I'm studying fluid mechanics in more depth during my Ph. D. and there is something related with the diffusive term that has been bothering me for a long time. Looking at the convection diffusion equation: $$ \frac{\partial v}{\partial t} + a\cdot\nabla v - \nabla(\nu \nabla v )=f, $$

and thinking in Fick's law, it's not hard for me thinking in diffusion as the process by which, for example, the particles of a solution with a concentration gradient get apart one from each other to "reduce" the energy of the system; being, each one of these particles, more "comfortable" inside the solvent, which is the same as saying "with the bigger free mean path possible".

But now when I look to the Navier Stokes equation (incompresible and viscous):

$$ \rho \left(\frac{\partial \mathbf{v}}{\partial t}+\mathbf{v}\cdot\nabla\mathbf{v}\right)=-\nabla p+\mu\nabla ^2\mathbf{v}+\mathbf{f} $$

I can easily see the viscous term as a diffusive one, but there's no way I can relate it with Fick's law. So, somebody can explain me how can I see viscosity as a diffusive process?

If you are interested in why I'm asking this, it's because in FEM there is a stabilization method called artificial viscosity that add some viscosity to increase diffusion and make the model more stable. So I understand a) why artificial viscosity increases the diffusion, b) why diffusion stabilizes the equation; but I don't understand why viscosity is diffusive (besides the fact that $\mu$ is multiplying $\nabla^2v$)

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    $\begingroup$ I'm unclear on what you're asking here... You say that you can see mathematically why "viscosity" is diffusive -- it is defined as $\mu \nabla^2 u$ so it is the mathematical definition of a diffusion term. Are you looking for a physical explanation of what viscosity is? What about linking it to Fick's Law is confusing you? Fick's Law describes the change in concentration of a quantity -- in the momentum equation, that quantity is velocity. $\endgroup$
    – tpg2114
    Commented Feb 23, 2014 at 12:38
  • $\begingroup$ Also, tangential comment -- artificial viscosity is a standard procedure in all numerical approaches, not just FEM. $\endgroup$
    – tpg2114
    Commented Feb 23, 2014 at 12:39

2 Answers 2


Yup, I wrote something like that in the wikipedia article...

The correspondence [between the Navier-Stokes equation and the convection-diffusion equation] is clearest in the case of an incompressible Newtonian fluid, in which case the Navier–Stokes equation is: $$\frac{\partial \mathbf{M}}{\partial t} = \frac{\mu}{\rho} \nabla^2\mathbf{M} -\mathbf{v} \cdot \nabla \mathbf{M} + (\mathbf{f}-\nabla\text{P})$$ where M is the momentum of the fluid (per unit volume) at each point (equal to the density $\rho$ multiplied by the velocity v), $\mu$ is viscosity, P is fluid pressure, and f is any other body force such as gravity. In this equation, the term on the left-hand side describes the change in momentum at a given point; the first term on the right describes viscosity, which is really the diffusion of momentum; the second term on the right describes the advective flow of momentum; and the last two terms on the right describes the external and internal forces which can act as sources or sinks of momentum.

So your question is: WHY is viscosity "really the diffusion of momentum"?

Well, think about what viscosity does. If you have two nearby regions of fluid with very different momenta, say a slow-moving region right next to a fast-moving region, viscosity describes the process whereby the slow region picks up some of the momentum from the fast region, so that the slow region moves faster and the fast region moves slower.

Basically, the momentum "diffuses" around, causing nearby regions to acquire similar momenta. That's the effect of high viscosity.

So, if you're thinking about it macroscopically and intuitively, it makes sense that momentum should satisfy a diffusion equation with diffusion coefficient proportional to the viscosity.

OK, that's my answer. If you're looking instead for a mathematical proof that starts from the definition of viscosity and ends with a diffusion-like term, I don't know that proof off-hand (although I'm sure it exists). Maybe someone else will give you an answer along those lines. :-D

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    $\begingroup$ The proof you are looking for is from statistical mechanics where you consider molecules from the slow moving stream being deflected by random thermal motion into the fast moving stream and vice versa $\endgroup$
    – tpg2114
    Commented Feb 23, 2014 at 23:07
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    $\begingroup$ Nono, I don´t want any mathematical proof, just a simple explanation of why viscosity is the diffusion of the momentum. Great answer, just what I was looking for. $\endgroup$ Commented Feb 24, 2014 at 10:28

I think Chapter 1-3 of the book Transport Phenomena by Bird et al. might help in understanding why.

In short, consider the stress tensor $\pmb{\tau}$. We have in Navier Stokes equation $\nabla \cdot \pmb{\tau}$ as the viscous term. It's not hard to see that this is also a diffusion term where you can define an effective flux tensor as $$\mathrm{\pmb{J_{eff}}} = \pmb{\tau}.$$

A nice summary of the momentum transport of fluids (i.e. the Navier-Stokes Equation) is in Equation 3.2-9. Diffusion is a synonym for molecular, or fluid particles, transport.

Reference: Bird, R.B., Stewart, W.E. and Lightfoot, E.N. (August 2001). Transport Phenomena (Second ed.). John Wiley & Sons. ISBN 0-471-41077-2.

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