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Gauss's law in 2D

  1. Mar 5, 2008 #1
    1. The problem statement, all variables and given/known data
    Is there an analog to Gauss's theorem (sorry about the title) in 2D.
    The reason is to develop the continuity equation in 2D.


    2. Relevant equations
    In 3D it is.
    [tex]\int_{S}{\bf A}. d{\bf S}=\int_{V}(\nabla . {\bf A})dV[/tex]
    (please forgive my latex. A and dS should be vectors).

    3. The attempt at a solution
    I suppose we may extend the situation to 3d by allowing a little slice "out of the page" (so that our volume looks like a pancake) and just use the theorem as usual, while imposing that A does not have any component "out of the page", then afterwards take the thickness of the slice to go to zero.
     
    Last edited: Mar 5, 2008
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  3. Mar 5, 2008 #2

    Dick

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    Sure. That's just the divergence theorem. Replace S by boundary(V). In fact there is a 1D version. It's called the fundamental theorem of calculus.
     
  4. Mar 5, 2008 #3

    HallsofIvy

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    Gauss' Law is also known as the "divergence theorem". You can get [itex]\vec{A}[/itex] and [itex]\d\vec{S}[/itex] by [ itex ]\vec{A}[ /itex ] and [ itex ]d\vec{A}[ /itex ] without the spaces. In general you can see the latex code by clicking on the LaTex.

    The closest analogue to Gauss' law in 2 dimensions is Stokes Theorem:
    [tex]\int_C \vec{v}\cdot ds= \int\int_S (\del\cdot d\vec{S}[/tex]
    where C is the boundary of the surface S. If S is in the xy-plane, that is Green's Theorem.

    All of those are special cases of the generalized Stoke's theorem:
    [tex]\int_M d\omega= \int_{\partial M} \omega[/tex]
    where [itex]\omega[/itex] is a differential form on the simply connected manifold M, [itex]d\omega[/itex] is the differential of [itex]\omega[/itex], and [itex]\partial M[/itex] is the boundary of M.
     
  5. Mar 5, 2008 #4
    So in 2D it would look like
    [tex]\int_{L}{\bf A}. d{\bf L}=\int_{V}(\nabla . {\bf A})dS[/tex]
    Where S is now the area (taking the place of the volume in the 3D version), and L is the contour enclosing the area (taking the place of the surface in the 3D version), and [tex]\nabla[/tex] is the 2D gradient operator. Is that right?
     
  6. Mar 5, 2008 #5
    Looking good. What I'm getting is that the continuity equation looks like
    [tex]\nabla.(\rho u)+{{\partial \rho}\over{\partial t}}=0[/tex]
    where rho is the density and u is the 2D velocity field.
    i.e. exactly the same as the 3D continuity equation, but now the [tex]\nabla[/tex] is the 2D gradient operator.
    Is all this correct?
     
  7. Mar 5, 2008 #6

    Dick

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    Yes. The continuity equation looks the same in all dimensions.
     
  8. Mar 5, 2008 #7
    Great. Thank you both very much.
     
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