# How can I derive the stress tensor for a Newtonian fluid in more physical terms?

The question is quite fundamental and more on a beginner's level (not sure if good in this high-level-forum, but I try): I have big problems in understanding the stress tensor for Newtonian fluids in terms of velocities u.

The result (assuming $\mu'=0$) is (according to my text book)

$\tau_{ij} = \mu (\partial _j u_i + \partial _i u_j -\frac{2}{3}\delta_{ij} \partial _k u_k)$

I know the derivation of that, but it is not very intuitive from a physical perspective. I would like to understand in particular, why we have the term $\partial _j u_i$

From the following image and the definition of viscous stresses I would naively expect the stress in 1-direction on surface-2 just to be

$\sigma_{12} = \mu \partial _2 u_1$

Why is there also the contribution $\partial _1 u_2$ ?

And where does the symmetrical additional 2/3-term come from?

I understand readily the derivation of my textbook, but this is rather mathematically and I cannot see physics behind (yet). EDIT:

I have seen that asymmetry is a consequence of having zero momentum along all axes. I didn't recognize that, but now its clear.

• The reason for the complicated mathematics is to guarantee that the state of stress that the law describes must be independent of the motion of the observer (which, of course, can not affect the state of stress in a fluid), even including rigid body rotations of the observer. Basically, Newton's empirical law of viscosity says that the state of stress in a fluid must be a linear function of the components of the velocity gradient tensor. The constant of proportionality is backed out by considering the case of only a single component to the velocity gradient, and matching to that case. – Chet Miller May 16 '18 at 11:56
• This might or might not help. Also, this might or might not be correct. At the end of the day, the reason why things get confusing is because of the philosophical differences between "continuous mathematics" and "discrete mathematics." In physics, we start off studying discrete systems (i.e. particles). Applying "continuous math" (read derivatives/integrals/continuous number lines) to "discrete particles" is much much easier than applying "continuous math" to "continuous systems." The fundamental difference is the difference between points in 3D space and infinitesimal regions in 3D space – DWade64 May 16 '18 at 20:43
• With points, you can say "there exists a force at this point in a given direction." With continuous matter, we have to get used to 1) force densities (read forces per unit area) because that's the only thing that makes sense in this context (force at points in continuum don't make sense. Define force density just as mass density makes more sense than mass for a continuum. How much mass is located at a point in a fluid?) and 2) that these force densities vary with direction. Point 2 is harder to understand intuitively. Especially if you try relating this to introductory physics – DWade64 May 16 '18 at 21:00
• because there is no comparison. It's a completely different way of thinking – DWade64 May 16 '18 at 21:00
• Related, if not dupe of, physics.stackexchange.com/q/152927/25301 – Kyle Kanos Dec 21 '18 at 23:28

## 1 Answer

It's subjective to say what's "intuitive" for one person, but here's one way to think about it based on three physical assumptions.

You could a priori say that the stress tensor $$\bar{\bar{\sigma}}$$ is a linear function of the velocity gradient tensor $$\nabla\vec{v}$$. There are a couple of physical ways to justify this: the current state of stress is independent of the previous states of stress, the relationship is invariant with respect to classical inertial frames so no dependence on $$\vec{v}$$, etcetra.

This means you can represent the stress-strain relationship as

$$\bar{\bar{\sigma}} = \mathcal{C}\nabla\vec{v}$$

where $$\mathcal{C}$$ is a fourth-order tensor. Now it's a question of figuring out what elements of the tensor $$\mathcal{C}$$ are zero.

We can then assume that the stress-strain relationship doesn't change with respect to the orientation of the way you look at it, which mathematically translates to the $$\mathcal{C}$$ tensor being isotropic. This is certainly true for simple liquids, where we assume the constituent atoms/molecules are approximately spherical for all intended purposes, but isn't true if we have rod-like polymers/molecules making up our fluid. (Doing the math, this kills off a great deal of elements in the above tensor.)

This isotropy, combined with the assumption that the fluid does not carry surface couples or internal torques, will kill off any remaining extra terms and immediately generates the Newtonian fluid model. (This assumption is almost always true expect for materials that have nontrivial magnetic coupling or some other exotic effects.)

Hope this helps!

• If you think that the notation is terrible, why mark the order at all? You could write just $\sigma = C \, \nabla \vec{v}$ and then say that $C$ is a fourth-order tensor (you say that anyway, even with all the bars). – md2perpe Dec 21 '18 at 20:00
• Very true; I’ll update accordingly! – aghostinthefigures Dec 21 '18 at 20:38