# SR - Time dilation, space-time diagram, and radio signals

## Homework Statement

P.S.: I'm not sure if it is allowed to ask multi-part questions.

Two equally old sisters Alice and Barbara leave Earth simultaneously in opposite directions. The following velocities and distances have been measured in the Earth system. Alice travels with a speed of ##v_A = 4/5c## to a star α, 4 light-years away, and Barbara with a speed ##v_B = 3/5c## to a star β, 3 light-years away. The moment the sisters arrive at their destinations, they turn around and head back to Earth with the same speeds they left Earth.

Question A:
After how many years (as seen by the observer from Earth) do both sisters return? Draw a space-time diagram of the Earth system of the trips of Alice and Barbara.

Question B:
How long does the entire trip take in Alice's frame? Similarly for Barbara? Who is older when returning? Also calculate the distance travelled.

Question C:
The sisters have agreed to sent each other a monthly radio signal. Exactly when Alice wants to turn around at star ##\alpha##, she receives a radio signal. When has Barbara sent this signal (as measured by her clock) and how many messages has Alice already received?

## Homework Equations

Lorentz-boost
$$\begin{pmatrix} \gamma & -\beta\gamma & 0 & 0\\ -\beta\gamma & \gamma & 0 & 0\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 &1 \end{pmatrix}$$

Length contraction
##L' = \gamma^{-1}L##

Distance formula
##\Delta v = \frac{\Delta s}{\Delta t}##

Relative velocity
##w = \frac{|v_A - v_B|}{1-(v_A\cdot v_B)/c^2}##

## The Attempt at a Solution

Part A: [/B]I believe this can simply be down using Newtonian mechanics. So ##s_B/v_B = t_B = \frac{3}{3/5c} = 5## years. Times two to get the distance and back: so ##10## years. Similarly, ##\frac{4}{4/5c} = 5##, hence, also ##10## years in total for Alice.

Part B:
Using the Lorentz-boost we find that Alice measures

$$\begin{pmatrix} \gamma & -\beta\gamma & 0 & 0\\ -\beta\gamma & \gamma & 0 & 0\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 &1 \end{pmatrix} \begin{pmatrix} 10\\ 8\\ 0\\ 0 \end{pmatrix} = \begin{pmatrix} 10\gamma - 8\beta\gamma\\ -10\beta\gamma + 8\gamma\\ 0\\ 0 \end{pmatrix} = \begin{pmatrix} 6\\ 0\\ 0\\ 0 \end{pmatrix}$$

Hence, Alice experiences ##6## years having gone by, but she hasn't travelled any distance. Because she "sees" herself standing still, and the star moving towards her? However, she undergoes an acceleration, so should realise it is the her moving? Shouldn't it be, using length-contraction ##L' = L\sqrt{1-\beta^2} = 8\sqrt{9/25} = 24/5 = 4.8## ly. What am I doing wrong with using the boost?

Barbara measures (using the same method) ##8## years having gone by, and also ##0## light-years travelled. Also, with length-contraction, shouldn't this be ##6\sqrt{16/25} = 4.8## ly? Also, Barbara would be older, as her trip took ##8## years, compared to Alice's ##6## years. This result confuses me, since in part A we have seen that they should arrive (and left) simultaneously: same time, same place. So no shenanigans should happen with relativity of simultaneity? Hence they should be equally old?

Part C:

We first need to find their relative velocity. Which is
##w = \frac{|v_A - v_B|}{1-(v_A\cdot v_B)/c^2} = \frac{4/5c-(-3/5c)}{1 - \frac{-\frac{4c}{5}\frac{3c}{5}}{c^2}} = \frac{\frac{7c}{5}}{\frac{37}{25}} = \frac{35}{37}c##
Is there a way to do this using Lorentz-boosts?

Since Alice is the one who receives the signals, we shall look at it from her frame of reference.
It takes her ##3## years to reach her destination. Hence, the signal had to travel ##\frac{35}{37}c\cdot 3 = 2.8..## ly. So the signal had to have been sent approximately ##2.8## years ago -both for Alice and Barbara?-. I can't figure out how to determine the amount of signals already sent..