# non-vanishing magnetic helicty density

+ 6 like - 0 dislike
1043 views

Suppose you are given a nowhere-vanishing exact 2-form $B=dA$ on an open, connected domain $D\subset\mathbb{R}^3$. I'd like to think of $B$ as a magnetic field.

Consider the product $H(A)=A\wedge dA$. At least in the plasma physics literature, $H(A)$ is known as the magnetic helicity density.

How can one determine if there is a closed one-form $\mathbf{s}$ such that $H(A+\mathbf{s})$ is non-zero at all points in $D$?

The reason I am interested in this question is that if you can find such an $\mathbf{s}$, then $A+\mathbf{s}$ will define a contact structure on $D$ whose Reeb vector field gives the magnetic field lines. Thus, the question is closely related to the Hamiltonian structure of magnetic field line dynamics.

I'll elaborate on this last point a bit. If there is a vector potential $A$ such that $A∧dA$ is non-zero everywhere, then the distribution $ξ=\text{ker}(A)$ is nowhere integrable, meaning $ξ$ defines a contact structure on $D$ with a global contact 1-form $A$. The Reeb vector field of this contact structure relative to the contact form $A$ is the unique vector field $X$ that satisfies $A(X)=1$ and $i_XdA=0$. Using the standard volume form $μ_o$, $dA$ can be expressed as $i_B\mu_o$ for a unique divergence-free vector field $\mathbf{B}$ (I'm having trouble typing $\mathbf{B}$ as a subscript). Thus, the second condition on the Reeb vector field can be expressed as $\mathbf{B}×X=0$, which implies the integral curves of X coincide with the magnetic field lines.

More generally, suppose $M$ is an orientable odd-dimensional manifold equipped with an exact 2-form $\omega$ of maximal rank. Also assume that the characteristic line bundle associated to $\omega$ admits a non-vanishing section $b:M\rightarrow \text{ker}(\omega)$. What is the obstruction to the existence of a 1-form $\vartheta$ with $d\vartheta=\omega$ and $\vartheta(b)>0$?

1) If $A(\mathbf{B})$ is bounded above and below on $D$, then a sufficient condition for there to be an $\mathbf{s}$ that gives a nowhere-vanishing helicity density is the existence of a closed one-form $\alpha$ with $\alpha(\mathbf{B})$ nowhere vanishing. In that case, $\mathbf{s}=\lambda \alpha$, where $\lambda$ is some large real number (with appropriate sign), would work.

If there is such an $\alpha$, then, being closed, it defines a foliation whose leaves are transverse to the divergence-free field $\mathbf{B}$. I suspect the question that asks whether a given non-vanishing divergence-free vector field admits a transverse co-dimension one foliation has been studied before, but I am not familiar with any work of this type.

An example where $D=$3-ball and helicity density must have a zero:

Let $D$ consist of those points in $\mathbb{R}^3$ with $x^2+y^2 < a^2$ for a real number $a>1$. Note that all closed 1-forms are exact in this case. Let $f:[0,\infty)\rightarrow\mathbb{R}$ be a smooth, non-decreasing function such that $f(r)=0$ for $r<1/10$ and $f(r)=1$ for $r\ge1/2$. Let $g:\mathbb{R}\rightarrow \mathbb{R}$ be the polynomial $g(r)=1-3r+2r^2$. Define the 2-form $B$ using the divergence free vector field $\mathbf{B}(x,y,z)=f(\sqrt{x^2+y^2})e_\phi(x,y,z)+g(\sqrt{x^2+y^2})e_z$. Here $e_\phi$ is the azimuthal unit vector and $e_z$ is the $z$-directed unit vector. It is easy to verify that $B$, thus defined, is an exact 2-form that is nowhere vanishing.

Because $g(1)=0$ and $f(1)=1$, the circle, $C$, in the $z=0$-plane, $x^2+y^2=1$, is an integral curve for the vector field $\mathbf{B}$. I will use this fact to prove that the helicity density must have a zero for any choice of gauge. Let $A$ satisfy $dA=B$ and suppose $A\wedge B$ is non-zero at all points in $D$. Note that $A\wedge B=A(\mathbf{B})\mu_o$, meaning $h=A(\mathbf{B})$ is a nowhere vanishing function. Without loss of generality, I will assume $h>0$. Thus, the line integral $I=\oint_C h\frac{dl}{|\mathbf{B}|}$ satisfies $I>0$. But, by Stoke's theorem, $I=2\pi\int_0^1g(r)rdr=0$, as is readily verified by directly evaluating the integral. Thus, there can be no such $A$.

An example where $D=T^2\times (0,2\pi)$ and helicity density must have a zero:

Set $D=S^1\times S^1\times(0,2\pi)$ and let $(\theta,\zeta,r)$ be the obvious coordinate system. Set $B=f(r) dr\wedge d\theta+g(r) dr\wedge d\zeta$ where $$f(r)=\cos(2r),$$ and $$g(r)=\sin(r).$$ Clearly, $A=\frac{1}{2}\sin(2r)d\theta-\cos(r)d\zeta$ satisfies $B=dA$ and $B$ is nowhere vanishing. A quick calculation shows that $\int_D A\wedge B=0$.

Now suppose that $\mathbf{s}$ is an arbitrary closed 1-form. Either by using Stoke's theorem or by direct calculation, the fact that the total toroidal and poloidal fluxes, $2\pi\int_0^{2\pi}f(r)dr$ and $2\pi\int_0^{2\pi}g(r)dr$, are zero implies that $\int_D(A+\mathbf{s})\wedge B=0$. Thus, the helicity density must always have a zero.

This post imported from StackExchange MathOverflow at 2015-05-27 22:10 (UTC), posted by SE-user Josh Burby
retagged May 27, 2015
Can you elaborate a little on your motivation? -- it's interesting and so am wondering if there is a reference.

This post imported from StackExchange MathOverflow at 2015-05-27 22:10 (UTC), posted by SE-user Chris Gerig
I edited the problem statement to give a bit more explanation of the connection between magnetic fields and contact geometry.

This post imported from StackExchange MathOverflow at 2015-05-27 22:10 (UTC), posted by SE-user Josh Burby
I have some questions. 1) What is $A(B)$ ? 2) Do you have an example when such form $s$ does not exist? 3) Can you prove that such $s$ exists if $D$ is a 3-ball? 4) What role plays for you that $B$ can be unbounded?

This post imported from StackExchange MathOverflow at 2015-05-27 22:10 (UTC), posted by SE-user Dmitri
1) $A(\mathbf{B})$ is just the contraction of the 1-form $A$ with the vector field $\mathbf{B}$, $A(B)=\text{i}_{\mathbf{B}} A$. 2+3) I have an example where $D$ is a 3-ball and such an $s$ cannot exist. I've added this above. 4) I am not terribly interested in unbounded $B$.

This post imported from StackExchange MathOverflow at 2015-05-27 22:10 (UTC), posted by SE-user Josh Burby

 Please use answers only to (at least partly) answer questions. To comment, discuss, or ask for clarification, leave a comment instead. To mask links under text, please type your text, highlight it, and click the "link" button. You can then enter your link URL. Please consult the FAQ for as to how to format your post. This is the answer box; if you want to write a comment instead, please use the 'add comment' button. Live preview (may slow down editor)   Preview Your name to display (optional): Email me at this address if my answer is selected or commented on: Privacy: Your email address will only be used for sending these notifications. Anti-spam verification: If you are a human please identify the position of the character covered by the symbol $\varnothing$ in the following word:p$\hbar$ysicsOverf$\varnothing$owThen drag the red bullet below over the corresponding character of our banner. When you drop it there, the bullet changes to green (on slow internet connections after a few seconds). Please complete the anti-spam verification