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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 $A\wedge dA$ is non-zero everywhere, then the distribution $\xi=\text{ker}(A)$ is nowhere integrable, meaning $\xi$ 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 $\text{i}_XdA=0$. Using the standard volume form $\mu_o$, $dA$ can be expressed as $\text{i}_{\mathbf{B}}\mu_o$ for a unique divergence-free vector field $\mathbf{B}$. Thus, the second condition on the Reeb vector field can be expressed as $\mathbf{B}\times X=0$, which implies the integral curves of $X$ coincide with the magnetic field lines.

An example where $D=$3-ball and no $\mathbf{s}$ can exist:

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 $A$.

Josh Burby
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  • Can you explain this relation with the contact geometry and B-fields? – Chris Gerig Oct 10 '12 at 17:42
  • Just for the record, this was crossposted at MO as well: http://mathoverflow.net/questions/109237/non-vanishing-magnetic-helicty-density – Neal Oct 14 '12 at 00:16

2 Answers2

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I believe that the answer, in generality, is that you cannot find $\mathbf s$. As an example, consider a field configuration that is translationally symmetric in, say, the $z$ direction (for example, if there is a smooth current density that directed parallel to the $z$ axis). Speaking in physics terms, if we assume that the helicity is non-zero in some gauge $\mathbf s=\boldsymbol\nabla \phi$, then without loss of generality, we can assume that $H > 0$ everywhere in the domain $D$. However, consider the line integral of $H$ over any closed field loop: $$0<\oint H\frac{\,dl}{|\mathbf B|}=\oint (\mathbf A + \boldsymbol\nabla\phi) \cdot\,d\mathbf l=\iint (\boldsymbol\nabla\times \mathbf A)\cdot\,d\boldsymbol\sigma= \iint \mathbf B\cdot \,d\boldsymbol\sigma=0$$ Here, $\,dl$ is a line measure, and $\,d\boldsymbol\sigma$ is the surface measure. The first integral is well defined since $\mathbf B$ is nowhere vanishing.

Ivan
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  • You should be careful about defining $D$. In your example, $D$ cannot contain the the zeros of the field. If you suppose $D$ is ${(x,y,z)|x^2+y^2>0}$, then you will not be justified in changing your line integral into a surface integral using Stoke's theorem. In particular, suppose you start with a vector potential that is smoothly defined on all of $\mathbb{R}^3$ and then restrict it to this modified $D$. Now change gauge using $\lambda\nabla\phi$, where $\phi$ is the azimuthal angle and $\lambda$ is any constant. Then $\oint A+\lambda d\phi = 2\pi\lambda$. – Josh Burby Oct 13 '12 at 21:49
  • I managed to find a different example, similar to yours in spirit. See my edit. – Josh Burby Oct 14 '12 at 00:06
  • Unless there is a topological defect, the argument stands as it is. Just consider a current density $j(\rho)=\rho\exp(-1/|\rho-a|^2)$ in the domain $\rho>a$. Either way, I don't know of a general argument or condition pro or contrary to your question. – Ivan Oct 18 '12 at 18:51
  • The issue is $\mathbf{s}$ might not be the gradient of a scalar; it only need satisfy $\nabla\times\mathbf{s}=0$, which certainly has non-gradient solutions in your domain $\rho > a$. You won't come to the contradiction in your argument if you take into consideration these non-exact gauge changes. But thanks for your help. – Josh Burby Oct 19 '12 at 00:36
  • The example I gave allows extension to the entire space, which is unique if we fix the boundary conditions at infinity. Furthermore, under the same boundary conditions every $s$ such that $\nabla\times s=0$, has to be a gradient. Again, unless you have a topological defect (which can ruin your boundary conditions), you are pretty safe. – Ivan Oct 23 '12 at 20:14
  • Sorry, I think I'm missing your point. If $j(\rho)=\rho\exp(-1/|\rho-a|^2)$ when $\rho>a$ and you want to extend your example to all of space, then what were you thinking of setting $j(\rho)$ in the part of space with $\rho<a$ for your example to work? It's hard for me to imagine a nowhere vanishing magnetic field defined on all of space arising from a purely z-directed current density. Also, I don't understand why you bring up boundary conditions on $\mathbf{s}$. Is there a good reason that $s$ should satisfy the same boundary conditions as $B$ in the context of my question? – Josh Burby Oct 26 '12 at 06:36
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Since $s$ is closed: \begin{eqnarray}H(A+s) &=& (A+s) \wedge d(A+s)\\ & =& (A+s) \wedge (dA + ds) = (A+s) \wedge dA \\ &=& (A \wedge dA) + (s \wedge dA)\end{eqnarray} If $dA$ is nowhere-vanishing on your domain, $|dA|$ should have a maximum value on the closure $\overline{D}$.

So you should be able to add a sufficiently large constant 1-form, so that $H$ is also nowhere vanishing.

cactus314
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  • You're assuming $dA$ can be continuously extended to $\overline{D}$, which is not necessarily the case. Consider $A = \frac{1}{1-r} d\theta$ on the open unit disc (minus the origin). Then $dA = \frac{1}{(1-r)^2} dr\wedge d\theta$ which cannot be continuously extended to the boundary. (There may be physical considerations which imply every such physically possible $A$ can be extended continuously to the boundary - I have no idea). – Jason DeVito - on hiatus Oct 09 '12 at 17:26
  • (My example works in $\mathbb{R}^2$, not $\mathbb{R}^3$. To fix it, instead of $d\theta$, use $d\omega_{S^2}$, the area form on $S^2$.) – Jason DeVito - on hiatus Oct 09 '12 at 17:33
  • I don't think it's physical. To have magnetic field $H = dA$ be infinite, you must be very close to the magnetic pole. – cactus314 Oct 09 '12 at 18:12
  • There is a more serious issue than extending $dA$. Simply choosing $\mathbf{s}$ to be large and constant will not work in general. This is because $\mathbf{s}\wedge dA$ will typically have zeros for $\mathbf{s}$ constant. This is obvious if you identify $dA$ with a divergence-free vector field $\mathbf{B}$ using the standard volume form $\mu_o$ on $\mathbb{R}^3$. Then $\mathbf{s}\wedge dA=\mathbf{s}(\mathbf{B})\mu_o$. Near one of these zeros, the term $\mathbf{s}\wedge dA$ will not automatically dominate $A\wedge dA$, even if $|dA|$ is takes a maximum value on $D$. – Josh Burby Oct 13 '12 at 21:52