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Polar coordinates?

  1. May 18, 2005 #1

    quasar987

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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?
     
  2. jcsd
  3. May 18, 2005 #2

    arildno

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    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]
     
  4. May 18, 2005 #3

    quasar987

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    Oh damn!

    But more like 'phew!' really. :smile:
     
  5. May 18, 2005 #4

    dextercioby

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    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.
     
  6. May 18, 2005 #5

    quasar987

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    What does that mean Daniel?
     
  7. May 18, 2005 #6

    dextercioby

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    Invertible [itex] C^{\infty} [/itex] maps from one space to another.

    U'll have to exclude the Oy axis.

    Daniel.
     
    Last edited: May 18, 2005
  8. May 18, 2005 #7

    quasar987

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    Well I don't see your point. Why must the transformation from rectangular to polar coord. be a diffeomorphism absolutely?!
     
  9. May 18, 2005 #8

    dextercioby

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

    Daniel.
     
  10. May 18, 2005 #9

    quasar987

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    No. So how do you suggest we avoid this problem?
     
  11. May 18, 2005 #10

    arildno

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    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]
     
    Last edited: May 18, 2005
  12. May 18, 2005 #11

    quasar987

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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?
     
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