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Homework Help: Can Green's Theorem disagree with itself sometimes?

  1. Jun 29, 2015 #1
    1. The problem statement, all variables and given/known data

    Firstly, I was seeking any clarification on whether I've made any mistakes. Secondly, further insight into Green's Theorem, if my working is all good.

    I've been reading the mathinsight.org on Subtleties about curl: http://mathinsight.org/curl_subtleties

    Regarding the vector field [itex]\vec{F} = \frac{1}{x^2+y^2} \langle -y,x,0 \rangle[/itex] I decided to test Green's Theorem out.

    This page talks about how this vector field has macroscopic circulation but no microscopic circulation.

    It would seem that for this vector field [itex]\oint_C \vec{F}\cdot d\vec{r} \neq \int\int_D (\nabla \times \vec{F})\cdot \hat{k} \ dA[/itex]

    2. Relevant equations

    3. The attempt at a solution

    So [itex]curl \ \vec{F} = \vec{0} [/itex] so [itex]\int\int_D (\vec{0})\cdot \hat{k} \ dA = 0[/itex] regardless of our path C or domain D.

    So let's consider the unit circle as our path

    [itex]C: \ x^2+y^2=1[/itex] for [itex]\oint_C \vec{F} \cdot d\vec{r}[/itex]

    Standard polar form conversion [itex]\vec{r}(t) = \langle cos(t), sin(t), 0 \rangle[/itex]

    So [itex]\oint_C \vec{F} \cdot d\vec{r} = \int_0^{2\pi}\vec{F}(\vec{r}(t)) \cdot \vec{r}'(t) dt = 2\pi[/itex]

    So Green's Theorem doesn't work for all vector fields, even if it's a simple enclosed region...is this due to a discontinuity at (x,y) = (0,0)?
  2. jcsd
  3. Jun 29, 2015 #2


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    The point is that the computation of curl F is only valid away from r = 0, where the vector field is singular. If you include the appropriate delta function at r = 0, you will find that Green's theorem holds.
  4. Jun 29, 2015 #3


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    No one (including Green) has ever claimed that Green's theorem works for all vector fields. A correct statement of Green's theorem is
    "If L and M have continuous partial derivatives inside a simple closed curve, then [tex]\oint_C (Ldx+ Mdy)= \int\int_D\left(\frac{\partial M}{\partial x}- \frac{\partial L}{\partial y}\right) dxdy[/tex].

    Your example does not have "continuous partial derivatives" at the origin. (And if you try to exclude (0, 0) by adding a small circle around it as part of the boundary, the boundary is no longer "simple".)
  5. Jun 29, 2015 #4
    ... :) So what does all that mean?
  6. Jun 29, 2015 #5
    Does this mean a region that does not include the origin will be fine for Green's Theorem?
  7. Jun 29, 2015 #6


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    This seems too restrictive. It should be straight forward to generalise it to arbitrary distributions.
  8. Jun 30, 2015 #7


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    Yes, if you are willing to go to "homotopies", "homologies", and "algebraic topology".
  9. Jun 30, 2015 #8


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    Green's theorem as you know it has been proven when the region ##D## is simple (type I or type II):

    $$\oint_C \vec F \cdot d \vec r = \iint_D Q_x - P_y \space dA$$

    The vector field is given as:

    $$\vec F = \frac{-y \hat i + x \hat j}{x^2 + y^2}$$

    For the given vector field, there is an issue at the origin because the vector field is undefined at the origin and it doesn't have continuous partial derivatives there either. Any attempt to apply Green's theorem directly will fail for any region containing the origin.

    This poses a problem for the general version of Green's theorem, but Green's theorem can be extended to apply to regions that are not simple, i.e the region will have a hole somewhere. We do this by dividing the larger region into two simpler regions where we can apply Green's theorem and superimpose the results.

    So taking the curve as the positively oriented unit circle ##C_1: x^2 + y^2 = 1##, it encloses the region ##D: x^2 + y^2 \leq 1##. This is going to cause a problem because it contains the origin, and as mentioned before, we can't do that for this particular vector field.

    By enclosing the origin in a negatively oriented, smaller circle ##C_2: x^2 + y^2 = r, \space 0 < r < 1##, we can extend Green's theorem to show that:

    $$\oint_C \vec F \cdot d \vec r = 2 \pi$$

    Where ##C = C_1 \cup C_2##.

    We do this by dividing ##D## into two simple regions ##D'## and ##D''## such that ##D = D' \cup D''## and we apply Green's theorem to the two simple regions like so:

    $$\iint_D Q_x - P_y \space dA = \iint_{D'} Q_x - P_y \space dA + \iint_{D''} Q_x - P_y \space dA = \oint_{\partial D'} Pdx + Qdy + \oint_{\partial D''} Pdx + Qdy$$

    The line integrals are along common boundary lines and are opposite in direction, so they cancel and we get:

    $$\oint_{\partial D'} Pdx + Qdy + \oint_{\partial D''} Pdx + Qdy = \oint_{C_1} P dx + Q dy + \oint_{C_2} P dx + Q dy = \oint_C P dx + Q dy$$

    This is Green's theorem as we know it because we have shown:

    $$\iint_D Q_x - P_y \space dA = \oint_C P dx + Q dy$$

    Even though we divided the region ##D## into two simpler regions before applying Green's theorem. This allows you to get around that pesky origin problem.

    So for the problem at hand, we have specifically shown:

    $$\oint_{C_1} \vec F \cdot d \vec r = \oint_{C_2} \vec F \cdot d \vec r = \int_0^{2 \pi} d \theta = 2 \pi$$
  10. Jul 10, 2015 #9
    You're trying to use Green's Theorem without the field being of C1 class, it's not continuously differentiable.
    As the previous poster mentioned you need to use the second form of Green's theorem if you wish to use it at all in this problem, you'll end up integrating by definition regardless, just in a smaller circle
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