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Thread: parametrization

  1. #1
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    parametrization

    The parametrization of the sphere:

    $\displaystyle {\bf x}\,(u^1,u^2) = (r \cos u^2 \cos u^1,\ r \cos u^2 \sin u^1,\ r \sin u^2)$
    $\displaystyle 0 \leq u^1 < 2\pi,\ \ \ -\frac{\pi}2 \leq u^2 \leq \frac{\pi}2$

    appears in my book without explanation. Any help on how to derive it? (I played with it and found the coordinates correspond to latitude and longitude).
    Last edited by myrerer; May 15th 2009 at 03:30 PM.
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  2. #2
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    What do you know about "spherical coordinates"? In standard spherical coordinates, we have $\displaystyle \rho$, the straight line distance from the origin to the point, $\displaystyle \theta$, the angle a line from the origin to the point on the xy-plane directly below the given point makes with the x-axis (the same angle as in two-dimensional polar coordinates, and [itex]\phi[/itex] the line from the origin to the point makes with the xy-plane. You can think of the two angles as being "longitude" and "co-latitude" (since it is measured from the "north pole" rather than the "equator").

    In spherical coordinates, $\displaystyle x= \rho cos(\theta)sin(\phi)$, $\displaystyle y= \rho sin(\theta) sin(\phi)$, and $\displaystyle z= \rho cos(\phi)$.

    Here, I notice we have "sin" in z rather than "cos". Okay, they are using "latitude" rather than "co-latitude". That is, they are using $\displaystyle \psi= \pi/2- \phi$ so that
    $\displaystyle x= \rho cos(\theta)cos(\psi)$
    $\displaystyle y= \rho sin(\theta)cos(\psi)$
    $\displaystyle z= \rho sin(\psi)$

    Finally, note the changes in notation: r instead of $\displaystyle \rho$, $\displaystyle u
    ^1$ instead of $\displaystyle \theta$, and $\displaystyle u^2$ instead of $\displaystyle \psi$.
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  3. #3
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    Quote Originally Posted by HallsofIvy View Post
    In spherical coordinates, $\displaystyle x= \rho cos(\theta)sin(\phi)$, $\displaystyle y= \rho sin(\theta) sin(\phi)$, and $\displaystyle z= \rho cos(\phi)$.
    How did you get this? Thanks.
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