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Inverse of jacubian

  1. May 1, 2012 #1
    Hi all,

    When there are two variables x, and p related via any differentiable function, we have the following identity:

    [itex]\frac{dx}{dp}\frac{dp}{dx}=1[/itex]

    When it comes to partial derivatives we can NOT say [itex]\frac{\partial x}{ \partial t}\frac{\partial t}{\partial x}=1[/itex] . However I have seen the following identity which seems to be the generalization of the above identity for variables (x,y,z) and (p,q,r):

    \begin{equation}
    \left( \begin{array}{ccc}
    \frac{\partial x}{ \partial p} & \frac{\partial y}{ \partial p} & \frac{\partial z}{ \partial p}\\
    \frac{\partial x}{ \partial q} & \frac{\partial y}{ \partial q} & \frac{\partial z}{ \partial q}\\
    \frac{\partial x}{ \partial r} & \frac{\partial y}{ \partial r} & \frac{\partial z}{ \partial r}\end{array} \right)
    \left( \begin{array}{ccc}
    \frac{\partial p}{ \partial x} & \frac{\partial q}{ \partial x} & \frac{\partial r}{ \partial x}\\
    \frac{\partial p}{ \partial y} & \frac{\partial q}{ \partial y} & \frac{\partial r}{ \partial y}\\
    \frac{\partial p}{ \partial z} & \frac{\partial q}{ \partial z} & \frac{\partial r}{ \partial z}\end{array} \right)=
    \left( \begin{array}{ccc}
    1 & 0 & 0\\
    0 & 1 & 0\\
    0 & 0 & 1\end{array} \right)
    \end{equation}

    Is the above identity true in general? If so, does anyone know of an elegant proof for that? I have two proofs but they I am looking for better ones. One mentions a function f(p,q,r) for the proof which I think it is unnecessary and also confusing, Another proof ( my own) is that we write

    [itex]\frac{\partial p}{\partial p}=1 \rightarrow \frac{\partial p}{\partial x}\frac{\partial x}{\partial p}+\frac{\partial p}{\partial y}\frac{\partial y}{\partial p}+\frac{\partial p}{\partial z}\frac{\partial z}{\partial p}=1[/itex]

    [itex]\frac{\partial p}{\partial q}=0 \rightarrow \frac{\partial p}{\partial x}\frac{\partial x}{\partial q}+\frac{\partial p}{\partial y}\frac{\partial y}{\partial q}+\frac{\partial p}{\partial z}\frac{\partial z}{\partial q}=0[/itex]

    and so on. From nine equations we can get the identity but the proof is insulting!

    I would be grateful if you share a better proof.

    P.S. The left matrix is called the jacubian. Does the other matrix, which in fact is the inverse if jacubian, have a particular name?
     
  2. jcsd
  3. May 1, 2012 #2

    chiro

    User Avatar
    Science Advisor

    Hey Hassan2.

    If the Jacobian is indeed non-zero then the inverse will exist and relate to volume change for the inverse situation just as you have conjectured.

    Are you familiar with tensors at all?
     
  4. May 1, 2012 #3
    Thanks for the reply.

    I'm afraid I'm alien to tensors. A proof based on tensors might be difficult for me to understand.
     
  5. May 1, 2012 #4
    Here is my confusion about the other proof. Say we have
    [itex]f(x,y,z)[/itex]
    then we have
    \begin{equation}
    \left( \begin{array}{ccc}
    \frac{\partial f}{ \partial p} \\
    \frac{\partial f}{ \partial q} \\
    \frac{\partial f}{ \partial r}\end{array} \right)=

    \left( \begin{array}{ccc}
    \frac{\partial x}{ \partial p} & \frac{\partial y}{ \partial p} & \frac{\partial z}{ \partial p}\\
    \frac{\partial x}{ \partial q} & \frac{\partial y}{ \partial q} & \frac{\partial z}{ \partial q}\\
    \frac{\partial x}{ \partial r} & \frac{\partial y}{ \partial r} & \frac{\partial z}{ \partial r}\end{array} \right)
    \left( \begin{array}{ccc}
    \frac{\partial f}{ \partial x} \\
    \frac{\partial f}{ \partial y} \\
    \frac{\partial f}{ \partial z}\end{array} \right)
    \end{equation}


    on the other hand, it says we have
    \begin{equation}
    \left( \begin{array}{ccc}
    \frac{\partial f}{ \partial x} \\
    \frac{\partial f}{ \partial y} \\
    \frac{\partial f}{ \partial z}\end{array} \right)=

    \left( \begin{array}{ccc}
    \frac{\partial p}{ \partial x} & \frac{\partial q}{ \partial x} & \frac{\partial r}{ \partial x}\\
    \frac{\partial p}{ \partial y} & \frac{\partial q}{ \partial y} & \frac{\partial r}{ \partial y}\\
    \frac{\partial p}{ \partial z} & \frac{\partial q}{ \partial z} & \frac{\partial r}{ \partial z}\end{array} \right)
    \left( \begin{array}{ccc}
    \frac{\partial f}{ \partial p} \\
    \frac{\partial f}{ \partial q} \\
    \frac{\partial f}{ \partial r}\end{array} \right)
    \end{equation}

    Although it makes sense but I'm unsure about the notations. I doubt we can replace f(x,y,z) with f(p,q,r).
     
  6. May 2, 2012 #5
    Yes, this identity is true in general, it follows immediately from the chain rule in several variables. The proof of the chain rule in several variables is quite simple, if one does it in the coordinate-free form (this is the way it is done in most analysis, but not calculus, textbooks)

    You can look for example here
    http://www.trillia.com/zakon-analysisII.html
     
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