Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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


    User Avatar
    Science Advisor
    Gold Member

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


    User Avatar
    Science Advisor
    Gold Member

    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


    User Avatar
    Science Advisor
    Gold Member

    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


    Staff: Mentor

    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.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook