Divergence Free But Not the Curl of Any Vector

[TranscendArcu]

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.

LetF =\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^-

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 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^2Could 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.

 

[Dickfore]

Suppose \vec{F} = \nabla \times \vec{G}. But, then, every other \vec{G} + \nabla \chi. If you apply this integral theorem:
<br /> \int{d^3 x \, (\nabla \times \vec{G})} = \oint{d a \, (\hat{n} \times \vec{G})}<br />
to the surface mentioned, what does the left side equal to? How about the integral:
<br /> \oint{d a \, (\hat{n} \times \nabla \chi)}<br />

 

[TranscendArcu]

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

<br /> \int

 

[Dickfore]

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

<br /> \int

<br /> <br /> i edited my response.

 

[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.

 

[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

 

[HallsofIvy]

That's the three dimensional version of Stoke's theorem.

 

[Dickfore]

You can derive it starting from Gauss's Law:
<br /> \int{d^3 x \, (\nabla \cdot \vec{A})} = \oint{da \, (\hat{n} \cdot \vec{A})}<br />
if you consider a vector field:
<br /> \vec{A} = \vec{C} \times \vec{G}<br />
where \vec{C} is a constant, but otherwise arbitrary vector. Then:
<br /> \nabla \cdot \vec{A} = \nabla \cdot (\vec{C} \times \stackrel{\downarrow}{\vec{G}}) = -\vec{C} \cdot (\nabla \times \vec{G})<br />
and
<br /> \hat{n} \cdot \vec{A} = \hat{n} \cdot (\vec{C} \times \vec{G}) = -\vec{C} \cdot (\hat{n} \times \vec{G})<br />
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:
<br /> \vec{C} \cdot \left( \int{d^3 x \, (\nabla \times \vec{G})} - \oint{da \, (\hat{n} \times \vec{G})} \right) = 0<br />
Because this has to hold for an arbitrary vector \vec{C}, we deduce that the term in the parentheses is identically equal to zero.

 

[TranscendArcu]

So I think \hat{n} = \frac{&lt;2x,2y,2z&gt;}{\sqrt{4x^2 + 4y^2 +4z^2}}Next, we have

\oint d aOkay. 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.

 

[Dickfore]

In spherical coordinates:

<br /> da = R^2 \, \sin{\theta} \, d\theta \, d\phi<br />

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:

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

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

 

[TranscendArcu]

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

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 &lt;\frac{f_1(\theta, \phi)}{r},\frac{f_2(\theta, \phi)}{r},0&gt;
= \frac{32π}{3} &lt;\frac{f_1(\theta, \phi)}{r},\frac{f_2(\theta, \phi)}{r},0&gt;Hmm. I don't know how this simplifies to 4πBut 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.