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Metric Tensor in Spherical Coordinates

  1. Jun 28, 2014 #1
    I recently derived a matrix which I believe to be the metric tensor in spherical polar coordinates in 3-D. Here were the components of the tensor that I derived. I will show my work afterwards:

    g11 = sin2(ø) + cos2(θ)

    g12 = -rsin(θ)cos(θ)

    g13 = rsin(ø)cos(ø)

    g21 = -rsin(θ)cos(θ)

    g22 = r2sin2(ø) + r2sin2(θ)

    g23 = 0

    g31 = rsin(ø)cos(ø)

    g32 = 0

    g33 = r2cos2(ø)

    The above is what I derived, but when I tried to verify to see if my answer was correct by checking various websites, I did not see any site have what I derived.

    Here is my work:

    The axes were:

    x1 = r

    x2 = θ

    x3 = ø

    The vector that I differentiated was:

    <rcos(θ)sin(ø) , rsin(θ)sin(ø) , rcos(θ)>

    I then differentiated the vector with respect to the various axes in order to derive my tangential basis vectors.

    Here were my basis vectors:

    er = <cos(θ)sin(ø) , sin(θ)sin(ø), cos(θ)>

    eθ = <-rsin(θ)sin(ø), rcos(θ)sin(ø) , -rsin(θ)>

    eø = <rcos(θ)cos(ø), rsin(θ)cos(ø) , 0>

    Finally, I did the dot product with these basis vectors to derive the components of my metric tensor.

    That is how I got what I derived, but I don't see any confirmation of this online.

    Can anyone please either verify if I am right with this metric tensor or tell me where I went wrong?
     
  2. jcsd
  3. Jun 28, 2014 #2

    Bill_K

    User Avatar
    Science Advisor

    The transformation to spherical polar coordinates is

    x = r sin(θ) cos(ø)
    y = r sin(θ) sin(ø)
    z = r cos(θ)

    r = <r sin(θ) cos(ø), r sin(θ) sin(ø), r cos(θ)>
     
  4. Jun 28, 2014 #3
    So I made a careless mistake it seems. Thank you Bill K.
     
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