 Solution

Let $d(\mathbf{y},t)=[u^{[n]}]$ denote the signed opening, positive when the faces separate, and let $S$ be the crack surface. Its displacement jump is $\mathbf{b}=d\mathbf{n}$. The singular strain associated with this jump has moment density $c_{ijkl}{b}_k{n}_l$. For the isotropic elastic stiffness tensor, contraction givesThe equivalent distributional body force is the negative divergence of this prescribed moment density:The resulting forced elastic wave in an isotropic solid equation is $\rho u_{i,tt}={\partial }_j{\sigma }_{ij}+f_i$. This is a tensile-crack seismic source; it need not be represented as an ordinary smooth volume force. For a small nearly planar patch of area $S_0$, uniform normal $\mathbf{n}$ and opening $d(t)$, define its volume change $V(t)=S_{0}d(t)$. Its seismic moment tensor and point-source force areThe sign convention makes a positive opening a positive extensional moment. Reversing the convention for the displacement jump changes both signs together; the force is independent of reversing the orientation of the crack normal.