The Coefficient $6\pi$ in Stokes' Drag Law

Contents

The stream function and the equation it obeys
Solution for the sphere
The pressure, and a check
The traction and its integral

This appendix supplies the one thing Phenomenon 35.45 of Fluid Dynamics does not derive. The chapter's inline argument settles the form of Equation (35.39) completely: linearity of Equation (35.38) makes the drag exactly proportional to $U$, and Theorem 5.2 then leaves $\mu a U$ as the only force that can be built, so $F=C\mu aU$ with $C$ a pure number. What is owed is $C$. It is obtained here by solving the creeping-flow problem for the sphere in closed form and integrating the surface traction, and the calculation returns rather more than a number: it says how the drag divides between pressure and friction, which is the physically informative part and the part the chapter has no room for.

The stream function and the equation it obeys

Work in the frame of the sphere, spherical coordinates $(r,\theta,\varphi)$ with the polar axis along the stream, and take the flow axisymmetric and without swirl, $v_{\varphi}=0$. Boundary conditions: no slip on the sphere, and a uniform stream at infinity,

\begin{equation}\tag{A60.1} v_{r}=v_{\theta}=0\ \text{at }r=a\ec \qquad \vect{v}\longrightarrow U\hat{\vect{z}}\ \text{as } r\longrightarrow\infty\ep \end{equation}

Definition A60.1 (Stokes stream function).

The Stokes stream function $\psi(r,\theta)$, in $\mathrm{m}^{3}/\mathrm{s}$, is defined by

\begin{equation}\tag{A60.2} v_{r}=\frac{1}{r^{2}\sin\theta}\pdv{\psi}{\theta}\ec \qquad v_{\theta}=-\frac{1}{r\sin\theta}\pdv{\psi}{r}\ep \end{equation}

Rests on Example 35.15 and Equation (11.131).

Incompressibility is then an identity rather than a constraint: in spherical coordinates $\vect{\nabla}\cdot\vect{v} =r^{-2}\pp_{r}(r^{2}v_{r}) +\left(r\sin\theta\right)^{-1}\pp_{\theta}(\sin\theta\,v_{\theta})$, and substituting Equation (A60.2) gives $\left(r^{2}\sin\theta\right)^{-1} \left(\pp_{r}\pp_{\theta}\psi-\pp_{\theta}\pp_{r}\psi\right)=0$. This is the axisymmetric counterpart of the plane construction of Example 35.15, and it is why the whole problem collapses to one scalar equation.

Lemma A60.2 (The creeping-flow equation for $\psi$).

Under Definition A60.1 the Stokes equations Equation (35.38) are equivalent to

\begin{equation}\tag{A60.3} E^{4}\psi=0\ec \qquad E^{2}:=\pp_{r}^{2} +\frac{\sin\theta}{r^{2}}\,\pp_{\theta} \left(\frac{1}{\sin\theta}\,\pp_{\theta}\right)\ec \end{equation}

together with a pressure recovered by quadrature. Rests on Definitions 35.43 and A60.1.

Proof.

Derives Lemma A60.2. Take the curl of $\vect{0}=-\vect{\nabla}p+\mu\nabla^{2}\vect{v}$. The pressure disappears because the curl of a gradient vanishes (Equation (11.167)), leaving $\nabla^{2}\vect{\omega}=\vect{0}$ with $\vect{\omega}=\vect{\nabla}\times\vect{v}$. In an axisymmetric flow without swirl the vorticity has only an azimuthal component, and substituting Equation (A60.2) into $\omega_{\varphi}=r^{-1}\left[\pp_{r}(rv_{\theta}) -\pp_{\theta}v_{r}\right]$ gives

\begin{equation}\tag{A60.4} \omega_{\varphi}=-\frac{1}{r\sin\theta}\,E^{2}\psi\ep \end{equation}

The vector Laplacian acting on a purely azimuthal field of this form reduces, term by term in the spherical expression of Equation (11.133), to $\left(\nabla^{2}\vect{\omega}\right)_{\varphi} =-\left(r\sin\theta\right)^{-1}E^{2}\left(E^{2}\psi\right)$, which is Equation (A60.3). Conversely, given $\psi$ with $E^{4}\psi=0$, the momentum equation reads $\vect{\nabla}p=\mu\nabla^{2}\vect{v} =-\mu\vect{\nabla}\times\vect{\omega}$ by Equation (35.6) and $\vect{\nabla}\cdot\vect{v}=0$, and its right-hand side is then curl-free, so $p$ exists and is obtained by integrating along any path.

∎

Solution for the sphere

The condition at infinity fixes the angular dependence once and for all. A uniform stream has $\psi\longrightarrow\tfrac{1}{2}Ur^{2}\sin^{2}\theta$ — read Equation (A60.2) backwards — and both the equation and the boundary conditions are then satisfied by the single separated form

\begin{equation}\tag{A60.5} \psi(r,\theta)=f(r)\sin^{2}\theta\ep \end{equation}

Acting with $E^{2}$ on Equation (A60.5),

\[ E^{2}\psi =\left[f''-\frac{2f}{r^{2}}\right]\sin^{2}\theta\ec \]

since $\pp_{\theta}\left(\sin^{-1}\theta\,\pp_{\theta}\sin^{2}\theta \right)=\pp_{\theta}\left(2\cos\theta\right)=-2\sin\theta$. Applying $E^{2}$ a second time to $g:=f''-2f/r^{2}$ in the same way and collecting,

\begin{equation}\tag{A60.6} f''''-\frac{4f''}{r^{2}}+\frac{8f'}{r^{3}}-\frac{8f}{r^{4}}=0\ep \end{equation}

Lemma A60.3 (The radial solution).

The general solution of Equation (A60.6) is $f=A/r+Br+Cr^{2}+Dr^{4}$, and the conditions Equation (A60.1) select

\begin{equation}\tag{A60.7} D=0\ec\qquad C=\tfrac{1}{2}U\ec\qquad A=\tfrac{1}{4}Ua^{3}\ec\qquad B=-\tfrac{3}{4}Ua\ep \end{equation}

Rests on Lemma A60.2 and Equation (A60.1).

Proof.

Derives Lemma A60.3. Equation (A60.6) is equidimensional: every term scales as $r^{n-4}$ under $f=r^{n}$, so the substitution turns it into the algebraic equation

\[ n(n-1)(n-2)(n-3)-4n(n-1)+8n-8=0\ec \]

whose left-hand side factorizes as $\left(n+1\right)\left(n-1\right)\left(n-2\right)\left(n-4\right)$; one checks the four roots directly, $n=-1$ giving $24-8-16=0$, $n=1$ giving $0-0+8-8=0$, $n=2$ giving $0-8+16-8=0$ and $n=4$ giving $24-48+32-8=0$. The four powers are independent, so they span the solution space.

The condition at infinity requires $f\longrightarrow\tfrac{1}{2}Ur^{2}$, which forbids $r^{4}$ and fixes $C=U/2$. No slip requires both components of Equation (A60.2) to vanish at $r=a$, that is $f(a)=0$ and $f'(a)=0$:

\[ \frac{A}{a}+Ba+\frac{Ua^{2}}{2}=0\ec \qquad -\frac{A}{a^{2}}+B+Ua=0\ep \]

Multiplying the second by $a$ and subtracting it from the first eliminates $B$ and leaves $2A/a-Ua^{2}/2=0$, whence $A=Ua^{3}/4$; back-substitution gives $B=A/a^{2}-Ua=-3Ua/4$.

∎

Reading the velocity field off Equations (A60.2) and (A60.7),

\begin{equation}\tag{A60.8} v_{r}=U\cos\theta\left(1-\frac{3a}{2r}+\frac{a^{3}}{2r^{3}}\right)\ec \qquad v_{\theta}=-U\sin\theta \left(1-\frac{3a}{4r}-\frac{a^{3}}{4r^{3}}\right)\ep \end{equation}

Both vanish at $r=a$ and both tend to the free stream, as they must.

The pressure, and a check

From Equation (A60.7), $f''-2f/r^{2}=3Ua/2r$, so by Equation (A60.4)

\begin{equation}\tag{A60.9} \omega_{\varphi}=-\frac{3aU\sin\theta}{2r^{2}}\ep \end{equation}

The radial component of $\vect{\nabla}p=-\mu\vect{\nabla}\times\vect{\omega}$ is then

\[ \pdv{p}{r} =-\frac{\mu}{r\sin\theta}\, \pp_{\theta}\left(\sin\theta\,\omega_{\varphi}\right) =\frac{3\mu aU}{r\sin\theta}\, \pp_{\theta}\left(\frac{\sin^{2}\theta}{2r^{2}}\right) =\frac{3\mu aU\cos\theta}{r^{3}}\ec \]

and integrating from infinity,

\begin{equation}\tag{A60.10} p=p_{\infty}-\frac{3\mu aU\cos\theta}{2r^{2}}\ep \end{equation}

It is worth confirming that this is harmonic, since $\nabla^{2}p=0$ follows from taking the divergence of Equation (35.38): the function $\cos\theta/r^{2}$ is the $l=1$ exterior solid harmonic of Example 14.33, and it is. The pressure is high in front ($\theta=\pi$, taking the stream along $+\hat{\vect{z}}$ so that $\theta=\pi$ faces upstream) and low behind, by equal amounts — the fore-and-aft antisymmetry that would give zero drag in an ideal fluid (Phenomenon 35.30). Here it does not, because the pressure is no longer the whole traction.

The traction and its integral

For a Newtonian fluid Equation (35.30) gives, in spherical coordinates,

\begin{equation}\tag{A60.11} \sigma_{rr}=-p+2\mu\,\pdv{v_{r}}{r}\ec \qquad \sigma_{r\theta}=\mu\left[ r\,\pp_{r}\!\left(\frac{v_{\theta}}{r}\right) +\frac{1}{r}\,\pdv{v_{r}}{\theta}\right]\ep \end{equation}

Evaluate both at $r=a$ from Equations (A60.8) and (A60.10). For the normal component, $\pp_{r}v_{r} =U\cos\theta\left(3a/2r^{2}-3a^{3}/2r^{4}\right)$, which vanishes at $r=a$; the entire normal traction is therefore the pressure,

\begin{equation}\tag{A60.12} \left.\sigma_{rr}\right|_{r=a} =-p_{\infty}+\frac{3\mu U\cos\theta}{2a}\ep \end{equation}

For the tangential component, $v_{r}$ vanishes identically on $r=a$ so its $\theta$ derivative does too, and $v_{\theta}/r=-U\sin\theta\left(1/r-3a/4r^{2}-a^{3}/4r^{4}\right)$ gives $\pp_{r}\left(v_{\theta}/r\right) =-U\sin\theta\left(-1/r^{2}+3a/2r^{3}+a^{3}/r^{5}\right)$, equal to $-3U\sin\theta/2a^{2}$ at $r=a$; hence

\begin{equation}\tag{A60.13} \left.\sigma_{r\theta}\right|_{r=a} =-\frac{3\mu U\sin\theta}{2a}\ep \end{equation}

Both are in $\mathrm{Pa}$, as $\mu U/a$ has the dimensions $\mathsf{M}\mathsf{L}^{-1}\mathsf{T}^{-2}$.

Theorem A60.4 (Stokes' drag law with its coefficient).

The force exerted on the sphere by the fluid is directed along the stream and has magnitude

\begin{equation}\tag{A60.14} F=6\pi\mu aU\ec \end{equation}

of which $2\pi\mu aU$ — one third — comes from the pressure and $4\pi\mu aU$ — two thirds — from the tangential viscous traction. Rests on Lemma A60.3 and Theorem 35.36.

Proof.

Derives Theorem A60.4. The component along the stream of the traction on the surface element $\dd S$ is $\left(\sigma_{rr}\cos\theta-\sigma_{r\theta}\sin\theta\right)\dd S$, the projections of the radial and meridional directions on $\hat{\vect{z}}$. Substituting Equations (A60.12) and (A60.13),

\[ \sigma_{rr}\cos\theta-\sigma_{r\theta}\sin\theta =-p_{\infty}\cos\theta +\frac{3\mu U}{2a}\left(\cos^{2}\theta+\sin^{2}\theta\right) =-p_{\infty}\cos\theta+\frac{3\mu U}{2a}\ep \]

The uniform pressure integrates to zero over the closed surface, and what remains is a constant traction $3\mu U/2a$ over the whole sphere:

\[ F=\frac{3\mu U}{2a}\oint\dd S =\frac{3\mu U}{2a}\cdot4\pi a^{2}=6\pi\mu aU\ep \]

For the split, keep the two sources separate. The pressure contributes

\[ F^{(p)}=\oint\left(-p\right)\cos\theta\,\dd S =\frac{3\mu U}{2a}\,2\pi a^{2} \int_{0}^{\pi}\cos^{2}\theta\sin\theta\,\dd\theta =\frac{3\mu U}{2a}\,2\pi a^{2}\cdot\frac{2}{3} =2\pi\mu aU\ec \]

and the tangential traction contributes

\[ F^{(\tau)}=\oint\left(-\sigma_{r\theta}\right)\sin\theta\,\dd S =\frac{3\mu U}{2a}\,2\pi a^{2} \int_{0}^{\pi}\sin^{3}\theta\,\dd\theta =\frac{3\mu U}{2a}\,2\pi a^{2}\cdot\frac{4}{3} =4\pi\mu aU\ec \]

using $\int_{0}^{\pi}\cos^{2}\theta\sin\theta\,\dd\theta=2/3$ and $\int_{0}^{\pi}\sin^{3}\theta\,\dd\theta=4/3$. Their sum is Equation (A60.14), which is Equation (35.39) with $C=6\pi$ [Stokes:1851].

∎

Remark A60.5 (Why the split is worth printing).

The one-third–two-thirds division is a fact about creeping flow that survives no other regime and is easy to get backwards. At high Reynolds number the drag of a bluff body is almost entirely pressure drag, because the wake destroys the rear pressure recovery (Phenomenon 35.55); at low Reynolds number the pressure distribution Equation (A60.10) is exactly the fore-and-aft antisymmetric one, and it contributes only because it is multiplied by $\cos\theta$ and integrated — it is not the residue of a broken symmetry but the symmetric distribution itself doing work against the motion. The larger part of the resistance is skin friction, which is the sense in which Stokes flow is the opposite of the ideal flow of Section 35.3.3: there the friction was absent and the pressure did nothing, here the friction dominates and the pressure does a third.

Remark A60.6 (The slow decay, and what it costs).

Equation (A60.8) decays only as $a/r$: the term $-3aU\cos\theta/2r$ in $v_{r}$ is the leading disturbance, in contrast with the $a^{3}/r^{3}$ dipole of the ideal-flow solution Equation (35.27). A sphere in creeping flow drags a very large volume of fluid with it, which is why suspensions interact hydrodynamically at many diameters' separation, and why the drag is proportional to $a$ rather than to $a^{2}$: the resisted region has the size of the disturbance, not the size of the body. The same slow decay is the origin of the non-uniformity discussed at Remark 35.47 — the neglected inertial term falls off as $r^{-3}$ while the retained viscous term falls off as $r^{-4}$, so beyond a distance of order $a/\mathrm{Re}$ the approximation inverts itself. Whitehead's paradox is the failure of the naive second approximation to satisfy the condition at infinity, and Oseen's repair, Equation (35.41), keeps the advection by the uniform stream in the outer region from the start.

Remark A60.7.

The Coefficient $6\pi$ in Stokes' Drag Law discharges the numerical coefficient of Phenomenon 35.45 in Fluid Dynamics, where Equation (35.39) is derived up to the pure number $C$ from linearity and Theorem 5.2 and the calculation of $C$ is deferred here. With $C=6\pi$ in hand, Example 35.46 converts an observed fall speed into a radius, which is how the oil droplets of [Millikan:1913] were sized, and Remark 35.47 states the first correction. Nothing in this section is valid outside $\mathrm{Re}\ll1$ (Definition 35.43), and the honest statement of its range is Equation (35.41): the leading correction to Equation (A60.14) is a relative $\tfrac{3}{16}\mathrm{Re}$, so the law is good to one percent only up to $\mathrm{Re}\approx0.05$.