Continuous 2nd Partials a Substantial Requirement for Conservative Field?

1. Jun 2, 2008

breez

It can be shown that if F has continuous 2nd partials, then div curl F = 0. According to Stoke's Theorem, the work done around a closed path C is equal to the flux integral of curl F on a surface sigma that has C as its boundary in positive orientation. However, this integral is equal to the volume triple integral of div curl F. But if F has continuous 2nd partials, then div curl F = 0, and hence the work around C must be 0. Doesn't this show that F is conservative if F has continuous 2nd partials?

2. Jun 2, 2008

HallsofIvy

Staff Emeritus
It isn't "F" that is "conservative"!

A vector field $\vec{f}$ is said to be "conservative" (that's physics terminology; I prefer "is an exact derivative") if there exist a scalar function F, having continuous partials, such that $\vec{f}= \nabla F$. IF there exist such an F, then yes, $\vec{f}$ is "conservative"!

3. Jun 2, 2008

nicksauce

So you're saying that because

$$\int_{\partial S}\vec{F}\cdot d \vec{l} = \int_{S} \nabla \times \vec{F} \cdot d \vec{S}$$

and because

$$\int_{\partial V}\vec{F} \cdot d \vec{S} = \int_{V} \nabla \cdot \vec{F} dV$$

it follows that

$$\int_{\partial V}\vec{F}\cdot d \vec{l} = \int_{V} \nabla \cdot \nabla \times \vec{F} dV = 0$$

assuming F is smooth enough? The problem is that you can't go from the first theorem to the second. In the first S is an open surface (which has a boundary curve) and in the second S = delV is a closed surface (which does not have a boundary curve)

4. Jun 2, 2008

nicksauce

The first integral in the third equation should be around delS not delV.