To solve problems in fluid dynamics one states often the assumption that the flow is vortex free i.e.

$rot(u) = 0$

It is a basic assumption which is needed for potential flow problems etc.. My question is, how do you know that a flow is vortex free? When are you allowed to assume that?

In literature $rot(u) = 0$ often occurs, when you are dealing with inviscid fluids but I haven't found a source yet, which makes precise correlation between those two situations like (inviscid fluid => $rot(u) = 0$ or inviscid fluid <= $rot(u) = 0$ or inviscid fluid <=> $rot(u) = 0$).


First important thing to understand is that vortices and vorticity are not the same thing, despite the similarity of the words. A vortex is a region in a flow with spinning features (at a rather large scale if you wish), but it may be irrotational (zero vorticity). Vorticity is a local property of the fluid, the rate of rotation of an imaginary particle fluid at that point. A viscous flow between two plates is actually rotational (non-zero vorticity), although it is laminar (layered).

To answer your question regarding potential flow. A potential flow is a flow where the velocity field derives from a potential. In 2D for example we'd have, given the velocity potential $\phi$: $$ u = \frac{\partial}{\partial x} \phi\\ v = \frac{\partial}{\partial y} \phi$$ Such a flow is necessarily irrotational ($\nabla \wedge \vec{u} = 0$), since $$ -\frac{\partial}{\partial y}u+\frac{\partial}{\partial x}v = -\frac{\partial}{\partial y}\frac{\partial}{\partial x}\phi + \frac{\partial}{\partial x}\frac{\partial}{\partial y}\phi = 0$$ Note that it doesn't tell us anything directly about vortices.

But how do we know if we're supposed to suspect a potential flow in the first place? A clear no-go are viscous flows with no-slip boundaries. A no-slip boundary (in a reference frame where the boundary is at rest) is characterized by $$ \vec{u}\vert_\text{boundary} = 0$$ The fluid in contact with the no-slip boundary (hereafter just "wall") does not move, but the one slightly above does. If the wall is parallel to the $x$-axis, then this means that the variation of the flow in the $y$-axis must be non-zero: $$ \frac{\partial}{\partial y} u \neq 0$$ else the fluid would never move. But as soon as you have that, then of course $\omega= \nabla \wedge \vec{u} \neq 0$ too! For a viscous fluid, there is always vorticity generated at a wall.

For non-viscous flows, walls are given by the condition $$\vec{u}\cdot\vec{n}\vert_\text{boundary} = 0$$ i.e., no flow may go through the wall, but it might well slip tangentially without any loss of speed. Given a reasonable geometry the flow can move around, there should be no vorticity. I wouldn't bet on what happens if you blow a perfect fluid into a closed box: hard to believe that the flow impacting with box walls (what with the corners and all) wouldn't be rotational, but I'd have to check.

There's still another mechanism that can introduce $\partial_y u\neq0$ or $\partial_x v \neq 0$: very particular body forces. That's actually one of the means by which you can study turbulence in a periodic box of fluid (i.e., without walls): you generate shear by a random body force. But you won't see those in most cases.

In applications, potential flow is particularly used in the inviscid flow approximation around profiles, with the flow extending to infinity all around (avoiding the "box scenario" I mentioned above). At most you'd have the gravity force, but that's a nice uniform field in most approximations.

Moreover, in 2D, an irrotational inviscid flow stays irrotational in time by the Kelvin circulation theorem. Good stuff. Ever noticed that we always study airfoils in 2D? Interesting.

(There's another reason for studying airfoils in 2D: the lift of a wing section per unit length is actually quite well predicted, whereas in 3D inviscid flows, the lift is always zero! See this other question. That's where more advanced methods kick in if you need to calculate stuff in 3D anyway.)

If I missed something, ask me some questions through the comments and I'll try to update the answer.

  • $\begingroup$ We encounter free-vortices in viscous flow as well. in many confined-vortex flows I have studied, this free vortex exists Rankine vortex . although this is not a completely free vortex, note that a totally free vortex cannot exist in a real scenario. $\endgroup$ – aditya kp Apr 29 '13 at 4:21
  • $\begingroup$ Hello @aditya kp. Good example of a vortex model for viscous flow. The vorticity profile is however particular: it's constant in the core, zero outside (discontinuous). $\endgroup$ – Christoph B. Apr 29 '13 at 8:16
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    $\begingroup$ yes. that's true. but I did not understand why you are pointing that out. Could you please explain? $\endgroup$ – aditya kp May 1 '13 at 6:53
  • $\begingroup$ My answer tries to distinguish the concepts of vortex and that of vorticity. The Rankine vortex describes a rotating motion of fluid, but the vorticity is zero outside the core. The remark about the vorticity being discontinuous at the core edge points out its model-like nature. The pressure associated with this vortex is discontinuous at the core edge too. One would expect smooth solutions from the Navier-Stokes equations (at least for incompressible flows and low velocities). $\endgroup$ – Christoph B. May 1 '13 at 8:53

You are confusing terms. It is never assumed that a flow is vortex free. A vortex is a generic term used to describe fluid that is spinning about some axis. Potential flow uses the assumption that a flow is irrotational or $$ \vec{\omega} =\vec{\nabla}\times\vec{U} = 0$$ Irrotational flow can in fact be composed from vorticies. What is even more confusing to some people is that a simple shear flow is not irrotational despite the fact that their is no physical rotation involved.

The justification behind the validity of the irrotational assumption for some incompressible flows can be seen through the vorticity evolution equation $$\frac{D\vec{\omega}}{Dt}=\nu\nabla^2\vec{\omega} + \vec{\omega}\cdot\nabla\vec{U}$$This equation demonstrates that regions that start out irrotational tend to remain so in the absence of external interaction that leads to diffusion of vorticity into irrotational zones.


Viscosity opposes adjacent layers of fluid moving past each other. If you had a vortex, then at different radial distance, the rotating layers of fluid should have different velocities in order to avoid them moving past each other. Imagine twisting the fluid, to make a vortex... for a vortex to be formed, you want the fluid at a smaller radius to drag along the fluid at a larger radius, and make it rotate. Else, you cannot create vortices. So, if you don't have viscosity, you can't create vortices.

Viscosity doesn't always imply vortices, since you could subject the fluid to a nice laminar flow. Typically, only if you stir, or somehow "force" a fluid, will you form vortices. However, things might get very different at higher Reynolds number, where the flow will be turbulent and I don't have a handle on that.

I propose a (thought) experiment. You could actually perform this and check, if you wish. I just came up with this idea and have not done this experiment. So if someone does try performing this, I'd like to hear about it.

Out of density, pressure and viscosity, you can generate a time-scale by dimensional analysis. If you try to stir a fluid faster than this time scale, you will (presumably) notice that it starts forming a vortex, but not if you stirred slower. Alternatively, it should be "easier" (from the point of view of stirring rate) to form vortices in more viscous liquids. Note that "easier" in the sense of rate of rotation while stirring. For a denser liquid, you will have to put in more effort for the same qotational speed while stirring.


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