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%TCIDATA{Created=Wed Oct 30 23:13:37 1996}
%TCIDATA{LastRevised=Sat Nov 02 15:17:05 1996}
%TCIDATA{Language=American English}

\input tcilatex
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\markright{{\sf  Lorentz matrices}}
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\begin{document}


\section{11.7\quad Lorentz Matrices.}

\subsection{11.7.1}

\quad \thinspace \textbf{a. }Formally, the Lorentz transformations $%
x^{\prime \,\alpha }=\Lambda _{\,\,\,\beta }^{\alpha }\,x^{\beta }$
(summation implied) is characterized by the invariance $\mathsf{x}^{\prime }
\cdot\, \mathsf{x}^{\prime }=\mathsf{x}\cdot \mathsf{x}$ for any \textsf{x}
(more generally, $\mathsf{A}^{\prime }\cdot \mathsf{B}^{\prime }=\mathsf{A}%
\cdot \mathsf{B}$ for any two vectors). In matrix notation, $\underline{x}%
^{\prime }=\underline{\Lambda }\,\underline{x}$ and $\mathsf{x}\cdot \mathsf{%
x}=\underline{\widetilde{x}}\,\underline{g}\,\underline{x},$ where 
\[
\underline{\widetilde{x}}=(x^{0}\,\,x^{1}\,\,x^{2}\,\,x^{3})\quad \quad
\quad \underline{x}=\left( 
\begin{tabular}{l}
$x^{0}$ \\ 
$x^{1}$ \\ 
$x^{2}$ \\ 
$x^{3}$%
\end{tabular}
\right) \quad \quad \quad \underline{g}=\left( 
\begin{tabular}{llll}
$+1$ & $\,\,\,\,0$ & $\,\,\,\,0$ & $\,\,\,\,0$ \\ 
$\,\,\,\,0$ & $-1$ & $\,\,\,\,0$ & $\,\,\,\,0$ \\ 
$\,\,\,\,0$ & $\,\,\,\,0$ & $-1$ & $\,\,\,\,0$ \\ 
$\,\,\,\,0$ & $\,\,\,\,0$ & $\,\,\,\,0$ & $-1$%
\end{tabular}
\right) \,. 
\]
Then $\mathsf{x}^{\prime }\cdot \mathsf{x}^{\prime } = \widetilde{\,(%
\underline{\Lambda }\,\underline{x})}\,\underline{g}\,\underline{\Lambda }\,%
\underline{x},$ and equating this to $\widetilde{\underline{x}}\,\underline{g%
}\,\underline{x}$ we obtain 
\begin{equation}
\widetilde{\underline{\Lambda }}\,\underline{g}\,\underline{\Lambda }=%
\underline{g}  \label{1}
\end{equation}
or, with $\widetilde{\Lambda }_{\alpha }^{\,\,\,\beta }=\Lambda
_{\,\,\,\alpha }^{\beta },$%
\[
\Lambda _{\,\,\,\alpha }^{\beta }\,g_{\beta \gamma }\,\Lambda _{\,\,\,\delta
}^{\gamma }=g_{\alpha \delta }. 
\]
Compare $\widetilde{\underline{R}}\,\underline{R}=\underline{1}$, or $%
R_{\,\,a}^{b}\,\delta _{bc}\,R_{\,\,d}^{c}=\delta _{ad},$ for
three-dimensional orthogonal matrices $\,\underline{R}$ (rotations) that
leave $\mathbf{x}^{\prime }\cdot\mathbf{x}^{\prime }=\mathbf{x\cdot x.}$

\smallskip\ 

\textbf{b}. Taking the determinant of both sides of (\ref{1}) we find 
\[
\det \,\underline{\Lambda }=\pm 1. 
\]
Also, the time-time component of (\ref{1}) gives 
\[
\left( \Lambda _{\,\,\,0}^{0}\right) ^{2}=1+\sum_{i}\left( \Lambda
_{\,\,\,0}^{i}\right) ^{2}, 
\]
so that $\Lambda _{\,\,\,0}^{0}$ is either $\geq 1$ or $\leq -1.$

The $\,\underline{\Lambda }$ matrices can then be classified according to
the signs of $\det \,\underline{\Lambda }$ and of $\Lambda _{\,\,\,0}^{0}.$
The \textsl{proper} Lorentz transformations have $\det \,\underline{\Lambda }%
=+1$ and $\Lambda _{\,\,\,0}^{0}$ $\geq 1$; they form a continuous group,
the proper Lorentz group, since the identity and the infinitesimal $\,%
\underline{\Lambda }$'s are proper.

From now on, $\,\underline{\Lambda }$ will denote a proper matrix; all the
other (improper) Lorentz matrices are of the forms $\underline{P}\,%
\underline{\Lambda }$, $\underline{T}\,\underline{\Lambda }$ or $\underline{P%
}\,\underline{T}\,\underline{\Lambda },$ where $\underline{P}$ is the parity
operation $(\mathbf{x}^{\prime }=-\mathbf{x}) $ and $\underline{T}$ the time
reversal ($t^{\prime }=-t$). Note that $\det \,\underline{P}=-1$, $%
P_{\,\,\,0}^{0}=1$ and that $\det \underline{T}=-1$, $T\,_{\,\,\,0}^{0}=-1$. 
$\underline{T}$ turns the light cone ``upside down''. Compare the
classification of rotations $\,\underline{R}$ into proper ($\det \,%
\underline{R}=1$) and improper ($\det \,\underline{R}=-1$), with $\,%
\underline{R}_{improper}= \underline{P}\,\underline{R}_{proper}.$

\newpage

\subsection{11.7.2}

\quad \quad \textbf{a. }The proper Lorentz group is a six-parameter Lie
group. It takes six real numbers to identify any matrix $\,\underline{%
\Lambda },$ because (\ref{1}) gives $10$ independent relations among the $16$
$(=4\times 4)$ elements of any $\,\underline{\Lambda }.$ By comparison, the
rotation group is a three-parameter Lie group, because each $\,\underline{R}$
has $9$ $(=3\times 3)$ elements connected by six relations. The three
parameters can be taken to be the Euler angles, or the three components of
the ``rotation vector'' $\mathbf{\omega }$ This is a vector directed along
the rotation axis and having magnitude $\omega $ equal to the rotation angle.

\smallskip\ 

\textbf{b.} We shall next find a representation of $\,\underline{\Lambda }$
that generalizes the well known representation of $\,\underline{R}$%
\[
\,\underline{R}=\exp \left[ -i\mathbf{\omega }\cdot \underline{\mathbf{%
\sigma }}\right] =\exp \left[ -i\omega ^{1}\,\underline{\sigma }_{1}-i\omega
^{2}\,\underline{\sigma }_{2}-i\omega ^{3}\,\underline{\sigma }_{3}\right] 
\]
with 
\[
i\,\underline{\sigma }_{1}=\left( 
\begin{tabular}{lll}
0 & 0 & \thinspace \thinspace \thinspace \thinspace 0 \\ 
0 & 0 & $-1$ \\ 
0 & 1 & \thinspace \thinspace \thinspace \thinspace 0
\end{tabular}
\right) \quad \quad i\,\underline{\sigma }_{2}=\left( 
\begin{tabular}{lll}
\thinspace \thinspace \thinspace \thinspace 0 & 0 & 1 \\ 
\thinspace \thinspace \thinspace \thinspace 0 & 0 & 0 \\ 
$-1$ & 0 & 0
\end{tabular}
\right) \quad \quad i\,\underline{\sigma }_{3}=\left( 
\begin{tabular}{lll}
0 & $-1$ & 0 \\ 
1 & \thinspace \thinspace \thinspace \thinspace 0 & 0 \\ 
0 & \thinspace \thinspace \thinspace \thinspace 0 & 0
\end{tabular}
\right) . 
\]
Note that $\underline{1}+i\omega ^{3}\underline{\sigma }_{3},$ for instance,
represents an infinitesimal rotation by $\omega ^{3}$ (counterclockwise)
about the $x^{3}$ axis. Note also that $i\,\underline{\sigma }_{3}$ has the
same effect as $\partial /\partial \varphi ,$ or $x^{1}\partial /\partial
x^{2}-x^{2}\partial /\partial x^{1},$ in the sense that 
\[
\frac{\partial }{\partial \varphi }\left( 
\begin{tabular}{l}
$x^{1}$ \\ 
$x^{2}$ \\ 
$x^{3}$%
\end{tabular}
\right) =\frac{\partial }{\partial \varphi }\left( 
\begin{tabular}{l}
$r\sin \vartheta \cos \varphi $ \\ 
$r\sin \vartheta \sin \varphi $ \\ 
$r\cos \vartheta $%
\end{tabular}
\right) =\left( 
\begin{tabular}{l}
$-\,x^{2}$ \\ 
$\,\,\,\,x^{1}$ \\ 
$\,\,\,\,0$%
\end{tabular}
\right) =i\,\underline{\sigma }_{3}\left( 
\begin{tabular}{l}
$x^{1}$ \\ 
$x^{2}$ \\ 
$x^{3}$%
\end{tabular}
\right) . 
\]
Thus $\hbar \underline{\mathbf{\sigma }}$ is the angular momentum, and we
can take over many results from quantum mechanics, for instance 
\[
\left[ \,\underline{\sigma }_{i},\,\underline{\sigma }_{j}\right]
=i\,\epsilon _{ijk}\,\underline{\sigma }_{k}\quad \text{or\quad \quad }%
\mathbf{\,\underline{\mathbf{\sigma }}\times \underline{\mathbf{\sigma }}}=i%
\mathbf{\,\underline{\mathbf{\sigma }}.} 
\]
The eigenvalues of $\,\underline{\sigma }_{3}$ are $0$ and $\pm 1$, with
eigenvectors $x^{3}$ and $(x^{1}\pm ix^{2})/\sqrt{2}$. Thus we are in the $%
l=1$ representation of angular momentum. It can be checked that, for
instane, 
\[
\,\left( \underline{\sigma }_{3}\right) _{diagonalized}=\left( 
\begin{tabular}{lll}
1 & \thinspace \thinspace \thinspace \thinspace 0 & 0 \\ 
0 & $-1$ & 0 \\ 
0 & \thinspace \thinspace \thinspace \thinspace 0 & 0
\end{tabular}
\right) \quad \quad \text{so that\quad }\left( \underline{\sigma }%
_{3}\right) ^{2}=\left( 
\begin{tabular}{lll}
1 & 0 & 0 \\ 
0 & 1 & 0 \\ 
0 & 0 & 0
\end{tabular}
\right) 
\]
so that $\underline{\mathbf{\sigma }}\,\cdot \mathbf{\underline{\sigma }}%
=l(l+1)\,\underline{1}=\underline{2}$ (twice the unit matrix) and $\left( 
\underline{\sigma }_{i}\right) ^{3}=\underline{\sigma }_{i}\,.$

\newpage

\subsection{11.7.3}

\quad \quad \textbf{a. }By analogy with $\,\underline{R}=\exp (i\mathbf{%
\omega \cdot \underline{\sigma }})$ we put $\,\underline{\Lambda }=\exp (%
\underline{L})$ and we determine $\,\underline{L}$ by imposing the condition 
$\widetilde{\,\underline{\Lambda }}\,\underline{g}\,\underline{\Lambda }=%
\underline{g}.$ This is equivalent to $\underline{g}\,\widetilde{\,%
\underline{\Lambda }}\,\underline{g}=\underline{\Lambda }^{-1},$ or $\exp (\,%
\underline{g}\widetilde{\,\underline{L}}\,\underline{g})=\exp (-\,\underline{%
L}),$ which gives $\underline{g}\widetilde{\,\underline{L}}\,\underline{g}%
=-\,\underline{L},$ or $\widetilde{\,\underline{g}\,\underline{L}}=-\,%
\underline{g}\,\underline{L}.$ Explicitly, 
\[
\underline{g}\,\underline{L}=\left( 
\begin{tabular}{llll}
$-\,L_{\,\,\,0}^{0}$ & $-\,L_{\,\,\,1}^{0}$ & $-\,L_{\,\,\,2}^{0}$ & $%
-\,L_{\,\,\,3}^{0}$ \\ 
$\,\,\,\,L_{\,\,\,0}^{1}$ & $\,\,\,\,\,L_{\,\,\,1}^{1}$ & $%
\,\,\,\,\,L_{\,\,\,2}^{1}$ & $\,\,\,\,L_{\,\,\,3}^{1}\,$ \\ 
$\,\,\,\,L_{\,\,\,0}^{2}$ & $\,\,\,\,L_{\,\,\,1}^{2}$ & $\,\,\,\,L_{\,\,%
\,2}^{2}$ & $\,\,\,\,L_{\,\,\,2}^{2}$ \\ 
$\,\,\,\,L_{\,\,\,0}^{3}$ & $\,\,\,\,L_{\,\,\,1}^{3}$ & $\,\,\,\,L_{\,\,%
\,2}^{3}$ & $\,\,\,L_{\,\,\,3}^{3}$%
\end{tabular}
\right) 
\]
and the condition $\widetilde{\underline{g}\underline{L}}=-\,\underline{g}\,%
\underline{L}$ requires that: the diagonal elements $L_{\,\,\,\alpha
}^{\alpha }$ must vanish, $L_{\,\,\,a}^{0}=L_{\,\,\,0}^{a},$ and $%
L_{\,\,\,b}^{a}=-\,L_{\,\,\,a\,}^{b}.$ Thus $\,\underline{L}$ can be written
in the form 
\[
\underline{L}=-\mathbf{\omega \cdot \underline{\mathbf{S}}}-\mathbf{\zeta
\cdot }\,\underline{\mathbf{K}}\,, 
\]
where 
\[
\underline{K}_{1}=\left( 
\begin{tabular}{llll}
0 & 1 & 0 & 0 \\ 
1 & 0 & 0 & 0 \\ 
0 & 0 & 0 & 0 \\ 
0 & 0 & 0 & 0
\end{tabular}
\right) \quad \quad \underline{K}_{2}=\left( 
\begin{tabular}{llll}
0 & 0 & 1 & 0 \\ 
0 & 0 & 0 & 0 \\ 
1 & 0 & 0 & 0 \\ 
0 & 0 & 0 & 0
\end{tabular}
\right) \quad \quad \underline{K}_{3}=\left( 
\begin{tabular}{llll}
0 & 0 & 0 & 1 \\ 
0 & 0 & 0 & 0 \\ 
0 & 0 & 0 & 0 \\ 
1 & 0 & 0 & 0
\end{tabular}
\right) 
\]
and $\underline{S}_{j}$ are the angular momentum matrices $i\underline{%
\sigma }_{j}$, bordered by a row and a column of zeros to make them $4\times
4:$ 
\[
\underline{S}_{1}=\left( 
\begin{tabular}{llll}
0 & 0 & \thinspace 0 & \thinspace \thinspace \thinspace 0 \\ 
0 & 0 & 0 & \thinspace \thinspace \thinspace 0 \\ 
0 & 1 & $\,$0 & $-1$ \\ 
0 & 0 & \thinspace 1 & \thinspace \thinspace \thinspace \thinspace 0
\end{tabular}
\right) \quad \quad \underline{S}_{2}=\left( 
\begin{tabular}{llll}
0 & \thinspace \thinspace \thinspace \thinspace 0 & 0 & 0 \\ 
0 & \thinspace \thinspace \thinspace 0 & 0 & 1 \\ 
0 & $\,\,\,$0 & $\,$0 & 0 \\ 
0 & $-1$ & \thinspace 0 & 0
\end{tabular}
\right) \quad \quad \underline{S}_{3}=\left( 
\begin{tabular}{llll}
0 & 0 & \thinspace \thinspace \thinspace \thinspace 0 & 0 \\ 
0 & 0 & $-1$ & 0 \\ 
0 & 1 & $\,\,\,\,0$ & 0 \\ 
0 & 0 & \thinspace \thinspace \thinspace \thinspace 0 & 0
\end{tabular}
\right) . 
\]

\smallskip

\textbf{b.} The matrices $\underline{K}_{j}$ have eigenvalues $1,-\,1$ and $0
$ (doubly degenerate). It is easy to check that $\left( \underline{K}%
_{1}\right) ^{3}=\underline{K}_{1},$ or more generally 
\begin{equation}
\left( \,\underline{\mathbf{K}}\mathbf{\cdot \epsilon }\right) ^{3}=%
\underline{\mathbf{K}}\mathbf{\cdot \epsilon \quad \quad \quad }\left( \,%
\underline{\mathbf{S}}\mathbf{\cdot \epsilon }\right) ^{3}=-\,\,\underline{%
\mathbf{S}}\mathbf{\cdot \epsilon }  \label{2}
\end{equation}
for any unit vector $\mathbf{\epsilon }$.

We also note that all the $\underline{S}$ and $\underline{K}$ matrices are
traceless, hence $\limfunc{Tr}\,\underline{L}=0,$as it must be to satisfy $%
\det \,\underline{\Lambda }=\exp (\limfunc{Tr}\,\underline{L})$ with $\det \,%
\underline{\Lambda }=1.$

\newpage

\subsection{11.7.4}

\quad \quad \textbf{a. }Expanding $\exp (-\,\zeta \underline{K}_{1})$ and
using (\ref{2}) one finds 
\[
\underline{\Lambda }=\exp (-\,\zeta \underline{K}_{1})=(1-\underline{K}%
_{1}^{2})-\underline{K}_{1}\sinh \zeta +\underline{K}_{1}^{2}\cosh \zeta 
\]
or explicitly, 
\[
\underline{\Lambda }=\left( 
\begin{tabular}{llll}
$\,\,\,\,\cosh \zeta $ & $-\sinh \zeta $ & 0 & 0 \\ 
$-\sinh \zeta $ & $\,\,\,\,\cosh \zeta $ & 0 & 0 \\ 
\quad \quad 0 & \quad \quad 0 & 1 & 0 \\ 
\quad \quad 0 & \quad \quad 0 & 0 & 1
\end{tabular}
\right) . 
\]

\smallskip

\textbf{b.} Comparison with the well-known $x^{\prime \,0}=\gamma
(x^{0}-vx^{1}/c^{2}),$ $x^{\prime \,1}=\gamma (x^{1}-vx^{0})$ gives $\tanh
\,\zeta =v/c,$ $\cosh \,\zeta =\gamma .$

\smallskip

\textbf{c.} Rotations and boots do \textsl{not} commute in general. The
basic commutators for the six infinitesimal generators of the group, i.e.
the matrices $\,\underline{K}_{i}$, $\,\underline{S}_{i},$ are

\begin{eqnarray*}
\left[ \,\,\,\underline{S}_{i}\,,\,\underline{S}_{j}\right]
&=&\,\,\,\,\epsilon _{ijk}\,\,\underline{S}_{k} \\
\left[ \,\,\,\underline{S}_{i}\,,\,K_{j}\right] &=&\,\,\,\,\epsilon _{ijk}\,%
\underline{K}_{k} \\
\left[ \underline{K}_{i}\,,\,\underline{K}_{j}\right] &=&-\,\epsilon
_{ijk}\,\,\underline{S}_{k}
\end{eqnarray*}
(all summations reduce to just one term).

\smallskip

\textbf{d.} Note that $\,\underline{\Lambda }=\exp (-\,\mathbf{\omega \cdot }%
\,\,\,\underline{\mathbf{S}}\mathbf{\,-}$\textbf{$\,$}$\mathbf{\zeta \cdot }%
\,\,\,\underline{\mathbf{K}})$ does \textsl{not} express $\,\underline{%
\Lambda }$ as the product of a rotation times a boost, because in general $%
\exp (-\,\mathbf{\omega \cdot }\,\,\,\underline{\mathbf{S}}\mathbf{\,-}$%
\textbf{$\,$}$\mathbf{\zeta \cdot }\,\,\,\underline{\mathbf{K}})$ is \textsl{%
not} equal to $\exp (-\,\mathbf{\omega \cdot }\,\,\,\underline{\mathbf{S}}%
)\exp (\mathbf{-}$\textbf{$\,$}$\mathbf{\zeta \cdot }\,\,\,\underline{%
\mathbf{K}}).$

\smallskip

\textbf{e.} Given a general $\,\underline{\Lambda },$ how does one recognize
if it is a pure boost \underline{$B$} or a pure rotation $\,\underline{R}$
(or neither)?

\begin{itemize}
\item  A pure boost \underline{$B$}$\,=\exp (\mathbf{-\,\zeta \cdot }\,\,%
\underline{\mathbf{K}})$ corresponds to a \textsl{symmetric} matrix, $%
\widetilde{\underline{B}}=\underline{B}.$ The first row, $B_{\,\,\,\alpha
}^{0}$ is actually equal to $u_{\alpha }$, or more explicitly, $%
B_{\,\,\,\,0}^{0}=\gamma =u_{0},$ $B_{\,\,\,\,a}^{0}=-\gamma v^{a}=u_{a}.$
More about this on next page.

\item  A pure rotation $\,\underline{R}=\exp (-\mathbf{\omega \cdot }\,\,%
\underline{\mathbf{S}})$ is of the form 
\[
\left( 
\begin{tabular}{llll}
1 & 0 & 0 & 0 \\ 
0 &  &  &  \\ 
0 &  & $R_{\,\,\,\,b}^{a}$ &  \\ 
0 &  &  & 
\end{tabular}
\right) 
\]
where $R_{\,\,\,\,b}^{a}$ is a three dimensional orthogonal matrix, $%
R_{\,\,a}^{b}\,\delta _{bc}\,R_{\,\,d}^{c}=\delta _{ad}$.
\end{itemize}

\newpage

\subsection{11.7.5 Boost and rotation}

\quad \quad \textbf{a. }We can write a generic Lorentz matrix \underline{$%
\Lambda $} as the product \underline{$R$}$\,\underline{B},$ where \underline{%
$R$} is a rotation and \underline{$B$} a boost. The generic boost matrix can
be deduced from Section 3.3a (or Jackson, eq. (11.98)): 
\[
B_{\;\beta }^{\alpha }=\left( 
\begin{array}{cc}
\gamma  & -\gamma v^{b} \\ 
-\gamma v^{a} & \delta _{\;b}^{a}+\gamma ^{2}v^{a}v^{b}/(\gamma +1)
\end{array}
\right) 
\]
where $v^{a}$ are the Cartesian components of the boost velocity $\mathbf{v}$
and $\gamma =\sqrt{1-v^{2}/c^{2}}$. On the other hand, \underline{$R$} is of
the form 
\[
R_{\;\beta }^{\alpha }=\left( 
\begin{array}{cc}
1 & 0 \\ 
0 & R_{\;b}^{a}
\end{array}
\right) 
\]
where $R_{\;b}^{a}$ is a three-dimensional rotation matrix. The problem is
to determine \underline{$B$} and $\underline{R},$ given \underline{$\Lambda $%
}$.$

We can keep the indices consistently placed (up or down) and rewrite 
\underline{$B$} more compactly by introducing the 4-velocity $u^{\alpha
}=dx^{\alpha }/d\tau :$%
\[
B_{\;\beta }^{\alpha }=\left( 
\begin{array}{cc}
\gamma  & u_{b} \\ 
-\,u^{a} & \delta _{\;b}^{a}-u^{a}u_{b}/(\gamma +1)
\end{array}
\right) 
\]
$u_{a}=-u^{a}$ are covariant components for $a=1,2,3$ (and of course $%
u_{0}=u^{0}=\gamma $). We find 
\[
\left( 
\begin{array}{cc}
\Lambda _{\;0}^{0} & \Lambda _{\;b}^{0} \\ 
\Lambda _{\;0}^{a} & \Lambda _{\;b}^{a}
\end{array}
\right) =R_{\;c}^{\alpha }B_{\;\beta }^{c}=\left( 
\begin{array}{cc}
\gamma  & u_{b} \\ 
-R_{\;c}^{a}u^{c} & R_{\;b}^{a}-R_{\;c}^{a}u^{c}u_{b}/(\gamma +1)
\end{array}
\right) 
\]
hence $\gamma =\Lambda _{\;0}^{0},\;u_{b}=\Lambda
_{\;b}^{0},\;R_{\;c}^{a}u^{c}=-\,\Lambda _{\;0}^{a},$ and $%
R_{\;b}^{a}=\Lambda _{\;b}^{a}+R_{\;c}^{a}u^{c}u_{b}/(\gamma +1).$ From
these we get the explicit formulas:
\[
v^{b}=\frac{u^{b}}{\gamma }=-\frac{\Lambda _{\;b}^{0}}{\Lambda _{\;0}^{0}}
\]
\[
R_{\;b}^{a}=\Lambda _{\;b}^{a}-\Lambda _{\;0}^{a}\Lambda _{\;b}^{0}/(\Lambda
_{\;0}^{0}+1)
\]

\end{document}
