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Curl of gradient

  1. Apr 5, 2012 #1
    Under what circumstances can a conservative field NOT be irrotational?
     
  2. jcsd
  3. Sep 17, 2012 #2
    According to Wikipedia:

    "Conservative vector fields are also irrotational, meaning that (in three-dimensions) they have vanishing curl. In fact, an irrotational vector field is necessarily conservative provided that a certain condition on the geometry of the domain holds: it must be simply connected.
    An irrotational vector field which is also solenoidal is called a Laplacian vector field because it is the gradient of a solution of Laplace's equation."

    We say that the field is irrotational if its curl is zero.

    See more: http://en.wikipedia.org/wiki/Conservative_vector_field

    Good luck!
     
  4. Oct 13, 2012 #3

    Simon Bridge

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    eg.
    there are no conditions in which a conservative field is irrotational.
    Did you have a particular situation in mind?
     
  5. Oct 18, 2012 #4
    TrickyDicky,
    Do you have any situations in your mind for existence of the curl of a gradient or conservative field being rotational? Honestly I am looking forward to hear that..
    (and that's why after waiting for a while I wrote this.. )
     
  6. Feb 21, 2013 #5
    When the vector field is on a surface that is not simply connected. How is this so?
    Have a look at the curl operator in matrix form:

    NumberedEquation4.gif

    If you look closely you can see that the curl operator, which I will refer to from now on as ∇× is a skew-symmetric matrix. It turns out that the exponential map of a skew-symmetric matrix is an orthogonal matrix(its transpose is equal to its inverse). This can be shown by defining the matrix S = e∇× and ST=e-∇×. Now we take the matrix product S×ST = e∇××e-∇× = e(∇×)-(∇×) = e[0] = I3 = e-∇×+∇× = e-∇××e∇× = ST×S = I3 where [0] is the zero matrix(a 3×3 matrix of all zero entries) and I3 is the square identity matrix of dimension 3.

    Now that we have shown that the exponential of ∇× is an orthogonal matrix, there's an even more detailed proof here showing that the exponential of a skew symmetric matrix like the Curl operator is a rotation matrix.:smile:

    So if G is the gradient of a 3 dimensional scalar field and ∇×G = 0, then e∇×G = I3 which is the identity element of SO(3)( the group of rotations about the origin of vectors in Euclidean 3 space R3) which is itself not simply connected. So the map of all simply connected scalar fields in R3 maps to the Kernel{SO(3)} as is not 1-to-1. Whereas scalar fields in
    R3 that are not simply connected map into SO(3)~I3. This map is injective but not surjective. So the curl of the gradient of a scalar field that is not simply connected corresponds to a rotation operator.
     
  7. Feb 21, 2013 #6

    micromass

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    You're mixing things up. The curl of a gradient is always zero. No matter what the domain of the vector field is. The "converse" is not generally true. That is: if the curl of a vector field is 0, then it is the gradient of something. That requires the domain to be simply connected. This last statement is essentially the Poincaré lemma.
     
    Last edited: Feb 21, 2013
  8. Feb 21, 2013 #7

    WannabeNewton

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    It is much more appropriate to talk about conservative co - vector fields NOT vector fields because conservative has line integral definitions for which you need co - vector fields but if you care only about euclidean 3 - space then it's not really a big deal. A conservative vector field corresponds to an exact one form and an irrotational vector field corresponds to a closed one - form in the context of euclidean 3 - space. Every exact form is closed. On the other hand, a closed form is exact on a contractible domain. All these things are MUCH MUCH more elegant and natural in the more general context of differential forms.
     
  9. Feb 21, 2013 #8

    mathwonk

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    there might be a continuity assumption. otherwise, always.
     
  10. Feb 21, 2013 #9

    UGH.....Ya got me! iStand corrected. :frown:

    But I'd like to see the proof of this if you will.
     
  11. Feb 21, 2013 #10

    WannabeNewton

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    [itex]\omega = df[/itex] hence [itex]d\omega = d(df) = d^{2}f = 0[/itex]. See how pretty it is with differential forms =D. Or you know, if you're a masochist, go ahead and work out the curl of a gradient in coordinates xD.
     
  12. Feb 23, 2013 #11
    micromass, upon looking at some old notes of mine attempting to disprove that ∇×∇f = 0
    for any scalar function f mapping R2 into R3 which is continuous and is differentiable up to at least order 2, I found a page which actually proves it to be correct.

    One can write δ2f/δyδx = lim(Δy→0) {(f'(x,y+Δy,z) -f'(x,y,z))/Δy} and the quantity in the brackets can be re-written as lim(Δx→0){ (f(x+Δx,y,z) -f(x,y,z))/Δx}. Substituting we get:

    δ2f/δyδx = lim(ΔyΔx→0) { (f(x+Δx,y+Δy,z) - f(x+Δx,y,z) + f(x,y,z))/ΔyΔx}.

    and

    δ2f/δxδy = lim(ΔxΔy→0) { (f(x+Δx,y+Δy,z) - f(x,y+Δy,z) + f(x,y,z))/ΔxΔy}.

    Since multiplication of real numbers is commutative, ΔyΔx = ΔxΔy and if we subtract the 2 quantities in brackets we now have:

    2f/δyδx -δ2f/δxδy] = lim(ΔxΔy→0){ (f(x,y+Δy,z)-f(x+Δx,y,z))/ΔxΔy}

    By the continuity of f: R2→R3, there is a point a = (x,y) in R2 whose image is f(x,y,z) and any open δ-neighborhood has an ε-neighborhood on G centered at Img(a) in G. So if we consider the points ax =(x+Δx,y) and ay=(x,y+Δy) in R2 and the vector X connecting them, then |X| = √(Δx2+Δy2). Now I claim that
    lim(ΔxΔy→0){ (f(x,y+Δy,z)-f(x+Δx,y,z))/ΔxΔy} = 0. Which implies that for any ε > 0, there exists a δ > 0 such that:

    |(f(x,y+Δy,z)-f(x+Δx,y,z))|/|ΔxΔy| < ε whenever the distance d(ax,ay) < δ. So what is the delta? Well since the vectors from a→ax and a→ay are orthogonal(to see this just set (x,y)=0), if you were to move one of those points towards the other along the X which connects them such that d(ax,ay) < √(Δx2+Δy2) = |X|, then we can choose δ = |X| and place our δ-disk at the midpoint of X between ax and y. So shrinking ε means we shrink the magnitude of Δy and Δx to move the 2 a-points even closer together along X to re-size our δ-disk. And as ΔxΔy→0(along with |X|), then the |Img(X)|→0 and the proof is complete. [δxiδxj]°δf = 0 whenever i≠j.
     
    Last edited: Feb 23, 2013
  13. Feb 23, 2013 #12

    micromass

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    Your proof looks more like a proof for the theorem of Clairaut-Schwarz... http://en.wikipedia.org/wiki/Symmetry_of_second_derivatives
     
  14. Feb 23, 2013 #13
    Well I'm not surprised that someone didn't construct a theorem about mixed partial derivatives(and prove it too). My proof was a specific case of n=3. And I can see that the Clairaut-Schwartz theorem is an induction of this concept to n dimensions. :wink: The commutativity of mixed partials for a continuous function f of differential class C2 that defines a (2-D) surface in Euclidean 3 space is a consequence of continuity as well as the Cauchy-Schwartz inequality used for defining ε-δ neighborhoods.
     
  15. Feb 25, 2013 #14
    Might I add, the fact that the Curl of the Gradient of a scalar function is always zero is a direct consequence of Clairaut's theorem since the theorem states that mixed partials commute.

    But what about a continuous well defined function of differential class C2+ on real projective space? Or more specifically a scalar function f in R3 that defines a non-orientable surface?
     
  16. Feb 25, 2013 #15

    micromass

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    Of course you can calculate partial derivatives on weird manifolds. Then the partial derivatives don't necessarily commute since you can have a weird chart. You can define a Lie-bracket to see how much the partial derivatives commute.

    But curl and gradient are defined completely different on manifolds. They are not longer vector fields but differential forms. The curl and gradient correspond to the exterior derivative and you can show that applying the exterior derivative twice gives 0.
     
  17. Feb 26, 2013 #16
    A conservative vector field for this purpose is simply the gradient of a function that defines a 2-dimensional manifold embedded in R3. So when you speak of manifolds you are speaking of a completely separate topological space whereas what I'm talking about is restricted to (2D)surfaces in Euclidean 3D space. The CURL operator is a differential 2-form on such an object.

    Now consider the scalar function ψ: (x,y,z) → (x/√(x2+y2+z2),y/√(x2+y2+z2),z/√(x2+y2+z2)) = (ψ(x),ψ(z),ψ(z))

    This is a mapping of R3 into Real projective space(of dimension 2). If you take ∇ψ = (δψ(x)/δx, δψ(y)/δy,δψ(z)/δz) and then apply the CURL operator, you will find that ∇×∇ψ ≠ 0 ! So any scalar field f(x,y,z) that is defined by a continuous mapping of the Real Projective Plane will have asymmetric mixed partial derivatives and a non-vanishing CURL. And while I haven't proved it just yet; I'm going to conjecture that:

    Let f is a continuous scalar function of 3 real variables and differential class C2+ that defines a surface in R3. If f is orientable then ∇f is a conservative vector field.
     
    Last edited: Feb 26, 2013
  18. Feb 26, 2013 #17

    micromass

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    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 [itex]F:\mathbb{R}^3\rightarrow \mathbb{R}[/itex], 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

    [tex]\psi(x)=\frac{x}{\sqrt{x^2 + y^2 + z^2}}[/tex]

    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 [itex]f:\mathbb{R}^3\rightarrow \mathbb{R}[/itex]. You seem to be taking a gradient for something whose codomain is not [itex]\mathbb{R}[/itex]. I don't know how that is defined.
     
  19. Feb 26, 2013 #18

    lavinia

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    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[itex]_{x}[/itex] - u[itex]_{y}[/itex]

    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.
     
    Last edited: Feb 26, 2013
  20. Feb 26, 2013 #19

    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:


    [tex]\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}}) [/tex]

    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.
     
    Last edited: Feb 26, 2013
  21. Feb 27, 2013 #20

    micromass

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    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.
     
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