How do you find the scalar field of the vector field $F=-y\hat{i}+x\hat{j}$

Question

I know that:

$F=\nabla f = \frac{\partial f}{\partial x}\hat{i}+\frac{\partial f}{\partial y}\hat{j}+\frac{\partial f}{\partial z}\hat{k}$

So in cases like:

$F=2x\hat{i}+2y\hat{j}$

I just integrate each term to get:

$f=x^2+y^2$

But when I do the same process for this:

$F=-y\hat{i}+x\hat{j}$

I get:

$f=-xy+xy = 0$

Which seems wrong.

I know that $F=-y\hat{i}+x\hat{j}$ is the vector field of a counter clockwise rotating vortex but I just can't seem to get its scalar field.

Answer 1

If $\hat F=-y\,\hat\imath+x\,\hat\jmath$ is a gradient field, then the corresponding scalar function $f$ satisfies

$$\frac{\partial f}{\partial x} = -y \\ \frac{\partial f}{\partial y}=x$$

Integrating both sides of the first equation with respect to $x$ gives

$$f(x,y) = -xy + g(y)$$

and differentiating this with respect to $y$ gives

$$\frac{\partial f}{\partial y} = -x +\frac{dg}{dy} = x$$

We assumed $g$ is independent of $x$, so its derivative should also be independent of $x$. Yet here $\frac{dg}{dy}=2x$ so no such scalar $g$ exists. $\hat F$ is not a gradient field.

In contrast to the first example, if $\hat F=2x\,\hat\imath+2y\,\hat\jmath$, the same procedure yields

$$\frac{\partial f}{\partial x} = 2x \implies f(x,y)=x^2+g(y) \implies \frac{\partial f}{\partial y}=\frac{dg}{dy}=2y \implies g(y)=y^2+C \\ \implies f(x,y)=x^2+y^2+C$$

Answer 2

Sometimes there is no scalar field. We have a scalar field V if $\nabla V=\vec{F}$. It can be shown the curl of any gradient is 0.

So if $\nabla \times \vec{F} \neq 0, \vec{F}$ has no scalar field.

This implies a test for whether $\vec{F}$ has a scalar field. In two dimensions, that means find $\partial F_y/\partial x-\partial F_x/\partial y$. If that is non-zero then there is no scalar field.