Eshelby energy-momentum tensor

Remarks

If $y$ is a smooth equilibrium deformation without body forces, so that $\operatorname{Div} P = 0$, then $\operatorname{Div} b = \partial W / \partial X$ at fixed $F$. Check: by the chain rule, $\partial_L (W \delta_{KL}) = \partial W/\partial X_K + P_{iL} \, \partial_K F_{iL}$, and, by $\operatorname{Div} P = 0$, $\partial_L (F_{iK} P_{iL}) = \partial_L F_{iK} \, P_{iL}$, and these two second terms agree because $\partial_K F_{iL} = \partial_L F_{iK} = \partial_K \partial_L y_i$. Hence $b$ is divergence-free in a homogeneous material, where $W$ does not depend on $X$. Example: for a uniform deformation $F$ of a homogeneous body, $b$ is constant, so each of its partial derivatives is zero and $\operatorname{Div} b = 0$. Non-example: the Cauchy stress $\sigma$ itself is divergence-free at equilibrium, but it is not an energy-momentum tensor: its flux gives the force on the material, not on a defect.