Recognitions:

In the planar case one can think of the curl of a vector field as the divergence of its rotation by -90 degrees.

∇xV = ∇.A(V) where a is the -90 degree rotation of V(x) around x.

For if V = (u,w) then A(v) = (w,-u) so ∇.A(V) = w$_{x}$ - u$_{y}$

The divergence gives the infinitesimal flux density of the vector field away or towards the point. It is a simple limiting arguement to show this.

So one can think of the curl of the vector field as the infinitestimal rotation density either clockwise or counter clockwise around the point.

 Quote by micromass I am very sorry to say that I could make very little sense of your post. I'm not following. If you define your manifold as a level set of $F:\mathbb{R}^3\rightarrow \mathbb{R}$, then the gradient is a normal field on F. I don't get why you call this a conservative vector field at all. The curl operator is not a 2-form. You say that $$\psi(x)=\frac{x}{\sqrt{x^2 + y^2 + z^2}}$$ But the right hand side depends clearly on x, y and z. And the left-hand side depends on x. So the notation is not very legal. In my knowledge, a gradient is only defined for a scalar field, that is for a sufficiently smooth function $f:\mathbb{R}^3\rightarrow \mathbb{R}$. You seem to be taking a gradient for something whose codomain is not $\mathbb{R}$. I don't know how that is defined.

I do realize that my notation for ψ equations was not legal. However, regarding the final paragraph, the Möbius strip is a surface whose codomain is R3 but is homeomorphic to the real projective plane. However, it is not an embedding.

Let's try this again:

$$\psi(x,y,z)=(\frac{x}{\sqrt{x^2 + y^2 + z^2}},\frac{y}{\sqrt{x^2 + y^2 + z^2}},\frac{z}{\sqrt{x^2 + y^2 + z^2}})$$

What am suggesting is that rule that ∇×∇f = 0 for any scalar function with domain R3 does not hold when f defines a non-orientable surface.

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 Quote by Zelyucha I do realize that my notation for ψ equations was not legal. However, regarding the final paragraph, the Möbius strip is a surface whose codomain is R3 but is homeomorphic to the real projective plane. However, it is not an embedding.
What do you mean with the "codomain of a surface"??

Furthermore, do you have any reference that says that the Mobius strip is homeomorphic to the projective plane?? I find this very difficult to believe.

 Let's try this again: $$\psi(x,y,z)=(\frac{x}{\sqrt{x^2 + y^2 + z^2}},\frac{y}{\sqrt{x^2 + y^2 + z^2}},\frac{z}{\sqrt{x^2 + y^2 + z^2}})$$ What am suggesting is that rule that ∇×∇f = 0 for any scalar function with domain R3 does not hold when f defines a non-orientable surface.
It's simply not true. You can actually prove that the curl of the gradient is zero for surfaces (provided we express everything as forms). And as of now, you have still not given a valid counterexample to that claim.
 Recognitions: Science Advisor In a conservative force field, the work done on a particle moving between two points is independent of the path. In an open neighborhood such as a ball in Euclidean space, one can show that this independence of path implies that there is a potential function (scalar field) whose gradient is the field. Conversely, the ngradient of a function must be conservative. This follows directly from the Fundamental Theorem of calculus. The total work done against a conservative field on a closed path is zero. Therefore the field can not circulate around a closed path . Otherwise put, its circulation around any closed path is zero. Therefore it's circulation density, its curl, must be zero

 Quote by micromass What do you mean with the "codomain of a surface"?? Furthermore, do you have any reference that says that the Mobius strip is homeomorphic to the projective plane?? I find this very difficult to believe. It's simply not true. You can actually prove that the curl of the gradient is zero for surfaces (provided we express everything as forms). And as of now, you have still not given a valid counterexample to that claim.

I actually worked out the ∇×∇ψ and it does not vanish. For the homeomorphism between the Mobius strip and the (Real)projective plane, look here.

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 Quote by Zelyucha I actually worked out the ∇×∇ψ and it does not vanish. For the homeomorphism between the Mobius strip and the (Real)projective plane, look here.
Your example literally makes no sense because your map doesn't even map to the reals so what does the gradient even mean in that case? Secondly, the real projective plane is compact but the Mobius strip isn't so your claim regarding them being homeomorphic is already false. Your link also never claims they are homeomorphic. I have no idea what is being attempted to be shown here.
 Recognitions: Science Advisor The Mobius strip is not homeomorphic to the real projective plane.You can construct the projective plane from the Mobius band by attaching a disk to its bounding circle. The projective plane has no boundary. The Mobius band's boundary is a circle. The Mobius band can be embedded in R^3. The Projective plane can not. In fact no non-orientable closed surface without boundary can be embedded in 3 space. The fundamental group of the Mobius band is the integers. The fundamental group of the projective plane is Z/2Z. The Mobius band can be given a flat metric. The Projective plane can not.

 Quote by WannabeNewton Your example literally makes no sense because your map doesn't even map to the reals so what does the gradient even mean in that case? Secondly, the real projective plane is compact but the Mobius strip isn't so your claim regarding them being homeomorphic is already false. Your link also never claims they are homeomorphic. I have no idea what is being attempted to be shown here.
$$\psi(x,y,z)=\frac{x+y+z}{\sqrt{x^2 + y^2 + z^2}}$$

This is a scalar function that describes a non-orientable manifold of dimension 2. The non-orientable comes from that fact that:

$$\psi(-x,-y,-z)=\frac{-x-y-z}{\sqrt{(-1)^2( x^2 + y^2 + z^2)}}$$ = $$\frac{-(x+y+z)}{-\sqrt{x^2 + y^2 + z^2}}$$ = $$\frac{x+y+z}{\sqrt{x^2 + y^2 + z^2}}$$

And furthermore, for any real valued scalar λ in R1~{0} it is easy to check that ψ(λx,λy,λz) = ψ(x,y,z). Now I just finished working out the ∇ψ( which consists of fractions with some big ol' cubic polynomials in the numerators). And as it turns out, if you compute the ∇×∇ψ you can see that it does not vanish. That is, ∇×∇ψ≠ {0}(the zero vector in R3). This function does not actually violate Clairaut's theorem(symmetry of partial derivatives) since it's partials are not continuous at the origin(0,0,0).

Part of the reason that this interests me so much is due to Gauss's law for magnetism( ∇°B=0) which follows from the fact that ∇×∇F = 0 by virtue of the symmetry of mixed partials for a scalar function F whose 1st and 2nd derivatives are continuous. In the case of a magnetic field which violates Gauss's law( ∇°B ≠ 0), such as a magnetic monopole, then we have a scalar potential whose gradient has non-zero curl.

Lavinia: You are correct about the topology of Möbius strip.
 Recognitions: Gold Member Science Advisor You made a calculation error somewhere along the road. Just plug it into wolfram. I got 0.

 Quote by WannabeNewton You made a calculation error somewhere along the road. Just plug it into wolfram. I got 0.

I think that Wolfram Alpha may have biffed here.

I plugged the Grad ψ and got the following:

So bear with me and let's compute the first component(ex) of ∇×∇ψ.....

Now since all components of ∇ψ have teh same denominator we can use the quotient rule which is $$(\frac{f}{g})'=\frac{f'g-g'f}{g^2}$$

Let$$r={\sqrt{x^2+y^2+z^2}}$$ and let c = r3.

Now let

a = x2+z2-xy-zy (∇ψy numerator)

b = x2+y2-zy-xz (∇ψz numerator)

So the 1st component of ∇x∇ψ = dy^dz(ψ) =

$$\frac{b'_{y}c -bc'_{y}-a'_{z}c+ac'_{z}}{c^2}$$

Now for a moment let's first focus on terms in the numerator where a and b are differentiated. It is easy to check that $$\frac{δb}{δy}=2y-z$$ and $$\frac{δa}{δz}=2z-y$$

Therefore, (b'yc-a'zc)c-2 = ((2y-z-2z+y)r3)r-6 = $$\frac{(3y-3z)r^3}{r^6} = \frac{3(y-z)}{r^3}$$
Now this expression is only equal to zero when y=z; but everywhere else it is non-zero.

Now for the other 2 terms. Since r is a square root, we use the chain rule to get

ac'z -bc'y = (3r/2)*2z(x2+z2-xy-zy)-(3/2)*2y(x2+y2-zy-xz) =

3r*(z3+zx2-z2y-yx2-y3+zy2+(xyz-xyz)) =

3r*(z3+zx2-z2y-yx2-y3+zy2)

Now when we factor in the denominator we get $$\frac{3(z^3+zx^2-z^2y-yx^2+zy^2-y^3)}{r^5}$$

We can now write the complete 1st component of ∇×∇ψ:

$$(\frac{3(y-z)}{\sqrt{x^2+y^2+z^2}^3}-\frac{3(z^3+zx^2-z^2y-yx^2+zy^2-y^3)}{\sqrt{x^2+y^2+z^2}^5}){\bf e_{x}}$$

So now you can see for yourself that ∇×∇ψ≠(0,0,0).

 Quote by WannabeNewton Your example literally makes no sense because your map doesn't even map to the reals so what does the gradient even mean in that case? Secondly, the real projective plane is compact but the Mobius strip isn't so your claim regarding them being homeomorphic is already false. Your link also never claims they are homeomorphic. I have no idea what is being attempted to be shown here.

FYI: The Möbius strip is compact. However, it is not homeomorphic to the projective plane because the Möbius strip has 2 closed boundary curves whereas RP2 does not.

 Quote by micromass The curl operator is not a 2-form.

It actually is. Look here at the wikipedia article regarding differential forms and if you scroll down to section 4.1 there is the definition of the wedge product for a differential 2-form which is precisely what the CURL operator is for surfaces in R3. So let F be a vector field in R3. ∇×F = [(dx^dy)+(dx^dz)+(dy^dz)]°F. You can write out the CURL in matrix form and see that it is a skew-symmetric 3×3 matrix with Trace=0.

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 Quote by Zelyucha FYI: The Möbius strip is compact.
Depends entirely on how you define the Mobius strip. It makes perfect sense to define it as not containing the boundary. In fact, since you were talking about manifolds, this is the only way to define it. If you add the boundary, then you're not longer a manifold (but rather a manifold with boundary).

 However, it is not homeomorphic to the projective plane because the Möbius strip has 2 closed boundary curves whereas RP2 does not.
Sigh... The Mobius strip only has one boundary curve.

 It actually is. Look here at the wikipedia article regarding differential forms and if you scroll down to section 4.1 there is the definition of the wedge product for a differential 2-form which is precisely what the CURL operator is for surfaces in R3. So let F be a vector field in R3. ∇×F = [(dx^dy)+(dx^dz)+(dy^dz)]°F. You can write out the CURL in matrix form and see that it is a skew-symmetric 3×3 matrix with Trace=0.
The curl is not a 2-form. It is something that acts on one-forms and that yields a 2-form (and actually, the curl is the dual of that, but that's not important here). The curl is not a two-form by itself.

I think you should get a good book on differential geometry and work through that. I highly recommend Lee's "Introduction to Smooth Manifolds". I hope it will clear up many misconceptions.

I'm not seeing that anybody is using this thread to learn. There are only arguments. Thus I'm locking this.

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