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\begin{document}

\LARGE FTL leads to going backward in time \normalsize
% \author{Christopher Neufeld} %
% \date{18 April 1995} %

Now that we've covered the lead up, let us consider a general FTL drive
in the context of special relativity. In special relativity we must
assume that there is no preferred reference frame, so the FTL drive
behaves identically (from the viewpoint of its passengers) no matter what
its absolute speed is in the universe, assuming one could even define
such a speed in a meaningful way.

We will use the Lorentz transformations derived earlier to calculate how
long an FTL ship is away from Earth for a particular journey. The journey
will consist of the following steps:
\begin{list}
{\roman{steps}}{\usecounter{steps}\setlength{\rightmargin}{\leftmargin}}
\item At time $t = 0$ the ship leaves Earth (located at $x = 0$) for
a stellar system \emph{Hyperion} located a distance $d_{Hyp}$ away,
in the $+x$ direction, with
its FTL drive. \setcounter{depart}{\value{steps}}
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\item The ship arrives at \emph{Hyperion} at time $t_{\roman{steps}}$.
\setcounter{arrive}{\value{steps}}
\item The ship spends a certain amount of time,
$\Delta \! t_{\roman{steps}}$ (as measured by an observer at rest on Earth),
taking photographs and then
accelerating to a slower-than-light speed $v$ relative to the Earth,
in the $x$ direction (away from the Earth).
\setcounter{pause}{\value{steps}}
\item The ship leaves \emph{Hyperion} for Earth using the same FTL drive it used
earlier. \setcounter{headback}{\value{steps}}
\item The ship arrives back at Earth at time $t_{\roman{steps}}$.
\setcounter{return}{\value{steps}}
\end{list}

Now, assume that the FTL drive has an effective normal space velocity of
$u$. That is, the time it takes to cover a distance $d$ is $d / u$, as
measured by a clock which stayed behind and was at rest relative to the
ship before the ship activated its drive. This time is not the time it
takes for light from the arriving ship to arrive back at the clock
which stayed behind, it's the true time taken by subtracting out the
known time delay caused by the finite speed of light.

So, what is the total elapsed time,
$t_{\roman{return}} - t_{\roman{depart}}$? Let's set
$t_{\roman{depart}} = 0$ for simplicity.
We can set a couple of these other
variables fairly easily:
\begin{equation}
\begin{array}{l}
t_{\roman{arrive}} = d_{Hyp} / u \\
t_{\roman{headback}} = t_{\roman{arrive}} + \Delta \! t_{\roman{pause}}
\nonumber
\end{array}
\end{equation}

The next step is not so simple. The ship is now moving under FTL drive
with a speed $u$ as measured in the \emph{new} frame, the one moving at
speed $v$ relative to the Earth. In this frame, the distance from
\emph{Hyperion} to Earth is given by $d'_{Hyp} = d_{Hyp} / \gamma$ where
$\gamma$ was defined in the section about Lorentz transformations. So, in
this new frame the ship will cover a distance and time interval:
\begin{equation}
\begin{array}{l}
\Delta \! x' = - d_{Hyp} / \gamma \\
\Delta \! t' = d_{Hyp} / ( \gamma u ) \nonumber
\end{array}
\end{equation}

We now apply the Lorentz transformation to find the time elapsed in the
frame of the Earth.
\begin{equation}
\Delta \! t = \gamma ( \Delta \! t' + \frac{v \Delta \! x'}{c^{2}} )
\end{equation}

This yields an elapsed time for the return journey in the Earth's frame
of $\Delta \! t = d_{Hyp} ( \frac{1}{u} - \frac{v}{c^{2}} )$. Now, here's
the crucial point. If $u > c$, this $\Delta \! t$ can be negative for
values of $v < c$. This
can't happen if $u \leq c$. So, the time taken for the return journey
can take negative time as measured by people on the Earth. The total time
of the trip is then:
\begin{equation}
t_{\roman{return}} = \frac{2 d_{Hyp}}{u} + \Delta \! t_{\roman{pause}} -
\frac{d_{Hyp} v}{c^{2}} \nonumber
\end{equation}

For $v > 0$ and sufficiently large $d$
it is possible for this sum to be negative, but only if $u > c$. So, we
have derived the following result: if an FTL ship goes from point A to B,
then accelerates away from A once it arrives at B, and activates the FTL
drive again to return to A, it is entirely possible that it will arrive
at A before it ever left.

Now, a \emph{caveat}. This analysis is not valid if the FTL drive
violates the ``no preferred frame'' postulate of special relativity. If
the FTL drive behaves differently if moving than if at rest, it is
possible to arrange things so that a time machine cannot be produced.
For instance, if the FTL drive always causes the person using it to move
at a speed $u$ relative to the stationary frame of the cosmic microwave
background, these time loops will not appear. This would violate the
postulates of special relativity, but there is really no reason to
believe that this violation would have been seen by experiments to date.
In other words, there is room to fit such an FTL drive which remains
consistent with all experiments performed to date.

\end{document}
