Question about computing Jacobians of transformationsby Boorglar Tags: computing, jacobians, transformations 

#1
Dec1812, 08:20 PM

PF Gold
P: 161

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
Dec1912, 03:54 AM

HW Helper
P: 2,168

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} aU & V & W \\ U & bV & W \\ U & V & cW \end{array} \right =abcbc Uac VabW[/tex] 



#3
Dec1912, 07:25 PM

HW Helper
P: 2,168

^The inverse is right.




#4
Dec2012, 08:11 AM

Sci Advisor
P: 1,716

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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