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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.

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  • $\begingroup$ 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. $\endgroup$
    – m4r35n357
    Commented Dec 28, 2023 at 9:05
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    $\begingroup$ @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. $\endgroup$
    – cplindem
    Commented Dec 29, 2023 at 3:30
  • $\begingroup$ 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 . . . $\endgroup$
    – m4r35n357
    Commented Dec 29, 2023 at 9:49
  • $\begingroup$ 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). $\endgroup$
    – m4r35n357
    Commented Dec 29, 2023 at 9:57
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    $\begingroup$ @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). $\endgroup$
    – cplindem
    Commented Dec 29, 2023 at 18:49

1 Answer 1

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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.

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  • $\begingroup$ 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. $\endgroup$
    – cplindem
    Commented Dec 30, 2023 at 2:34
  • $\begingroup$ @cplindem Try -(3/5)c , 0, +(3/5)c. $\endgroup$
    – robphy
    Commented Dec 30, 2023 at 2:43
  • $\begingroup$ @cplindem Look at my answer to physics.stackexchange.com/questions/434060/… and physics.stackexchange.com/a/383363/148184 $\endgroup$
    – robphy
    Commented Dec 30, 2023 at 2:46

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