'Constant' vector field is equivalent to some scalar field

[JanEnClaesen]

To every scalar field s(x,y) there corresponds a 'constant' vector field x = A s(x,y) and y = B s(x,y), where A,B are direction cosines. The vector field is only partially constant since only the directions, and not the magnitudes, which are equal to |f(x,y)|, of the field vectors are constant.

The scalar field corresponds to the magnitudes of a vector field, that can be specified by only the magnitudes of the field vectors since the directions are constants A,B.

Is this correct?

This came up when evaluating a line integral f . dr , and splitting it up in the dx,dy components of dr. Can the dot product of a scalar field and a vector field define a scalar field as well as a vector field (since a(x,y)=(ax,ay))?

 

[JanEnClaesen]

In more concrete terms: does it make sense to speak of the potential of a scalar field?

 

[UltrafastPED]

Can the dot product of a scalar field and a vector field define a scalar field as well as a vector field (since a(x,y)=(ax,ay))?

The dot product is only defined for a pair of vectors, and the result is a scalar.

 

[pasmith]

In more concrete terms: does it make sense to speak of the potential of a scalar field?

Yes, but it's of no practical use. The only scalar-valued first-order differential operator is divergence, so the potential would be a vector. But then if \phi = \nabla \cdot \mathbf{f} for some vector potential \mathbf{f} and the original scalar PDE for \phi is \mathcal{L}(\phi) = 0 then we now have \mathcal{L}(\nabla \cdot \mathbf{f}) = 0 which is replacing one equation in one unknown with one equation in three unknowns, which seems pointless. We can impose the condition \nabla \times \mathbf{f} = 0 by replacing \mathbf{f} with \mathbf{f}' = \mathbf{f} + \mathbf{g} where \mathbf{g} satisfies \nabla \cdot \mathbf{g} = 0 \\ \nabla \times \mathbf{g} = -\nabla \times \mathbf{f} so that \nabla \cdot \mathbf{f}' = \nabla \cdot \mathbf{f} and \nabla \times \mathbf{f}' = 0, but then we immediately have \mathbf{f}' = \nabla \theta and we end up with the scalar equation \mathcal{L}(\nabla^2 \theta) = 0 and the obvious way to solve that is to set \phi = \nabla^2 \theta and first solve for \phi.

 

[JanEnClaesen]

It's interesting that you call divergence the only scalar-valued first-order differential operator. This means that divergence is a necessary consequence, as opposed to an arbitary construct. Is there a way to see why divergence is the only possible first-order differential operator? Vector differential operators always seemed a bit arbitrary to me.

The vector potential of a scalar field isn't by any chance the scalar potential of that vector field (duality)?

 

[JanEnClaesen]

So it doesn't make sense to think of a scalar field as being 'conservative', in that the line integral between two points is path-independent?
EDIT: a scalar field is probably almost never path-independent, unless it zero everywhere.

 

[UltrafastPED]

Path independence is a vector field property - and requires curl F = 0.

See Stokes' theorem: http://mathinsight.org/stokes_theorem_idea