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Background: In classical electrodynamics, given the shape of a wire carrying electric current, it is possible to obtain the magnetic field $\mathbf{B}$ via the Biot-savart law. If the wire is a curve $\gamma$ parametrized as $\mathbf{y}(s)$, where $s$ is the arc-length, then

$$ \mathbf{B}(\mathbf{x}) = \beta \int_\gamma \dfrac{ d\mathbf{y}\times(\mathbf{x} - \mathbf{y}) }{|\mathbf{x} - \mathbf{y}|^3} = \beta \int_\gamma ds \dfrac{ \mathbf{y}'(s) \times(\mathbf{x} - \mathbf{y}(s)) }{|\mathbf{x} - \mathbf{y}(s)|^3} \, , $$

where $\beta $ is just a physical constant proportional to the current in the wire.

Another application of the Biot-Savart law is to find the velocity field $\mathbf{v}$ around a bent vortex line in a fluid, in the approximation of incompressible and irrotational fluid flow (i.e. $\nabla \cdot \mathbf{v} =0$ and $\nabla \times \mathbf{v} =0$ almost everywhere) and very-thin diameter of the vortex core. In fact, by demanding that the vorticity of the fluid is concentrated on the vortex core (i.e. it is distributed as a Dirac delta peaked on the vortex core),

$$ \mathbf{w}(\mathbf{x}) = \nabla \times \mathbf{v}(\mathbf{x})= c \int_\gamma ds \, \mathbf{y}'(s)\, \delta( \mathbf{x} - \mathbf{y}(s)) \, , $$

we have that the Helmholtz decomposition and the fact $\delta(\mathbf{y}-\mathbf{x} ) = -\nabla^2 \, (4 \pi |\mathbf{y}-\mathbf{x}|)^{-1}$ tell us that

$$ \mathbf{v}(\mathbf{x})= \frac{c}{4 \pi} \int_\gamma ds \dfrac{ \mathbf{y}'(s) \times(\mathbf{x} - \mathbf{y}(s)) }{|\mathbf{x} - \mathbf{y}(s)|^3}$$

Again, the constant $c$ is just a physical constant that sets the value of the circulation of the field $\mathbf{v}$ around the vortex.

Question: The above construction works in $\mathbb{R}^3$. Imagine now that the wire (or, equivalently, the irrotational vortex) is a curve in the three-dimensional torus $\mathbb{T}^3 = S^1 \times S^1 \times S^1$. How to obtain the equivalent of the Biot-Savart law?

Note on topology: We are changing the base manifold from $\mathbb{R}^3$ to $\mathbb{T}^3 $ but the local differential relations should be unchanged (i.e. the definition of the vorticity 2-form as the external derivative of the velocity 1-form, or the local form of Maxwell equations $dF = J$). The problem is that the Biot-Savart law is non-local, so it is a global problem that "feels" the topology of the manifold. Maybe in the end the question is related to how the Helmholtz decomposition works on a torus.

Important (possible answer): See the following closely related questions in Physics SE: Rotating away a constant gauge field, Periodic boundary conditions for vortex in a square lattice. There is a topological problem in considering Biot-Savart in a torus!

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    $\begingroup$ What do you mean by "the equivalent of the Biot-Savart law", if I may ask? What kind of "equivalent" equality are you looking for in the case of $\mathbb{T}^3$? $\endgroup$ Jun 29, 2020 at 16:20
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    $\begingroup$ Thank you for the comment, I have been a bit sloppy. I mean an integral formula over the curve $\gamma$ that gives you the value of $\mathbf{v}$ in the whole torus, exactly as the BS law gives you the value of $\mathbf{v}$ in the whole $R^3$ space. $\endgroup$
    – Quillo
    Jun 29, 2020 at 16:50
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    $\begingroup$ In the Wikipedia page about Helmholtz decomposition, this is stated for a general bounded domain, you then have an extra boundary term. Is it not what you want? Is your issue the regularity of your domain? Or the regularity of your distributions of charge? $\endgroup$
    – LL 3.14
    Jun 30, 2020 at 18:06
  • $\begingroup$ @LL the issue is to find the field $\mathbf{v}$ on the torus given a curve, just this. The point is how (if possible) to modify the BS law so that it gives the velocity/magnetic field given the vortex/wire. It seems to me that (at least part of) the difficulty is in the fact that the trick of writing the Dirac delta as the laplacian of $1/r $ cannot work on the torus because $1/r$ does not have the periodicity of the torus (i.e. $1/r$ is not a function that naturally lives on $T^3$). $\endgroup$
    – Quillo
    Jun 30, 2020 at 23:26
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    $\begingroup$ Oh, I see, from the comments below, that your problem is to have periodic solutions. But I think it should diverge then, unless perhaps you have some special cancellations? $\endgroup$
    – LL 3.14
    Jul 1, 2020 at 5:00

2 Answers 2

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The Biot-Savart law is essentially a case of a Green's function. If we want to solve a scalar, linear, inhomogeneous PDE of the form $$ Df=g $$ Where $D$ is a linear differential operator, $f$ is the unknown function, and $g$ is a "source" function, we can split this into two steps: first, we may find a family of Green's functions $G$ satisfing $$ DG(x,y)=\delta(x-y) $$ where $\delta$ is a Dirac function, and $D$ is understood to treat $G$ as a function of the first argument only. Then, by using the linearity of $D$, we can "split up" the source $g$ into an integral of delta functions and write the solution in terms of the Green's function. $$ f(x)=\int_Mg(y)G(x,y)dy $$ The upshot is in many cases $G(x,y)$ has a very simple form due to the symmetries of the underlying space. One complication is that this problem is inherently global, and there are often tricky questions of existence and uniqueness. If it's a vector valued PDE, so we can choose a basis $e_i$ and think of $G$ as a "matrix" valued function satisfying $D(G^i_j(x,y)e_i)=\delta(x-y)e_j$.

For you're specific problem, the Biot-Savart law can in principal be generalied by obtaining a Green's function of the magnetostatic PDE $\nabla\times B=J$, $\nabla\cdot B=0$ on the torus, with an appropriate set of boundary conditions. I don't happen to know a closed form for such a Green's function, but using Fourier decomposition it should be possible to find at least a series solution.

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  • $\begingroup$ Yes, the problem is global, otherwise it would be more easy. Using some reasoning in Fourier space could be actually useful, because on the torus you should need only a countable number of Fourier modes. On the torus the Dirac delta is a "Dirac comb" and you just need a countable number of modes. Maybe this helps (thank you for your answer!). $\endgroup$
    – Quillo
    Jul 1, 2020 at 18:58
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    $\begingroup$ The Fourier approach isn't strictly necessary, of course. See this physics SE post for a computation of a similar Green's function in two dimensions. (For the Laplacian rather than curl, but that could be applied here by rewriting the problem in terms of the magnetic vector potential in Coulomb gauge.) $\endgroup$
    – Kajelad
    Jul 1, 2020 at 20:22
  • $\begingroup$ Interesting reference, linked from the physics SE post you mentioned: arxiv.org/abs/math/0608358 $\endgroup$
    – Quillo
    Jul 1, 2020 at 20:38
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I will record some considerations of which I am not 100% sure, hence the community wiki.

I agree that the Maxwell equations remain the same on the torus. (What changes are the boundary conditions, but this is not important here, I think). The Biot-Savart law is the solution to the equations of Maxwell with a filiform source term $\gamma$.

Now, identifying the torus with $[0, 1]^3$ with appropriate identification of the boundary points, we see by compactness that $\gamma$ is the sum of a finite number of filaments $\gamma_j$ that are contained in $(0, 1)^3$. For each one of those, the Biot-Savart law is exactly the same. Thus, the Biot-Savart law is the same for $\gamma$, too.

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  • $\begingroup$ We have a closed curve wrapping the torus and the field $\mathbf{v}$ (or $\mathbf{B}$) should be "periodic". Take the torus $[-1/2,1/2]^3$ and a vortex/wire (0,0,z) with $z$ in [-1/2,1/2]. In $R^3$ the BS law gives you a velocity field that is axially symmetric around the wire and decays as $1/r$, where $r$ is the distance from the z axis. On the torus the field cannot be axially symmetric, otherwise it is impossible to match the boundary conditions (that are "defined on a square"). Only close to the wire it could be similar to the one in $R^3$. $\endgroup$
    – Quillo
    Jun 29, 2020 at 15:53
  • $\begingroup$ Sorry, I hope my comment is clear despite the number of characters is limited. What I am saying is that a straight vortex in $R^3$ has a certain field that is axially symmetric around the vortex. If now we fold this vortex on a torus then the resulting velocity field cannot be axially symmetric: it is like considering many straight vortices in $R^3$ with the periodicity of the torus. $\endgroup$
    – Quillo
    Jun 29, 2020 at 15:56
  • $\begingroup$ So my feeling is that the BS can not be used as in $R^3$. Am I misunderstanding your answer? In any case, thank you for the hint, +1. $\endgroup$
    – Quillo
    Jun 29, 2020 at 16:00
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    $\begingroup$ @Quillo I think you're right. Instead of thinking about a cube with identified sides,maybe you should instead think about periodic solutions on $\mathbb R^3$: in your example you could lift your straight wire to a 2-dimensional lattice of infinite straight lines in $\mathbb R^3$, then try applying Biot-Savart to obtain a periodic field that descends to $\mathbb T^3$. I have a feeling you might end up with some divergent sums, though... $\endgroup$ Jun 30, 2020 at 14:29
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    $\begingroup$ @Quillo: I am sorry, I have nothing more to say. But Anthony is a great guy and he understands this stuff WAY better than I do, so if he says you're right and I'm wrong, that's because it's true $\endgroup$ Jun 30, 2020 at 21:01

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