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Explanation of Stokes' Theorem

  1. Sep 6, 2004 #1
    Would anyone be willing to explain Stokes' Theorem to me?

    I have managed to grasp the concepts of grad, div, curl, and what the text calls "green's theorem", but I cannot seem to grasp the geometric meaning of "stokes theorem." I've been trying to put the theorem together based on the following explanations from my electrodynamics text:

    [tex]\int_{S}(\nabla\times\nu)\cdot da = \oint_{p}v\cdot dl[/tex]

    "As always, the integral of a derivative (here, a curl) over a region (here, a patch of surface) is equal to the value of the function at the boundary (here, the perimeter of the patch). As in the case of green's theorem, the boundary term is itself an integral--specifically, a closed line integral."

    "Geometrical interpretation: Recall that the curl measures the "twist" of the vectors v; a region of high curl is a whirlpool--if you put a tiny paddlewheel there, it will rotate. Now, the integral of the curl over some surface (or, more precisely, the flux of the curl through the surface) represents the "total amount of swirl", and we can determine that swirl just as well by going around the edge and finding how much the flow is following the boundary. You may find this a rather forced interpretation of Stokes' theorem, but it's a helpful mnemonic, if nothing else."


    Thanks for any help,
    --Brian
     
    Last edited: Sep 6, 2004
  2. jcsd
  3. Sep 6, 2004 #2

    pervect

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    Well, let's look at the line integral of a small box in the x-y plane

    [tex]
    \longleftarrow
    [/tex]
    [tex]
    \downarrow \hspace{.24 in} \uparrow
    [/tex]
    [tex]
    \longrightarrow
    [/tex]

    Let the coordinates of the center of the box be (x,y). Let the lower left corner be at (x-dx/2,y-dy/2), the upper right corner be at (x+dx/2,y+dy/2).

    Let [tex] \vec{v} = [v_x,v_y,v_z] [/tex]. Then we can write the line integral [tex] \oint_{p}v\cdot dl[/tex] as follows, by approximating the vector function as the value at it's midpoint on each side.

    bottom line element: vx(x,y-dy/2) * dx
    top line element: -vx(x,y+dy/2) * dx

    left line element: -vy(x-dx/2,y) * dy
    right line element: vy(x+dx/2,y) * dy

    We can re-write this as

    [tex]-\frac{\partial v_x}{\partial y} dx dy + \frac{\partial v_y}{\partial x} dx dy [/tex]

    But [tex] \frac{\partial v_y}{\partial x} - \frac{\partial v_x}{\partial y} [/tex] is just the z component of the cross product [tex] \nabla \times \vec{v} [/tex].

    So we've established that the line integral over a small box is the surface integral over a small box.

    So we can replace the surface integral [tex] \int_{S}(\nabla\times\nu)\cdot da [/tex] by the sum of the line integrals over all the boxes in the surface.

    _______________
    |_|_|_|_|_|_|_|_|
    |_|_|_|_|_|_|_|_|
    |_|_|_|_|_|_|_|_|
    |_|_|_|_|_|_|_|_|
    |_|_|_|_|_|_|_|_|
    |_|_|_|_|_|_|_|_|
    |_|_|_|_|_|_|_|_|

    Now if you draw little arrows inside ach box, you'll see that most arrows occur once in each direction. Only the arrows around the outermost boundary don't cancel out. Thus the sum of the line integrals is the line integral around the whole box.

    Which is the result we wanted to prove.

    Note: actually, we've only shown Stokes theorem for the special case where we draw the surface on the x-y plane. The result generalizes, however, if you like you can do the same procedure for the xz and yz planes, I am not going to do that personally, though.
     
    Last edited: Sep 6, 2004
  4. Sep 6, 2004 #3
    Pervect,

    Thanks for your reply.

     
  5. Sep 6, 2004 #4
    Griffiths eh? It's a good book.
     
  6. Sep 6, 2004 #5

    pervect

    User Avatar
    Staff Emeritus
    Science Advisor

    OK, consider the the sum of the top and bottom elements

    vx(x,y-dy/2) * dx - vx(x,y+dy/2) * dx

    We can re-write this sum as

    -(vx(x,y+dy/2)-vx(x,y-dy/2))*dx

    We can re-write this as
    [tex]
    -(\frac{\partial v_x(x,y)}{\partial y} dy) dx
    [/tex]

    The justification for this is similar for saying that in the limt as dx->0

    [tex]f(x+dx)-f(x) = f(x+dx/2)-f(x-dx/2) = \frac{df}{dx} dx [/tex]

    but in the more comlicated case above we need to use partial derivatives because we have more than one variable.

    We can re-write the sum of the left and right elements similarly.
     
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