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Coördinate transformation

  1. Oct 5, 2003 #1
    Greeting

    A TA has got me very and utterly confused. He won't be avaible for a few days, so I'm asking you guys.

    Consider the transformation to cilindrical coord.

    x-->r.con[the]
    y-->r.sin[the]
    z-->z

    I have the Jabobian (no problems here).
    He then asks the differential da , where a is a vector.
    Enter the first confusion. I know the differential of the transformation (the linear function given by the Jacobian matrix), but what the hell is the differential of a vector?

    My guess is: da =dxx +dyy +dzz .

    I have the unity vectors of the new system.

    The trick is now what is dr in fuction of the new unity vectors?

    Then answer : da =drr +rd[the][the] +dzz

    How the hell is this determined?
     
  2. jcsd
  3. Oct 8, 2003 #2

    Tom Mattson

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    Staff Emeritus
    Science Advisor
    Gold Member

    It's not a whole lot different from the differential of a real-valued function. Surely you've seen them before. In Physics I, you learn that velocity v is related to displacement r by:

    v=dr/dt

    The dr in the derivative is nothing more than the differential of the vector r.

    That is true if a=xx+yy+zz. Is that what a is?
     
  4. Oct 8, 2003 #3
    Yep, that's exactly right. I just got a little confused. The things was part of a introduction to tensors, covariance and contravariance, and between the sea of indices I somewhat lost sight.

    It's all clear now, though. :smile:
     
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