Why does $\oint\mathbf{E}\cdot{d}\boldsymbol\ell=0$ imply $\nabla\times\mathbf{E}=\mathbf{0}$?

I am looking at Griffith's Electrodynamics textbook and on page 76 he is discussing the curl of electric field in electrostatics. He claims that since $$\oint_C\mathbf{E}\cdot{d}\boldsymbol\ell=0$$ then $$\nabla\times\mathbf{E}=\mathbf{0}$$ I don't follow this logic. Although I know that curl of $\mathbf{E}$ in statics is $\mathbf{0}$, I can't see how you can simply apply Stokes' theorem to equate the two statements.

If we take Stokes' original theorem, we have $\oint\mathbf{E}\cdot{d}\boldsymbol\ell=\int\nabla\times\mathbf{E}\cdot{d}\mathbf{a}=0$. How does this imply $\nabla\times\mathbf{E}=\mathbf{0}$? Griffiths seem to imply that this step is pretty easy, but I can't see it!

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What he is saying is: if the loop integral of every loop is zero then the curl must be zero everywhere. Convince yourself this is true. –  BebopButUnsteady Jul 1 '13 at 21:56
Hello Austin, you can write math symbols using LaTeX commands as the ones listed here: artofproblemsolving.com/Wiki/index.php/LaTeX:Symbols Just put the command(s) in between \$signs for inline math or between \$\$signs for display math. You'll need it later all the time if you are into physics, so give it a try! – pfold Jul 1 '13 at 22:17 5 Answers Suppose$\nabla \times {\bf E}$is a well-behaved function and$\nabla \times {\bf E}\neq 0$in some region. Then you could find a surface$S$through which$\int_S \nabla \times {\bf E} \cdot d{\bf a}\neq0$by making that surface very small and close to the aforementioned region. This contradicts Stokes's theorem, so it must be that$\nabla \times {\bf E}=0$everywhere. - Nice heuristic proof. _1!. – dimensio1n0 Aug 28 '13 at 18:03 $$\oint_C\mathbf{E}\cdot{d}\boldsymbol\ell=0$$ By 3-dimensional stokes's theorem applied to surfaces and their boundaries, $$\iint_S\left(\nabla\times\mathbf E\right)\cdot \hat{\mathbf{n}}\mbox{ d}S=0$$ For this to always be true, $$\left(\nabla\times\mathbf{E}\right)\cdot \hat{\mathbf{n}}=0$$ For the curl of this vector field to be orthogonal to the normal vector, the curl must be 0 (consider shifting your coordinate system to be such that$\hat{\mathbf{n}}$points in the$z$-direction if you want this to be more clear) and thus, $$\nabla\times\mathbf{E}=0$$ As required./Q.E.D.. - You can prove this by contradiction. Say that $$\oint_C\mathbf{E}\cdot{d}\boldsymbol\ell=0$$ everywhere and also that there exists some point such that $$\nabla \times \mathbf{E} \neq 0.$$ Since the electric field is continuous (a discontinuity of the electric field corresponds to an infinite charge density by Gauss' Law), it follows that the curl of the electric field must also be nonzero in some neighborhood of that point, with the same sign when dotted into an arbitrary unit vector. Then we can integrate over this neighborhood to get a nonzero value of $$\oint_C\mathbf{E}\cdot{d}\boldsymbol\ell$$ which is a contradiction. - The infinitesimal area$d\mathbf{a}$is arbitrary which means that$\nabla\times\mathbf{E}=0\$ must be true everywhere, and not just locally.

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The curl operator is defined as a limit of a line integral around the boundary of an oriented area as the area tends to zero:

$$(\nabla\times\mathbf{E})\cdot\mathbf{\hat{n}}=\;^{lim}_{A\rightarrow0}\left(\frac{1}{|A|}\oint_c\mathbf{E}\cdot d\mathbf{r}\right)$$

If this infinitesimal line integral tends to zero at a point, then so must the curl of the field there.

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