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.

Let us consider a 2D irrotational flow, such that $\nabla\times\boldsymbol u =\boldsymbol 0$. Defining the stream function such that $\boldsymbol u =\nabla\times\psi \boldsymbol n$ where $\boldsymbol n$ is the unit vector perpendicular to the plane of the motion. The incompressibility condition implies $\nabla^2\psi =0$, so $\nabla^2 \boldsymbol u= \nabla^2 \nabla\times\psi \boldsymbol n=\nabla\times\nabla^2\psi \boldsymbol n=\boldsymbol 0$. From this follows that in the Navier-Stokes equations the term $\nu\nabla^2\boldsymbol u=0$ vanishes, no matter the viscosity.

Now, does this allow us to say that the fluid is inviscid? Or, after considering the nature of the inertia, we can say that:

  • if inertia can be neglected, the equation of the motion is $\nabla p=\boldsymbol 0$.

  • if inertia is important, the flow is at high Reynolds number.

My problem is: Is every irrotational flow inviscid? This is kind of counter-intuitive. But I think my error is in the "no matter the viscosity"...

share|improve this question
add comment

3 Answers

up vote 4 down vote accepted

A version of your proof without a stream function:

The Laplace acting on the velocity may be expressed via the curl of the curl identity and aside from the $\nabla\times (\nabla\times \vec u)$ which vanishes, you also get another term $\nabla\cdot (\nabla\cdot \vec u)$ which vanishes (only) if one assumes incompressibility (it's the conservation of the mass).

So yes, one may neglect the viscous term if the flow is irrotational. In this sense, the irrational flows are automatically inviscid, too. I believe that this is the right solution to the problem 2 on page 143 of this Chapter

http://www.lcs.syr.edu/faculty/lewalle/FluidDynamics/fluidsCh5.pdf

that is entirely dedicated to the inviscid and irrotational flows in this very combination. The converse isn't true. Inviscid flows may refuse to be irrotational: they may have vorticity.

However, I would still mention that the implication proved above isn't necessarily conceptually important. It's because while people sometimes discuss inviscid flows in which the whole viscous ($\mu \Delta \vec u$) term is zero or negligible, they're more likely to discuss "inviscid fluids". When we talk about the inviscid flows, we really want to claim that this term vanishes not because $\Delta\vec u=0$ but because $\mu$ is zero (or negligible). Even though the adjective "negligible" depends on the precise condition, typical speeds and Reynold's number etc., it's still meant to be a general property of a fluid rather than a property of some particular solutions. And of course, the fact that a fluid is able to exhibit irrotational flows does not mean that it is an inviscid fluid.

share|improve this answer
    
Yes, clearly. My doubt was about the possibility to have an irrotational flow at low Reynolds numbers. This is possible only if the inertia is neglettable, and this can be only by means of lenght and density, because the viscosity doesn't enter in the eq- of motion for an irrotational flow. Is that true? –  usumdelphini Jul 3 '12 at 7:32
    
Dear usumdelphini, irrotational incompressible flow may still have both low and high Reynolds number. For low Re, you may neglect the inertia term $\rho\vec u\cdot \nabla \vec u$. However, in that case you typically want to preserve the viscous term. The loophole is that $\nabla\cdot\vec v =0$ doesn't hold because one must include temperature-related and other effects on $\rho$. You must also be careful on whether the "irrotational character" survives at later times when it holds at $t=0$. It's not guaranteed. –  Luboš Motl Jul 3 '12 at 7:44
    
Kelvin's circulation theorem en.wikipedia.org/wiki/Kelvin%27s_circulation_theorem says that for inviscid fluids, the irrotationality is conserved if it holds at $t=0$. ... Irrotational incompressible flows may also have a high Reynolds number. In that case you keep the inertia term and ignore the viscous term. The solutions to these equations are then generally turbulent i.e. not unique. –  Luboš Motl Jul 3 '12 at 7:44
    
At first, thanks for the answers. Actually, this was the reason I took the 2D case (K.theorem) "However, in that case you typically want to preserve the viscous term." I can't, because the laplacian of $u$ vanishes, and I usually work with low Re flows with the incompressible condition satisfied. –  usumdelphini Jul 3 '12 at 7:54
    
I retry to formulate my question and give a possible answer : for irrotationality I have laplacian of the velocity equal to zero, so that I can just write the Euler equation, no matter the viscosity. If the fluid is even quasi-inertia-less, I neglet the inertia term (this is an approximation that must pass through the density and characteristic lenght and velocities) , and I get $\nabla p = \boldsymbol 0$, which is the actual low-Re irrotational equation. –  usumdelphini Jul 3 '12 at 7:56
show 3 more comments

(I try to answer to my own question, after some reflections made with the help of Luboš.)

For an incompressible and irrotational flow, the conditions $\nabla\times \boldsymbol u=\boldsymbol 0$ and $\nabla\cdot \boldsymbol u = 0$ imply, $\nabla^2\boldsymbol u =\boldsymbol 0$. Indeed:

$$\nabla^2\boldsymbol u = \nabla(\nabla\cdot\boldsymbol u)-\nabla\times(\nabla\times \boldsymbol u) = \boldsymbol 0$$

This forces us to write down the Navier-Stokes equation for the motion of the fluid without the viscous term $\mu\nabla^2\boldsymbol u$, no matter the viscosity:

$$ \rho(\partial_t \boldsymbol u + u\cdot\nabla \boldsymbol u) = -\nabla p \ \ \ ,\ \ \ \nabla\cdot \boldsymbol u = 0$$

Now, it could seem that this implies the flow is automatically a high-Reynolds number flow (for which we could have wrote down the same equation, but for a different reason: $\mu=\rho\nu\simeq 0$, and this would have been an approximation). But, even if the viscosity is far from neglectable, we can make another kind of approximation, saying that the inertia, represented by the left-hand-side terms in N-S equation, can be neglected because of $Re=UL\rho/\mu\ll 1$ (this can happen in a lot of situations: microobjects, extra-slow flows, and - of course - high viscosity. In this case, the equations of motion become: $$-\nabla p = \boldsymbol 0\ \ \ ,\ \ \ \nabla\cdot \boldsymbol u = 0$$ which are, in fact, the equations we would arrive at if we started by the Stokes equation (for inertia-less flows) for irrotational flows.

Then, an irrotational flow is not necessarily governed by the Euler equation, i.e., it's not necessarily inviscid.

share|improve this answer
add comment

I would like to add a few comments:

1) The form $\mu\nabla^2 {\boldsymbol u}$ for the viscous forces is based on the assumption that $\mu$ is constant. The general result is $\nabla_i (\mu \sigma_{ij})$ with $$\sigma_{ij}=\nabla_i u_j+\nabla_ju_i-\frac{2}{3}\delta_{ij}\nabla_ku_k.$$ This is important in a variety of cases. For example, the famous minimum viscosity fluid described by AdS/CFT has $\mu\sim T^3$, so $\nabla_i\mu=0$ only if the flow is isothermal.

2) Even if the viscous forces vanish the viscous stress $\sigma_{ij}$ may not be zero. This will lead to i) viscous forces on the bounbdary, and ii) dissipation of heat. The most famous example is sheared flow between parallel plates. The equation of motion is $\nabla^2{\boldsymbol u}=0$, solved by a linear flow profile $u_x\sim u_0 z/d$ ($d$ is the distance between the plates). This is not inviscid flow, because there is a force $$ F/A =\mu\nabla_z u_x $$ on the plates. This force does work on the fluid, and leads to viscous heating.

3) In a viscous fluid, even if the vorticity is zero initially, non-zero vorticity can be generated, typically by diffusion from a boundary. This is what happens in the viscous decay of a line vortex (vorticity initially concentrated at the origin).

share|improve this answer
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.