On the electrodynamics of moving bodies
3. Theory of the Transformation of Co-ordinates and Times from a Stationary Sytem to another System in Uniform Motion of Translation Relatively to the Former
- $(x, y, z)$, $t$,
- stationary system
- $(\xi , \eta, \zeta)$, $\tau$,
- moving system
Assumptions
(1)
There exists $f$ such that
\[\tau = f(x, y, z, t)\](2)
$f$ is linear
\[\tau =\](3)
\[x^{\prime} := x - vt .\] \[\tau_{1} - \tau_{0} := \tau_{2} - \tau_{1} .\] \[\tau = a \left( t - \frac{ v }{ c^{2} - v^{2} }x^{\prime} \right)\]where $a$ is a function $\phi(v)$. For brevity, we assume that $\tau = 0$ at $(0, 0, 0)$ in $k$, when $t = 0$.
For a ray of light emitted from the origin of the moving sytem $k$ at the time $\tau = 0$ in the direction of the increasing $\xi$,
\[\begin{eqnarray} \xi & = & c \tau \nonumber \\ & = & ac \left( t - \frac{ v }{ c^{2} - v^{2} }x^{\prime} \right) . \nonumber \end{eqnarray}\]When it is measured in the stationary system,
\[\frac{ x^{\prime} }{ c - v } = t.\]If we insert the value of $t$,
\[\begin{eqnarray} \xi & = & ac \left( t - \frac{ v }{ c^{2} - v^{2} }x^{\prime} \right) \nonumber \\ & = & ac \left( \frac{ x^{\prime} }{ c - v }x^{\prime} - \frac{ v }{ c^{2} - v^{2} }x^{\prime} \right) \nonumber \\ & = & ac \left( \frac{ 1 }{ c - v } - \frac{ v }{ c^{2} - v^{2} } \right) x^{\prime} \nonumber \\ & = & ac \left( \frac{ c }{ c^{2} - v^{2} } \right) x^{\prime} . \nonumber \end{eqnarray}\]In an analogous manner by considering rays moing along the two other axes, when
\[\frac{ y }{ \sqrt{ c^{2} - v^{2} } } = t, \ x^{\prime} = 0,\]then
\[\begin{eqnarray} \eta & = & c \tau \nonumber \\ & = & ac \left( t - \frac{ v }{ c^{2} - v^{2} } x^{\prime} \right) \nonumber \end{eqnarray}\]Hence,
\[\begin{eqnarray} \eta & = & ac t \nonumber \\ & = & ac \frac{ y }{ \sqrt{ c^{2} - v^{2} } } \nonumber . \end{eqnarray}\]Similary,
\[\begin{eqnarray} \zeta & = & ac \frac{ z }{ \sqrt{ c^{2} - v^{2} } } \nonumber . \end{eqnarray}\]Substituting for $x^{\prime} = x - vt$,
\[\begin{eqnarray} \beta & := & \frac{ 1 }{ \sqrt{ 1 - v^{2} / c^{2} } } \nonumber \\ & = & \frac{ c }{ \sqrt{ c^{2} - v^{2} } } \nonumber \\ \phi(v) & := & a \beta, \nonumber \end{eqnarray}\] \[\begin{eqnarray} \tau & = & a \left( t - \frac{ v }{ c^{2} - v^{2} } x^{\prime} \right) \nonumber \\ & = & a \left( t - \frac{ v }{ c^{2} - v^{2} } (x - vt) \right) \nonumber \\ & = & a \left( t + \frac{ v^{2} }{ c^{2} - v^{2} } t - \frac{ v }{ c^{2} - v^{2} } x \right) \nonumber \\ & = & a \left( \beta^{2}t - \frac{ v }{ c^{2} - v^{2} } x \right) \nonumber \\ & = & a \left( \beta^{2}t - v \beta^{2} \frac{1}{c^{2}} x \right) \nonumber \\ & = & \phi(v) \beta \left( t - \frac{v}{c^{2}} x \right) \label{equation_relation_tau} . \end{eqnarray}\] \[\begin{eqnarray} \xi & = & a \frac{ c^{2} }{ c^{2} - v^{2} } x^{\prime} \nonumber \\ & = & a \beta^{2} (x - vt) \nonumber \\ & = & \phi(v) \beta (x - vt) . \label{equation_relation_xi} \end{eqnarray}\] \[\begin{eqnarray} \eta & = & \frac{ a c }{ \sqrt{c^{2} - v^{2}} }y \nonumber \\ & = & a \beta y \nonumber \\ & = & \phi(v) y . \label{equation_relation_eta} \end{eqnarray}\] \[\begin{eqnarray} \zeta & = & \phi(v) z . \label{equation_relation_zeta} \end{eqnarray}\]We have to prove that the principle of the constancy of the velocity of light iscompatible iwth the principle of relativity.
Let a spherical wave be emitted at $t = \tau = 0$ with the velocity $c$ in stationary system $K$. $(x, y, z)$ is a point attained by this wave.
\[\begin{eqnarray} \sqrt{x^{2} + y^{2} + z^{2}} = ct \nonumber & \Leftrightarrow & x^{2} + y^{2} + z^{2} = c^{2} t^{2} \nonumber . \end{eqnarray}\]From the equations above,
\[\begin{eqnarray} v \tau + \xi & = & \phi(v) \beta \left( vt - \frac{v^{2}}{c^{2}}x +x - vt \right) \nonumber \\ & = & \phi(v) \beta x \left( 1 - \frac{v^{2}}{c^{2}} \right) \nonumber \\ & = & \phi(v) \beta x \frac{1}{\beta^{2}} \nonumber \\ \Leftrightarrow \quad x & = & \frac{\beta}{\phi(x)} \left( v \tau + \xi \right) \nonumber \end{eqnarray}\]Similary,
\[\begin{eqnarray} \tau + \frac{v}{c^{2}}\xi & = & \phi(v) \beta \left( t - \frac{v}{c^{2}}x + \frac{v}{c^{2}}x - t \frac{v^{2}}{c^{2}} \right) \nonumber \\ & = & \phi(v) \beta t \left( 1 - \frac{v^{2}}{c^{2}} \right) \nonumber \\ & = & \phi(v) \beta t \frac{1}{\beta^{2}} \nonumber \\ \Leftrightarrow \quad t & = & \frac{\beta}{\phi(x)} \left( \tau + \frac{v}{c^{2}}\xi \right) \nonumber \end{eqnarray}\]Other equations are as follows.
\[\begin{eqnarray} y & = & \frac{y}{\phi(v)}, \nonumber \\ \zeta & = & \frac{z}{\phi(v)}. \nonumber \end{eqnarray}\]Substituting the equations into the above equation,
\[\begin{eqnarray} & & x^{2} +y^{2} +z^{2} = c^{2} t^{2} \nonumber \\ & \Leftrightarrow & \frac{\beta^{2}}{\phi^{2}(v)} \left( v \tau + \xi \right)^{2} + \frac{\eta^{2}}{\phi^{2}(v)} + \frac{\zeta^{2}}{\phi^{2}(v)} = c^{2} \beta^{2} \frac{\beta^{2}}{\phi^{2}(v)} \left( \tau + \frac{v}{c^{2}} \xi \right)^{2} \nonumber \\ & \Leftrightarrow & \beta^{2} \left( v \tau + \xi \right)^{2} - c^{2} \beta^{2} \left( \tau + \frac{v}{c^{2}} \xi \right)^{2} + \eta^{2} + \zeta^{2} = 0 \nonumber \\ & \Leftrightarrow & \beta^{2} \left( v \tau + \xi + c \left( \tau + \frac{v}{c^{2}} \xi \right) \right) \left( v \tau + \xi - c \left( \tau + \frac{v}{c^{2}} \xi \right) \right) + \eta^{2} + \zeta^{2} = 0 \nonumber \end{eqnarray} .\]Arraging the first term further, we will obtain
\[\begin{eqnarray} & & \beta^{2} \left( v \tau + \xi + c \tau + \frac{v}{c} \xi \right) \left( v \tau + \xi - c \tau - \frac{v}{c} \xi \right) \nonumber \\ & = & \beta^{2} \left( (v + c) \tau + \left( 1 + \frac{v}{c} \right) \xi \right) \left( (v - c)\tau + \left( 1 - \frac{v}{c} \right) \xi \right) \nonumber \\ & = & \beta^{2} \left( (v + c)(v - c) \tau ^{2} + (v + c) \left( 1 - \frac{v}{c} \right) \tau \xi + (v - c) \left( 1 + \frac{v}{c} \right) \xi \tau + \left( 1 + \frac{v}{c} \right) \left( 1 - \frac{v}{c} \right) \xi^{2} \right) \nonumber \\ & = & \beta^{2} \left( (v^{2} - c^{2}) \tau ^{2} + \left( v + c - \frac{v^{2}}{c} - v \right) \tau \xi + \left( v - c + \frac{v^{2}}{c} - v \right) \xi \tau + \left( 1 - \frac{v^{2}}{c^{2}} \right) \xi^{2} \right) \nonumber \\ & = & \beta^{2} \frac{c^{2}}{\beta^{2}} \tau ^{2} + \beta^{2} \left( \left( c - \frac{v^{2}}{c} \right) \tau \xi + \left( \frac{v^{2}}{c} - c \right) \xi \tau \right) + \beta^{2} \frac{1}{\beta^{2}} \xi^{2} \nonumber \\ & = & c^{2} \tau ^{2} + \xi^{2} \nonumber \end{eqnarray}\]Combining the results,
\[\begin{eqnarray} x^{2} + y^{2} + z^{2} = c^{2} t^{2} \Leftrightarrow \xi^{2} + \eta^{2} + \zeta^{2} = c^{2} \tau^{2} . \nonumber \end{eqnarray}\]Now we will determine an unkown function $\phi$ of $v$.
- $K^{\prime}$ be a third system of co-ordinates
- coordinates in $K^{\prime}$ is $(\Xi, H, Z)$,
- relatively to the moving system $k$ is in a state of parallel translatory motion parallel to the axis of $\Xi$
- moves with velocity $-v$
- $x^{\prime}, y^{\prime}, z^{\prime}$,
By using \(\eqref{equation_relation_tau}\),
\[\begin{eqnarray} t^{\prime} & = & \phi(-v) \beta(-v) (\tau + v \frac{\xi}{c^{2}}) \nonumber \\ & = & \phi(-v) \beta(-v) \left( \phi(v) \beta(v) \left( t - \frac{v}{c^{2}}x \right) + \frac{v}{c^{2}} \left( \phi(v) \beta(v) (x - vt) \right) \right) \nonumber \\ & = & \phi(-v) \phi(v) \beta^{2} \left( t - \frac{v}{c^{2}}x + \frac{v}{c^{2}} x - \frac{v^{2}}{c^{2}} t \right) \nonumber \\ & = & \phi(-v) \phi(v) \beta^{2} t \frac{1}{\beta^{2}} \nonumber \\ & = & \phi(-v) \phi(v) t . \end{eqnarray}\]By using \(\eqref{equation_relation_xi}\) and \(\eqref{equation_relation_tau}\),
\[\begin{eqnarray} x^{\prime} & = & \phi(-v) \beta(-v) (\xi + v \tau) \nonumber \\ & = & \phi(-v) \beta(-v) \left( \phi(v)\beta(v) (x - vt) + v \phi(v)\beta(v) \left( t - \frac{v}{c^{2}}x \right) \right) \nonumber \\ & = & \phi(-v) \phi(v) \beta^{2} \left( x - vt + v t - \frac{v^{2}}{c^{2}}x \right) \nonumber \\ & = & \phi(-v) \phi(v) \beta^{2} x \frac{1}{\beta^{2}} \nonumber \\ & = & \phi(-v) \phi(v) x . \nonumber \end{eqnarray}\]By using \(\eqref{equation_relation_eta}\) and \(\eqref{equation_relation_zeta}\), ,
\[\begin{eqnarray} y^{\prime} & = & \phi(-v) \eta \nonumber \\ & = & \phi(-v) \phi(v) y \nonumber \\ z^{\prime} & = & \phi(-v) \zeta \nonumber \\ & = & \phi(-v) \phi(v) z \nonumber . \end{eqnarray}\]Since the relations between $x^{\prime}, y^{\prime}, z^{\prime}$ and $x, y, z$ do not contain the time $t$, it is clear that the transformation from $K$ to $K^{\prime}$ must be the identical transformation.
Thus,
\[\begin{equation} \phi(v)\phi(-v) = 1 \label{equation_relation_phi_1} \end{equation} .\]Let $l$ be the length of a rigid rod. To determine $\phi(v)$, we will consider a ridid rod moving parallel to $x$ axis with velocity $v$ relatively to stationary system $K$ lying between $(\xi, \eta, \zeta) = (0, 0, 0)$ and $(\xi, \eta, \zeta) = (0, l, 0)$. The coordinates of the end of the rod in $K$ is $(x_{1}, y_{1}, z_{1})$
By using \(\eqref{equation_relation_xi}\), \(\eqref{equation_relation_eta}\), \(\eqref{equation_relation_zeta}\), the coordinates of moving system $k$ can be transformed into $K$ by the following relations.
\[\begin{eqnarray} & & \xi = \phi(v) \beta (x - vt) \nonumber \\ & \Leftrightarrow & x = \frac{ \xi }{ \phi(v) \beta } + vt \nonumber \end{eqnarray}\] \[\begin{eqnarray} & & \eta = \phi(v) y \nonumber \\ & \Leftrightarrow & y = \frac{ \eta }{ \phi(v) } . \nonumber \end{eqnarray}\] \[\begin{eqnarray} & & \zeta = \phi(v) z \nonumber \\ & \Leftrightarrow & z = \frac{ \zeta }{ \phi(v) } . \nonumber \end{eqnarray}\]Using the relations, we can determine the coordinates of the rods in stationary system $K$. For $(\xi, \eta, \zeta) = (0, l, 0)$,
\[\begin{eqnarray} x_{1} & = & vt, \nonumber \\ y_{1} & = & \frac{l}{\phi(v)} \nonumber \\ z_{1} & = & 0. \nonumber \end{eqnarray}\]For $(\xi, \eta, \zeta) = (0, 0, 0)$,
\[\begin{eqnarray} x_{2} & = & vt, \nonumber \\ y_{2} & = & 0 \nonumber \\ z_{2} & = & 0 . \nonumber \end{eqnarray}\]The lenght of the lod in stationary system $K$ is $l / \phi(v)$. From reasons of symmetry, the length of the moving rod measured in the stationary ssytem $K$ does not change if $v$ and $-v$ are interchanged.
\[\begin{eqnarray} & & \frac{l}{\phi(v)} = \frac{l}{\phi(-v)} \nonumber \\ & \Leftrightarrow & \phi(v) = \phi(-v) . \label{equation_relation_phi_2} \end{eqnarray}\]It follows that $\phi(v) = 1$ from \(\eqref{equation_relation_phi_1}\) and \(\eqref{equation_relation_phi_2}\). Therefore,
\[\begin{eqnarray} \tau & = & \beta (t - vx / c^{2}), \\ \xi & = & \beta (x - vt) \\ \eta & = & y \\ \zeta & = & z . \end{eqnarray}\]