in the diagram (Fig \ref{fig:2}), this problem is solved by means of the
equations:
-\begin{eqnarray*}
-x' &=& \frac{x-vt}{\sqrt{I-\frac{v^2}{c^2}}} \\
-y' &=& y \\
-z' &=& z \\
-t' &=& \frac{t-\frac{v}{c^2}x}{\sqrt{I-\frac{v^2}{c^2}}} \\
-\end{eqnarray*}
+\begin{align*}
+x' &= \frac{x-vt}{\sqrt{I-\frac{v^2}{c^2}}} \\
+y' &= y \\
+z' &= z \\
+t' &= \frac{t-\frac{v}{c^2}x}{\sqrt{I-\frac{v^2}{c^2}}} \\
+\end{align*}
\noindent This system of equations is known as the ``Lorentz transformation."\footnotemark
character of times and lengths, then instead of the above we should
have obtained the following equations:
-\begin{eqnarray*}
-x' &=& x - vt \\
-y' &=& y \\
-z' &=& z \\
-t' &=& t \\
-\end{eqnarray*}
+\begin{align*}
+x' &= x - vt \\
+y' &= y \\
+z' &= z \\
+t' &= t \\
+\end{align*}
\noindent This system of equations is often termed the ``Galilei
transformation." The Galilei transformation can be obtained from the
the value $ct$ in the first and fourth equations of the Lorentz
transformation, we obtain:
-\begin{eqnarray*}
-x' &=& \frac{(c-v)t}{\sqrt{I-\frac{v^2}{c^2}}} \\
-t' &=& \frac{(I-\frac{v}{c})t}{\sqrt{I-\frac{v^2}{c^2}}}
-\end{eqnarray*}
+\begin{align*}
+x' &= \frac{(c-v)t}{\sqrt{I-\frac{v^2}{c^2}}} \\
+t' &= \frac{(I-\frac{v}{c})t}{\sqrt{I-\frac{v^2}{c^2}}}
+\end{align*}
\noindent from which, by division, the expression
of the first equation of the Lorentz transformation the values of
these two points at the time $t = 0$ can be shown to be
-\begin{eqnarray*}
-x_{\mbox{(begining of rod)}} &=& 0 \sqrt{I-\frac{v^2}{c^2}} \\
-x_{\mbox{(end of rod)}} &=& 1 \sqrt{I-\frac{v^2}{c^2}}
-\end{eqnarray*}
+\begin{align*}
+x_{\mbox{(begining of rod)}} &= 0 \sqrt{I-\frac{v^2}{c^2}} \\
+x_{\mbox{(end of rod)}} &= 1 \sqrt{I-\frac{v^2}{c^2}}
+\end{align*}
~
\noindent the distance between the points being $\sqrt{I-v^2/c^2}$.
of the table (Gaussian co-ordinates). For example, the point $P$ in the
diagram has the Gaussian co-ordinates $u=3$, $v=1$. Two neighbouring
points $P$ and $P_1$ on the surface then correspond to the co-ordinates
-\begin{eqnarray*}
-P: & u ~~,~~v \\
-P': & u + du , v + dv
-\end{eqnarray*}
+
+\begin{align*}
+P:&~u,~v \\
+P':&~u + du,~v + dv
+\end{align*}
+
where $du$ and $dv$ signify very small numbers. In a similar manner we may
indicate the distance (line-interval) between $P$ and $P_1$, as measured
with a little rod, by means of the very small number $ds$. Then
We can characterise the Lorentz transformation still more simply if we
introduce the imaginary $\sqrt{-I} \cdot ct$ in place of $t$, as time-variable. If, in
accordance with this, we insert
-\begin{eqnarray*}
- x_1 & = & x \\
- x_2 & = & y \\
- x_3 & = & z \\
- x_4 & = & \sqrt{-I} \cdot ct
-\end{eqnarray*}
+
+\begin{align*}
+ x_1 & = x \\
+ x_2 & = y \\
+ x_3 & = z \\
+ x_4 & = \sqrt{-I} \cdot ct
+\end{align*}
+
and similarly for the accented system $K^1$, then the condition which is
identically satisfied by the transformation can be expressed thus: