Results 1 to 5 of 5

Math Help - Derivation of Jacobian Determinant

  1. #1
    Newbie
    Joined
    Sep 2009
    Posts
    3

    Derivation of Jacobian Determinant

    Hi,

    I'm having some problems with the derivation of the Jacobian determinant when used to describe co-ordinate transformations. As I understand it, the Jacobian determinant should relate the areas defined by two vectors in both co-ordinate systems. As the vectors are not necessarily perpendicular, the area is calculated using the cross-product, giving:

    \lvert \mathrm{d}x \times \mathrm{d}y \rvert = J \mathrm{d}u\mathrm{d}v

    (Examples of the derivation can be found here and here.)

    My problem is that \mathrm{d}u\mathrm{d}v is not a cross product and so doesn't describe the area of a parallelogram in the u-v co-ordinate system. So, as far as I can see it, one of two things is happening:

    1) du and dv are assumed to be perpendicular, and so the area is just the product of the sides of the rectange, dudv.

    2) du and dv are assumed to be very small, so that the area approximates a rectangle

    So yeah, any ideas? What am I missing?

    Thanks

    James
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Jester's Avatar
    Joined
    Dec 2008
    From
    Conway AR
    Posts
    2,392
    Thanks
    56
    Quote Originally Posted by alt20 View Post
    Hi,

    I'm having some problems with the derivation of the Jacobian determinant when used to describe co-ordinate transformations. As I understand it, the Jacobian determinant should relate the areas defined by two vectors in both co-ordinate systems. As the vectors are not necessarily perpendicular, the area is calculated using the cross-product, giving:

    \lvert \mathrm{d}x \times \mathrm{d}y \rvert = J \mathrm{d}u\mathrm{d}v

    (Examples of the derivation can be found here and here.)

    My problem is that \mathrm{d}u\mathrm{d}v is not a cross product and so doesn't describe the area of a parallelogram in the u-v co-ordinate system. So, as far as I can see it, one of two things is happening:

    1) du and dv are assumed to be perpendicular, and so the area is just the product of the sides of the rectange, dudv.

    2) du and dv are assumed to be very small, so that the area approximates a rectangle

    So yeah, any ideas? What am I missing?

    Thanks

    James
    If x and y are given parametrically, say x = x(u,v), y=y(u,v) then holding v constant and varying u will give a trace in the region from which we can find the tangent vector <x_u,y_u> to this curve. Similarly, hold u constant and varying v gives another trace and another tangent vector <x_v,y_v>. Now a vary small area here is given by the parallogram with these vectors as there sides, i.e.

     \bold{T}_u = <x_u,y_u>du and \bold{T}_v = <x_v,y_v>dv

    and the area of this small parallogram is given by

     <br />
\bold{T}_u \times \bold{T}_v = \left|\begin{array}{cc}x_u&y_u\\x_v&y_v\end{array} \right| dudv<br />
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Sep 2009
    Posts
    3
    So, if you're varying u and v to form a parallelogram in the x-y space, then I guess you're making the assumption that du and dv are perpendicular to each other (or at least will be for small du and dv)?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    Jester's Avatar
    Joined
    Dec 2008
    From
    Conway AR
    Posts
    2,392
    Thanks
    56
    Quote Originally Posted by alt20 View Post
    So, if you're varying u and v to form a parallelogram in the x-y space, then I guess you're making the assumption that du and dv are perpendicular to each other (or at least will be for small du and dv)?
    I'm not assuming that. They may or they may not be perpendicular. It doesn't matter.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Newbie
    Joined
    Sep 2009
    Posts
    3
    Ah. Sorry, I'm obviously being really thick here, but I don't see how dudv can represent an area without the vectors being perpendicular, as only a rectangle has an area defined by length x height? Unless however, I'm not supposed to be thinking of dudv as an area at all, and the cross product just defines the area produced on the x-y surface by small deviations on the u-v surface. But if that's the case, then if you assume u and v are perpendicular, then dudv does become an area...? No?
    Last edited by alt20; September 8th 2009 at 01:51 PM.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Derivation and Jordan derivation
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: May 8th 2011, 10:22 PM
  2. Jacobian determinant question
    Posted in the Calculus Forum
    Replies: 1
    Last Post: October 23rd 2010, 01:32 AM
  3. Jacobian determinant
    Posted in the Calculus Forum
    Replies: 3
    Last Post: May 7th 2010, 07:07 AM
  4. The problem of Jacobian determinant
    Posted in the Calculus Forum
    Replies: 0
    Last Post: May 1st 2009, 05:36 AM
  5. Jacobian
    Posted in the Calculus Forum
    Replies: 4
    Last Post: October 23rd 2007, 10:49 AM

Search Tags


/mathhelpforum @mathhelpforum