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Symmetry of the Stress Tensor

  1. Nov 11, 2006 #1
    Hello !
    I'm having trouble with the symmetry of the stress tensor.
    What it means physically?
    In the demostration of the symmetry my book applies the angular momentum equation to a third material volume (a cube, lenght L). In the final step we have:

    [tex]
    0 = \mathop {\lim }\limits_{L \to 0} {1 \over {L^3 }}\int\limits_{Am(t)} {(\vec r \times \vec t_{(n)} )dA}
    [/tex]

    solving the cross product, forming the scalar product with base vector "k" (to observe the z-component of the torque). Dividing the total Area in 6 integrals (one for each face of the cube), and the we apply the mean value, when we take the lim L->0, we find that:
    Tyx = Txy
    And if we do something similar with the other components, we will find:
    Tzy = Tyz
    Txz = Tzx

    But, what it means at really?
    What if the stress tensor is not symmetric?

    I don't know, the prior integral says something like the torque is zero whe the material volume is a point, but, isn't this obvious? Even more, is not zero the angular moment too? (I can't imagine the angular moment of a point without quantum)

    Well... i hope some pious souls come to clarify the little things.

    Greets !
    jenri.
    PD: sorry for my ugly english, I'll copy the above in spanish too.

    -------------------------------------------------------------------------------------

    Hola a todos !
    Estoy teniendo problemitas con la smetría del tensor de tensiones (stress tensor).
    Cuál es el significado físico?
    En la demostración se suele aplicar el principio del momento angular a un cubo (lado L) y se termina llegando a:

    [tex]
    0 = \mathop {\lim }\limits_{L \to 0} {1 \over {L^3 }}\int\limits_{Am(t)} {(\vec r \times \vec t_{(n)} )dA}
    [/tex]

    Si a su vez resolvemos el producto vectorial, luego analizamos una sola componente (por ejemplo la "z", multiplicando por el vector k), dividimos el área total del cubo en 6 integrales (una para cada cara) y las resolvemos aplicando el valor medio. Resulta que cuando tomamos el lím L->0 nos encontramos con que Tyx = Txy. Si hacemos lo mismo para las otras componentes encontramos las otras dos igualdades.

    Pero qué significa esto?
    Podría no haber sido así?
    Que consecuencias tendría que el tensor no sea simétrico?

    La integral de más arriba parece indicar que el momento de torsión al tomar un volumen puntual es cero, pero este no tiene ni sentido, es claro que debe ser cero, no hay brazo de palanca...
    De hecho el momento angular deberia ser cero también para un punto, por lo menos no me imagino como podría tener un punto momento angular, solo en cuántica creo.

    Bueno, ahí dejo, espero algún alma piadosa se anime a aclarar estas cositas.

    Saludos !
    enri.
     
  2. jcsd
  3. Nov 13, 2006 #2
    A symmetric stress tensor means that there is no torque (moment, couple, angular force, or whatever) on the mass at that point. It simply means that the point in question will not rotate. A symmetrical stress tensor is a simplification for fairly static objects.

    It is not obvious, nor is it in general true. In general, there will be torque on the material "at a point", and the total stress tensor will not be symmetric anymore. In this case, what is often done is to split the stress tensor into symmetric and anti-symmetric parts (You can do this for any matrix I believe). The symmetric part is then sometimes referred to as the "stress tensor"(It is only a part of that), and the anti-symmetric part as the rotation tensor.
     
  4. Nov 13, 2006 #3
    Thanks for your reply !
    Now... if the stress tensor it's not symmetric in general... what are we trying to prove when we prove that T is symmetric? and why this demostration is correct, ... when this demostration is correct?

    can you explain it with an example? maybe it's easier so.

    enri.
     
  5. Nov 13, 2006 #4
    The "proof" that T is symmetric, involves the assumption that there is no torque on the particle at that point. To say that T is symmetric is only correct when there is no torque, i.e. no change in rotation. (Note that there can still be steady rotation, for example a particle spinning in space, but no change in rotation.)

    For most solid materials under consideration, the bodies are not even rotating, let alone changing their rotation. In fact in many cases the bodies in question may not even be moving at all. Under such cases, it is valid to assume a symmetric stress tensor. Or if you like, it is valid to assume that the anti-symmetric part of the tensor(the torque) is zero.
     
  6. Nov 13, 2006 #5
    OK, I think I see your point.
    But, I can't see where in the proof of the symmetry of T we assume that there is no torque.
    My book ("Introduction to fluid Mechanics" - STEPHEN WHITAKER) starts with:

    [tex]
    {D \over {Dt}}\int\limits_{{\rm{Vm(t)}}} {(\vec r \times \rho \vec v)} dV = \int\limits_{{\rm{Vm(t)}}} {(\vec r \times \rho \vec g)} dV + \int\limits_{{\rm{Am(t)}}} {(\vec r \times \vec t_{(\vec n)} )} dA
    [/tex]
     
  7. Nov 13, 2006 #6
    There doesn't appear to be an assumption of no torque in that equation. In fact, he seems to explicitly include it. I think.
     
  8. Nov 13, 2006 #7
    and now?
    Why we can prove the symmetry?
    I'll put the the proof, textually:

    For an arbitrary material volume, the angular momentum equation is:
    [tex]
    {D \over {Dt}}\int\limits_{Vm(t)} {(\vec r \times \rho \vec v)dV} = \int\limits_{Vm(t)} {(\vec r \times \rho \vec g)dV} + \int\limits_{Am(t)} {(\vec r \times \vec t_{(n)} )dA}
    [/tex]

    provided all torques are the moments of forces -i.e, there are no local stress couples(*) acting on the surface Am(t).
    (*) D. W. Condiff and J.S. Dahler, "Fluid Mechanical Aspects of Antisymmetrical Stress", Phys. Fluids, 1964, 7:842:54
    //I couldn't find it in google...

    We present the material volume as

    [tex]
    V_m (t) = \alpha (t)L^3
    [/tex]

    and apply the mean value theorem to the volume integrals

    [tex]
    {D \over {Dt}}[ < \vec r \times \rho \vec v > \alpha (t)L^3 ] = < \vec r \times \rho \vec g > \alpha (t)L^3 + \int\limits_{Am(t)} {(\vec r \times \vec t_{(n)} )dA}
    [/tex]

    For a cube, the characteristic length is taken as [tex]L=\delta x=\delta y=\delta z[/tex]. Since the position vector [tex]\vec r[/tex] is a directed line segment from the origin to any point on the cube, it tends to zero as L->0. Dividing by L^3 and taking the limit L->0, we reach the result that torques are in local equilibrium.

    [tex]
    0 = \mathop {\lim }\limits_{L \to 0} {1 \over {L^3 }}\int\limits_{Am(t)} {(\vec r \times \vec t_{(n)} )dA}
    [/tex]

    solving this for the z-component (for a cube), will result in:

    Txy = Tyx

    and so on...
     
  9. Nov 13, 2006 #8
    That proof looks incredibly fishy to me. The stress tensor is not in general symmetric. Take a ruler, make it horizonal. The top side faces directly up. Now apply pressure to the ends and observe the ruler bend. Most of the top side does not now face directly up. The elements have been moved and rotated. The external forces were applied only at the ends, so the only thing creating torque in the inside was the stress!

    I've seen a lot of arguments for the symmetry of the stress tensor. None of them are very good. Does anyone know any better ones? My expierience of all this begins from the point of view of the strain tensor, which is certainly not symmetric. I find it almost incredulous that application of a general Hooke's tensor to an unsymmetric strain tensor produces a symmetric stress tensor.
     
  10. Nov 13, 2006 #9
    Here is another "proof":

    [tex]
    {{{\rm{D\vec L}}} \over {{\rm{Dt}}}}{\rm{ = \vec N}}
    [/tex]

    ---

    [tex]
    {\rm{L = }}\int\limits_{{\rm{Vm(t)}}} {\rho (\vec r \times \vec v)} dV
    [/tex]

    ---

    [tex]
    {D \over {Dt}}\int\limits_{{\rm{Vm(t)}}} {\rho (\vec r \times \vec v)} dV = \vec N
    [/tex]

    ---

    [tex]
    \int\limits_{{\rm{Vm(t)}}} {\rho {D \over {Dt}}(\vec r \times \vec v)} dV = \vec N
    [/tex]

    ---

    [tex]
    {D \over {Dt}}(\vec r \times \vec v) = {{D\vec r} \over {Dt}} \times \vec v + \vec r \times {{D\vec v} \over {Dt}} = \vec v \times \vec v + \vec r \times \vec a = \vec r \times \vec a
    [/tex]

    ---

    [tex]
    \int\limits_{{\rm{Vm(t)}}} {\rho (\vec r \times \vec a)} dV = \vec N
    [/tex]

    ---

    [tex]
    \vec N = \int\limits_{{\rm{Vm(t)}}} {\rho (\vec r \times \vec g)} dV + \int\limits_{{\rm{Am(t)}}} {(\vec r \times \vec t_{(\vec n)} )} dA
    [/tex]

    ---

    [tex]
    \int\limits_{{\rm{Vm(t)}}} {\rho (\vec r \times \vec a)} dV = \int\limits_{{\rm{Vm(t)}}} {\rho (\vec r \times \vec g)} dV + \int\limits_{{\rm{Am(t)}}} {(\vec r \times \vec t_{(\vec n)} )} dA
    [/tex]


    Gauss (divergence theorem):

    [tex]
    \int\limits_{{\rm{Vm(t)}}} {\rho (\vec r \times \vec a)} dV = \int\limits_{{\rm{Vm(t)}}} {\rho (\vec r \times \vec g)} dV + \int\limits_{{\rm{Vm(t)}}} {DIV[ (\vec r \times \vec t_{(\vec n)} )} ]dV
    [/tex]

    indicial notation, and remembering:
    [tex]
    \vec t_{(\vec n)} = \vec n \cdot [T]
    [/tex]

    ---

    [tex]
    \int\limits_{{\rm{Vm(t)}}} {\rho \in _{ijk} r_j a_k } dV = \int\limits_{{\rm{Vm(t)}}} {\rho \in _{ijk} r_j g_k } dV + \int\limits_{{\rm{Vm(t)}}} {{\partial \over {\partial x_l }}( \in _{ijk} r_j T_{kl} )} dV
    [/tex]

    ---

    [tex]
    {\partial \over {\partial x_l }}( \in _{ijk} r_j T_{kl} ) = \in _{ijk} {{\partial r_j } \over {\partial x_l }}T_{kl} + \in _{ijk} {{\partial T_{kl} } \over {\partial x_l }}r_j = \in _{ijk} \delta _{jl} T_{kl} + \in _{ijk} {{\partial T_{kl} } \over {\partial x_l }}r_j
    [/tex]

    ---

    [tex]
    {\partial \over {\partial x_l }}( \in _{ijk} r_j T_{kl} ) = \in _{ijk} T_{kj} + \in _{ijk} r_j {{\partial T_{kl} } \over {\partial x_l }}
    [/tex]

    together:

    [tex]
    \int\limits_{{\rm{Vm(t)}}} {\rho \in _{ijk} r_j a_k } dV = \int\limits_{{\rm{Vm(t)}}} {\rho \in _{ijk} r_j g_k } dV + \int\limits_{{\rm{Vm(t)}}} { \in _{ijk} r_j {{\partial T_{kl} } \over {\partial x_l }}} dV + \int\limits_{{\rm{Vm(t)}}} { \in _{ijk} T_{kj} } dV
    [/tex]

    Applying the Lim Vm(t)->0 (volume -> 0)

    [tex]
    \in _{ijk} T_{kj} = 0
    [/tex]

    because the first three terms goes to zero like r^4, and the last term like r^3...

    Then, we have our symmetry proved. (we have?)

    And another one (over my limits...)

    The Symmetry of the Stress-Tensor Obtained by Schroedinger's Rule:
    http://www.pubmedcentral.nih.gov/picrender.fcgi?artid=1085229&blobtype=pdf
     
  11. Nov 13, 2006 #10
    If he's doing what I think he's doing, this step is wrong.

    The phrase "handwaving" comes to mind here.


    This guys assumes the tensor is symmetric to begin with!

    Still no good proofs. Frankly I'm skeptical.
     
  12. Nov 13, 2006 #11
    I think it's Reynolds Theorem. An special form... but I was uncertain about it too... because I did'nt knew that it can be applyed to a vector field.
    Why do you think it's wrong?


    Why handwaving? I don't know... maybe... but how would you prove the symmetry so?

    bye !
    enri.
     
  13. Nov 15, 2006 #12
    Because the boundary of the integral is also a function of time. It cannot simple be disregarded in this way.

    I wouldn't, because it is not in general true. If I was considering a material in which no particle expieriences any overall torque, then I would be able to say that the stress tensor is symettric. Otherwise, it simply isn't.
     
  14. Nov 15, 2006 #13
    I was checking it and I think I have the annswer.
    The Stress Tensor is simmetric because we must obey the principle of momemnt of momentum.
    The torque is proportional to the mass, i.e to the density.volume. When we take the limit to Volume->0, the Volume = dx.dy.dz tends to zero (together with the total torque) faster than the Area (dx.dy), therefore, we need a symmetric T at that limit.
    It has no other interpretation, only the principle of conservation of angular momentum.
    Now, it's the tensor always symmetric? NO ! because if we have some type of force not proportional to the volume, then that force will not tend to zero whe V->0, therefore, in that limit, we have torque=0, and 2 types of forces that can produce torque, so, we don't need to do the tensor symmetric to accomplish the principle of conservation, the two types of forces can be canceled to do it.

    bye ! and thanks !
    enri.
    PD: it is possible to "put" the D/Dt inside the integral because of the special form of the Reynolds Transport Theorem (which uses the Transport Theorem along with the continuity equation to prove it, at least for a scalar function...)
     
    Last edited: Nov 15, 2006
  15. Nov 15, 2006 #14

    PerennialII

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    Gold Member

    Symmetry of the Cauchy stress tensor requires its definition and laws of balance of momentum & balance of moment of momentum. It doesn't introduce the nature of the field but considers a property of the tensor, so we should be careful not to mix the property of the tensor to the nature of a specific field. I'm not sure whether you're talking about a system or a point, or a property of a tensor or a field (state)?
     
  16. Nov 16, 2006 #15
    We are trying to prove the simmetry of the stress tensor.
    To do it we just make use of the pricniple of conservation of angular momentum. Then, in the limit when V->0 the torque (which dependes on mass => proportional to volume) tends to cero faster than the surface forces (stress tensor), which not depends on mass (=> proportional to area).
    Then the stress tensor MUST be symmetric, but ONLY if it's the only force that "survives" when V->0, if we have another type of forces, that not depends on mass, then the tensor is not symmetric.
    A typical problem is:
    "Show that the symmetry of the stress tensor is not valid if there are body moments per unit volume, as in the case of a polarized anisotropic dielectric solid."

    bye !
    enri.
     
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