# Divergence Free But Not the Curl of Any Vector

1. Dec 5, 2011

### TranscendArcu

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$$F =\frac{<x,y,z>}{(x^2 + y^2 + z^2)^{3/2}}$$
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 $$1 < x^2 + y^2 + z^2 < 9$$ that is, the region between two concentric spheres of radii 1, 3 centered at the origin.

Let S be the surface $$x^2 +y^2 +z^2 = 4,z ≤ a$$ 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

$$\int _C G • dr = 4π$$ as $$a → 2^-$$

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 $$a → 2^-$$ 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

$$x^2 +y^2 = 4-a^2$$Could C then be parametrized as $$C = r(t) = <(4-a^2)cost,(4-a^2)sint>$$Anyway, I'm not entirely sure how to begin.

Last edited: Dec 5, 2011
2. Dec 5, 2011

### Dickfore

Suppose $\vec{F} = \nabla \times \vec{G}$. But, then, every other $\vec{G} + \nabla \chi$. 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
3. Dec 5, 2011

### TranscendArcu

Hmm. I don't see an integral theorem. I just see:

$$\int 4. Dec 5, 2011 ### Dickfore i edited my response. 5. Dec 5, 2011 ### TranscendArcu Okay. I don't think I've ever seen that theorem before. Also, some of the notation is unfamiliar to me. Specifically, what are $\nabla \chi$, $d^3 x$, $da$, and $\hat{n}$? Does this theorem have a name? I think I'll probably have to investigate it more since, as of right now, it looks so peculiar. 6. Dec 5, 2011 ### Dickfore $\nabla \chi$ - gradient of a scalar function $d^3 x$ - 3 dimensional volume element $da$ - surface element $\hat{n}$ - unit vector perpendicular to the surface 7. Dec 5, 2011 ### HallsofIvy That's the three dimensional version of Stoke's theorem. 8. Dec 5, 2011 ### Dickfore You can derive it starting from Gauss's Law: [tex] \int{d^3 x \, (\nabla \cdot \vec{A})} = \oint{da \, (\hat{n} \cdot \vec{A})}$$
if you consider a vector field:
$$\vec{A} = \vec{C} \times \vec{G}$$
where $\vec{C}$ is a constant, but otherwise arbitrary vector. Then:
$$\nabla \cdot \vec{A} = \nabla \cdot (\vec{C} \times \stackrel{\downarrow}{\vec{G}}) = -\vec{C} \cdot (\nabla \times \vec{G})$$
and
$$\hat{n} \cdot \vec{A} = \hat{n} \cdot (\vec{C} \times \vec{G}) = -\vec{C} \cdot (\hat{n} \times \vec{G})$$
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 $\vec{C}$ 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 $\vec{C}$, we deduce that the term in the parentheses is identically equal to zero.

9. Dec 5, 2011

### TranscendArcu

So I think $$\hat{n} = \frac{<2x,2y,2z>}{\sqrt{4x^2 + 4y^2 +4z^2}}$$Next, we have

$$\oint d a$$Okay. So this bit is still a little confusing to me. You say $da$ is the surface element. This looks like it might be written as $\oint 1 d a$, 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, $\int \int_S 1 dS$ is the surface area of S.) What this property might be, however, is still a mystery to me.

Finally, does $\hat{n} \times \vec{G}$ 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. Dec 5, 2011

### Dickfore

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 $\vec{G}$ that would satisfy:

$$\nabla \times \vec{G} = \frac{\vec{r}}{r^3}$$
In spherical coordinates, this would amount to:
$$\begin{array}{l} \frac{1}{r \, \sin \theta} \, \left[ \frac{\partial}{\partial \theta} \left( \sin \theta \, G_\phi \right) - \frac{\partial G_\theta}{\partial \phi} \right] = \frac{1}{r^2} \\ \frac{1}{\sin \theta} \frac{\partial G_r}{\partial \phi} - \frac{\partial}{\partial r} \left( r \, G_\phi \right) = 0 \\ \frac{\partial}{\partial r} \left( r \, G_\theta \right) - \frac{\partial G_r}{\partial \theta} = 0 \end{array}$$

For example, you may take $G_r = 0$. 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:
$$\frac{\partial}{\partial \theta} \left( \sin \theta \, f_2 \right) - \frac{\partial f_1}{\partial \phi} = \sin \theta$$

11. Dec 5, 2011

### TranscendArcu

So if I'm understanding this, we can say $\vec{F} = curl \vec{G} = <\frac{f_1(\theta, \phi)}{r},\frac{f_2(\theta, \phi)}{r},0>$?

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

$$\int_0^{2π} \int_0^π \int_0^2 p^2sin{\phi} drd{\phi}dθ <\frac{f_1(\theta, \phi)}{r},\frac{f_2(\theta, \phi)}{r},0>$$
$$= \frac{32π}{3} <\frac{f_1(\theta, \phi)}{r},\frac{f_2(\theta, \phi)}{r},0>$$Hmm. I don't know how this simplifies to $4π$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.