37-13

In the figure below (fig.7) the $ct$ axis represents the earth. The line at $x=26$ ly represents vega (it is a distance $x=26$ ly away from earth at all times $t$). The line from the origin to vega is the travelers path to vega. Once at vega the traveler send a (light) signal back to earth which travels with $\beta=1$ (always the case for light).

Figure 7: Space-time diagram for 37-13.

Since we know the speed of the traveler and the distance to vega as measured by earth we can calculate the time of travel as measured by earth by dividing as usual

$$t=\frac{x}{v}=\frac{26\;{\rm ly}}{.99c}=\frac{26\;{c\cdot \rm 1\;yr}}{.99c} = \frac{26\;{\rm yr}}{.99}=26.26\;{\rm yr},$$

where we used the fact that a light-year is the distance traveled by light in one year, which is the speed of light times one year.

Now, the light sent by the traveler back to earth travels at the speed of light, so it takes 26 years to make the trip since it travels 26 ly. So the total time is $26\;{\rm ly}+26.26\;{\rm ly}=52.26\;{\rm ly}$.

Finally, to find how long the traveler judges the trip to have taken, we use the time dilation formula from problem 4 to get

$$t'=\frac{t}{\gamma}=\sqrt{1-\beta^2}t=\sqrt{1-.99^2}(26\;{\rm ly})=3.70\;{\rm yr}.$$