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Shortest arc between two points in polar coordinates

  1. Apr 14, 2010 #1
    If we consider a Euclidean plane [tex]\mathbb{R}^2[/tex] with the ordinary inner product, and we "distort" it through a cartesian->polar transformation, how should I compute the shortest arc between two points [tex](r,\theta)[/tex] and [tex](r',\theta')[/tex] ?
  2. jcsd
  3. Apr 14, 2010 #2
    What metric are you using on the polar plane?
  4. Apr 14, 2010 #3
    I am using the metric I derived from the equations
    [tex]x=r cos(\theta)[/tex]
    [tex]y=r sin(\theta)[/tex]

    From those I got:

    [tex]M = diag(1,r^2)[/tex]
  5. Apr 14, 2010 #4
    I guess I'm not sure what you're looking for. The shortest arc will be the image of an honest straight line under the isometry.
  6. Apr 14, 2010 #5
    Ok. I guess my original question was meaningless.
    As far as I could understand, computing a shortest-arc length makes sense only on surfaces whose curvature changes locally. The [itex]\mathbb{R}^2[/itex] plane is flat, so the shortest arcs between two points are always straight lines.

    Basically, all I have to do is to consider the straight line connecting the two points (in cartesian coordinates), and convert its parametric representation into polar coordinates.

    Is this correct?
  7. Apr 15, 2010 #6
    Computing geodesics is the same no matter what the metric does. It's just particularly easy here.

    So yes, you're correct.
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