# Cartesian Coordinates to Polar Coordinates

I apologize if this question is trivial, but I am new to physics and am struggling with some of the basic concepts.

Working in $\mathbb{R}^2$ with standard coordinates $(x,y)$, suppose we have a particle of mass $m$ moving on a curve $(x(t),y(t))\in\mathbb{R}^2$. It's tangent vector (velocity vector) is $$x^\prime(t)\frac{\partial}{\partial x}+y^\prime(t)\frac{\partial}{\partial y} \ \ \ \ \ \ \ \ \ \ (1)$$This particle's kinetic energy is $\frac{1}{2}m\left((x^\prime(t))^2+(y^\prime(t))^2\right)$. Also, suppose we have some conservative force $F$ so that $F=\left(\frac{\partial U}{\partial x},\frac{\partial U}{\partial y}\right)$ where $U$ is some smooth potential $U:\mathbb{R}^2\to\mathbb{R}$.

Anything I've read says the kinetic energy in polar coordinates is $$\frac{1}{2}m\left((\dot r)^2+(r\dot\theta)^2\right)$$ and the force in the $r$ and $\theta$ directions are $$F_r=-\frac{\partial U}{\partial r} \ \ \ \ \text{ and } \ \ \ \ F_\theta=\frac{1}{r}\frac{\partial U}{\partial \theta}$$

For the second point, I don't understand what it means to say force in the $r$ or $\theta$-direction. It's clear the force in the $x$-direction is just the first component of $F$, but is the force in the $r$-direction just the first component of $F$ in polar coordinates? I don't see how that really makes sense. Also, computing $\frac{\partial U}{\partial x}=\frac{\partial U}{\partial r}\frac{\partial r}{\partial x}+\frac{\partial U}{\partial \theta}\frac{\partial \theta}{\partial x}$ (and similarily $\frac{\partial U}{\partial y}$) I can see where these terms pop up, but don't get how to put the concepts together.

For the first point, I don't understand how they are getting these equations, and especially how they get them so fast! If you use the change of variables formula (i.e. $\frac{\partial}{\partial x}=\frac{\partial r}{\partial x}\frac{\partial}{\partial r}+\frac{\partial\theta}{\partial x}\frac{\partial}{\partial \theta}$ and so on) on equation $(1)$, compute $x^\prime , y^\prime$, and collect like terms you get that the velocity vector above is $\dot r\frac{\partial}{\partial r}+r\dot\theta\frac{\partial}{\partial\theta}$. This takes some work but in this form it makes sense, to me, to say that the kinetic energy in polar coordinates is $\frac{1}{2}m\left((\dot r)^2+(r\dot\theta)^2\right)$. But any book I've read just computes this extremely quick by saying $$x^\prime(t)=\dot r\cos\theta +r\dot\theta\sin\theta \ \ \ \ \text{ and } \ \ \ \ y^\prime(t)=\dot r\sin\theta-r\dot\theta\cos\theta$$I see sometimes $\hat r=(\cos\theta,\sin\theta)$ and $\hat\theta=(-\sin\theta,\cos\theta)$ but how can you have a "basis" that changes at every point?

Any help would be greatly appreciated!!

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## 2 Answers

"but how can you have a "basis" that changes at every point?"

This is really the root of your problem. Mathematically (and physically) speaking, such a basis works fine. You just have a (hopefully temporary) conceptual problem.

Maybe try thinking of it this way: How does $\hat{x}$ know to point "to the right" in cartesian coordinates, at an arbitrary point in the coordinate system? You could define it as "parallel to the $x$-axis". Similarly for $\hat{y}$. Now how would you define $\hat{r}$? Parallel to a line joining the current point and the origin. $\hat{\theta}$? Perpendicular to $\hat{r}$. Not that different, right?

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Thank you for your answer! I know that on the tangent bundle, for each point you have a basis of the tangent space and so the basis changes in that sense. But isn't $\hat x$ defined to be $(1,0)$ and $\hat y=(0,1)$? I figured, $\frac{\partial}{\partial x}$ is being associated with $(1,0)$ and $\frac{\partial}{\partial y}$ with $(0,1)$. Is $\frac{\partial}{\partial r}$ being associated with $(\cos\theta,\sin\theta)$? I think this is what is bothering me? – JonHerman Jul 29 '14 at 23:32
Defining $\hat{x}$ as $(1,0)$ is one (very convenient, widely used) way, but not the only way of course. $\frac{\partial}{\partial r}$ is the derivative along the direction $(\cos\theta,\sin\theta)$, yes. – Kyle Oman Jul 29 '14 at 23:36

In case anyone is interested I'll post here what I learnt. If anyone has any corrections or comments please let me know.

The thing that I was missing was that the metric makes the kinetic energy well defined, i.e. independent of coordinates. Working in $\mathbb{R}^2$ with standard coordinates $(x^1,x^2)$ we have the induced coordinates $(x^1,x^2,v^1,v^2)$ on $T\mathbb{R}^2=\mathbb{R}^4$. Let $g$ be an arbitrary metric on $\mathbb{R}^2$. The kinetic energy is a function $$K:\mathbb{R}^4\to\mathbb{R} \ , \ (x^1,x^2,v^1,v^2)\mapsto \frac{1}{2}mg_{ij}v^i\cdot v^j$$We also have the definition of the gradient of a function $U$, which is $$\nabla U=\frac{\partial U}{\partial x^j}g^{ij}\frac{\partial}{\partial x^i}$$In the case that the metric is the standard one, we have $g_{ij}=\delta_{ij}$ and so this reduces to the standard definition of kinetic energy and the usual notion of the gradient. With this metric, but in polar coordinates a quick computation shows that $$g=\left[\begin{array}{c c}1&0\\ 0&r^2\end{array}\right]$$ and so with standard induced coordinates $r,\theta,\widetilde r,\widetilde\theta$, since the velocity vector of a curve is $\dot r\frac{\partial}{\partial r}+\dot\theta\frac{\partial}{\partial\theta}$ the kinetic energy is just $\frac{1}{2}m(\dot r)^2+r^2(\dot\theta)^2$. We also have that $$g^{-1}= \left[\begin{array}{c c}1&0\\ 0&\frac{1}{r^2}\end{array}\right]$$so that the gradient is $(\frac{\partial U}{\partial r},\frac{1}{r^2}\frac{\partial U}{\partial\theta})$. It then makes sense to define the force in the $r$-direction as the projection of $F$ onto $\frac{\partial}{\partial r}$ which is associated with $(\cos\theta,\sin\theta)$ and the force in the $\theta$-direction as the projection of $F$ onto $\frac{\partial}{\partial \theta}$ which is associated with $(-\sin\theta,\cos\theta)$.

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