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Cartesian coordinates to Polar coordinates (dx,dy question)

  1. Feb 3, 2015 #1
    The usual change of variables in this case (mentioned in the title of this topic) is this:
    ##x = rcos(t)##
    ##y = rsin(t)##

    When I rewrite (say my integral) in polar coordinates I have to change ##dxdy## to ##rdrdt##

    My question is why can't I just compute dx and dy the usual way (the already mentioned change of variables) meaning finding the complete differentials dx and dy, multiplying them together, etc.?
     
  2. jcsd
  3. Feb 3, 2015 #2

    jambaugh

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    Try it, just remember that the dr^2 and dt^2 terms must be crossed out (they are Grassmann variables and nilpotent or equivalently they represent differential area of parallelogram with both sides in the dr direction (or both in the dt direction) which is thus zero.).
     
  4. Feb 3, 2015 #3
    Mathematically speaking the differentials ##dx## and ##dy## are not ordinary numbers that you can simply divide/multply since they are defined through a limit process. However, you can use your "multiplication thumb rule" only when the new coordinate are independent from each other, i.e. when you have something like ##x=f(x^\prime)##, ##y=g(y^\prime)##, without mixing ##x^\prime## and ##y^\prime##. This is simply because in this case the Jacobian matrix is diagonal. Otherwise you have to include the determinant of the Jacobian.
     
  5. Feb 3, 2015 #4

    Stephen Tashi

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    Consider a rectangle with corners (x,y),(x+h,y),(x+h,y+k),(x,y+k).

    If you make a change of variables x = f(u), y = g(v) then in the u,v coordinate system, the rectangle is still a rectangle of some sort although the length of its sides may be different. The area of the transformed figure is still its width times its height.

    However if you make a change of variables of the form x = p(s,t), y = q(s,t) then in the s,t coordinate system the figure may no longer be nearly a rectangle. It could be a quadrilateral with no side parallel to either axis. Hence it may not even have a "width" and "height". The procedure that uses the Jacobian determinant uses the idea of approximating the quadrilateral by a parallelogram.
     
    Last edited: Feb 3, 2015
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