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A question about Jacobian when doing coordinates transformation

  1. Apr 7, 2013 #1
    Hi,

    When I do the following transformation:

    $$
    X_1=x_1+x_2 \\
    X_2=x_2
    $$

    It turns out that the Jacobian ##\partial (X_1,X_2)/\partial (x_1,x_2)## is 1. But we have:

    $$
    dx_1dx_1+dx_1dx_2=d(x_1+x_2)dx_2=dX_1dX_2=|\partial (X_1,X_2)/\partial (x_1,x_2)|dx_1dx_2=dx_1dx_2
    $$

    So we have ##dx_1dx_1=0##. Is this kind of weird? Why does ##(dx_1)^2## have to be 0?

    Thank you!
     
    Last edited: Apr 7, 2013
  2. jcsd
  3. Apr 7, 2013 #2

    jambaugh

    User Avatar
    Science Advisor
    Gold Member

    It represents the differential area for the parallelogram formed by varying by dx1 and then dx1. Since both sides are the same direction, the area is zero.

    At a higher level the differentials are treated as Grassmann variables (like cross product but yielding tensor instead of vector). Then the Jacobian is built into the algebra.
     
  4. Apr 8, 2013 #3

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    Strictly speaking "[itex]dx_1dx_1[/itex]" has no meaning! How did it get in that problem?
     
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