Thermodynamic Substate Variables for a Solid

A set of tensorial thermodynamic substate variables $V_{\mathrm{ij}}$ has been found such that under hydrostatic pressure the enthalpy $H$ of a solid, defined by $H\equiv U-{\tau }_{\mathrm{ij}}{V}_{\mathrm{ij}}$, where ${\tau }_{\mathrm{ij}}\equiv {\left(\frac{\partial U}{\partial V_{\mathrm{ij}}}\right)}_S$, reduces to the conventional enthalpy $U+pV$. $U$ is the internal energy per unit mass, $p$ the pressure, $V$ the specific volume, and $S$ the entropy per unit mass. Previously used tensorial variables (e.g., the Lagrangian strain components) lead to a different enthalpy. The new variables are defined by $V_{\mathrm{ij}}=\left(\frac{1}{3\overline{\rho }}\right){D^{\left(\frac{3}{2}\right)}}_{\mathrm{ij}}$, $D_{\mathrm{ij}}=\left(\frac{\partial x_k}{\partial X_i}\right)\left(\frac{\partial x_k}{\partial X_j}\right)$. Here the $X_i$ are the Cartesian coordinates of the particles of the body in some arbitrarily stressed reference configuration of density $\overline{\rho }$ and stresses ${\overline{T}}_{\mathrm{ij}}$, $x_i$ are the present coordinates, and ${D^{\left(\frac{3}{2}\right)}}_{\mathrm{ij}}$ is a symbol for the $\mathrm{ij}$ element of the positive real 3/2 power of the tensor $D_{\mathrm{ij}}$. Under these definitions, the enthalpy of a solid of arbitrary symmetry has been proved to reduce to $U+pV$ whenever x and X both correspond to states in which the stress in hydrostatic pressure. As a special case, X may of course be an unstressed configuration. The above choice of variables is not unique. In fact, the enthalpy reduces to $U+pV$ if $V_{\mathrm{ij}}$ is any matrix function of $D_{\mathrm{ij}}$ whose determinant is proportional to the 3/2 power of the determinant of the matrix $D_{\mathrm{ij}}$. However, the above choice of $V_{\mathrm{ij}}$ has the additional desirable property that when evaluated at x = X, the thermodynamic tensions ${\tau }_{\mathrm{ij}}$ equal the stresses $T_{\mathrm{ij}}$. In the case of cubic crystals and isotropic media under hydrostatic pressure, the present $V_{\mathrm{ij}}$ and ${\tau }_{\mathrm{ij}}$ reduce to $V_{\mathrm{ij}}=\frac{1}{3}V{\delta }_{\mathrm{ij}}$, ${\tau }_{\mathrm{ij}}=-p{\delta }_{\mathrm{ij}}$.