# Polar coordinates?

1. May 18, 2005

### quasar987

As I understand it, the polar coordinates of a point is defined by the rectangular coordinates of that point according to the transformation T from R² to R² defined by

$$T:(x,y)\mapsto (\sqrt{x^2+y^2},tan\left(\frac{y}{x}\right))$$

But this definition fails for y=pi and x=2 because tan(pi/2) is not defined.

We could defined it this way for $(x,y)\in \mathbb{R}^2 \backslash \{(x,y) \ \vert \ y/x = (n+1/2)\pi, \ n\in \mathbb{Z}\}$, and by

$$T:(x,y)\mapsto (\sqrt{x^2+y^2}, Arccos_n \left(\frac{x}{\sqrt{x^2+y^2}}\right))$$

for $(x,y) \in \{(x,y) \ \vert \ y/x = (n+1/2)\pi, \ n\in \mathbb{Z}\}$ and where Arccos_n is the inverse function of cos in the interval containing (n+1/2)pi.. i.e. $Arccos_n(z): [-1,1]\rightarrow [n\pi, (n+1)\pi]$.

This is phenomenally ugly. Is there a nicer way to define the polar coordinates?

2. May 18, 2005

### arildno

That has never been correct, this is somewhat more correct:
$$T:(x,y)\mapsto (\sqrt{x^2+y^2},arctan\left(\frac{y}{x}\right))$$

3. May 18, 2005

### quasar987

Oh damn!

But more like 'phew!' really.

4. May 18, 2005

### dextercioby

These basis changes should be made through diffeomorphisms.Obviously

$$\hat{T}:(x,y)\mapsto \left(\sqrt{x^{2}+y^{2}},\arctan \frac{y}{x}\right)$$

is not a diffeomorphism from $$\mathbb{R}^2}\rightarrow \mathbb{R}^{2}$$.

Daniel.

5. May 18, 2005

### quasar987

What does that mean Daniel?

6. May 18, 2005

### dextercioby

Invertible $C^{\infty}$ maps from one space to another.

U'll have to exclude the Oy axis.

Daniel.

Last edited: May 18, 2005
7. May 18, 2005

### quasar987

Well I don't see your point. Why must the transformation from rectangular to polar coord. be a diffeomorphism absolutely?!

8. May 18, 2005

### dextercioby

Pick the point $(0,2)$ in cartesian.Can your mapping send it to polar coords...?

Daniel.

9. May 18, 2005

### quasar987

No. So how do you suggest we avoid this problem?

10. May 18, 2005

### arildno

It's simplest to define the mapping by through:
$$x=r\cos\theta, y=r\sin\theta$$

This sets up a bijection almost everywhere between (x,y) and $$(r,\theta)$$.
(That is with (x,y) on the plane, r on the non-negative half-axis, and $$\theta$$ on the half-open interval $$[0,2\pi)$$

Last edited: May 18, 2005
11. May 18, 2005

### quasar987

If theta is traped in the interval [0, 2pi), then surely when we say that the potential energy of an object free to rotate around the z axis is

$$V(\theta) = -\int_{\theta _s}^{\theta}N\z(\theta)d\theta$$ (Symon pp.212)

the theta involved in this equation is not the theta of polar/cylindrical coordinates (i.e. constrained in [0, 2pi)), is it?

I had succeeded in proving this equation but it involved treating the polar angle has being free to take any value in $(-\infty, \infty)$. I was trying to justify that it was justified to do that. But now I'm a little confused. Is it justified?