Question about computing Jacobians of transformations


by Boorglar
Tags: computing, jacobians, transformations
Boorglar
Boorglar is offline
#1
Dec18-12, 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?
Phys.Org News Partner Science news on Phys.org
NASA's space station Robonaut finally getting legs
Free the seed: OSSI nurtures growing plants without patent barriers
Going nuts? Turkey looks to pistachios to heat new eco-city
lurflurf
lurflurf is offline
#2
Dec19-12, 03:54 AM
HW Helper
P: 2,151
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
lurflurf is offline
#3
Dec19-12, 07:25 PM
HW Helper
P: 2,151
^The inverse is right.

lavinia
lavinia is offline
#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.


Register to reply

Related Discussions
Question on Jacobians Calculus 3
Optical Computing: special issue - Natural Computing, Springer General Physics 1
Optical Computing: special issue - Natural Computing, Springer General Physics 0
Optical Computing: special issue - Natural Computing, Springer General Physics 0
Optical Computing: special issue - Natural Computing, Springer General Physics 0