Converting this vector into polar form

[Rodger125]

TL;DR

I'd like to convert this surface velocity vector into the form I described. It might be just a matter of converting it into polar coords

In the following paper, the surface velocity for a moving, spherical particle is given as (eq 1):

$$\textbf{v}_s(\hat{\textbf{r}}) = \sum {2\over{{n(n+1)}}} B_n (\hat{\textbf{e}} \cdot \hat{\textbf{r}} \hat{\textbf{r}} - \hat{\textbf{e}}) P_n'( \hat{\textbf{e}} \cdot \hat{\textbf{r}})$$

where $\hat{\textbf{r}}$ is the preferred swimming axis (we consider that the sphere carries with it a fixed coordinate system that determines its preferred moving direction at each instant). $\hat{\textbf{r}}$ is a unit vector from the particle center to a point on the surface, we have the Legendre polynomials with $$P_n'$$ being the derivative of the n-th order Legendre polynomial, and $$B_n$$ is the amplitude of the corresponding mode.

They then (up to N=2) write the following expression for the surface tangential velocity, as a function of theta
$$\textbf{v}_s(\theta) = B_1 [sin(\theta) + {\alpha\over{2}} sin(2\theta)] \hat{\theta}$$
where $$\beta = B_2 / B_1$$.

How does one arrive at the second equation? Do you convert the first vector into polar coordinates? If so, how do you do this?

Thank you

 

[anuttarasammyak]

I do not go into the mathematics but I am convinced that velocity of fluid at particle surface has no radial component. In fact
$$\mathbf{\hat{r}} \cdot \textbf{v}_s(\hat{\textbf{r}}) = \sum {2\over{{n(n+1)}}} B_n (\mathbf{\hat{r}}\cdot (\hat{\textbf{e}}\cdot \hat{\textbf{r}}) \hat{\textbf{r}} - \mathbf{\hat{r}}\cdot\hat{\textbf{e}}) P_n'( \hat{\textbf{e}} \cdot \hat{\textbf{r}}) =0$$
where
$$\mathbf{\hat{r}}\cdot \mathbf{\hat{r}}=1$$

 

[Rodger125]

I do not go into the mathematics but I am convinced that velocity of fluid at particle surface has no radial component. In fact
$$\mathbf{\hat{r}} \cdot \textbf{v}_s(\hat{\textbf{r}}) = \sum {2\over{{n(n+1)}}} B_n (\mathbf{\hat{r}}\cdot (\hat{\textbf{e}}\cdot \hat{\textbf{r}}) \hat{\textbf{r}} - \mathbf{\hat{r}}\cdot\hat{\textbf{e}}) P_n'( \hat{\textbf{e}} \cdot \hat{\textbf{r}}) =0$$
where
$$\mathbf{\hat{r}}\cdot \mathbf{\hat{r}}=1$$

That would make sense, I think. If we have a coordinate system moving with the sphere, and the sphere does not change in radius, surely the radial velocity would be zero

 

[Fred Wright]

How does one arrive at the second equation? Do you convert the first vector into polar coordinates? If so, how do you do this?

The authors describe the polar angle as
$$
\hat r \cdot \hat e = \cos(\theta)
$$
Therefore taking the sum,
$$
\mathbf u^s (\hat r)=B_1(\cos(\theta)\hat r - \hat e)P_1^{'}(\cos(\theta))
$$
$$
+ \frac{1}{3} B_2(\cos(\theta)\hat r - \hat e)P_1^{'}(\cos(\theta))
$$
with
$$
P_1^{'}(\cos(\theta))=-\sin(\theta)
$$
$$
P_2^{'}(\cos(\theta))=-\frac{3}{2}\sin(2\theta)
$$
The sum becomes,
$$
-B_1(\cos(\theta)\hat r -\hat e)\sin(\theta) - \frac{1}{2} B_2(\cos(\theta)\hat r -\hat e)\sin(2\theta)
$$
The result follows if
$$
\cos(\theta)\hat r -\hat e=-\hat \theta
$$

 

[Rodger125]

The authors describe the polar angle as
$$
\hat r \cdot \hat e = \cos(\theta)
$$
Therefore taking the sum,
$$
\mathbf u^s (\hat r)=B_1(\cos(\theta)\hat r - \hat e)P_1^{'}(\cos(\theta))
$$
$$
+ \frac{1}{3} B_2(\cos(\theta)\hat r - \hat e)P_1^{'}(\cos(\theta))
$$
with
$$
P_1^{'}(\cos(\theta))=-\sin(\theta)
$$
$$
P_2^{'}(\cos(\theta))=-\frac{3}{2}\sin(2\theta)
$$
The sum becomes,
$$
-B_1(\cos(\theta)\hat r -\hat e)\sin(\theta) - \frac{1}{2} B_2(\cos(\theta)\hat r -\hat e)\sin(2\theta)
$$
The result follows if
$$
\cos(\theta)\hat r -\hat e=-\hat \theta
$$

Thank you!