# Notes: Magnetostatic multipole expansion using STF tensors

$\newcommand{\pd}{\partial} \newcommand{\cd}{\nabla} \newcommand{\bs}{\boldsymbol} \newcommand{\nn}{\nonumber}$

These notes are intended for students (or profs) aware of the multipole expansion for electrostatics in terms of symmetric tracefree (STF) tensors. Standard texts on electrodynamics (like Jackson) hardly mention the STF version, though it is extremely well-known to researchers in GR. After I developed these notes, Julio Parra-Martinez pointed me to a paper by Andreas Ross which implicitly includes these results, though I want to explain a bit more slowly.

# Refresher: Electrostatic STF multipole expansion

Before getting to magnetostatics, we’ll start with electrostatics. This is easier since we only need to solve for the scalar potential, which satisfies

\begin{align} \cd^2 \Phi = - \frac{\rho}{\epsilon_0} . \end{align}

We’re interested in the case where $\rho$ vanishes outside of a compact region. The most efficient way to get to the STF version of the multipole expansion is to start from the Green’s function solution,

\begin{align} \Phi(\bs{x}) = \frac{1}{4\pi\epsilon_0} \int \frac{\rho(\bs{x}')}{|\bs{x}-\bs{x}'|} d^3\bs{x}' . \end{align}

We then take the function $1/|\bs{x}-\bs{x}'|$ and perform a multivariate Taylor series expansion about the point $\bs{x}'=0$, since far away from the source, $|\bs{x}'| \ll |\bs{x}|$. This expansion is

\begin{align} \frac{1}{|\bs{x}-\bs{x}'|} &= \sum_{\ell=0}^\infty \frac{(-1)^\ell}{\ell!} x^{\prime j_1} x^{\prime j_2} \cdots x^{\prime j_\ell} \pd_{j_1} \pd_{j_2} \cdots \pd_{j_\ell} \frac{1}{r} \,, \\ &= \sum_{\ell=0}^\infty \frac{(-1)^\ell}{\ell!} (r')^\ell n'^{j_1} n'^{j_2}\cdots n'^{j_\ell} \pd_{j_1} \pd_{j_2} \cdots \pd_{j_\ell} \frac{1}{r} . \end{align}

In the $\ell$ index tensor $\pd_{j_1} \pd_{j_2} \cdots \pd_{j_\ell} (1/r)$, all indices are obviously symmetric; they are also tracefree away from the origin, where we get a delta function, owing to $\cd^2 (1/r) = -4\pi \delta_{(3)}(\bs{x})$. Since this tensor is symmetric and tracefree (STF), we are free to take only the STF part of the product of $\bs{x}'$ direction vectors. We denote this with angle brackets around the relevant indices,

\begin{align} \frac{1}{|\bs{x}-\bs{x}'|} &= \sum_{\ell=0}^\infty \frac{(-1)^\ell}{\ell!} x'^{\langle j_1} x'^{j_2}\cdots x'^{j_\ell\rangle} \pd_{j_1} \pd_{j_2} \cdots \pd_{j_\ell} \frac{1}{r} . \end{align}

Plugging this in to the Green’s function integral, we get

\begin{align} \Phi(\bs{x}) &= \frac{1}{4\pi\epsilon_{0}} \sum_{\ell=0}^{\infty} \frac{(-1)^\ell}{\ell!} \left( \pd_{j_1} \pd_{j_2} \cdots \pd_{j_\ell} \frac{1}{r} \right) M^{j_{1}j_{2}\cdots j_{\ell}} , \end{align}

where we have defined the $\ell$th STF multipole tensor of the source as

\begin{align} M^{j_{1}j_{2}\cdots j_{\ell}} \equiv \int \rho(\bs{x}) x^{\langle j_1} x^{j_2} \cdots x^{j_\ell \rangle} \ d^{3} \bs{x} . \end{align}

# Magnetostatic multipole expansion

We can apply our results from the electrostatic multipole expansion to magnetostatics, using the potential formulation. In magnetostatics, we are trying to find a magnetic field $\bs{B}(\bs{x})$ satisfying

\begin{align} \cd\times\bs{B} &= \mu_{0} \bs{J}\,, & \text{(static)} \end{align}

where as usual $\cd\cdot\bs{B}=0$, and in statics, conservation of charge demands that $\cd\cdot\bs{J}=0$. Now we go to the potential formulation, $\bs{B}=\cd\times\bs{A}$, and use our gauge freedom to go to Coulomb gauge, $\cd\cdot\bs{A}=0$. Plugging in, we are now trying to solve

\begin{align} \cd^{2} A^{i} = - \mu_{0} J^{i} \,. \end{align}

In Cartesian coordinates, this is just three independent copies of the Poisson equation, one for each Cartesian component $A^{i}$. Therefore we can use the Green’s function for the scalar Laplacian’s for each component,

\begin{align} A^{i}(\bs{x}) = \frac{\mu_{0}}{4\pi} \int \frac{J^{i}(\bs{x}')}{|\bs{x}-\bs{x}'|} d^{3}\bs{x}' \,. \end{align}

Just like in the electrostatic case, we Taylor expand $\tfrac{1}{|\bs{x}-\bs{x}'|}$, pull things out of the integrals, etc. Essentially, we are just making a replacement in the electrostatic case: $\Phi \to A^{i}, \tfrac{\rho}{\epsilon_{0}} \to \mu_{0} J^{i}$. This means our multipole moments get an extra index that does not participate in the STF operation. As it stands, our solution is

\begin{align} \label{eq:A-mpole-external} A^{i}(\bs{x}) &= \frac{\mu_{0}}{4\pi} \sum_{\ell=0}^{\infty} \frac{(-1)^\ell}{\ell!} \left( \pd_{j_1} \pd_{j_2} \cdots \pd_{j_\ell} \frac{1}{r} \right) \mathcal{M}^{i;j_{1}j_{2}\cdots j_{\ell}} \,, \end{align}

where the magnetic multipole moments are defined as

\begin{align} \label{eq:Bstatic-mpole-tensor-def} \mathcal{M}^{i;j_{1}j_{2}\cdots j_{\ell}} \equiv \int J^{i}(\bs{x}) x^{\langle j_1} x^{j_2} \cdots x^{j_\ell \rangle} \ d^{3} \bs{x} \,, \end{align}

which are STF only on the $j_{1}\cdots j_{\ell}$ indices after the semicolon.

Notice that the $\ell=0$ term vanishes – no magnetic monopoles! – by conservation of charge. Integrate $(\cd\cdot\bs{J})x^{j}$ and use integration by parts:

\begin{align} \int (\pd_{i}J^{i})x^{j} \ d^{3} \bs{x} = -\int J^{i}\pd_{i}x^{j} \ d^{3} \bs{x} = -\int J^{i}\delta_{i}^{j} \ d^{3} \bs{x} = - \mathcal{M}^{i} \,. \end{align}

The left hand side vanishes since in magnetostatics, $\cd\cdot\bs{J}=0$. Therefore the $\ell=0$ magnetic monopole moment vanishes, $\mathcal{M}^{i}=0$.

Before handling the arbitrary $\ell$ term, let’s write the dipole in the traditional form seen in e.g. Griffiths. The traditional form for a magnetic dipole is

\begin{align} \bs{A}_{\text{dip}} &= \frac{\mu_{0}}{4\pi} \frac{\bs{m}\times\bs{n}}{r^{2}} \,,\\ A^{i}_{\text{dip}} &= \frac{\mu_{0}}{4\pi} \epsilon^{ijk} m_{j} \pd_{k} \frac{-1}{r} \,. \end{align}

Here the magnetic dipole pseudo-vector is related to the 2-index magnetic dipole tensor,

\begin{align} m^{i} &= \frac{1}{2} \epsilon^{ijk} \mathcal{M}_{k;j} \,, & \mathcal{M}^{k;j} &= \epsilon^{jki} m_{i} \,,\\ \bs{m} &= \frac{1}{2} \int \bs{x} \times \bs{J}(\bs{x}) \ d^{3}\bs{x} \,. \end{align}

This gives the ideal dipole magnetic field

\begin{align} \bs{B}_{\text{dip}} &= \cd\times\bs{A}_{\text{dip}} \,, \\ B^{i}_{\text{dip}} &= \frac{\mu_{0}}{4\pi} m^{j} \pd_{i}\pd_{j}\frac{1}{r} \,, \end{align}

except that we have dropped the singular $\mu_{0}m^{i}\delta_{(3)}(\bs{x})$ term.

It seems like we’ve discarded some information — only the antisymmetric part of $\mathcal{M}^{i;j}$ contributed to $m^{i}$. What about the symmetric part? This exactly vanishes, and that generalizes to all higher $\ell$. The proof follows similarly to why $\mathcal{M}^{i}$ vanished above. Use that $\cd\cdot\bs{J}=0$, and integrate this divergence against $x^{j_{1}}x^{j_{2}}\cdots x^{j_{\ell}}$,

\begin{align} 0 &= -\int (\pd_{i}J^{i})x^{j_{1}}x^{j_{2}}\cdots x^{j_{\ell}} \ d^{3}\bs{x} \\ &= +\int J^{i} \pd_{i} \left( x^{j_{1}}x^{j_{2}}\cdots x^{j_{\ell}} \right) \ d^{3}\bs{x} \\ &= \int J^{i} \left[ \delta_{i}^{j_{1}}x^{j_{2}}\cdots x^{j_{\ell}} + x^{j_{1}}\delta_{i}^{j_{2}}x^{j_{3}}\cdots x^{j_{\ell}} \right. \nn\\ &\qquad\qquad \left. + \ldots + x^{j_{1}}x^{j_{2}}\cdots x^{j_{\ell-1}} \delta_{i}^{j_{\ell}} \right]\ d^{3}\bs{x} \\ &= \ell \int J^{(j_{1}} x^{j_{2}}\cdots x^{j_{\ell})} \ d^{3}\bs{x} \,. \end{align}

What we found is that the completely symmetric part of $\mathcal{M}^{i;j_{1}\cdots j_{\ell-1}}$ vanishes (in fact it vanishes even before removing the traces on the indices after the semicolon).

The next step for understanding these magnetic multipole tensors requires a little knowledge of how Young diagrams classify the index symmetries of tensors (ok, maybe not strictly necessary, but this was how I first realized what to do). We know that the tensor $x^{\langle j_{1}}x^{j_{2}}\cdots x^{j_{\ell}\rangle}$ lives in the representation labeled by the diagram of shape $(\ell)$, Now recall that when we tensor-product a vector with some tensor in a diagram with shape $\lambda$, we generate tensors in irreps related by adding one box at the end of any allowed row or as a new row underneath (the decomposition of tensor products into irreps is determined by the Littlewood–Richardson rule; adding one box where allowed is the simplest case. This is encapsulated in a Hasse diagram called Young’s lattice, which gives a partial order on Young diagrams, seen in here: Now, since $x^{\langle j_{1}}x^{j_{2}}\cdots x^{j_{\ell}\rangle}$ lives in the $(\ell)$ representation, we know that tensoring with $J^{i}$ can produce content in exactly two representations: the $(\ell+1)$ diagram, and the $(\ell,1)$ diagram, having shapes However, above we showed that the completely symmetric part, labeled by $(\ell+1)$, vanishes. Therefore, we have shown that each $\mathcal{M}^{i;j_{1}\cdots j_{\ell}}$ lives in the $(\ell,1)$ diagram, and this means the $i$ index is antisymmetric with each $j$ index.

Because of the antisymmetry between $i$ and any one of the $j$’s, we are free to insert a projector in the space of 2-forms,

\begin{align} \mathcal{M}^{i;j_{1}j_{2}\cdots j_{\ell}} = \delta^{i}_{[k}\delta^{j_{1}}_{p]} \mathcal{M}^{k;pj_{2}\cdots j_{\ell}} = \tfrac{1}{2}\epsilon^{ij_{1}q}\epsilon_{qkp} \mathcal{M}^{k;pj_{2}\cdots j_{\ell}} \,. \end{align}

This motivates defining an auxiliary tensor $m$, like in the dipole case,

\begin{align} \label{eq:mstatic-mpole-and-dual-rels} m^{k j_{2}j_{3}\cdots j_{\ell}} &\equiv -\tfrac{1}{2} \epsilon^{k}{}_{i j_{1}} \mathcal{M}^{i;j_{1}j_{2}\cdots j_{\ell}} \,, & \mathcal{M}^{i;j_{1}j_{2}\cdots j_{\ell}} &= -\epsilon^{ij_{1}}{}_{k} m^{k j_{2}j_{3}\cdots j_{\ell}} \,. \end{align}

The two minus signs (which cancel) are here to agree with the traditional notation for the magnetic dipole vector. We can insert the integral expression,

\begin{align} m^{k j_{2}j_{3}\cdots j_{\ell}} = \int -\tfrac{1}{2} \epsilon^{k}{}_{i j_{1}} J^{i} x^{\langle j_{1}} x^{j_{2}} \cdots x^{j_{\ell}\rangle} d^{3}x \,. \end{align}

What are the symmetries of $m^{kj_{2}\cdots j_{\ell}}$? It is obviously symmetric and tracefree on the $j$’s. It is also easy to see that tracing $k$ with any of the $j$’s would result in a symmetric pair of indices contracting with the $\epsilon$ tensor in the definition of $m$, so by symmetry-antisymmetry, $m^{kj_{2}\cdots j_{\ell}}$ is tracefree on all indices.

Now we will show that $m^{kj_{2}\cdots j_{\ell}}$ is symmetric on $(k,j_{2})$ and thus on $k$ with any of the $j$’s. Suppose we split the tensor into parts that are symmetric and antisymmetric on these two indices, $m^{kj_{2}\cdots j_{\ell}} = m^{(kj_{2})\cdots j_{\ell}} + m^{[kj_{2}]\cdots j_{\ell}}$. For the antisymmetric part, we could again insert a projector in the space of 2-forms. While evaluating this projector, we have the dual on $[k j_{2}]$. But this is simply a trace of $\mathcal{M}$: from Eq. \eqref{eq:mstatic-mpole-and-dual-rels},

\begin{align} \epsilon^{p}{}_{kj_{2}}m^{kj_{2}\cdots j_{\ell}} = \mathcal{M}^{p;j_{1}j_{2}j_{3}\cdots j_{\ell}}\delta_{j_{1}j_{2}} = 0 \,, \end{align}

which vanishes since $\mathcal{M}$ is tracefree on all the $j$’s. Since this antisymmetric part of $m^{kj_{2}\cdots j_{\ell}}$ vanished, we found that $m$ is STF on all indices.

We can finally restate $A^{k}$ and $B^{i}$ in terms of these magnetic STF moments, after a bit of algebra:

\begin{align} A^{k}(\bs{x}) &= \frac{\mu_{0}}{4\pi} \sum_{\ell=0}^{\infty} \frac{(-1)^{\ell}}{\ell!} \left( \pd_{j_1} \pd_{j_2} \cdots \pd_{j_\ell} \frac{1}{r} \right) \epsilon^{kp j_{1}}m^{p j_{2}\cdots j_{\ell}} \,, \\ B^{i} = \epsilon^{ijk}\pd_{j}A_{k} &= \frac{\mu_{0}}{4\pi} \sum_{\ell=0}^{\infty} \frac{(-1)^{\ell+1}}{\ell!} \left( \pd_{i} \pd_{j_{1}}\pd_{j_2} \cdots \pd_{j_\ell} \frac{1}{r} \right) m^{j_{1} j_{2}\cdots j_{\ell}} \,. \end{align}

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