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Stokes theorem

Hi, i don't really know how to start to show this using stokes theorem.
In similar questions i have done i have used F=aϕ \textbf{F} = \textbf{a}\phi, where a \textbf{a} is a constant vector. But here we don't have a scalar in what we are required to show, so i'm not sure how to start. Any help would be great.
(edited 10 years ago)
Original post by John taylor
Hi, i don't really know how to start to show this using stokes theorem.
In similar questions i have done i have used F=aϕ \textbf{F} = \textbf{a}\phi, where a \textbf{a} is a constant vector. But here we don't have a scalar in what we are required to show, so i'm not sure how to start. Any help would be great.

You want a cross to appear in a line integral where a dot would normally be - you should also know that the scalar triple product is invariant under cyclic permutation of the vectors involved, and is often a good way to get "dots and crosses" to swap over. How might you choose F now so the STP arises in the line integral that you wish to apply Stokes' to?

Think about it before opening the spoiler.

Spoiler

(edited 10 years ago)
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John taylor
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Original post by Farhan.Hanif93
You want a cross to appear in the a line integral where a dot would normally be - you should also know that the scalar triple product is invariant under cyclic permutation of the vectors involved, and is often a good way to get "dots and crosses" to swap over. How might you choose F now so the STP arises in the line integral that you wish to apply Stokes' to?

Think about it before opening the spoiler.

Spoiler



I'm trying to evaluate
∇×(a×r)=2a(∇⋅r)−2r(∇⋅a) \nabla \times (a \times r) = 2a(\nabla \cdot r) - 2r(\nabla \cdot a), and i know ∇⋅a=0 \nabla \cdot a = 0 , but what about ∇⋅r \nabla \cdot r, how would i evaluate this. It clearly has to be one, but i can't see how to get that, l tried r=xi+yj+zkr = xi + yj + zk and so ∇⋅r=3\nabla \cdot r = 3?
(edited 10 years ago)
Original post by John taylor
I'm trying to evaluate
∇×(a×r)=2a(∇⋅r)−2r(∇⋅a) \nabla \times (a \times r) = 2a(\nabla \cdot r) - 2r(\nabla \cdot a), and i know ∇⋅a=0 \nabla \cdot a = 0 , but what about ∇⋅r \nabla \cdot r, how would i evaluate this.

I'm not entirely convinced that your RHS is correct, how did you obtain that?

For the sake of answering your question, r=(x1x2x3)\mathbf{r} = \begin{pmatrix} x_1 \\ x_2 \\ x_3 \end{pmatrix} is simply the radial position vector, so it's divergence should be apparent if you work it out explicitly component-by-component.
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John taylor
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Original post by Farhan.Hanif93
I'm not entirely convinced that your RHS is correct, how did you obtain that?

For the sake of answering your question, r=(x1x2x3)\mathbf{r} = \begin{pmatrix} x_1 \\ x_2 \\ x_3 \end{pmatrix} is simply the radial position vector, so it's divergence should be apparent if you work it out explicitly component-by-component.


Its an identity on wikipedia, http://en.wikipedia.org/wiki/Vector_calculus_identities.
I made a mistake i think as ∇⋅r≠r⋅∇\nabla \cdot r \neq r \cdot \nabla

I think i finally managed it, ∇×(a×r)=3a−a=2a \nabla \times (a \times r) = 3a - a = 2a .
And so it all works out, thanks so much for your help!
(edited 10 years ago)
Original post by John taylor
Its an identity on wikipedia, http://en.wikipedia.org/wiki/Vector_calculus_identities.
I made a mistake i think as ∇⋅r≠r⋅∇\nabla \cdot r \neq r \cdot \nabla

Yeah, that was the mistake. Can you see how to proceed without making that incorrect assumption?

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