Proof that $\nabla \times E = 0$ using Stokes's theorem Firstly, I know that this can be proved by showing the curl of a gradient is $0$. I am not interested in that. I am interested in the validity of using Stokes's theorem. One way that Jackson proves that $\nabla \times E = 0$ is the following:
$$ F = q E $$
$$ W = - \int_A^B F \cdot dl = - q \int_A^B E \cdot dl = q \int_A^B \nabla \phi \cdot dl = 
 q \int_A^B d \phi = q(\phi_B - \phi_A) $$
so $\int_A^B E \cdot dl = -(\phi_B - \phi_A)$ therefore $\oint_A^B E \cdot dl = 0$
Then using Stokes's theorem $\int (\nabla \times E) \cdot \hat{n} da = 0$ which implies $\nabla \times E=0$.
Isn't the last line true only if $\nabla \times E \geq 0$? It seems like Jackson is instantiating $\int f(x) dx = 0 \implies f(x) = 0$, but this isn't guaranteed unless we already know $f(x) \geq 0$.
 A: Jackson is using the fact that closed curve line integrals of conservative vector fields are equal to $0$.
By Stokes' Theorem we have;
$$\int_{S}(\nabla \times F)\cdot dS=\oint_{\partial S}F\cdot ds$$
But for any $F=\nabla\phi$ where $\phi$ is a scalar field, we have;
$$\oint_{\partial S}F\cdot ds=0$$
So by Stokes' theorem it follows, by necessity, that;
$$\int_{S}(\nabla \times F)\cdot dS=0$$
$\textbf{EDIT}:$
If $(\nabla \times F)$ were not equal to zero everywhere in the interior of our closed contour then it would be possible to restrict our open surface to the neighborhood $M \subset S$ in which $|(\nabla \times F)|\geq 0$. This would lead to;
$$\oint_{\partial M}F\cdot dm \not = 0$$
Which is not possible for $F= \nabla\phi$.
A: He's using that the line integral vanishes for all nice curves, so the surface integral vanishes for all "nice surfaces" (whose boundaries are said curves) and this is what allows us to prove the integrand vanishes. If we only have an integral equal to $0$ for one surface, then of course we can't say anything about the integrand; but here it's $0$ for "all nice surfaces".
A: The key thing is that the above argument is true for all possible regions, which implies the claim. Here is a rigorous version of this:
Suppose $f: \mathbb{R} \to \mathbb{R}$ is measurable and that $$\int_E f=0 $$
for all measurable sets $E$, then $f=0$ almost everywhere
Proof:
Let $ U = \{x \in \mathbb{R} \text{ s.t } f(x)>0 \}$
Suppose $\mu(U)>0$ then there exist an $n \in \mathbb{N}$ such that $U_n = \{x \in \mathbb{R} \text{ s.t } f(x)>1/n \}$ and $\mu(U_n)>0$ hence
$$\int_U f \ge \int_{U_n}f \ge \dfrac{1}{n} \mu(U_n) \gt0$$
Contradicting our assumption.
In general you decompose $f=f_+ +f_-$ and apply the above argument to each function sparately. So in conclusion you'll have that:
$$ U_+ = \{x \in \mathbb{R} \text{ s.t } f(x)>0 \}$$
$$ U_- = \{x \in \mathbb{R} \text{ s.t } f(x)<0 \}$$
Both have zero measure hence $f=0$ almost everywhere. Since in physics we assume at least continuity then $f=0$ everywhere.
The above can be easily generalized to $\mathbb{R}^n$; in fact, the argument looks exactly the same with the only difference being the domains.
A: Firstly, it doesn't really make sense to say that $\nabla \times E \ge 0$, since $\nabla \times E$ isn't a real number (it's an element of $\mathbb R^3)$.
What Jackson's using here is continuity of $\nabla \times E$.
Let's suppose there's some point $p$ at which $F:=\nabla \times E \ne 0$. WLOG, we can assume $p$ is the origin and that the x-component of $F$ is positive. Then, by continuity, there is a disk $D$ about $0$ in the $x=0$ plane on which the x-component of $F$ is positive. We see that $\int_{D} F.n \cdot \mathrm{dA} > 0$ since the integrand is positive over this region. But this is a contradiction to what you've already proven.
