Why are stress components on opposite faces identical?

Question

The stress tensor stores information about the stress on each of the faces of an infinitesimal volume of material. But I am confused as to why the stress components on opposite faces should be equal. The textbook I am reading (Fluid Mechanics by Kundu and Cohen) gives the following explanation:

On the opposite face EFGH the stress components have the same value as on ABCD, but their directions are reversed. This is because Figure 2.4 shows the stress at a point. The cube shown is supposed to be of “zero” size, so that the faces ABCD and EFGH are just opposite faces of a plane perpendicular to the x2-axis. That is why the stresses on the opposite faces are equal and opposite.

I am not sure why the cube being infinitesimal size implies that the stress components on opposite faces should be the same. Could someone please explain?

Answer 1

The argument in your text is a bit muddled.

The usual argument is the following: If the sides of the cube have length $L$ the the net force $F_x$ in the $x$ direction, say, is equal to $L^2\Delta \tau_{xx}$ where $\Delta \tau_{xx}$ is the difference in the stress components on the opposite faces. Meanwhile the mass of the cube is $m=\rho L^3$ where $\rho$ is the density. Thus, from $F=ma$ the acceleration is $$ a =\frac 1{\rho} \Delta \tau_{xx}L^{-1}. $$ If $a$ is to stay finite as $L\to 0$, then $\Delta \tau_{xx}$ must go to zero $\propto L$. Or, to put it another way, if $\Delta \tau_{xx}$ were not zero in the limit, the material would move infinitely fast so as make the two stresses equal. A similar argument applied to torque and moments of intertia shows that $\tau_{ij}=\tau_{ji}$.

This argument is similar to the one that says that the tension in a massless string has to be the constant along the length of the string.