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The force density due to viscosity in an incompressible fluid is $-\mu \nabla^2 \mathbf{u}$ where $\mathbf{u}$ is the velocity and $\mu$ is the dynamic viscosity.

Let's suppose some particular small bit of fluid has volume $dV$. Then the $x$-component of force it feels is

$$f_x = -\mu dV \left( \frac{\partial^2 u_x}{\partial x^2} + \frac{\partial^2 u_x}{\partial y^2} + \frac{\partial^2 u_x}{\partial z^2}\right)$$

To understand the term $\partial^2 u_x/\partial y^2$, imagine that the entire fluid is moving in the $x$-direction only, and that $u_x$ is a function of $y$ only. Then $\partial u_x/\partial y$ describes the momentum flux in the $y$ direction, and the derivative of this is the net momentum building up at some point, and so that's a net force. The situation is like this picture from Wikipedia. $x$-momentum flows from the top of this picture to the bottom. enter image description here

The same thing applies for the $\partial^2 u_x/\partial z^2$ term, just relabel the axes.

How should I understand the $\partial^2 u_x/\partial x^2$ term? I can't have flow purely in the $x$ direction where $u_x$ depends solely on $x$ (and varies with $x$); that would violate incompressibility. Is there some simple picture that demonstrates what this term is, or if not, some simple argument that shows that this term is right and should exist, based on the understanding of viscosity you get from the picture above?

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    $\begingroup$ It is my understanding that the force density only appears in compressible fluids en.wikipedia.org/wiki/… $\endgroup$
    – user83548
    Jun 3, 2016 at 17:20
  • $\begingroup$ In a Newtonian viscous fluid, the force depends on the gradient of the velocity. There is a similar term with the so called kinematic viscosity in the incompressible navier stokes equation, are we talking about this one? $\endgroup$
    – Sanya
    Jun 7, 2016 at 21:49
  • $\begingroup$ Excuse me sir. I am not expert in fluid mechanics so much. So, please don't get uncomfortable if I am wrong. I think force is derivative of momentum with respect to the time, but you have said "derivative of this is the net momentum building up at some point, and so that's a net force" $\endgroup$
    – lucas
    Jun 28, 2016 at 16:23
  • $\begingroup$ @lucas Not sure what your point is. I am aware of the relation between momentum, force, and time. I don't see anything in what you quoted that contradicts it. $\endgroup$ Jun 28, 2016 at 18:03
  • $\begingroup$ @bruce No, the term I wrote is for incompressible fluids. See any exposition of Navier-Stokes for incompressible fluids. $\endgroup$ Jun 28, 2016 at 18:08

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The basic point is that the picture in wikipedia shows one very particular flow, a sheared laminar flow, but viscosity also acts in more complicated situations. Viscosity arises from diffusion of momentum, which acts to equalize the fluid velocity in adjacent fluid elements, no matter what the geometry of the flow.

Purely based on rotational symmetry and Galilean invariance the stress tensor must have the form $$ \tau_{ij} = -\mu (\nabla_iu_j+\nabla_j u_i) -\nu \delta_{ij} \nabla\cdot u $$ which we conventionally combine into a trace-less and a trace term $$ \tau_{ij} = -\mu \left(\nabla_iu_j+\nabla_j u_i-\frac{2}{3}\delta_{ij} \nabla\cdot u\right) -\zeta\delta_{ij} \nabla\cdot u. $$ You see that we can have a fluid for which the trace of the stress tensor vanishes, but not one for which the diagonal components are all zero.

This form of the stress tensor also arises in kinetic theory, where $$ \tau_{ij} \sim \int d^3p \, p_i p_j \, \delta f(x,p) $$ where $\delta f$ is a slight deviation of the distribution function from thermal equilibrium. For an inhomogeneous flow $\delta f\sim f_0(p_a p_b\nabla_a u_b)$, which comes from the streaming term in the Boltzmann equation acting on the distribution function $f_0$. The stress tensor can be traceless, but the diagonal components do not vanish.

All of this is true for both compressible and incompressible fluids, but obviously for a flow of the form $u_x=u_x(x,t)$, the gradient $\nabla_xu_x$ can only be non-zero is the fluid is compressible.

Postscript: A general problem with viscosity, reinforced by the wikipedia picture, is that we tend to have the wrong intuition for how viscosity operates. We tend to think/argue/assume that viscosity is friction between fluid layers, somehow caused by "stickiness" of fluid molecules (indeed, the name viscosity comes from the latin word for mistletoe). But this is not how Newtonian fluids work: There is no stickiness between fluid or gas molecules, a much better picture is provided by colliding billiard balls. In a system of colliding billiard balls, viscosity is due to momentum diffusion. This may happen between layers moving at different speed, but it also happens in more complicated situations that do not look anything like sliding layers, for example fluids undergoing anisotropic expansion.

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  • $\begingroup$ I am familiar with these sorts of arguments, but I cannot possibly use them to teach biology majors. Hence the question: "Is there some simple picture that demonstrates what this term is, or if not, some simple argument that shows that this term is right and should exist, based on the understanding of viscosity you get from the picture above?" $\endgroup$ Jun 28, 2016 at 18:41
  • $\begingroup$ I think part of the problem is that we lie when we describe the wikipedia picture. Viscosity in the Newtonian fluid is not due to friction between layers, it is due to momentum to fusion which is isotropic. $\endgroup$
    – Thomas
    Jun 28, 2016 at 19:02
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To counter one source of puzzlement, if you did have one-dimensional flow with $\partial_{x} u_x=0$, you would also have $\partial_{xx} u_x=0$ and that funny term would go away. This may still not reach your biology students, but since they do understand partial derivatives, note that in an incompressible fluid $$\partial_x u_x+\partial_y u_y+\partial_z u_z=0$$ so the viscous term can be rewritten $$-\partial_{xy}u_y-\partial_{xz}u_z+\partial_{yy}u_x+\partial_{zz}u_z =-\partial_{y}(\partial_xu_y-\partial_y{u_x}) -\partial_z(\partial_zu_x-\partial_xu_z)=-\partial_y\omega_z-\partial_z\omega_y$$ and the force in the $x$-direction is due to lateral gradients of vorticity. I believe that this might generate some illusion of understanding

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