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+---
+title: "Stokes' law"
+date: 2021-05-04
+categories:
+- Physics
+- Fluid mechanics
+- Fluid dynamics
+layout: "concept"
+---
+
+**Stokes' law** describes the size of the drag force $D$
+at low [Reynolds number](/know/concept/reynolds-number/) $\mathrm{Re} \ll 1$
+experienced by a spherical object in a steady, uniform flow at velocity $U$.
+
+
+## Flow field
+
+Imagine a sphere with radius $a$ sinking in a viscous liquid.
+To model this situation, let us pretend that the sphere is fixed instead,
+and the fluid comes from infinity at velocity $U$ along the $z$-axis,
+flows past the sphere, and continues to infinity at the same $U$.
+The Reynolds number is:
+
+$$\begin{aligned}
+    \mathrm{Re}
+    = \frac{2 a U}{\nu}
+\end{aligned}$$
+
+We assume that $\mathrm{Re} \ll 1$, in which case
+the incompressible [Navier-Stokes equations](/know/concept/navier-stokes-equations/)
+are reduced to the **steady Stokes equations**:
+
+$$\begin{aligned}
+    \nabla p
+    = \eta \nabla^2 \va{v}
+    \qquad \quad
+    \nabla \cdot \va{v}
+    = 0
+\end{aligned}$$
+
+The goal is to solve for $p$ and $\va{v}$.
+We make the following ansatz in
+[spherical coordinates](/know/concept/spherical-coordinates/) $(r, \theta, \phi)$,
+where $q(r)$, $f(r)$ and $g(r)$ are unknown functions:
+
+$$\begin{gathered}
+    p
+    = \eta U q(r) \cos\theta
+    \\
+    v_r
+    = U f(r) \cos\theta
+    \qquad
+    v_\theta
+    = - U g(r) \sin\theta
+    \qquad
+    v_\phi
+    = 0
+\end{gathered}$$
+
+The fluid hits the sphere head on,
+so the solution is taken to be $\phi$-independent due to symmetry.
+Note that $\theta$ is the angle to the positive $z$-axis,
+which is the direction of $\va{U} = U \vu{e}_z$.
+Moreover, note that $\va{U} \cdot \vu{e}_r = U \cos\theta$
+and $\va{U} \cdot \vu{e}_\theta = - U \sin\theta$,
+where $\vu{e}_r$ and $\vu{e}_\theta$ are basis vectors.
+
+To begin with, we insert this ansatz into the incompressibility condition,
+yielding:
+
+$$\begin{aligned}
+    0
+    = \nabla \cdot \va{v}
+    &= \pdv{v_r}{r} + \frac{1}{r} \pdv{v_\theta}{\theta} + \frac{2 v_r}{r} + \frac{v_\theta}{r \tan \theta}
+    \\
+    &= U \dv{f}{r} \cos\theta - \frac{U g}{r} \cos\theta + \frac{2 U f}{r} \cos\theta - \frac{U g}{r} \cos\theta
+    \\
+    &= U \cos\theta \Big( \dv{f}{r} + \frac{2}{r} f - \frac{2}{r} g \Big)
+\end{aligned}$$
+
+The parenthesized expression must be zero for all $r$,
+leading us to the following relation:
+
+$$\begin{aligned}
+    g(r)
+    = f + \frac{r}{2} \dv{f}{r}
+\end{aligned}$$
+
+Next, we take the divergence of the first Stokes equation,
+and insert incompressibility:
+
+$$\begin{aligned}
+    \nabla^2 p
+    = \eta \nabla \cdot (\nabla^2 \va{v})
+    = \eta \nabla^2 (\nabla \cdot \va{v})
+    = 0
+\end{aligned}$$
+
+This is simply the Laplace equation,
+which is as follows for our ansatz $p(r, \theta)$:
+
+$$\begin{aligned}
+    0
+    = \nabla^2 p
+    &= \frac{1}{r^2} \pdv{}{r}\Big( r^2 \pdv{p}{r} \Big) + \frac{1}{r^2 \sin\theta} \pdv{}{\theta}\Big( \sin\theta \pdv{p}{\theta} \Big)
+    \\
+    0
+    &= \frac{\eta U \cos\theta}{r^2} \dv{}{r}\Big( r^2 \dv{q}{r} \Big)
+    - \frac{\eta U q}{r^2 \sin\theta} \pdv{}{\theta}\Big( \sin^2\theta \Big)
+    \\
+    &= \frac{\eta U \cos\theta}{r^2} \dv{}{r}\Big( r^2 \dv{q}{r} \Big)
+    - \frac{2 \eta U q}{r^2 \sin\theta} \sin\theta \cos\theta
+    \\
+    &= \eta U \cos\theta \Big( \dvn{2}{q}{r} + \frac{2}{r} \dv{q}{r} - \frac{2}{r^2} q \Big)
+\end{aligned}$$
+
+Again, the parenthesized expression must be zero for all $r$,
+meaning it is an ODE for $q(r)$,
+whose solution is straightforwardly found to be:
+
+$$\begin{aligned}
+    q(r)
+    = \frac{C_3}{r^2} + C_4 r
+\end{aligned}$$
+
+Where $C_3$ and $C_4$ are linearity constants ($C_1$ and $C_2$ appear later).
+The pressure is therefore:
+
+$$\begin{aligned}
+    p
+    = \eta U \cos\theta \Big( \frac{C_3}{r^2} + C_4 r \Big)
+\end{aligned}$$
+
+Consequently, its gradient $\nabla p$ in spherical coordinates is as follows:
+
+$$\begin{aligned}
+    \nabla p
+    = \vu{e}_r \pdv{p}{r} + \vu{e}_\theta \frac{1}{r} \pdv{p}{\theta}
+    = \vu{e}_r \Big( \eta U \cos\theta \dv{q}{r} \Big) - \vu{e}_\theta \Big( \eta U \sin\theta \frac{q}{r} \Big)
+\end{aligned}$$
+
+According to the Stokes equation, this equals $\eta \nabla^2 \va{v}$.
+Let us look at the $r$-component of $\nabla^2 \va{v}$:
+
+$$\begin{aligned}
+    (\nabla^2 \va{v})_r
+    &= \pdvn{2}{v_r}{r} + \frac{1}{r^2} \pdvn{2}{v_r}{\theta} + \frac{2}{r} \pdv{v_r}{r}
+    + \frac{\cot\theta}{r^2} \pdv{v_r}{\theta} - \frac{2}{r^2} \pdv{v_\theta}{\theta} - \frac{2}{r^2} v_r - \frac{2 \cot\theta}{r^2} v_\theta
+    \\
+    &= U \cos\theta \Big( \dvn{2}{f}{r} - \frac{1}{r^2} f + \frac{2}{r} \dv{f}{r} - \frac{1}{r^2} f
+    + \frac{2}{r^2} g - \frac{2}{r^2} f + \frac{2}{r^2} g \Big)
+    \\
+    &= U \cos\theta \Big( \dvn{2}{f}{r} + \frac{2}{r} \dv{f}{r} - \frac{4}{r^2} f + \frac{4}{r^2} g \Big)
+\end{aligned}$$
+
+Substituting $g$ for the expression we found from incompressibility lets us simplify this:
+
+$$\begin{aligned}
+    \eta (\nabla^2 \va{v})_r
+    &= \eta U \cos\theta \Big( \dvn{2}{f}{r} + \frac{4}{r} \dv{f}{r} \Big)
+\end{aligned}$$
+
+The Stokes equation says that this must be equal to the $r$-component of $\nabla p$:
+
+$$\begin{aligned}
+    \eta U \cos\theta \Big( \dvn{2}{f}{r} + \frac{4}{r} \dv{f}{r} \Big)
+    = \eta U \cos\theta \Big( \!-\! \frac{2 C_3}{r^3} + C_4 \Big)
+\end{aligned}$$
+
+Where we have inserted $\idv{q}{r}$.
+Dividing out $\eta U \cos\theta$ leaves an ODE for $f(r)$,
+satisfied by:
+
+$$\begin{aligned}
+    f(r)
+    = C_1 + \frac{C_2}{r^3} + \frac{C_3}{r} + \frac{C_4 r^2}{10}
+\end{aligned}$$
+
+Then, thanks to our earlier relation again,
+we know that $g(r)$ is as follows:
+
+$$\begin{aligned}
+    g(r)
+    = C_1 - \frac{C_2}{2 r^3} + \frac{C_3}{2 r} + \frac{C_4 r^2}{5}
+\end{aligned}$$
+
+So what about $C_1$, $C_2$, $C_3$ and $C_4$?
+For $r\!\to\!\infty$, we expect that $\va{v}\!\to\!\va{U}$,
+meaning that $f(r)\!\to\!1$ and $g(r)\!\to\!1$.
+This implies that $C_4 = 0$ and $C_1 = 1$, leaving:
+
+$$\begin{aligned}
+    f(r)
+    = 1 + \frac{C_2}{r^3} + \frac{C_3}{r}
+    \qquad \quad
+    g(r)
+    = 1 - \frac{C_2}{2 r^3} + \frac{C_3}{2 r}
+\end{aligned}$$
+
+Furthermore, the viscous *no-slip* condition demands
+that $\va{v} = 0$ at the sphere's surface $r = a$, so $f(a) = g(a) = 0$ there.
+Inserting $a$ into $f$ and $g$, setting them to zero,
+and solving the resulting system of equations
+yields $C_2 = a^3 / 2$ and $C_3 = -3 a / 2$.
+Therefore the full solution is:
+
+$$\begin{gathered}
+    \boxed{
+        p
+        = - \frac{3 \eta U a}{2 r^2} \cos\theta
+    }
+    \\
+    \boxed{
+        v_r
+        = U \cos\theta \Big( 1 + \frac{a^3}{2 r^3} - \frac{3 a}{2 r} \Big)
+        \qquad
+        v_\theta
+        = - U \sin\theta \Big( 1 - \frac{a^3}{4 r^3} - \frac{3 a}{4 r} \Big)
+    }
+\end{gathered}$$
+
+
+## Drag force
+
+From the definition of [viscosity](/know/concept/viscosity/),
+we know that there must be shear stresses at the sphere surface,
+described by the fluid's [Cauchy stress tensor](/know/concept/cauchy-stress-tensor/) $\hat{\sigma}$.
+The drag force $\va{D}$ on the surface is:
+
+$$\begin{aligned}
+    \va{D}
+    = \oint \hat{\sigma} \cdot \dd{\va{S}}
+    = \int_0^{2\pi} \!\!\!\! \int_0^\pi \big( \hat{\sigma} \cdot \vu{e}_r \big) \:a^2 \sin\theta \dd{\theta} \dd{\phi}
+\end{aligned}$$
+
+Where $\vu{e}_r$ is the sphere's surface normal vector.
+The integrand can be expanded as follows:
+
+$$\begin{aligned}
+    \hat{\sigma} \cdot \vu{e}_r
+    = \vu{e}_r \sigma_{rr} + \vu{e}_\theta \sigma_{\theta r}
+\end{aligned}$$
+
+To calculate this, we start by taking the gradient of the velocity field $\va{v}$:
+
+$$\begin{aligned}
+    \nabla\va{v}
+    &= \vu{e}_r \vu{e}_r \pdv{v_r}{r} + \vu{e}_r \vu{e}_\theta \pdv{v_\theta}{r}
+    + \vu{e}_\theta \vu{e}_r \Big( \frac{1}{r} \pdv{v_r}{\theta} - \frac{v_\theta}{r} \Big)
+    \\
+    &\qquad + \vu{e}_\theta \vu{e}_\theta \Big( \frac{1}{r} \pdv{v_\theta}{\theta} - \frac{v_r}{r} \Big)
+    + \vu{e}_\phi \vu{e}_\phi \Big( \frac{v_\theta}{r \tan\theta} + \frac{v_r}{r} \Big)
+\end{aligned}$$
+
+Some of these terms are necessary to calculate the stress elements $\sigma_{rr}$ and $\sigma_{\theta r}$:
+
+$$\begin{aligned}
+    \sigma_{rr}
+    &= - p + 2 \eta (\nabla\va{v})_{rr}
+    = - p + 2 \eta \pdv{v_r}{r}
+    \\
+    &= \frac{3 \eta U a}{2 r^2} \cos\theta + 2 \eta U \cos\theta \: \Big( \!-\! \frac{3 a^3}{2 r^4} + \frac{3 a}{2 r^2} \Big)
+    \\
+    &= \frac{3 \eta U a}{2 r^2} \cos\theta \: \Big( 3 - 2 \frac{a^2}{r^2} \Big)
+\end{aligned}$$
+$$\begin{aligned}
+    \sigma_{\theta r}
+    &= \eta \big( (\nabla\va{v})_{\theta r} + (\nabla\va{v})_{r \theta} \big)
+    = \eta \: \Big( \pdv{v_\theta}{r} + \frac{1}{r} \pdv{v_r}{\theta} - \frac{v_\theta}{r} \Big)
+    \\
+    &= \eta U \sin\theta \: \Big( \!-\! \frac{3 a^3}{4 r^4} - \frac{3 a}{4 r^2}
+    - \frac{1}{r} - \frac{a^3}{2 r^4} + \frac{3 a}{2 r^2}
+    + \frac{1}{r} - \frac{a^3}{4 r^4} - \frac{3 a}{4 r^2} \Big)
+    \\
+    &= - \frac{3 \eta U a^3}{2 r^4} \sin\theta
+\end{aligned}$$
+
+At the sphere's surface we set $r = a$, so these expressions reduce to the following:
+
+$$\begin{aligned}
+    \sigma_{rr}
+    = \frac{3 \eta U}{2 a} \cos\theta
+    \qquad \quad
+    \sigma_{\theta r}
+    = - \frac{3 \eta U}{2 a} \sin\theta
+\end{aligned}$$
+
+Now we can finally calculate the effective stress on the surface,
+by converting the basis vectors $\vu{e}_r$ and $\vu{e}_\theta$ to Cartesian coordinates:
+
+$$\begin{aligned}
+    \hat{\sigma} \cdot \vu{e}_r
+    &= \vu{e}_r \frac{3 \eta U}{2 a} \cos\theta - \vu{e}_\theta \frac{3 \eta U}{2 a} \sin\theta
+    \\
+    &= \Big( \vu{e}_x \sin\theta \cos\phi + \vu{e}_y \sin\theta \sin\phi + \vu{e}_z \cos\theta \Big) \frac{3 \eta U}{2 a} \cos\theta
+    \\
+    &\qquad - \Big( \vu{e}_x \cos\theta \cos\phi + \vu{e}_y \cos\theta \sin\phi - \vu{e}_z \sin\theta \Big) \frac{3 \eta U}{2 a} \sin\theta
+    \\
+    &= \Big( \vu{e}_z \cos^2\theta + \vu{e}_z \sin^2\theta \Big) \frac{3 \eta U}{2 a}
+    = \vu{e}_z \frac{3 \eta U}{2 a}
+\end{aligned}$$
+
+Remarkably, the stress at every point on the sphere is purely in the $z$-direction!
+This is not entirely unexpected though: symmetry cancels out all other components.
+
+With this, we can do the integrals for $\va{D}$,
+which reduce to a surface area factor $4 \pi a^2$:
+
+$$\begin{aligned}
+    \va{D}
+    = \vu{e}_z \frac{3 \eta U}{2 a} \int_0^{2\pi} \!\!\!\! \int_0^\pi a^2 \sin\theta \dd{\theta} \dd{\phi}
+    = \vu{e}_z \frac{3 \eta U}{2 a} 2 \pi a^2 \int_0^\pi \sin\theta \dd{\theta}
+    = \vu{e}_z \: 6 \pi \eta U a
+\end{aligned}$$
+
+At last, we arrive at Stokes' law,
+which simply expresses the magnitude of $\va{D}$:
+
+$$\begin{aligned}
+    \boxed{
+        D
+        = 6 \pi \eta U a
+    }
+\end{aligned}$$
+
+To arrive at this result,
+we assumed that the sphere was fixed, and the fluid was flowing past it.
+We can equally well let the fluid be at rest,
+with the sphere falling through it at $U$.
+The force of gravity then exerts the following force $G$ on it,
+subtracting [buoyancy](/know/concept/archimedes-principle/):
+
+$$\begin{aligned}
+    G
+    = \frac{4 \pi a^3}{3} (\rho_s - \rho_f) g_0
+\end{aligned}$$
+
+Where $\rho_s$ and $\rho_f$ are the sphere's and fluid's densities,
+and $g_0$ is the gravitational acceleration.
+Since $D$ acts in the opposite sense of $G$,
+after some time, they cancel out:
+
+$$\begin{aligned}
+    6 \pi \eta U a
+    = \frac{4 \pi a^3}{3} (\rho_s - \rho_f) g_0
+\end{aligned}$$
+
+This is an equation for the **terminal velocity** $U_t$,
+which we find to be as follows:
+
+$$\begin{aligned}
+    \boxed{
+        U_t
+        = \frac{2 a^2 (\rho_s - \rho_f) g_0}{9 \eta}
+    }
+\end{aligned}$$
+
+The falling sphere will accelerate until $U_t$,
+and then continue falling at constant speed.
+
+
+
+## References
+1.  B. Lautrup,
+    *Physics of continuous matter: exotic and everyday phenomena in the macroscopic world*, 2nd edition,
+    CRC Press.
-- 
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