How can I measure the space time interval between two events?

Question

This is my last resort for a frustrating problem that I am certainly overcomplicating, but I don't have enough understanding to complete alone.

The problem goes like this:

Consider a space station immobile in a given $(t, x, y, z)$ coordinate system. At $t=0$, three spaceships leave a space station with velocities $\bar{v}_A = 0.8c\hat{x}$, $\bar{v}_B = -0.6c\hat{x}$, and $\bar{v}_C = 0.7c\hat{y}$. Write the worldline of the station and each spaceship in the $(t, x, y, z)$ coordinates, as a function of parameters $\lambda_{S,A,B,C}$. Choose your $\lambda_X$ parameters so that they are all $0$ at $t = 0$.

Spaceship A waits for a time $\Delta t'_A$ after launch (measured in its own reference frame) before emitting a brief pulse of light. Consider the events corresponding to spaceships B and C receiving the pulse of light. How are they causally related (i.e. is one in the future of the other, or are they causally disconnected)? What is the proper distance (if spacelike) or proper time (if timelike) between them?

Normally calculating the interval $(\Delta s)^2$ between two events isn't difficult, but the worldlines for each ship are parameterized with different $\lambda$s and I know that in order to determine whether they are time/spacelike I need to calculate the interval but this single problem is enough to make calculating it very difficult.

For reference, I have the worldlines in the frame of the station S $(t,x,y,z)$:

$W_S = (\lambda_S,0,0,0)$

$W_A = (\lambda_A, 0.8c\lambda_A,0,0)$

$W_B = (\lambda_B, -0.6c\lambda_B,0,0)$

$W_C = (\lambda_C, 0,0.7c\lambda_C,0)$

and (after a Lorentz boost) in the frame of A $(t',x',y',z')$:

$W_S' = (\frac{5}{3}\lambda_S,-\frac{4}{3}c\lambda_S,0,0)$

$W_A' = (\frac{3}{5}\lambda_A,0,0,0)$

$W_B' = (\frac{37}{15}\lambda_B, -\frac{7}{3}c\lambda_B,0,0)$

$W_C' = (\frac{5}{3}\lambda_C,-\frac{4}{3}c\lambda_C,0.7c\lambda_C,0)$

I also calculated the time it takes for the pulse to reach B and C in A's frame as well as how long it takes for B and C to receive the pulse in their own frames (using time dilation):

$\Delta t_{BA}' = \Delta t_A' + \frac{|W_B'^x|}{c} = \Delta t_A' + \frac{7}{3}\lambda_B$

$\Delta t_{CA}' = \Delta t_A' + \frac{|W_C'^x + W_C'^y|}{c} \approx \Delta t_A' + \frac{3}{2}\lambda_B$

In B's frame: $\tau_{BA} = \frac{\Delta t_{BA}'}{\gamma_{BA}}$

In C's frame: $\tau_{CA} = \frac{\Delta t_{CA}'}{\gamma_{CA}}$

I have a feeling that I am messing something up by still having the $\lambda$s in my time calculations, but I'm not sure if I should substitute a time (maybe $\Delta t_A'$?) into each worldline to compute the time in A's frame or if I've missed something else.

Any help on how to calculate the interval between B and C would be greatly appreciated! I believe I can calculate proper time/distance after that on my own.

Answer 1

wait--:

$$ W_S = (t, 0, 0, 0)_S $$ $$ W_A = (t, v_At, 0, 0)_S $$ $$ W_B = (t, v_Bt, 0, 0)_S $$ $$ W_C = (t, 0, v_Ct, 0)_S $$

The pulse is emitted at $\Delta t'_A$ -- which occurs in $S$ at:

$$ \Delta t_A = \Delta t'_A/\gamma_A \equiv a $$

which is:

$$ E_A = (a, v_A a,0,0)$$

To find where that intersects $B$, which occurs at $\Delta t_B\equiv b$ (in S), so the event is:

$$ E_B = (b, v_B b, 0 , 0) $$

such that:

$$ (E_A-E_B)^2 = (E_{A, t} - E_{B, t})^2 - (E_{A, x} - E_{B, x})^2 - (E_{A, y} - E_{B, y})^2-(E_{A, z} - E_{B, z})^2 = 0 $$

since the emission and detection are light-like separated in all frames. So:

$$ (E_A-E_B)^2 = (a-b)^2 - (v_A a-v_B b)^2 = 0 $$

Solve $b$ in terms of $a$, and then do the same for $W_C$ to get $E_C$, then work with $E_B$ and $E_C$ in frame $S$ to get $\Delta s_{BC}$, which is a Lorentz invariant.