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Dot product for vectors in spherical coordinates

  1. Jul 31, 2015 #1
    Hi all.

    I'm struggling with taking dot products between vectors in spherical coordinates. I just cannot figure out how to take the dot product between two arbitrary spherical-coordinate vectors ##\bf{v_1}## centered in ##(r_1,\theta_1,\phi_1)## and ##\bf{v_2}## centered in ##(r_2,\theta_2,\phi_2)## without converting them to cartesian coordinates first.

    Could you guys please help me? The main issue is that the basis for ##v_1## and ##v_2## are different so everything becomes super complicated.

    Thanks
     
    Last edited: Jul 31, 2015
  2. jcsd
  3. Jul 31, 2015 #2
    By the way, if this requires advanced mathematics and is very complicated, please just tell me that's the case instead of spending your time writing a long post. I don't really have the time to dwell too long on this issue.
     
  4. Jul 31, 2015 #3

    mathman

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    Your last remark is confusing. What do you mean by "basis for v1 and v2 are different"?
     
  5. Jul 31, 2015 #4

    SteamKing

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    I don't think there is a definition of the dot (scalar) product which involves using vectors expressed in spherical coordinates, at least, not one which makes sense.

    What started you down this dark road in the first place?
     
  6. Jul 31, 2015 #5
    The basis vectors ##\hat{\phi}##, ##\hat{\theta}## depend on the angles ##\phi##,##\theta##. This is what I meant with the basis are different for vectors centered in different points of spherical space. Sorry for the confusion.

    I was derailed while doing an assignment :( . I ended up wanting to calculate ##\vec{L} \cdot \vec{L}## where ##\vec{L} = -i \hbar \vec{r} \times \nabla## is the angular momentum operator in spherical coordinates.
     
  7. Aug 1, 2015 #6
    The main problem is that in spherical coordinates, the local orthonormal basis are not the global coordinate basis, and hence you cannot obtain a 'neat' expression for the dot product using them. You can obtain an expression in terms of them using Cartesian conversions, but the expression is long, and it would be better to simply change coordinates first and then perform the dot product.
     
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