Countless arguments between highly intelligent people have been waged (on this very site in fact) as to exactly how lift can be explained in an experimentally and mathematically rigorous way. Taking the potential flow approximation and invoking the experimentally-observed Kutta condition provides a fairly accurate model. A majority of explanations for the Kutta condition involve Nature avoiding the infinite velocities implied by potential flow around a corner of zero radius. This, however, is where the problem arises. No man-made object has a zero radius of curvature. We cannot manufacture perfectly sharp corners in the same way that we cannot manufacture perfectly straight edges; all real objects have a nonzero radius of curvature. Thus, no potential flow would actually require an infinite velocity to properly flow around it. By this reasoning, claiming that Nature "enforces the Kutta condition to avoid infinite velocities" has to be false, because infinite velocities are not needed to flow around any real geometry. Moreover, we know that the Kutta Condition is actually not upheld for very low Reynolds numbers (see here and below). Is there a better explanation for the Kutta Condition than this spurious reference to infinite velocities? I know the potential flow model is just an approximation, but why does a real viscous flow force the rear stagnation point to the trailing edge?

From the MIT 16.100 Lecture Notes:

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Hele-Shaw Flow Around an Airfoil (note that the rear stagnation point is not at the trailing edge):

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A video of the above experiment can be seen here.

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    $\begingroup$ I'd like to point out that your examples where "we know the KC is not upheld for very low Reynolds numbers" is not a proof that the KC is wrong or baseless. Reynolds number is the ratio of inertial to viscous forces, so very low Re means viscous forces dominate. Meanwhile, the Kutta condition is something we impose in potential flow equations, which assume infinitely large Re (negligible viscosity). So of course it doesn't hold, it's not supposed to. The equations that use it aren't valid. $\endgroup$
    – tpg2114
    Sep 16, 2014 at 2:14
  • $\begingroup$ What is a more physically accurate explanation for the stagnation point moving to the trailing edge? I know it involves vortex shedding, but can you add some detail in your answer? $\endgroup$
    – Bryson S.
    Sep 16, 2014 at 2:32
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    $\begingroup$ Also -- and this is just being pedantic because we can be once in awhile -- you state that no man-made object can be perfectly sharp (zero radius of curvature). And while true, we are capable of creating knife edges that get down to nanometers in thickness at the very edge. Since this is well below the mean-free-path of air at atmospheric conditions, this is effectively zero radius of curvature over the length scales of the flow. $\endgroup$
    – tpg2114
    Sep 16, 2014 at 2:45
  • $\begingroup$ What I need is specific detail on exactly how viscosity moves the rear stagnation point to the trailing edge at high Reynolds numbers. Again, we know that at low enough Re it doesn't do this and I am looking for why. Any explanation for lift involving viscosity must deal with why the rear stagnation point does not migrate to the trailing edge at low enough Reynolds numbers. $\endgroup$
    – Bryson S.
    Sep 16, 2014 at 20:32
  • $\begingroup$ Can you give a citation that says there is no stagnation point at the rear point in low Re flow? Stokes flow over a cylinder and over airfoils both have rear stagnation points. In fact, the solutions to those flows look identical to the potential flow solutions, confusing many people about why the potential solutions don't hold at low Re. $\endgroup$
    – tpg2114
    Sep 16, 2014 at 21:24

1 Answer 1


The Kutta condition is completely artificial.

The potential equations are completely artificial.

The potential equations are a mathematical construct we use because it's much simpler than the full Navier-Stokes set of equations. We know the Kutta condition is never actually upheld in any real flow ever. However, when we perform all of our mathematical trickery to get to the potential equations, the very nature of the equations is now changed.

In the full Navier-Stokes, we have a second order PDE. This requires 2 boundary conditions. The first is that there is no flow through the body. The second is that the tangential velocity is zero along the body (and note -- this is also not true in real life either, there is some slip velocity along bodies in real flow under some conditions). When we get the potential equations, we have a first order PDE and now we can only impose a single boundary condition -- no flow through the body.

However, lift in real life is due to viscosity. The following explanation is from the linked answer:

The reason we need the Kutta condition is purely mathematical. When the inviscid assumption is made, the order of the governing equations drops and we can no longer enforce two boundary conditions. If we look at the incompressible, viscous momentum equation:

$\frac{\partial u_i}{\partial t} + u_i\frac{\partial u_i}{\partial x_j} = -\frac{1}{\rho}\frac{\partial P}{\partial x_i} + \nu \frac{\partial^2 u_i}{\partial x_j \partial x_i}$

we can enforce two boundary conditions because we have a second derivative in $u$. We typically set these to be $u_n = 0$ and $u_t = 0$, implying no flux through the surface and no velocity along the surface.

Dropping the viscous term results in only having the first derivative in $u$ and so we can only enforce one boundary condition. Since flow through the body is impossible, we drop the requirement that tangential velocity be zero -- this results in the slip boundary condition. However, it is not physically correct to let this slip line persist downstream of the trailing edge. So, the Kutta condition is needed to force the velocities to match at the trailing edge, eliminating the discontinuous velocity jump downstream.

John Anderson Jr explains in Fundamentals of Aerodynamics (emphasis in text):

... in real life, the way that nature insures the that the flow will leave smoothly at the trailing edge, that is, the mechanism that nature uses to choose the flow... is that the viscous boundary layer remains attached all the way to the trailing edge. Nature enforces the Kutta condition by means of friction. If there were no boundary layer (i.e. no friction), there would be no physical mechanism in the real world to achieve the Kutta condition.

He chooses to explain that nature found a way to enforce the Kutta condition. I prefer to think of it the other way around -- the Kutta condition is a mathematical construction we use to enforce nature in our mathematical approximation.

  • $\begingroup$ So you agree that the explanation given in the MIT 16.10 course notes is completely baseless? $\endgroup$
    – Bryson S.
    Sep 16, 2014 at 2:08
  • $\begingroup$ @BrysonS. No, not at all. In fact, I think it explains it quite well. The only way to get a solution to the potential equations that has the flow leaving the trailing edge smoothly (like what is observed in physical situations) is if the Kutta condition is imposed. What in there do you see as baseless or misleading? $\endgroup$
    – tpg2114
    Sep 16, 2014 at 2:11
  • $\begingroup$ The claim is made within the notes (pg.2) that an infinite velocity is implied by the sharp trailing edge, and so—to avoid this—the rear stagnation point must be at the trailing edge. $\endgroup$
    – Bryson S.
    Sep 16, 2014 at 2:21
  • $\begingroup$ @BrysonS. Remember that they are solving the potential equations over a simplified model of the airfoil. In the simplified problem, the trailing edge is infinitely sharp. It is because the equations to generate it will generate an infinitely sharp edge. There is nothing bogus or baseless about that either. I think part of your confusion is connecting the mathematical approximation to real life. They are two distinct things, and if we're really lucky, the former will describe the latter pretty well. $\endgroup$
    – tpg2114
    Sep 16, 2014 at 2:23
  • $\begingroup$ Can you provide a more physically accurate explanation (based on the Navier-Stokes equations) in your answer? That is really what I am after. In other words, what's the real explanation? $\endgroup$
    – Bryson S.
    Sep 16, 2014 at 2:29

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