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Apologies if its the wrong section, encountered this while studying mechanics :|

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- Thread starter dreamLord
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- #1

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Apologies if its the wrong section, encountered this while studying mechanics :|

- #2

Simon Bridge

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What makes you think curl is supposed to be a gradient of something?

Surely grad is the gradient?

Surely grad is the gradient?

- #3

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"If [itex] \displaystyle{\nabla \times \vec{A} = \vec{0}} [/itex] why does it follow that there must be a [itex] \phi(x) [/itex] such as [itex] \displaystyle{\vec{A}=\nabla \phi} [/itex] ? "

- #4

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Do you know the ∇ operator ?

∇ is a sort of notational "trick" that let you think about it as a vector

(the vector (∂/∂x, ∂/∂y, ∂/∂z))

the gradient of a scalar field is simply multiplying ∇ by the scalar

the curl of a vector field is simply the vector cross product of ∇ and this vector field

the cross product of two collinear vectors is always 0.

if you take the curl of something which is a grad of some field ψ, it looks like this:

∇x(∇ψ), but ∇ψ is clearly collinear to ∇, since it is just ∇ 'times' ψ

therefore, ∇x(∇ψ)=0

for the same reason, if ∇xA is 0, then A must be collinear to ∇, which means it can be written as ∇ 'times' some ψ, and therefore A is the gradient of some ψ

- #5

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@Dexter ; yes, that is what I meant, thank you for rephrasing it.

@Oli4 ; I'm a little uncomfortable with that method. How would you link your explanation to say a conservative force (A is conservative in this case, right?)

- #6

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@Oli4 ; I'm a little uncomfortable with that method. How would you link your explanation to say a conservative force (A is conservative in this case, right?)

Hmm, I wouldn't link it I guess.

A is conservative yes, we have shown that since rot(A)=0 then A must be the gradient of some scalar field, this in turns means that the force is conservative because it can be proved (gradient theorem) that therefore the line integral through this gradient is path independent.

(for a force, it means the work done by this force over any path does not depend on the path itself but only of the end points)

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- #8

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- #9

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curl (

zero is a scalar and curl (

curl (

- #10

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curl (v) is never zero.

zero is a scalar and curl (v) is a vector.

curl (v) can be the zero vector (0,0,0). This is still valid vector, but it does not possess a unique direction, any direction will do.

It is very customary to call the zero vector (0,0,0) (or whichever null vector) just zero, just as it is customary to call 1 the identity matrix and so on. unless there is risk of confusion, the context is generally enough

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Dextercioby has already drawn attention to this formally, but his point seems not to have been heard.

The distinction is very important in this question.

- #12

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May I know why the distinction is so important in this question ?

- #13

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May I know why the distinction is so important in this question ?

Of course, but I was trying to determine your maths level. You asked for a physical explanation - to do with forces.

Do you know what a vector field and partial derivatives and the curl itself are? (It's easy to explain if you don't)

- #14

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First let's look at a vector field that is the gradient of a scalar field,

[tex]\vec{V}=-\vec{\nabla} \phi.[/tex]

In cartesian coordinates you have

[tex]V_j=-\partial_j \phi.[/tex]

Suppose now that [itex]V_j[/itex] are continuously differentiable wrt. all three coordinates, then you necessarily have

[tex]\partial_k V_j-\partial_j V_k=-(\partial_k \partial_j-\partial_j \partial_k) \phi=0,[/tex]

because under these assumptions the partial derivatives commute. The hodge dual of this antisymmetric tensor is the curl of the vector field, i.e., we have (coordinate independently!)

[tex]\vec{\nabla} \times \vec{V}=0.[/tex]

Here I use, as usual in the physics literature, the symbol "0" for both scalar and vector quantities. There shouldn't be any trouble with this, for mathematicians somewhat sloppy, notation. It's clear from the context, that here we mean the zero vector of Euclidean vector space. I wouldn't like the notation (0,0,0), because that are the components of the zero vector wrt. to a basis but not the coordinate independent zero vector itself.

Now the question is the opposite, i.e., given that for a continuously differentiable vector field

[tex]\vec{\nabla} \times \vec{V}=0[/tex]

holds in some region of space, does there always exist a scalar field such that the vector field is given by the gradient.

The answer is often yes, but not always! It depends on the region of space, where the curl vanishes and where the vector field is well defined. The answer is that in any open region that is simply connected the assertion is correct. Simply connected means that any closed curve within this region can be continuously contracted to a single point within this region.

Then it is easy to show that for any curve [itex]\mathcal{C}(\vec{x}_1,\vec{x})[/itex], connecting a fixed point [itex]\vec{x}_1[/itex] with [itex]\vec{x}[/itex] fully contained in that region, the integral

[tex]\phi(\vec{x})=-\int_{\mathcal{C}(\vec{x}_1,\vec{x})} \mathrm{d} \vec{y} \cdot \vec{V}(\vec{y})[/tex]

is independent of the particular choice of the curve. Since the region has been assumed to be open, one can also take the gradient of this field, and it is easy to show that indeed

[itex]\vec{\nabla} \phi=-\vec{V}.[/itex]

There's another theorem of this kind. If for a continuously differentiable vector field [itex]\vec{\nabla} \cdot \vec{V}=0[/itex] in a simply connected open region, then there exists a vector potential [itex]\vec{A}[/itex], such that

[tex]\vec{V}=\vec{\nabla} \times \vec{A}[/tex]

in this region.

These are special cases of Poincare's lemma, specialized to three-dimensional space. You find the proof in any textbook on vector calculus (I know only German ones, so that I can't give a particular reference).

Then there is Helmholtz decomposition theorem, stating that any "sufficiently nice" vector field can be decomposed into a gradient and a vector field. The former contains all sources and the second all vortices of the field. I.e. there is always a scalar and a vector potential such that

[tex]\vec{V}=-\vec{\nabla} \phi+\vec{\nabla} \times \vec{A}.[/tex]

- #15

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Vj=−∂jϕ.

What if they don't exist?

For example the force example asked for by the OP.

- #16

mfb

Mentor

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That is an important requirement, I just want to highlight it here.The answer is that in any open regionthat is simply connectedthe assertion is correct.

A quite intuitive (imo) approach: Just construct your ϕ: Fix an arbitrary point x

- #17

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Kleppner and Kolenkow 5.8 says it can be proved by using physical arguments - how would one do that?

- #18

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If you have understood the mathematical derivation consider the following vector fields

Edit : F_{y} edited sorry.

1) Force Field A has F_{x} = -by and F_{y} = +bx

2) Force Field A has F_{x} = +by and F_{y} = +bx

Are the fields conservative or not?

3) Lewis Hamilton drives around the track at Monza.

Associating (a) A scalar field and (b) A vector field with one single circuit does the line integral around the circuit exist. What does it mean? Are the fields differentiable?

4) A point load F_{p} acts at the centre of a beam. Does this force have a curl? If so what is it?

Edit : F

1) Force Field A has F

2) Force Field A has F

Are the fields conservative or not?

3) Lewis Hamilton drives around the track at Monza.

Associating (a) A scalar field and (b) A vector field with one single circuit does the line integral around the circuit exist. What does it mean? Are the fields differentiable?

4) A point load F

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