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Cross Product in Spherical Coordinates - Getting conflicting oppinions

  1. Mar 3, 2009 #1
    Hey all,
    I really need some clarification here.

    I've seen problems dealing with the Angular Momentum of a particle, working in spherical coordinates. Wolfram says that there is no simple way to perform this and do the determinant, and you will find many people and other websites claiming this. i.e. you must convert to Cartesian, or I've also seen some kind of operator.

    However,
    Some of my own professors have said you MAY do this. There are also E-mag textbooks, etc. that do the cross product in spherical coords!

    In my own meddling, it seems like it works fine, as long as you look at an instant in time, since the directions are not constant, but they are however, all mutually perpendicular to each other. When I do this, it is identical to what I would get in Cartesian.

    Also, if this is allowed, can someone tell me in what order the directions should be in the determinant? (r, theta, phi)
    (phi being in the x,y plane)
     
  2. jcsd
  3. Mar 3, 2009 #2

    clem

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    Science Advisor

    Maybe W and others were referring to Del X something. The cyclic order r, theta, phi is correct, with any one of them going first.
     
  4. Mar 4, 2009 #3
    Hmm ... I did a back-of-the-envelope caliculation and got myself

    L = m r 2 * (d[tex]\theta[/tex]/dt * phi-direction - sin [tex]\theta[/tex] * d [tex]\phi[/tex]/dt * theta-direction )

    Of course, you could convert it into Cartesian, but it's easiest to do it in the coordinate system in which your trajectory is given.
     
  5. Mar 4, 2009 #4
    What you wrote, xlines, is your result from the cross-product?
     
  6. Mar 5, 2009 #5
    Yes, that is [tex]\vec{L}[/tex] = m [tex]\vec{r}[/tex] x [tex]\vec{v}[/tex] in spherical coordinates, angular momentum of a pointlike particle.
     
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