# Calculation of polar velocity components given cartesian counterparts

I would like to calculate the polar velocity components given the position $(x,y)$ and velocity $(u_x,u_y)$ in Cartesian coordinates. First of all, $$r=\sqrt{x^2+y^2}\text{ and }\theta=\tan^{-1}\left(\frac yx\right).$$ By now, I know the angle and radius in the global cylindrical coordinate system. I assume that $\mathbf u=u_r\,\mathrm e_r+u_\theta\,\mathrm e_\theta$. Is it correct to write, \begin{align} u_r&=\cos(\theta)u_x + \sin(\theta)u_y \\ u_\theta&=-\sin(\theta)u_x + \cos(\theta)u_y? \end{align} The problem is that I would like to calculate the components in polar coordinates (not angular velocity though) given that a particle is moving from $(x_1,y_1)$ to $(x_2,y_2)$.

• I am unsure of your question. Can you be more specific? What you have written is correct, but the components of the velocity in moving from point 1 to point 2 completely depends on how you actually move between those points. – Aaron Stevens Jun 13 '18 at 18:09
• Those points are discrete, thus I assume I am moving linearly. The relations above are related to the cyclotron motion of an electron in a magnetic field. I know the velocity and position in cartesian coordinate but I would like to translate them in a global cylindrical system (not the local one of the electron) – dimpep Jun 14 '18 at 7:41
• Motion on a straight line still does not uniquely define the velocity. It still matters how you move from the first point to the second point. If you are working on a problem about electron motion in a magnetic field then I would suggest actually putting more detail of your problem into your question. That way you can get more specific and directed help on what you are really asking about. Right now it is still unclear. – Aaron Stevens Jun 14 '18 at 14:00
• I do not know how moving from one point to another. Even if I have to deal with a cyclotron motion, I do not know a priori that it is such kind of motion. Therefore I have only $u_x$ and $u_y$ for points defined at $x_1,x_2$ and $y_1,y_2$. I have then to translate those characteristics to linear and radial velocity in a global cylindrical coordinate system. Not the electrons. The system has origin at $x,y=0,0$ – dimpep Jun 19 '18 at 11:00

From what I can gather you just want to replace the trigonometric functions with their Cartesian counterparts

$$cos(\theta)=\frac xr=\frac{x}{\sqrt {x^2+y^2}}$$

$$sin(\theta)=\frac yr=\frac{y}{\sqrt {x^2+y^2}}$$

So what you have for $u_r$ becomes

$$u_r=\frac{xu_x}{\sqrt {x^2+y^2}}+\frac{yu_y}{\sqrt {x^2+y^2}}=\frac{xu_x+yu_y}{\sqrt {x^2+y^2}}$$

And $u_\theta$ becomes

$$u_\theta=-\frac{yu_x}{\sqrt {x^2+y^2}}+\frac{xu_y}{\sqrt {x^2+y^2}}=\frac{xu_y-yu_x}{\sqrt {x^2+y^2}}$$

Therefore, the velocity $\vec v$ with polar components expressed in Cartesian coordinates is

$$\mathbf{\vec v=\frac{xu_x+yu_y}{\sqrt {x^2+y^2}}e_r+\frac{xu_y-yu_x}{\sqrt {x^2+y^2}}e_\theta}$$

If you want to explicitly substitute your expressions for $u_x$ and $u_y$ as expressed in your comments, then at any position $(x,y)$ the velocity will be given by

$$\mathbf{\vec v=\frac{x(x_2-x_1)+y(y_2-y_1)}{\delta t\sqrt {x^2+y^2}}e_r+\frac{x(y_2-y_1)-y(x_2-x_1)}{\delta t\sqrt {x^2+y^2}}e_\theta}$$

• To be more precise: I have been given $(x_1,y_1)$ and $(x_2,y_2)$ for a given time interval $\delta t$ as well as velocities $u_x$ and $u_y$, defined as $u_x=\frac{x_2-x_1}{\delta_t}$ and $u_y=\frac{y_2-y_1}{\delta_t}$. I can calculate $\theta$ and $r$ (as well as $\theta_1,\theta_2,r_1,r_2$) and I would like to calculate the radial and linear components of the corresponding velocity. – dimpep Jun 20 '18 at 8:47
• @dimpep So we are just assuming we are moving at a constant speed in a straight line from point 1 to point 2? – Aaron Stevens Jun 20 '18 at 18:49
• Yes, exactly. Constant speed. – dimpep Jun 21 '18 at 10:31
• @dimpep Ok, then the answer I have provided is fine. $u_x$ and $u_y$ are constant. Then you can plug any $(x,y)$ coordinate into the final expression to determine the components of the velocity at that point. – Aaron Stevens Jun 21 '18 at 13:08
• @dimpep I have added what I said in my previous comment to my answer. – Aaron Stevens Jun 21 '18 at 21:12