In Fluid Mechanics textbook, I found the following example:

In the rotating sphere viscometer, a solid sphere of radius $R$ is suspended from a wire and rotates slowly at constant angular velocity $\Omega$ about the long axis of the wire in an incompressible Newtonian fluid. The fluid is quiescent far from the sphere.

(a) Use the no-slip boundary condition on the surface of the rotating sphere to postulate the functional form of the fluid velocity profile when rotation is slow enough and centrifugal forces can be neglected.

Answer: Consider rigid-body rotation of a solid sphere about the $z$ axis of a Cartesian coordinate system and calculate the velocity vector at the fluid–solid interface by invoking the no-slip condition: $$\vec{v}=\left(\left .\vec{\Omega} \times \vec{r} \right) \right|_{r=R} \tag{1}$$

The angular velocity vector is oriented in the $z$ direction (i.e., = $\Omega \vec{\delta}_{z}$), and the position vector from the axis of rotation (i.e., along the wire) to any point on the surface of the solid sphere is: $$\vec{r}=R\sin(\theta)\left(\vec{\delta}_{r}\sin(\theta)+\vec{\delta}_{\theta}\cos(\theta) \right) \tag{2}$$

where $\theta$ is the polar angle measured from the $z$ axis.[...]

My questions are:

1) Why the position vector is not defined by the corresponding expresion in spherical coordinates:


Here, $\vec{\delta}_{i}$ is the unit vector in the $i$ direction.

2) How deduce the expresion in equation $(2)$


I made some advances since the last time I asked this question. I found, based on the coment of @npojo that there is a relation between the $\vec{r}$ and the $\vec{\delta}_{r}^{C}$ unit vector (the unit vector in the $r$ direction in cylindrical coordinates).


$\vec{\delta}_{r}^{C}$ in cartesian coordinate system is:


where $\theta^{C}$ is the polar angle in cylindrical coordiantes and equals to azimuthal angle in spherical coordinates $\phi$, thus: $$\vec{\delta}_{r}^{C}=\cos(\phi)\vec{\delta}_{x}+\sin(\phi)\vec{\delta}_{y}$$

The relation of spherical and cartesian unit vectors are:

$$ \vec{\delta}_{x}=(\sin\theta\cos\phi)\vec{\delta}_{r}+(\cos\theta\cos\phi)\vec{\delta}_{\theta}+(-\sin\phi)\vec{\delta}_{\phi}$$ $$\vec{\delta}_{y}=(\sin\theta\sin\phi)\vec{\delta}_{r}+(\cos\theta\sin\phi)\vec{\delta}_{\theta}+(\cos\phi)\vec{\delta}_{\phi}$$

finally we get:

$$\vec{\delta}_{r}^{C}=\sin\theta(\cos^2\phi+\sin^2\phi)\vec{\delta}_{r}+\cos\theta(\cos^2\phi+\sin^2\phi)\vec{\delta}_{\theta}$$ $$\boxed{\vec{\delta}_{r}^{C}=\left(\vec{\delta}_{r}\sin(\theta)+\vec{\delta}_{\theta}\cos(\theta) \right)}$$

  • $\begingroup$ What does your book say about the angle $\theta$? How is it defined? $\endgroup$
    – flaudemus
    Commented Mar 10, 2019 at 21:06
  • 2
    $\begingroup$ As for (1), r is defined from the axis of rotation, not from center of sphere. So these are cylindrical coordinates. Regarding (2), I am guessing the leftmost $\theta$ is wrong and should be say, $\phi$, defining latitude on the sphere. $\endgroup$
    – npojo
    Commented Mar 10, 2019 at 21:14
  • $\begingroup$ @flaudemus the textbook says: "where $\theta$ is the polar angle measured from the $z$ axis." $\endgroup$
    – camd92
    Commented Mar 10, 2019 at 21:22

1 Answer 1


I have problem with your calculations ?

enter image description here you can "create" a sphere by rotating (about the $z$ axis) a half circle in the ($x',z$) plane .

first i calculate the components of the vector $\vec{r}$


transformed to cartesian coordinates (sphere parameters) :

$$\frac{1}{R}\,\vec{r}= \begin{bmatrix} \cos(\phi) & -\sin(\phi) & 0 \\ \sin(\phi) & \cos(\phi) & 0 \\ 0 & 0 & 0 \\ \end{bmatrix}\begin{bmatrix} r_{x'} \\ 0 \\ 0 \\ \end{bmatrix}= \begin{bmatrix} \cos(\phi)\,\sin(\theta) \\ \sin(\phi)\,\sin(\theta) \\ 0 \\ \end{bmatrix}=\sin(\theta)\,\begin{bmatrix} \cos(\phi) \\ \sin(\phi) \\ 0 \\ \end{bmatrix} $$

so the components of the vector $\vec{r}$ are: $$\vec{r}=\begin{bmatrix} x \\ y \\ z \\ \end{bmatrix} = R\,\sin(\theta)\begin{bmatrix} \cos(\phi) \\ \sin(\phi) \\ 0 \\ \end{bmatrix}$$

  • $\begingroup$ the vector in the last line: \begin{bmatrix} \cos(\phi) \\ \sin(\phi) \\ 0 \\ \end{bmatrix} is in cartesian coordinates $xyz$? $\endgroup$
    – camd92
    Commented Mar 11, 2019 at 20:44
  • $\begingroup$ @camd92 the components are in cartesian coordinte $\endgroup$
    – Eli
    Commented Mar 11, 2019 at 22:30

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