I'm reading in the textbook Electromagnetics with Applications and it says (page 18) that if the line integral around a closed path is zero, $$\oint_C\vec{F}.d\vec{L}=0$$
then the vector field $\vec{F}$ is lamellar (conservative). Specifically, the book says, "Any field for which the line integral around a closed path is zero is called a conservative or lamellar field."
I read this wiki page and it makes sense. Here is my question: Is the above quote strictly correct (specifically for physical systems like electromagnetic fields)? In other words, is it possible to find a path such that a line integral is zero for a field that is not a conservative field?
Just trying to get this locked down. Thanks.
Best Answer
Given a path integral of the form:
$$\oint_C\vec{F}\cdot d\vec{L}$$
Letting $\vec{F}=\langle P(x,y),Q(x,y) \rangle$, we can rewrite as follows: $$\oint_C Pdx +Qdy$$ We can then invoke green's theorem:
$$\oint_C Pdx +Qdy=\iint_R\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}dA$$
As you can see if the vector field is conservative, that is if $\frac{\partial Q}{\partial x}=\frac{\partial P}{\partial y}$ the integral is clearly zero. However we want to know if there exists a vector field that is not conservative, where the above integral is zero. Define a function:
$$f(x,y)=\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}$$ And then plug that into the integral and set it equal zero: $$\iint_Rf(x,y)dA=0$$ If we let $f(x,y)=x^3y^2$ and the region $R$ be a square of side length 2 centered at the origin it is clear that this integral goes to zero. But can we write that function in terms of the partial derivatives? Yes, say we let our $Q$ and $P$ functions equal the following: $$P(x,y)=x$$ $$\frac{\partial P}{y}=0$$ $$Q(x,y)=\frac{x^4y^2}{4}$$ $$\frac{\partial Q}{\partial X}=x^3y^2$$ Therefore we have found a vector field $\vec{F}=\langle x,\frac{x^4y^2}{4} \rangle $ that is not conservative, which can yield a closed path integral equal to $0$.