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Curvilinear basis in spherical polar coordinates

  1. Sep 5, 2014 #1
    1. The problem statement, all variables and given/known data
    As a part of my self study I am trying to find the covariant basis vectors in the spherical polar coordinates. Since I have never done anything like this before I would appreciate if someone could tell me whether I am on the rigth track.


    2. Relevant equations
    [itex]r=\sqrt(x^2+y^2+z^2)[/itex]
    [itex]\theta=\arctan(\frac{y}{x})[/itex]
    [itex]\psi=\arccos(\frac{z}{\sqrt(x^2+y^2+z^2)})[/itex]

    I used the equation; [itex]\overline{\epsilon}^{i}_r = \frac{\nabla x_i}{\left| \nabla x_i \right|}[/itex]

    and also

    [itex]x=r\sin(\theta)\cos(\psi)[/itex]
    [itex]y=r\sin(\theta)\sin(\psi)[/itex]
    [itex]z=r\cos(\theta)[/itex]


    3. The attempt at a solution

    I took the grad of the first three equations and then divided that by the magnitude of the vector. Then I substituted x,y,z with the last equation. I got [itex] \overline{\epsilon}^{i}= (\sin(\theta)\cos(\psi), \sin(\theta)\sin(\psi), \cos(\theta))[/itex]. This seem correct to me? I got lost evaluating the other two vectors, but provided that this is the correct method then I am confident that I will get the correct solution sooner or latter.

    I would also appreciate it if someone could tell me whether there is some easier way to find the covariant basis vectors in spherical polar coordinates?

    Thank you very much for all the help

    P.S. I was also wondering whether the covariant and contravariant basis vectors are the same in the if the metric coefficient is 1? (sorry for the very vague wording)
     
    Last edited: Sep 5, 2014
  2. jcsd
  3. Sep 7, 2014 #2
    The correct way is indeed to calculate the normalized grads of the three equations defining the sets of coordinate planes and then eliminate x, y and z from the results by substituting the expressions for x, y and z as functions of r, θ, ψ. For the two sets you mention to be identical, you first need one of them to consist of mutually perpendicular vectors. In this case, the vectors in the other set will have the same directions as the vectors in the first set, but not necessarily the same moduli. If, in addition, the vectors on one of the sets have all modulus one, so do the vectors on the other set and so the two sets are identical in this case.
     
  4. Sep 9, 2014 #3
    Great thank you for the clarification.
     
    Last edited: Sep 9, 2014
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