Is it generally true that $\nabla\times\vec{n}=0$ for any surface or is this only true for a simply connected domain? (see ftp://ftp.math.ucla.edu/pub/camreport/cam12-18.pdf) and discussion here (Curl of unit normal vector on a surface is zero?)
I think Stoke’s theorem implies that $\vec{n}\cdot\nabla\times\vec{n}=0$ but this isn’t quite $\nabla\times\vec{n}=0$.
In particular, it doesn't seem like the unit vector for u in toroidal coordinates $(u,v,\phi)$ satisfies this (http://mathworld.wolfram.com/ToroidalCoordinates.html) Yet constant u corresponds to toroidal surfaces.
So what I'm pondering is how to translate $\nabla\times\vec{n}=0$ into
practice. For example, the curl in general orthogonal curvilinear coordinates
is
\begin{align}
\nabla\times\vec{f}=&\frac{1}{h_2\,h_3}\,\left[\frac{\partial}{\partial
x_2}\left(h_3\,f_3\right)-\frac{\partial}{\partial
x_3}\left(h_2\,f_2\right)\right]\,\vec{e}_1+\frac{1}{h_3\,h_1}\,\left[\frac{\partial}{\partial
x_3}\left(h_1\,f_1\right)-\frac{\partial}{\partial
x_1}\left(h_3\,f_3\right)\right]\,\vec{e}_2\nonumber\\
&\qquad+\frac{1}{h_1\,h_2}\,\left[\frac{\partial}{\partial x_1}\left(h_2\,f_2\right)-\frac{\partial}{\partial x_2}\left(h_1\,f_1\right)\right]\,\vec{e}_3
\end{align}
where $h_i$ are the scale factors and $\vec{e}_i$ are the unit vectors along
coordinate lines $x_i$. For spherical coordinates
$(x_1,x_2,x_3)=(r,\theta,\phi)$ the scale factors are $h_1=1$, $h_2=r$, and
$h_3=r\,\sin\theta$. If I take surfaces normal to the sphere $\vec{n}=\vec{e}_1$ with $f_1=1$,
clearly $\frac{\partial}{\partial
x_3}\left(h_1\,f_1\right)=\frac{\partial}{\partial
x_2}\left(h_1\,f_1\right)=0$ and $\nabla\times\vec{n}=0$.
If I take toroidal coordinates $(x_1,x_2,x_3)=(u,v,\phi)$ with scale factors
$h_1=h_2=a/\left(\cosh{u}-\cos{v}\right)$ and
$h_3=a\,\sinh{u}/\left(\cosh{u}-\cos{v}\right)$ where $a$ is a parameter and
$u=\mathrm{constant}$ are toroidal surfaces. Then naively I would take the
unit normal to the toroidal surfaces to be $\vec{n}=\vec{e}_1$ again
corresponding to $f_1=1$. However this leads to
\begin{equation}
\nabla\times\vec{f}=\frac{1}{h_3\,h_1}\,\left(\frac{\partial{h}_1}{\partial
x_3}\right)\,\vec{e}_2-\frac{1}{h_1\,h_2}\,\left(\frac{\partial{h}_1}{\partial x_2}\right)\,\vec{e}_3=\frac{\sin{v}}{a}\,\vec{e}_3
\end{equation}
So it seems that the naive assumption $\vec{n}=\vec{e}_1$ isn't correct for
this coordinate system? So how do I find the proper normal that satisfies
$\nabla\times{n}=0$ and $\vec{n}\cdot\vec{n}=1$ in an general orthogonal
curvilinear coordinate system? Obviously, from definition of the $\nabla\times$
above $f_1=1/h_1$ satisfies the former condition but not the latter
condition.
Thanks
This is entirely a local result. Only if you want to go from $\text{curl }\vec F = \vec 0$ to $\vec F = \nabla f$ do you need some global topological restrictions.
The only satisfactory proof I see, after a bit of thought, is to use a basic fact from differential geometry. Curl of a vector field is the antisymmetric part of its derivative. (That is, write $D\vec F = \frac12\big(D\vec F + (D\vec F)^\top\big) + \frac12\big(D\vec F - (D\vec F)^\top\big)$, and curl is identified with the latter term.) Now, the shape operator, which gives the derivative of the map $\vec n$ from the surface to the unit sphere, is symmetric. (See, for example, pp. 45-46 of my differential geometry text.) Thus, the matrix representation of $D\vec n$ with respect to an orthonormal basis for the tangent space of the surface will give a symmetric matrix, and the curl term is $0$.
A different argument would come from extending the map $\vec n$ to a neighborhood of the surface (say by having it stay constant along normal lines) and then differentiating the resulting vector field on an open set in $\Bbb R^3$ with the usual rules for curl. It would still be beneficial to work with a basis adapted to the geometry, however.
EDIT: There's nothing wrong with your computations. But let's look at something simpler. What if you look at the level surface $f_2=\pi/4$—a cone—in spherical coordinates? Then you get a nonzero curl computation, as well. You just happened to luck out when you looked at the spheres.
So what's the explanation? The result you quoted (and which I proved) applies to the computation of an "intrinsic" curl living only in the surface, not in 3-D. Let me explain this using the language of differential forms. You have a 3-D coordinate system with $\vec e_1,\vec e_2,\vec e_3$ an orthonormal frame along the appropriate coordinate curves. Let $\omega_1,\omega_2,\omega_3$ be the dual coframe — so these are $1$-forms satisfying $\omega_i(\vec e_j) = \delta_{ij}$. The level surfaces with unit normal $\vec e_1$ are given by the differential equation $\omega_1 = 0$. Now, you're wanting curl $0$ to mean $d\omega_1 = 0$. This needn't be true (as in our examples). What is true is that $d\omega_1 = \omega_1\wedge\eta$ for some $1$-form $\eta$; this will tell us that when we evaluate $d\omega_1$ on any surface $\omega_1=0$ we do get $0$, of course, but it needn't be $0$ as a $2$-form on all of 3-space.
This is what's going on in my spherical coordinates example (and, in more complicated fashion, in your example). Explicitly, we have $\omega_1= dr$, $\omega_2 = r\,d\theta$, $\omega_3 = r\sin\theta\,d\phi$. You can compute, using your formulas, that $\text{curl }\vec e_2$ has a non-zero $\vec e_3$ component. I see that with differential forms by taking $d\omega_2 = dr\wedge d\theta$; applying the Hodge star to get back a $1$-form, I get $\star (dr\wedge d\theta) = \star (\frac 1r\omega_1\wedge\omega_2) = \frac 1r\omega_3$, corresponding to the vector field $\frac1r\vec e_3$. Indeed, as a 2-form on the cone, $d\omega_2$ does vanish identically; but as a 2-form on $\Bbb R^3$ it most definitely does not. (The moral of the story is that to understand deeply what's going on here one has to use differential forms; staying in the land of vectors won't cut it.)