(75) 
$X_{x}={\frac {1}{2}}({\mathfrak {m}}_{x}{\mathfrak {M}}_{x}{\mathfrak {m}}_{y}{\mathfrak {M}}_{y}{\mathfrak {m}}_{z}{\mathfrak {M}}_{z}+{\mathfrak {e}}_{x}{\mathfrak {E}}_{x}{\mathfrak {e}}_{y}{\mathfrak {E}}_{y}{\mathfrak {e}}_{z}{\mathfrak {E}}_{z})$,
$X_{y}={\mathfrak {m}}_{x}{\mathfrak {M}}_{y}+{\mathfrak {e}}_{y}{\mathfrak {E}}_{x},\ Y_{x}={\mathfrak {m}}_{y}{\mathfrak {M}}_{x}+{\mathfrak {e}}_{x}{\mathfrak {E}}_{y}$, u.s.f.
$X_{t}={\mathfrak {e}}_{y}{\mathfrak {M}}_{z}{\mathfrak {e}}_{z}{\mathfrak {M}}_{y}$,
$T_{x}={\mathfrak {m}}_{z}{\mathfrak {E}}_{y}{\mathfrak {m}}_{y}{\mathfrak {E}}_{z}$, u.s.f.
$T_{t}={\frac {1}{2}}({\mathfrak {m}}_{x}{\mathfrak {M}}_{x}+{\mathfrak {m}}_{y}{\mathfrak {M}}_{y}+{\mathfrak {m}}_{z}{\mathfrak {M}}_{z}+{\mathfrak {e}}_{x}{\mathfrak {E}}_{x}+{\mathfrak {e}}_{y}{\mathfrak {E}}_{y}+{\mathfrak {e}}_{z}{\mathfrak {E}}_{z})$,

(76) 
$L={\frac {1}{2}}({\mathfrak {m}}_{x}{\mathfrak {M}}_{x}+{\mathfrak {m}}_{y}{\mathfrak {M}}_{y}+{\mathfrak {m}}_{z}{\mathfrak {M}}_{z}{\mathfrak {e}}_{x}{\mathfrak {E}}_{x}{\mathfrak {e}}_{y}{\mathfrak {E}}_{y}{\mathfrak {e}}_{z}{\mathfrak {E}}_{z})$, 
These quantities are all real. In the theory for bodies at rest, the combinations ($X_{x},\ X_{y},\ X_{z},\ Y_{x},\ Y_{y},\ Y_{z},\ Z_{x},\ Z_{y},\ Z_{z}$ are known as Maxwell's Stresses", $T_{x},\ T_{y},\ T_{z}$ are known as the Poynting's Vector, $T_{t}$ as the electromagnetic energydensity, and L as the Langrangian function.
On the other hand, by multiplying the alternating matrices of f and F, we obtain
(77) 
$F^{*}f^{*}=\left{\begin{array}{llll}S_{11}L,&S_{12},&S_{13},&S_{14}\\S_{21},&S_{22}L,&S_{23},&S_{23}\\S_{31},&S_{32},&S_{33}L,&S_{34}\\S_{41},&S_{42},&S_{43},&S_{44}L\end{array}}\right$ 
and hence, we can put
(78) 
$fF=SL,\ F^{*}f^{*}=SL$, 
where by L, we mean Ltimes the unit matrix, i.e. the matrix with elements
$\leftLe_{hk}\right\ \left({\begin{array}{c}e_{hh}=1,\ e_{hk}=0,\ h\gtrless k\\h,k=1,2,3,4\end{array}}\right)$
Since here $SL=LS$, we deduce that,
$F^{*}f^{*}fF=(SL)(SL)=SS+L^{2}$,
and find, since $f^{*}f=Det^{\frac {1}{2}}f,\ F^{*}F=Det{\frac {1}{2}}F$, we arrive at the interesting conclusion
(79) 
$SS=L^{2}Det^{\frac {1}{2}}fDet^{\frac {1}{2}}F$, 
i.e. the product of the matrix S into itself can be expressed as the multiple of a unit matrix — a matrix in which all the elements except those in the principal diagonal are zero, the elements in the principal diagonal are all equal