Laplacian Vector: Definition & Derivation

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SUMMARY

The discussion centers on the derivation and definition of the Laplacian vector in various coordinate systems, particularly curvilinear coordinates. It highlights that while the Laplacian can be expressed in Cartesian coordinates, its definition in non-Cartesian systems is not straightforward due to the dependence on the coordinate system used. The forum participants emphasize that defining the Laplacian vector as the scalar Laplacian of each component would yield results that vary with the coordinate system, which is undesirable. The conversation concludes that the established vector derivative identity from MathWorld is applicable across different coordinate systems, provided the components are treated appropriately.

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  • Understanding of vector calculus, specifically the Laplacian operator.
  • Familiarity with curvilinear coordinate systems and their properties.
  • Knowledge of the vector derivative identity and its applications.
  • Basic concepts of differential operators in multiple dimensions.
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  • Study the derivation of the Laplacian operator in curvilinear coordinates.
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  • Investigate the differences between scalar and vector Laplacians in different coordinate systems.
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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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If (u,v,w) is an arbitrary curvilinear coordinate system, then:

\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 )

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

\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

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

\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)

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.
 
StatusX said:
If (u,v,w) is an arbitrary curvilinear coordinate system, then:

\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 )

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