# Question about computing Jacobians of transformations

1. Dec 18, 2012

### Boorglar

Suppose I have the following transformation:

$$u = \frac{x}{x^2+y^2+z^2}$$
$$v = \frac{y}{x^2+y^2+z^2}$$
$$w = \frac{z}{x^2+y^2+z^2}$$

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. Dec 19, 2012

### lurflurf

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

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

The determinant is closely related to

$$\left| \begin{array}{ccc} a-U & -V & -W \\ -U & b-V & -W \\ -U & -V & c-W \end{array} \right| =abc-bc U-ac V-abW$$

Last edited: Dec 19, 2012
3. Dec 19, 2012

### lurflurf

^The inverse is right.

4. Dec 20, 2012

### 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.