## 4 Results and Discussions
### Propagation of Elasto-Plastic Wave
In the present context, the free charge (\(\
abla\cdot\mathbf{E}\)) corresponds to \(\
abla\cdot\mathbf{v}\), which, in the one-dimensional form in a shear band, is \(\partial v_{s}/\partial x_{s}\). The disappearance of a shear band corresponds to the decrease in free charges, and is accompanied by a stress rise that corresponds to the voltage rise. The above-discussed narrowing of the shear band corresponds to the increase in the electric charge density, which triggers the positive feedback (self-enhancing) mechanism (an increase in conductivity increases the charge density that, in turn, increases the conductivity) that leads to the final arc discharge (See Appendix E). The final sharp shear band narrowing corresponds to this self-enhancing effect. If a deforming solid enters this stage, there is no more recovery; the solid will break. Stage 3 (\(18.69\%<\epsilon_{xx}<18.91\%\))  
This is the stage when the stress is at the peak on the stress-strain curve. From this point onward, the stress decreases monotonically until the specimen breaks. In this stage, both the \(u\) and \(v\) fringe patterns are stationary. At the same time, the shear band pattern becomes more prominent, and the \(u\) and \(v\) fringes start to be similar to each other. The similarity between the \(u\) and \(v\) fringe patterns is very interesting from the field theoretical viewpoint. This condition indicates that \(u\) and \(v\) can be expressed with the same function \(f(x,y)\). Consider the volume expansion and shear strain under this condition. \[\
abla\cdot\mathbf{\xi} = \frac{\partial u}{\partial x}+\frac{\partial v}{\partial y}= \frac{\partial f}{\partial x}+\frac{\partial f}{\partial y} \tag{40}\] \[\epsilon_{xy} = \frac{\partial v}{\partial x}+\frac{\partial u}{\partial y}= \frac{\partial f}{\partial x}+\frac{\partial f}{\partial y} \tag{41}\]  
Apparently, the two quantities take the identical form. The \((\
abla\cdot\mathbf{v})\) term in the plastic energy dissipative term is the time derivative of the volume expansion. \[\
abla\cdot\mathbf{v}=\frac{\partial}{\partial t}(\
abla\cdot\mathbf{\xi}) \tag{42}\]  
This means that, under this condition, the shear strain causes energy dissipation. This is interesting, because it is consistent with the concept of plastic deformation proposed by [PERSON] [58]; plastic deformation is shear instability. It is also consistent with the the field theoretical interpretation of the onset of plastic deformation discussed above; the deformation transitions from elastic to plastic when the strain tensor expressed in the principal coordinate system starts to have shear components. In addition, \(u\) and \(v\) constitute the \(x\) and \(y\) components of the differential displacement vector at a given point. Having the same \(x\) and \(y\) components, the differential displacement vector is directed at \(45^{\circ}\) to the \(x\) axis. This is why the linear fringe patterns run in \(45^{\circ}\) to the tensile axis, or along the maximum shear stress. Figure 11 shows three stationary \(u\) and \(v\) patterns observed in Stage 3. Note that the shear band pattern runs from the bottom left to the top right when \(\epsilon_{xx}=18.88\%\), it disappears and, consequently, the half-circle patterns become symmetric when \(\epsilon_{xx}=18.92\%\), and it runs from the bottom right to the top left when \(\epsilon_{xx}=18.94\%\). It seems that approximately \(45^{\circ}\) is the preferred orientation for the linear fringes to run. Notice that the shear band does not run vertically when the two half-circle patterns are symmetric at \(\epsilon_{xx}=18.92\%\). This is consistent with the above argument regarding the similarity of the \(u\) and \(v\) fringes (Equations (40) and (41)).