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Solenoidal and conservative fields 
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#1
Mar2912, 07:53 AM

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I had doubts whether to post this here or in in the physics section but I did here because I'm more interested in a purely mathematical view in this case.
I understand a solenoidal vector field implies the existence of another vector field, of wich it is the curl: [tex]S=\nabla X A[/tex] because the divergence of the curl of any vector field is zero. But what if the vector field is conservative instead? I guess in this case it is not necessarly implied the existence of a vector potential. How about in the case of a laplacian vector field, that is both conservative and solenoidal, does it imply the existence of a vector potential? 


#2
Mar3012, 03:34 AM

P: 3,063

no takers?
Is this question more appropriate maybe for the physics section? 


#3
Apr512, 10:56 PM

P: 1,309

What do you mean by solenoidal?
Cartan's exterior calculus and De Rham cohomology sheds a lot of light on these things. Locally, every closed form is exact. Closed means its exterior derivative is zero. Exact means it is the exterior derivative of something. Taking the gradient, curl, divergence can be done by taking some sort of exterior derivative. In Euclidean space, the exterior derivative of a function is the gradient, the exterior derivative of the dual of a vector field the hodge dual of the curl, and the exterior derivative of the Hodge dual of the dual to the vector field is divergence. If you believe that this is giving you some sort of cohomology, if you know what that is, it's obvious that, locally, a potential should exist if your curl is zero, and a vector potential should exist if your divergence is zero because cohomology is a topological invariant and a manifold has trivial local topology (this fact is essential what is known as the Poincare Lemma). Globally, these things actually depend on the topology of the space you are living in. A conservative vector field is one that is curlfree. That doesn't tell you anything about a vector potential, just a potential. You can easily have a vector field that is curlfree, but has some divergence. That's basically what happens in electostatics all the time. If there are no timevarying magnetic fields, Maxwell's equations tell you that the electric field is curlfree. But it can easily have divergence. Just put down some charges wherever you like. However, if a vector field has a vector potential, then it must be divergence free because it is the curl of something. So, vanishing divergence is a requirement to have a vector potential. Luckily, one of Maxwell's equations, again, tells you that that's always true of magnetic fields. 


#4
Apr512, 11:05 PM

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Solenoidal and conservative fields
In not so complicated words, working in Euclidean space, a curl free vector field is the gradient of a scalar field and a divergence free vector field is the curl of a vector field.



#5
Apr612, 02:15 AM

P: 3,063

Thanks both for your answers.



#6
Apr612, 02:45 AM

P: 3,063

Also my question referred to vector fields like the magnetic field that seem to be both divergence free and curlfree, 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? 


#7
Apr612, 02:59 AM

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#8
Apr612, 05:07 AM

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P: 2,953

If you have a general manifold on which to work, you have to deal with a somewhat nontrivial extension of what you mean by "curl". I don't want to get too technical, since I'm bound to mess up.
The general result is that the exterior derivative (off of which curl, gradient and divergence are based) satisfies the property that dd=0 (e.g. curl of gradient=0 or divergence of curl =0). The converse, that given df=0 implies that f=dg locally, arises from Poincare's lemma, and that is true in general. If you want to remove the requirement of "locally" and have it hold globally, then your manifold must be contactable (i.e. smoothly deformable to a point  it can't have any holes or w/e). Flatness is not a requirement. In fact, an exterior algebra can be defined on a manifold without defining a connection (and therefore any sense of curvature). 


#9
Apr612, 05:11 AM

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hi homeomorphic!
i assume you don't have the ∂ or Λ symbols? let's translate … ∂ Λ is the exterior derivative (also called the boundary, for an intuitively obvious reason which i can't remember ): it converts a scalar to a 1form (a vector), a 1form to a 2form, and so on ∂ Λ ∂ Λ anything is zero ("the boundary of a boundary is 0") in n dimensions, a kform has ^{n}C_{k} components (so an nform is effectively a scalar, and higher forms do not exist) * is the hodge dual, converting kforms to (nk)forms ** is the identity F is closed if ∂ Λ F = 0. F is exact if F = ∂ Λ G (and since ∂ Λ ∂ Λ G has to be 0, that means that any exact form is obviously closed). Any closed form is exact. in ℝ^{3}, a 1form is a vector, and the dual of a 2form is a 1form (in the dual space), and so is a pesudovector or crossproductvector (a vector in the dual space), and … ∂ Λ f = ∇f = grad(f)in spacetime, a 1form is a 4vector, the dual of a 3form is a (pseudo?)4vector, and a 2form is a new sort of thing with 6 components, such as the electromagnetic field (technically, the faraday 2form, whose dual is the maxwell 2form), and … an electromagnetic field is ∂ Λ (the electromagnetic potential 4vector) … ∂ Λ (φ,A) = (E;B) = (∇φ  ∂A/∂t;∇xA)so ∂ Λ (electromagnetic field) = 0 (this is 2 of maxwell's equations) … ∂ Λ (E;B) = *(∇.B, ∂B/∂t + ∇xE) = 0and *∂ Λ *(electromagnetic field) = a 1form, the 4vector (ρ,J) (this is the other 2) … ∂ Λ *(E;B) = *(∇.E, ∂E/∂t + ∇xB) = *(ρ,J) 


#10
Apr612, 08:17 AM

P: 3,063

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 oneform related by the metric, that in this case would not be trivially (1,1,1) as in the Euclidean case. 


#11
Apr612, 01:58 PM

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How are you generalizing curl to a curved manifold? And are you only considering manifolds of dimension 3 with a connection? I don't know of any "standard" generalizations.



#12
Apr612, 02:07 PM

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I can't find any online or in my books, R^3 is always assumed. 


#13
Apr612, 02:30 PM

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P: 2,953

Well, one way would be to just define curl as the same thing as it was in the Euclidean case, since the exterior derivative is still defined in a general manifold. Only in the 3D case though can you identify a 2form with a 1form (and viceversa) through the Hodge duality operator (and if you have a metric, a 1form can be identified with a vector).
So define [tex]\nabla \times \mathbf{A}\equiv [*(d\mathbf{A}^\flat)]^\sharp[/tex] The gradient is then simply: [tex]\nabla f\equiv (df)^\sharp[/tex] In this case then, we are asking is the following equality true: [tex]0=\nabla \times (\nabla f)=[*(d((df^\sharp)^\flat))]^\sharp=[*(ddf)]^\sharp[/tex] This is true in general by the definition of the exterior derivative (and the fact that the musical isomorphisms are indeed isomorphisms). 


#14
Apr612, 02:44 PM

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PF Gold
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With a riemannian metric g on a 3manifold M, all kinds of natural isomorphisms pop up. For one thing, the tangent and cotangent bundles TM and T*M get identified via the musical isomorphisms. Also, the Hodgestar * identifies 2forms to 1forms and 3forms to smooth functions. Equivalently, this last identification can also be expressed by saying that every 3form can be written as fv_{g}, where v_{g} is the riemannian volume form. The identification is then f <> fv_{g}. To sum up, we have natural identifications
[itex]C^{\infty}(M)[/itex] = [itex]\Omega^0(M)[/itex] [itex]\Gamma (TM)[/itex] <> [itex]\Omega^1(M)[/itex] [itex]\Gamma (TM)[/itex] <> [itex]\Omega^2(M)[/itex] [itex]C^{\infty}(M)[/itex] <> [itex]\Omega^3(M)[/itex] And there is an obvious way to go "differentially" from [itex]\Omega^i(M)[/itex] to [itex]\Omega^{i+1}(M)[/itex], namely the exterior derivative d. So by using the above identifications, you can define grad, curv, and div as the map between functions and vector fields corrersponding to d at the level of differential forms. In particular, when M=R³ and g = δ_{ij}dx^{i}dx^{j} is the standard metric, this construction gives the usual grad, curv, div. Edit: I am saying the same thing as Matterwave but without the nice explicit formulae. 


#15
Apr612, 03:13 PM

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#16
Apr612, 03:26 PM

P: 3,063

I mean that the curl involves the covariant derivative and therefore takes into account not only the covariant oneform but the contravariant vector.
The exterior derivative of the gradient oneform in this case is already zero due to the d^2=0 rule so taking the hodge dual of it here doesn't help us solve the curl in the curved case. 


#17
Apr712, 10:00 AM

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