# Conservative Force problem that has been worked but I still have a question

1. Jul 12, 2005

### karen03grae

My question: Show that $$\vec{F}$$ is a conservative vector field then find a potential function "f" such that $$\vec{F} =\nabla f$$.

$$\vec{F} (x,y) = sin(y)\vec{i} + (xcos(y) + sin(y))\vec{j}$$

I worked the problem and found out that the force was conservative and I found the potential function. Okay, I want to know why it is considered conservative if
$$\frac{\partial Q} {\partial x} = \frac {\partial P} {\partial y}$$

is true.

where P is the scalar function sin(y) and Q is the scalar function xcos(y) + sin(y);

Now in order to work this problem this way I had to assume that the force was conservative and then imply that all conservative forces are equal to their potential function. So I had $$\frac{\partial f} {\partial x} = sin(y)$$ and $$\frac{\partial f} {\partial y} = xcos(y) + sin(y)$$

now if I take the partial derivative of each of these with respect to the other variable then I can show that
$$\frac{\partial Q} {\partial x} = \frac {\partial P} {\partial y}$$
by Clairaut's theorem. Well that's great that they are equal but why does that show conservatism?

2. Jul 12, 2005

### dextercioby

Well, what is the definition of a conservative force ? To give you a hint, i'd say you might consider the theorem of Stokes to be useful

Daniel.

3. Jul 12, 2005

### karen03grae

A conservative force is independant of the path taken

4. Jul 12, 2005

### Nylex

No, the work done by the force is independent of the path taken.

5. Jul 15, 2005

### karen03grae

Opps, okay the work done is independent of the path taken. And Stokes' theorem says that $$\int \int_S \ curl\vec{F}\cdot d\vec{S}$$ =
$$\int_c\vec{F}\cdot d\vec{r}$$

So if we take the cross product of $$\nabla$$ and $$\vec{F}$$ (Which by the way, isn't "del" an operator? How is it that we can cross and operator with a vector? I thought we could only take the cross prod. of vectors) we are left with $$\frac {\partial Q} {\partial x} - \frac {\partial P} {\partial y}$$

And if these are equal then the force is conservative. But I still don't know why.

longshot:
Maybe it's because in order to do a surface integral one must have a closed path. And if we use Stokes' theorem and integrate along the boundary of the surface...and that integral just happens to be zero...because $$\frac {\partial Q} {\partial x} = \frac {\partial P} {\partial y}$$
and the curl of $$\vec{F}$$ just happens to lead to
$$\frac {\partial Q} {\partial x} - \frac {\partial P} {\partial y}$$
which equal zero then we can say that the force is conservative because no matter what the boundary of the surface is (path) the work done will still be zero because the particle started and ended in the same place.

6. Jul 15, 2005

### dextercioby

That's right. You're on the right track. A conservative force should send you to the closed curvilinear integral of the second kind equal to zero, for ANY PATH. Applying Stokes' theorem, you automatically get the curl of the force equal to zero. Mathematically and logically speaking, it is an equivalence.

It's a notation abuse this $\nabla\times\vec{F}$. Why? Well, on one hand we have that for 3D euclidean vectors

$$\mbox{curl} \ \vec{F}=:\epsilon_{ijk}\partial_{i}F_{j}\vec{e}_{k}$$

and on the other hand, for the same class of vectors

$$\vec{F}\times\vec{G}=:\epsilon_{ijk}F_{i}G_{j}\vec{e}_{k}$$

Daniel.

Last edited: Jul 15, 2005