We will develop a spacetime diagram for the scenario you described.
(This method is based on my PF Insights:
https://www.physicsforums.com/insights/spacetime-diagrams-light-clocks/
https://www.physicsforums.com/insights/relativity-rotated-graph-paper/
https://www.physicsforums.com/insights/relativity-using-bondi-k-calculus/
https://www.physicsforums.com/insights/relativity-on-rotated-graph-paper-a-graphical-motivation/
)
Steps 0 through 3 provide a general setup.
Steps 4 and 5 focus on the specifics of this problem.
Step 0:
We will work in 1+1 dimensions (for 1 dimension of space).
Start with a large sheet of graph paper rotated by 45-degrees.
This is a grid of
"light-clock diamonds" modeled on
"the light-rays in one tick of a light-clock at rest in this [lab] frame".
(Time runs upwards as in the typical spacetime diagram.)
Use this grid to locate events and make measurements according to the "lab frame".
Step 1:
Use the diamonds of the lab frame to construct the
slopes (velocities) for
the worldlines of the two incoming ships that meet the lab frame at event Z (the origin).
Note that SZ has velocity ##\frac{\Delta x}{\Delta t}=\frac{0-6}{0-(-10)}c=\frac{-6}{10}c##, and, similarly, PZ has velocity (6/10)c.
We also display a
radar measurement done by the lab frame.
To reach distant event S, the lab frame sends a light-signal at event Q and it receives a radar echo at event U.
(It might help to think of Q and U as the "intersection of the light-cone of (target) event S and the lab-frame (measurer's) worldline".)
So, the lab records ##t_Q=-16## and ##t_U=-4##.
The lab says that event M is simultaneous with the distant event S,
and the
lab assigns time coordinate ##t_S=t_M=\frac{1}{2}(t_U + t_Q)=-10## (the mid-time [half the sum]).
Next, the
lab assigns "space" coordinate ##x_S/c=\frac{1}{2}(t_U - t_Q)=6## (half the round trip time).
This is in agreement with the diagram.
The lab frame can also make a measurement of event Z.
The radar measurement in that case has equal time-readings for the emission and the reception.
So, ##t_Z=0## and ##x_Z/c=0##.
In preparation for the next step, we note the following constructions involving the
"square-interval".
Form the "causal diamond of MZ" by intersecting the future light-cone of M with the past light-cone of Z.
The parallelogram formed has the area of 100 light-clock diamonds.
The square root gives the magnitudes along the diagonal: thus 10.
That is, MZ (along the timelike diagonal) has magnitude 10.
In an analogous construction, MS is the spacelike diagonal of a causal diamond with area 36 light-clock diamonds. [Technically, this can be assigned the
signed-area -36.] Thus MS (along the spacelike diagonal) has magnitude 6. QS (along a lightlike direction) is the diagonal of causal diamond with area 0 light-clock diamonds. Thus QS has magnitude 0.
Step 2:
We know the lab frame says that the duration of SZ takes 10 ticks (a temporal displacement).
We also know the spatial displacement is (-6) space-ticks ["sticks"].
What does the Green-ship measure for the duration of SZ (using Green's wristwatch along SZ)?
We can use the formula ##(SZ)_t{}^2 - (SZ)_x{}^2 =(10)^2-(-6)^2=(8)^2## to find "8 ticks" along SZ.
But let's try to use the diagram.
Construct the causal diamond of SZ, which has area ##uv=(4)(16)=(8)^2##. (Here, ##v## is not velocity ##V##.)
So, there are 8 of Green's light-clock diamonds along the diagonal SZ, similar in shape to the causal diamond of SZ. Green's light-clock diamonds have the same area as those of the lab-frame (as expected since a boost has determinant one). [The edge-sizes 4 and 16 here are related to the earlier radar-measurement time-readings.]
Step 3:
So, we can construct all of Green's ticks.
The ratio ##\frac{MZ}{SZ}=\frac{10}{8}## is the
time-dilation factor ##\gamma=\frac{1}{\sqrt{1-(v/c)^2}}##.
Side note: (3/5)c is an
arithmetically-nice value for velocity V that leads to calculations involving rational numbers [fractions]. (4/5)c is nice, but (1/2)c is not nice. It turns out that the Doppler factor ##k=\sqrt{\frac{1+V/c}{1-V/c}}## satisfies ##k^2=\frac{u}{v}##.
When ##k## is rational, then the associated ##V## is arithmetically nice. Here, ##k=\sqrt{\frac{4}{16}}=\frac{1}{2}##. So, Green's diamonds are scaled down by 2 along the "u" direction and scaled up by 2 along the "v" direction.
(Check for yourself that ##u/v=1## for the causal diamond of MZ.)
By the way, since ##k=\sqrt{\frac{1+V/c}{1-V/c}}##, then ##(V/c)=\frac{k^2-1}{k^2+1}##.
If ##V## were not arithmetically-nice, the diamonds along the diagonal would not be as easy to draw by hand on the rotated graph paper.
Checkpoint:
By the
principle of relativity, Green should be able to say similar things about the lab-frame.
If Green followed the lab frame's procedure for making a radar measurement,
Green would get radar-times -16 and -4, as the lab frame did.
Green would make the same coordinate assignments [which could be adjusted for signs when they argee on what is "the positive direction"].
In particular, taking "to the right as positive",
Green says that
the lab-frame has velocity +(3/5)c according to Green and has time-dilation factor ##\gamma=\frac{10}{8}## and (compared to Green's grid of diamonds) ##k=2##.
Further,
note that while the events on segment MS are
simultaneous for the lab-frame,
they are
not simultaneous for the Green frame.
Step 4:
Green's measurement of Blue's velocity
By trial and error, maybe guided by the fact that "composition of velocities" is encoded in the multiplication of Doppler factors (so ##k_{\scriptsize{\rm Blue, wrt\ Green}}=(2)(2)=4## and thus ##V_{\scriptsize{\rm Blue, wrt\ Green}}=\frac{4^2-1}{4^2+1}=\frac{15}{17}\approx 0.882##), we find that
Green can do a radar measurement of Blue using signals JP and PY.
Thus, Green uses radar times ##t_J=-32## and ##t_Y=-2##.
Thus, ##x_P=\frac{1}{2}((-2)-(-32))=15##(which must be adjusted to ##-15## for the common direction of positive) and ##t_P=\frac{1}{2}((-2)+(-32))=-17##.
In the end,
Green measures the velocity of Blue to be ##V_{\scriptsize{\rm Blue, wrt\ Green}}=+\frac{15}{17}##.
(As a check, the time-dilation factor should be ##\frac{17}{8}= 2.125##. Try the textbook time-dilation formula. Also check that ##(17)^2-(15)^2=(8)^2##.)
Furthermore,
Green says Blue will travel from P to Z with velocity (15/17)c in 17-Green-ticks, covering the 15 space-ticks from event K on Green's worldline and event P on Blue's worldline, which Green says are simultaneous.
Step 5:
Blue's measurement of Green's velocity
You can follow the procedure in the preceding step to see that
Blue says essentially the same thing about Green and about the lab-frame.
Note that there are three reference frames with their tickmarks shown on this one spacetime diagram...
and you really don't need a calculator (given the arithmetically-nice velocities).
You just need a sheet of graph paper and practice using and interpreting the methods shown here.
