Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Divergence Free But Not the Curl of Any Vector

  1. Dec 5, 2011 #1
    1. The problem statement, all variables and given/known data
    So this is part of a problem set in which I have to show that a vector field is divergence free but not the curl of any vector field.

    Let[tex]F =\frac{<x,y,z>}{(x^2 + y^2 + z^2)^{3/2}}[/tex]
    Then F is smooth at every point of R3 except the origin, where it is not defined. (This vector field is identical, up to a constant multiple, to the electric field generated by a point charge at the origin.) Let E be the region [tex]1 < x^2 + y^2 + z^2 < 9[/tex] that is, the region between two concentric spheres of radii 1, 3 centered at the origin.

    Let S be the surface [tex]x^2 +y^2 +z^2 = 4,z ≤ a[/tex] where a is a number slightly smaller than 2. (S is the sphere of radius 2 with its top sliced off.) Let C be the boundary of this surface. Show that if F = ∇ × G for some vector field G defined on E, then

    [tex]\int _C G • dr = 4π[/tex] as [tex]a → 2^-[/tex]

    3. The attempt at a solutionSo I have confirmed that ∇ • F = 0. I have also found that the region E is not a simply connected since any sphere contained in E will form the boundary of a region that contains points not in E. Therefore, such a sphere could not be shrunk to a point without leaving E. Establishing these two points were the first two parts of the problem.

    This third part, however, seems much more difficult. First, just a bit of clarification, what does [tex]a → 2^-[/tex] mean? Is it like "as a approaches 2 from the negative direction?"

    I think I probably need to define C in some way. Supposing I let z=a, then

    [tex]x^2 +y^2 = 4-a^2[/tex]Could C then be parametrized as [tex]C = r(t) = <(4-a^2)cost,(4-a^2)sint>[/tex]Anyway, I'm not entirely sure how to begin.
    Last edited: Dec 5, 2011
  2. jcsd
  3. Dec 5, 2011 #2
    Suppose [itex]\vec{F} = \nabla \times \vec{G}[/itex]. But, then, every other [itex]\vec{G} + \nabla \chi[/itex]. If you apply this integral theorem:
    \int{d^3 x \, (\nabla \times \vec{G})} = \oint{d a \, (\hat{n} \times \vec{G})}
    to the surface mentioned, what does the left side equal to? How about the integral:
    \oint{d a \, (\hat{n} \times \nabla \chi)}
    Last edited: Dec 5, 2011
  4. Dec 5, 2011 #3
    Hmm. I don't see an integral theorem. I just see:

  5. Dec 5, 2011 #4
    i edited my response.
  6. Dec 5, 2011 #5
    Okay. I don't think I've ever seen that theorem before. Also, some of the notation is unfamiliar to me. Specifically, what are [itex]\nabla \chi[/itex], [itex]d^3 x[/itex], [itex]da[/itex], and [itex]\hat{n}[/itex]?

    Does this theorem have a name? I think I'll probably have to investigate it more since, as of right now, it looks so peculiar.
  7. Dec 5, 2011 #6
    [itex]\nabla \chi[/itex] - gradient of a scalar function

    [itex]d^3 x[/itex] - 3 dimensional volume element

    [itex]da[/itex] - surface element

    [itex]\hat{n}[/itex] - unit vector perpendicular to the surface
  8. Dec 5, 2011 #7


    User Avatar
    Science Advisor

    That's the three dimensional version of Stoke's theorem.
  9. Dec 5, 2011 #8
    You can derive it starting from Gauss's Law:
    \int{d^3 x \, (\nabla \cdot \vec{A})} = \oint{da \, (\hat{n} \cdot \vec{A})}
    if you consider a vector field:
    \vec{A} = \vec{C} \times \vec{G}
    where [itex]\vec{C}[/itex] is a constant, but otherwise arbitrary vector. Then:
    \nabla \cdot \vec{A} = \nabla \cdot (\vec{C} \times \stackrel{\downarrow}{\vec{G}}) = -\vec{C} \cdot (\nabla \times \vec{G})
    \hat{n} \cdot \vec{A} = \hat{n} \cdot (\vec{C} \times \vec{G}) = -\vec{C} \cdot (\hat{n} \times \vec{G})
    where we used the properties of the nabla operator and the mixed product of three vectors. Substituting these expressions into Gauss's Law and taking [itex]\vec{C}[/itex] in front of the integrals because it is a constant vector, we get:
    \vec{C} \cdot \left( \int{d^3 x \, (\nabla \times \vec{G})} - \oint{da \, (\hat{n} \times \vec{G})} \right) = 0
    Because this has to hold for an arbitrary vector [itex]\vec{C}[/itex], we deduce that the term in the parentheses is identically equal to zero.
  10. Dec 5, 2011 #9
    So I think [tex]\hat{n} = \frac{<2x,2y,2z>}{\sqrt{4x^2 + 4y^2 +4z^2}}[/tex]Next, we have

    [tex]\oint d a[/tex]Okay. So this bit is still a little confusing to me. You say [itex]da[/itex] is the surface element. This looks like it might be written as [itex]\oint 1 d a[/itex], which leads me to believe the result is some property of the surface. (I'm basing my reasoning here off of the fact that, for example, [itex]\int \int_S 1 dS[/itex] is the surface area of S.) What this property might be, however, is still a mystery to me.

    Finally, does [itex]\hat{n} \times \vec{G}[/itex] have some type of identity that I'm not aware of? Since I don't know a definition of G, I'm having a hard time evaluating this.
  11. Dec 5, 2011 #10
    In spherical coordinates:

    da = R^2 \, \sin{\theta} \, d\theta \, d\phi

    where R is the radius of the spherical surface over which you are integrating. The surface is 2-dimensional, so you have two integration variables.

    Did you find some [itex]\vec{G}[/itex] that would satisfy:

    \nabla \times \vec{G} = \frac{\vec{r}}{r^3}
    In spherical coordinates, this would amount to:
    \frac{1}{r \, \sin \theta} \, \left[ \frac{\partial}{\partial \theta} \left( \sin \theta \, G_\phi \right) - \frac{\partial G_\theta}{\partial \phi} \right] = \frac{1}{r^2} \\

    \frac{1}{\sin \theta} \frac{\partial G_r}{\partial \phi} - \frac{\partial}{\partial r} \left( r \, G_\phi \right) = 0 \\

    \frac{\partial}{\partial r} \left( r \, G_\theta \right) - \frac{\partial G_r}{\partial \theta} = 0

    For example, you may take [itex]G_r = 0[/itex]. Then the last two equations would imply that:
    G_\theta = \frac{f_1(\theta, \phi)}{r}, G_\phi = \frac{f_2(\theta, \phi)}{r}
    Inserting this into the first condition gives:
    \frac{\partial}{\partial \theta} \left( \sin \theta \, f_2 \right) - \frac{\partial f_1}{\partial \phi} = \sin \theta
  12. Dec 5, 2011 #11
    So if I'm understanding this, we can say [itex]\vec{F} = curl \vec{G} = <\frac{f_1(\theta, \phi)}{r},\frac{f_2(\theta, \phi)}{r},0> [/itex]?

    You said [itex]d^{3}x[/itex] was the 3-D volume element. Is this ever written as [itex]dV[/itex]? If so, then would I have,

    [tex]\int_0^{2π} \int_0^π \int_0^2 p^2sin{\phi} drd{\phi}dθ <\frac{f_1(\theta, \phi)}{r},\frac{f_2(\theta, \phi)}{r},0>[/tex]
    [tex]= \frac{32π}{3} <\frac{f_1(\theta, \phi)}{r},\frac{f_2(\theta, \phi)}{r},0>[/tex]Hmm. I don't know how this simplifies to [itex]4π[/itex]But looking at the theorem again, I see you only have one integral (and of the right, a closed line integral), whereas I have written three integrals for the volume. This leads me to think I've done something wrong.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook