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Homework Help: Difference between Two Vectors, Spherical Coordinates

  1. Nov 8, 2008 #1
    1. The problem statement, all variables and given/known data
    I'm doing a problem that involves expressing, for two arbitrary vectors [itex]\vec{x}[/itex] and [itex]\vec{x'}[/itex],

    [tex]|\vec{x}-\vec{x'}|[/tex]

    in spherical coordinates ([itex]\rho,\theta,\phi[/itex]).

    2. Relevant equations

    Law of Cosines:

    [tex]c^{2}=a^{2}+b^{2}-2ab\cos\gamma[/tex]

    where [itex]\gamma[/itex] is the angle between a and b.

    3. The attempt at a solution

    Using the law of cosines, we can write

    [tex]|\vec{x}-\vec{x'}|=(\rho^{2}+\rho'^{2}-2\rho\rho'\cos\gamma)^{\frac{1}{2}}[/tex]

    but I can't figure out what the angle between the vectors [itex]\gamma[/itex] would be. I imagine a plane formed by the two vectors, and the angle being between the two vectors in that plane. How I describe this mathematically, though, is confusing. Can anyone push me in the right direction?

    EDIT:
    Alright, I took a new approach. I decided to express the difference in Cartesian coordinates, and then convert to spherical coordinates. In doing so, I get:

    [tex]|\vec{x}-\vec{x'}|=(\rho^{2}+\rho'^{2}-2\rho\rho'[\sin\theta\sin\theta'(\cos\phi\cos\phi'+\sin\phi\sin\phi')+\cos\theta\cos\theta'])^{\frac{1}{2}}[/tex]

    which gives me the [itex]\gamma[/itex] I was looking for. It would be nice if there were an easy way to simply that, though...
     
    Last edited: Nov 8, 2008
  2. jcsd
  3. Nov 8, 2008 #2

    gabbagabbahey

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    Usually, you are free to orient your coordinate system as you choose, and so a good choice is often to orient it such that [itex]\vec{x'}[/itex] points in the z-direction and hence [itex]\theta'=0[/itex] and you get a much simpler expression:

    [tex]|\vec{x}-\vec{x'}|=\rho^{2}+(\rho ')^{2}-2\rho\rho ' cos \theta[/tex]
     
  4. Nov 8, 2008 #3
    That's useful a point.

    In the larger problem that this is for, a rigid body (described by f(x')) is rotating around a particular axis with angular velocity [itex]\omega[/itex] - so I am restricted to making the z-axis parallel to [itex]\vec{\omega}[/itex]. Hence, the most general expression is required.
     
  5. Nov 8, 2008 #4

    gabbagabbahey

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    In that case, I think the most you can do to simplify is to use the trig Identity [itex]\cos (\phi -\phi ')=\cos \phi \cos \phi '+\sin \phi \sin\phi '[/itex]
     
  6. Nov 8, 2008 #5
    Right! Working it out right now. Thanks.
     
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