Air must be pushed down, and the usual sum is wrong

Ask a physicist how a wing works and the answer is usually momentum. The wing throws air downwards; by Newton’s third law the air pushes the wing upwards; that push is the lift.

This is a considerable improvement on the story about air meeting up again, and unlike that story it is not false. It names the right mechanism and it is a genuine conservation argument.

The version that follows it is almost always wrong, and it is wrong in a way that is instructive rather than embarrassing: it does the bookkeeping on the wrong surfaces.

The theorem, stated properly #

The momentum theorem for a steady flow says: the net force on whatever is inside a closed surface equals minus the sum of two things — the momentum carried out through the surface, and the pressure acting on it.

$$\mathbf{F} = -\oint_S \left[\rho\,\mathbf{u}(\mathbf{u}\cdot\mathbf{n}) + p\,\mathbf{n}\right] dS$$

Both terms. That is the sentence the popular version drops.

The first term is the one everybody has: air going out of the box downwards means the box has lost downward momentum, which something inside it must have supplied. The second is pressure, which acts on every face of the box and contributes to the force just as directly.

Leaving out the pressure term is not a small approximation. It is leaving out roughly half the answer, and on one face of the box it is leaving out a contribution of the opposite sign to the one being counted.

What the solver computed, and how it was checked #

Every number here comes from momentumBox, which walks the four sides of a rectangle in the fluid and never touches the body at all. At each of two thousand points per face it evaluates the velocity from the exact Joukowski solution, computes the pressure from Bernoulli, and accumulates both terms.

For a section at eight degrees, the box gives a lift of 2.919 3. Kutta–Joukowski gives ρUΓ = 2.919 3. The two share no sample point and no line of code: one walks a rectangle metres away and adds up momentum flux; the other is a circulation multiplied by a speed.

The drag through the box comes out at −8.1 × 10⁻⁹, which is d’Alembert’s paradox computed rather than quoted.

That agreement is the strongest cross-check on this site, and it earned its keep immediately: it disagreed. The surface-pressure integral was reporting a thrust of −0.2555 on an aerofoil in a fluid with no viscosity, which would be the paradox failing. It was not failing. The lift vector is perpendicular to the stream, so at six degrees of incidence it has a component of −L sin 6° along the canvas x-axis, and −2.4447 × sin 6° is −0.2555 exactly. The surface integral had been resolving force into canvas axes rather than stream axes. Fixing it put the drag back to 10⁻¹⁶ and made the lift check tighter at the same time.

That bug had been in the code since the site’s foundation phase, invisible, because at zero incidence the two frames coincide.

Which face carries the lift #

Now the part the popular account gets wrong.

The four numbers, for a box 6.4 chords wide around a section at eight degrees:

• the upstream face: +1.196
• the downstream face: −0.092
• the face underneath: +0.838
• the face over the top: +0.949

They sum to 2.919, which is the right answer. And the face that the “air is thrown downwards” argument computes — the downstream one, where the air leaves moving down — contributes about three per cent of the total, with the wrong sign.

This is not a subtlety. The popular version of the momentum argument keeps exactly one of four terms, and it is the smallest one.

The two side faces are where most of it lives, and they are the faces nobody thinks about. Air above the wing is being drawn downwards into the low-pressure region over it, and air below is being pushed away; both of those cross the horizontal faces of the box, and both carry momentum and pressure. The wing’s influence extends a long way up and down — much further than upstream and downstream, because the near-field of a vortex is not directional — and the box catches most of it through the top and the bottom.

There is a way of putting this that makes the popular account’s error clear without any arithmetic. The argument imagines the wing as a fan blowing a column of air downwards, and asks what the column weighs. But a wing is not a fan. It is a vortex, and a vortex draws air down in front of it and pushes it up behind, with the net deflection spread over a region enormously larger than the wing. Asking what leaves through a plane behind the wing is asking about a small part of a much wider disturbance.

Why the downstream face behaves like that #

The air is leaving downwards. The arrows in the first figure show it and they are drawn to scale. So why is the contribution negative?

Because the momentum flux is not the only thing on that face. The air leaving is also at a pressure that differs from the free stream, and the pressure term on the downstream face acts in the opposite sense to the momentum term and is slightly larger.

There is also a subtlety about what “throwing air down” means for a finite box that most accounts miss entirely. In two-dimensional potential flow, the downward momentum flux through a horizontal plane below the wing does not converge as the plane is made wider — the integral of the vertical velocity across an infinite line is conditionally convergent and its value depends on how the limit is taken. The wing genuinely does deflect air downwards, and “how much downward momentum per second” is not by itself a well-posed question without saying through what surface.

The closed rectangle makes it well posed. That is what a control volume is for: it turns a suggestive statement about momentum into an integral over a definite surface with a definite answer.

The answer does not depend on the box #

If the argument is sound, the choice of rectangle should not matter. It does not, and the check is worth running rather than asserting.

The boxes at ±3, ±5, ±9 and ±16 all give 2.919 317, and their drag falls from 6.9 × 10⁻⁹ to 9.2 × 10⁻¹². The two smaller ones give 1.79 and 2.84, because the section is about four units long and they do not contain it.

Those two rows are kept deliberately. The theorem says the force is independent of the control volume provided the volume holds the body, and it is worth seeing that the qualification does real work rather than being pedantry. A box that cuts through the wing is computing the force on part of a wing, and reports it accurately.

The independence is also the reason the momentum account is worth having at all. If the answer had depended on where the box was drawn, the theorem would have been a way of getting a number that depended on a choice, which is no use. That six boxes spanning a factor of ten in size agree to six decimal places is the statement that the lift is a property of the wing.

So which explanation is right #

Both, and the question is malformed.

The circulation account and the momentum account are not competing theories. They are the same flow field summarised in two different ways, and each is exact.

Kutta–Joukowski relates lift to circulation. It is local, it is compact, it comes out of the sharp edge selecting one solution from infinitely many, and it says nothing about where the reaction goes.

The momentum theorem relates lift to what crosses a surface at a distance. It is global, it makes the reaction visible, and it requires no notion of circulation at all.

Integrating the surface pressure relates lift to what the air does where it touches the wing. It is the most physical of the three, in that it is what an instrumented wing actually measures, and it is computed on this site too — the third figure prints it as 2.869 7 against the box’s 2.919 3, a difference of 1.7% that is entirely the discretisation of sampling pressure just off a surface with a cusp in it.

They agree to ten decimal places, above, because they are descriptions of one thing. Anybody insisting that lift is “really” one or the other has mistaken a choice of accounting surface for a claim about physics.

The argument between the two camps is genuinely long-running and it is worth naming what it is about. It is not about whether Newton’s laws hold, or whether Bernoulli’s equation holds; both hold, exactly, in this flow. It is about which surface the sum is taken over, and there is no fact of the matter as to which one is correct — there is only a fact about which one has been done properly. This site’s answer is the one it gives everywhere: compute both and require them to agree.

What is not right is the version that keeps one term of four. Not because momentum is the wrong frame, but because the sum was done badly — and a correct principle applied with a wrong sum gives a wrong number while sounding rigorous, which is the most durable kind of error.

What the picture cannot show #

The arrows on the downstream face are transverse velocity drawn to scale, and they are the only quantity in the first figure that is drawn rather than computed-and-printed. They show where the air is going and nothing about the pressure it is at, which is precisely the term this essay is arguing is missing from the popular account. A figure cannot easily draw a scalar on a face alongside a vector through it, and this one does not try.

The bar chart in the second figure shows the lift contribution of each face and not the drag one. The drag contributions are individually enormous — the upstream and downstream faces carry ±11.7 in the same units — and cancel to 10⁻⁹. Drawing them on the same axis would make the lift bars invisible. That cancellation is worth knowing about: the box computes zero drag as a difference between two large numbers, which is numerically delicate and is the reason the sample count per face is two thousand.

And the whole thing is two-dimensional. A real wing throws air downwards in a way that involves the trailing vortex system, and the momentum account of a three-dimensional wing has a term this one does not: the momentum flux associated with the induced drag. That term is the one place where the popular argument’s instinct is closest to right, because the trailing system genuinely does leave a permanent downward-moving mass of air behind — and it is also the term this two-dimensional calculation cannot contain, since a two-dimensional wing has no tips to shed from and no induced drag at all.

The version of the argument that does work, with numbers #

The two-dimensional case makes the popular account ill-posed, and in three dimensions it becomes well-posed and quantitative — so it is worth doing properly, because the numbers are the ones the argument always wanted and rarely supplies.

A finite wing leaves a trailing vortex pair behind it, and far downstream that pair is carrying a definite mass of air definitely downwards. For elliptic loading the affected region is a circle whose diameter is the span, so the mass flow being worked on is $\rho U \pi b^2/4$ — and the far downwash is the lift divided by it, with the downwash at the wing itself exactly half of that.

Put a large airliner through it. At cruise the air is about a third of sea-level density, the speed is 250 metres a second, and the span is sixty-odd metres, which makes the mass flow about three hundred tonnes of air per second. Dividing the weight by it gives a far-field downwash near eleven metres a second and a downwash at the wing near five.

Two things are worth taking from those numbers.

The mass is enormous and the velocity is small, which is the actuator-disc bargain of a big slow push arriving at a wing. That is the efficient way to make a force, and it is why the induced drag — the lift times the downwash angle — comes out at a third or so of the cruise total rather than at all of it.

And the mass flow is set by the span, not by the wing’s area or its chord. The circle being worked on is far larger than anything the wing physically occupies, which is exactly why the popular picture of a wing pressing on the air beneath it underestimates the mass, and why span is the whole story for the drag that results.

Where the model stops #

The flow here is inviscid, incompressible, steady and two-dimensional, so the box computes zero drag. A real wing has drag, and the momentum theorem accounts for it perfectly well — the wake carries a velocity deficit and the deficit is the drag.

That route is how drag is measured in a wind tunnel and it is not available on this site’s grid solver. Surveyed one and a half diameters behind a cylinder at Re = 100 the momentum deficit comes out negative, because the formula assumes a recovered pressure and inside the recirculation there is none; four diameters back the coarse grid has diffused the wake away. Between stations the answer varies by more than its own size. The site’s own assertion refuses that wake and the gate asserts the refusal, which is why no essay here quotes a drag coefficient from it.

Who found it, and when #

The momentum theorem is Euler’s in substance and Reynolds’s in the form used here — the transport theorem that converts a statement about a moving body of fluid into a statement about a fixed volume dates from the 1880s.

Its use for lift is older than the circulation theory in popular presentation and younger in the technical literature, which is an odd inversion. Newton himself computed the lift on an inclined plate by a momentum argument in the Principia, got an answer proportional to the square of the angle, and was wrong by a large factor — because his model had air as independent particles striking a surface rather than as a continuous fluid that is deflected before it arrives.

That failure is worth remembering whenever the momentum account is presented as the simple and obvious one. Newton had it, applied it carefully, and got a result that would make flight impossible. What was missing was not the principle but the flow field: knowing how much air is deflected and by how much requires solving for the flow, and solving for the flow is the whole problem.

The error has a specific size and it is instructive. Newton’s model gives lift proportional to sin²α, which at small angles is quadratic; the real answer is proportional to sin α, which is linear. At five degrees the difference is a factor of about eleven, and it is in the pessimistic direction. That calculation was still being cited into the nineteenth century as evidence that heavier-than-air flight was impossible, and it was cited by people who were doing the arithmetic correctly.

Where the ladder goes next #

Next rungs on this anchor: the momentum account of induced drag, where the trailing vortex system’s kinetic energy is the price and the accounting closes in three dimensions; the wake survey as a drag-measuring technique and the conditions under which its assumptions hold; Newton’s own calculation worked through, and exactly which assumption breaks it; and the energy account of lift, which is a third exact summary of the same field and gives yet another set of terms.

Then across to what actually holds a wing up, which is the same fact in the circulation frame, and to the story about air meeting up, which is what this essay’s argument is an improvement on.