Question about computing Jacobians of transformations

[Boorglar]

Suppose I have the following transformation:

<br /> u = \frac{x}{x^2+y^2+z^2}<br />
<br /> v = \frac{y}{x^2+y^2+z^2}<br />
<br /> w = \frac{z}{x^2+y^2+z^2}<br />

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?

 

[lurflurf]

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

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

The determinant is closely related to

\left|<br /> \begin{array}{ccc}<br /> a-U &amp; -V &amp; -W \\<br /> -U &amp; b-V &amp; -W \\<br /> -U &amp; -V &amp; c-W<br /> \end{array} \right| =abc-bc U-ac V-abW

 

[lurflurf]

^The inverse is right.

 

[lavinia]

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.