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Is stress tensor symmetric in Navier-Stokes Equation?

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  1. Jun 26, 2014 #1
    Hello,

    In CFD computation of the Navier-Stokes Equation, is stress tensor assumed to be symmetric?
    We know that in NS equation only linear momentum is considered, and the general form of NS equation does not assume that stress tensor is symmetric. Physically, if the tensor is asymmetric then there is torque on the microscopic volume m, and within its streamline in general it is subjected to the influence of:
    1. gravity
    2. normal stress
    3. shear force ⇒ torque ⇒ angular acceleration mra (r assumed to be mean radius of m)
    attachment.php?attachmentid=70898&stc=1&d=1403765913.jpg
    I think this understand is basically correct regarding torque: visualize a crowed square in which thousands of people, shoulder-to-shoulder, are moving toward a same direction (for example pilgrims in Mecca), then each body is subjected to torque from the neighbor, and in general he will not always orient toward the same direction during his course along the streamline. The orientation change is therefore due to the angular a above, and his movement can always be decomposed into
    1. linear translation ⇐ normal stress
    2. rotation about his own axis ⇐ torque
    therefore the assumption seems perfectly valid, especially when liquid swirl is encountered.
    And although in general is a present, the accumulated mr⋅v(rotational) is small because temporarily (time) microscopically, and individual “rotation” not in-phase/aligned with its neighbors increases stress which soon acts to decelerate that.

    However, if many places I read that the symmetry of stress is assumed, such as with Rutherford Aris’s Vectors, Tensors, and the Basic Equations of Fluid Mechanics pp102, in section 6.41 and he also used a somewhat questionable term “polar fluid” to refer to fluid with torque/asymmetric stress.

    I would like to know that in popular CFD packages Fluent, CFX , FloEFD and Solidworks Flow Simulation, is stress tensor assumed to be symmetric or not? I wonder if they do assume symmetric, then for viscous fluid symmetric shear stress implies a symmetric velocity field which is unrealistic for example for swirls, so the software will fail.

    I also read that the time-averaged Reynolds Stress in RANS (Reynolds-averaged Navier–Stokes equations) is probably symmetric, and is used for steady-state solutions. However, there are situations where particle/parcel rotation is involved, particularly in cases where convection is heavy like for combustion. Therefore, any assumption of symmetric stress tensor might seem to be an unrealistic oversimplification.

    Because, even for steady-state flow, particles (or larger "parcel"s) still flow and most likely change orientation (the pilgrim queue can reach steady state, but people sitll chang orientation along the streamline). So what makes the symmetric stress tensor assumption valid?

    I just started learning this subject for a few days so the question might seem very basic to experienced users. Sincerely hope someone could resolve my puzzle!


    Andy
     

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    Last edited: Jun 26, 2014
  2. jcsd
  3. Jun 26, 2014 #2
    The CFD packages all assume that the stress tensor is symmetric. Symmetry of the stress tensor does not imply symmetry of the velocity field; it implies symmetry of the rate of deformation tensor. The velocity gradient tensor can be resolved into a symmetric part (the rate of deformation tensor) and an antisymmetric part (the vorticity tensor). The vorticity tensor corresponds to rotation of the fluid elements, which does not contribute to deformational stress, so only the rate of deformation tensor (which is symmetric) determines the stress tensor. And the stress tensor for a Newtonian fluid is linear in the rate of deformation tensor, so it too must be symmetric.

    Chet
     
  4. Jun 27, 2014 #3

    vanhees71

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    Well, there are cases where the stress tensor is not symmetric. That goes under the name "polar fluids". It occurs when the forces at work are not of the simple Newtonian type.

    One must distinguish between a microscopic and a macroscopic notions of polarity here, however. E.g., water consists of molecules with an electric dipole moment and in this sense it's polar on a microscopic level, giving rise to all kinds of "anomalies", e.g., it has a pretty large static dielectricity constant compared to other materials, due to the strong electric dipole moment of its molecules.

    Macroscopically, however water is unpolar, because averaging (coarse graining) over volume elements large on a microscopic scale, i.e., containg a large number of molecules, but small on a microscopic scale (i.e., small compared to the typical spatial scales upon which the macroscopic properties like density, flow-velocity field, etc. change considerably) the polarization averages out. This macroscopic description is the point of hydrodynamics vs. more microscopic descriptions as the Boltzmann transport equation or the Boltzmann-Uehling-Uhlenbeck equation, which incorporates also quantum effects. A full quantum treatment leads to the Kadanoff-Baym equations which only very recently have been able to study for simple systems (like relativistic [itex]\phi^4[/itex] theory), but that's not the subject of this thread.

    A fluid is polar in the macroscopic sense, if it has for some reason intrinsic macroscopic spin on the level of the fluid cells (small compared to the macroscopic scale but lage on the microscopic scale as detalied above). This is the case if you have magnetic fluids, where you have macroscopic domains of ordered magnetic moments. Then in addition to the usual "body forces" you have to consider also "body torque", and then the usual Cauchy proof for the symmetry of the stress tensor doesn't hold anymore. This is all nicely demonstrated in Asis's book, but only on a quite mathematical level. He doesn't give much details about the physics. That's more detailed in the famous treatise by Truesdal in Flügge's physics encyclopedia, cited by Asis. But this is awfully difficult to read for us nowadays, since he uses totally different conventions in naming his variables (e.g., he uses [itex]\vec{p}[/itex] instead of [itex]\vec{x}[/itex] for the Euler coordinates, which drives me nuts, because I'm used to label momenta with [itex]\vec{p}[/itex]; it's however still a brillant exhibition of fluid dynamics).

    To express it shortly: A fluid is nonpolar, if in the macroscopic sense of hydrodynamics, the total torque of a fluid element is described as arising from usual force, i.e.,
    [tex]\rho \vec{m}=\vec{x} \times (\rho \vec{f} + \vec{\nabla} \cdot \hat{T}.[/tex]
    Then the total angular momentum density is given by
    [tex]\vec{J}=\rho \vec{v} \times \vec{v}.[/tex]
    Then the balance equation for total angular momentum of any fluid volume, i.e., the equation expressing angular-momentum conservation leads to the conclusion that the stress tensor is symmetric, and we call such a fluid nonpolar.

    If the fluid is polar, then there is an additional intrinsic torque, and you have body torque and couple stress
    [tex]\rho \vec{m}=\vec{x} \times (\rho \vec{f} + \vec{\nabla} \cdot \hat{T}) + \rho \vec{g} + \vec{\nabla} \cdot \hat{C},[/tex]
    and the angular momentum contains an intrinsic part of the fluid, i.e., a kind of macroscopic "spin"
    [tex]\vec{J}=\rho \vec{v} \times \vec{v}+\rho \vec{S}.[/tex]
    Then the balance equation for angular momentum of a fluid element modifies, and the stress tensor is not necessarily symmetric anymore. In the angular-momentum balance it occurs in the in the intrinsic angular momentum (spin) compensating its contribution to the balance of orbital angular momentum such that the total angular momentum is conserved as it must be. This is nicely explained in detail in Aris's book (Sect. 5.13).
     
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