D'Alembert's Paradox for a Body of Arbitrary Shape

Contents

Statement and hypotheses
The far field carries no source
The momentum theorem on a large sphere
Every surviving term integrates to zero

This appendix proves Phenomenon 35.30 of Fluid Dynamics in the generality in which it was stated. The chapter derives the result for a sphere, where the fore-and-aft symmetry of Equation (35.28) makes the cancellation visible in one line; what is owed, and is discharged here, is the statement for a body of any shape, where no such symmetry is available and the vanishing of the force has to be extracted from the structure of the far field instead. The conclusion is stronger than the chapter's phrasing: not only the drag but the whole resultant vanishes, so an arbitrary body in a steady, simply connected, irrotational flow of an ideal fluid feels no lift either. That is what makes Remark 35.31 exhaustive — the only way to buy a transverse force is to break the hypotheses, which is what Theorem 35.68 does.

Statement and hypotheses

Theorem A59.1 (D'Alembert's paradox).

Let $S$ be the smooth closed surface of a finite rigid body held fixed in an unbounded incompressible ideal fluid of uniform density $\rho$, in $\mathrm{kg}/\mathrm{m}^{3}$, occupying the exterior region $\Omega$, and let the motion be steady and irrotational with a single-valued potential $\phi$ on $\Omega$, satisfying

\begin{equation}\tag{A59.1} \nabla^{2}\phi=0\ \text{in }\Omega\ec \qquad \pp_{n}\phi=0\ \text{on }S\ec \qquad \vect{\nabla}\phi\longrightarrow U\hat{\vect{z}}\ \text{as } \abs{\vect{x}}\longrightarrow\infty\ep \end{equation}

Then the resultant force exerted on the body by the fluid pressure vanishes identically,

\begin{equation}\tag{A59.2} \vect{F}=-\oint_{S}p\,\vect{N}\,\dd S=\vect{0}\ec \end{equation}

$\vect{N}$ being the unit normal on $S$ directed into the fluid. The statement holds for every shape of $S$, for every value of $U$ in $\mathrm{m}/\mathrm{s}$, and componentwise: there is no drag and no lift. Rests on Proposition 35.28 and Theorem 35.24.

Three hypotheses carry the whole argument and each is named in Remark 35.31 as a way out. Steadiness is what licenses the momentum balance of The momentum theorem on a large sphere. Single-valuedness of $\phi$ is the simple-connectivity assumption; it is what forbids a circulation, and dropping it is the plane-flow escape of Section 35.7.1. Attachment — that $\Omega$ really is the whole exterior, with the fluid following the surface everywhere — is what nature declines to provide, and its failure is Section 35.7.2.

The far field carries no source

Write

\begin{equation}\tag{A59.3} \phi=Uz+\chi\ec \qquad \vect{u}':=\vect{\nabla}\chi\ec \end{equation}

so that $\chi$ is harmonic on $\Omega$ and $\vect{u}'$, the disturbance velocity, tends to zero at infinity. The first step is that $\chi$ has no monopole term, and this is a consequence of impermeability alone.

Lemma A59.2 (The disturbance flux vanishes).

Let $S_{R}$ be a sphere of radius $R$ large enough to contain $S$, with outward unit normal $\vect{n}$. Then

\begin{equation}\tag{A59.4} \oint_{S_{R}}\vect{u}'\cdot\vect{n}\,\dd S=0 \end{equation}

for every such $R$. Rests on Theorem 11.133 and Equation (A59.3).

Proof.

Derives Lemma A59.2. Apply the divergence theorem (Theorem 11.133) to $\vect{u}'$ on the region bounded by $S$ and $S_{R}$. Since $\vect{\nabla}\cdot\vect{u}'=\nabla^{2}\chi=0$ there,

\[ \oint_{S_{R}}\vect{u}'\cdot\vect{n}\,\dd S =\oint_{S}\vect{u}'\cdot\vect{N}\,\dd S =\oint_{S}\vect{\nabla}\phi\cdot\vect{N}\,\dd S -U\oint_{S}\hat{\vect{z}}\cdot\vect{N}\,\dd S\ep \]

The first integral is zero because $\pp_{n}\phi=0$ on $S$: the body is impermeable. The second is zero because the integral of the unit normal over any closed surface vanishes — take $\vect{a}$ constant in the divergence theorem and read $\oint_{S}\vect{a}\cdot\vect{N}\,\dd S =\int\vect{\nabla}\cdot\vect{a}\,\dd V=0$.

∎

Theorem A59.3 (Exterior expansion in solid harmonics; quoted).

Let $\chi$ be harmonic in the exterior of a ball of radius $R_{0}$ in three dimensions and tend to zero at infinity. Then, in spherical coordinates $(r,\theta,\varphi)$ centred in that ball,

\begin{equation}\tag{A59.5} \chi(\vect{x})=\sum_{l=0}^{\infty}\sum_{m=-l}^{l} \frac{c_{lm}}{r^{\,l+1}}\,Y_{lm}(\theta,\varphi)\ec \end{equation}

the series converging uniformly, together with the series of its term-by-term derivatives of every order, on $r\ge R_{1}$ for each $R_{1}>R_{0}$. Here $Y_{lm}$ is the spherical harmonic Equation (13.163), and the powers $r^{-l-1}$ are the decaying radial solutions of Equation (14.34) at $k=0$ (Example 14.33). Rests on Equations (13.163) and (14.34).

Remark A59.4 (What is quoted here).

Theorem A59.3 is the only statement in this section that is not derived. Part II supplies one half of it and not the other. The half it supplies is the basis: separation of Laplace's equation in spherical coordinates, carried out at Example 14.33, produces exactly the radial factors $r^{l}$ and $r^{-l-1}$ against the spherical harmonics of Section 13.9.3, and the decay condition selects the second family. The half it does not supply is completeness with convergence — that every exterior harmonic function decaying at infinity is the sum of such a series, uniformly and differentiably. That is a theorem of potential theory, resting on the mean-value property (The Mean-Value Property Characterizes Harmonic Functions) and on the analyticity of harmonic functions, and this treatise does not build it; it is recorded here as a debt against Partial Differential Equations, in the same spirit as the vector identities recorded at Remark 35.5. Its classical use in exactly the present setting — the far field of a body in a stream as a source, a dipole and higher terms — is Lamb's [Lamb:1932]. Nothing below uses more of it than the leading two terms.

Corollary A59.5 (The disturbance is a dipole at leading order).

There is a constant vector $\vect{A}$, of dimension $\mathsf{L}^{4}\mathsf{T}^{-1}$, such that

\begin{equation}\tag{A59.6} \chi=\frac{\vect{A}\cdot\vect{x}}{r^{3}}+O\!\left(r^{-3}\right)\ec \qquad u'_{i}=\frac{1}{r^{3}} \left(A_{i}-3\left(\vect{A}\cdot\vect{n}\right)n_{i}\right) +O\!\left(r^{-4}\right)\ec \end{equation}

with $\vect{n}=\vect{x}/r$. Rests on Lemma A59.2 and Theorem A59.3.

Proof.

Derives Corollary A59.5. In Equation (A59.5) the $l=0$ term is $c_{00}Y_{00}/r$, a point source of strength $-4\pi c_{00}Y_{00}$; integrating its gradient over $S_{R}$ gives that strength, and every $l\ge1$ term integrates to zero over the sphere by the orthogonality of the spherical harmonics (Theorem 13.119). So Lemma A59.2 forces $c_{00}=0$. The three $l=1$ terms are linear combinations of $x_{i}/r^{3}$, which is the first expression in Equation (A59.6); the remainder is $l\ge2$, i.e. $O(r^{-3})$. Differentiating,

\[ \pp_{i}\frac{A_{j}x_{j}}{r^{3}} =\frac{A_{i}}{r^{3}}-\frac{3A_{j}x_{j}x_{i}}{r^{5}} =\frac{1}{r^{3}}\left(A_{i}-3\left(\vect{A}\cdot\vect{n}\right) n_{i}\right)\ec \]

and term-by-term differentiation of the remainder, licensed by Theorem A59.3, gives $O(r^{-4})$.

∎

The physical reading of Lemma A59.2 is worth one sentence, because it is the only place where the body enters at all: a rigid impermeable body neither creates nor destroys fluid, so it cannot look like a source from far away, and the slowest decay it can produce is that of a dipole. The sphere of Example 35.29 is the worked case, $\chi=Ua^{3}\cos\theta/2r^{2}$, with $\vect{A}=\tfrac{1}{2}Ua^{3}\hat{\vect{z}}$.

The momentum theorem on a large sphere

Lemma A59.6 (The force is a far-field integral).

For every $R$ large enough that $S_{R}$ encloses $S$,

\begin{equation}\tag{A59.7} F_{i}=-\oint_{S_{R}}\left[p\,n_{i} +\rho v_{i}\left(\vect{v}\cdot\vect{n}\right)\right]\dd S\ec \end{equation}

with $\vect{v}=\vect{\nabla}\phi$ and $p$ the pressure. In particular the right-hand side is independent of $R$. Rests on Theorems 11.133 and 35.22.

Proof.

Derives Lemma A59.6. Let $V$ be the fluid region between $S$ and $S_{R}$ and let $\vect{m}$ be the outward normal of $V$, equal to $\vect{n}$ on $S_{R}$ and to $-\vect{N}$ on $S$. In steady flow of an ideal fluid with no body force, the Euler equation (Theorem 35.22) combined with $\vect{\nabla}\cdot\vect{v}=0$ reads $\pp_{j}\left(\rho v_{i}v_{j}+p\,\delta_{ij}\right)=0$, since $\rho v_{j}\pp_{j}v_{i}=\pp_{j}(\rho v_{i}v_{j})$. Integrating this over $V$ and applying the divergence theorem (Theorem 11.133) to each $i$,

\[ 0=\oint_{\pp V}\left[p\,m_{i} +\rho v_{i}\left(\vect{v}\cdot\vect{m}\right)\right]\dd S =\oint_{S_{R}}\left[p\,n_{i} +\rho v_{i}\left(\vect{v}\cdot\vect{n}\right)\right]\dd S -\oint_{S}p\,N_{i}\,\dd S\ec \]

where the momentum flux over $S$ dropped because $\vect{v}\cdot\vect{N}=0$ there. The last integral is $-F_{i}$ by Equation (A59.2), which is the assertion.

∎

The pressure is supplied by Bernoulli's theorem (Theorem 35.24) at constant height. With $q^{2}=\abs{\vect{v}}^{2} =U^{2}+2Uu'_{z}+\abs{\vect{u}'}^{2}$,

\begin{equation}\tag{A59.8} p=p_{\infty}+\tfrac{1}{2}\rho U^{2}-\tfrac{1}{2}\rho q^{2} =p_{\infty}-\rho Uu'_{z}-\tfrac{1}{2}\rho\abs{\vect{u}'}^{2}\ep \end{equation}

By Corollary A59.5 the last term is $O(R^{-6})$ on $S_{R}$; multiplied by the area element it contributes $O(R^{-4})$ to Equation (A59.7) and vanishes in the limit. The same estimate disposes of the quadratic part of the momentum flux:

\begin{equation}\tag{A59.9} \rho v_{i}\left(\vect{v}\cdot\vect{n}\right) =\rho U^{2}\delta_{i3}n_{z} +\rho U\delta_{i3}\left(\vect{u}'\cdot\vect{n}\right) +\rho Uu'_{i}n_{z} +O\!\left(R^{-6}\right)\ep \end{equation}

Every surviving term integrates to zero

Assemble Equations (A59.7), (A59.8) and (A59.9) and take the four contributions in turn.

The uniform terms.

$\oint_{S_{R}}p_{\infty}n_{i}\,\dd S=0$ and $\rho U^{2}\delta_{i3}\oint_{S_{R}}n_{z}\,\dd S=0$, both because the integral of the unit normal over a closed surface vanishes. A constant pressure and a uniform stream, taken alone, push a closed surface nowhere.

The disturbance flux.

$\rho U\delta_{i3}\oint_{S_{R}}\vect{u}'\cdot\vect{n}\,\dd S=0$ by Lemma A59.2 — exactly, at every $R$, not merely in the limit.

The two surviving dipole terms.

What is left is

\begin{equation}\tag{A59.10} F_{i}=\rho U\oint_{S_{R}} \left[u'_{z}n_{i}-u'_{i}n_{z}\right]\dd S +O\!\left(R^{-2}\right)\ec \end{equation}

the first term coming from the pressure through Equation (A59.8) and the second from the momentum flux, with the signs of Equation (A59.7) already applied. Insert Equation (A59.6) and write $\dd S=R^{2}\dd\Omega$ with $\dd\Omega$ the solid angle:

\begin{align} \oint_{S_{R}}\left[u'_{z}n_{i}-u'_{i}n_{z}\right]\dd S &=\frac{1}{R}\oint\Bigl[ \left(A_{3}-3\left(\vect{A}\cdot\vect{n}\right)n_{3}\right)n_{i} -\left(A_{i}-3\left(\vect{A}\cdot\vect{n}\right)n_{i}\right)n_{3} \Bigr]\dd\Omega \nonumber\\ &=\frac{1}{R}\oint\left[A_{3}n_{i}-A_{i}n_{3}\right]\dd\Omega \nonumber\\ &=\frac{1}{R}\left[A_{3}\oint n_{i}\,\dd\Omega -A_{i}\oint n_{3}\,\dd\Omega\right]=0\ec \tag{A59.11} \end{align}

because the two terms in $\vect{A}\cdot\vect{n}$ cancel identically — they enter the two integrands as $-3(\vect{A}\cdot\vect{n})n_{3}n_{i}$ and $+3(\vect{A}\cdot\vect{n})n_{i}n_{3}$, the same quantity with opposite signs — and because $\oint n_{i}\,\dd\Omega=0$ over the whole sphere.

That cancellation is the mechanism, and it deserves to be read rather than merely verified. The dipole disturbance contributes to the pressure integral and to the momentum-flux integral separately, and neither contribution is zero; what is zero is their difference, at every $R$ and for every orientation of $\vect{A}$, hence for every shape of body.

Proof.

Derives Theorem A59.1. By Lemma A59.6 the quantity $F_{i}$ defined by Equation (A59.7) does not depend on $R$. By Equations (A59.10) and (A59.11) it equals $O(R^{-2})$. A constant that is $O(R^{-2})$ for arbitrarily large $R$ is zero, so $\vect{F}=\vect{0}$ — exactly, and for all three components at once.

∎

Remark A59.7 (Where the sphere's fore-and-aft symmetry went).

The chapter's proof for the sphere runs on the symmetry of Equation (35.28) about $\theta=\pi/2$, and a general body has no such symmetry: the pressure distribution over an aerofoil at incidence is grossly asymmetric, and every individual surface element carries a real force. What survives in general is weaker and sufficient — the resultant is fixed by the far field alone, and the far field of any impermeable body is a dipole, whose two contributions to Equation (A59.7) cancel. The near field, where all the physics of shape lives, never enters the calculation. This is also why the theorem is so unforgiving: nothing about the body is used except that it is closed and that the fluid does not pass through it.

Remark A59.8.

D'Alembert's Paradox for a Body of Arbitrary Shape discharges the general case of Phenomenon 35.30 in Fluid Dynamics, where the sphere is done inline and the arbitrary body is deferred here. Read it next to Remark 35.31, whose claim that there are exactly two escapes is now exact rather than rhetorical, because Theorem A59.1 kills the resultant and not only its streamwise component. The first escape is multiple connectivity: in the plane, dropping single-valuedness of $\phi$ admits a circulation and produces the transverse force of Theorem 35.68 — which still leaves the drag zero, so the paradox survives its own resolution for lift, and the appendix computing the circulation itself is The Joukowski Map, the Kutta Condition and the $2\pi$ Lift Slope. The second is that real flow is neither steady nor attached, which is Section 35.7.2, quantified by Theorem 35.71 and measured in Section 40.5.