(54) 
$\Psi _{4}=i({\mathfrak {w}}_{x}\Psi _{1}+{\mathfrak {w}}_{y}\Psi _{2}+{\mathfrak {w}}_{z}\Psi _{3})$ 
The vector $\Psi$ is perpendicular to w; we can call it the Magnetic restforce.
Relations analogous to these hold among the quantities $iwF^{*},{\mathfrak {M,E,w}}$ and Relation (D) can be replaced by the formula
{D} 
$wF^{*}=\mu wf^{*}$ 
We can use the relations (C) and (D) to calculate F and f from $\Phi$ and $\Psi$, we have
$wF=\Phi ,\ wF^{*}=i\mu \Psi ,\ wf=\epsilon \Phi ,\ wf^{*}=i\Psi$
and applying the relation (45) and (46), we have
(55) 
$F=[w,\Phi ]+i\mu [w,\Psi ]^{*}$, 
(56) 
$f=\epsilon [w,\Phi ]+i[w,\Psi ]^{*}$, 
i.e.

$F_{12}=(w_{1}\Phi _{2}w_{2}\Phi _{1})+i\mu (w_{3}\Psi _{4}w_{4}\Psi _{3})$, etc.
$f_{12}=\epsilon (w_{1}\Phi _{2}w_{2}\Phi _{1})+i(w_{3}\Psi _{4}w_{4}\Psi _{3})$, etc.

Let us now consider the spacetime vector of the second kind $[\Phi \Psi ]$, with the components

$\Phi _{2}\Psi _{3}\Phi _{3}\Psi _{2},\ \Phi _{3}\Psi _{1}\Phi _{1}\Psi _{3},\ \Phi _{1}\Psi _{2}\Phi _{2}\Psi _{1}$,
$\Phi _{1}\Psi _{4}\Phi _{4}\Psi _{1},\ \Phi _{2}\Psi _{4}\Phi _{4}\Psi _{2},\ \Phi _{3}\Psi _{4}\Phi _{4}\Psi _{3}$,

Then the corresponding spacetime vector of the first kind
$w[\Phi ,\Psi ]=(w{\overline {\Psi }})\Phi +w({\overline {\Phi }})\Psi$
vanishes identically owing to equations 49) and 53).
Let us now take the vector of the 1st kind
(57) 
$\Omega =iw[\Phi ,\ \Psi ]^{*}$ 
with the components
$\Omega _{1}=i\left{\begin{array}{ccc}w_{2},&w_{3},&w_{4}\\\Phi _{2},&\Phi _{3},&\Phi _{4}\\\Psi _{2},&\Psi _{3},&\Psi _{4}\end{array}}\right$, etc.
Then by applying rule (45), we have