This is taken from Pg 84 of Kundu-Cohen, 3rd Edition:
We shall now prove that the stress tensor is symmetric.
Consider the torque on an element about a centroid axis parallel to $x_3$. This torque is generated only by the shear stresses in the $x_1 x_2$-plane and is (assuming $dx_3 = 1$):
$T = [\tau_{12}+ \frac{1}{2}\frac{\partial \tau_{12}}{\partial x_1} dx_1]dx_{2}\frac{dx_1}{2}+[\tau_{12}- \frac{1}{2}\frac{\partial \tau_{12}}{\partial x_1} dx_1]dx_{2}\frac{dx_1}{2}-[\tau_{21}+ \frac{1}{2}\frac{\partial \tau_{21}}{\partial x_2} dx_2]dx_{1}\frac{dx_2}{2}+[\tau_{12}- \frac{1}{2}\frac{\partial \tau_{21}}{\partial x_2} dx_2]dx_{1}\frac{dx_2}{2}$This simplifies to $T= (\tau_{12}-\tau_{21})dx_1dx_2$.
This is straightforward.
The rotational equilibrium of the element requires that $T = I \dot{ω}_{3}$, where $\dot{ω}_{3}$ is the angular acceleration of the element and $I$ is its moment of inertia.
For the rectangular element considered, it is easy to show that $I = dx_1 dx_2 (dx_1 ^2+dx_2 ^2)\rho/12$.
The rotational equilibrium then requires $(\tau_{12}-\tau_{21})dx_1dx_2 = \frac{\rho}{12}dx_1 dx_2 (dx_1 ^2+dx_2 ^2)\dot{ω}_{3}$.
For $dx_, dx_2 \rightarrow 0$, $\tau_{12}=\tau_{21}$ must hold.
I don't follow as to why it must be specified this is under rotational equilibrium. If the line elements tend to zero, the RHS will tend to zero, and so should the LHS. But LHS is independent of the area element considered, so the LHS is strictly zero. But that directly implies that torque must be zero.
Thus, shouldn't it be the other way around in the proof then; that we obtain rotational equilibrium as a byproduct of the proof?
PS. Rotational equilibrium already means that sum of torques is zero, so it does not make sense to do the scaling analysis, since we have already established that LHS must be zero for all area elements.
