
#1
Apr512, 06:16 AM

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Under what circumstances can a conservative field NOT be irrotational?




#2
Sep1712, 03:09 PM

P: 325

"Conservative vector fields are also irrotational, meaning that (in threedimensions) 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! 



#3
Oct1312, 08:38 PM

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eg.
there are no conditions in which a conservative field is irrotational. Did you have a particular situation in mind? 



#4
Oct1812, 08:30 AM

P: 196

curl of gradient
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.. ) 



#5
Feb2113, 05:27 PM

P: 25

Have a look at the curl operator in matrix form: If you look closely you can see that the curl operator, which I will refer to from now on as ∇× is a skewsymmetric matrix. It turns out that the exponential map of a skewsymmetric matrix is an orthogonal matrix(its transpose is equal to its inverse). This can be shown by defining the matrix S = e^{∇×} and S^{T}=e^{∇×}. Now we take the matrix product S×S^{T} = e^{∇×}×e^{∇×} = e^{(∇×)(∇×)} = e^{[0]} = I_{3} = e^{∇×+∇×} = e^{∇×}×e^{∇×} = S^{T}×S = I_{3} where [0] is the zero matrix(a 3×3 matrix of all zero entries) and I_{3} 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. So if G is the gradient of a 3 dimensional scalar field and ∇×G = 0, then e^{∇×G} = I_{3} which is the identity element of SO(3)( the group of rotations about the origin of vectors in Euclidean 3 space R^{3}) which is itself not simply connected. So the map of all simply connected scalar fields in R^{3} maps to the Kernel{SO(3)} as is not 1to1. Whereas scalar fields in R^{3} that are not simply connected map into SO(3)~I_{3}. 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. 



#6
Feb2113, 06:04 PM

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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. 



#7
Feb2113, 06:04 PM

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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.




#8
Feb2113, 09:11 PM

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there might be a continuity assumption. otherwise, always.




#9
Feb2113, 10:21 PM

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UGH.....Ya got me! iStand corrected. But I'd like to see the proof of this if you will. 



#10
Feb2113, 10:35 PM

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#11
Feb2313, 12:57 PM

P: 25

micromass, upon looking at some old notes of mine attempting to disprove that ∇×∇f = 0
for any scalar function f mapping R^{2} into R^{3} 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 δ^{2}f/δyδx = lim(Δy→0) {(f'(x,y+Δy,z) f'(x,y,z))/Δy} and the quantity in the brackets can be rewritten as lim(Δx→0){ (f(x+Δx,y,z) f(x,y,z))/Δx}. Substituting we get: δ^{2}f/δyδx = lim(ΔyΔx→0) { (f(x+Δx,y+Δy,z)  f(x+Δx,y,z) + f(x,y,z))/ΔyΔx}. and δ^{2}f/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: [δ^{2}f/δyδx δ^{2}f/δxδy] = lim(ΔxΔy→0){ (f(x,y+Δy,z)f(x+Δx,y,z))/ΔxΔy} By the continuity of f: R^{2}→R^{3}, there is a point a = (x,y) in R^{2} whose image is f(x,y,z) and any open δneighborhood has an εneighborhood on G centered at Img(a) in G. So if we consider the points a_{x} =(x+Δx,y) and a_{y}=(x,y+Δy) in R^{2} and the vector X connecting them, then X = √(Δx^{2}+Δy^{2}). 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(a_{x},a_{y}) < δ. So what is the delta? Well since the vectors from a→a_{x} and a→a_{y} 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(a_{x},a_{y}) < √(Δx^{2}+Δy^{2}) = X, then we can choose δ = X and place our δdisk at the midpoint of X between a_{x} and _{y}. So shrinking ε means we shrink the magnitude of Δy and Δx to move the 2 apoints even closer together along X to resize our δdisk. And as ΔxΔy→0(along with X), then the Img(X)→0 and the proof is complete. [δx_{i}δx_{j}]°δf = 0 whenever i≠j. 



#12
Feb2313, 01:14 PM

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#13
Feb2313, 05:33 PM

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#14
Feb2513, 12:04 AM

P: 25

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 C^{2+} on real projective space? Or more specifically a scalar function f in R^{3} that defines a nonorientable surface? 



#15
Feb2513, 03:49 AM

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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. 



#16
Feb2613, 01:11 PM

P: 25

Now consider the scalar function ψ: (x,y,z) → (x/√(x^{2}+y^{2}+z^{2}),y/√(x^{2}+y^{2}+z^{2}),z/√(x^{2}+y^{2}+z^{2})) = (ψ(x),ψ(z),ψ(z)) This is a mapping of R^{3} 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 nonvanishing 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 C^{2+} that defines a surface in R^{3}. If f is orientable then ∇f is a conservative vector field. 



#17
Feb2613, 06:25 PM

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I am very sorry to say that I could make very little sense of your post.
[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 lefthand side depends on x. So the notation is not very legal. 



#18
Feb2613, 07:20 PM

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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. 


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