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Gradient theorem for time-dependent vector field

  1. May 29, 2013 #1
    Let's say we have some time-independent scalar field [itex]\phi[/itex]. Obviously [tex]\phi\left(\mathbf{q}\right)-\phi\left(\mathbf{p}\right) = \int_{\gamma[\mathbf{p},\,\mathbf{q}]} \nabla\phi(\mathbf{x})\cdot d\mathbf{x}.[/tex]
    This is of course still true if the path [itex]\gamma[/itex] is the trajectory of a particle moving through space. But let's say we have a time-dependent field instead, with [itex]\gamma[/itex] still being the trajectory of the particle. Will
    [tex]\phi(\mathbf{q},t_2)-\phi(\mathbf{p},t_1)=\int_{\gamma[\mathbf{p},\,\mathbf{q},t]} \nabla\phi(\mathbf{x} (t))\cdot d\mathbf{x}?[/tex]
     
  2. jcsd
  3. May 29, 2013 #2
    Consider that the gradient theorem is basically a thinly disguised form of Stokes' Theorem, id est,

    $$\int\limits_{\partial\gamma} \phi = \int\limits_{\gamma} d\phi$$

    So, let's consider evaluating the exterior derivative of ##\phi(\vec{x}(t))##.

    $$d\left(\phi(\vec{x}(t))\right) = \sum_{i=1}^{n}\frac{\partial\phi}{\partial x^i}dx^i = \langle\nabla\phi(\vec{x}(t)), \cdot\rangle$$
    where the angle brackets denote the inner product of phi with something.

    Do you agree with this? There are other ways to do this, but this method seems easiest.

    (I'm fairly sure I did this right, but if someone from the upper math-y places wants to correct me, they should.)

    If this works how I thinks it does,

    $$\int\limits_{\partial\gamma}\phi(\vec{x}(t)) = \phi(\vec{q}(t))-\phi(\vec{p}(t)) = \int\limits_{\gamma} \langle\nabla\phi(\vec{x}(t)), d\vec{x}\rangle$$

    where ##\vec{q}(t)## is the vector q at time t.
     
    Last edited: May 29, 2013
  4. May 30, 2013 #3
    I'm sorry, but you're using a lot of terminology and notation that I'm unfamiliar with. What does [tex]\int_{\partial \gamma}\phi[/tex] mean? What's an exterior derivative?
     
  5. May 30, 2013 #4
    Oh...sorry. That may be out of what you've learned before. My apologies.

    ##\int\limits_{\partial\gamma}\phi## would be the oriented sum of phi at the endpoints of ##\gamma##. Because ##\gamma## is 1-dimensional (it's a curve), its boundary is a finite set of points (namely, its endpoints).

    The exterior derivative is a a generalization of the differential. See here.
     
  6. May 31, 2013 #5

    HallsofIvy

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    The left side is evaluated at two different values of t. What value of t are you using on the right?
     
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