Vtyjkyuk

Is it generally true that $nablatimesvecn=0$ 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 $vecncdotnablatimesvecn=0$ but this isn’t quite $nablatimesvecn=0$.

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 $nablatimesvecn=0$ into
practice. For example, the curl in general orthogonal curvilinear coordinates
is
beginalign
nablatimesvecf=&frac1h_2,h_3,left[fracpartialpartial
x_2left(h_3,f_3right)-fracpartialpartial
x_3left(h_2,f_2right)right],vece_1+frac1h_3,h_1,left[fracpartialpartial
x_3left(h_1,f_1right)-fracpartialpartial
x_1left(h_3,f_3right)right],vece_2nonumber\
&qquad+frac1h_1,h_2,left[fracpartialpartial x_1left(h_2,f_2right)-fracpartialpartial x_2left(h_1,f_1right)right],vece_3
endalign

where $h_i$ are the scale factors and $vece_i$ 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
$h_3=r,sintheta$. If I take surfaces normal to the sphere $vecn=vece_1$ with $f_1=1$,
clearly $fracpartialpartial
x_3left(h_1,f_1right)=fracpartialpartial
x_2left(h_1,f_1right)=0$ and $nablatimesvecn=0$.

If I take toroidal coordinates $(x_1,x_2,x_3)=(u,v,phi)$ with scale factors
$h_1=h_2=a/left(coshu-cosvright)$ and
$h_3=a,sinhu/left(coshu-cosvright)$ where $a$ is a parameter and
$u=mathrmconstant$ are toroidal surfaces. Then naively I would take the
unit normal to the toroidal surfaces to be $vecn=vece_1$ again
corresponding to $f_1=1$. However this leads to
beginequation
nablatimesvecf=frac1h_3,h_1,left(fracpartialh_1partial
x_3right),vece_2-frac1h_1,h_2,left(fracpartialh_1partial x_2right),vece_3=fracsinva,vece_3
endequation

So it seems that the naive assumption $vecn=vece_1$ isn't correct for
this coordinate system? So how do I find the proper normal that satisfies
$nablatimesn=0$ and $vecncdotvecn=1$ in an general orthogonal
curvilinear coordinate system? Obviously, from definition of the $nablatimes$
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 $textcurl vec F = vec 0$ 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 $Dvec F = frac12big(Dvec F + (Dvec F)^topbig) + frac12big(Dvec F - (Dvec F)^topbig)$, 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 $Dvec n$ 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 $domega_1 = 0$. This needn't be true (as in our examples). What is true is that $domega_1 = omega_1wedgeeta$ for some $1$-form $eta$; this will tell us that when we evaluate $domega_1$ 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 $omega_1= dr$, $omega_2 = r,dtheta$, $omega_3 = rsintheta,dphi$. You can compute, using your formulas, that $textcurl vec e_2$ has a non-zero $vec e_3$ component. I see that with differential forms by taking $domega_2 = drwedge dtheta$; applying the Hodge star to get back a $1$-form, I get $star (drwedge dtheta) = star (frac 1romega_1wedgeomega_2) = frac 1romega_3$, corresponding to the vector field $frac1rvec e_3$. Indeed, as a 2-form on the cone, $domega_2$ 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.)