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Transformation matrix derivation problem

  1. Sep 1, 2009 #1
    1. The problem statement, all variables and given/known data

    As part of an assignment, I need to derive a transformation matrix to convert a vector in cartesian coordinates to spherical coordinates.

    2. Relevant equations

    What I've got so far is:

    For an arbitrary vector V,

    [tex]
    \textbf{V}=\left[\begin {array}{ccc}V_{x}&V_{y}&V_{z}\end{array}\right]\left[\begin{array}{c}\textbf{i}\\\textbf{j}\\\textbf{k}\end{array}\right] = \left[\begin{array}{ccc}V_{R}&V_{\theta}&V_{\phi}\end{array}\right]\left[\begin{array}{c}\textbf{e}_{R}\\\textbf{e}_{\theta}\\\textbf{e}_{\phi}\end{array}\right]
    [/tex]

    Knowing that VR is the component of V in the eR direction, then:

    [tex]
    V_{R}=aV_{x}+bV_{y}+cV_{z}
    [/tex]

    where a, b and c are some transformation coefficients to be derived from geometry.

    Similarly,

    [tex]
    V_{\theta}=dV_{x}+eV_{y}+fV_{z}
    [/tex]
    and
    [tex]
    V_{\phi}=gV_{x}+hV_{y}+iV_{z}
    [/tex]

    From this, a transformation matrix T exists such that:

    [tex]
    \left[\begin{array}{ccc}V_{x}&V_{y}&V_{z}\end{array}\right]\textit{T}=\left[\begin{array}{ccc}aV_{x}+bV_{y}+cV_{z}&dV_{x}+eV_{y}+fV_{z}&gV_{x}+hV_{y}+iV_{z}\end{array}\right]=\left[\begin{array}{ccc}V_{R}&V_{\theta}&V_{\phi}\end{array}\right]
    [/tex]

    3. The attempt at a solution

    My problem is finding this matrix T.
    It's obvious from inspection that:

    [tex]
    T=\left[\begin{array}{ccc}a&d&g\\b&e&h\\c&f&i\end{array}\right]
    [/tex]

    But, I'm not sure that just stating 'it's obvious that' will be sufficient.
    I'm wondering if and how the below equation can be rearranged and solved to give T.

    [tex]
    \left[\begin{array}{ccc}V_{x}&V_{y}&V_{z}\end{array}\right]\textit{T}=\left[\begin{array}{ccc}V_{R}&V_{\theta}&V_{\phi}\end{array}\right]
    [/tex]

    After this point, finding the coefficients shouldn't be a problem.

    Thank you
     
  2. jcsd
  3. Sep 2, 2009 #2

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    Welcome to PF!

    Hi Fryderyk! Welcome to PF! :smile:

    (have a theta: θ and a phi: φ :wink:)
    Yes, you're really just stating the obvious …

    Finding T and finding a b c etc are the same thing.

    Try it this way … you have two sets of axes, with angles between them, and you first need to find the angles …

    do the two-dimensional case first (it's easier! :wink:) …

    er and eθ are at angle θ to i and j, so the matrix is … ? :smile:
    :rofl: :rofl:
     
  4. Sep 4, 2009 #3
    hi, im new to PF and i have a very similer problem to this. I am also having trouble finding the T matrix. so far i hav found:
    R= (x^2+y^2+z^2)^1/2
    theta= sin-1(y/(x^2+y^2)^1/2)
    phi =??

    hope some one can help...
    thanks
     
  5. Sep 4, 2009 #4

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    Welcome to PF!

    Hi demha! Welcome to PF! :smile:

    (have a square-root: √ and try using the X2 tag just above the Reply box :wink:)
    You seem to be trying to find a matrix that transforms "global" coordinates.

    A matrix is linear, and only gives linear transformations, but Cartesian to polar vectors isn't linear.

    The matrix T isn't intended to convert vectors which start at the origin … it's only for converting "local" vectors at a particular point (x,y,z) = (r,θ,φ) … so you're converting from i j and k at that point to er eθ and eφ at that point.

    For that, you don't need √ or sin-1, you just use the angles. :wink:

    As I said to Fryderyk, try the two-dimensional case first. :smile:
     
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