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I Lie derivative of a differential form

  1. Sep 17, 2016 #1
    Hello,

    I have a maybe unusual question. In a paper, I recently found the equation $$\mathcal{L}_v(v_i dx^i) = (v^j \partial_j v_i + v_j \partial_i v^j) dx^i$$
    Where [itex]v[/itex] denotes velocity, [itex]x[/itex] spatial coordinates and [itex]\mathcal{L}_v[/itex] the Lie derivative with respect to [itex]v[/itex]. Now I'm an undergraduate who understands very little of differential geometry. Besides this, my paper does not require any knowledge of this discipline.

    Is there someone who could help me out with some rough explanation or short derivation of this formula?
     
  2. jcsd
  3. Sep 17, 2016 #2

    fresh_42

    Staff: Mentor

    This is basically the product rule, Leibniz's rule resp. of differentiation: the directional derivative of ##v_i## along ##v## performed in ##x-##coordinates. You may search the web for "Lie derivative product rule" for a proof that fits the terminology you are used to.
    I assume it will be one of the first properties stated in every book that defines the Lie derivative. I first thought it would even be part of the definition before I looked it up to be sure.
     
  4. Sep 17, 2016 #3
    Thanks, I think this has put me on the right track. My current attempt:
    Using the product rule, we get
    $$\mathcal{L}_v(v_i dx^i) = \mathcal{L}_v (v_i) dx^i + v_i \mathcal{L}_v (dx^i)$$
    Interpreting [itex] \mathcal{L}_v [/itex] as a directional derivative the first term equals [itex] v^j \partial_j v_i dx^i [/itex]. For the second one we get [itex] v_i d \mathcal{L}_v (x^i) [/itex], since the Lie derivative is said to commute with the exterior derivative. Furthermore:
    $$ \begin{align*}
    v_i d \mathcal{L}_v (x^i) & = v_i d(v^j \partial_j x^i) \\
    & = v_i d(v^i) \\
    & = v_i \partial_j v^i dx^j
    \end{align*}
    $$
    Relabeling the indices results in $$\mathcal{L}_v(v_i dx^i) = v^j \partial_j v_i dx^i + v_j \partial_i v^j dx^i$$ Please correct me if I'm wrong!
     
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