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Circumference of a circle (in strange coordinates)

  1. Sep 10, 2012 #1
    1. The problem statement, all variables and given/known data

    We are given a function defined by
    [itex] x = uv,
    y = 1/2 (u^2-v^2)[/itex]


    2. Relevant equations

    I derived the line element [itex] ds^2 = (u^2+v^2) dv^2 + (u^2+v^2) du^2 [/itex]

    However I decided this was to unwieldy to derive our circumference where

    C = [itex]2*{R}\oint_{-R}^{R} ds[/itex]

    So I decided to try to convert my coordinats to polar. I realized the equation of a 2-d circle in these coordinates is
    [itex] 4(y^2+x^2) = (u^2+v^2) = R^2[/itex]

    so then R = 2r in polar coordinates.. right? I think I can use this information to construct a simple integral in polar coordinates but i don't know exactly what to do.

    Thanks in advance
     
    Last edited: Sep 11, 2012
  2. jcsd
  3. Sep 10, 2012 #2
    Perhaps I'm missing the point of the problem; why would the circumference be any different in these coordinates?
     
  4. Sep 11, 2012 #3
    I believe its a non-euclidean geometry problem (could be wrong). For example if I take the area of a circle on the face of a sphere using the area element in spherical coordinates you will get

    [itex] \int_{\vartheta=0}^{2\pi} \int_{
    \varphi=0}^{\pi/2}a^2 sin(\phi) d\phi d\rho d\theta [/itex]

    Which is the not the same as the area of a circle in euclidean (plane) geometry.

    edit: you will run into the same problem calculating the circumference also.
     
  5. Sep 11, 2012 #4
    I don't think this is correct :frown:

    Well all you're doing is changing coordinates. The names have been changed, but the geometry is the same. Things like lengths should be coordinate independent.
     
  6. Sep 11, 2012 #5
    Yeah, the metric was ultimately flat, so actual lengths are still Euclidean. In the case of a sphere, when you write out the metric in terms of two coordinates, the metric is not flat. No coordinate transformations can change that.
     
  7. Sep 11, 2012 #6
    Ok your right i screwed up the circle equation it should be 4y^2 + x^2.

    I am still trying to figure out how to calculate the area and circumference for a circle in this coordinate system though. I'm not arguing that the circumference isn't changed, but how would you show that?

    after all spherical coordinates are just a coordinate transformation where your mapping
    [itex] x = \rho cos\phi sin\theta , y = \rho sin\phi sin\theta , z = r cos\theta [/itex]

    I don't see how that is so much different from what is being performed here. And i already demonstrated that the area of a circle on the 2-d surface of a sphere isn't equal to the area of a circle on a plane.

    However I think i titled this post poorly and should have outlined the question more accurately. This is how it is stated in my book.

    Calculate the ratio of the circumference to the diameter of a circle using (u,v)
    coordinates. Do you get the correct answer?
     
  8. Sep 11, 2012 #7
    No that's still not right - I got
    [tex] x^2 + y^2 = \frac{1}{4}(u^2 + v^2)^2.[/tex]
    See if you agree.

    So it is true that circles in the x-y plane are mapped onto circles in the u-v plane - but see the hint below.

    Ok I worked the problem myself and it's actually quite cute.

    To find the radius and circumference, just work in the (u,v) coordinates using your line element - which amounts to multiplying the usual lengths by R, with

    [tex]R^2 = u^2 + v^2.[/tex]

    You'll have to do a little integral to get the radius, and you can just use c = 2 pi R to get the circumference - but (hint) watch out for a subtle factor of 2.

    The whole thing has a nice interpretation as a conformal map - if you know a bit of complex analysis - I don't know if that's the point of view from which you're approaching this.

    Yeah I think this turned out to be a bit of a red herring.

    As you rightly point out, there are two different areas here - the area of the (flat) disc bounded by the circle and the area of the bit of the 2-sphere bounded by the circle. But note that they're different areas because they're different surfaces, not because of any choice of coordinate system. You can work out either area in the coordinate system of your choice.
     
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