Laplancian vector?


by pivoxa15
Tags: laplancian, vector
pivoxa15
pivoxa15 is offline
#1
Oct28-06, 09:02 PM
P: 2,268
In http://mathworld.wolfram.com/Laplacian.html under

'Using the vector derivative identity' It has that forumula for the laplacian of a vector. In cartesian coords it can be derived but I read in books that the this laplacian on vector fields that are not cartesian can only be defined. But why don't they define it in other systems as the scalar laplacian (wrt to their own coords systems) of each components of their vector field? It would make more sense wouldn't it?
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StatusX
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#2
Oct29-06, 10:51 AM
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P: 2,566
If (u,v,w) is an arbitrary curvilinear coordinate system, then:

[tex]\nabla^2 \vec A = \nabla^2 (\hat e_u A_u + \hat e_v A_v + \hat e_w A_w ) = \nabla^2 (\hat e_u A_u) + \nabla^2 ( \hat e_v A_v) + \nabla^2 ( \hat e_w A_w )[/tex]

In a system where the basis vectors depended on position, there will be extra terms, eg:

[tex] \nabla^2 (\hat e_u A_u) = \hat e_u \nabla^2 A + 2 (\nabla \cdot \hat e_u) \nabla A_u + A_u \nabla^2 \hat e_u [/tex]

If you're asking why we don't just define the Laplacian so that:

[tex]\nabla^2 \vec A = \hat e_u \nabla^2 (A_u) + \hat e_v \nabla^2 (A_v)+ \hat e_w \nabla^2 (A_w)[/tex]

the reason is that the answer we'd get would depend on the coordinate system we're using, which is something we don't want.
pivoxa15
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#3
Oct30-06, 04:59 AM
P: 2,268
Quote Quote by StatusX
If (u,v,w) is an arbitrary curvilinear coordinate system, then:

[tex]\nabla^2 \vec A = \nabla^2 (\hat e_u A_u + \hat e_v A_v + \hat e_w A_w ) = \nabla^2 (\hat e_u A_u) + \nabla^2 ( \hat e_v A_v) + \nabla^2 ( \hat e_w A_w )[/tex]

So is this equation equivalent to the (established) vector derivative identity given in MathWorld for any coordinate system?

StatusX
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#4
Oct30-06, 10:21 AM
HW Helper
P: 2,566

Laplancian vector?


I don't know what you mean. All that equation used was the expansion of A in the (u,v,w) coordinate system and the linearity of the laplacian.
pivoxa15
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#5
Oct30-06, 04:53 PM
P: 2,268
I think my question arose from a misunderstanding on my behalf. Is it the case that the equation (highlighted in post 3) is the first step in solving the vector Laplacian of A. The next step is to apply the definition of a Laplacian vector on each component, like breaking a complicated problem into smaller bits but applying the same principle, which in this case is the definition of a Laplacian vector.

When you say the answer would depend on the coord system, do you mean that if I got in spherical coords, A=r(wrt r coord) than in cartesian, it should be A=(x,y,z). But this doesen't always happen if you (hypothetically) define the Laplacian as you have done in the last equation of post 2.


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