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Solenoidal and conservative fields

  1. Mar 29, 2012 #1
    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. jcsd
  3. Mar 30, 2012 #2
    no takers?
    Is this question more appropriate maybe for the physics section?
  4. Apr 5, 2012 #3
    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 curl-free. That doesn't tell you anything about a vector potential, just a potential. You can easily have a vector field that is curl-free, but has some divergence. That's basically what happens in electostatics all the time. If there are no time-varying magnetic fields, Maxwell's equations tell you that the electric field is curl-free. 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.
  5. Apr 5, 2012 #4


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    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.
  6. Apr 6, 2012 #5
    Thanks both for your answers.
  7. Apr 6, 2012 #6
    Just divergence-free

    Yes, but is it a sufficient condition too?
    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?
  8. Apr 6, 2012 #7
    Good that you use the qualifier "in Euclidean space"( I have a question about this in another subforum that got no answers). The "gradient of a scalar field has vanishing curl" rule is not necessarily the case if the space is not flat, right?
  9. Apr 6, 2012 #8


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    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).
  10. Apr 6, 2012 #9


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    hi homeomorphic! :smile:

    i assume you don't have the ∂ or Λ symbols? :wink:

    let's translate …
    Λ is the wedge product (or exterior product)

    ∂ Λ is the exterior derivative (also called the boundary, for an intuitively obvious reason which i can't remember :rolleyes:): it converts a scalar to a 1-form (a vector), a 1-form to a 2-form, and so on

    ∂ Λ ∂ Λ anything is zero ("the boundary of a boundary is 0")

    in n dimensions, a k-form has nCk components (so an n-form is effectively a scalar, and higher forms do not exist)

    * is the hodge dual, converting k-forms to (n-k)-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 ℝ3, a 1-form is a vector, and the dual of a 2-form is a 1-form (in the dual space), and so is a pesudovector or cross-product-vector (a vector in the dual space), and
    ∂ Λ f = f = grad(f)
    *∂ Λ A = x A = curl(A)
    *(A Λ B) = A x B
    *(∂ Λ *A) = .A = div(A)​

    in space-time, a 1-form is a 4-vector, the dual of a 3-form is a (pseudo?)4-vector, and a 2-form is a new sort of thing with 6 components, such as the electromagnetic field (technically, the faraday 2-form, whose dual is the maxwell 2-form), and

    an electromagnetic field is ∂ Λ (the electromagnetic potential 4-vector) …
    ∂ 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) = 0​

    and *∂ Λ *(electromagnetic field) = a 1-form, the 4-vector (ρ,J) (this is the other 2) …
    ∂ Λ *(E;B) = *(.E, -∂E/∂t + xB) = *(ρ,J)​
    the ∂ Λ stuff works in any space, but it doesn't necessarily translate nicely into ∇ stuff :redface:
    Last edited by a moderator: Apr 8, 2012
  11. Apr 6, 2012 #10
    Yes, I'm aware of the topological requirements about the space being contractible and simply-connected. And also that in the exterior algebra-differential forms context flatness would not be a requirement. But as tiny tim pointed out the covariant derivative doesn't translate well into the exterior derivative and in the case a curved connection is needed (in the presence of curvature of the space) the "exterior covariant derivative" allows D^2 different than zero.
    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.
    Last edited: Apr 6, 2012
  12. Apr 6, 2012 #11


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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.
  13. Apr 6, 2012 #12
    Well, that is precisely what I'm asking help here for. :smile:
    I can't find any online or in my books, R^3 is always assumed.

    Unlike divergence, curl doesn't generalize well to dimension other than three,(maybe that is why neither you nor me know any "standard" generalization) so I'm only considering those for the moment.
  14. Apr 6, 2012 #13


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    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 3-D case though can you identify a 2-form with a 1-form (and vice-versa) through the Hodge duality operator (and if you have a metric, a 1-form 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).
  15. Apr 6, 2012 #14


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    With a riemannian metric g on a 3-manifold 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 Hodge-star * identifies 2-forms to 1-forms and 3-forms to smooth functions. Equivalently, this last identification can also be expressed by saying that every 3-form can be written as fvg, where vg is the riemannian volume form. The identification is then f <--> fvg. 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 = δijdxidxj 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.
  16. Apr 6, 2012 #15
    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).
  17. Apr 6, 2012 #16
    I mean that the curl involves the covariant derivative and therefore takes into account not only the covariant one-form but the contravariant vector.
    The exterior derivative of the gradient one-form 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.
  18. Apr 7, 2012 #17
    A minor correction, in terms of dif. forms the divergence is "*d*", and the curl is "*d". With "d" being the exterior derivative.
  19. Apr 7, 2012 #18


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    Thanks :smile:

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

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

    But my div equation is ok, isn't it?
  20. Apr 7, 2012 #19
    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.
  21. Apr 7, 2012 #20


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    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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