# 'Constant' vector field is equivalent to some scalar field

1. May 14, 2014

### 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))?

Last edited: May 14, 2014
2. May 14, 2014

### JanEnClaesen

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

3. May 14, 2014

### UltrafastPED

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

4. May 14, 2014

### pasmith

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 $\phi$.

5. May 14, 2014

### 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)?

6. May 14, 2014

### 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.

Last edited: May 14, 2014
7. May 14, 2014