37-33

This problem is very difficult without the aid of a space-time diagram to keeps things straight. In the diagram (fig.9), the $S$ frame is the stationary space station, $S'$ is the armada, and $S''$ is the messenger ship. The two events of interest are when the messenger leaves the armada (we choose the origin) and when it reaches the front of the armada. These are labeled as (1) and (2). The rear and front of the armada are the $ct'$ axis and a line parallel to that axis through $x'=L'=1\;{\rm ly}$, the rest length of the armada.

Figure 9: Space-time diagram for 37-33.

Now, what we need to calculate is the time coordinate in all three reference systems of when the messenger reaches the front of the armada. Since this is an event in space-time, all we need to do is calculate the time in one coordinate system and then convert it to the others using Lorentz transformations. I will do the calculations in the stationary $S$ frame an then transform to the others.

As seen by $S$, the messenger starts at the origin and then moves with $\beta_M=.95$ until it reaches the front of the armada. But the armada is also seen to be moving with speed $\beta_A=.8$. So the ship has to move the length of the armada and move an extra distance since the armada is moving. $S$ will judge this to take a time $t$. Let us denote the total distance moved as $x$ (as judged by $S$). Then we have

$$x=\beta_M(ct)=L+\beta_A(ct),$$

where $L$ is the length of the armada as judged by $S$. Make sure this expression makes sense to you as it is the crux of the problem.

Now that we've established that relationship we can solve for $t$ once we know the length $L$ of the armada as measured by $S$. The armada will be length contracted (it is moving as seen by $S$), so we have

$$L=\frac{L'}{\gamma_A},$$

where $\gamma_A=(1-\beta_A^2)^{-1/2}=(1-.8^2)^{-1/2}=1.667$. Be careful to use the right $\gamma$'s and $\beta$'s! Then

$$L=\frac{1\;{\rm ly}}{1.667}=.6\;{\rm ly}.$$

Finally we have

$$\begin{eqnarray*} \beta_M(ct) &=& L+\beta_A(ct)\\ ct &=& \frac{L}{\beta_M-\beta... ...{.6\;{\rm ly}}{.95-.8}\\ ct&=& 4\;{\rm ly}\\ t&=& 4\;{\rm yr}. \end{eqnarray*}$$

Now we need to know the $x$ coordinate of this event before we can transform it to the other systems. We have (substituting), that

$$x=\beta_M(ct)=(.95)(4\;{\rm ly})=3.8\;{\rm ly}.$$

So the hard work is done. Now we use the Lorentz transformations (again, being careful to use the right $\beta$'s)

$$\begin{eqnarray*} ct' &=& \gamma_A(ct-\beta_Ax)\\ ct'&=& 1.667(4\;{\rm ly}-.8(3.8\;{\rm ly}))\\ ct'&=&1.6\;{\rm ly}\\ t' &=&1.6\;{\rm yr}, \end{eqnarray*}$$

and

$$\begin{eqnarray*} ct''&=& \gamma_M(ct-\beta_Mx)\\ ct''&=& 3.2(4\;{\rm ly}-.95(3.8\;{\rm ly}))\\ ct''&=&1.25\;{\rm ly}\\ t'' &=&1.25\;{\rm yr}. \end{eqnarray*}$$

As an exercise to the reader, find $x'$ and $x''$. What do these mean?