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\markright{{\sf  The Lorentz Transformation}}
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\begin{document}


\section{11.3\quad The Lorentz Transformation.}

\subsection{11.3.1}

\quad \thinspace \textbf{a. }The coordinate transformations that leave
invariant the form of the wave equations, $\square \psi =0,$ form a group.
We begin by discussing some simple examples of such transformations in this
section and defer a fuller account to Sections. 11.6 and 11.7.

\smallskip\ 

\textbf{b}. It is convenient to proceed by analogy with the familiar groups
of $3$-dimensional Euclidean space. It is well known that the square of the
distance between two points, $\left| \mathbf{x-y}\right|
^{2}=(x_{1}-y_{1})^{2}+(x_{2}-y_{2})^{2}+(x_{3}-y_{3})^{2}$ has the same
invariance group as the equation $\nabla ^{2}u=0.$ This group includes
translations, proper rotations and the inversion, or parity, operation. If
we fix $\mathbf{y,}$ say $\mathbf{y}=0,$ only rotations around axes through
point $\mathbf{y,}$ say the origin, are left. (The word ``rotations''
includes proper rotations and their products with the inversion). Taking the
rotation axis to be $x_{3},$ a proper rotation through an angle $\vartheta $
is described by 
\begin{eqnarray}
x_{1}^{\prime } &=&x_{1}\cos \vartheta +x_{2}\sin \vartheta   \label{1} \\
x_{2}^{\prime } &=&-x_{1}\sin \vartheta +x_{2}\cos \vartheta   \nonumber
\end{eqnarray}

\smallskip\ 

\textbf{c.} In the four-dimensional space spanned by $x_{0}=ct,$ $x_{1},$ $%
x_{2}$ and $x_{3},$ the transformations that leave $\square \psi $ invariant
are, analogously, those that leave invariant the quadratic form 
\begin{equation}
(x_{0}-y_{0})^{2}-\left[
(x_{1}-y_{1})^{2}+(x_{2}-y_{2})^{2}+(x_{3}-y_{3})^{2}\right] .  \label{2}
\end{equation}
These transformations include space and time translations, space and time
inversion and the generalization to space-time of proper rotations.

\newpage

\subsection{11.3.2}

\quad \quad \textbf{a. }Space and time inversion (parity and time reversal)
will be discussed later. Translations in $4$-d correspond to a shift in
origin and can be eliminated from consideration by fixing $y_i=0$ $%
(i=0,1,2,3).$ Among the transformations that leave $x_0^2-\left| \mathbf{x}%
\right| ^2$ invariant are obviously all the rotations, with $t^{\prime }=t,$
and in particular (\ref{1}) supplemented by $x_3^{\prime }=x_3,$ $%
x_0^{\prime }=x_0.$

\smallskip\ 

\textbf{b.} The simplest ``new'' transformations are those that leave two
space-like coordinates, say $x_{2}$ and $x_{3},$ unchanged. To find these
transformations, and for other formal manipulations, it is convenient to
define, following Minkowski, an \textsl{imaginary} fourth coordinate $%
x_{4}=ix_{0}=ict.$ We are then looking for transformations that leave
invariant $x_{1}^{2}+x_{2}^{2}+x_{3}^{2}+x_{4}^{2},$ or just $%
x_{1}^{2}+x_{4}^{2}$ in the particular case $x_{2}^{\prime
}=x_{2},x_{3}^{\prime }=x_{3}.$ Such a transformation is clearly a
``rotation'' of the form (\ref{2}) 
\begin{eqnarray}
\quad \quad x_{1}^{\prime } &=&\quad \quad x_{1}\cos \vartheta +x_{4}\sin
\vartheta   \label{3} \\
ict^{\prime } &\equiv &x_{4}^{\prime }=-x_{1}\sin \vartheta +x_{4}\cos
\vartheta   \nonumber
\end{eqnarray}

\textbf{c. }The novelty comes from the requirement that $x_{1}^{\prime }$
and $x_{0}^{\prime },$ or $ct^{\prime },$ must of course be real. This
implies that $\vartheta $ must be purely imaginary: $\vartheta =i\zeta $
with real $\zeta $. We have then 
\begin{eqnarray}
ct^{\prime } &=&x_{0}^{\prime }=x_{0}\cosh \zeta -x_{1}\sinh \zeta  \label{4}
\\
x_{1}^{\prime } &=&-x_{0}\sinh \zeta +x_{1}\cosh \zeta  \nonumber
\end{eqnarray}

\textbf{d.} This simple transformation is called a \textsl{Lorentz boost }in
the $x_{1}$ direction, with $\zeta $ the \textsl{boost} \textsl{parameter}
or \textsl{rapidity}. These names are suggested by the physical meaning of
the transformation, which becomes clear if we note that $x^{\prime }=0,$ the
origin of the $K^{\prime }$ system, corresponds to the relation $x_{0}\sinh
\zeta =x_{1}\cosh \zeta $ in the $K$ system. It follows that the $K^{\prime
} $ system is moving with constant velocity $v=c\tanh \zeta $ in the $x_{1}$
direction, as seen from the $K$ system. This conclusion is confirmed by the
fact that, for small $\zeta ,$ (\ref{4}) becomes the Galilean transformation 
$t^{\prime }=t,$ $x_{1}^{\prime }=x_{1}-vt,$ up to terms of order $%
v^{2}/c^{2}.$

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\newpage

\subsection{11.3.3}

\quad \quad \textbf{a. }Following common use (and Jackson) we define $\beta
=v/c=\tanh \zeta $ and $\gamma =(1-\beta ^{2})^{-1/2}=\cosh \zeta $ (hence $%
\gamma \beta =\sinh \zeta )$ and rewrite (\ref{4}) as 
\begin{eqnarray}
x_{0}^{\prime } &=&\gamma (x_{0}-\beta x_{/\!/})  \label{5} \\
x_{/\!/}^{\prime } &=&\gamma (x_{/\!/}-\beta x_{0})  \nonumber \\
\mathbf{x}_{\perp }^{\prime } &=&\mathbf{x}_{\perp }.  \nonumber
\end{eqnarray}
The subscripts $//$ and $\perp $ refer to the direction of the velocity.
This change of notation shows how to write a Lorentz boost in a general
direction. If we want to write (\ref{5}) in pure vector notation, we can
introduce $\mathbf{\beta =v/}c$ and use $\beta x_{/\!/}=\mathbf{\beta \cdot x%
}$ and $x_{\perp }=\mathbf{x-}\left( \mathbf{\beta \cdot x}\right) \mathbf{%
\beta /}\beta ^{2}.$ The result is Jackson's (11.19). 
\begin{eqnarray}
x_{0}^{\prime } &=&\gamma (x_{0}-\mathbf{\beta \cdot x})  \label{5b} \\
\mathbf{x}^{\prime } &=&\mathbf{x}+\frac{\gamma -1}{\beta ^{2}}\,(\mathbf{%
\beta \cdot x})\mathbf{\beta }-\gamma \mathbf{\beta }x_{0}  \nonumber
\end{eqnarray}
Using the identity $\gamma ^{2}-1=\beta ^{2},$ we can write $\mathbf{x}%
^{\prime }=\mathbf{x}+\dfrac{1}{\gamma +1}\,(\mathbf{\beta \cdot x})\mathbf{%
\beta }-\gamma \mathbf{\beta }x_{0}$.

\smallskip \ 

\textbf{b. }The inverse transformation is obtained by changing the sign of 
\textbf{v}, or $\mathbf{\beta }.$ It is a good exercise to verify
algebraically that the inverse of (\ref{5}) is 
\begin{eqnarray}
x_{0} &=&\gamma (x_{0}^{\prime }+\beta x_{/\!/})  \label{5c} \\
x_{/\!/} &=&\gamma (x_{/\!/}^{\prime }+\beta x_{0})  \nonumber \\
\mathbf{x}_{\perp } &=&\mathbf{x}_{\perp }^{\prime }.  \nonumber
\end{eqnarray}

\textbf{c. }Simple as it is, the Lorentz transformation is hard to grasp
intuitively. As an example, suppose that a light signal is flashed at a time 
$t=t^{\prime }=0$ at a common origin of $t$ and $t^{\prime }$. The relation $%
\left| \mathbf{x}\right| ^{2}-c^{2}t^{2}=\left| \mathbf{x}^{\prime }\right|
^{2}-c^{2}t^{\prime 2}$ implies that at later times both $K$ and $K^{\prime
},$ in their own coordinates, see themselves at the center of an expanding
spherical light front. The sketch shows what one would expect classically if
the signal is emitted in $K$ (solid circle) or in $K^{\prime }$ (dashed
circle) and travels with speed $c$ with respect to the emitter.

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\newpage

\subsection{11.3.4}

\quad \quad \textbf{a. }\textsl{The Fitzgerald-Lorentz} contraction is seen
to follow from a proper interpretation of the Lorentz transformation (\ref{5}%
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Lengths measured in $K$ are by definition values of $\left| s\right| $ at a
given time $t,$ so that $s^{2}=-\left| \Delta \mathbf{x}\right| ^{2}.$ If
the origins of $K$ and $K^{\prime }$ coincide at $t=t^{\prime }=0,$ the end
of the bar in $K^{\prime }$ (see sketch) is not coincident with the end of
an identical bar in $K$ at $t=0.$ Further, and this is crucial, a clock
attached to $K^{\prime }$ at $A^{\prime }$ will not show $t^{\prime }=0.$
Rather, according to (\ref{5})\ 
\begin{eqnarray*}
t^{\prime } &=&-v\gamma x/c^{2} \\
x^{\prime } &=&\gamma x
\end{eqnarray*}
hence $x=\gamma ^{-1}x^{\prime }=\sqrt{1-\beta ^{2}}x^{\prime }.$ If $%
x^{\prime }=1$ m and $\beta =0.6,$ $x=80$ cm. That means that a goat $1$ m
long can be trapped in a stall $1$ m long by shutting both ends of the stall
at the same time, if the goat goes fast enough. But the shutting must be
done at the same time according to the stall's clocks. The passing goat sees
the stall shortened by $\sqrt{1-\beta ^{2}}$ and will get hit on the head
and in the tail if he shuts the gates at the same time according to \textsl{%
his} clock. Note that $t=0$ corresponds to $t^{\prime }=-vx^{\prime }/c^{2}.$
Since $\left| x^{\prime }/t^{\prime }\right| =c^{2}/v>c,$ the shutting of
one gate is unaffected by the shutting of the other at an earlier time. Put
it another way, it takes at least a time $x^{\prime }/c$ for the tail of the
goat to know that a gate has been shut on the goat's nose.

\smallskip\ 

\textbf{b. }The \textsl{time dilation} also follows from (\ref{5}), with
proper interpretation. Now the moving clock has a fixed coordinate (say $%
x^{\prime }=0)$ in the \textsl{moving} system. One must solve

\begin{tabular}[b]{l}
$t^{\prime }=\gamma \left( t-vx/c^{2}\right) $ \\ 
$0=\gamma \left( x-vt\right) $ \\ 
\\ 
\end{tabular}
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\ 

or use (\ref{3}) with $x^{\prime }=0.$ Either way, $t^{\prime }=t/\gamma ,$
or $t=\gamma t^{\prime }.$

\newpage

\subsection{11.3.5}

\quad \quad \textbf{a. }By analogy with three-dimensional rotations, the
general Lorentz transformation is of the form $x_{\alpha }^{\prime
}=R_{\alpha \beta }\,x_{\beta }$ ($\alpha ,\beta =1,2,3,4$), with $R_{\alpha
\beta }$ a 4-dimensional orthogonal matrix: $R_{\alpha \beta }R_{\beta
\gamma }=\delta _{\alpha \gamma }$ (repeated indices are summed). These
transformations form the Lorentz group, which is discussed in more detail in
Section 11.7.

\ 

\textbf{b.} Any group of four quantities $A_{\alpha }$ that transform like
the cartesian coordinates and $ict$ is by definition a vector in the
4-dimensional \textsl{Minkowski space}. The three spatial components $%
A_{1},\,A_{2},\,A_{3}$ transform like $x_{1},\,x_{2},\,x_{3}$ under ordinary
(3-d) rotations and identify an ordinary 3-vector $\mathbf{A}$. In order to
keep the formal analogy with the 3-dimensional euclidean space, one must use
a fourth component $A_{4}$ that is imaginary when it corresponds to a real
physical quantity. The modern preference is to use a real fourth component $%
A_{0},$ with $A_{4}=iA_{0}$.

\ 

\textbf{c.} The ``squared length'' of a 4-vector is naturally defined by $%
\sum_{\alpha }(A_{\alpha })^{2}=\left| \mathbf{A}\right| ^{2}-\left(
A_{0}\right) ^{2}.$ This implies that the scalar product of two 4-vectors is 
$\sum_{\alpha }A_{\alpha }B_{\alpha }=\mathbf{A\cdot B-}A_{0}B_{0}.$ One can
directly prove that this is a \textsl{Lorentz invariant}: by using the
orthogonality property $R_{\alpha \beta }R_{\alpha \gamma }=\delta \,_{\beta
\gamma }\,$(sum implied): 
\[
A_{\alpha }^{\prime }B_{\alpha }^{\prime }=R_{\alpha \beta }A_{\beta
}R_{\alpha \gamma }B_{\gamma }=\delta \,_{\beta \gamma }A_{\beta }B_{\gamma
}=A_{\beta }B_{\beta } 
\]

\ 

\textbf{d.} In Minkowski space, the quantity $\sum_{\alpha }(A_{\alpha
})^{2}=\left| \mathbf{A}\right| ^{2}-\left( A_{0}\right) ^{2}$ is not
positive definite. We are thus free to fix a sign in making a reasonable
definition of the ``length'' of a 4-vector. Perversely, Jackson (along with
most particle physicists) picks $\left( A_{0}\right) ^{2}-\left| \mathbf{A}%
\right| ^{2}$ as the definition, so that for instance the ``distance
squared'' between two space-time points is defined as $s^{2}=\left(
x_{0}\right) ^{2}-\left| \mathbf{x}\right| ^{2}=c^{2}t^{2}-r^{2}.$ The
scalar product of two 4-vectors \textsf{A }and \textsf{B }is then 
\[
\mathsf{A}\cdot \mathsf{B}=A_{0}B_{0}-\mathbf{A\cdot B} 
\]

\newpage

\subsection{11.3.6}

\quad \quad \textbf{a. }\textsl{The addition of velocities} is given by the
formulae 
\begin{equation}
u_{/\!/}=\frac{u_{/\!/}^{\prime }+v}{1+\dfrac{\mathbf{v\cdot u}^{\prime }}{%
c^{2}}}\quad \quad \quad \quad \quad u_{\perp }=\frac{u_{\perp }^{\prime }}{%
\gamma _{v}\left( 1+\dfrac{\mathbf{v\cdot u}^{\prime }}{c^{2}}\right) }
\label{4a}
\end{equation}
with $u_{\alpha }=dx_{\alpha }/dt,$ $u_{\alpha }^{\prime }=dx_{\alpha
}^{\prime }/dt^{\prime },\gamma _{v}=(1-v^{2}/c^{2})^{-1/2}.$ This is easily
obtained by differentiating (\ref{5c}). For small speeds, (\ref{4a}) reduces
to the Galilean $\mathbf{u}^{\prime }=\mathbf{u+v.}$

\smallskip\ 

\textbf{b.} In accord with Einstein's postulates, $u$ can never exceed $c$
if $u^{\prime }$ and $v$ are $\leq c.$ More than that, $u=c$ if $u^{\prime }$
\textsl{or} $v$ equal $c.$ This is particularly easy to see for parallel
velocities: 
\[
u=\frac{u^{\prime }+v}{1+\dfrac{vu^{\prime }}{c^{2}}}\,.
\]
For perpendicular velocities 
\[
u_{\perp }=u^{\prime }\sqrt{1-v^{2}/c^{2}},\quad \quad u_{/\!/}=v.
\]

\smallskip\ 

\textbf{c. }If $u_{\perp }/u_{/\!/}=\tan \vartheta $ and $u_{\perp }^{\prime
}/u_{/\!/}^{\prime }=\tan \vartheta ^{\prime },$ we have from (\ref{4a}) 
\begin{equation}
\tan \vartheta =\frac{u^{\prime }\sin \vartheta ^{\prime }}{\gamma
_{v}(u^{\prime }\cos \vartheta ^{\prime }+v)}  \label{5a}
\end{equation}

\smallskip\ 

\textbf{d.} The relativistic explanation for stellar aberrations is
contained in (\ref{5a}). For a star at the zenith ($\vartheta ^{\prime }=\pi
/2),$ $\tan \vartheta =(c/v)\sqrt{1-v^{2}/c^{2}}.$ The aberration angle is $%
\alpha =\pi /2-\vartheta .$ Thus $\tan \alpha \simeq \alpha =v/c$ to leading
order in $v/c,$ in agreement with observations (see 1.2a).

\smallskip\ 

\textbf{e. }The Fizeau experiment (see 1.2b) is explained by the addition
formula for parallel velocities: 
\[
u=\frac{\dfrac{c}{n}\pm v}{1\pm \dfrac{vc}{nc^{2}}}\simeq \left( \frac{c}{n}%
\pm v\right) \left( 1\mp \frac{v}{nc}\right) =\frac{c}{n}\pm v\left( 1-\frac{%
1}{n^{2}}\right) .
\]

\newpage

\subsection{11.3.7}

\quad \quad \textbf{a. }The \textsl{relativistic Doppler shift formulae} are
simply derived from the requirement that the phase $\omega t-\mathbf{k\cdot x%
}$ must be a Lorentz invariant, since it counts the number of crests in the
wave. It follows that $(\omega /c,\,k_{1,}\,k_{2}\,,k_{3})$ must be the
components of a $4$-vector $k_{\alpha },$ so that the phase can be written
as the invariant scalar product $k_{\alpha }x_{\alpha }=\mathsf{k\cdot x\,}.$

\smallskip\ 

\textbf{b.} The transformation properties of $k_{\alpha }$ are then, with $%
k_{0}=\omega /c$%
\begin{eqnarray}
k_{0}^{\prime } &=&\frac{\omega ^{\prime }}{c}=\gamma \left( k_{0}-\beta
k_{/\!/}\right)   \label{6} \\
k_{/\!/}^{\prime } &=&\gamma \left( k_{/\!/}-\beta k_{0}\right)   \nonumber
\\
\mathbf{k}_{\perp }^{\prime } &=&\mathbf{k}_{\perp }\quad \quad \quad \text{%
(see (\ref{5}))}  \nonumber
\end{eqnarray}
For light, $\omega =c\left| \mathbf{k}\right| ,$ or $k_{/\!/}=k_{0}\cos
\vartheta ,$ $k_{\perp }=k_{0}\sin \vartheta ,$ so that 
\begin{equation}
\omega ^{\prime }=\gamma \omega \left( 1-\frac{v}{c}\cos \vartheta \right)
,\quad \quad \quad \quad \quad \tan \vartheta ^{\prime }=\frac{k_{\perp
}^{\prime }}{k_{/\!/}^{\prime }}=\frac{\sin \vartheta }{\gamma \left( \cos
\vartheta -v/c\right) }.  \label{7}
\end{equation}

\smallskip\ 

\textbf{c.} We see from (\ref{7}) that there is a longitudinal Doppler shift
($\vartheta =0)$%
\[
\omega ^{\prime }=\omega \frac{1-v/c}{\sqrt{1-v^{2}/c^{2}}}=\omega \sqrt{%
\frac{1-v/c}{1+v/c}}
\]
as well as a transverse Doppler shift ($\vartheta ^{\prime }=\pi /2)$%
\[
\omega ^{\prime }=\gamma \omega \left( 1-\frac{v^{2}}{c^{2}}\right) =\omega 
\sqrt{1-\frac{v^{2}}{c^{2}}}
\]

\smallskip

\textbf{d.} For wave propagation in a classical medium, there is no
transverse Doppler shift. The longitudinal shift depends on whether the
source $S,$ the medium $M,$ or the observer $O$ move (with respect to the
other two), according to the scheme: \smallskip 

\begin{tabular}{ccccccc}
&  & $\omega ^{\prime }$ &  & $k^{\prime }$ &  & $\omega ^{\prime
}/k^{\prime }$ \\ 
&  &  &  &  &  &  \\ 
$O\text{ moving (away from }S)$ &  & $\omega (1-v/c)$ &  & $k$ &  & $c-v$ \\ 
&  &  &  &  &  &  \\ 
$S\text{ moving (away from }O)$ &  & $\omega /(1+v/c)$ &  & $k/(1+v/c)$ &  & 
$c$ \\ 
&  &  &  &  &  &  \\ 
$M\text{ moving (from }O\text{ to }S)$ &  & $\omega $ &  & $k/(1-v/c)$ &  & $%
c-v$ \\ 
&  &  &  &  &  &  \\ 
$\text{%
\begin{tabular}{c}
$\text{Relativistic result}$ \\ 
$O\text{ moving or }S$ moving
\end{tabular}
}$ &  & $\omega \sqrt{\dfrac{1-v/c}{1+v/c}}$ &  & $k\sqrt{\dfrac{1-v/c}{1+v/c%
}}$ &  & $c$%
\end{tabular}

\newpage

\subsection{11.3.8}

\quad \quad \textbf{a. }\textsl{The Michelson-Morley experiment} is
explained trivially by the fact that source and observer are at rest with
respect to each other. The use of starlight is irrelevant, due to the
considerations of 1.6a.

\smallskip

\textbf{b. }The aberration of starlight is explained (again) by (\ref{7}),
which is identical to the inverse of (\ref{5a}) with $u=c.$

\smallskip

\textbf{c. }In solving problems of relativistic kinematics, it is often
convenient to introduce a frame where the system (or at least its center,
properly defined in next section) is instantaneously at rest. This is called
the rest frame.

\smallskip

\textbf{d.} Denoting this comoving system by $K^{\prime }$, we see at once,
from $ds^{2}=c^{2}dt^{\prime \,2}-d\mathbf{x}^{\prime \,2}$ and $d\mathbf{x}%
^{\prime }=0,$ that $dt^{\prime }=ds/c$ is a Lorentz invariant quantity,
called the \textsl{proper time}. Since $ds$ is also equal to $cdt/\gamma $ $%
\left( \text{from }ds^{2}=c^{2}dt^{2}-d\mathbf{x}^{2}=c^{2}\left(
1-v^{2}/c^{2}\right) dt^{2}\right) $ it follows that the time intervals $%
\Delta t$ are related to proper time intervals $\Delta t^{\prime }$ by 
\begin{equation}
\Delta t=\int_{0}^{\Delta t^{\prime }}\gamma dt^{\prime }=\int_{0}^{\Delta
t^{\prime }}\frac{dt^{\prime }}{\sqrt{1-\dfrac{v^{2}}{c^{2}}}}.  \label{8}
\end{equation}

\smallskip

\textbf{e. }This formula is the generalization of the time-dilation formula
of 3.4b to arbitrary (non-uniform) relative motion. It is particularly
useful, of course, if the velocity $v$ (of $K^{\prime }$ relative to $K)$ is
known as a function of $t^{\prime },$ i.e. if the kinematic problem has been
solved in the rest frame of the moving system.

\smallskip

\textbf{f. }The above considerations are sufficient for a full discussion of
the twin paradox. The twin that travels (non-uniformly) is younger when he
rejoins his inertial twin. General relativity is not needed to explain the
effect.

\end{document}
