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"...

  • $\begingroup$ Behaving in an inviscid way and actually being inviscid are technically two different things. One could argue that—even though the fluid behaves inviscidly—it is still a viscous flow because the kinematic viscosity itself is nonzero. $\endgroup$
    – Bryson S.
    Commented Jan 31, 2015 at 15:43
  • $\begingroup$ (Irrotational: $\nabla\times v=0$) + (incompressible: $\nabla\cdot v=0$) = ($\nabla^2v=0$) and this "looks like" an inviscid flow. $\endgroup$
    – Quillo
    Commented Oct 10, 2022 at 12:23

4 Answers 4


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


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.

  • $\begingroup$ 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? $\endgroup$ Commented Jul 3, 2012 at 7:32
  • $\begingroup$ 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. $\endgroup$ Commented Jul 3, 2012 at 7:44
  • $\begingroup$ 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. $\endgroup$ Commented Jul 3, 2012 at 7:44
  • $\begingroup$ 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. $\endgroup$ Commented Jul 3, 2012 at 7:54
  • $\begingroup$ 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. $\endgroup$ Commented Jul 3, 2012 at 7:56

(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 written 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.


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).


The person who posted the comment about an inviscid flow being able to exhibit rotation is so out to lunch. He stated an inviscid flow experiencing vorticity can be considered a rotational flow. This is simply not true. I hope the person who posted that is still an undergrad like myself.

There is a huge difference between vorticity and rotation in a fluid. vorticity is fluid traveling in a circle around a point without rotation. To imagine vorticity, imagine that you have a giant vat of water that is moving in a circle around the center point, similar to water moving in a toilet, but without it moving downwards. Now, place a small t in that fluid. put a dot on the top part of the t. as the t moves around with the swirling fluid, the t will always face the same direction. This is an example of a fluid exhibiting vorticity. It is very tough (impossible) to create a flow that only exhibits vorticity. a rotational flow is a flow where that same t, spins around its own axis.

I hope my explanation was clear, it can be tough to explain descriptively, it is more of a visual thing, but my point is that vorticity and rotation in a fluid are two hugely different things. I really hope that person was an undergrad. also, saying that the Euler equations do not apply to inviscid flows is incorrect. They apply to every fluid flow, both Newtonian and non Newtonian, inviscid and viscid.

  • 2
    $\begingroup$ You should check your definitions. What you describe as rotation is also called vorticity. Also, the Euler equations most definitely only apply to inviscid flow. What was being said in the other answers is that some solutions (flows) to the viscous fluid equations can be irrotational, in which case viscosity happens to not matter. $\endgroup$
    – user10851
    Commented Jan 31, 2015 at 4:17
  • 1
    $\begingroup$ This is a relevant pair of definitions... $\endgroup$
    – Mike
    Commented Jan 31, 2015 at 4:26

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