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Inverse Jacobian

  1. Jan 22, 2015 #1
    1. The problem statement, all variables and given/known data
    Don't understand why the inverse jacobian has the form that it does.

    2. Relevant equations
    $$ J = \begin{pmatrix} \frac{\partial{x}}{\partial{u}} & \frac{\partial{y}}{\partial{u}} \\ \frac{\partial{x}}{\partial{v}} & \frac{\partial{y}}{\partial{v}} \end{pmatrix} $$

    $$ J^{-1}= \begin{pmatrix} \frac{\partial{u}}{\partial{x}} & \frac{\partial{v}}{\partial{x}} \\ \frac{\partial{u}}{\partial{y}} & \frac{\partial{v}}{\partial{y}} \end{pmatrix} $$

    3. The attempt at a solution
    This flow of logic makes sense to me:
    $$ \begin{pmatrix} \frac{\partial{f}}{\partial{u}} \\ \frac{\partial{f}}{\partial{v}} \end{pmatrix} = J \begin{pmatrix} \frac{\partial{f}}{\partial{x}} \\ \frac{\partial{f}}{\partial{y}} \end{pmatrix} $$
    (due to the equation for change of variables in two dimensions).

    Similarly, looking at the inverse transformation, if we say that there is some matrix ## M ## that represents the inverse of ## J ##, we know:

    $$ \begin{pmatrix} \frac{\partial{f}}{\partial{x}} \\ \frac{\partial{f}}{\partial{y}} \end{pmatrix} =M \begin{pmatrix} \frac{\partial{f}}{\partial{u}} \\ \frac{\partial{f}}{\partial{v}} \end{pmatrix} $$

    Using the equation of the change of variables in 2D it's clear that ##J^{-1} ## should look as shown.

    I'm confused however, because if I apply the general equation for the inverse of a ##2X2 ## matrix to ## J##, we get a matrix, ## j^{-1} ##, like this:

    $$ j^{-1} = \frac{1}{|J|} \begin{pmatrix} \frac{\partial{y}}{\partial{v}} & - \frac{\partial{y}}{\partial{u}} \\ -\frac{\partial{x}}{\partial{v}} & \frac{\partial{x}}{\partial{u}} \end{pmatrix} $$

    But this would then imply that
    $$ \frac{1}{|J|} \frac{\partial{y}}{\partial{v}} = \frac{\partial{u}}{\partial{x}} $$ etc .. .
    Which wouldn't seem to be generally true.

    Similarly, if we multiply ## J ## and ## J^{-1} ##, using the standard rules for matrix multiplication, we get:

    $$ JJ^{-1} = \begin{pmatrix} \frac{\partial{x}}{\partial{u}}\frac{\partial{u}}{\partial{x}}+\frac{\partial{y}}{
    \partial{u}}\frac{\partial{u}}{\partial{y}} & \frac{\partial{x}}{\partial{u}}\frac{\partial{v}}{\partial{x}}+\frac{\partial{y}}{\partial{u}}\frac{\partial{v}}{\partial{y}} \\ \frac{\partial{x}}{\partial{v}}\frac{\partial{u}}{\partial{x}}+\frac{\partial{y}}{\partial{v}}\frac{\partial{u}}{\partial{y}} & \frac{\partial{x}}{\partial{v}}\frac{\partial{v}}{\partial{x}}+\frac{\partial{y}}{\partial{v}}\frac{\partial{v}}{\partial{y}} \end{pmatrix} $$

    If the above ##'J^{-1} ## is a true inverse of ## J ## then clearly: ## JJ^{-1}=I ##, where ## I ## is the identity matrix. But I don't understand how the terms in the above expression equate to the corresponding terms in the identity matrix! :(

    Can somebody please help?

    Thank you!
     
  2. jcsd
  3. Jan 27, 2015 #2
    Thanks for the post! This is an automated courtesy bump. Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post?
     
  4. Jan 28, 2015 #3

    haruspex

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    I think you have some signs wrong in your JJ-1 matrix. After fixing that, how about try it out with a nontrivial transform, e.g. polar/Cartesian?
     
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