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Can i do cross product in spherical coordinates as the same way i do in cartesian coordinates?

  1. Sep 21, 2014 #1
    Is A x B = | i j k | also true for Spherical Coordinates?
    | r1 theta1 phi1 |
    | r2 theta2 phi2 |

    Or I have to convert them to Cartesian Coordinates and do the cross product and then convert them back?
     
  2. jcsd
  3. Sep 21, 2014 #2

    kreil

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    Gold Member

    The short answer: just convert to Cartesian, perform the cross product, then convert back. That's probably the easiest way to go in most cases.

    The reason the rules change are because in Cartesian coordinates, the unit vectors are all linear and perpendicular to each other,

    [tex]
    A = a_1 \hat{e}_x + a_2 \hat{e}_y + a_3 \hat{e}_z
    [/tex]

    But in spherical coordinates, just one of the unit vectors is linear ([itex]\hat{e}_r[/itex]) and the other two are spherical ([itex]\hat{e}_{\theta}[/itex] and [itex]\hat{e}_{\phi}[/itex]). Of course the cross product is independent of any coordinate system you choose, but it's considerably more difficult to do it in [itex](r,\theta,\phi)[/itex]. See http://en.wikipedia.org/wiki/Cross_product#Computing_the_cross_product for more information about the basis vectors.

    There is also a short, informative discussion here:
    http://math.stackexchange.com/quest...lculating-dot-and-cross-products-in-spherical
     
    Last edited: Sep 21, 2014
  4. Sep 22, 2014 #3

    pasmith

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    Homework Helper

    Spherical polar coordinates [itex](r, \theta, \phi)[/itex] are an orthogonal right-handed system: [tex]
    \mathbf{e}_r \times \mathbf{e}_\theta = \mathbf{e}_\phi \\
    \mathbf{e}_\theta \times \mathbf{e}_\phi = \mathbf{e}_r \\
    \mathbf{e}_\phi \times \mathbf{e}_r = \mathbf{e}_\theta,[/tex] as are cartesian coordinates (x,y,z):[tex]
    \mathbf{i} \times \mathbf{j} = \mathbf{k} \\
    \mathbf{j} \times \mathbf{k} = \mathbf{i} \\
    \mathbf{k} \times \mathbf{i} = \mathbf{j}
    [/tex] Learning these results and using distributivity of the cross product over vector addition results in better conceptual understanding than using determinants.
     
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