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Cylindrical vs. spherical coordinates

  1. Aug 24, 2009 #1
    Hi everyone!

    There's a question bothering me about the two coordinate systems - cylindrical and spherical:

    Consider the two systems, i.e. [tex](r, \theta, \phi)\rightarrow\left(\begin{array}{c}r\sin\theta\cos\phi\\r\sin\theta\sin\phi\\r\cos\theta\end{array}\right)[/tex] and [tex](r,\phi,z)\rightarrow\left(\begin{array}{c}r\cos\phi\\r\sin\phi\\z\end{array}\right)[/tex]

    Now, restricting us to only 2 dimensions and setting z=0 it automatically follows that [tex]\theta=\frac{\pi}{2}[/tex]

    The big question is: WHY are the differential operators (div and Laplacian) different within this restriction. I mean, we are describing the plane via the usual polar coordinates. Thus the operators have to produce the same results.


    Here are some defs:

    Let U be a scalar and F be a vector field:

    [tex]\nabla U = \frac{\partial f}{\partial r}\vec r+\frac{1}{r}\frac{\partial f}{\partial\phi}\vec\phi[/tex] for cylindrical coordinates and

    [tex]\nabla U=\frac{\partial f}{\partial r}\vec r+\frac{1}{r\sin\theta}\frac{\partial f}{\partial\phi}=\frac{\partial f}{\partial r}\vec r+\frac{1}{r}\frac{\partial f}{\partial\phi}\vec\phi[/tex] for spherical coordinates, since z=0 and [tex]\theta=\pi/2[/tex]

    It's ok for now, just as it should be. But here comes the interesting part, considering the divergence:

    [tex]\nabla\vec F= \frac{1}{r}\frac{ \partial (r F_r)}{\partial r}+\frac{1}{r}\frac{\partial F_{\phi}}{\partial\phi}[/tex] for cylindrical coordinates and

    [tex]\nabla\vec F=\frac{1}{r^2}\frac{\partial (r^2F_r)}{\partial r}+\frac{1}{r\sin\theta}\frac{\partial F_{\phi}}{\partial\phi}=\frac{1}{r^2}\frac{\partial (r^2F_r)}{\partial r}+\frac{1}{r}\frac{\partial F_{\phi}}{\partial\phi}[/tex] for spherical coordinates

    (all the \theta-terms do not contribute since the fields U and F have no \theta-coordinate)

    Now I calculated that:

    [tex]\frac{1}{r}\frac{ \partial (r F_r)}{\partial r}\neq\frac{1}{r^2}\frac{\partial (r^2F_r)}{\partial r} [/tex]

    but How am I supposed to explain this to me?

    I thought that the two coordinate systems are equivalent, when reduced to the 2-dim. polar coordinate form, but apperantly they're not.

    What is more, the Laplacians are also different, which leads to different equations for the radial part of, let's say, the 2D wave equation. But the coordinate system has (physically) not been changed - just the angle \phi and the radius r. This would suggest that I could describe a 2-dimensional problem with a point-symmetry via both coordinate system restrictions and obtain different differential equations respectively, which have to describe the same problem in the same coordinate system?!
    (it is the circular membrane wave problem (drum) described both ways via the cylindrical and spherical Bessel functions that is meant here in particular)


    Any help is much appreciated!

    Thanks in advance!

    marin
     
  2. jcsd
  3. Aug 24, 2009 #2

    arildno

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    Quite simply in that what you regard as the radial component of the divergence from the spherical coordinates DO involve theta-differentiations of the radial unit vector.
     
  4. Aug 25, 2009 #3
    Ok, so it follows that spherical coordinates are inappropriate for dealing with 2D point-symmetry physical problems (like the vibrating cicular membrane), as the formulae for div and thus for the Laplacian do not correctly describe the plane using the angle phi and the radius, as they do in cylindrical?

    But if this is so, then there must be a mistake in what is stated in this Wikipedia article, section 'Vibrating Membrane':

    http://en.wikipedia.org/wiki/Helmholtz_equation

    since they are applying spherical coordinates there, and not cylindrical!

    What is more, the resulting equation is then NOT the original Bessel equation (as they call it), since the factor 2 in front of the first derivative, coming form the spherical divergence, should not be there!

    ...or I didn't quite get your idea?
     
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