Talk:PlanetPhysics/Simple Derivation of the Lorentz Transformation

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\begin{document}

 \subsection{Simple Derivation of the Lorentz Transformation
(Supplementary to Section 11)}
From \htmladdnormallink{Relativity: The Special and General Theory}{http://planetphysics.us/encyclopedia/SpecialTheoryOfRelativity.html} by \htmladdnormallink{Albert Einstein}{http://planetphysics.us/encyclopedia/AlbertEinstein.html}
For the relative orientation of the co-ordinate \htmladdnormallink{systems}{http://planetphysics.us/encyclopedia/GenericityInOpenSystems.html} indicated in
Fig. 2, the x-axes of both systems pernumently coincide. In the
present case we can divide the problem into parts by considering first
only events which are localised on the $x$-axis. Any such event is
represented with respect to the co-ordinate system $K$ by the abscissa $x$
and the time $t$, and with respect to the system $K^1$ by the abscissa $x'$
and the time $t'$. We require to find $x'$ and $t'$ when $x$ and $t$ are given.

A light-signal, which is proceeding along the positive axis of $x$, is
transmitted according to the equation

$$x = ct$$
or
\begin{equation}
\label{eqn:a1}
x - ct = 0
\end{equation}

Since the same light-signal has to be transmitted relative to $K^1$ with
the \htmladdnormallink{velocity}{http://planetphysics.us/encyclopedia/Velocity.html} $c$, the propagation relative to the system $K^1$ will be
represented by the analogous \htmladdnormallink{formula}{http://planetphysics.us/encyclopedia/Formula.html}
\begin{equation}
\label{eqn:a2}
x' - ct' = 0
\end{equation}

Those \htmladdnormallink{space-time}{http://planetphysics.us/encyclopedia/SR.html} points (events) which satisfy (\ref{eqn:a1}) must also satisfy
(\ref{eqn:a2}). Obviously this will be the case when the \htmladdnormallink{relation}{http://planetphysics.us/encyclopedia/Bijective.html}
\begin{equation}
\label{eqn:a3}
(x' - ct') = \lambda (x - ct)
\end{equation}

\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
$(x' - ct')$.

If we apply quite similar considerations to light rays which are being
transmitted along the negative x-axis, we obtain the condition

\begin{equation}
\label{eqn:a4}
(x' + ct') = \mu (x + ct)
\end{equation}

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 $\mu$,
where

$$a = \frac{\lambda+\mu}{2}$$

\noindent and

$$a = \frac{\lambda-\mu}{2}$$ % ??

\noindent we obtain the equations

\begin{equation}
\label{eqn:a5}
\left. \begin{array}{rcl} x' &=& ax-bct \\ ct' &=& act-bx \end{array} \right\}
\end{equation}

We should thus have the solution of our problem, if the constants $a$
and $b$ were known. These result from the following discussion.

For the origin of $K^1$ we have permanently $x' = 0$, and hence according
to the first of the equations (\ref{eqn:a5})

$$x = \frac{bc}{a}t$$

If we call $v$ the velocity with which the origin of $K^1$ is moving
relative to $K$, we then have

\begin{equation}
\label{eqn:a6}
v=\frac{bc}{a}
\end{equation}

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
(directed towards the negative $x$-axis) of a point of $K$ with respect to
$K'$. In short, we can designate $v$ as the relative velocity of the two
systems.

Furthermore, the principle of relativity teaches us that, as judged
from $K$, the length of a unit measuring-rod which is at rest with
reference to $K^1$ must be exactly the same as the length, as judged from
$K'$, of a unit measuring-rod which is at rest relative to $K$. In order
to see how the points of the $x$-axis appear as viewed from $K$, we only
require to take a ``snapshot'' of $K^1$ from $K$; this means that we have
to insert a particular value of $t$ (time of $K$), {\it e.g.} $t = 0$. For this
value of $t$ we then obtain from the first of the equations (5)

$$x' = ax$$

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}
\label{eqn:a7}
\Delta x = \frac{I}{a}
\end{equation}

\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
obtain

$$x' = a \left( I - \frac{v^2}{c^2} \right) x$$

\noindent From this we conclude that two points on the $x$-axis separated by the
distance $I$ (relative to $K$) will be represented on our snapshot by the
distance

$$\Delta x' = a \left( I - \frac{v^2}{c^2} \right) \quad . \quad . \quad . \quad \mbox{(7a)}$$

But from what has been said, the two snapshots must be identical;
hence $\Delta x$ in (7) must be equal to $\Delta x'$ in (7a), so that we obtain

$$a = \frac{I}{I-\frac{v^2}{c^2}} \quad . \quad . \quad . \quad \mbox{(7b)} $$

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 \htmladdnormallink{section}{http://planetphysics.us/encyclopedia/IsomorphicObjectsUnderAnIsomorphism.html} 11.

\begin{equation}
\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}

Thus we have obtained \htmladdnormallink{The Lorentz transformation}{http://planetphysics.us/encyclopedia/LorentzTransformation.html} for events on the
$x$-axis. It satisfies the condition

$$x'^2 - c^2t'^2 = x^2 - c^2t^2 \quad . \quad . \quad . \quad \mbox{(8a)} $$

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}
\label{eqn:a9}
\left. \begin{array}{rcl} y' &=& y \\ z' &=& z \end{array} \right\}
\end{equation}

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
the system $K$ and for the system $K'$. This may be shown in the following
manner.

We suppose a light-signal sent out from the origin of $K$ at the time $t
= 0$. It will be propagated according to the equation

$$r = \sqrt{x^2+y^2+z^2} = ct$$

\noindent or, if we \htmladdnormallink{square}{http://planetphysics.us/encyclopedia/PiecewiseLinear.html} this equation, according to the equation

\begin{equation}
\label{eqn:a10}
x^2 + y^2 + z^2 = c^2t^2 = 0
\end{equation}

It is required by the law of propagation of light, in conjunction with
the postulate of relativity, that the \htmladdnormallink{transmission}{http://planetphysics.us/encyclopedia/FluorescenceCrossCorrelationSpectroscopy.html} of the signal in
question should take place---as judged from $K^1$---in accordance with
the corresponding formula

$$r' = ct'$$

\noindent or,

$$x'^2 + y'^2 + z'^2 - c^2t'^2 = 0  \quad . \quad . \quad . \quad \mbox{(10a)} $$

In order that equation (10a) may be a consequence of equation (\ref{eqn:a10}), we
must have

\begin{equation}
\label{eqn:a11}
x'^2 + y'^2 + z'^2 - c^2t'^2 = \sigma (x^2 + y^2 + z^2 - c^2t^2)
\end{equation}


Since equation (8a) must hold for points on the $x$-axis, we thus have $\sigma
= I$. It is easily seen that the Lorentz transformation really
satisfies equation (\ref{eqn:a11}) for $\sigma = I$; for (\ref{eqn:a11}) is a consequence of (8a)
and (\ref{eqn:a9}), and hence also of (\ref{eqn:a8}) and (\ref{eqn:a9}). We have thus derived the
Lorentz transformation.

The Lorentz transformation represented by (\ref{eqn:a8}) and (\ref{eqn:a9}) still requires
to be generalised. Obviously it is immaterial whether the axes of $K^1$
be chosen so that they are spatially parallel to those of $K$. It is
also not essential that the velocity of translation of $K^1$ with respect
to $K$ should be in the direction of the $x$-axis. A simple consideration
shows that we are able to construct the Lorentz transformation in this
general sense from two kinds of transformations, {\it viz.} from \htmladdnormallink{Lorentz transformations}{http://planetphysics.us/encyclopedia/CosmologicalConstant.html} in the special sense and from purely spatial
transformations. which corresponds to the replacement of the
rectangular co-ordinate system by a new system with its axes pointing
in other directions.

Mathematically, we can characterise the generalised Lorentz
transformation thus:

It expresses $x', y', x', t'$, in terms of linear homogeneous \htmladdnormallink{functions}{http://planetphysics.us/encyclopedia/Bijective.html} of $x, y, x, t$, of such a kind that the relation

$$x'^2 + y'^2 + z'^2 - c^2t'^2 = x^2 + y^2 + z^2 - c^2t^2  \quad . \quad . \quad . \quad \mbox{(11a)} $$

\noindent is satisficd identically. That is to say: If we substitute their
expressions in $x, y, x, t$, in place of $x', y', x', t'$, on the
left-hand side, then the left-hand side of (11a) agrees with the
right-hand side.

\subsection{References}
This article is derived from the Einstein Reference Archive (marxists.org) 1999, 2002. \htmladdnormallink{Einstein Reference Archive}{http://www.marxists.org/reference/archive/einstein/index.htm} which is under the FDL copyright.

\end{document}