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Understanding geometrically the equivalence between path independence and exactness

  1. Oct 31, 2011 #1
    Hi friends, sorry that i have posted so many threads recently regarding complex analysis. i am trying hard to understand as much as possible.

    anyway i was wondering if anyone had any good geometric interpretation for the equivalence between a differential being exact and it being path independent. I understand the concepts of exactness and path independence but i have a difficult time relating these two

    A differential Pdx + Qdy is exact if Pdx+Qdy=dh for some function h. Where a differential of h is = (∂h/∂x)dx + (∂h/∂y)dy

    And a function is path independent on a domain D if given two points A and B, we can integrate the function through any path and get the same value.

    I can see the connection between exactness and the fundamental theorem of calculus part I (h is like the antiderivative so we can apply the fundamental theorem) but I guess i have a hard to just equating these two concepts because they seem different. so... if a function isn't exact/path independent then we have to be careful about what curve/path we integrate over. But if path independent we can choose anything. I have a hard time grasping exactly (no pun intended) why that is obvious from the exactness condition.

  2. jcsd
  3. Nov 1, 2011 #2
    Re: Understanding geometrically the equivalence between path independence and exactne

    If a differential is exact, what that is telling you is that it's the gradient of a function (or the "dual" of the gradient vector field, to be more precise). But gradients are curl-free. And Green's theorem says that the integral around a simple closed curve is equal to the integral of the curl over the region inside the curve. But it's curl free, so the curl is zero, so that integral is zero. That gives you path-independence.

    So, that's the outline. It's pretty geometric--provided you understand everything involved there.

    Actually, there's a sort of deep argument that the curl of a gradient is zero involved in making it more geometric, but it can be verified fairly easily by calculation. This is the tip of a big iceberg that has to do with differential forms and de Rham cohomology.
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