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A Coordinate transformation: derivative of spherical coordinate with cartesian coordinate

  1. Mar 23, 2017 #1
    I have the following equations:
    [tex]\left\{ \begin{array}{l}
    x = \sin \theta \cos \varphi \\
    y = \sin \theta \cos \varphi \\
    z = \cos \theta
    \end{array} \right.[/tex]

    Assume [tex]\vec r = (x,y,z)[/tex], which is a 1*3 vector. Obviously, x, y, and z are related to each other. Now I want to calculate [tex]\frac{{\partial \vec r}}{{\partial z}}[/tex], could you please tell me if you have any hint?

    I have Googled the questions a lot with different terms but can't find an answer that I am sure of. Many thanks for your attention!

    Attempt 1: The problem seems related to coordinate transformation between spherical and cartesian coordinates.

    Attempt 2: The problem seems related to "The Cartesian partial derivatives in spherical coordinates" shown here: http://mathworld.wolfram.com/SphericalCoordinates.html.
     
    Last edited: Mar 23, 2017
  2. jcsd
  3. Mar 23, 2017 #2

    PeroK

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    First, you have:

    ##\vec{r} = (x, y, z)##

    And

    ##r = \sqrt{x^2 + y^2 + z^2}##

    ##x, y, z## are independent variables, so are unrelated to each other. But, both ##\vec{r}## and ##r## can be treated as functions of ##x, y, z## and differentiated.
     
  4. Mar 23, 2017 #3
    1, Thank you for reminding me to write [tex]{\vec r}[/tex] as vector, I have made revisions above.
    However, x, y, and z are NOT independent variables, for example [tex]{x^2} + {y^2} + {z^2} = 1[/tex]. If they are independent, the problem is very easy because [tex]\frac{{d{\vec r}}}{{dz}} = \frac{{d(x,y,z)}}{{dz}} = (0,0,1)[/tex]
    2, the physical problem at my hand tells me they are NOT independent.
     
    Last edited: Mar 23, 2017
  5. Mar 23, 2017 #4
    which problem do you really solve? I am afraid the current statement merely does not make sense
     
  6. Mar 23, 2017 #5
    I am trying to solve a ray tracing problem. That is all I can say. Other information may not appropriate on this forum. Could you please tell me which part doesn't make sense? Thank you very much!
     
  7. Mar 23, 2017 #6
    the partial derivative by its definition uses some coordinate of a coordinate system. Coordinate system consists of independent coordinates
     
  8. Mar 23, 2017 #7

    PeroK

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    You effectively have a function of two independent variables, with ##r=1##. But, to get a partial derivative wrt ##z## you must be explicit about what the other variable is. For example, you could have:

    ##\vec{r} = (x, \sqrt{1-x^2-z^2}, z)## or ##\vec{r} = (\sqrt{1-y^2-z^2}, y, z)## or ##\vec{r} = (\sin \theta \cos \phi, \sin \theta \sin \phi, \cos \theta)## where ##z = \cos{\theta}##
     
  9. Mar 23, 2017 #8
    Thanks indeed! You clear up my mind!
     
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