Alpha Centauri is roughly 4 lightyears from Earth. Dr. X travels (at constant velocity) from Earth to Alpha Centauri in 3 years. Immediately upon her arrival at Alpha Centauri, she turns on a powerful laser aimed at the Earth. (Ignore the motion of the Earth!)

1. How fast did she travel?
2. How far did she travel? (That is, how far does she think Alpha Centauri is from Earth?)
3. How long after her departure does the Earth find out she arrived safely?
4. Draw a spacetime diagram showing the worldlines of Dr. X, the Earth, and the laser beam. Clearly indicate which events correspond to departure, arrival/turning on the laser, and the receipt of the laser signal.

Figure 7.10: The spacetime diagram for Dr. X.

See the spacetime diagram in Figure \ref{travel}, which answers Question 4.

1. The known data is that Alpha Centauri is 4 lightyears from Earth, and that Dr. X travels for 3 years as measured by her own clock. The lower triangle in Figure \ref{travel} therefore has horizontal leg $\Delta x=4 \hbox{lightyears}$ and hypotenuse $c\Delta\tau=3 \hbox{lightyears}$, as shown. The remaining (vertical) leg of this triangle therefore has length $5$ lightyears, from which we can read off \begin{equation} \frac{v}{c} = \tanh\beta = \frac45 \end{equation}

2. This same triangle gives us a Lorentz contraction factor of \begin{equation} \cosh\beta = \frac53 \end{equation} from which it follows that \begin{equation} \ell' = \frac{\ell}{\cosh\beta} = \frac{4}{\frac53} = \frac{12}{5} \end{equation} Alternatively, if Dr. X travels for 3 years at $\frac45$ the speed of light, she must have traveled $\frac{12}{5}$ lightyears.

3. We already know that the vertical leg of the lower triangle has length $5$, indicating that observers on Earth believe Dr. X arrives 5 years after departure. The laser signal will take 4 years to travel 4 lightyears, for a total time difference of 9 years.