Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

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: $$ \frac{\partial u}{\partial t} + a\cdot\nabla v - \nabla(\nu \nabla v )=f $$

, 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):

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

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 $\nabla^2v$)

EDIT1: Added MathJax Equations.

share|improve this question
    
You can use MathJaX here. –  Dimensio1n0 Feb 23 at 11:55
    
Thank you, didn't knew that. Much better this way. –  Alfonso Santiago Feb 23 at 12:12
    
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. –  tpg2114 Feb 23 at 12:38
    
Also, tangential comment -- artificial viscosity is a standard procedure in all numerical approaches, not just FEM. –  tpg2114 Feb 23 at 12:39
add comment

1 Answer 1

up vote 1 down vote accepted

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

share|improve this answer
1  
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 –  tpg2114 Feb 23 at 23:07
    
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. –  Alfonso Santiago Feb 24 at 10:28
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.