Spacetime diagram - Twin paradox

[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.

 

[ghwellsjr]

Jaumzaum's link shows an aborted attempt to combine the last two diagrams from the last post. Here I will show a better way to do this but they still have problems and they cannot show the paths of the message for both twins on the same diagram. First is the combined diagram in which Pam is always at rest and in which she is receiving the messages from Jim. Note that everything she sees is accurate:

Next is the combined diagram in which Pam is always at rest and in which she is sending messages to Jim. Although it correctly shows when she sent the messages, she cannot tell the path they take to Jim.

This final non-inertial drawing in which Pam is always at rest correctly shows the timings for both Pam and Jim in terms of when they send and receive all the messages:

Note that Pam could always use a radar method to determine how far away Jim was and this diagram takes advantage of that information.

 

[PAllen]

Thanks, ghwellsjr, the last is radar coordinates I've referred to, in simplest form (another form is to scale the horizontal - stretching to the left, so the coordinate distance along a horizontal coordinate line corresponds to proper distance along the simultaneity line it represents).

 

[phyti]

Fig. 1 is the typical 'twin' drawing with simultaneity axis, from static twin
Bert's view.
Fig. 2 is traveling twin Bart's view, with distorted space and time
coordinates resulting from simultaneity convention. The 'convention' is a
mathematical device/stipulation and as such cannot relocate events (events
do not move!).
Fig. 3 is fig. 2 without the 'convention' and using relative light speeds.
Fig. 4 is Bart's view with an equivalent G-field as his reversal. Bart will see
Bert curve back toward him as a result of the G-field.


https://www.physicsforums.com/attachment.php?attachmentid=55394&d=1360089535

 

[PAllen]

Fig. 1 is the typical 'twin' drawing with simultaneity axis, from static twin
Bert's view.
Fig. 2 is traveling twin Bart's view, with distorted space and time
coordinates resulting from simultaneity convention. The 'convention' is a
mathematical device/stipulation and as such cannot relocate events (events
do not move!).
Fig. 3 is fig. 2 without the 'convention' and using relative light speeds.
Fig. 4 is Bart's view with an equivalent G-field as his reversal. Bart will see
Bert curve back toward him as a result of the G-field.


https://www.physicsforums.com/attachment.php?attachmentid=55394&d=1360089535

Fig. 4 is ok, but you should be aware that drawn as you have, x coordinated distance is wildly different from proper distance computed along horizontal line in the coordinates. Often, when using some simultaneity convention, you want spatial coordinate differences to reflect proper distances computed on those surfaces. If you do this, the curved path gets highly stretched to the left in your fig. 4.

 

[phyti]

Fig. 4 is ok, but you should be aware that drawn as you have, x coordinated distance is wildly different from proper distance computed along horizontal line in the coordinates. Often, when using some simultaneity convention, you want spatial coordinate differences to reflect proper distances computed on those surfaces. If you do this, the curved path gets highly stretched to the left in your fig. 4.

The stretching/distortion, etc., as in fig. 2, results from the simultaneity convention.
It is is not a deduction using physics, it's, as Einstein states, a definition to assign time and position to the remote reflection events. He also states in a different souce, his definition has nothing to do with physical light propagation. It's to support the pseudo-rest frame that the observer thinks he occupies, with equal light paths out and return, so divide the round trip time in half.
Any locations and times for the reflection events are speculation and inverifiable by the inertially moving observer, until someone can time the light propagation along a 1-way path.

 

[PAllen]

The stretching/distortion, etc., as in fig. 2, results from the simultaneity convention.
It is is not a deduction using physics, it's, as Einstein states, a definition to assign time and position to the remote reflection events. He also states in a different souce, his definition has nothing to do with physical light propagation. It's to support the pseudo-rest frame that the observer thinks he occupies, with equal light paths out and return, so divide the round trip time in half.
Any locations and times for the reflection events are speculation and inverifiable by the inertially moving observer, until someone can time the light propagation along a 1-way path.

You drew a picture in coordinates. I am simply stating a mathematical fact about those coordinates: x coordinate differences are not proportional to proper distance computed along horizontal coordinate lines. There is no requirement that this be so (the metric will contain the scaling, when you transform the metric correctly). However, it is an important feature to know about the coordinates.

Another possible way to get your diagram 4 is if the coordinates are far from orthonormal near the vertical line. This is also ok, as long as you understand its implications. It means that even in the inertial parts of the path, and right near the time axis, the metric will not be close to the Minkowski metric.

 

[phyti]

Fig. 1 shows B accelerating away from A as observed by A.
Fig. 2 shows A accelerating away from B as observed by B.
The curve in fig.2 is a reflection of the curve in fig.1 except fig.2 is scaled down,
due to time dilation and length contraction.
There is no cusp (as shown in post 30) in either curve because the td and lc begin at zero and increase as the speed of B increases. At deceleration, follow the curve backward until B speed equals A speed, with the same results, no cusp. As B decelerates toward A, the G-field points away from A. A cusp at reversal would indicate a repulsive gravitational effect, and occur at minimal relative speed!
A discontinuity is a flag indicating something is not according to physical laws/rules. The instantaneous reversal is such an instance, therefore the results are fiction. In post 38, event A4 in fig.1 does not jump from 16 ly distance to 48 ly in fig.2 unless there is some new and weird undiscovered physics. As stated there, the extreme time and space excursions are math computations using a convention, and not consequences of physical phenomena.
There have been other threads showing that the simplest ‘twin’ case, restricting all acceleration to one twin will appear to be explained by the non inertial asymmetry, but in general cases involving both twins accelerating, the aging question is decided by which twin loses the most time, which is path dependent.

https://www.physicsforums.com/attachments/55914