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Spherical Coordinates

  1. Sep 3, 2004 #1
    I'm trying to find the line element in spherical coordinates as well as a velocity element. I know that they are (ds)^2=(dr)^2+r^2(sin(theta))^2(dtheta)^2+r^2(dphi)^2 and sqrt[(dr/dt)^2+r^2(sin theta)^2(dtheta/dt)^2+r^2(dphi/dt)^2].

    I know that this should be a quick and easy problem, but I simply can not figure it out. I would really appreciate some help on this one.
     
  2. jcsd
  3. Sep 3, 2004 #2

    Tide

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    Start from Cartesian coordinates in which [itex]ds^2 = dx^2+dy^2+dz^2[/itex] then calculate the differentials dx, dy and dz using:

    [tex]x = r \sin \theta \cos \phi[/tex]
    [tex]y = r \sin \theta \sin \phi[/tex]
    [tex]z = r \cos \theta[/tex]

    Substitute for dx, dy and dz in [itex]ds^2 = dx^2+dy^2+dz^2[/itex] and after a bit of algebra you should get the desired result.
     
  4. Sep 3, 2004 #3

    robphy

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    To find the velocity vector, write the position vector as [itex] r(t) \hat r [/itex].
    Then [itex]v(t)=\frac{d}{dt} \left[ r(t) \hat r \right] [/itex].
    Use the product rule.
    You'll have to compute [itex]\frac{d}{dt} \hat r [/itex],
    where [tex]\hat r= \sin\theta\cos\phi \hat\imath + \sin\theta\sin\phi \hat\jmath + \cos\theta \hat k[/tex].

    To simplify what you get, you might find it useful to know that
    [tex]\hat \theta= \cos\theta\cos\phi \hat\imath + \cos\theta\sin\phi \hat\jmath - \sin\theta \hat k[/tex]
    and [tex]\hat \phi= -\sin\phi \hat\imath + \cos\phi \hat\jmath [/tex]

    You can derive these expressions for the spherical-polar unit vectors if you calculate the vectorial element
    [tex]d \vec s = (dx)\hat \imath + (dy)\hat \jmath + (dz)\hat k [/tex]
    using Tide's expressions for x, y, and z. [The strategy is to group the terms in [itex] dr[/itex], [itex]d\theta[/itex], and [itex]d\phi[/itex].]
     
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