I have a professor that is fond of saying that vorticity cannot be destroyed. I see how this is true for inviscid flows, but is this also true for viscous flow? The vorticity equation is shown below for reference. From this equation, it looks as if vorticity only convects and diffuses. This would suggest that it can't be destroyed.

$$\frac{D\boldsymbol{\omega}}{Dt} = (\boldsymbol{\omega}\cdot\nabla)\boldsymbol{V} + \nu\nabla^2\boldsymbol{\omega}$$

However, consider this thought experiment:

Suppose we have a closed container filled with water with initial vorticity field $\boldsymbol{\omega}_0$ at time $t_0$. If the container is allowed to sit undisturbed, as $t\to\infty$ the water will become stationary ($\boldsymbol{V}\to 0$) with zero vorticity ($\boldsymbol{\omega}\to 0$).

This suggests vorticity can be destroyed.

My professor claims the boundary-layer vorticity at the sides of the container is equal and opposite in sign to the bulk vorticity. If this is the case, the vorticity cancels out after a long time resulting in the stationary fluid and vorticity is not destroy (just cancelled out).

EDIT: I'm looking for either a proof that the boundary-layer vorticity is equal and opposite to the bulk vorticity or a counter-explanation or proof. (I'm using proof in a very loose hand-wavy sense)

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    $\begingroup$ It sounds like you've answered your own question. What, exactly, would you still like clarified? $\endgroup$ Apr 23, 2013 at 21:07
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    $\begingroup$ @IlmariKaronen I'm not entirely convinced the boundary-layer vorticity is always equal and opposite in sign to the bulk vorticity. What if the closed container was filled with superfluid helium. Initially there would be no boundary layer (no boundary layer vorticity) but suppose the initial vorticity field is the same as in the previous case. After a while suppose the helium is slightly heated just enough that a boundary-layer can form but not so much that the container blows up. I don't know if boundary-layer vorticity explains this. $\endgroup$
    – OSE
    Apr 23, 2013 at 21:17
  • $\begingroup$ Admittedly, the superfluid helium situation is a bit of a stretch and most likely introduces many more complications. It just seems like that does not always have to be the case. I guess the answer that I am looking for is a proof that boundary-layer vorticity is equal and opposite to bulk vorticity. $\endgroup$
    – OSE
    Apr 23, 2013 at 21:20
  • $\begingroup$ Let's say we have a glass of water sitting still and I start stirring it to form a vortex. Since this is the first time I've come across the term "boundary layer vorticity", I assume you mean a layer of water along the walls of the glass, spinning in the opposite direction (Correct me if I'm wrong). Then I stop stirring (take my spoon out) and let the fluid flow come to rest. $\endgroup$
    – Siva
    Apr 25, 2013 at 2:57
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    $\begingroup$ Vorticity can certainly be destroyed, this is the basis of the energy cascade in 3D turbulence where energy is channeled across wavenumber space from large to small scales all the way down to the Kolmogorov scale at which point it dissipates into heat. Of course in order to do that you need to be looking at the complete equations... this link should answer all your questions. eng.utah.edu/~mcmurtry/Turbulence/turbvort.pdf $\endgroup$ Apr 25, 2013 at 4:21

5 Answers 5


Your professor is correct, but I agree with you that the statement “vorticity can’t be destroyed or created” seems jarring - I would prefer to think of this as “vorticity is conserved” because the conservation of vorticity derives from the Navier-Stokes Eq and the conservation of angular momentum. I confess this is splitting terminology hairs (don’t push it with your professor) but I think it helped me.

So, I think, maybe I can understand this as an analogy with linear momentum, because linear momentum is conserved too. I remember the problem of a car of mass m, traveling toward the right at velocity v, and on the same road an identical car traveling to the left at velocity –v. They collide head-on and smash and stick together. Velocities after the crash – zero. Momentum after the crash – zero and of course, momentum is conserved. The total momentum of the system was zero before and after.

Let say your container filled with water is a long annulus with thick steel walls. The flow is initially a circular flow around the axis (i.e. 2D flow.) What is initial total angular momentum of the system? Eventually the fluid stops moving, so the final total angular momentum of the system must be zero. How do we show that the initial angular momentum is zero too?

At this point you need to recognize that the vorticity vector in the moving fluid is everywhere parallel to the axis of the container. And you need to use Stokes’ theorem to write an integral equation with a line integral on the LHS (the circulation) and a surface integral on the RHS (vorticity integrated over the container cross section. )

\begin{align*} \oint_{C} v \cdot dl = \int_{S} w \cdot dS\ \end{align*}

Take your integration path (the closed path C) entirely inside the steel wall of your container. The velocity inside the container wall is always zero, and so the circulation along the path is always zero, and so the total vorticity across the cross-sectional area (the area S) of the container and fluid is always zero too.

You can calculate for yourself that the boundary layer vorticity is equal and opposite to the bulk viscosity using a similar approach. Imagine spinning the container about its axis at a constant angular velocity. Eventually the entire viscous-fluid and container system will be rotating like a rigid body around the axis. Every point has the same angular velocity, and there is now a vortex located at the center. Compute the circulation for any closed path that includes the vortex inside it – this will be the strength of the vortex, and the magnitude and sign of vorticity in the bulk fluid. You can show yourself that the strength of this vortex is the total vorticity. Compute the circulation around any path that does not include the vortex, this will always come out to be zero. Pick a path near the fluid-container boundary, s’, so half of it is in fluid and half is inside the container wall, as long as the container and fluid are still rotating together the circulation around this path will be zero too. Now stop the container’s rotation. The fluid continues to move. Compute the circulation around the path s’ again, it is no longer zero and is the vorticity at the boundary layer. It’s sign is opposite that of the vortex at the center. Every point along the fluid-boundary can be associated with path like s’ and a small amount of boundary layer vorticity. Integrate around the entire boundary and the sum will be equal in magnitude and opposite in sign to the strength of the vortex at the center. Eventually, boundary layer vorticity will diffuse towards the center and annihilate the center vortex.

@Isopycnal_Oscillation is correct to point out that in 3D, and particularly near turbulent conditions, vorticity is not conserved. The second term on the RHS of your ‘transport equation’ says that the stretching and tilting of vortex tubes can change vorticity too. However, I expect that in the classes where your professor is fond of saying that “vorticity cannot be destroyed” turbulent flow is seldom if ever encountered.

Finally, assuming that the LHS of your ‘transport equation’ equals zero does not necessarily require that the fluid be inviscid or that the problem be 2D – you are assuming that the terms on the RHS happen to cancel exactly and the vorticity is fortuitously ‘steady-state.’ So yes, that is a very strong assumption to accept.


Vorticity can certainly be destroyed, this is the basis of the energy cascade in 3D turbulence where energy is channeled across wavenumber space from large to small scales all the way down to the Kolmogorov scale at which point it dissipates into heat. Of course in order to do that you need to be looking at the complete equations... this link should answer all your questions.

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    $\begingroup$ This is interesting, but I think it doesn't quite answer the central question, which is about the equations for an incompressible (but viscous) fluid, in which case vorticity seems to be conserved. Is it necessary to consider a compressible fluid in order to see how a glass of water with initial vorticity settles? $\endgroup$
    – AJK
    Apr 25, 2013 at 18:49
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    $\begingroup$ It is conserved in the 2D case and for high Reynolds numbers where diffusion is negligible. In that case vorticity is just advected within the fluid as a scalar just like a particle would. Moreover, since the flow is confined to a plane, vortex stretching is not allowed. That is why in 2D you get the inverse energy cascade where energy flows from smaller to larger scales - there is no mechanism to break up large eddies into smaller eddies. $\endgroup$ Apr 25, 2013 at 21:52

Vorticity is indeed a linear transformation of momentum, which is conserved in isolated systems. So if your system is isolated (no transfer of mass and energy) momentum,mass, energy and vorticity are conserved. Otherwise you can transfer vorticity through the boundaries. This is what happens to your tank with initial vorticity. Because it is not an isolated system (it is transferring momentum to the the tank walls). If you put your tank on a rotating plate of zero mass and with no friction and exclude air friction on the free surface, your tank will rotate forever with the fluid and the initial vorticity of the fluid will be equal to the final vorticity of the fluid+tank.

The vorticity at the boundary layer is not always equal to the vorticity at the bulk. Your professor is wrong. Imagine the following with your tank: you extract some of the initial vorticity to spin a wheel with no friction. With the logic of your professor, ( bulk and boundary vorticity cancel out). But your wheel is still spinning! Where does this energy come from?

In fact, the vorticity in the fluid is drained slowly by transfering momentum to the walls of the tank through the boundary layer and thus to the Earth, if the tank is not on our frictionless plate.

PS: No real system is truly isolated. But you can imagine your tank with free-slip condition at the walls. In this case there is no momentum transfer through the boundaries and the fluid keeps diffusing its vorticity until it reaches a state of solid body rotation where the surface stress field is identically zero and thus the state of the fluid is only balanced by pressure-gradients and centrifugal forces. At this point, diffusion of vorticity stops as you can see from your equation. But the total amount had always been the same.

Confusion with the energy cascade: Many think vorticity can be destroyed or created because of the energy cascade and dissipation or to the vortex stretching term $(\omega \cdot \nabla)\vec V$. This because some authors call it as a vortex production term. This is misleading. What can be produced or destroyed is not vorticity but enstrophy, which are the equivalent of momentum and kinetic energy. Momentum is conserved and kinetic energy is dissipated through viscosity. Similarly, vorticity is conserved and enstrophy is dissipated trough viscosity. Another point critical to understand is the difference of the equation for the evolution of vorticity in a material particle, or loop as circulation, and the integral of vorticity over the whole domain. Of course in a particle it will in general change (will be diffused, stretched, tilted), but the integral of an isolated system remais the same.


I just stumbled across this one, and since no clear answer has been given so far, I'll throw in my two cents:

Your professor is wrong, plain and simple.

The fundamental equations clearly show that vorticity is not conserved in viscous flows, neither in the compressible nor the incompressible case. In the 2-D incompressible case, the vorticity transport equation looks like a convection-diffusion equation, but this is misleading. There will, in general, be vorticity production on the boundary, and vorticity is lost through dissipation in the interior of the domain. In the 3-D case, vorticity can also be generated through nonlinear effects (vortex-stretching).

I notice that the vorticity equation posted in the original question is correct, and thus is (a) clearly not just a convection-diffusion equation, and (b) that a quantity described by a convection diffusion equation is in general not conserved.

I note in passing that some of the material posted previously confuses vorticity with circulation, or the existence of vortices. Those are different concepts.

P.S.: The statement "I'm looking for either a proof that the boundary-layer vorticity is equal and opposite to the bulk vorticity or a counter-explanation or proof." really makes no sense. Vorticity is a continuous vector field, and there is only one such field. The term "bulk viscosity" is undefined.


I agree with the comments made by the individual just above and would like to make few observations. Circulation is the quantity derived from vorticity which is conserved for barotropic ($\rho = \rho(P)$), inviscid and in the absence of non conservative forces. This does not hold for vorticity which can be altered by compressibility and non linear effects within the fluid. Your professor might have been talking about Circulation under special circumstances.

Also remember vorticity is more of a mathematical concept. The most physical sense to vorticity would come from its magnitude which is twice that of angular velocity and that is jus that. So it should not be confused with vortex shedding or generation, dissipation etc. all being physical aspects of the flow.


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