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Jacobian for kronecker delta

  1. Aug 16, 2014 #1
    Dear all,
    I was revising on a bit of tensor calculus, when I stumbled upon this:

    $$\delta^i_j = \frac{\partial y^i}{\partial x^\alpha} \frac{\partial x^\alpha}{\partial y^j}$$

    And the next statement reads,

    "this expression yields:

    $$|\frac{\partial y^i}{\partial x^j}| |\frac{\partial x^\alpha}{\partial y^\beta}|= 1$$, ..............................(1)

    With $ |\frac{\partial y^i}{\partial x^j}|$ being the jacobian for transformation $$y^i=y^i(x^1.....x^n)$$, and $$|\frac{\partial x^\alpha}{\partial y^\beta}|$$ being the Jacobian of the INVERSE transformation."

    My question is, how do you get eq. 1 from the Kronecker delta, as they are merely jacobians of coordinate transformatios, and being inverse of each other, are 1. But how do they follow from $$\delta^i_j$$'s expansion? *(i.e. DERIVE eq. 1 using the expansion for kronecker delta)*? I am most probably making a conceptual error, but this is the first time I have seen such a representation of the kronecker delta. Thanks in advance!!!!
  2. jcsd
  3. Aug 16, 2014 #2
    It's just matrix multiplication. You have the two Jacobian matrices
    [tex] T_{i\alpha}=\frac{\partial y^i}{\partial x^\alpha} ,~~~~
    S_{\alpha j}=\frac{\partial x^\alpha}{\partial y^j} . [/tex]
    for the coordinate transformation and its inverse. When you multiply them together and use the first formula in your post, you get the matrix
    (TS)_{ij}=\frac{\partial y^i}{\partial x^\alpha}\frac{\partial x^\alpha}{\partial x^j}=\delta^i_j
    so [itex]TS=\mathrm{id} [/itex]. By multiplicativity of the determinant the result now follows.
  4. Aug 16, 2014 #3


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    Recall the equality ##\det(AB)=\det(A)\det(B)## for ##A,B## square matrices of equal size (as coordinate transformations and their inverses have to be). Equation one will follow directly when you take the determinant of both sides of the top equation. The determinant of the Kronecker delta is obviously 1 since it is the identity matrix.
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