\documentclass{letter}

\usepackage{graphics,latexsym}

\setlength{\topmargin}{0cm}
\setlength{\oddsidemargin}{0cm}
\setlength{\footskip}{0cm}
\setlength{\marginparsep}{1cm}
\setlength{\headheight}{0cm}
\setlength{\headsep}{0cm}
\setlength{\textwidth}{16cm}

\begin{document}

\LARGE Derivation of the Lorentz transformations \normalsize
% \author{Christopher Neufeld} %
% \date{18 April 1995} %

First, we need some definitions. We will define an inertial reference
frame as a coordinate system associated with an observer moving at some
constant velocity. For simplicity we will assume that this observer is at
the origin, \emph{i.e.} $x = y = z = 0$. In this coordinate system there
is also a time variable which measures the elapsed time as measured on a
clock at rest in this coordinate system.

Consider two frames, $S$ and $S'$, such that an observer in $S$ sees
the $S'$ frame as moving in the direction of positive $x$ with a
speed $v$. Prior to special relativity one transformed between these
frames via the Galilean transformations:

\begin{equation}
\begin{array}{ll}
t' = t & \\
x' = x - v t & x = x' + v t' \\
y' = y & z' = z
\end{array}
\label{eq:galilean}
\end{equation}

We saw earlier that the speed of light is required to be preserved
between frames. Clearly the Galilean transformations do not preserve the
speed of light, a light ray moving with speed $c$ in the positive $x$
direction in
frame $S$
will appear in frame $S'$ to be moving with speed $c - v$ in the
positive $x'$ direction.

What we want is a relation between the $S$ and $S'$ frames which
preserves the speed of light. Consider a spherically symmetric expanding
wave which starts at the origin when the frames are overlaid so that
$x = x' = y = y' = z = z' = t = t' = 0$.
In frame $S$, at a given time $t$ it is found on the spherical shell given by:
\begin{equation}
x^{2} + y^{2} + z^{2} - c^{2} t^{2} = 0
\label{eq:sphere1}
\end{equation}
and in the frame $S'$ it is on the shell given by:
\begin{equation}
x'^{2} + y'^{2} + z'^{2} - c^{2} t'^{2} = 0
\label{eq:sphere2}
\end{equation}

Now, since the frame $S'$ is moving in the $x$ direction relative to $S$
it is clear by symmetry that $y' = y$ and $z' = z$. Also, the equation we
get must not change if the origin of time or space is arbitrarily defined
to be somewhere else, so $x'$ and $t'$ must be linear functions of $x$
and $t$. The general form of this is:
\begin{equation}
x' = A x + B t
\label{eq:linear1}
\end{equation}
\begin{equation}
t' = D x + E t
\label{eq:linear2}
\end{equation}
where $A$, $B$, $D$, and $E$ are some constants.

Now, a point at rest in $S'$ at $x' = 0$ must appear in $S$ to be moving
with speed $+v$, so it follows immediately that $-B / A = v$. We can
rewrite equation~\ref{eq:linear1}:
\begin{equation}
x' = A ( x - v t )
\label{eq:linear3}
\end{equation}

We now substitute equation~\ref{eq:linear3} into equation~\ref{eq:sphere2}:
\begin{equation}
\{ A ( x - v t ) \} ^{2} + y^{2} + z^{2} - c^{2} ( D x + E t )^{2} = 0
\nonumber
\end{equation}

Next, perform a bit of algebra on this. After some thrashing, we produce:
\begin{equation}
(A^{2} - c^{2} D^{2}) x^{2} + y^{2} + z^{2} +
(A^{2} v^{2} - E^{2} c^{2}) t^{2} - 2 (A^{2} v + D E c^{2}) x t = 0
\end{equation}

This equation must be identical with equation~\ref{eq:sphere1}. Comparing
terms from the two expressions we obtain the following:
\begin{equation}
\begin{array}{l}
A^{2} - c^{2} D^{2} = 1 \\
A^{2} v^{2} - E^{2} c^{2} = -c^{2} \\
A^{2} v + D E c^{2} = 0 \nonumber
\end{array}
\end{equation}

We can solve these equations for $A$, $D$, and $E$, and can discard
extraneous solutions by requiring that the valid solution has the $x$ and
$x'$ pointing in the same direction, and $t$ and $t'$ both increase as
time passes into the future. This gives us:
\begin{equation}
\begin{array}{l}
A = (1 - \frac{v^{2}}{c^{2}})^{-\frac{1}{2}} \\
B = - v A \\
D = - \frac{v}{c^{2}} (1 - \frac{v^{2}}{c^{2}})^{-\frac{1}{2}} \\
E = (1 - \frac{v^{2}}{c^{2}})^{-\frac{1}{2}} \nonumber
\end{array}
\end{equation}

If we now adopt the notation $\gamma = (1 - v^{2} / c^{2})^{-1/2}$, we can
write the Lorentz transformations as they usually appear:

\begin{equation}
\begin{array}{ll}
x' = \gamma ( x - v t ) & x = \gamma ( x' + v t' ) \\
y' = y & y = y' \\
z' = z & z = z' \\
t' = \gamma ( t - \frac{v x}{c^{2}} ) &
t = \gamma ( t' + \frac{v x'}{c^{2}} )
\end{array}
\end{equation}

Of course, the Galilean transformations in Eq.~\ref{eq:galilean} make intuitive
sense, the Lorentz transformations seem ridiculous. Notice that the
Lorentz transformations reduce to the Galilean transformations in the
limit of small velocities. Expanding $\gamma$ in a McLauren series in
$v / c$ produces:

\begin{equation}
\begin{array}{ll}
x' = (x - v t ) (1 + \frac{v^{2}}{2 c^{2}}) &
x = ( x' + v t' ) (1 + \frac{v^{2}}{2 c^{2}}) \\
y' = y & y = y' \\
z' = z & z = z' \\
t' = (t - \frac{x}{c} \frac{v}{c}) (1 + \frac{v^{2}}{2 c^{2}}) &
t = (t' + \frac{x}{c} \frac{v}{c}) (1 + \frac{v^{2}}{2 c^{2}}) \\
\end{array}
\end{equation}

For small values of $v / c$ the Lorentz transformations reduce to the
Galilean transformations. The departure is proportional to
$(\frac{v}{c})^{2}$ for the spatial coordinate. The departure for the
time coordinate depends a bit on the system in question. If $\Delta \! x$ is
zero in one frame then the departure of the time coordinate from the
Galilean transformation value is again proportional to the square of
$v / c$.

\end{document}
