Page:Scientific Papers of Josiah Willard Gibbs.djvu/246

enlargement of volume. Since the position of the co-ordinate axes is arbitrary, it follows that the sum of the squares of the ratios of elongation or enlargement of three lines or surfaces which in the unstrained state are at right angles to one another, is otherwise independent of the direction of the lines or surfaces. Hence, ${\tfrac {1}{3}}E$ and ${\tfrac {1}{3}}F$ are the mean squares of the ratios of linear elongation and of superficial enlargement, for all possible directions in the unstrained solid.

There is not only a practical advantage in regarding the strain as determined by $E,F,$ and $H$, instead of $E,F,$ and $G$, because $H$ is more simply expressed in terms of ${\frac {dx}{dx'}},...{\frac {dz}{dz'}}$, but there is also a certain theoretical advantage on the side of $E,F,H$. If the systems of co-ordinate axes $X,Y,Z$, and $X',Y',Z'$, are either identical or such as are capable of superposition, which it will always be convenient to suppose, the determinant H will always have a positive value for any strain of which a body can be capable. But it is possible to give to $x,y,z$ such values as functions of $x',y',z'$ that $H$ shall have a negative value. For example, we may make

$x=x',$$y=y',$$z=-z'.$

(440)

This will give $H=-1$, while

$x=x',$$y=y',$$z=z'.$

(441)

will give #=1. Both (440) and (441) give $H=1$. Now although such a change in the position of the particles of a body as is represented by (440) cannot take place while the body remains solid, yet a method of representing strains may be considered incomplete, which confuses the cases represented by (440) and (441).

We may avoid all such confusion by using $E,F,$ and $H$ to represent a strain. Let us consider an element of the body strained which in the state $(x',y',z')$ is a cube with its edges parallel to the axes of $X',Y',Z'$, and call the edges $dx',dy',dz'$ according to the axes to which they are parallel, and consider the ends of the edges as positive for which the values of $x',y',$ or $z'$ are the greater. Whatever may be the nature of the parallelepiped in the state $(x,y,z)$ which corresponds to the cube $dx',dy',dz'$ and is determined by the quantities ${\frac {dx}{dx'}},...{\frac {dz}{dz'}}$, it may always be brought by continuous changes to the form of a cube and to a position in which the edges $dx',dy'$ shall be parallel to the axes of $X$ and $Y$, the positive ends of the edges toward the positive directions of the axes, and this may be done without giving the volume of the parallelepiped the value zero, and therefore without changing the sign of $H$. Now two cases are possible;—the positive end of the edge $dz'$ may be turned toward