Divergence Free But Not the Curl of Any Vector

  • #1

Homework Statement


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]

The Attempt at a Solution

So 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:

Answers and Replies

  • #2
2,967
5
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:
[tex]
\int{d^3 x \, (\nabla \times \vec{G})} = \oint{d a \, (\hat{n} \times \vec{G})}
[/tex]
to the surface mentioned, what does the left side equal to? How about the integral:
[tex]
\oint{d a \, (\hat{n} \times \nabla \chi)}
[/tex]
 
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  • #3
Hmm. I don't see an integral theorem. I just see:

[tex]
\int
 
  • #4
2,967
5
Hmm. I don't see an integral theorem. I just see:

[tex]
\int
i edited my response.
 
  • #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.
 
  • #6
2,967
5
[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
 
  • #7
HallsofIvy
Science Advisor
Homework Helper
41,833
956
That's the three dimensional version of Stoke's theorem.
 
  • #8
2,967
5
You can derive it starting from Gauss's Law:
[tex]
\int{d^3 x \, (\nabla \cdot \vec{A})} = \oint{da \, (\hat{n} \cdot \vec{A})}
[/tex]
if you consider a vector field:
[tex]
\vec{A} = \vec{C} \times \vec{G}
[/tex]
where [itex]\vec{C}[/itex] is a constant, but otherwise arbitrary vector. Then:
[tex]
\nabla \cdot \vec{A} = \nabla \cdot (\vec{C} \times \stackrel{\downarrow}{\vec{G}}) = -\vec{C} \cdot (\nabla \times \vec{G})
[/tex]
and
[tex]
\hat{n} \cdot \vec{A} = \hat{n} \cdot (\vec{C} \times \vec{G}) = -\vec{C} \cdot (\hat{n} \times \vec{G})
[/tex]
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:
[tex]
\vec{C} \cdot \left( \int{d^3 x \, (\nabla \times \vec{G})} - \oint{da \, (\hat{n} \times \vec{G})} \right) = 0
[/tex]
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.
 
  • #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.
 
  • #10
2,967
5
In spherical coordinates:

[tex]
da = R^2 \, \sin{\theta} \, d\theta \, d\phi
[/tex]

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:

[tex]
\nabla \times \vec{G} = \frac{\vec{r}}{r^3}
[/tex]
In spherical coordinates, this would amount to:
[tex]
\begin{array}{l}
\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
\end{array}
[/tex]

For example, you may take [itex]G_r = 0[/itex]. Then the last two equations would imply that:
[tex]
G_\theta = \frac{f_1(\theta, \phi)}{r}, G_\phi = \frac{f_2(\theta, \phi)}{r}
[/tex]
Inserting this into the first condition gives:
[tex]
\frac{\partial}{\partial \theta} \left( \sin \theta \, f_2 \right) - \frac{\partial f_1}{\partial \phi} = \sin \theta
[/tex]
 
  • #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.
 

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