\begin{problem*}{9.18}
An observer in reference frame $S$ measures two events as
simultaneous.  Event $A$ occurs at the point $(50.0\U{m},0,0)$ at the
instant 9:00:00 Universal time on January 15, 2005.  Event $B$ occurs
at the point $(150\U{m},0,0)$ at the same moment.  A second observer,
moving past with a velocity of $0.800c\ihat$, also observes the two
events.  In her reference frame $S'$, which event occured first and
what time interval elapsed between the events?
\end{problem*} % problem 9.18

\begin{solution}
The $(x,t)$ coordinates in the $S$ frame are $(50.0\U{m},t_0)$ and
$(150\U{m},t_0)$.  In the $S'$ frame, the coordinates are given by
the Lorentz transformations
\begin{align}
  \gamma &= \frac{1}{\sqrt{1-v^2/c^2}}
    = \frac{1}{\sqrt{1-0.800^2}}
    = 1.67 \\
  t' &= \gamma \p({t - v x/c^2}) \\
  x' &= \gamma \p({x - v t}) \\
  y' &= y \\
  z' &= z
\end{align}
We see from the formula for $t'$ that the event with the largest $x$
value will have the earliest time.  Therefore, \ans{event $B$ happens
first in $S'$}.  The time interval is
\begin{align}
  \Delta t = t_A' - t_B'
    = \gamma \p({\Delta t - v \Delta x/c^2})
    = \gamma \p({0 - 0.800 \cdot (-100\U{m}) / 3.00\E{8}\U{m/s}})
    = \ans{444\U{ns}}
\end{align}
\end{solution}
