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\begin{document}

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\centerline{\bf PROBLEM 47}
\medskip

\smallskip
The Matrix $L$ is defined by:

\begin{center}
$L=\left( 
\begin{array}{cccc}
l_{\ 0}^{0} & l_{\ 1}^{0} & l_{\ 2}^{0} & l_{\ 3}^{0} \\ 
l_{\ 0}^{1} & l_{\ 1}^{1} & l_{\ 2}^{1} & l_{\ 3}^{1} \\ 
l_{\ 0}^{2} & l_{\ 1}^{2} & l_{\ 2}^{2} & l_{\ 3}^{2} \\ 
l_{\ 0}^{3} & l_{\ 1}^{3} & l_{\ 2}^{3} & l_{\ 3}^{3}
\end{array}
\right) .$

\bigskip 
\end{center}

(a)--Calculate $-gL$. \ (b)--Write down the transpose matrix
$\widetilde{L}$.\ (c)--Calculate$\widetilde{L}g$. \ (d)--Compare
(a) and (c) to find the general form of $L$ ({\it i.e.} use
$\widetilde{L}g=-gL$).

\bigskip 

(a)--The matrix g is defined by:

\begin{center}
$g=\left( 
\begin{array}{cccc}
1 & 0 & 0 & 0 \\ 
0 & -1 & 0 & 0 \\ 
0 & 0 & -1 & 0 \\ 
0 & 0 & 0 & -1
\end{array}
\right) .$
\end{center}

\bigskip 

\bigskip --By matrix multiplication:

\begin{center}
$-gL=\left( 
\begin{array}{cccc}
-l_{\ 0}^{0} & -l_{\ 1}^{0} & -l_{\ 2}^{0} & -l_{\ 3}^{0} \\ 
l_{\ 0}^{1} & l_{\ 1}^{1} & l_{\ 2}^{1} & l_{\ 3}^{1} \\ 
l_{\ 0}^{2} & l_{\ 1}^{2} & l_{\ 2}^{2} & l_{\ 3}^{2} \\ 
l_{\ 0}^{3} & l_{\ 1}^{3} & l_{\ 2}^{3} & l_{\ 3}^{3}
\end{array}
\right) .$

\bigskip 
\end{center}

(b)--The transpose of $L$ is:

\begin{center}
$\widetilde{L}=\left( 
\begin{array}{cccc}
l_{\ 0}^{0} & l_{\ 0}^{1} & l_{\ 0}^{2} & l_{\ 0}^{3} \\ 
l_{\ 1}^{0} & l_{\ 1}^{1} & l_{\ 1}^{2} & l_{\ 1}^{3} \\ 
l_{\ 2}^{0} & l_{\ 2}^{1} & l_{\ 2}^{2} & l_{\ 2}^{3} \\ 
l_{\ 3}^{0} & l_{\ 3}^{1} & l_{\ 3}^{2} & l_{\ 3}^{3}
\end{array}
\right) .$

\bigskip 
\end{center}

(c)--By matrix multiplication:

\begin{center}
$\widetilde{L}g=\left( 
\begin{array}{cccc}
l_{\ 0}^{0} & -l_{\ 0}^{1} & -l_{\ 0}^{2} & -l_{\ 0}^{3} \\ 
l_{\ 1}^{0} & -l_{\ 1}^{1} & -l_{\ 1}^{2} & -l_{\ 1}^{3} \\ 
l_{\ 2}^{0} & -l_{\ 2}^{1} & -l_{\ 2}^{2} & -l_{\ 2}^{3} \\ 
l_{\ 3}^{0} & -l_{\ 3}^{1} & -l_{\ 3}^{2} & -l_{\ 3}^{3}
\end{array}
\right) .$

\bigskip 
\end{center}

(d)--Set $-gL=\widetilde{L}g$ and compare the components:

\begin{equation} \label{matrix_l_equations}
\left( \begin{array}{cccc}
-l_{\ 0}^{0}=l_{\ 0}^{0} & -l_{\ 1}^{0}=-l_{\ 0}^{1} & -l_{\ 2}^{0}=-l_{\
0}^{2} & -l_{\ 3}^{0}=-l_{\ 0}^{3} \\ 
l_{\ 0}^{1}=l_{\ 1}^{0} & l_{\ 1}^{1}=-l_{\ 1}^{1} & l_{\ 2}^{1}=-l_{\ 1}^{2}
& l_{\ 3}^{1}=-l_{\ 1}^{3} \\ 
l_{\ 0}^{2}=l_{\ 2}^{0} & l_{\ 1}^{2}=-l_{\ 2}^{1} & l_{\ 2}^{2}=-l_{\ 2}^{2}
& l_{\ 3}^{2}=-l_{\ 2}^{3} \\ 
l_{\ 0}^{3}=l_{\ 3}^{0} & l_{\ 1}^{3}=-l_{\ 3}^{1} & l_{\ 2}^{3}=-l_{\ 3}^{2}
& l_{\ 3}^{3}=-l_{\ 3}^{3}
\end{array} \right)\ .
\end{equation}

\smallskip

--The resulting matrix is:

\begin{equation} \label{matrix_l_result}
\left( \begin{array}{cccc}
0 & l_{\ 1}^{0} & l_{\ 2}^{0} & l_{\ 3}^{0} \\ 
l_{\ 1}^{0} & 0 & l_{\ 2}^{1} & l_{\ 3}^{1} \\ 
l_{\ 2}^{0} & -l_{\ 2}^{1} & 0 & l_{\ 3}^{2} \\ 
l_{\ 3}^{0} & -l_{\ 3}^{1} & -l_{\ 3}^{2} & 0
\end{array} \right)\ .
\end{equation}

\smallskip

(e)--Obtain the same result by discussing the elements of
$ g^{\alpha\beta}\, \tilde{l}_{\beta}^{\ \gamma}\,
   g_{\gamma\delta} = - l^{\alpha}_{\ \delta}$.
Performing the contractions we get
$\tilde{l}^{\alpha}_{\ \delta}=-l^{\alpha}_{\ \delta}$. Using the 
definition of $\tilde{l}^{\alpha}_{\ \delta}$ we have the equations
$$ \tilde{l}^{\alpha}_{\ \delta} = l_{\delta}^{\ \alpha} 
                                 = -l^{\alpha}_{\ \delta}\,, $$ 
which are identical with the equations (\ref{matrix_l_equations}), 
namely (no summations, $a=0,1,2,3$,
$i=1,2,3$ and $j=1,2,3$):
$$ \tilde{l}^a_{\ a} = l_a^{\ a} = + l^a_{\ a} \Rightarrow
   l^a_{\ a} = - l^a_{\ a} = 0\ , $$
$$ \tilde{l}^i_{\ 0} = l_0^{\ i} = - l^0_{\ i} \Rightarrow
   l^0_{\ i} = + l^i_{\ 0}\ , $$
$$ \tilde{l}^i_{\ j} = l_j^{\ i} = + l^j_{\ i} \Rightarrow
   l^j_{\ i} = - l^i_{\ j} $$
and the resulting matrix is again (\ref{matrix_l_result}).

\vfil
\end{document}

