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'Constant' vector field is equivalent to some scalar field

  1. May 14, 2014 #1
    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))?
     
    Last edited: May 14, 2014
  2. jcsd
  3. May 14, 2014 #2
    In more concrete terms: does it make sense to speak of the potential of a scalar field?
     
  4. May 14, 2014 #3

    UltrafastPED

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    The dot product is only defined for a pair of vectors, and the result is a scalar.
     
  5. May 14, 2014 #4

    pasmith

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    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 [itex]\phi = \nabla \cdot \mathbf{f}[/itex] for some vector potential [itex]\mathbf{f}[/itex] and the original scalar PDE for [itex]\phi[/itex] is [tex]\mathcal{L}(\phi) = 0[/tex] then we now have [tex]\mathcal{L}(\nabla \cdot \mathbf{f}) = 0[/tex] which is replacing one equation in one unknown with one equation in three unknowns, which seems pointless. We can impose the condition [tex]\nabla \times \mathbf{f} = 0[/tex] by replacing [itex]\mathbf{f}[/itex] with [itex]\mathbf{f}' = \mathbf{f} + \mathbf{g}[/itex] where [itex]\mathbf{g}[/itex] satisfies [tex]\nabla \cdot \mathbf{g} = 0 \\ \nabla \times \mathbf{g} = -\nabla \times \mathbf{f}[/tex] so that [itex]\nabla \cdot \mathbf{f}' = \nabla \cdot \mathbf{f}[/itex] and [itex]\nabla \times \mathbf{f}' = 0[/itex], but then we immediately have [itex]\mathbf{f}' = \nabla \theta[/itex] and we end up with the scalar equation [tex]\mathcal{L}(\nabla^2 \theta) = 0[/tex] and the obvious way to solve that is to set [itex]\phi = \nabla^2 \theta[/itex] and first solve for [itex]\phi[/itex].
     
  6. May 14, 2014 #5
    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)?
     
  7. May 14, 2014 #6
    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.
     
    Last edited: May 14, 2014
  8. May 14, 2014 #7

    UltrafastPED

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