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Question about computing Jacobians of transformations

by Boorglar
Tags: computing, jacobians, transformations
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Boorglar
#1
Dec18-12, 08:20 PM
P: 179
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?
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lurflurf
#2
Dec19-12, 03:54 AM
HW Helper
P: 2,264
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]
lurflurf
#3
Dec19-12, 07:25 PM
HW Helper
P: 2,264
^The inverse is right.

lavinia
#4
Dec20-12, 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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