Is the vector field generated by $\langle-y,x \rangle$ a gradient vector field?

Question

Let's do a two dimensional case please: I guess what I am asking is this: Are there any gradients that are not generated by a potential function and what do these gradients generated by a potential function have to do with path independence ?

A gradient of a function generates many vectors, it is a place holder for values of $x$ and $y$ and when various values are inserted into $x$ and $y$ you generate a vector field. Now if there is a potential function that means there is an antiderivative for each of those place holders correct?

It also means the vector field is conservative which in turn means the path you take from one point in the field to another point in the field doesn't make any difference ... the energy you would end up gaining or loosing is the same.

But why does the existence of antiderivatives and potential functions mean that energy is conserved following a path through the field.? I used the example $\langle-y, x\rangle$ because what ever this link is then it would be missing for $\langle -y,x\rangle$ and perhaps I can see the difference such as $\langle x, y \rangle$ .

There is a fundamental link I am missing. I need some help.

Answer 1

A gradient field is, by definition, the gradient of some function (usually called a potential function), so the answer to your first question is no. What I believe you're after is this: Can you rescale the vector field $\langle -y,x\rangle$ to make it conservative (or a gradient)? That depends on your domain. If you allow me to remove a ray containing the origin, then on that domain the vector field $\langle -\frac y{x^2+y^2},\frac x{x^2+y^2}\rangle$ is a gradient field. (One can actually show it's impossible on the whole plane.)

Energy is conserved in the presence of a gradient field because of the fundamental theorem of calculus for line integrals: $$\int_A^B \overrightarrow{\nabla f}\cdot d\vec r = f(B)-f(A)$$ for any path from $A$ to $B$. In particular, the net work done around any closed path is $0$, so energy is conserved.

Answer 2

As long as $\underline{F}$ is conservative, there is a $\nabla f$ that is $\underline{F}$

Path independence has to do with the fundamental theorem of line integrals.

If $\underline{\gamma}(t)$, $a\leq t\leq b$, traces a path of integration, $C$, for the vector field $\underline{F}$, then

$$I=\int_C\underline{F}\cdot d\underline{\gamma}=\int_a^b\underline{F}(\underline{\gamma}(t))\cdot\underline{\gamma}'dt$$

We don't know what this integral is unless we know $\underline{F}$ and $\underline{\gamma}$. But suppose $\underline{F}$ is conservative, then we know $\underline{F}=\nabla f$ from potential theory, so

$$I=\int_a^b\nabla f(\underline{\gamma}(t))\cdot d\underline{\gamma}'dt=\int_a^b\left(\frac{\partial f}{\partial x}\Bigg{\vert}_{\underline{\gamma}(t)}\frac{\partial x}{\partial t}dt+\frac{\partial f}{\partial y}\Bigg{\vert}_{\underline{\gamma}(t)}\frac{\partial y}{\partial t}dt\right)=\int_a^b\left(\frac{d}{dt}f(\underline{\gamma}(t))dt\right)$$

So

$$I=\int_a^bd\left(f(\underline{\gamma}(t))\right)=f(\underline{\gamma}(b))-f(\underline{\gamma}(a))$$

This says the integral is only dependent on the endpoints of $\gamma$ and nothing inbetween

In particular, if $C$ is closed so that $\underline{\gamma}(a)=\underline{\gamma}(b)$, then $I=0$

As for $\underline{F}=-y\underline{i}+x\underline{j}$, this field is not conservative because

$$\displaystyle\frac{\partial}{\partial y}\left(-y\right)\neq\displaystyle\frac{\partial}{\partial x}\left(x\right)$$

Whereas $\underline{F}=x\underline{i}+y\underline{j}$ is conservative because

$$\displaystyle\frac{\partial}{\partial y}\left(x\right)= \displaystyle\frac{\partial}{\partial x}\left(y\right)$$

Answer 3

The problem is asking if there exist a function, F(x,y), such that $\nabla F= <-y, x>$. That requires that $\frac{\partial F}{\partial x}= -y$ and $\frac{\partial F}{\partial y}= x$. Integrating $\frac{\partial F}{\partial x}= -y$ with respect to x, $F(x, y)= -xy+ u(y)$ (since the partial derivative with respect to x treats y as a constant, the "constant of integration" may be a function of y only). Differentiating that with respect to y, $\frac{\partial (-xy+ u)}{\partial y}= -x+ u'= x$. But that would mean that u'= 2x which contradicts the fact that u is a function of y only. So, no, this is not a gradient.