- #1

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Is it? If so, can you show how? Thanks

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- Thread starter Tosh5457
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- #1

- 130

- 28

Is it? If so, can you show how? Thanks

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- #2

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It is possible, what have you attempted?

- #3

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It is possible, what have you attempted?

I don't know where to start, unfortunately the mathematics disciplines in physics, in my university, are a bit superficial, with no theoretical exercises (only exercises to apply the theory). So I'm not trained for this, I want to see someone doing it so I can learn

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- #5

lavinia

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I don't know where to start, unfortunately the mathematics disciplines in physics, in my university, are a bit superficial, with no theoretical exercises (only exercises to apply the theory). So I'm not trained for this, I want to see someone doing it so I can learn

in three space there is a natural duality between vector fields and 2 forms.

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I'm wondering what deluks917 had in mind...

- #7

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I'm referring to the Kelvin-Stokes theorem

- #8

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div (A x B) = B . curl A - A . curl B

is useful. Also, on the volume V you have curl N = 0, where N is the normal vector field from S extended along itself a small distance h. If you can do that, then maybe you will see what I've been missing and do the derivation in the other direction, too. Note that what I'm talking about is a very physics-book style "proof"... it would make mathematicians giggle.

- #9

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Tosh5457, any luck deriving Kelvin-Stokes from the Divergence theorem or vice-versa?

- #10

lavinia

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I'm referring to the Kelvin-Stokes theorem

The Divergence theorem does not refer to line integrals. However the generalized Stokes theorem implies the Divergence Theorem.

the duality I referred to above is (a,b,c) -> ady^dz - bdx^dz + cdx^dy

The exterior derivative of this 2 form is the divergence of the vector field (a,b,c) times the Euclidean volume element. Have fun figuring out the rest.

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- #11

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[tex]\int_{\partial V} (\mathbf{F} \times \mathbf{N})\cdot \mathbf{n} dS = \int_{V} \nabla \cdot (\mathbf{F} \times \mathbf{N}) dV =

\int_{V} \mathbf{N} \cdot (\nabla \times \mathbf{F}) dV =

h\int_{S} \mathbf{N} \cdot (\nabla \times \mathbf{F}) dS + o(h) [/tex]

Also,

[tex]\int_{\partial V} (\mathbf{F} \times \mathbf{N})\cdot \mathbf{n} dS =

\int_{\mbox{top surface}} (\mathbf{F} \times \mathbf{N})\cdot \mathbf{N} dS +

\int_{\mbox{bottom surface}} (\mathbf{F} \times \mathbf{N})\cdot \mathbf{N} dS +

\int_{\mbox{edge}} (\mathbf{F} \times \mathbf{N})\cdot (\mathbf{t} \times \mathbf{N}) dS

[/tex]

The first two integrals on the right hand side are obviously zero. The last one is

[tex]

\int_{\mbox{edge}} (\mathbf{F} \times \mathbf{N})\cdot (\mathbf{t} \times \mathbf{N}) dS =

\int_{\mbox{edge}} \mathbf{F}\cdot \mathbf{t} dS =

h\int_{C} \mathbf{F}\cdot \mathbf{t} dl + o(h)

[/tex]

I wish I could see how to do it in the other direction.

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