# Converting this vector into polar form

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• Rodger125
In summary, the paper discusses the surface velocity of a moving, spherical particle, which is given by a summation of Legendre polynomials. They then provide an expression for the surface tangential velocity as a function of theta, which can be derived by converting the first vector into polar coordinates and taking the sum.

#### Rodger125

TL;DR Summary
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

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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
$$\mathbf{\hat{r}}\cdot \mathbf{\hat{r}}=1$$

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Rodger125
anuttarasammyak said:
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

Rodger125 said:
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
Fred Wright said:
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!

## What is polar form?

Polar form is a way of representing a vector in terms of its magnitude (or length) and direction. It is typically written as (r, θ), where r is the magnitude and θ is the angle between the vector and the positive x-axis.

## Why would I need to convert a vector into polar form?

Converting a vector into polar form can be useful when dealing with complex mathematical operations, such as finding the sum or difference of two vectors. It also allows for a more intuitive understanding of the vector's direction and magnitude.

## What are the steps to convert a vector into polar form?

The first step is to graph the vector on a coordinate plane. Then, use the Pythagorean theorem to find the magnitude of the vector. Next, use trigonometric functions to find the angle between the vector and the positive x-axis. Finally, write the vector in the form (r, θ), where r is the magnitude and θ is the angle in radians.

## Can any vector be converted into polar form?

Yes, any vector can be converted into polar form. However, it is important to note that polar form is most commonly used with two-dimensional vectors in the Cartesian coordinate system.

## Is there a difference between polar form and polar coordinates?

Yes, there is a difference between polar form and polar coordinates. Polar form refers to the representation of a vector in terms of its magnitude and direction, while polar coordinates refer to a specific point on a coordinate plane with coordinates (r, θ). However, the two are often used interchangeably in mathematics.