### Asymmetric Stress Tensor?

Suppose that I remove a truncated cylinder from some large stress free solid.
The cylinder is at {(x, y, z)| x^{2} + y^{2}<1 ∧ z < 1 ∧ z > −1}.
Its density is 0.
After the removal it is still under no stress.
Then I twist the cylinder by small amount t and replace it, glueing the ends to the cavity ends.
Have I produced a plausible volume where the stress tensor is asymmetric?
Before the removal the classic stress tensor (3×3) was all zero.
After twist and replacement the following tensor terms immediately appear resisting the twist:

T^{02} = rty

T^{12} = −rtx

where (x, y, z) = (x^{0}, x^{1}, x^{2})

t = amount of twist

r = √(x^{0}x^{0} + x^{1}x^{1}).

That terms T^{20} and T^{21} also appear long confused me.
It is because of conservation of angular momentum.
Without those extra symmetrizing terms we would have a field of local generators of angular momentum and the differential elements would spin ever faster.
The conclusion is that the classic stress tensor is symmetric.
A corollary of this is that in some frame the tensor is diagonal.
This is covered more convincingly in the late 19th century book on Elasticity by Love.

Searching the web finds a description of the Cauchy Stress Tensor and the same sort of symmetry argument.