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## solenoidal and conservative fields

 Quote by TrickyDicky A minor correction, in terms of dif. forms the divergence is "*d*", and the curl is "*d". With "d" being the exterior derivative.
Thanks

The curl equation should be *∂ Λ A = x A = curl(A) ?

(I don't know how that extra * got in there )

But my div equation is ok, isn't it?

 Quote by tiny-tim Thanks The curl equation should be *∂ Λ A = ∇ x A = curl(A) ? (I don't know how that extra * got in there ) But my div equation is ok, isn't it?
Yes about the curl, and about the div I have always seen it without the parenthesis but I don't think the way is set changes the result at all.

 Quote by TrickyDicky Yes, I have seen the curl generalized as the hodge dual of the exterior derivative before, however, in a curved manifold the vector field and the one-form are different wrt their covariant derivative unlike the case in R^3 (where the covariant derivative is trivially just the partial derivative without the need for additional connection terms).
This is why I asked you how you were defining curl. The way that I defined it previously is perfectly valid even for a curved 3-manifold since nothing in the definitions depended on curvature. If you want to change the derivatives to co-variant derivatives rather than exterior derivatives, then you have to tell me your definition.

If you want to change it to a covariant exterior derivative, I won't be able to help you much since I myself am not very familiar, at this point, with that operation.

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 Quote by TrickyDicky Yes about the curl, and about the div I have always seen it without the parenthesis but I don't think the way is set changes the result at all.
Thanks.

I've asked one of the mentors to change the curl, to correct it.

 Quote by Matterwave This is why I asked you how you were defining curl. The way that I defined it previously is perfectly valid even for a curved 3-manifold since nothing in the definitions depended on curvature. If you want to change the derivatives to co-variant derivatives rather than exterior derivatives, then you have to tell me your definition.
After a little research online it turns out the reason the curl of the grad is not zero in curved spaces lies at the heart of Riemannian geometry and is a way to get to the Riemann curvature tensor. It goes like thist: the proof that curl (grad f)=0 in R^3 goes like this:
$\nabla$ X$(\nabla f)=0$
Since: $\nabla f=(\frac{∂f}{∂x},\frac{∂f}{∂y},\frac{∂f}{∂z})=F$

$\nabla$X $F=(\frac{∂F_3}{∂y}-\frac{∂F_2}{∂z},\frac{∂F_1}{∂z}-\frac{∂F_3}{∂x},\frac{∂F_2}{∂x}-\frac{∂F_1}{∂y})$
and therefore curl (grad f)=:
$(\frac{∂}{∂y}\frac{∂f}{∂z}-\frac{∂}{∂z}\frac{∂f}{∂y},\frac{∂}{∂z}\frac{∂f}{∂x}-\frac{∂}{∂x}\frac{∂f}{∂z},\frac{∂}{∂x}\frac{∂f}{∂y}-\frac{∂}{∂y}\frac{∂f}{∂x})$
And by the equality of mixed partials everyone knows about from multivariate calculus the result is (0,0,0).
Well according to Riemann's generalization of calculus to manifolds, space is flat if and only if mixed partials are equal, like above. That is the equality of mixed partials implies the Riemann curvature tensor must vanish. And conversely if the Riemann curvature tensor doesn't vanish the mixed partials are not equal and therefore:curl (grad f)≠0
Christoffel formalized this result (due to early death of Riemann) in the form of the well known Christoffel symbols of second class around 1868.

 The Riemann tensor is usually given as the values of the curvature operator: $$R(X,Y)=\nabla_X\nabla_Y-\nabla_Y\nabla_X-\nabla_{[X,Y]}$$ If you want to call that a curl of a gradient, that's fine with me. I still don't see you giving a definition of a curl in a curved 3-space.

 Quote by Matterwave The Riemann tensor is usually given as the values of the curvature operator: $$R(X,Y)=\nabla_X\nabla_Y-\nabla_Y\nabla_X-\nabla_{[X,Y]}$$ If you want to call that a curl of a gradient, that's fine with me. I still don't see you giving a definition of a curl in a curved 3-space.
No, I woudn't call that curl of a gradient. I call it a Riemann tensor math definition.
I said a result of Riemannian geometry is that when the Riemann tensor vanishes in a space (meaning no curvature or flatness), it implies the equality of mixed partials holds, and that is what makes the curl of the grad be zero in R^3.

 Precisely the curvature opearator measures the failure of commutativity of mixed partials. The definition of curl in curved 3-space just would need to take into account all this thru the corresponding christoffel symbols from the specific metric of the specific curved space that were the case when applying the nabla operator to the specific (grad f), I think.
 The equality of mixed partials always holds (as long as there are no holes in the manifold at least). I think what you mean is the equality of mixed covariant derivatives. But if you want to formulate a curl using covariant derivatives, then you have to tell me how because I see no obvious way to do it.

 Yes, but is it a sufficient condition too?
Yes, that's what I was saying about the Poincare Lemma. In Euclidean space, closed forms are always exact, except 0-forms (which is the same thing as saying the De Rham cohomology vanishes, except in dimension 0 it's ℝ). That translates (in ℝ^3) to the fact that if you are divergence free, then you are the curl of something.

 Also my question referred to vector fields like the magnetic field that seem to be both divergence free and curl-free, that seems to require for that field to have both a vector potential and a scalar potential, like magnetic fields indeed have. That makes the magnetic field a Laplacian field(meaning the gradient of a Laplace equation solution), right?
The magnetic field is always divergence free (no magnetic monopoles), but it is only curl-free if there is no current or displacement current. It's only locally in the absence of current that the magnetic field has a scalar potential, but it always has a vector potential. Then, it's Laplacian, as you say.

 What I'm saying is that the proof of the vanishing curl of the gradient of a scalar field rests on the fact that the order that the partials are taken doesnt matter, so that every component of the resulting vector should be zero. But I'd say in the case the space has curvature the order of the partials does matter. One way gives the vector and the other the one-form related by the metric, that in this case would not be trivially (1,1,1) as in the Euclidean case.
Your reasoning is flawed in two ways. The first way is that you insist on not giving a definition of what you mean by curl. The second flaw is that you seem to imply that just because you USED a certain fact in a proof, it is actually necessary to the proof. That itself would require proof and is not true a priori.

If by curl, you mean the ordinary exterior derivative, then yes, the curl of a gradient is zero. It's a matter of definition.

By the way, this definition DOES involve the metric, if you carry it out. You have a vector field. To take the exterior derivative of it, what do you do? You take the dual with respect to the metric, aka, you raise indices, if you like. So, we used the metric. That means we are taking into account the geometry of the manifold. Then, you take the exterior derivative. Then you take the Hodge dual, which also involves the metric because the Hodge dual is the guy that wedges with stuff to give you the volume form that comes from the metric. Then, you take the dual again with respect to the metric.

Until you can give a coherent definition of what you mean by curl, I'm afraid we aren't going to get anywhere.

Edit: Also, we are not talking about the covariant exterior derivative here. On a Riemannian manifold, you still have the ordinary exterior derivative, so you may as well use it.

 Quote by homeomorphic Your reasoning is flawed in two ways. The first way is that you insist on not giving a definition of what you mean by curl. The second flaw is that you seem to imply that just because you USED a certain fact in a proof, it is actually necessary to the proof. That itself would require proof and is not true a priori. If by curl, you mean the ordinary exterior derivative, then yes, the curl of a gradient is zero. It's a matter of definition. Until you can give a coherent definition of what you mean by curl, I'm afraid we aren't going to get anywhere.
Well I tried to make my reasoning clearer in #22, #24 and #25.
The second flaw you mention I don't think that interprets correctly what I was trying to say. But I don't think it deserves further explaining here if you follow what I wrote in #22, #24 and #25.
About the curl definition, no , I don't mean "just" the exterior derivative, I mean the cross product of the $\nabla$ operator (I think that's what a curl is for everybody), in this case acting on a gradient field in a curved 3-space, say a hypersphere for instance.
Is that not possible to do? do you mean the curl is only defined in Euclidean spaces?

 Quote by Matterwave The equality of mixed partials always holds (as long as there are no holes in the manifold at least).
Yes in Euclidean spaces, see below
 Quote by Matterwave I think what you mean is the equality of mixed covariant derivatives. But if you want to formulate a curl using covariant derivatives, then you have to tell me how because I see no obvious way to do it.
Yes, the covariant derivative is the appropriate generalized notion for the partial derivatives in a curved space. See: http://www.math.ucsd.edu/~ctiee/math...rad_n_curl.pdf
in the first page in the note at the bottom.
About using the curl in general manifolds I ask you the same thing I did to homeomorphic, is it an operation not allowed for the curl? Can not the curl be generalized to general 3-manifolds then?

 I think it can be generalized like this (using the semicolon notation): curl of covector A= $$A_{i;j}-A_{j;i}$$ wich only commutes in the R^3 case and for a gradient field F: $$F_{i;j}=\partial_jF_i-\Gamma^k_{ij}F_k$$ Please correct if wrong.
 From wikipedia page on Vector calculus: "More generally, vector calculus can be defined on any 3-dimensional oriented Riemannian manifold, or more generally pseudo-Riemannian manifold. This structure simply means that the tangent space at each point has an inner product (more generally, a symmetric nondegenerate form) and an orientation, or more globally that there is a symmetric nondegenerate metric tensor and an orientation, and works because vector calculus is defined in terms of tangent vectors at each point." So to answer my previous questions the curl is perfectly generalizable to curved spaces. Also to avoid misunderstandings in what I wrote above, when I referred to mixed partials equality (or lack of) in the context of curved spaces I always meant its generalization for general manifolds (covariant derivative).
 Recognitions: Gold Member Homework Help Science Advisor From what I gather, you are unsatisfied with the generalization of grad, curl, div provided by Matterwave in post 13 to general riemannian oriented 3-manifolds. I don't understand what is unsatisfactory to you in this, but it seems to have to do with lack of covariant derivatives. So perhaps it would ease your mind to know that the exterior derivative and the covariant derivative are closely related by the formula $$d\omega(X_0,\ldots,X_p)=\sum_{i=0}^p(-1)^i(\nabla_{X_i}\omega)(X_0,\ldots,\hat{X_i}, \ldots ,X_p)$$ In other words, d is just the (normalized) anysymetrization of the covariant derivative. This actually holds not only for the Levi-Civita connexion of a riemannian structure but any torsion free connexion on TM.

 Quote by quasar987 From what I gather, you are unsatisfied with the generalization of grad, curl, div provided by Matterwave in post 13 to general riemannian oriented 3-manifolds. I don't understand what is unsatisfactory to you in this, but it seems to have to do with lack of covariant derivatives. So perhaps it would ease your mind to know that the exterior derivative and the covariant derivative are closely related by the formula $$d\omega(X_0,\ldots,X_p)=\sum_{i=0}^p(-1)^i(\nabla_{X_i}\omega)(X_0,\ldots,\hat{X_i}, \ldots ,X_p)$$ In other words, d is just the (normalized) anysymetrization of the covariant derivative. This actually holds not only for the Levi-Civita connexion of a riemannian structure but any torsion free connexion on TM.
But I'm not unsatisfied at all, how could I? I agreed with it (just like I agree with what you just posted about exterior derivatives) since I already knew about that generalization. Only problem was that it didn't actually addressed my specific question.
My mind is eased now that I think I found basically what I was looking for.
Finally I found online this very basic result from Riemannian geometry that says that commutativity of the appropriate generalization of mixed partials implies vanishing Riemann tensor, and this I would think that answers my question. My conclusion is that only if the 3-space is Euclidean must we expect the rule "curl of grad=0" to hold. Now do you disagree with this conclusion? If so, where does my logic fail here? Thanks.

 Tags exterior product, maxwell's equations