Page:Scientific Papers of Josiah Willard Gibbs.djvu/243

${\frac {dx}{dx'}},...{\frac {dz}{dz'}}$ under the most general supposition with respect to the position of the co-ordinate axes.

For any principal axis of strain we have

${\frac {d(r^{2})}{d\alpha '}}d\alpha '+{\frac {d(r^{2})}{d\beta '}}d\beta '+{\frac {d(r^{2})}{d\gamma '}}d\gamma '=0,$

when$\alpha 'd\alpha '+\beta '\beta '+\gamma 'd\gamma '=0,$

the differential coefficients in the first of these equations being determined from (420) as before. Therefore,

${\frac {1}{\alpha '}}{\frac {d(r^{2})}{d\alpha '}}={\frac {1}{\beta '}}{\frac {d(r^{2})}{d\beta '}}={\frac {1}{\gamma '}}{\frac {d(r^{2})}{d\gamma '}}\cdot$

(426)

From (420) we obtain directly

${\frac {\alpha '}{2}}{\frac {d(r^{2})}{d\alpha '}}+{\frac {\beta '}{2}}{\frac {d(r^{2})}{d\beta '}}+{\frac {\gamma '}{2}}{\frac {d(r^{2})}{d\gamma '}}=r^{2}.$

(427)

From the two last equations, in virtue of the necessary relation $\alpha ^{'2}+\beta ^{'2}+\gamma ^{'2}=1$, we obtain

${\tfrac {1}{2}}{\frac {d(r^{2})}{d\alpha '}}=\alpha 'r^{2},$${\tfrac {1}{2}}{\frac {d(r^{2})}{d\beta '}}=\beta 'r^{2},$${\tfrac {1}{2}}{\frac {d(r^{2})}{d\gamma '}}=\gamma 'r^{2},$

(428)

or, if we substitute the values of the differential coefficients taken from (420),

$\alpha '\textstyle \sum \displaystyle \left({\frac {dx}{dx'}}\right)^{2}+\beta '\textstyle \sum \displaystyle \left({\frac {dx}{dx'}}{\frac {dx}{dy'}}\right)+\gamma '\textstyle \sum \displaystyle \left({\frac {dx}{dx'}}{\frac {dx}{dz'}}\right)=\alpha 'r^{2},$	$\scriptstyle {\left.{\begin{matrix}\ \\\\\ \\\ \\\ \\\ \ \end{matrix}}\right\}\,}$ (429)
$\alpha '\textstyle \sum \displaystyle \left({\frac {dx}{dy'}}{\frac {dx}{dx'}}\right)+\beta '\textstyle \sum \displaystyle \left({\frac {dx}{dy'}}\right)^{2}+\gamma '\textstyle \sum \displaystyle \left({\frac {dx}{dx'}}{\frac {dx}{dz'}}\right)^{2}=\beta 'r^{2},$
$\alpha '\textstyle \sum \displaystyle \left({\frac {dx}{dz'}}{\frac {dx}{dx'}}\right)+\beta '\textstyle \sum \displaystyle \left({\frac {dx}{dz'}}{\frac {dx}{dy'}}\right)+\gamma '\textstyle \sum \displaystyle \left({\frac {dx}{dz'}}\right)^{2}=\gamma 'r^{2}.$

If we eliminate $\alpha ',\beta ',\gamma '$ from these equations, we may write the result in the form,

${\begin{vmatrix}\textstyle \sum \displaystyle \left({\frac {dx}{dx'}}\right)^{2}-r^{2}&\textstyle \sum \displaystyle \left({\frac {dx}{dx'}}{\frac {dx}{dy'}}\right)&\textstyle \sum \displaystyle \left({\frac {dx}{dx'}}{\frac {dx}{dz'}}\right)\\\textstyle \sum \displaystyle \left({\frac {dx}{dy'}}{\frac {dx}{dx'}}\right)&\textstyle \sum \displaystyle \left({\frac {dx}{dy'}}\right)^{2}-r^{2}&\textstyle \sum \displaystyle \left({\frac {dx}{dy'}}{\frac {dx}{dz'}}\right)\\\textstyle \sum \displaystyle \left({\frac {dx}{dz'}}{\frac {dx}{dx'}}\right)&\textstyle \sum \displaystyle \left({\frac {dx}{dz'}}{\frac {dx}{dy'}}\right)&\textstyle \sum \displaystyle \left({\frac {dx}{dz'}}\right)^{2}-r^{2}\end{vmatrix}}=0.$

(430)

We may write

$-r^{5}+Er^{4}-Fr^{2}+G=0.$

(431)

$E=\textstyle \sum '\sum \displaystyle \left({\frac {dx}{dx'}}\right)^{2}\cdot$

(432)