Defining polar coordinates without undefined tangent values

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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

[tex]T:(x,y)\mapsto (\sqrt{x^2+y^2},tan\left(\frac{y}{x}\right))[/tex]

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

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

[tex]T:(x,y)\mapsto (\sqrt{x^2+y^2}, Arccos_n \left(\frac{x}{\sqrt{x^2+y^2}}\right))[/tex]

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

This is phenomenally ugly. Is there a nicer way to define the polar coordinates?
 
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quasar987 said:
[tex]T:(x,y)\mapsto (\sqrt{x^2+y^2},tan\left(\frac{y}{x}\right))[/tex]
That has never been correct, this is somewhat more correct:
[tex]T:(x,y)\mapsto (\sqrt{x^2+y^2},arctan\left(\frac{y}{x}\right))[/tex]
 
Oh damn!

But more like 'phew!' really. :smile:
 
These basis changes should be made through diffeomorphisms.Obviously

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

is not a diffeomorphism from [tex]\mathbb{R}^2}\rightarrow \mathbb{R}^{2}[/tex].

Daniel.
 
What does that mean Daniel?
 
Invertible [itex]C^{\infty}[/itex] maps from one space to another.

U'll have to exclude the Oy axis.

Daniel.
 
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Well I don't see your point. Why must the transformation from rectangular to polar coord. be a diffeomorphism absolutely?!
 
No. So how do you suggest we avoid this problem?
 
It's simplest to define the mapping by through:
[tex]x=r\cos\theta, y=r\sin\theta[/tex]

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

[tex]V(\theta) = -\int_{\theta _s}^{\theta}N\z(\theta)d\theta[/tex] (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 [itex](-\infty, \infty)[/itex]. I was trying to justify that it was justified to do that. But now I'm a little confused. Is it justified?