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.
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.
Drawing worldline C such that it ends at the same point as the others, while maintaining the new slope, results in the diagram below.
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:
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.
Best Answer
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:
so that the future-forward light-signal (along +45 degrees) is now horizontal (0 degrees).
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].)
Each step is a particular area-preserving transformation. However, there may be round-off errors in how precisely the stretching factors can be entered.
Using your final diagram,
note that, according to A,
so C is traveling with speed (-14/15)c=(-0.933)c [as expected?]
so B is traveling with speed (-7/10)c=(-0.7)c [as expected?]
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.