Page:A history of the theories of aether and electricity. Whittacker E.T. (1910).pdf/175

known that the vector curl e denotes twice the rotation of the part of the solid in the neighbourhood of the point (x, y, z) from its equilibrium orientation. In an ordinary elastic solid, the potential energy of strain depends only on the change of size and shape of the volume-elements; on their compression and distortion, in fact. For MacCullagh's new medium, on the other hand, the potential energy depends only on the rotation of the volume-elements.

Since the medium is not supposed to be in a state of stress in its undisturbed condition, the potential energy per unit volume must be a quadratic function of the derivates of e; so that in an isotropic medium this quantity φ must be formed from the only in variant which depends solely on the rotation and is quadratic in the derivates, that is from (curl e)²; thus we may write

$\phi ={\tfrac {1}{2}}\left\{\left({\frac {\partial e_{z}}{\partial y}}-{\frac {\partial e_{y}}{\partial z}}\right)^{2}+\left({\frac {\partial e_{x}}{\partial z}}-{\frac {\partial e_{z}}{\partial x}}\right)^{2}+\left({\frac {\partial e_{y}}{\partial x}}-{\frac {\partial e_{x}}{\partial y}}\right)^{2}\right\}$.

The equation of motion is now to be determined, as in the case of Green's aether, from the variational equation

$\iiint \rho \left\{{\frac {\partial ^{2}e_{x}}{\partial t^{2}}}\delta e_{x}{\frac {\partial ^{2}e_{y}}{\partial t^{2}}}\delta e_{y}{\frac {\partial ^{2}e_{z}}{\partial t^{2}}}\delta e_{z}\right\}\ dx\ dy\ dz=\iiint \delta \phi \ dx\ dy\ dz$;

the result is

$\rho {\frac {\partial ^{2}\mathbf {e} }{\partial t^{2}}}=-\mu {\text{ curl curl }}\mathbf {e}$.

It is evident from this equation that if div e is initially zero it will always be zero: we shall suppose this to be the case, so that no longitudinal waves exist at any time in the medium. One of the greatest difficulties which beset elastic-solid theories is thus completely removed.

The equation of motion may now be written

$\rho {\frac {\partial ^{2}\mathbf {e} }{\partial t^{2}}}=\mu \nabla ^{2}\mathbf {e}$,