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Is the curl of a field a vector or a vector density?

  1. May 8, 2013 #1
    Hi All,

    I'm trying to figure out how the components of the curl transform upon changing the coordinate system. In general coordinates, the contravariant components of the curl (if applied to the velocity field; then the curl is known as vorticity) are defined as
    [tex]\omega^k = \frac{\epsilon^{ijk}}{\sqrt{g}} \partial_i u _j, [/tex]
    where [itex]\omega^k[/itex] is the k-th component of the vorticity, and [itex]u_i[/itex] represents the velocity field.

    The permutation symbol, [tex]\epsilon^{ijk} = \mathbf{g}^i \cdot \mathbf{g}^j \times \mathbf{g}^k,[/tex] is a tensor density. But if we define the permutation symbol as [itex]\frac{\epsilon^{ijk}}{\sqrt{g}}[/itex], it is an honest tensor. If we use this tensor in the definition of the curl, as done above, shouldn't the vorticity be a true vector as well (rather than a vector density)? I guess I'm not really sure how to show that, however. Simply applying the transformation matrix to the free index doesn't seem give the right answer, because I'd ignore the contracted indices (i.e., the epsilon tensor and the determinant of the metric tensor also need to be transformed, right?). Perhaps someone can point me in the right direction. (I'm confused because I keep reading that the curl is a vector density.)

    There is a related "philosphical" question: I used to think that only true vectors (or tensors/spinors in general) represent physically meaningful quantities. If an object doesn't transform like a tensor, doesn't that mean that it isn't independent of the coordinate bases?
    The local rotation of a fluid is something that exists independently of our choice of a coordinate basis, so shouldn't the vorticity be a true vector? Or am I misinterpreting what a vector density really is. Perhaps the additional factor that apparently appears during the transformation is *required* to ensure that the components indeed represent the same object in every coordinate system?

    Thanks for helping out!

    Johannes
     
  2. jcsd
  3. May 8, 2013 #2

    dextercioby

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    I won't address the <philosophical> issue, just state that the sqrt of det of the metric tensor does just that: in a riemannian geometry converts the vorticity into a vector, rather than pseudovector. The <curl>'s general definition should be ammended with the 1/sqrt.
     
  4. May 11, 2013 #3
    Thanks! I think there's an error in the way I defined the permutation symbol. It should be {+1, -1, 0}. If defined in terms of the basis vectors as done above, we actually wouldn't need the sqrt-factor in the definition of the curl. But my main confusion has been settled...
     
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