockets never meet at all could be indistinguishable. ]

That is not quite true. Since the Euclidean length of the worldlines in the space-propertime corresponds to coordinate time, you can tell that objects meet: When they arrive at the same space coordinate, after the same Euclidean path distance.

But I agree that they should be used together with Minkowski diagrams, because both have their weak points:
- Space-propertime diagrams don't show meetings directly as intersections of world lines
- Minkowski diagrams don't show poper-time directly as a length

 

[ghwellsjr]

If you've ignored the paths that light takes to travel, then you haven't addressed jaumzaum's issue, have you? But while we're waiting for your diagram that does that, maybe you could answer Peter's questions:

Jaunzaum's second diagram is incorect. It ignores the length contraction that will make Pam see a much shorter distance between the start and end of her journey while she is traveling. Only as they match speeds at the midpoint of the journey does this distance expand again to match the distance seen by Jim. So Pam sees Jim shoot out further out at turnaround, as has been mentioned before (by Peter, I think).

Could you be a little more specific with which diagram you are referring to? I couldn't find one that was incorrect.

In the meantime, I decided to redraw the diagrams from the link that Jaumzaum provided specifically to combine the signals for both Jim and Pam in each drawing. Here is the first one for the Inertial Reference Frame (IRF) of Jim (shown in blue--Pam is in black):

Next is the last diagram shown in the link which is the IRF in which Pam is at rest during the return part of the trip:

I don't know why they only showed the messages going from Pam to Jim. It's just as easy to show Jim's messages going to Pam.

And here is a diagram they didn't show which is the IRF in which Pam is at rest during the first part of the trip when she is traveling away:

Please note that each drawing illustrates exactly the same information. You can follow any message being sent by either twin, noting the year it was sent, and track how it was received by the other twin in which year it was received.

On the next post, I will show another aborted attempt to marry portions of these last two diagrams together in which Pam is always at rest and then I will show a successful way to depict a non-inertial diagram in which Pam is always at rest and it also correctly describes the paths of the messages.