[fluid dynamics] why are there no shear forces in a fluid at rest?

by nonequilibrium
Tags: isotropic
 P: 1,412 It is said that the stress tensor $\mathbf T$ for a fluid at rest is diagonal. Why is this? The fact that the fluid is at rest means there is no net force, i.e. the body and surface forces cancel each other: $0 = \mathbf F + \nabla \cdot \mathbf T$, but from this condition is does not follow that the shear forces (i.e. off diagonal element of T) are zero. Then again, that condition only formulates that $\frac{\partial \mathbf v}{\partial t} = 0$, not that $\mathbf v=0$. I'm not sure how the statement follows. Perhaps it is not a mathematical statement but more of a physical statement: shear forces can only be generated by the presence of layers with differing velocities (?), hence if v = 0 everywhere, no shear forces can be physically generated.
 Sci Advisor HW Helper Thanks P: 26,157 hi mr. vodka! at any particular point, the pressure in a fluid at rest is the same in every direction to prove this, place a constant-diameter pipe (wholly within the fluid) with both open ends at the same point, both perpendicular to the pipe itself, but facing different directions … the fluid inside the pipe can only move along the pipe, and has only two external forces along the pipe (the pressure in different directions at the two ends) … so those two forces must be equal! (for a solid, or for a viscous fluid not at rest, a similar tube could have forces along the pipe from the wall of the tube, so the argument doesn't apply)
 P: 1,412 But how do you know that there are no forces along the pipe from the wall of the tube? (as you say is true for solids) EDIT: why was this moved to the physics thread? Fluid dynamics is also a specific field of applied math... oh well. Also, tiny-tim, why are forces perpendicular to the pipe (also from the wall of the tube) no problem? (since you leave them out of your discussion)
 PF Gold P: 1,490 [fluid dynamics] why are there no shear forces in a fluid at rest? You have to look at the definition of the stress tensor in a fluid $$T_{ij} = -p\delta_{ij} + T^{\prime}_{ij}$$ where $p$ is the pressure, $\delta_{ij}$ is the Kronecker delta, and $T^{\prime}_{ij}$ is the viscous stress tensor and is dependent only on the rate of deformation of the fluid. For a Newtonian fluid, the actual definition of $T^{\prime}_{ij}$ is as follows: $$D_{ij} = \frac{1}{2}\left( \frac{\partial v_i}{\partial x_j} + \frac{\partial v_j}{\partial x_i} \right)$$ $$T^{\prime}_{ij} = \lambda D_{kk} \delta_{ij} + 2 \mu D_{ij}$$ where $D_{ij}$ is known as the rate of deformation tensor, $\lambda$ and $\mu$ are Lamé's constants. $\mu$ is simply the dynamic viscosity and $\lambda$ relates to compressibility. Clearly, for a fluid at rest, $D_{ij} \equiv 0$ and therefore $$T_{ij} = -p\delta_{ij}$$ which is a diagonal tensor.
 P: 1,412 That's not how we introduced the stress tensor, we introduced it simply as a surface force, without specifying the cause. It seems what you are writing down is the physical cause of the stress tensor, i.e. my original guess (in my OP) that the reason is physical rather than mathematical seems correct (?) More specifically it's because a strain force can only be generated by layers of fluid with differing velocities, as your formula indeed suggests. Also oddly I didn't like the proof that the pressure is isotropic that was given in my book (after stating that the stress tensor is diagonal, which I can now accept). I also didn't like tiny-tim's proof *for now* due to the objections I raised in my last post. And now I see that according to you it's simply by definition. Hm, I can't seem to find a satisfying proof that pressure is isotropic (in a fluid at rest).
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P: 26,157
 Quote by mr. vodka But how do you know that there are no forces along the pipe from the wall of the tube? (as you say is true for solids)
because if there were (forces along the wall in a fluid at rest) …
we wouldn't call it a fluid!
 Also, tiny-tim, why are forces perpendicular to the pipe (also from the wall of the tube) no problem? (since you leave them out of your discussion)
ah, because i'm only considering movement along the tube, since that's the only movement that's possible …
forces perpendicular to the tube wall have no component in that direction!
 P: 1,412 Now that I think about it, I think the real answer is: by definition of "fluid", since a fluid is a substance that responds to any shear stress applied, hence if v = 0 there can be no shear stress. I suppose your deformation tensor confirms this since it was derived in the assumption that the behaviour is linear for a small neighbourhoud, which -although mathematically natural- is a result of the definition of fluid since that tells you the behaviour of the fluid is in a sense continuous(?) If somebody thinks I'm speaking nonsense, let me know. EDIT: ah, tiny-tim. Yes, your first segment is what I had just realized, that the definition of fluid plays a fundamental role here. As for your second segment: aha, I thought you were talking about an imaginary tube. Shouldn't the argument also work for an imaginary tube? If the argument depends on there physically being a tube in the water, I don't know, it feels like cheating in some way...
PF Gold
P: 1,490
Well, I don't know how familiar you are with continuum mechanics, but the definitions are not arbitrary. Most fluid mechanics books approach the subject on a physical basis, and indeed you can derive most of it from a force balance (which is quite easy is you assume incompressible flow). For a more mathematical treatment, the viewpoint of continuum mechanics may hold more appeal. The book I learned from was Introduction to Continuum Mechanics by Lai, Rubin and Krempl. It is a decent book. There are a few things I didn't like about it but at one point I sat down and read it cover-to-cover (in studying for a crazy final exam) and it was pretty clear, at least to me.

As for proving that pressure is a scalar (i.e. that its tensor is isotropic), you could just sum up the forces you have on a differential fluid element. What it ends up coming down to is that sure, pressure can vary significantly over even small distances, but the equations are valid at a point, so if you take the limit as the size of the differential element goes to zero, the pressures on each side of the element approach the same value.

 Quote by mr. vodka Now that I think about it, I think the real answer is: by definition of "fluid", since a fluid is a substance that responds to any shear stress applied, hence if v = 0 there can be no shear stress. I suppose your deformation tensor confirms this since it was derived in the assumption that the behaviour is linear for a small neighbourhoud, which -although mathematically natural- is a result of the definition of fluid since that tells you the behaviour of the fluid is in a sense continuous(?) If somebody thinks I'm speaking nonsense, let me know.
True, the definition of a fluid is a material which deforms continuously under the influence of an applied shear stress. Therefore, if the fluid isn't moving, there is no shear stress. I suppose we could have stated that and left it there, but I figured you wanted something more... mathematical based on your original post. The rate of deformation tensor I mentioned is linear, though I should caution you that this is only necessarily the case for Newtonian fluids. Other non-Newtonian fluids can have some funky stress tensors, though all have a shear stress that goes to zero when the fluid is at rest.
 P: 1,412 How do you conclude that the pressure on each side of the element approach the same value? (without assuming it's a scalar...)
 PF Gold P: 1,490 Hmm... perhaps a different approach would interest you: http://stommel.tamu.edu/~baum/reid/b...ok/node55.html
 P: 1,412 Perhaps a trivial proof that pressure is isotropic: suppose $p_x \neq p_y$, then there will be a shear force for the surface with the normal $\frac{1}{\sqrt{2}} \left( \mathbf e_x + \mathbf e_y \right)$ (Intuitively I had first assumed that a diagonal stress tensor implied no shear forces, but as this proof shows that obviously wrong!)
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P: 26,157
 Quote by boneh3ad Hmm... perhaps a different approach would interest you: http://stommel.tamu.edu/~baum/reid/b...ok/node55.html
it's very difficult to follow it without seeing the diagram
 Quote by mr. vodka suppose $p_x \neq p_y$, then there will be a shear force for the surface with the normal $\frac{1}{\sqrt{2}} \left( \mathbf e_x + \mathbf e_y \right)$
why?
P: 1,412
 Quote by tiny-tim why?
Darn, am I overlooking something?

I had first written it out in latex but for some reason the specific codes won't compile, but the reasoning is simple enough. So suppose p_x is different from p_y, then the stress tensor applied to the normal vector (1 1 0) won't be parallel to (1 1 0), hence there is a force on the surface parallel to the surface, which is a shear force (?)
 Sci Advisor HW Helper Thanks P: 26,157 i don't even understand the relevance of px and py (to that surface)
 P: 1,412 They're the elements of the (diagonal) stress tensor.
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