Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Question about computing Jacobians of transformations

  1. Dec 18, 2012 #1
    Suppose I have the following transformation:

    [tex]
    u = \frac{x}{x^2+y^2+z^2}
    [/tex]
    [tex]
    v = \frac{y}{x^2+y^2+z^2}
    [/tex]
    [tex]
    w = \frac{z}{x^2+y^2+z^2}
    [/tex]

    Is there a fast way to calculate the determinant jacobian without having to deal with the whole 3x3 determinant?

    I noticed that the inverse transformation is the same (switching x,y,z with u,v,w gives the equality again) but the determinant is not 1, so I don't really know if this can help.

    Or would I really have to do it the long and boring way?
     
  2. jcsd
  3. Dec 19, 2012 #2

    lurflurf

    User Avatar
    Homework Helper

    Your inverse is wrong.

    Sure there is a fast way, consider the slight generalization where the denominator is a function R so your case is R=x^2+y^2+z^2

    then we have

    [tex]J=
    \left|
    \begin{array}{ccc}
    u_x & u_y & u_z \\
    v_x & v_y & v_z \\
    w_x & w_y & w_z
    \end{array} \right|
    =\frac{1}{R^3}
    \left|
    \begin{array}{ccc}
    1 - x R_x/R & -x R_y/R & -x R_z/R \\
    -y R_x/R & 1 - y R_y/R & -y R_z/R \\
    -z R_x/R & -z R_y/R & 1 - z R_z/R
    \end{array} \right|
    =\frac{1-(x R_x+y R_y+z R_z)/R}{R^3}[/tex]

    The determinant is closely related to

    [tex]\left|
    \begin{array}{ccc}
    a-U & -V & -W \\
    -U & b-V & -W \\
    -U & -V & c-W
    \end{array} \right| =abc-bc U-ac V-abW[/tex]
     
    Last edited: Dec 19, 2012
  4. Dec 19, 2012 #3

    lurflurf

    User Avatar
    Homework Helper

    ^The inverse is right.
     
  5. Dec 20, 2012 #4

    lavinia

    User Avatar
    Science Advisor
    Gold Member

    This function just divides each vector by the square of its length. A small cube alligned so that its base is tangent to a sphere centered at the origin will be shrunk in volume by a factor of the sixth power of the radius of the sphere. So the magnitude of the determinant is the reciprocal of the sixth power of the radius of the sphere.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Question about computing Jacobians of transformations
  1. Question on Jacobians (Replies: 3)

Loading...