# Page:BatemanElectrodynamical.djvu/11

1909.]
283
The transformations of the electrodynamical equations.

have $\theta ^{2}=1$ and eighteen relations of the types

 ${\frac {\partial (y',z')}{\partial (x,y)}}=\theta {\frac {\partial (x',t')}{\partial (z,t)}}$ ${\frac {\partial (y',z')}{\partial (x,t)}}=-\theta {\frac {\partial (x',t')}{\partial (y,z)}}$ There are clearly nine relations of the first type and nine relations of the second type. We shall now show that when $\theta ^{2}=1$ these relations imply that there is a relation of the form

$dx'^{2}+dy'^{2}+dz'^{2}-dt'^{2}=\lambda ^{2}\left(dx^{2}+dy^{2}+dz^{2}-dt^{2}\right);$ for this purpose we shall require the following lemma.

Lemma. — Let the sixteen quantities

$\left(\alpha _{1}\alpha _{2}\alpha _{3}\alpha _{4}\right),\ \left(\beta _{1}\beta _{2}\beta _{3}\beta _{4}\right),\ \left(\gamma _{1}\gamma _{2}\gamma _{3}\gamma _{4}\right),\ \left(\delta _{1}\delta _{2}\delta _{3}\delta _{4}\right)$ be connected by the eighteen relations of type

 $\beta _{2}\gamma _{3}-\beta _{3}\gamma _{2}=\alpha _{1}\delta _{4}-\alpha _{4}\delta _{1},\ \gamma _{2}\alpha _{3}-\gamma _{3}\alpha _{2}=\beta _{1}\delta _{4}-\beta _{4}\delta _{1},$ $\alpha _{2}\beta _{3}-\alpha _{3}\beta _{2}=\gamma _{1}\delta _{4}-\gamma _{4}\delta _{1},$ which imply that conjugate minors of the determinant $\left[\alpha _{1}\beta _{2}\gamma _{3}\delta _{4}\right]$ are equal. The identity

$\alpha _{1}\left(\alpha _{2}\beta _{3}-\alpha _{3}\beta _{2}\right)+\alpha _{2}\left(\alpha _{3}\beta _{1}-\alpha _{1}\beta _{3}\right)+\alpha _{3}\left(\alpha _{1}\beta _{2}-\alpha _{2}\beta _{1}\right)=0$ then gives

$\alpha _{1}\left(\gamma _{1}\delta _{4}-\gamma _{4}\delta _{1}\right)+\alpha _{2}\left(\gamma _{2}\delta _{4}-\gamma _{4}\delta _{2}\right)+\alpha _{3}\left(\gamma _{3}\delta _{4}-\gamma _{4}\delta _{3}\right)=0$ or

$\delta _{4}\left(\alpha _{1}\gamma _{1}+\alpha _{2}\gamma _{2}+\alpha _{3}\gamma _{3}+\alpha _{4}\gamma _{4}\right)=\gamma _{4}\left(\alpha _{1}\delta _{1}+\alpha _{2}\delta _{2}+\alpha _{3}\delta _{3}+\alpha _{4}\delta _{4}\right).$ Introducing the notation

$(\alpha \gamma )\equiv \alpha _{1}\gamma _{1}+\alpha _{2}\gamma _{2}+\alpha _{3}\gamma _{3}+\alpha _{4}\gamma _{4},$ we may obtain in the above way the equations

${\begin{array}{ccc}\delta _{1}(\alpha \gamma )=\gamma _{1}(\alpha \delta ),&&\delta _{3}(\alpha \gamma )=\gamma _{3}(\alpha \delta ),\\\\\delta _{2}(\alpha \gamma )=\gamma _{2}(\alpha \delta ),&&\delta _{4}(\alpha \gamma )=\gamma _{4}(\alpha \delta ),\\\\\delta _{1}(\beta \gamma )=\gamma _{1}(\beta \delta ),&&\delta _{3}(\beta \gamma )=\gamma _{3}(\beta \delta ),\\\\\delta _{2}(\beta \gamma )=\gamma _{2}(\beta \delta ),&&\delta _{4}(\beta \gamma )=\gamma _{4}(\beta \delta ).\end{array}}$ Hence, either

$(\alpha \gamma )=(\alpha \delta )=(\beta \gamma )=(\beta \delta )=0$ or all the quantities of type $\gamma _{1}\delta _{2}-\gamma _{2}\delta _{1}$ are zero. In the same way we can prove that either

$(\alpha \beta )=(\alpha \delta )=(\gamma \beta )=(\gamma \delta )=0$  