D'Alembert's Paradox

Derivation

D'Alembert's Paradox

D-324 Home PU-307 Threads force · fields · symmetry Depends on Potential Flow and Laplace's Equation, Bernoulli's Theorem, Steady and Unsteady

Statement

For the steady, incompressible, inviscid, irrotational flow of an unbounded fluid past a rigid body of finite extent held in a uniform stream \(\vec U = U\hat x\), the net hydrodynamic force exerted on the body vanishes identically: \(\vec F = -\oint_{S_b} p\,\hat n\,dA = \vec 0\). In particular the drag \(F_x = 0\), independent of the body's shape, size, or the free-stream speed.

Why it matters

A ship, a wing, or a sphere plainly experiences drag, yet the most natural continuum model of a fluid — potential flow — predicts none. This clash, sharpened by d'Alembert in 1752, is the cleanest demonstration in physics that a term deliberately omitted from the equations (viscosity) is nonetheless responsible for a leading-order effect. It tells us that no amount of care with the ideal-flow solution can recover drag; the resolution is qualitative, not quantitative.

The paradox is the historical seed of boundary-layer theory (Prandtl, 1904): viscosity matters not because it is large, but because it changes the topology of the flow — it permits separation and a wake, which carry away momentum. Understanding precisely why the ideal integral collapses shows exactly which physical ingredient must be restored.

Assumptions

Steady flow, \(\partial_t = 0\).Unsteady potential flow carries an added-mass reaction force \(\vec F = -m_{ij}\dot U_j\); the cancellation is specific to the steady state.

Incompressible, \(\nabla\cdot\vec u = 0\).Compressibility admits acoustic radiation and, above \(M=1\), wave drag — a genuine non-zero force even without viscosity.

Irrotational, \(\vec\omega = \nabla\times\vec u = \vec 0\), so \(\vec u = \nabla\phi\).Vorticity shed into a wake produces a downstream momentum deficit; the far-field integral no longer vanishes and drag survives.

Inviscid, \(\mu = 0\).The surface stress reduces to pressure alone, \(-p\hat n\); a viscous shear stress \(\boldsymbol\tau\cdot\hat n\) would add skin friction directly.

Unbounded fluid; body impermeable and of finite extent.Impermeability kills the source (monopole) term, so the disturbance decays as a dipole \(\phi' = O(r^{-2})\). A net mass flux, a free surface, or solid walls introduce slower-decaying fields and the far-field estimate fails.

Single-valued potential (no net circulation in 3D).In 2D a circulation \(\Gamma\) is admissible and produces lift by Kutta–Joukowski; it still contributes zero drag, but the flow is no longer simply connected.

Derivation

\[ \vec F = -\oint_{S_b} p\,\hat n\,dA \]

With \(\mu=0\) the only surface traction is the normal pressure; \(\hat n\) is the outward normal of the fluid at the body. A

\[ \oint_{S_b}\!\big(\rho\,\vec u(\vec u\!\cdot\!\hat n)+p\hat n\big)dA + \oint_{S_R}\!\big(\rho\,\vec u(\vec u\!\cdot\!\hat n)+p\hat n\big)dA = \vec 0 \]

Steady momentum theorem \(\oint_S(\rho\,\vec u\,u_n+p\hat n)\,dA=\vec 0\) applied to the fluid volume bounded internally by \(S_b\) and externally by a large sphere \(S_R\) of radius \(R\); no body force, no viscous flux. B

\[ \vec F = -\oint_{S_b} p\hat n\,dA = -\oint_{S_R}\big(\rho\,\vec u\,(\vec u\!\cdot\!\hat n)+p\hat n\big)dA \]

On \(S_b\) the impermeability condition \(\vec u\!\cdot\!\hat n=0\) kills the convective term, leaving exactly \(-\vec F\); rearrange step 2. The force is now an integral over the far sphere only. B

\[ \phi = Ux + \phi',\qquad \phi' = -\frac{\vec p_d\cdot\vec r}{4\pi r^{3}}+O(r^{-3}),\qquad \vec u' = \nabla\phi' = O(r^{-3}) \]

A finite impermeable body emits no net volume flux, so \(\oint u'_n\,dA=0\) on any enclosing surface: the monopole (source) vanishes and the leading disturbance is a dipole. Hence \(\phi'=O(r^{-2})\) and its gradient \(\vec u'=O(r^{-3})\). C

\[ p = p_\infty + \tfrac12\rho U^2 - \tfrac12\rho\,|\vec u|^2 = p_\infty - \rho\,\vec U\!\cdot\!\vec u' - \tfrac12\rho\,|\vec u'|^2 \]

Bernoulli's theorem for steady irrotational flow, \(p+\tfrac12\rho|\vec u|^2=\text{const}\), with \(\vec u=\vec U+\vec u'\) and \(|\vec u|^2=U^2+2\vec U\!\cdot\!\vec u'+|\vec u'|^2\). The constant is fixed at infinity. B

\[ \vec F = -\oint_{S_R}\!\Big[\rho(\vec U+\vec u')\big((\vec U+\vec u')\!\cdot\!\hat n\big)+\big(p_\infty-\rho\vec U\!\cdot\!\vec u'-\tfrac12\rho|\vec u'|^2\big)\hat n\Big]dA \]

Substitute steps 4–5. The pure free-stream part \(\oint_{S_R}(\rho\vec U(\vec U\!\cdot\!\hat n)+p_\infty\hat n)\,dA=\vec 0\) on a closed surface (constant vectors, \(\oint\hat n\,dA=\vec 0\)). B

\[ \vec F = -\oint_{S_R}\!\big[\underbrace{\rho\vec U(\vec u'\!\cdot\!\hat n)+\rho\vec u'(\vec U\!\cdot\!\hat n)-\rho(\vec U\!\cdot\!\vec u')\hat n}_{\text{linear: }O(R^{-3})}+\underbrace{\rho\vec u'(\vec u'\!\cdot\!\hat n)-\tfrac12\rho|\vec u'|^2\hat n}_{\text{quadratic: }O(R^{-6})}\big]dA \]

Collect terms by order in \(\vec u'\). Every linear term is \(O(R^{-3})\); over area \(4\pi R^2\) it is \(O(R^{-1})\). (More sharply, the linear integral is exactly zero for a pure dipole by angular parity — the surviving pieces integrate to zero over the sphere.) The quadratic terms are \(O(R^{-6})\cdot R^2=O(R^{-4})\). C

\[ \vec F = \lim_{R\to\infty}\Big[O(R^{-1})+O(R^{-4})\Big] = \vec 0 \]

The force is independent of \(R\) (it equals the fixed surface integral over \(S_b\)), so it must equal its own limit as \(R\to\infty\), which is zero. A

Result

\[ \boxed{\;\vec F = -\oint_{S_b} p\,\hat n\,dA = \vec 0\;}\qquad\Longrightarrow\qquad F_{\text{drag}} = 0 \]

Reading. In steady, incompressible, irrotational, inviscid flow the pressure field around any finite closed body is fore–aft balanced: whatever the front pushes back, the rear pushes forward by exactly as much. The high pressure at the forward stagnation point is mirrored by an equal high pressure at the rear stagnation point, and the suction over the shoulders is symmetric. No net force of any kind acts on the body — no drag, and (absent circulation) no lift.

Units check. \([p][A]=(\mathrm{Pa})(\mathrm{m^2})=(\mathrm{N\,m^{-2}})(\mathrm{m^2})=\mathrm{N}\). The far-field integrand \(\rho U u'\) has units \((\mathrm{kg\,m^{-3}})(\mathrm{m\,s^{-1}})(\mathrm{m\,s^{-1}})=\mathrm{kg\,m^{-1}s^{-2}}=\mathrm{Pa}\), and \(\times\,\mathrm{m^2}=\mathrm{N}\). Consistent; the "answer" \(\vec 0\) carries newtons.

Limiting cases

\(U\to 0\): trivially \(\vec F\to\vec 0\); pressure reduces to hydrostatic and cancels around a closed surface regardless.
Sphere of radius \(a\): surface speed \(\tfrac32 U\sin\theta\), \(C_p = 1-\tfrac94\sin^2\theta\); the pressure integral gives exactly zero drag (see Worked Example 1).
2D cylinder with circulation \(\Gamma\): drag still \(0\), but lift \(L=\rho U\Gamma\ne 0\) — d'Alembert kills drag, not lift.
Slender body, thickness \(t\to 0\): the dipole strength \(\to 0\) and the whole disturbance vanishes; zero drag is recovered as the trivial \(\vec F=\vec 0\).
Real fluid, \(\mathrm{Re}\to\infty\): naive intuition says viscosity "disappears" and drag should \(\to 0\), yet measured \(C_D\) approaches a finite constant — the sharpest statement of the paradox.

Breaks when

Viscosity and separation (real flows). A boundary layer forms; at finite Reynolds number it separates on the rear shoulder, destroying the fore–aft pressure symmetry. The rear stagnation pressure is never recovered, leaving a low-pressure wake and a finite pressure (form) drag — the actual origin of drag on a sphere.
Vorticity / lifting bodies. Any shed vorticity (a starting vortex, a trailing vortex sheet behind a finite wing) leaves a downstream momentum deficit or downwash. The far-field flux integral no longer vanishes: induced drag \(D_i = L^2/(\tfrac12\rho U^2\pi b^2 e)\) appears even in the inviscid limit.
Compressibility, \(M\gtrsim 0.3\). Bernoulli's incompressible form fails; above \(M=1\) shock waves create wave drag with no viscosity required.
Unsteadiness / acceleration. \(\partial_t\phi\ne 0\) restores an added-mass force \(F=-m_a\dot U\); the cancellation is a steady-state artifact.
Confined or free-surface flows. Walls, a nearby free surface (wave drag on ships), or cavitation introduce slower-decaying far fields that break the \(O(R^{-1})\) estimate.

Failure modes

"Bernoulli explains drag." Bernoulli is exactly what forbids drag here; students invoke it to produce drag, not seeing that it enforces fore–aft symmetry.
Integrating pressure over only the front half. The forward stagnation pressure looks like a net push; forgetting the identical rear stagnation pressure gives a spurious drag.
Believing viscosity "adds a little skin friction" to fix the paradox. On a bluff body the dominant real drag is pressure drag from separation, not skin friction — a topological change, not a small correction.
Using \(C_p = 1-4\sin^2\theta\) (cylinder) for a sphere. The sphere factor is \(\tfrac94\), not \(4\); the surface-speed multiplier is \(\tfrac32\), not \(2\).
Assuming zero drag means zero force everywhere. Local pressures are large; only the integral cancels. The body is genuinely squeezed.
Confusing zero drag with zero lift. With circulation, lift survives; d'Alembert's statement is strictly about the streamwise component in an irrotational field.

Discussion

The physical heart of the paradox is momentum bookkeeping. Drag on a body equals the rate at which the fluid removes streamwise momentum from the flow and carries it downstream. In steady potential flow the disturbance is a dipole that decays as \(r^{-3}\): the fluid far downstream is indistinguishable from the fluid far upstream — there is no wake, no velocity defect, no momentum deficit. With nothing carried away, Newton's third law leaves nothing on the body. Drag is not "cancelled by a lucky pressure symmetry"; the symmetry is the shadow of a deeper fact — an irrotational steady flow cannot transport net momentum to infinity.

Prandtl's resolution is subtle: viscosity is confined to a thin boundary layer of thickness \(\delta\sim L/\sqrt{\mathrm{Re}}\), which shrinks as \(\mathrm{Re}\to\infty\). One might expect its effect to vanish with it. But the layer's function — allowing the no-slip condition and, crucially, permitting the flow to separate — does not vanish. Separation replaces the smooth rear closure of the streamlines with a broad wake, and the wake carries the momentum deficit that the potential solution lacked. Thus an arbitrarily thin viscous region produces an \(O(1)\) drag. This is a singular limit: \(\mu=0\) is not the same as \(\mu\to 0^+\).

Sharper still: the far-field integral in steps 6–8 is an exact conservation statement. Restoring vorticity replaces the dipole tail with a rotational wake whose transverse velocities decay only as \(r^{-1}\) (a trailing vortex) or leave a momentum-flux slab of fixed cross-section. Either way the \(O(R^{-1})\) estimate that sent \(\vec F\to\vec 0\) is destroyed at its root — the surviving flux integral is the drag, and it equals \(\rho\iint(U-u)\,u\,dA\) over a downstream plane (the wake-integral method used in wind tunnels). D'Alembert's paradox and modern drag measurement are two readings of the same momentum theorem.

Common misconceptions. The paradox is often stated as "potential flow gives zero drag and zero lift." That is false in general: in two dimensions a bound circulation gives lift \(L=\rho U\Gamma\) with still-zero drag. The correct statement isolates the streamwise force. Equally, the paradox is not a failure of mathematics or a sign potential flow is "useless" — potential flow predicts the pressure distribution on the front of a streamlined body extremely well; it fails precisely where separation occurs.

Worked examples

Example 1 — Zero drag on a sphere (direct pressure integral). Sphere of radius \(a=0.10\ \mathrm{m}\) in water (\(\rho=1000\ \mathrm{kg\,m^{-3}}\)) with free stream \(U=2.0\ \mathrm{m\,s^{-1}}\). Compute the potential-flow drag by integrating the surface pressure.

\[ u_\theta(a,\theta) = -\tfrac32 U\sin\theta,\qquad |\vec u|=\tfrac32 U|\sin\theta| \]

Tangential surface velocity of the classic sphere solution \(\phi=U\cos\theta\,(r+a^3/2r^2)\), evaluated at \(r=a\). A

\[ p(\theta) = p_\infty + \tfrac12\rho U^2\!\left(1-\tfrac94\sin^2\theta\right),\qquad C_p = 1-\tfrac94\sin^2\theta \]

Bernoulli with \(|\vec u|^2=\tfrac94U^2\sin^2\theta\). B

\[ F_x = -\oint p\,n_x\,dA = -\int_0^{2\pi}\!\!\int_0^{\pi} p(\theta)\cos\theta\,\big(a^2\sin\theta\big)\,d\theta\,d\varphi \]

Drag = streamwise projection of pressure; \(n_x=\cos\theta\), \(dA=a^2\sin\theta\,d\theta\,d\varphi\). B

\[ \int_0^\pi \cos\theta\sin\theta\,d\theta = 0,\qquad \int_0^\pi \sin^2\theta\cos\theta\sin\theta\,d\theta = \left[\tfrac{\sin^4\theta}{4}\right]_0^\pi = 0 \]

Both angular integrals — from the constant part of \(C_p\) and from the \(\sin^2\theta\) part — vanish by fore–aft antisymmetry in \(\cos\theta\). A

\[ F_x = 0\ \mathrm{N} \]

Reading. Despite a stagnation over-pressure of \(\tfrac12\rho U^2=2.0\times10^{3}\ \mathrm{Pa}\) at the nose, the identical over-pressure at the tail and the symmetric shoulder suction cancel exactly. The real drag at this \(\mathrm{Re}=U(2a)/\nu\approx4\times10^{5}\) is roughly \(\tfrac12\rho U^2 C_D\pi a^2\approx\tfrac12(1000)(4)(0.4)\pi(0.01)\approx 25\ \mathrm{N}\) — the entire measured force is the "missing" physics.

Units check. \(\rho U^2 a^2 = (1000)(4)(0.01)=40\), units \(\mathrm{kg\,m^{-3}\cdot m^2 s^{-2}\cdot m^2}=\mathrm{N}\); the integral of it is \(0\ \mathrm{N}\).

Example 2 — Cylinder with circulation: zero drag, finite lift. Circular cylinder of radius \(a=0.050\ \mathrm{m}\) in air (\(\rho=1.2\ \mathrm{kg\,m^{-3}}\)), \(U=3.0\ \mathrm{m\,s^{-1}}\), bound circulation \(\Gamma=2.0\ \mathrm{m^2\,s^{-1}}\). Find the force per unit span.

\[ u_\theta(a,\theta) = -2U\sin\theta - \frac{\Gamma}{2\pi a} \]

Surface velocity of the cylinder-plus-vortex potential flow. A

\[ p(\theta) = p_\infty + \tfrac12\rho U^2 - \tfrac12\rho\Big(2U\sin\theta+\tfrac{\Gamma}{2\pi a}\Big)^2 \]

Bernoulli on the surface. B

\[ D = -\!\int_0^{2\pi}\! p\cos\theta\,a\,d\theta = 0 \]

Every term of \(p(\theta)\) is a constant or a multiple of \(\sin\theta\) or \(\sin^2\theta\); each integrates against \(\cos\theta\) over \([0,2\pi]\) to zero — d'Alembert for the cylinder. B

\[ L = -\!\int_0^{2\pi}\! p\sin\theta\,a\,d\theta = \rho\,\frac{2U\Gamma}{2\pi a}\!\int_0^{2\pi}\!\sin^2\theta\,a\,d\theta\cdot\!\Big.\Big/\!\ldots = \rho U\Gamma \]

Only the cross term \(2\rho U\sin\theta\cdot\tfrac{\Gamma}{2\pi a}\) survives (via \(\int\sin^2\theta\,d\theta=\pi\)); this is the Kutta–Joukowski theorem. C

\[ D = 0\ \mathrm{N\,m^{-1}},\qquad L = \rho U\Gamma = (1.2)(3.0)(2.0) = 7.2\ \mathrm{N\,m^{-1}} \]

Reading. Circulation breaks the up–down symmetry to give lift, but leaves the fore–aft symmetry — and hence zero drag — completely intact. This is why thin-airfoil theory predicts lift well yet gives no profile drag.

Units check. \(\rho U\Gamma=(\mathrm{kg\,m^{-3}})(\mathrm{m\,s^{-1}})(\mathrm{m^2\,s^{-1}})=\mathrm{kg\,s^{-2}}=\mathrm{N\,m^{-1}}\), a force per unit span. Correct.

Problems

State the four independent physical assumptions behind d'Alembert's paradox and, for each, name one physical effect that is restored when it is dropped.
Solution
(i) Steady — added-mass reaction force under acceleration, \(F=-m_a\dot U\). (ii) Incompressible — wave (shock) drag above \(M=1\). (iii) Irrotational — induced drag / wake momentum deficit from shed vorticity. (iv) Inviscid — skin friction and, decisively, boundary-layer separation giving form (pressure) drag. Impermeability/finiteness of the body is the extra structural assumption ensuring dipole decay.
For the sphere of Worked Example 1, compute the pressure at (a) the forward stagnation point \(\theta=0\), (b) the equator \(\theta=\pi/2\), taking \(p_\infty=1.01\times10^{5}\ \mathrm{Pa}\). Comment on the sign at the equator.
Solution
\(\tfrac12\rho U^2=\tfrac12(1000)(2^2)=2000\ \mathrm{Pa}\). (a) \(C_p(0)=1\Rightarrow p=p_\infty+2000=1.030\times10^{5}\ \mathrm{Pa}\). (b) \(C_p(\pi/2)=1-\tfrac94=-\tfrac54\Rightarrow p=p_\infty+2000(-1.25)=p_\infty-2500=0.985\times10^{5}\ \mathrm{Pa}\). The equatorial pressure is below ambient (suction) because the flow accelerates to \(\tfrac32U=3\ \mathrm{m\,s^{-1}}\) there. The identical values at \(\theta=0\) and \(\theta=\pi\) are what force zero drag.
Show explicitly that for the 2D cylinder without circulation, \(C_p=1-4\sin^2\theta\), the drag integral \(\int_0^{2\pi}C_p\cos\theta\,d\theta\) vanishes.
Solution
\(\int_0^{2\pi}\cos\theta\,d\theta=0\) and \(\int_0^{2\pi}\sin^2\theta\cos\theta\,d\theta=\big[\tfrac{\sin^3\theta}{3}\big]_0^{2\pi}=0\). Hence \(\int_0^{2\pi}(1-4\sin^2\theta)\cos\theta\,d\theta=0-4\cdot0=0\), so \(D=-\tfrac12\rho U^2 a\!\int C_p\cos\theta\,d\theta=0\).
A body accelerates from rest along \(\hat x\) at \(\dot U=5\ \mathrm{m\,s^{-2}}\) in water; its added mass is \(m_a=2.6\ \mathrm{kg}\) (a sphere of radius \(a=0.10\ \mathrm{m}\), \(m_a=\tfrac23\pi\rho a^3\)). Find the instantaneous inviscid force and reconcile with d'Alembert.
Solution
\(m_a=\tfrac23\pi(1000)(0.1)^3=2.09\ \mathrm{kg}\) (the quoted \(2.6\) rounds a slightly larger body; use the formula value). Force to hold/drive it: \(F=-m_a\dot U=-2.09\times5\approx-10.5\ \mathrm{N}\) opposing motion. This is not drag — it is a reactive inertial force that reverses sign on deceleration and vanishes at constant speed. D'Alembert's paradox concerns the steady state \(\dot U=0\), where this term is absent and \(F=0\).
In a wind tunnel the wake behind a 2D body is measured on a downstream plane: the streamwise velocity is \(u=U-\Delta u\) with \(\Delta u=6\ \mathrm{m\,s^{-1}}\) over a slab of height \(h=0.04\ \mathrm{m}\), in air with \(U=30\ \mathrm{m\,s^{-1}}\), \(\rho=1.2\ \mathrm{kg\,m^{-3}}\) (treat \(\Delta u\) as uniform across \(h\), zero outside). Estimate the drag per unit span and explain why this is non-zero while potential flow gave zero.
Solution
Wake-integral drag \(D=\rho\!\int u(U-u)\,dy\approx\rho\,(U-\Delta u)\,\Delta u\,h=1.2\,(30-6)(6)(0.04)=1.2(24)(6)(0.04)=6.9\ \mathrm{N\,m^{-1}}\). It is non-zero because the real (viscous, separated) flow leaves a genuine momentum deficit \(\Delta u\ne0\) in the wake — exactly the far-downstream defect that the \(r^{-3}\) potential dipole did not produce. The drag is that momentum flux.