Page:Scientific Papers of Josiah Willard Gibbs.djvu/242

Moreover, if we write ${\frac {d(r^{2})}{d\alpha '}},{\frac {d(r^{2})}{d\beta '}},{\frac {d(r^{2})}{d\gamma '}}$ for the differential coefficients obtained from (420) by treating $\alpha ',\beta ',\gamma '$ as independent variables,

	${\frac {d(r^{2})}{d\alpha '}}+{\frac {d(r^{2})}{d\beta '}}+{\frac {d(r^{2})}{d\gamma '}}=0,$
when	$\alpha 'd\alpha '+\beta 'd\beta '+\gamma 'd\gamma '=0,$
and	$\alpha '=1,$$\beta '=0,$$\gamma '=0.$
That is,	${\frac {d(r^{2})}{d\beta '}}=0,$and${\frac {d(r^{2})}{d\gamma '}}=0,$
when	$\alpha '=1,$$\beta '=0,$$\gamma '=0.$
Hence,	${\frac {dx}{dy'}}=0,$${\frac {dx}{dz'}}=0.$

Therefore a line of the element which in the unstrained state is perpendicular to $X'$ is perpendicular to $X$ in the strained state. Of all such lines we may choose one for which the value of $r$ is at least as great as for any other, and make the axes of $Y'$ and $Y$ parallel to this line in the unstrained and in the strained state respectively. Then

${\frac {dz}{dy'}}=0;$

(424)

and it may easily be shown by reasoning similar to that which has just been employed that

${\frac {dy}{dz'}}=0.$

(425)

Lines parallel to the axes of $X',Y'$, and $Z'$ in the unstrained body will therefore be parallel to $X,Y$, and $Z$ in the strained body, and the ratios of elongation for such lines will be

${\frac {dx}{dx'}},{\frac {dy}{dy'}},{\frac {dz}{dz'}}\cdot$

These lines have the common property of a stationary value of the ratio of elongation for varying directions of the line. This appears from the form to which the general value of $r^{2}$ is reduced by the positions of the co-ordinate axes, viz.,

$r^{2}={\frac {dx}{dx'}}\alpha ^{'2}+{\frac {dy}{dy'}}\beta ^{'2}+{\frac {dz}{dz'}}\gamma ^{'2}.$

Having thus proved the existence of lines, with reference to any particular strain, which have the properties mentioned, let us proceed to find the relations between the ratios of elongation for these lines (the principal axes of strain) and the quantities