A moving sphere in a quiescent fluid

Consider a sphere of radius R, moving with velocity $\vec{v}_{0}$ in a quiescent fluid . Referring all coordinates and velocities to a frame which moves with velocity $\vec{v}_{0}$ relative to the fluid transforms the problem into one of a resting sphere in a fluid which, at large distances from the sphere, moves with constant velocity $-\vec{v}_{0}$.

Our velocity field will be stationary. The Stokes equations then read

  

$$\eta \nabla ^{2}\vec{v} \vec{\nabla}P$$ = (5.16)

$$\vec{\nabla}\cdot \vec{v}$$ = 0 (5.17)

with boundary conditions

$$\vec{v}(R) \vec{0}$$ = (5.18)

$$\vec{v}(\infty ) -\vec{v}_{0}.$$ = (5.19)

As a first step towards the solution, we take the divergence of Eq. (5.16). Making use of the incompressibility equation  (5.17) we then get

$$\nabla ^{2}P=0.$$ (5.20)

Now, notice that the flow field will be linear in $\vec{v}_{0}$. From Eq. (5.16) it follows that $\vec{\nabla}P$will then also be linear in $\vec{v}_{0}$. Moreover $P(\vec{r})$ must be invariant under a rotation of both the coordinate frame and $\vec{v}_{0}$. Therefore we write

$$P(\vec{r})=a+\vec{v}_{0}\cdot \vec{r}f(r)$$ (5.21)

Introducing this into Eq. (5.20) we get

$$2\vec{v}_{0}\cdot \vec{\nabla}f+\vec{v}_{0}\cdot \vec{r}\nabla ^{2}f$$ = 0

$$\frac{\partial ^{2}f}{\partial r^{2}}+\frac{4}{r}\frac{\partial f}{\partial r }$$ = 0 (5.22)

Solving this under the condition $f(\infty )=0$, we get f(r)=br^-3, and

$$P(\vec{r})=a+b\frac{\vec{v}_{0}\cdot \vec{r}}{r^{3}}.$$ (5.23)

Eq. (5.16) now reads

$$\eta \nabla ^{2}\vec{v}=\vec{v}_{0}\frac{b}{r^{3}}-\vec{r}(\vec{v}_{0}\cdot \vec{r})\frac{3b}{r^{5}}.$$ (5.24)

Now again, we notice that $\vec{v}(\vec{r})$ must be linear in $\vec{v}_{0}$. Moreover it must transform like $\vec{v}_{0}$ and $\vec{r}$ under a joint rotation of $\vec{v}_{0}$ and the coordinate frame. Therefore

$$\vec{v}(\vec{r})=\vec{v}_{0}F_{1}(r)+\vec{r}(\vec{v}_{0}\cdot \vec{r} )F_{2}(r).$$ (5.25)

Introducing this expression into Eq. (5.24), we get

$${\vec{v}_{0}\nabla ^{2}F_{1}+2\vec{\nabla}(\vec{v}_{0}\cdot \vec{r} F_{2})+\vec{r}\nabla ^{2}(\vec{v}_{0}\cdot \vec{r}F_{2})}$$

$$\vec{v}_{0}\{\nabla ^{2}F_{1}+2F_{2}\}+\vec{r}(\vec{v}_{0}\cdot \... ... )\left\{ \frac{4}{r}\frac{\partial F_{2}}{\partial r}+\nabla ^{2}F_{2}\right\}$$ =

$$\vec{v}_{0}\frac{b/\eta }{r^{3}}-\vec{r}(\vec{v}_{0}\cdot \vec{r}) \frac{3b/\eta }{r^{5}}$$ = (5.26)

from which

  

$$\nabla ^{2}F_{1}+2F_{2} \frac{b/\eta }{r^{3}}$$ = (5.27)

$$\nabla ^{2}F_{2}+\frac{4}{r}\frac{\partial F_{2}}{\partial r} -\frac{ 3b/\eta }{r^{5}}.$$ = (5.28)

The boundary conditions for these equations are

  

$$0,F_{1}(\infty )=-1$$ F₁(R) = (5.29)

$$0,F_{2}(\infty )=0.$$ F₂(R) = (5.30)

We first solve Eq. (5.28) under the condition Eq. (5.30). Solutions to the homogeneous equation are cr^-5; a particular solution is $(b/2\eta )r^{-3}$. According to the boundary condition Eq. (5.30) we then get

$$F_{2}(r)=\frac{b}{2\eta }\frac{1}{r^{3}}\left( 1-\frac{R^{2}}{r^{2}}\right) .$$ (5.31)

Equation (5.27) now reads

$$\nabla ^{2}F_{1}=\frac{bR^{2}}{\eta }\frac{1}{r^{5}}.$$ (5.32)

Solutions to the homogeneous equation are -1+cr^-1; a particular solution is $(bR^{2}/6\eta )r^{-3}$. The boundary conditions Eq. (5.29) then lead to

$$F_{1}(r)=\frac{bR^{2}}{6\eta }\frac{1}{r^{3}}+R\frac{1}{r}\left( 1-\frac{b}{ 6\eta R}\right) -1.$$ (5.33)

Finally we have to fulfill the incompressibility equation

$$\vec{\nabla}\cdot \vec{v} \vec{v}_{0}\cdot \vec{\nabla}F_{1}+3(\vec{v} _{0}\cdot \vec{r})F_{2}+\vec{r}\cdot \vec{\nabla}(\vec{v}_{0}\cdot \vec{r} F_{2})$$ =

$$(\vec{v}_{0}\cdot \vec{r})\left\{ \frac{1}{r}\frac{\partial F_{1}}{ \partial r}+4F_{2}+r\frac{\partial F_{2}}{\partial r}\right\} =0.$$ = (5.34)

Introducing F₁ and F₂ we find $b=\frac{3}{2}\eta R$.

We now introduce F₁ and F₂ from Eqs. (5.33) and (5.31), together with $b=\frac{3}{2}\eta R$ into Eq. (5.25), and add $\vec{v}_{0}$ to get the flow field around a sphere at the origin, moving with velocity $\vec{v}_{0}$ in a quiescent fluid.

$$\vec{v}(\vec{r})=\vec{v}_{0}\frac{3R}{4r}\left( 1+\frac{R^{2}... ... \vec{r})\frac{3R}{4r^{3}}\left( 1- \frac{R^{2}}{r^{2}}\right)$$ (5.35)

One easily checks the boundary conditions

$$\vec{v}(R) \vec{v}_{0}$$ = (5.36)

$$\vec{v}(\infty ) \vec{0}$$ = (5.37)

We shall now use this flow field  to calculate the force exerted by the fluid on the particle.

According to Eq. (5.8) the force exerted by the fluid on the sphere is given by

$$\vec{F}=\int d\Omega R^{2}S(\vec{R})\cdot \frac{\vec{R}}{R}$$ (5.38)

where $R^{2}d\Omega$ is a surface element on the sphere. The stress tensor is given by Eq. (5.13), which in the case of an incompressible fluid reads

$$S_{ij}=\eta \left\{ \frac{\partial v_{i}}{\partial x_{j}}+\frac{\partial v_{j}}{\partial x_{i}}\right\} -P\delta _{ij}.$$ (5.39)

Using the flow field given in Eq. (5.35) and the pressure given in Eq. (5.23), where $b=\frac{3}{2}\eta R$, we calculate

$$\bar{S}(\vec{R})\cdot \vec{R} -\frac{3}{2}\eta \vec{v}_{0}+\{\eta (\vec{v} _{0}\cdot \vec{R})\frac{3}{2R^{2}}-P(\vec{R})\}\vec{R}$$ =

$$-\frac{3}{2}\eta \vec{v}_{0}-a\vec{R}.$$ = (5.40)

Introducing this into Eq. (5.38) and performing the integral we obtain

$$\vec{F}=-6\pi \eta R\vec{v}_{0}$$ (5.41)

$\vec{F}$ is known as the Stokes friction .