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Divergence and Stoke's Theorems in 2D

  1. Jul 17, 2004 #1
    Could I get a demonstration of why they are the same? I have the two equations which the two theorems reduce to in two dimensions, and it's pretty tantalizing because they are virtually the same, but differ in a nice symmetrical way. But I can't for the life of me show that they are the same (I take that to mean that they state the same fact). Thanks.
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  3. Jul 17, 2004 #2
    Let's face it....it is in fact quite quite difficult to prove the Stoke's theorem in a non-ad-hoc way. I once saw a (non-rigorous) proof in a physics book on electromagnetism but it is messy.

    The general "physical" proof starts by dividing the region of integration (for example, a subset U of R^2) into many small "rectangles". Then a relation between the "flux" (of a vector field F) out of each rectangle and div(F)*(volume of rectangle) is established. Then one sums over all rectangles to obtain the ad-hoc result.

    The mathematical proof is neater, though it requires knowledge about integration over forms and some differential geometry. The fundamental result is [tex]\int_{\partial M} \omega = \int_{M} d\omega [/tex], where M is an oriented manifold and the result holds with some conditions on the form [tex] \omega [/tex]. You may want to try this http://mathworld.wolfram.com/StokesTheorem.html.

    Probably I'm messing things up for you....I think the best way to see why the theorems are true is to consult a good book, though it may require some heavy investment (i.e. time). Hope this helps.
  4. Jul 18, 2004 #3
    Without fancy language showing how general Stokes thm is the simplest answer to your question is that cosine (0) = 1.

    The angle whose cosine we need is that between the normal to the surface and the third coordinate. But these are identical in two dimesions for an orientable surface.
  5. Jul 18, 2004 #4
    Is what i wrote clear or do you need me to write the equations?
  6. Jul 18, 2004 #5
    Uhhh...I thought about your reply...are trying to explain why the surface integral of the curl is just a regular double integral over area in 2D? I got that bit. I have that:

    [tex] \oint_{C} F_y\,dx - F_x\,dy = \iint_{A} \left( \frac{ \partial F_{x}}{ \partial x } + \frac{ \partial F_{y}}{ \partial y } \right)\,dx\,dy [/tex]

    which is the divergence theorem in 2D, and that:

    [tex] \iint_A \left( \frac{ \partial F_{y}}{ \partial x} - \frac{ \partial F_{x}}{ \partial y} \right)\,dx\,dy = \oint_{C} F_x\,dx + F_y\,dy [/tex]

    which is Stoke's in 2D. Nice and symmetrical. But why the same? Thanks again for the replies.
    Last edited: Jul 18, 2004
  7. Jul 23, 2004 #6


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    well, this is going to be a bit sloppy, but the point is that when you rotate a vector which is tangent to a path by 90 degrees you get a vector which is perpendicular to the path. One theorem is about the component of your vector field which is tangent to the path and the other one is about a vector field which is perpendicular to the path.

    so all you have to do to get one theorem from the other is rotate the vector field 90 degrees.

    One (of the two possible) rotation of (A,B) by 90 degrees is (B,-A), which accounts for why one theorem has (df/dx,df/dy) in it and the other has (df/dy, -df/dx) in it.

    (At then end you will see I got a minus sign wrong so i chose the wrong rotation i guess.)

    Anyway, if you have a theorem (greens or stokes) that says to compute the tangential component of (A,B) around a path, you just integrate the curl of Adx +Bdy, i.e.

    (dB/dx - dA/dy) over the interior then if I want to calculate instead the normal component of (A,B) over the path, I just make a new vector field whose tangential component equals the other ones normal component.

    I.e. the outward normal component of (A,B) is just (A,B) dotted with an outward pointing vector to the path, i.e. with (-dy/dt,dx/dt). [here's my mistake, this is an inward pointing vector but i will not bother to correct it.]

    So I want to integrate (-Ady/dt + Bdx/dt). For that to be a tangential component, I use instead the vector field (B,-A). So the normal component of (A,B) integrated over the path, equals the tangential component of (B,-A) integrated over the path, which by greens thm equals the double integral of -dA/dx -dB/dy integrated over the interior, which seems to be the "convergence" of (A,B) rather than the divergence.

    Well I fell prey to the hardest thing in mathematics, namely geting the signs right, but i think you can see these theorems are equivalent in 2 dimensions because we can rotate there.

    (In fact the first person to publish these theorems, Maxwell, in his famous book on Electricity and Magnetism, used the opposite sign, as I did, and stated it in terms of "convergence". So at least I am in good company, although Maxwell meant to do this and i didn't. He got minus because instead of vectors he used quaternions, where minuses come in naturally when you square the basic quaternions i,j,k.)

    Does this help explain the mystery?
  8. Jul 23, 2004 #7
    Right, thank you. So Stokes in 2D on a field A is just the divergence, working with another field B such that the normal component of B is the tangential component of A.
  9. Jul 24, 2004 #8


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    that sounds right.
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