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Laplacian = tr(H)

  1. Feb 19, 2010 #1
    Stupid thing I noticed today:

    [tex]\nabla^2 U=tr(H(U))[/tex]

    Or, in other words, the Laplacian of a function is just the trace of its Hessian matrix. Whoop-de-frickin do, right? Is this useful knowledge or should I forget it immediately?


    N!
     
  2. jcsd
  3. Feb 19, 2010 #2

    Ben Niehoff

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    This is not true in curvilinear coordinates.
     
  4. Feb 19, 2010 #3
    But then isn't the 'Hessian' a tensor with co/contravariant components? is the trace even defined for tensors like that? And how is a differential operator like [tex]\nabla^2[/tex] even defined in a nonlinear coordinate system? (Genuine questions, I'm just starting to learn tensors and differential geometry and whatnot...)
     
  5. Feb 19, 2010 #4

    Ben Niehoff

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  6. Feb 19, 2010 #5
    okay, cool. thanks!!

    would the trace of an arbitrary tensor be [tex]\sum_{i}{A_{ii...i}}[/tex] i.e. summing over the elements of the tensor with identical indicies? For tensors rank>2 'diagonal' is sort of vague, or does 'diagonal' always mean 'elements with the same index'? I guess this doesn't have anything to do with the original question...thanks again Ben!
     
  7. Feb 19, 2010 #6

    Ben Niehoff

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    Tensor traces are taken by contracting any two indices with the metric tensor. The result will be a tensor whose rank has been reduced by 2. For example, the Ricci tensor is the trace of the Riemann tensor on the 1st and 3rd indices:

    [tex]R_{bd} = g^{ac} R_{abcd}[/tex]

    and then the scalar curvature is the trace of the Ricci tensor, or the double trace of the Riemann tensor:

    [tex]R = g^{bd} R_{bd} = g^{ac} g^{bd} R_{abcd}[/tex].

    In flat space, the metric tensor is just the identity matrix, so this reduces to the more familiar definition of the trace.
     
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