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Math Help - Please help in transforming rectangle coordinates to spherical coordinates.

  1. #1
    Junior Member
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    Please help in transforming rectangle coordinates to spherical coordinates.

    I am doing this as a practice of transforming  \vec B \;X\; \vec C = \vac A \; using two approach by a) performing the cross product first and then transform from rectangle to spherical coordinates and b) Transform both  \vec B \hbox { and }\vec C \; to spherical coordinates first before performing the cross product. I notice that I don’t get the same answer. I checked carefully and find nothing wrong that I can tell. In b), it will give  \vec A = 0 because being a position vector, the cross product is zero. Below is my work, please tell me what did I do wrong:


    I start out in rectangle coordinates with letting:

     \vec B_{(x,y,z)} = (2,4, 2\sqrt{(\frac 5 3)}) \;\;\hbox { and }\;\; \vec C = (4,2, 2\sqrt{(\frac 5 3)})


    a) Performing cross product first and transform:

     \vec A \;=\; \vec B \;X\; \vec C \;=\; \left| \begin{array}{ccc} \hat x & \hat y & \hat z \\ 2 & 4 & 2\sqrt{(\frac 5 3)} \\ 2 & 4 & 2\sqrt{(\frac 5 3)}\end{array}\right| \;=\; \hat x 4\sqrt{\frac 5 3} \;+\; \hat y 4\sqrt{\frac 5 3} -\hat z 12


     |\vec A_{(x,y,z)}| = \sqrt { x_A^2 +y_A^2 +z_A^2} = \sqrt{ 16\frac 5 3 + 16\frac 5 3 + 144} =14.04754

     \phi = tan^{-1} (\frac y x) = 45^o \;\hbox { and }\;\theta = cos ^{-1}(\frac { z}{|\vec A|}) = cos ^{-1}( \frac {-12}{14.04754})= 148.676^o

    \Rightarrow \; cos \phi = sin\phi = 0.7071 \;\;\hbox { and }\;\; cos \theta = -0.85424 \;\;\hbox { and }\;\; sin \theta = 0.5198



    Given equation of the  \hat {\theta} component:

    A_{\theta} \;=\; A_x cos \theta cos \phi \;+\; A_y cos \theta sin\phi \;-\; A_z sin \theta

    \Rightarrow A_{\theta} = 4\sqrt{\frac 5 3} X (-0.85824) X 0.7071 + 4\sqrt{\frac 5 3} X (-0.85824) X 0.7071 +12 X (0.5198) = -6.23843 -6.2376 = 0

    Same as  A_{\phi} \;=\;0




    b) Transform both  \vec B \;&\; \vec C to spherical coordinates first before performing the cross product:

    Using the same formula above, both  \vec B_R = \vec C_R = \hat R 5.163746 with different  \theta \;&\; \phi .

     \Rightarrow \;\vec B = \hat R B_R \;\;\hbox { and } \;\; \vec C = \hat R C_R


    As you can see, both only has the  \hat R components only as expected from a position vector.

    \vec B \;X\; \vec C \;=\; \left|\begin{array}{ccc}\hat R & \hat \theta & \hat \phi \\B_R & 0 & 0\\ C_R & 0 & 0\end{array}\right| = 0

    This imply you get a total different answer in two different procedures. It is my understanding you can perform the cross product either way but obviously it does not work. Please take a look and see what did I do wrong.

    Thanks
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  2. #2
    Senior Member
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    The problem here is that your \hat R vector is different for the points B and C. In the first case, it points in the direction \vec{B} and in the second it points in direction \vec{C}.

    Can you give a reference for this problem? It doesn't look standard.
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