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The below Minkowski spacetime diagram includes three worldlines, where B is the observer and has a rest frame. A and C both have a velocity of 0.71c.

initial spacetime diagram

I then created a second diagram where worldline A is the observer. I believe I calculated the new velocities correctly, where the velocity of A is 0c, the velocity of B is 0.71c (symmetrical to A's initial velocity), and the velocity of C is now 0.94c.

But drawing those new velocities/slopes produces the following diagram, which is obviously incorrect, because worldline C no longer meets the others.

spacetime diagram from worldline A

Drawing worldline C such that it ends at the same point as the others, while maintaining the new slope, results in the diagram below.

spacetime diagram with repositioned worldline C

The repositioning of the starting point of worldline C seems to imply that I was missing length contraction. It seems that the new starting point for C should be x = ~1.58. But that is simply an estimate based on the geometry of the diagram.

Am I correct that I need to account for length contraction to get the new spatial position for worldline C? If so, how do I accurately calculate that length contraction?

Edit: I discovered equations for Lorentz transformations of both space and time. Applying the equations for the velocity of worldline A (0.71c) means that three of the four points (technically six points, but three overlap) shift not only horizontally, but also vertically. I think this new diagram correctly shows the three transformed lines, relative to worldline A:

spacetime diagram with Lorentz transformations

So they all converge on t = 10 instead of 7. This seems a bit strange to me. Perhaps it is more appropriate to shift the y-axis itself such that they still converge on t = 7 and worldline B starts at t = -3 rather than 0? I'm not sure.

I was also expecting C's travel distance to get shorter (i.e., contracted due to its 0.94c speed according to A), rather than longer. According to A, C is now travelling over 14 rather than 5, according to B.

cplindem
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  • How the hell is this a "homework" question? How are people supposed to learn anything here at all? This is clearly someone attempting to learn by doing the right thing, and seeking a little basic assistance. – m4r35n357 Dec 28 '23 at 09:05
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    @m4r35n357 Thanks for your comments. I was also confused when my question was closed. Your (now deleted) comment also helped me understand that the points should shift vertically in time as well as horizontally in space. I think it's making more sense now. – cplindem Dec 29 '23 at 03:30
  • Thanks for pointing this out. I am in utter amazement that some joker has deleted my helpful comment, and left the complaining one in place! Can't get the staff, I suppose . . . – m4r35n357 Dec 29 '23 at 09:49
  • It is hard to tell from your diagram whether A's proper time is sqrt(24) or 5 ;) That "sloping" of simultaneity of start times is a hallmark of SR, and you should not try to explain it away, go with it! This example also demonstrates that although you can choose any frame to describe a situation, there is usually a "simplest" one (in this case B). – m4r35n357 Dec 29 '23 at 09:57
  • In the original diagram, your mistake was to label the time when all three meet as 7 rather than 0. This is not strictly wrong (you can choose any label you want) but it complicates the algebra and makes it easier to make mistakes. – WillO Dec 29 '23 at 11:46
  • @m4r35n357 : "How are people supposed to learn anything here at all?" One of the very best ways to learn is to tackle homework-like questions (such as this one), get screwy answers sometimes, and then think long and hard (on your own or with a study partner) about what went wrong. – WillO Dec 29 '23 at 11:51
  • @WillO in other words, don't expect any help from this site, work it out yourself! How do you justify all the truly idiotic & lazy questions that do get answered here? – m4r35n357 Dec 29 '23 at 13:19
  • @WillO to be absolutely clear, the OP has attempted to work it out, basically using the correct approach, realized they have made a mistake, and showed their working so far. A perfect candidate for a bit of gently nudging in the right direction, which I gave (then the comment was deleted!). As someone with no "study partner" myself, I am on the OP's side 100% here. – m4r35n357 Dec 29 '23 at 13:25
  • People can always go to physicsforums.com for homework help. Students have to show their work on that site, but they can get help. – Dale Dec 29 '23 at 13:28
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    @m4r35n357 In the final diagram, the points for worldline A are located at t = 5.041 and t = 9.94, which does indeed give a proper time of sqrt(24). – cplindem Dec 29 '23 at 18:49
  • @WillO That is an interesting point about setting the meeting point at t = 0. I'm actually building a series of diagrams and am trying to create some consistency between them, hence always starting at t = 0. I wonder if it would be better to move the origin to the starting point of the rest frame (e.g., A in the final diagram). I also wonder if it would be incorrect or misleading to crop the lower portion of the diagram, or if that would provide additional consistency. The weird part about this example is that A and B don't agree on the distance between them, even tho they both have v = 0.71c. – cplindem Dec 29 '23 at 18:57

1 Answers1

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Here's a way to check graphically.

(It uses the Lorentz transformation in light-cone coordinates.)

Using your first spacetime diagram, use a paint program like https://jspaint.app/ to do these steps:

  1. Rotate by 45 degrees (right 45)
    so that the future-forward light-signal (along +45 degrees) is now horizontal (0 degrees).

enter image description here (click for a full size image)

  1. To boost to make the v=+0.71 worldline vertical, (that is, make its new velocity zero)
    the first step is a reshaping of a "causal diamond":
    compute (the doppler factor k) =exp(arctanh(-0.71)) $\approx$ 0.412
    and its reciprocal is ( 1/exp(arctanh(-0.71)) ) = exp(arctanh(+0.71)) $\approx$ 2.428.
    https://www.wolframalpha.com/input?i=exp%28arctanh%28-0.71%29
    Alternatively, you can calculate sqrt((1-0.71)/(1+0.71)) $\approx$ 0.412 .

    Now, stretch 41.2% horizontally (width), 242.8% vertically (height).
    (These stretches preserve the speeds of light-signals, as well as preserves the area [within numerical errors].)

enter image description here (click for a full size image)

  1. Rotate by -45 degrees (left 45) so that the desired worldline is now vertical.

enter image description here (click for a full size image)

Each step is a particular area-preserving transformation. However, there may be round-off errors in how precisely the stretching factors can be entered.

I was also expecting C's travel distance to get shorter (i.e., contracted due to its 0.94c speed according to A), rather than longer. According to A, C is now travelling over 14 rather than 5, according to B.

Using your final diagram,
note that, according to A,

  • C must travel -14 space-ticks (sticks) in 15 ticks (as best as I can read),
    so C is traveling with speed (-14/15)c=(-0.933)c [as expected?]
  • B must travel -7 sticks in 10 ticks (as best as I can read),
    so B is traveling with speed (-7/10)c=(-0.7)c [as expected?]
  • By the way, "length contraction" involves two parallel worldlines. If you don't have them featured, then it probably isn't an issue to be concerned with.
  • Although it may not be obvious in this final diagram, the elapsed proper-time on the blue segment (as read on a blue wristwatch) is equal to the the elapsed proper-time on the green segment (as read on a green wristwatch)... since they were numerically equal in the original diagram (due to the symmetry of their motion about the orange worldline).
    As best as I can read off the final diagram,
    sqrt(15^2-14^2)$\approx$ 5.385 $\sim$ 5= sqrt(5^2-0^2).
    In the original diagram, sqrt(7^2-5^2)$\approx$ 4.898 $\sim$ 5.
    If you have more specific coordinates, repeat the calculation.

If you use arithmetically nice values (like v=(3/5)c or v=(4/5)c), you can end up with nice fractions... and can have coordinates that lie on rational values.

robphy
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  • I like the visual explanation. It makes it somewhat intuitive to see space and time stretching at those angles. I originally chose 0.71c because that resulted in the three worldlines converging at ~10, but that was before I really understood that they converge at another time altogether according to A and C. So I'll have to revisit those values. Maybe nicer velocity values will indeed be more intuitive. – cplindem Dec 30 '23 at 02:34
  • @cplindem Try -(3/5)c , 0, +(3/5)c. – robphy Dec 30 '23 at 02:43
  • @cplindem Look at my answer to https://physics.stackexchange.com/questions/434060/question-about-special-relativity-similar-to-twin-clock-experiment and https://physics.stackexchange.com/a/383363/148184 – robphy Dec 30 '23 at 02:46