2 Cross-overs

The derivation of the previous section, though giving the most direct evaluation of the stress, is too constraining in practice. While in a crystal structure it may be possible to define special planes $AA'$ that properly separate Lagrangian sets, for arbitrary solids and arbitrary planes at least some atoms will be close to the plane $AA'$ and heat motion will push them periodically across the plane. Note that from here on, set $I$ stands more precisely for the Lagrangian set of atoms whose nominal position is in region $I$ , below plane $AA'$ .

Figure 2: The cross-over problem.

Consider the example situation sketched in figure 2, in which a molecule from set $I$ crosses the plane $AA'$ into the region $J$ , bounces off the atoms there, and returns to region $I$ . The forces that bounce atom $i$ are clearly physical forces exerted by atoms in set $J$ on an atom in set $I$ . They should be included in our stress force: they add to the changes of momentum of Lagrangian set $I$ . Alternately, they add to the forces that must be exerted on set $I$ elsewhere to keep it at rest. Yet our sum of forces (1.2) completely ignores these forces as soon as the atom $i$ temporarily crosses the plane $AA'$ .

So, to get the proper sum of the physical forces of the set $J$ on the set $I$ , the sum of forces (1.2) has to be augmented by forces on atoms $i$ that are temporarily on the other side of the plane. To actually track atoms through the plane and back is of course very inconvenient, especially in a purely spatial average, and the standard trick is to instead substract the momentum of the atom $i$ when it leaves and add the new momentum back when it returns through the plane. The net effect is the same as including the integrated force on atom $i$ while it was at the other side.

Note that the atoms are not normally billiard balls as suggested above, and atom $i$ will probably be pulled back by its fellow atoms in $I$ as well as bounced back by the atoms in $J$ . Fortunately, due to Newton's third law, the part of the momentum change of atom $i$ due to its fellows in $I$ exactly cancels falsely counting the reaction forces by $i$ on its fellows as forces exerted by set $J$ on set $I$ . The same story holds in reverse for atoms $j$ wandering into region $I$ .

Thus, to get the correct stress, we need to add the net momentum flux of atoms through the surface $AA'$ to the sum of the forces (1.2). This dynamical term is not an artificial addition, but represents actual mechanical forces of the Lagrangian set $J$ on the set $I$ that are not correctly accounted for in the sum (1.2). (Compare Todd et al. 1995).

The above seems to be the key point missed by Zhou, who assumes the force term in the virial stress to be the correct evaluation of the forces and the dynamical term to be an addition (which cannot be zero since it involves squares.) In fact, as seen in the next section, the first term in the virial stress is exactly the correction for the physical forces on cross-overs missed by the force sum (1.2). This invalidates the central point made by Zhou; for example, in his abstract: ``The virial stress is not a measure for mechanical force between material points and cannot be regarded as a measure for mechanical stress in any sense.'' The discussion above shows that it does describe mechanical force, with the first dynamical term describing the mechanical force when atom $i$ is at the other side of plane $AA'$ , and the second force term describing the part when it is not.