Stress tensor in continuum mech.

[LC]

I recently began studying continuum mechanics, and I'm having a bit of
trouble trying to find a mathematically precise definition of the
stress tensor. Most sources define it as some sort of nonsensical
limit like

lim_(A -> 0) \delta(F) / \delta(A)

or by describing it as "the contact force per unit area" acting on an
area element dA. The only reasonable interpretation I can find for
this is to say that the stress tensor is a 9-component tensor T_ij
such that if we integrate TdA over the surface bounding a body of
material, then we get the total force acting on that body. By the
divergence theorem, this is equivalent to asserting that the vector
div T_ij is the total force per unit volume.

However, this definition also seems problematic to me, because the
tensor T_ij that results is not unique. In particular, let S_ij be
any tensor such that div S_ij = 0.
We would then have div(T_ij + S_ij) = force per unit volume. There
doesn't seem to be any reason to prefer T_ij over T_ij + S_ij. In
other words, the definition of T_ij involves a choice of gauge. This
seems irreconcilable with the notion of defining it once and for all
through the sort of limiting process described above.

Thus my question is, why do people refer to "the" stress tensor rather
than "a" stress tensor? Can someone either give me a precise
mathematical account of this discrepancy, point out where my analysis
went wrong, or explain the physical significance of the gauge? Are
there some implicit assumptions that I'm unaware of that allow us to
single out a particular choice of gauge, and therefore single out a
unique tensor T_ij?

Thanks
LC

[Bossavit]

>I'm having a bit of trouble trying to find a mathematically precise
>definition of the stress tensor.

You might find it helpful to read papers (easily available on line) by
Herrmann, from Karlsruhe, and his school: They view force as "momentum
flux", which allows for helpful analogies with other kinds of flow, like
that of electric charge for instance.

In this picture, momentum is mathematically represented by a covector,
whose time derivative is force (d_t p = f). Why force should be,
mathematically speakning, a covector, is easy to figure out: Force acts
on virtual displacement, a vector, to give a number, virtual power. So
force is a linear map of the VECTOR --> REAL kind, i.e., an element of
the dual vector space, aka covector.

Speaking physics, now, matter on one side of some fiducial surface
"sends momentum" to the other side, according to Herrmann's view. (See
e.g., F. Herrmann, G.B. Schmid: "Statics in the momentum current
picture", Am. J. Phys., 52, 2 (1984), pp. 146-52.) Looking at the
surface as an assembly of tiny patches, each of which can mathematically
be represented by a bivector, one gets the picture of stress as a
"covector-valued (twisted) differential form (DF) of degree 2". That's
the proper definition: "DF of degree 2" means linear map, at each
point, from 2-vectors to something; The something is, in that case, a
covector; And "twisted" means the surface over which the DF will be
integrated is equipped with a crossing direction. So integrating stress
over such a surface, we get a covector, which is the flow of momentum
passing through, in that direction.

It was not said that the surface is closed, not at all. Defining stress
by its integrals over *closed* surfaces only would carry less
information than defining stress as a 2-form, which is equivalent to
know its integrals over *all* surfaces, closed or not. This loss of
information is the source of the "gauge ambiguity" you mention, which is
a real issue, but *not* one inherent in the *definition* of stress.

The simpler case of electric charge may help clarify this point:
Electric current j is similar to stress, in being a twisted 2-form,
but a real-valued one: By integrating j over a surface S equipped
with a crossing direction, you get the flow of charge crossing S in
that direction, per unit of time--a real number. No one doubts that j
thus represents a well-defined physical entity (*the* current density).
But if j was defined by its integral over *closed* surfaces only, it
would be determined up to some "closed" 2-form only, and subject to the
gauge equivalence j --> j + dh, where h is a 1-form. Which means (in
standard, more familiar, vectorial notation), the vector field J would
only be defined up to the curl of some arbitrary field H. So that
would make it only *a* current density.

So I would argue, the same for stress: The gauge freedom you mention
does not deprive stress from its physical meaning. It's indeed *the*
stress tensor.

As for the representation of stress by a nine-component tensor, it's of
course correct, but it's just that, a *representation* of something more
fundamental. Just as the vector field J represents current density
j, a 2-form.

[Pierre Asselin]

LC <luccul@gmail.com> wrote:
> I recently began studying continuum mechanics, and I'm having a bit of
> trouble trying to find a mathematically precise definition of the
> stress tensor. Most sources define it as some sort of nonsensical
> limit like

> lim_(A -> 0) \delta(F) / \delta(A)

> or by describing it as "the contact force per unit area" acting on an
> area element dA. The only reasonable interpretation I can find for
> this is to say that the stress tensor is a 9-component tensor T_ij
> such that if we integrate TdA over the surface bounding a body of
> material, then we get the total force acting on that body.

It's more than that. If you do that on a surface element
in the interior of the body, you get the total force transmitted
*through* the surface element. The material on each side
exerts a traction on the other side; the two tractions
are equal and opposite.

It's easiest to see this in the one-dimensional case, where
the stress tensor has only one component, linear tension (or
compression). Think of a rope under tension. To measure
the tension you would have to cut the rope and apply equal
and opposite forces to the now free ends if you want to
keep them from flying apart. The required force is the
tension.

> By the
> divergence theorem, this is equivalent to asserting that the vector
> div T_ij is the total force per unit volume.

> However, this definition also seems problematic to me, because the
> tensor T_ij that results is not unique. In particular, let S_ij be
> any tensor such that div S_ij = 0.
> We would then have div(T_ij + S_ij) = force per unit volume. There
> doesn't seem to be any reason to prefer T_ij over T_ij + S_ij.

Not until you carve a free surface in the body's interior. This
is analoguous to cutting the rope in the one-dimensional case.
The material on each side of the surface will shift unless you
apply compensating forces.

> In other words, the definition of T_ij involves a choice of gauge.

which can only be resolved by an extra measurement.

--
pa at panix dot com

[Igor Khavkine]

On 2007-09-28, LC <luccul@gmail.com> wrote:
> I recently began studying continuum mechanics, and I'm having a bit of
> trouble trying to find a mathematically precise definition of the
> stress tensor. Most sources define it as some sort of nonsensical
> limit like
>
> lim_(A -> 0) \delta(F) / \delta(A)
>
> or by describing it as "the contact force per unit area" acting on an
> area element dA.

Why do you find this limit definition non-sensical?

> The only reasonable interpretation I can find for
> this is to say that the stress tensor is a 9-component tensor T_ij
> such that if we integrate TdA over the surface bounding a body of
> material, then we get the total force acting on that body. By the
> divergence theorem, this is equivalent to asserting that the vector
> div T_ij is the total force per unit volume.
>
> However, this definition also seems problematic to me, because the
> tensor T_ij that results is not unique. In particular, let S_ij be
> any tensor such that div S_ij = 0.
> We would then have div(T_ij + S_ij) = force per unit volume. There
> doesn't seem to be any reason to prefer T_ij over T_ij + S_ij. In
> other words, the definition of T_ij involves a choice of gauge. This
> seems irreconcilable with the notion of defining it once and for all
> through the sort of limiting process described above.

For a similar point of view and part of the answer to your question, see
Landau & Lifshitz, Theory of Elasticity (vol. 7 of their series),
Section 2. At the end of that section, you'll find the calculation of
the total torque on the body by adding up the torque acting on each
individual volume element. Just as with the total force on the body, the
total torque can be rewritten purely as a surface integral, and in a way
that is consistent with the interpretation of T_ij as the force per unit
area of the surface. But, this is possible only if T_ij is "almost"
symmetric.

Let me explain what I mean by "almost". If T_ij is symmetric, then this
condition is sufficient to express the total torque as a surface
integral. However, it is not necessary. The necessary condition is
weaker and requires that the antisymmetric part of T_ij be a total
divergence:

(T_ij - T_ji)/2 = U_ijk,k,

where ,k indicates divergence along the last index of U_ijk and clearly
U_ijk = -U_jik.

However, as we can use the freedom of adding, as you suggest, of
replacing T_ij by T_ij + S_ij, with S_ij,j = 0, to restore the symmetry
of T_ij. First, the divergence condition S_ij,j = 0 implies that S_ij is
necessarily of the form

S_ij = V_ijk,k,

for some V_ijk, such that V_ijk = -V_ikj. This is a kind of Helmholtz
decomposition for tensors. So, if you have some non-symmetric stress
tensor T_ij, you can always find V_ijk depending on U_ijk such that the
new stress tensor T_ij + S_ij is symmetric.

So, is the condition of symmetry sufficient to fix T_ij uniquely? The
answer is Yes. 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.
It's fairly easy to show that, under these conditions, V_ijk must
vanish. Thus T_ij is uniquely determined.

In more detail, we can combine the conditions on V to yield
V_ijk = -V_jki. Multiple applications of which finally yield
V_ijk = -V_ijk.

> Thus my question is, why do people refer to "the" stress tensor rather
> than "a" stress tensor? Can someone either give me a precise
> mathematical account of this discrepancy, point out where my analysis
> went wrong, or explain the physical significance of the gauge? Are
> there some implicit assumptions that I'm unaware of that allow us to
> single out a particular choice of gauge, and therefore single out a
> unique tensor T_ij?

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

Hope this helps.

Igor

[LC]

On Oct 2, 2:19 am, Igor Khavkine <igor...@gmail.com> 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

[Igor Khavkine]

On 2007-10-03, LC <luccul@gmail.com> wrote:
> On Oct 2, 2:19 am, Igor Khavkine <igor...@gmail.com> 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.

As you show below, my previous assertion is false.

> 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.

That should the be antisymmetric part of T_ij, not T_ij itself.

> 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.

Yes, this counterexample shows that symmetry is not yet a sufficient
condition for uniqueness of T_ij. Thanks for pointing this out.

> 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.

The appearance of the V_ijk tensor is easy to explain. Consider the
difference S_ij = T'_ij - T_ij, and for now ignore its first index, i.
In the second index, j, it is a vector field whose divergence vanishes.
This, using the Helmholtz decomposition, implies that it is the curl of
some other vector field, S_ij = curl W_ij, where the curl is only taken
over the second index. Recall that we can write the curl as

curl W_ij = e_jkl W_il,k = V_ijk,k,

where e_jkl is the completely antisymmetric Levi-Civita tensor and we've
defined V_ijk = e_jkl W_il. BTW, one can easily find a V_ijk whose
divergence reproduces S_ij in your counterexample.

That's simple enough. However, you have correctly pointed out the
location of the flaw in my uniqueness argument. The flaw lies in
assuming that V_ijk need be symmetric in the ij indices, since S_ij is
symmetric. What is correct is that V_ijk,k needs to be symmetric in ij,
but while V_ijk need not be symmetric itself. Without this extra
property, we can no longer show that V_ijk must vanish. Moreover, you
have already provided a counter example. BTW, this uniqueness assertion
did not come from Landau & Lifshitz, it was my own clumsy attempt to
demonstrate it. :-) Obviously it has failed.

My next thought was to consider bounday conditions. Is knowledge of the
forces applied to each surface element on the boundary of an elastic
medium, together with the knowledge of the forces applied to each volume
element of its interior, sufficient to specify the stress tensor
uniquely? The answer is still No.

Consider the example of a semi-infinite elastic medium, occupying the
z>0 half-space. Let f_i(x,y,z) be the force per unit volume acting on
its interior. Let F_i(x,y) be the force per unit area acting on the its
boundary at z=0. If you find a symmetric tensor T_ij such that
T_ij,j = f_i in the interior and T_iz = -F_i on the boundary, then any
tensor T'_ij = T_ij + S_ij satisfies the same conditions, provided that

[ g_,yy -g_,xy 0 ]
S_ij = [ -g_,xy g_,xx 0 ], where g=g(x,y,z) is any scalar function.
[ 0 0 0 ]

It's easy to show that S_ij,j = 0 and that S_iz = 0 at z=0. Also, g
could easily be picked to satisfy any desired boundary conditions at
infinity. Unfortunately, I suspect that, this parametrization is not
exhaustive. It can, however, be generalized to shapes other than the
half-space continuum.

> 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.

That opinion, I still have to disagree with. Consider a 1D example, as
suggested by another poster. For simplicity, consider a static
situation. The tension at any point of a stretched string can be
measured by cutting the string at that point and supplying just enough
force acting on the newly created free ends to keep the configuration of
the string unaltered. In other words, at the point of measurement, you
are eliminating part of the string and measuring the amount of force the
eliminated part of the string would have provided if it were still
attached. Conceptually similar, but practically more difficult
measurements can be carried out for the non-static case.

Coming back to 3D, consider a body under stress. You can slice away part
of the body and try to apply just enough force on the newly created
surface to compensate for the part of the body that was cut away. These
surface forces will in general nave both normal (pressure) and
tangential components (shear). If, at a particular point, such a surface
force were measured to be F_i, this measurement provides a direct
measurement of T_ij n_j, where n_j is the normal to the surface at the
point of measurement. This is precisely the definition of the the
stresses T_ij. Since you are allowed to slice at any angle, you can vary
n_j and thust obtain all components of T_ij.

I hope this argument is convincing enough to show that, in principle,
all components of T_ij are empirically accessible.

Here's another take on the apparent non-uniqueness of T_ij. Suppose for
a moment that we are dealing only with isotropic pressures (T_ij is
diagonal with each diagonal element equal to p). Only the gradient of
the pressure will enter the dynamical equations for the mechanical
contiuum in question. Is it fair to say that p is physically meaningful
only up to an additive constant? No, the absolute magnitude of p can be
measured, for example, at the boundary, fixing the additive constant to
some particular value. So, adding different constants to the pressure
produces different physical situations. The lesson here is that, in any
definite physical situation, the stress (including pressure as a special
case) will have to be specified through a constitutive relation
depending on other dynamical quantities (strains, temperature, etc.), in
addition to the equations of motion.

Hope this helps.

Igor