Replace uses of equation environment with align.

diff --git a/relat10.tex b/relat10.tex
index ceb5b32b4588918ed8fd89118b62a40e2e3a1ba2..63c5f1da0e02cb0dfb32c3c32d06ca1c28f04ca2 100644 (file)
--- a/relat10.tex
+++ b/relat10.tex
@@ -1282,17 +1282,17 @@ This equation expresses nothing else than the law of motion of the
 point with reference to the system $K$ (of the man with reference to the
 embankment). We denote this velocity by the symbol $W$, and we then
 obtain, as in Section 6,
-\begin{equation}
+\begin{align}
 W=v+w
 \label{eqnA}
-\end{equation}
+\end{align}
 
 But we can carry out this consideration just as well on the basis of
 the theory of relativity. In the equation
-\begin{equation}
+\begin{align}
 x'=wt'
 \label{eqnB}
-\end{equation}
+\end{align}
 
 \noindent we must then express $x'$and $t'$ in terms of $x$ and $t$, making use of the
 first and fourth equations of the Lorentz transformation. Instead of
@@ -3411,27 +3411,27 @@ transmitted according to the equation
 x = ct
 \end{align*}
 or
-\begin{equation}
+\begin{align}
 \label{eqn:a1}
                                    x - ct = 0
-\end{equation}
+\end{align}
  
 Since the same light-signal has to be transmitted relative to $K^1$ with
 the velocity $c$, the propagation relative to the system $K^1$ will be
 represented by the analogous formula
 
-\begin{equation}
+\begin{align}
 \label{eqn:a2}
                                    x' - ct' = 0
-\end{equation}
+\end{align}
 
 Those space-time points (events) which satisfy (\ref{eqn:a1}) must also satisfy
 (\ref{eqn:a2}). Obviously this will be the case when the relation
 
-\begin{equation}
+\begin{align}
 \label{eqn:a3}
           (x' - ct') = \lambda (x - ct)
-\end{equation}
+\end{align}
 
 \noindent is fulfilled in general, where $\lambda$ indicates a constant; for, according
 to (\ref{eqn:a3}), the disappearance of $(x - ct)$ involves the disappearance of
@@ -3440,10 +3440,10 @@ $(x' - ct')$.
 If we apply quite similar considerations to light rays which are being
 transmitted along the negative x-axis, we obtain the condition
 
-\begin{equation}
+\begin{align}
 \label{eqn:a4}
            (x' + ct') = \mu (x + ct)
-\end{equation}
+\end{align}
 
 By adding (or subtracting) equations (\ref{eqn:a3}) and (\ref{eqn:a4}), and introducing for
 convenience the constants $a$ and $b$ in place of the constants $\lambda$ and $\mu$,
@@ -3461,10 +3461,10 @@ a = \frac{\lambda-\mu}{2}
 
 \noindent we obtain the equations
 
-\begin{equation}
+\begin{align}
 \label{eqn:a5}
            \left. \begin{array}{rcl} x' &=& ax-bct \\ ct' &=& act-bx \end{array} \right\} 
-\end{equation}
+\end{align}
 
 We should thus have the solution of our problem, if the constants $a$
 and $b$ were known. These result from the following discussion.
@@ -3479,10 +3479,10 @@ x = \frac{bc}{a}t
 If we call $v$ the velocity with which the origin of $K^1$ is moving
 relative to $K$, we then have
 
-\begin{equation}
+\begin{align}
 \label{eqn:a6}
                         v=\frac{bc}{a}
-\end{equation}
+\end{align}
 
 The same value $v$ can be obtained from equations (\ref{eqn:a5}), if we calculate
 the velocity of another point of $K^1$ relative to $K$, or the velocity
@@ -3507,10 +3507,10 @@ Two points of the $x'$-axis which are separated by the distance $\Delta x' = I$
 when measured in the $K^1$ system are thus separated in our instantaneous
 photograph by the distance
 
-\begin{equation}
+\begin{align}
 \label{eqn:a7}
                         \Delta x = \frac{I}{a}
-\end{equation}
+\end{align}
 
 \noindent But if the snapshot be taken from $K'(t' = 0)$, and if we eliminate $t$
 from the equations (\ref{eqn:a5}), taking into account the expression (\ref{eqn:a6}), we
@@ -3539,13 +3539,13 @@ The equations (\ref{eqn:a6}) and (7b) determine the constants $a$ and $b$. By
 inserting the values of these constants in (\ref{eqn:a5}), we obtain the first
 and the fourth of the equations given in Section 11.
 
-\begin{equation}
+\begin{align}
 \label{eqn:a8}
            \left. \begin{array}{rcl} 
            x' &=& \frac{x-vt}{\sqrt{I-\frac{v^2}{c^2}}} \\ 
            ~ \\
            t' &=& \frac{t-\frac{v}{c^2}x}{\sqrt{I-\frac{v^2}{c^2}}} \end{array} \right\}  
-\end{equation}
+\end{align}
 
 Thus we have obtained the Lorentz transformation for events on the
 $x$-axis. It satisfies the condition
@@ -3558,10 +3558,10 @@ The extension of this result, to include events which take place
 outside the $x$-axis, is obtained by retaining equations (\ref{eqn:a8}) and
 supplementing them by the relations
 
-\begin{equation}
+\begin{align}
 \label{eqn:a9}
            \left. \begin{array}{rcl} y' &=& y \\ z' &=& z \end{array} \right\}
-\end{equation}
+\end{align}
 
 In this way we satisfy the postulate of the constancy of the velocity
 of light in vacuo for rays of light of arbitrary direction, both for
@@ -3577,10 +3577,10 @@ r = \sqrt{x^2+y^2+z^2} = ct
 
 \noindent or, if we square this equation, according to the equation
 
-\begin{equation}
+\begin{align}
 \label{eqn:a10}
           x^2 + y^2 + z^2 = c^2t^2 = 0
-\end{equation}
+\end{align}
 
 It is required by the law of propagation of light, in conjunction with
 the postulate of relativity, that the transmission of the signal in
@@ -3600,10 +3600,10 @@ r' = ct'
 In order that equation (10a) may be a consequence of equation (\ref{eqn:a10}), we
 must have
 
-\begin{equation}
+\begin{align}
 \label{eqn:a11}
              {x'}^2 + {y'}^2 + {z'}^2 - c^2{t'}^2 = \sigma (x^2 + y^2 + z^2 - c^2t^2)
-\end{equation}
+\end{align}
 
 
 Since equation (8a) must hold for points on the $x$-axis, we thus have $\sigma