From 8bed9b0564a0f2351676b74cce2dd4307dbe1914 Mon Sep 17 00:00:00 2001 From: Francis Russell Date: Thu, 1 Dec 2011 03:28:43 +0000 Subject: [PATCH] Replace uses of equation environment with align. --- relat10.tex | 52 ++++++++++++++++++++++++++-------------------------- 1 file changed, 26 insertions(+), 26 deletions(-) diff --git a/relat10.tex b/relat10.tex index ceb5b32..63c5f1d 100644 --- 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 -- 2.47.3