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Insights The Pantheon of Derivatives - Part II - Comments

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  1. Mar 17, 2017 #1

    fresh_42

    Staff: Mentor

  2. jcsd
  3. Mar 17, 2017 #2
    the space ##C^\infty_{\mathbb{R}}([0,1])## equipped with the uniform norm is not complete. That is not the point but above it is written that all the spaces are Banach

    The definition of a vector field on a manifold is very bad. It looks like one implicitly assumes that the manifold is embedded into ##\mathbb{R}^m##.

    the gradient of a function is not a vector field, it is an 1-form at least until the manifold is not equipped with a metric tensor

    the operation rotor as well as cross product give pseudo vectors
     
    Last edited: Mar 17, 2017
  4. Mar 17, 2017 #3

    fresh_42

    Staff: Mentor

    ... which (is not claimed and) emphasizes my point to watch out the definitions made in the relating contexts.
    True. I removed ##\mathbb{R}^n##.
    It is a line bundle, and the set ##\{(x,\nabla_xf)\,\vert \,x \in M\}## is a vector field.
    ##\vec{a} \times \vec{b}## is a vector in ##\mathbb{R}^3##.
     
  5. Mar 17, 2017 #4
    In principle I can dig textbooks up and show you corresponding paragraphs but what for?
    It would be great if you ask a specialist in differential geometry whom you trust in , to read your text carefully.
     
  6. Mar 17, 2017 #5

    fresh_42

    Staff: Mentor

    So can I. But what for? Your goal is apparently one which I'm not interested in.
     
  7. Mar 19, 2017 #6

    Stephen Tashi

    User Avatar
    Science Advisor

    Is the "material derivative" of fluid mechanics a special case of a Lie derivative? Or is it yet another kind of derivative?
     
  8. Mar 19, 2017 #7

    fresh_42

    Staff: Mentor

    I'll have it in the fourth part (where I also included an example from Wiki): The material derivative is a special case of a derivative in order to describe the flow of fluids or gases. It is more of a special tool for these currents rather than a special concept of a differentiation process.

    \begin{equation*}
    D_v\Phi = \frac{d_v}{dt} \Phi = \frac{\partial \Phi}{\partial t}+ (v \cdot \nabla)(\Phi) = \left(\,\frac{\partial}{\partial t}+v_x\,\frac{\partial}{\partial x}+v_x\,\frac{\partial}{\partial x}+v_x\,\frac{\partial}{\partial x}\,\right)\,(\Phi)
    \end{equation*}
    where ##v## represents the velocity of the flow at point ##x## and time ##t##. The first summand is the local behavior in time at a fixed point, the second is the convective change of a particle in the flow.

    I would rather call it by its name Euler operator (cp. http://www.math.nyu.edu/faculty/childres/fluidsbook.pdf ; p.8 f.) because the two summands are treated differently: the time dependent part keeps a particle location fixed, whereas the second is to analyze the velocity (flow, integral curve) of a fluid. So strictly speaking I'd say no, since I cannot imagine how to combine this in a single view of a vector field (spacetime aside), will say it's an operator that combines two Lie derivatives. It is also called a total derivative. As I understand it, are Lie derivatives directional derivatives.
     
    Last edited: Mar 19, 2017
  9. Mar 19, 2017 #8
    Good question.

    For simplicity sake let's consider a stationary flow of a fluid with velocity field ##v=v(x),\quad v=(v^1,\ldots,v^3)(x).##
    Let ##g_v^t(x)## stand for corresponding one-parametric group:
    $$\frac{d}{dt}g_v^t(x)=v(g_v^t(x)),\quad g_v^0(x)
    =x.$$
    Then take a smooth manifold ##S\subset \mathbb{R}^3,\quad \mathrm{dim}\,S=k##. The manifold ##S## can be a curve (k=1), a surface (k=2), or a domain of ##\mathbb{R}^3## (k=3) and let ##\omega## stand for a ##k-##form in ##\mathbb{R}^3##.
    Theorem. The following formula holds
    $$\frac{d}{dt}\Big|_{t=0}\int_{g^t_v(S)}\omega =\int_SL_v\omega,$$ here ##L_v## is the Lie derivative.

    This very simple theorem remains true in all dimensions, not only in ##\mathbb{R}^3##.
    For example, let ##\rho(x)## be a density of the fluid and let ##\Omega =\sqrt gdx^1\wedge dx^2\wedge dx^3,\quad g=\mathrm {det}(g_{ij}(x))## be the volume form; ##g_{ij}## is a metric tensor. Then the mass conservation law is written as follows
    $$\int_{g^t_v(D)}\rho\Omega =const,$$ here ##D## is an arbitrary volume and the constant does not depend on time but it surely depends on ##D##.
    Applying the above theorem we get
    $$\frac{d}{dt}\Big|_{t=0}\int_{g^t_v(D)}\rho\Omega=\int_DL_v(\rho\Omega)=0.\qquad (*)$$
    It is not hard to show that ##L_v(\rho\Omega)=\mathrm{div}\,(\rho v) \Omega## and sinse ##D## is an arbitrary volume, formula (*) gives the standard continuity equation: ##\mathrm{div}\,(\rho v)=0.##
    In the same way Helmholtz's theorems https://en.wikipedia.org/wiki/Helmholtz's_theorems and many other useful things from Hamiltonian dynamics follow.

    (I have dropped math details such as smoothness of ##v##, finite measure of ##S,D## etc)
     
    Last edited: Mar 19, 2017
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