# Question about computing Jacobians of transformations

by Boorglar
Tags: computing, jacobians, transformations
 P: 158 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?
 HW Helper P: 2,080 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 $$J= \left| \begin{array}{ccc} u_x & u_y & u_z \\ v_x & v_y & v_z \\ w_x & w_y & w_z \end{array} \right| =\frac{1}{R^3} \left| \begin{array}{ccc} 1 - x R_x/R & -x R_y/R & -x R_z/R \\ -y R_x/R & 1 - y R_y/R & -y R_z/R \\ -z R_x/R & -z R_y/R & 1 - z R_z/R \end{array} \right| =\frac{1-(x R_x+y R_y+z R_z)/R}{R^3}$$ The determinant is closely related to $$\left| \begin{array}{ccc} a-U & -V & -W \\ -U & b-V & -W \\ -U & -V & c-W \end{array} \right| =abc-bc U-ac V-abW$$
 HW Helper P: 2,080 ^The inverse is right.
P: 1,668

## Question about computing Jacobians of transformations

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.

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