Page:BatemanElectrodynamical.djvu/11

1909.] have $$\theta^{2}=1$$ and eighteen relations of the types

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

for this purpose we shall require the following lemma.

. — Let the sixteen quantities

be connected by the eighteen relations of type

which imply that conjugate minors of the determinant $$\left[\alpha_{1}\beta_{2}\gamma_{3}\delta_{4}\right]$$ are equal. The identity

then gives

or

Introducing the notation

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

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