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Diferentiation and differential operators

  1. Mar 30, 2014 #1
    If the gradient of f is equal to differential of f wrt s: [tex]\vec{\nabla}f=\frac{df}{d\vec{s}}[/tex] so, what is the curl of f and the gradient of f in terms of fractional differentiation?
     
  2. jcsd
  3. Apr 2, 2014 #2

    Matterwave

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    I'm not sure what you mean by "fractional differentiation" (wikipedia has a definition that seems very different than what you're talking about here), but these three operations are all related to the exterior derivative (if that's what you were talking about) [itex]\bf{d}[/itex].

    The gradient
    [tex]\nabla f = (\bf{d} f)^\sharp[/tex]
    The Curl
    [tex]\nabla\times \bf{f}=[\star(\bf{d}\bf{f}^\flat)]^\sharp[/tex]
    The Divergence
    [tex]\nabla\cdot \bf{f}=\star\bf{d}(\star \bf{f}^\flat)[/tex]

    Edit: I have no idea why the f is boldfaced in the gradient formula...and I don't know how to fix it. It should be non-boldfaced as it's a functinon.
     
    Last edited: Apr 2, 2014
  4. Apr 2, 2014 #3

    micromass

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    Hmm, this is weird. It looks like a bug, thanks for bringing this to our attention.
     
  5. Apr 3, 2014 #4

    Matterwave

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    Oh I should have explained my notation. The sharp and flat signs mean the musical isomorphisms which, given a metric, maps a one form to a vector and vice versa. The star is the Hodge dual operator which, again being metric dependent, maps a (k)-form to a (n-k) form where n is the dimension of the manifold.
     
  6. Apr 3, 2014 #5
    Fractional differentiation means, for me, express a derivative in the form of a fraction.
     
  7. Apr 3, 2014 #6

    Matterwave

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    Oh, so you mean like...

    $$\nabla\cdot\vec{f}=\frac{\partial f_x}{\partial x}+\frac{\partial f_y}{\partial y}+\frac{\partial f_z}{\partial z}$$

    ?
     
  8. Apr 5, 2014 #7
    Yes and not. Yes because you used fraction, and not because your fraction is a scalar.

    A Hessian of f, for example, can be write like:

    [tex]Hf = \frac{d^2f}{d\vec{r}^2}[/tex]
     
  9. Apr 5, 2014 #8

    Mark44

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    Then your definition is at odds with how the term is already used, which has nothing to do with the derivative appearing in the form of a fraction. See this article on fractional derivatives.
     
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