The momentum with no value

A cylinder sits at rest in still fluid. Something pushes it, and after a while it is moving at a steady speed $U$ with the whole flow settled into the familiar dipole pattern. The push did work. Energy went into the fluid — exactly $\tfrac12\rho\pi a^2 U^2$, a number that is finite, closed-form and checkable. The push also delivered an impulse, and the impulse had to go somewhere.

So: how much momentum is in the fluid?

The question sounds like arithmetic. The integral is $\rho\!\int\!\mathbf{u}\,dA$ over everything outside the body, the integrand is known in closed form, and the answer is finite. It is also different for every shape of region the sum is taken over, and the differences do not go away as the regions grow.

Why it does not converge absolutely #

The disturbance of a translating cylinder is a dipole, and a dipole’s velocity falls as $1/r^2$ in the plane. The area element grows as $r$. So the contribution of an annulus between $r$ and $r + dr$ goes as $r^{-2}\cdot r\,dr = dr/r$ — a logarithm’s worth of contribution from every decade of distance, in magnitude.

What saves the integral is that the contributions cancel in direction rather than decay in size. Around any circle the velocity’s $x$ component goes as $\cos 2\theta$, whose average over a full turn is zero. So a disc gives exactly nothing, and it gives nothing at every radius, by symmetry rather than by smallness.

Change the shape and the cancellation changes with it. A tall rectangle weights the parts of the annulus that lie above and below the body, where the fluid is being pushed backwards as the body goes by. A long one weights the parts ahead and behind, where the fluid is going forwards. The two answers have opposite signs, and both are of the same size as the impulse.

The zero over a disc is not an absence #

The disc’s answer of exactly zero is the one number here that looks like a resolution, and it is worth taking apart, because it is a cancellation between two finite quantities rather than a statement that nothing is there.

Split it with the boundary identity. Round a circle of radius $R$ the outer term is $-\rho\pi a^2 U$, independent of $R$ — the $1/R$ in the potential exactly cancelling the $R$ in the arc length, which is the conditional convergence showing up as a limit that exists and does not go to zero. The body’s term is $+\rho\pi a^2 U$. They sum to nothing.

So a computation that only ever looked at discs would return a tidy answer to a question that has none, and would return it at every radius, with beautiful consistency. A result that is independent of the cut-off is not the same as a result that is independent of the shape of the cut-off, and checking the first is the natural thing to do and proves the wrong thing.

The identity that makes it computable #

There is a piece of vector calculus that turns this from a two-dimensional quadrature into a pair of line integrals, and it is worth stating because it is where the well-defined quantity comes from.

For a potential flow, $\mathbf{u} = \nabla\phi$, and

$$\int_V \nabla\phi\,dV \;=\; \oint_{\partial V}\phi\,\mathbf{n}\,dS.$$

All of the momentum lives on the boundary and none of it in between. The fluid region has two boundaries — the outer surface and the body — so the total is the sum of two terms, with the body’s normal pointing the other way round.

That splits the answer cleanly. The outer term is where the ambiguity is: it depends on the shape of a surface that is supposed to be receding to infinity, and it does not settle down. The body’s term does not depend on the outer surface at all.

The impulse #

The body’s term, with a sign, is

$$\mathbf{I} \;=\; -\rho\oint_{\text{body}} \phi\,\mathbf{n}\,dS \;=\; \rho\pi a^2 U$$

for a cylinder — the mass of fluid it displaces, times its speed. This is Kelvin’s hydrodynamic impulse, and it is everything the fluid’s momentum is not: finite, unique, independent of any outer surface, and computable from the body alone.

What it means physically is the cleanest statement available. It is the impulse an external agency must supply to generate the motion from rest, and it is measurable: hit the fluid, and integrate the pressure over the instant. The pressure impulse that leaves is $-\rho\phi$, and integrating it over the surface gives this number back, by an entirely different route.

And its rate of change is the force #

The reason the impulse rather than the momentum is the useful object is that it behaves like a momentum in the one respect that matters.

$$\mathbf{F} = \frac{d\mathbf{I}}{dt} = \rho\pi a^2 \frac{dU}{dt},$$

which is the added-mass force, and is the whole content of the statement that accelerating a body in an ideal fluid costs something even though moving it steadily costs nothing. Newton’s second law survives, applied to a quantity that is not the fluid’s momentum.

The energy tells the same story from the other side. $E = \tfrac12\rho\pi a^2 U^2$, so $dE/dU = \rho\pi a^2 U = I$, exactly — the impulse is the derivative of the energy with respect to the speed, which is the relation a momentum has to a kinetic energy and which the fluid’s actual momentum does not have at all.

Two routes, and why both were run #

Everything above could be arithmetic error, and the arrangement of the computation is designed so that it would show.

The momentum inside a region is computed twice: once as an area integral of the velocity over a grid, and once as a line integral of the potential round the boundary. The two share no arithmetic — one samples a vector field on a lattice, the other evaluates a scalar on a curve — and they agree to a part in a billion of the impulse.

And the check that the whole essay rests on is written the other way round from every other check on this site: it requires the answers to disagree. If every shape returned the same momentum, the integral would be absolutely convergent, the ambiguity would be an illusion, and the argument would be false. So the check demands a spread of at least half the impulse and a change of sign across the family, and refuses a computation that is too well behaved.

The same question, asked about force #

There is a version of this question with a comfortable answer, and putting the two side by side is the fastest way to see what has gone wrong here.

Ask instead for the force on a body, by a momentum balance over a contour drawn round it. That answer is unique: it is the same on every contour, whatever its shape and however far out, because the integrand is a genuine flux of a conserved quantity through a closed surface and what is inside does not change. How much of it is pressure and how much is momentum flux depends on the contour — that split runs from three per cent to ninety-seven — but the total does not move.

The difference is that a force is a flux through a surface and a momentum is a density integrated over a volume. Fluxes through closed surfaces are well behaved when what they enclose is settled; volume integrals of slowly decaying densities are not. The lesson is about which questions a control volume can be asked, and it is one this collection has met before from the other direction, when a momentum deficit measured in a wake came out negative because the station was inside a region where the assumptions behind the formula did not hold.

Which integrals survive, and it depends on the multipole #

The trouble here is arithmetic about decay rates, so it can be sorted once for the whole subject rather than met again each time. Every disturbance in ideal flow is a multipole, each multipole has its own decay, and each integral has its own volume element to fight.

In the plane, a translating body is a dipole and its velocity falls as $1/r^2$. The area element grows as $r$, so a momentum density integrates as $dr/r$ — a logarithm’s worth from every decade, saved only by the angular cancellation this essay is about. The energy density falls as $1/r^4$, integrates as $dr/r^3$, and converges absolutely with nothing to argue about. That asymmetry is the whole reason the energy of a translating cylinder is a number anybody can quote and the momentum is not.

A body carrying circulation is a different multipole and it is one order worse in everything. A vortex’s velocity falls only as $1/r$, so its energy density integrates as $dr/r$ — and the kinetic energy of a two-dimensional flow with circulation is logarithmically divergent. Not conditionally convergent, not shape-dependent: infinite, for any aerofoil, at any speed, in an unbounded plane.

That is worth knowing because it quietly conditions a great deal of what this collection does. It is why the minimum-energy theorem is computed inside an outer circle a few radii out rather than over the plane, and why its energy figures are honest about that cut. It is why two-dimensional aerofoil theory is always written in terms of forces and circulations and never in terms of the energy of the field. And it is the sharpest possible statement of why induced drag is a three-dimensional quantity: a wing’s wake energy per unit length is finite only because the trailing vorticity comes in equal and opposite pairs whose fields cancel at large distance. A single-signed circulation in the plane has no energy to speak of, because it has too much.

Going up a dimension moves every case one step towards respectability without curing any of them. A sphere’s dipole falls as $1/r^3$ against a volume element growing as $r^2$, so the momentum integrates as $dr/r$ — the same conditional convergence as the plane’s, arrived at from different exponents, which is what the last section of this essay reports. Its energy density falls as $1/r^6$ and converges comfortably.

The pattern that emerges is worth stating as a rule, because it predicts the answer without doing the integral. A volume integral over an unbounded ideal flow is trustworthy when its integrand decays faster than the volume element grows, and every time it does not, the well-defined replacement is a surface integral over the body. Momentum gives way to impulse. In the plane, energy with circulation present gives way to a force computed on a contour. In each case the quantity that survives is the one written on the body, because the body is a finite surface and infinity is not.

Which is a more useful summary of this essay than the cylinder is. The cylinder is one case where the sum does not settle; the rule is that in this subject an integral spread over an unbounded fluid is guilty until its decay rate has been checked, and the check takes one line.

What the ambiguity is really about #

It is tempting to file this under mathematical pedantry. It is not, and the reason is that the missing information is physical and nameable.

An unbounded fluid is an idealisation, and every real fluid is in something. Put the cylinder in a long narrow tank and the fluid’s momentum is one number; put it in a wide shallow one and it is another; and the two differ by an amount of order the impulse, however large the tank. The momentum depends on the container, and the container was thrown away in the first line of the problem.

That is also the resolution of an old confusion about how a moving body sets fluid in motion far away. It does not: the disturbance dies as $1/r^2$ and there is nothing much happening out there. What happens is that a tiny velocity spread over an area growing as $r^2$ adds up to something of order one, and where that something ends up depends on the far boundary — which is a statement about the whole system rather than about the body.

What it is not #

Three misreadings, and the third is the useful one.

It is not a failure of the potential-flow model. The same conditional convergence appears in a viscous calculation, in a numerical simulation, and in a real tank. It is a property of a slowly-decaying disturbance in an unbounded domain, not of the approximation used to compute it.

It is not resolved by adding viscosity. A real fluid does eventually stop, and the momentum it had goes into the walls of whatever contains it — which is to say the container reappears in the answer, which was the point.

And it is not a licence to ignore the momentum entirely. The momentum flux through a surface — the thing that appears in a force balance — is perfectly well defined, and so is the impulse. What has no value is one particular volume integral, and knowing exactly which one is the difference between using a control volume correctly and getting a plausible number out of it.

What the picture cannot show #

Nothing at infinity is drawn, and infinity is where the whole difficulty lives. Every figure has a finite frame, the frames are chosen to make the near field legible, and the disagreement being described is invisible inside any of them.

The quadrature is over rectangles and the argument is about limits. A finite rectangle is not a limit; what the sweep shows is that the answer does not settle as the rectangle grows in one shape rather than another. Reading a limit off a table of finite regions is exactly the step that has to be done carefully, and the family is constructed at fixed area so that the only thing varying is shape.

And the three-dimensional case is a different problem. A sphere’s disturbance falls as $1/r^3$ against a volume element growing as $r^2$, so the integrand falls as $1/r$ — still not absolutely convergent, still shape-dependent, and with different coefficients. The plane case is drawn because it is the one that can be drawn.

Who found it, and when #

Kelvin introduced the impulse in the 1860s, in the same run of papers as the circulation theorem and the minimum-energy theorem, and he introduced it precisely because the momentum would not behave. Lamb gives the conditional convergence carefully in Hydrodynamics and most later textbooks do not, which is why the identification of added-mass-times-speed with “the momentum of the fluid” is so widespread.

The surprising connection is with a different subject entirely. The same conditional convergence, for the same reason, is why the electrostatic energy of a dipole lattice depends on the shape of the crystal — the Ewald summation problem, where a sum over a $1/r^3$ field in three dimensions gives an answer that depends on whether the sample is a needle or a plate. Physicists solved it by splitting the sum into a rapidly converging part and a shape-dependent surface term, which is precisely the split between the impulse and the outer integral above. Two fields, one mathematical structure, and in both of them the resolution is to find the quantity that does not depend on the boundary and use that instead.

Where the ladder goes next #

Below this rung is the added mass whose impulse this is, and mass conservation, which is the other integral over a control volume that this collection leans on.

Beside it are where the reaction to a wing’s lift actually is, which is the same question asked about force rather than momentum and has the more comfortable answer, and what survives the journey to infinity.

And above it, the one place the pressure impulse is not an abstraction: the flows that consist of nothing but the unsteady term.