# Stress tensor and viscosity relation

## Main Question or Discussion Point

Can anyone tell me why this is true? I can't find an explanation anywhere, and it doesn't make sense to me geometrically either, especially for i=j. Isn't viscous force, by definition, a shear force? How can it produce a normal stress? Also why would Txy=Tyx? Why can't they be different, producing a net moment on the volume element? What says that can't happen? I'm getting extremely frustrated.

$$\mathbb{T}_{ij} = \mu\left(\frac{\partial u_i}{\partial x_j} + \frac{\partial u_j}{\partial x_i}\right)$$

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## Answers and Replies

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Can anyone tell me why this is true?
Why what is true? Your expression for for the stress tensor is certainly not a general expression, if that's what you mean. It just describes the contribution of bulk (a.k.a volume or second) viscosity to the stress tensor. In general there may also be non-zero shear. It my help if you gave the context of where you found it.

I can't find an explanation anywhere, and it doesn't make sense to me geometrically either, especially for i=j. Isn't viscous force, by definition, a shear force? How can it produce a normal stress?
No, there can certainly be both shear and normal stresses on a system, and the bulk viscosity tells you the response of the system to a normal stress. If you have an incompressible fluid, the volume can't change and the divergence is always zero, so such an effect doesn't show up. That may be what you're thinking of.

Also why would Txy=Tyx? Why can't they be different, producing a net moment on the volume element? What says that can't happen? I'm getting extremely frustrated.
I was always under the impression that the stress tensor is always symmetric unless there is something external to your system that breaks conservation of angular momentum (or if you're considering extensions to general relativity with torsion). I could be wrong about this in general, but it's certainly true that in most cases the stress tensor will be symmetric. See
http://en.wikipedia.org/wiki/Normal_stress#Equilibrium_equations_and_symmetry_of_the_stress_tensor
and
http://en.wikipedia.org/wiki/Stress-energy_tensor

Andy Resnick
Can anyone tell me why this is true? I can't find an explanation anywhere, and it doesn't make sense to me geometrically either, especially for i=j. Isn't viscous force, by definition, a shear force? How can it produce a normal stress? Also why would Txy=Tyx? Why can't they be different, producing a net moment on the volume element? What says that can't happen? I'm getting extremely frustrated.

$$\mathbb{T}_{ij} = \mu\left(\frac{\partial u_i}{\partial x_j} + \frac{\partial u_j}{\partial x_i}\right)$$
As the_house said, this is not a general expression for the stress tensor. It's not even the whole expression for the stress in a perfect fluid.

Sorry, what I meant to ask is why is the stress tensor due to viscous forces equal to the expression I've posted.

The context is a Newtonian fluid with a velocity V=u(x,y,z,t)i + v(x,y,z,t)j + w(x,y,z,t)k

I thought by definition, the viscous shrear stress is the viscosity times the change of the parallel velocity component with respect to the orthogonal coordinate. I.e. τyx =μ du/dy.
τxy should be μ dv/dx, independent of how the x-velocity (u) changes with y.

To me it makes perfect sense to have τyx≠τxy, in which case you have a net moment on the volume element, meaning it has some angular acceleration. What's the problem with that, shouldn't that be allowed in fluids?

P.S. Got it from http://en.wikipedia.org/wiki/Deriva...es_equations#Application_to_different_fluids", and I'm ignoring the 2nd coefficient of viscosity terms.

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That will teach me not to read through a post so quickly. Ignore what I said about that term being a divergence, and the coefficient being bulk viscosity. Indeed, what you have written is a shear stress, whose coefficient is the shear, or "first", viscosity.

The rest of what I wrote should be correct. The stress tensor is almost always symmetric. As I mentioned, I believe it is because you won't conserve angular momentum otherwise (unless the antisymmetric part can be written as total divergence, in which case it has no effect). In the absence of "external" forces, this should be the case. I don't have time to look into it more carefully, though, so maybe I'm missing part of the argument for why it has to be symmetric.

Yes, apparently for Newtonian fluids, the stresses must be locally in equilibrium (hence the symmetry). I found one short paper on this, however I'm not following their explanation. Read on the 2nd page.

http://www.owlnet.rice.edu/~ceng501/Chap5.pdf

Does anyone know how to explain this? Why must the stresses be in equilibrium locally (as the volume element becomes infinitesimal)? I tried drawing blocks on paper and I still don't see it.

It's not just a property of Newtonian fluids or local equilibrium. It's much more general than that. See page 4 of the pdf you linked to "The symmetry of the stress tensor". As he says, it has to do with local conservation of angular momentum. Unless you have something external that applies a torque on your system, the stress tensor has to be symmetric.

Andy Resnick
Sorry, what I meant to ask is why is the stress tensor due to viscous forces equal to the expression I've posted.

<snip>
To me it makes perfect sense to have τyx≠τxy, in which case you have a net moment on the volume element, meaning it has some angular acceleration. What's the problem with that, shouldn't that be allowed in fluids?

P.S. Got it from http://en.wikipedia.org/wiki/Deriva...es_equations#Application_to_different_fluids", and I'm ignoring the 2nd coefficient of viscosity terms.
Yes, apparently for Newtonian fluids, the stresses must be locally in equilibrium (hence the symmetry). I found one short paper on this, however I'm not following their explanation. Read on the 2nd page.

<snip>
I'll chime in also- the symmetry of the stress tensor is true only for nonpolar materials- for example, fluids that cannot transmit or generate torques (Cauchy's second law). The most general relationship for the balance of stress (Cauchy's fundamental theorem) is:

mijq,q + $\rho$Lij = T[ij]

where the first term is the divergence of the couple-stress tensor, $\rho$ the mass density, L the applied couple field (body torques), and T[ij] means Tij-Tji.

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