# Page:BatemanElectrodynamical.djvu/12

284
[March 11,
Mr. H. Bateman

or all the quantities of type $\beta _{1}\delta _{2}-\beta _{2}\delta _{1}$ are zero, and that either

$(\beta \alpha )=(\beta \delta )=(\gamma \alpha )=(\gamma \delta )=0$ or all the quantities of type $\alpha _{1}\delta _{2}-\alpha _{2}\delta _{1}$ are zero. It follows from this that either (1) the six quantities (αβ) are zero or (2) that the thirty-six quantities $\left(\alpha _{1}\delta _{2}-\alpha _{2}\delta _{1}\right),\ \left(\beta _{1}\delta _{2}-\beta _{2}\delta _{1}\right),\ \left(\gamma _{1}\delta _{2}-\gamma _{2}\delta _{1}\right),\ \dots$ are all zero or (3) that there is a set of relations

 $(\alpha \gamma )=(\alpha \delta )=(\beta \gamma )=(\beta \delta )=0$ ${\frac {\beta _{1}}{\delta _{1}}}={\frac {\beta _{2}}{\delta _{2}}}={\frac {\beta _{3}}{\delta _{3}}}={\frac {\beta _{4}}{\delta _{4}}},\ {\frac {\alpha _{1}}{\delta _{1}}}={\frac {\alpha _{2}}{\delta _{2}}}={\frac {\alpha _{3}}{\delta _{3}}}={\frac {\alpha _{4}}{\delta _{4}}}.$ It is easy to see, however, that in the latter case we also have

$(\gamma \delta )=(\alpha \beta )=0.$ Hence, in all cases,

$(\alpha \gamma )=(\alpha \beta )=(\alpha \delta )=(\beta \gamma )=(\beta \delta )=(\gamma \delta )=0.$ Again, we have

$\left|{\begin{array}{ccc}\alpha _{1}&\beta _{1}&\gamma _{1}\\\alpha _{2}&\beta _{2}&\gamma _{2}\\\alpha _{3}&\beta _{3}&\gamma _{3}\end{array}}\right|{\begin{array}{l}=\alpha _{1}\left(\alpha _{1}\delta _{4}-\alpha _{4}\delta _{1}\right)+\alpha _{2}\left(\alpha _{2}\delta _{4}-\alpha _{4}\delta _{2}\right)+\alpha _{3}\left(\alpha _{3}\delta _{4}-\alpha _{4}\delta _{3}\right)\\=\delta _{4}\left(\alpha _{1}^{2}+\alpha _{2}^{2}+\alpha _{3}^{2}+\alpha _{4}^{2}\right)-\alpha _{4}\left(\alpha _{1}\delta _{1}+\alpha _{2}\delta _{2}+\alpha _{3}\delta _{3}+\alpha _{4}\delta _{4}\right);\end{array}}$ hence, since the last term is zero,

$\left|{\begin{array}{ccc}\alpha _{1}&\beta _{1}&\gamma _{1}\\\alpha _{2}&\beta _{2}&\gamma _{2}\\\alpha _{3}&\beta _{3}&\gamma _{3}\end{array}}\right|{\begin{array}{l}=\delta _{4}\left(\alpha _{1}^{2}+\alpha _{2}^{2}+\alpha _{3}^{2}+\alpha _{4}^{2}\right).\end{array}}$ This gives

${\begin{array}{l}\left(\alpha _{1}^{2}+\alpha _{2}^{2}+\alpha _{3}^{2}+\alpha _{4}^{2}\right)\left(\delta _{1}^{2}+\delta _{2}^{2}+\delta _{3}^{2}+\delta _{4}^{2}\right)=\end{array}}\left|{\begin{array}{cccc}\alpha _{1}&\beta _{1}&\gamma _{1}&\delta _{1}\\\alpha _{2}&\beta _{2}&\gamma _{2}&\delta _{2}\\\alpha _{3}&\beta _{3}&\gamma _{3}&\delta _{3}\\\alpha _{4}&\beta _{4}&\gamma _{4}&\delta _{4}\end{array}}\right|{\begin{array}{l}=\lambda ^{4},\end{array}}$ say, and there are similar equations in $\alpha ,\beta ,\alpha \gamma ,\beta \gamma ,\beta \delta ,\gamma \delta .$ . It follows that

${\begin{array}{cl}\alpha _{1}^{2}+\alpha _{2}^{2}+\alpha _{3}^{2}+\alpha _{4}^{2}&=\beta _{1}^{2}+\beta _{2}^{2}+\beta _{3}^{2}+\beta _{4}^{2}\\&=\gamma _{1}^{2}+\gamma _{2}^{2}+\gamma _{3}^{2}+\gamma _{4}^{2}=\delta _{1}^{2}+\delta _{2}^{2}+\delta _{3}^{2}+\delta _{4}^{2}=\pm \lambda ^{2}\end{array}}$ These conditions, combined with the previous set, imply that the sixteen quantities $\alpha _{1},\beta _{2},\dots$ are the elements of an orthogonal matrix. 