Page:Elektrische und Optische Erscheinungen (Lorentz) 083.jpg

and by

$Rot'{\mathfrak {A}}$

a vector with the components

$\left({\frac {\partial {\mathfrak {A}}_{z}}{\partial y}}\right)^{'}-\left({\frac {\partial {\mathfrak {A}}_{y}}{\partial z}}\right)^{'}$ etc.

The introduction of t' and ${\mathfrak {D}}'$ gives the advantage, that (as I will show now) the equations ($I_{c}$) — ($V_{c}$) assume the same form as the formulas that apply to ${\mathfrak {p}}=0$.

§ 57. At first we obtain, by consideration of formulas (35),

$Div\ {\mathfrak {D}}=Div'\ {\mathfrak {D}}-{\frac {1}{V^{2}}}\left({\mathfrak {p}}_{x}{\dot {\mathfrak {D}}}_{x}+{\mathfrak {p}}_{y}{\dot {\mathfrak {D}}}_{y}+{\mathfrak {p}}_{z}{\dot {\mathfrak {D}}}_{z}\right)$,

or by ($III_{c}$), if we replace (in the terms multiplied by ${\mathfrak {p}}_{x},{\mathfrak {p}}_{y},{\mathfrak {p}}_{z}$) ${\mathfrak {H}}'$ by ${\mathfrak {H}}$ and Div by $Div'$

$Div\ {\mathfrak {D}}=Div'\ {\mathfrak {D}}-{\frac {1}{4\pi V^{2}}}\left\{{\mathfrak {p}}_{x}[Rot\ {\mathfrak {H}}]_{x}+{\mathfrak {p}}_{y}[Rot\ {\mathfrak {H}}]_{y}+{\mathfrak {p}}_{z}[Rot\ {\mathfrak {H}}]_{z}\right\}=$ $=Div'\ {\mathfrak {D}}+{\frac {1}{4\pi V^{2}}}Div\ [{\mathfrak {p.H}}]=Div'\ {\mathfrak {D}}'$.

Hence the equation ($I_{c}$) becomes

$Div'\ {\mathfrak {D}}'=0$

$(I_{d})$

In a similar way

$Div\ {\mathfrak {H}}=Div'\ {\mathfrak {H}}-{\frac {1}{V^{2}}}\left({\mathfrak {p}}_{x}{\dot {\mathfrak {H}}}_{x}+{\mathfrak {p}}_{y}{\dot {\mathfrak {H}}}_{y}+{\mathfrak {p}}_{z}{\dot {\mathfrak {H}}}_{z}\right)$

i.e., by ($IV_{c}$),

$Div\ {\mathfrak {H}}=Div'\ {\mathfrak {H}}+{\frac {1}{V^{2}}}\left\{{\mathfrak {p}}_{x}[Rot\ {\mathfrak {E}}]_{x}+{\mathfrak {p}}_{y}[Rot\ {\mathfrak {E}}]_{y}+{\mathfrak {p}}_{z}[Rot\ {\mathfrak {E}}]_{z}\right\}=$ $=Div'\ {\mathfrak {H}}-{\frac {1}{V^{2}}}Div[{\mathfrak {p.E}}]=Div'\ {\mathfrak {H}}'$,

so that it can be written for ($II_{c}$)

$Div'\ {\mathfrak {H}}'=0$

($IId$)

Now let us turn to formula ($III_{c}$). In this one,