Re: Stress tensor in continuum mech.

On Oct 2, 2:19 am, Igor Khavkine <igor...@xxxxxxxxx> wrote:
So, is the condition of symmetry sufficient to fix T_ij uniquely? The
answer is Yes.

In short, the stress tensor is uniquely fixed by requiring that it
generates both the total force and torque due to contact forces.

This is by far the most helpful answer I've seen, because it explains
why symmetry of the stress tensor comes up so often, and gives a
reference (Landau-Lifshitz) that acknowledges this difficulty in
defining the stress tensor. I had been previously unable to find such
a reference. However, I am still not convinced that a unique stress
tensor can be defined by imposing the extra conditions you mentioned.
You brought up two such conditions we could impose on the stress
tensor, but I claim that neither of these are sufficient to gaurantee
a uniquely defined stress tensor.

Condition 0: The stress tensor can be integrated over the surface of a
body to give the total force acting on the body. Of course, we must
always assume this, and it is equivalent to saying T_ij,j = F

Condition 1: The moments of the stress tensor can be integrated over
the surface of a body to give the total torque acting on the body.
This is equivalent to the existence of a tensor U_ijk such that
U_ijk,k = T_ij and U_ijk = - U_jik.

Condition 2: The stress tensor is symmetric.

As both you and Landau-Lifshitz point out, condition 2 implies
condition 1. Furthermore, condition 1 only tells us that the
antisymmetric part of T_ij is a total divergence, and of course there
are many antisymmetric total divergences. Thus condition 1 cannot
possibly determine a unique tensor T_ij.

Now, you claim that condition 2 is enough to uniquely determine a
tensor T_ij. I claim that this is also false. As a counterexample,
consider the tensor:

S_11 = x_2, S_ij = 0 for all other i and j.

This is clearly a symmetric tensor. Its divergence F_i = Tij,j is
given by:

F_1 = T_11,1 + T_12,2 + T_13,3 = 0 + 0 + 0 = 0
F_2 = 0
F_3 = 0

In fact, the situation is even worse than this: I claim that any
tensor T_ij is equivalent to a **diagonal** tensor D_ij, (by
equivalent I mean that T_ij and D_ij give the same equations of
motion, or in other words T_ij,j = D_ij,j). Also, note that the
example above shows that even diagonal tensors are not uniquely
determined by their divergence, as we could always add S_ij and obtain
another diagonal tensor with the same divergence.

Here's the proof. Consider an arbitrary tensor T_ij. Define a
vector V_i by taking its divergence

V_i = T_ij,j

Integrating with respect to x_i, we can then find a vector W_i such
that:

d/d_{x_i} ( W_i ) = V_i

for all i. Now consider the diagonal tensor:

S_ij = 0 for i not equal to j
S_ii = W_i for all i

By definition, S_ij,j = V_i = T_ij,j. Thus (S_ij - T_ij),j = 0. But
this shows that we can turn T_ij into a **diagonal** tensor by adding
the divergenceless tensor R_ij = S_ij - T_ij.

It appears that your argument went wrong at this point:

If T'_ij is an equivalent symmetric stress tensor, then
we know that their difference is a total divergence

T'_ij - T_ij = V_ijk,k,

where the tensor V now satisfies both V_ijk = -V_ikj and V_ijk = V_jik.

The condition you state is not equivalent to saying that (T'_ij -
T_ij),j = 0. I'm not sure exactly where this tensor V came from, but
in any case the counterexample I gave proves that the argument is
invalid.

In summary, even if I am willing to accept that the stress tensor must
be symmetric (and thus must generate the total torque as well as the
total force), I still don't believe that a unique stress tensor can be
defined. Perhaps there are other conditions one can impose to
guarantee uniqueness, but I have yet to see any set forth.

Also, I have yet to see a concrete description of how one can measure
the components of the stress tensor, which leads me even more strongly
to believe that they are physically meaningless.

-LC

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