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Scalar product in spherical coordinates

  1. Nov 1, 2012 #1
    Hello!

    I seem to have a problem with spherical coordinates (they don't like me sadly) and I will try to explain it here. I need to calculate a scalar product of two vectors [itex]\vec{x},\vec{y}[/itex] from real 3d Euclidean space.
    If we make the standard coordinate change to spherical coordinates we can calculate it just fine in terms of [itex]\left( r, \theta, \phi \right)[/itex].
    However if we compute the metric tensor it depends on [itex]r[/itex] and [itex]\theta[/itex]. So we can't use the expression [itex]\vec{x}.\vec{y} = g_{ij} x^i y^j[/itex] (Einstein summation used). And here is my problem. It seems that our space has transformed from flat to curved.

    Now I think I understand on an intuitive level that this is like defining an atlas for a manifold but I would appreciate a little rigour. Also why one of the ways gives an answer not depending on the point we are calculating our scalar product at while the other depends on it? I suspect that it is just hidden in the bad notation and definition of the first one...

    I guess I haven't written my question very clearly but any answers will be appreciated and I will try to clear things a bit if needed :)
     
    Last edited: Nov 1, 2012
  2. jcsd
  3. Nov 1, 2012 #2
    In Cartesian coordinates, the basis vectors do not change with position. In spherical coordinates, they do, so to take a scalar product of two vectors, you must know what location this is taking place at.

    The expression for the scalar product in terms of the metric--##x \cdot y = g_{ij} x^i y^j##--takes this into account, as the metric components are functions of position.
     
  4. Nov 3, 2012 #3
    Yes. I also said that. However that doesn't make things more clear...
     
  5. Nov 3, 2012 #4

    chiro

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    Do you understand how g_ij is derived by relating tangential vectors of one basis to another?
     
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