Mesoscale Approximation of the Electromagnetic Fields

Abstract

The point-interaction approximation (or the Foldy–Lax approximation) of the electromagnetic fields generated by a cluster of small-scaled inhomogeneities is derived in the mesoscale (i.e., mesoscopic scale) regime, that is, when the minimum distance $\varvec{\delta }$ between the particles is proportional to their maximum diameter ${\varvec{a}}$ in the form $\varvec{\delta }={\varvec{c}}_r \, {\varvec{a}}$ with a positive constant ${\varvec{c}}_r$ that we call the dilution parameter. The small particles are modeled by anisotropic and variable electric permittivities and/or magnetic permeabilities with possibly complex values. We provide the dominating field (the so-called Foldy–Lax field) with explicit error estimates in terms of the dilution parameter ${\varvec{c}}_r$ uniformly in terms of the distribution of these inhomogeneities. Such approximations are key steps in different research areas as imaging and material sciences.

Appendix

1.1 Green Function Approximations

A simple application of mean value theorem gives

$$\begin{aligned} \begin{aligned} \bigl (\Phi _k(x,y)-\Phi _k({\varvec{z}}_m,y)\bigr )&=\int _{0}^{1}\nabla \Phi _k(tx+(1-t) {\varvec{z}}_m,y){\mathrm{d}}t\circ (x-{\varvec{z}}_m)\\&= O\left( \frac{1}{\varvec{\delta }_{mj}}\left( \frac{1}{\varvec{\delta }_{mj}}+|k|\right) {\varvec{a}}\right) ,\\ \nabla \bigl (\Phi _k(x,y)-\Phi _k({\varvec{z}}_m,y)\bigr )&=\int _{0}^{1}D^2\Phi _k(tx+(1-t) {\varvec{z}}_m,y){\mathrm{d}}t\circ (x-{\varvec{z}}_m)\\&=O\left( \frac{1}{\varvec{\delta }_{mj}}\left( \frac{1}{\varvec{\delta }_{mj}}+|k|\right) ^2{\varvec{a}}\right) ,\\ \nabla \nabla \left( \Phi _k(x,y)-\Phi _k({\varvec{z}}_m,y)\right)&= \int _{0}^{1}D^3\Phi _k(tx+(1-t){\varvec{z}}_m,y){\mathrm{d}}t\circ (x-{\varvec{z}}_m)\\&= O\left( \frac{1}{\varvec{\delta }_{mj}}\left( \frac{1}{\varvec{\delta }_{mj}}+|k|\right) ^3{\varvec{a}}\right) . \end{aligned} \end{aligned}$$

(A.1)

whenever $y\in D_j$ and $j\ne m.$ We also have the following first-order expansion of the Green’s function¹³

$$\begin{aligned}{}[\Phi _k-\Phi _{i\alpha }](x)&=\frac{(ik-\alpha )}{4\pi }\int _{[0,1]} e^{\bigl ((ik)t-(1-t)\alpha \bigr )|x|}~{\mathrm{d}}t, \end{aligned}$$

(A.2)

$$\begin{aligned} \nabla [\Phi _k-\Phi _{i\alpha }](x)&= \Bigr [\frac{(ik-\alpha )}{4\pi } \int _{[0,1]}{e^{(ikt-(1-t)\alpha )|x|}}~\bigl (ikt+(1-t)\alpha \bigr ){\mathrm{d}}t\Bigl ]~\frac{x}{|x|}, \end{aligned}$$

(A.3)

$$\begin{aligned}{}[\otimes \nabla ]^2[\Phi _k-\Phi _{i\alpha }](x)&= \int _{0}^1\frac{e^{(ikt-(1-t)\alpha )|x|} \bigl (ikt+(1-t)\alpha \bigr )}{4\pi ~(ik-\alpha )^{-1} |x|}\nonumber \\&\quad \left[ I+\left( ikt+(1-t)\alpha -\frac{1}{|x|} \right) \frac{[\otimes x]^2}{|x|}\right] {\mathrm{d}}t. \end{aligned}$$

(A.4)

1.2 Counting Lemma

Lemma A.1

For any non-negative function g, we have

$$\begin{aligned} \sum _{{j\ge 1},{j\ne m}}^{{{\aleph }}} g(\varvec{\delta }_{mj})\le 48\sum _{{\mathop {O\le i\le k\le l}\limits ^{1\le l\le {{\aleph }}}}} g\Bigl (\bigl [\bigl (l^2+k^2+i^2\Bigr )^\frac{1}{2}({\varvec{c}}_r+1)-1\bigr ]{\varvec{a}}\Bigr ), \end{aligned}$$

(A.5)

and for any non-negative sequence $(\alpha _j)_{j=1}^{{{\aleph }}}$, we have

$$\begin{aligned} \sum _{m=1}^{{{\aleph }}} \left( \sum _{{\mathop {j\ne m}\limits ^{j\ge 1}} }^{{{\aleph }}}\frac{\alpha _j}{\varvec{\delta }_{mj}^q}\right) ^2\le \left( \frac{{c_0}}{\varvec{\delta }^q}\sum _{l=1}^{{{\aleph }}^\frac{1}{3}}l^{2-q}\right) ^2 \sum _{j=1}^{{{\aleph }}}\alpha _j^2. \end{aligned}$$

(A.6)

where $c_0$ is a positive number.

Proof

We first address the following observation. The worst case (i.e., the maximum number of inhomogeneities around a fixed one, let’s say ${\varvec{z}}_m$) is described by the tessellation of a sphere of diameter $n\varvec{\delta },\,n\in {\mathbb {N}},$ by equilateral triangles of side $\varvec{\delta }+{\varvec{a}},$ where the vertices stand for the position of the inhomogeneities. Therefore, we will only consider a periodically distributed of these inhomogeneities, since the results will differ by a multiplicative constant. The reason is that there exists a bounded homeomorphism, which is a radial projection, between the two configurations. To justify this last statement, let CU(0, r) be a cube of fixed side r centered at the origin whose sides are parallel to the coordinates axis.

For $y=(y_1, y_2, y_3)$, we set $\tau _r(y):=\frac{r}{2|y|_{\infty }}y$ where $|y|_{\infty }:=\sup _{j=1,2,3}|y_j|$. Let B(0, r/2) be the ball of center the origin and radius r/2. It is obvious that $B(0,r/2)\subset CU(0,r)$. In addition, for any y we have $\tau _r(y)\in \partial CU(0,r).$ Indeed, as $\partial CU(0,r)$ is the union of six truncated plans, namely $\{x\in {\mathbb {R}}^3;\,\, x_i=\pm r/2, |x_j|_{j\ne i}\le r/2\}_{i=1,2,3},$ and $|y|_{\infty }=y_i,$ for some $i\in \{1,2,3\}$, then putting

$$\begin{aligned} x_i:=\frac{r}{2}\frac{y_i}{|y|_\infty }=\pm r/2,\,\,x_j:=\frac{r}{2}\frac{y_j}{|y|_\infty } \text { for }j\ne i \end{aligned}$$

guaranties us that $x:=\tau _r(y)\in \cup _{i=1}^3\{x\in {\mathbb {R}}^3 ;\,\, x_i=\pm r/2, |x_j|_{j\ne i}\le r/2\}$.

Let us prove that $\tau : \partial B(0,r/2) \rightarrow \partial CU(0,r)$ is a bounded homeomorphism.

To show the injectivity of $\tau$, let $y^1, y^2$ be any two points such that ${y^1}/{|y^1|_{\infty }}={y^2}/{|y^2|_{\infty }}$ then we have, for $i=1,2,3,$ ${y^1_i}/{|y^1|_\infty }={y^2_i}/{{|y^2|_\infty }},$ squaring and summing the previous identity gives $|y^2|_\infty /{|y^1|_\infty }=1$ which guaranties the injectivity $\tau _r.$

Concerning the surjectivity, it is sufficient to set, for x in the truncated plan $P_i^\pm :=\{x\in {\mathbb {R}}^3 ;\,\, x_i=\pm r/2, |x_j|_{j\ne i}\le r/2\}$,

$$\begin{aligned} y:=\pi (x):=\frac{r}{2|x|_2}x. \end{aligned}$$

Obviously $y \in B(0,r/2)$. Let us fix $x \in P^+_i$ for a certain i. As $|y|_\infty =r/2 |x_i|/|x|_2=r^2/(4|x|_2),$ we have

$$\begin{aligned} \tau (\pi (x))=\tau (y)=\frac{r}{2}\frac{y}{|y|_\infty }=x. \end{aligned}$$

Similar reasoning gives us $\pi \tau (y)=y$ for any $y\in \partial B(0,r/2),$ to conclude $\tau ^{-1}=\pi .$

It remains to show that $\tau$ and $\tau ^{-1}$ are continuous. For this purpose, noticing that for any $y \in \partial B(0,r/2)$ we have $|y|_\infty \ge r/2\sqrt{3}$, we obtain

$$\begin{aligned} |\tau (y^1)-\tau (y^2)|_2&= \frac{r}{2}|\frac{y^1}{|y^1|_\infty }-\frac{y^2}{|y^2|_\infty }|_2\le \frac{r}{2}|\frac{|y^2|_\infty y^1-|y^1|_\infty y^2}{|y^1|_\infty |y^2|_\infty }|_2 \\&\le 6/r\Bigl (|y^2|_\infty | y^1- y^2|_2+|y^2|_2| y^1- y^2|_\infty \Bigr )\\&\le 6/r\Bigl (\frac{r}{2}| y^1- y^2|_2+\frac{r}{2}| y^1- y^2|_2\Bigr ) \end{aligned}$$

which is $|\tau (y^1)-\tau (y^2)|_2\le 3| y^1- y^2|_2.$ With similar computations, for $x^1,x^2\in \partial CU(0,r),$ we get

$$\begin{aligned} |\tau ^{-1} (x^1)-\tau ^{-1}(x^2)|_2=|\pi (x^1)-\pi (x^2)|_2=\le \sqrt{3}| x^1- x^2|_2. \end{aligned}$$

Hence, $\tau : \partial B(0,r/2) \rightarrow \partial CU(0,r)$ is a bounded homeomorphism.

Now let ${\varvec{z}}_i, {\varvec{z}}_j$ be in $\partial B({\varvec{z}}_m,r)$ and set ${\varvec{z}}_j^{\prime }=\tau ({\varvec{z}}_j)$ and ${\varvec{z}}_i^{\prime }=\tau ({\varvec{z}}_i).$ Hence, as we have the continuity of $\tau ^{-1},$

$$\begin{aligned} d({\varvec{z}}_j,{\varvec{z}}_i)= & {} |\tau ^{-1}(\tau ({\varvec{z}}_j))-\tau ^{-1}(\tau ({\varvec{z}}_i))|_ 2\le \sqrt{3}|\tau ({\varvec{z}}_j)-\tau ({\varvec{z}}_i)|_2 =\sqrt{3}d({\varvec{z}}_j^{\prime },{\varvec{z}}_i^{\prime })\\\le & {} 3\sqrt{3} d({\varvec{z}}_j,{\varvec{z}}_i). \end{aligned}$$

Fig. 1

Disposition of the faces $F_l$

Fig. 2

Counting on the square $SQ_k$

From a given position ${\varvec{z}}_m,$ we split the space into equidistant cubes $(CU_l)$, centered at ${\varvec{z}}_m$, such that each of its faces support some of the $({\varvec{z}}_j)_{{j\ge 1,\,}{j\ne m}}^{{{\aleph }}}$ with $d(CU_l,CU_{l+1})=\varvec{\delta }+{\varvec{a}}$.

There is at most $O(\aleph ^\frac{1}{3})$ of such cubes. Indeed, let p denote the number of cubes, for a given $l\in \{1,\ldots ,p\},$ we have

$$\begin{aligned} d(CU_l,{\varvec{z}}_m)=l(\varvec{\delta }+{\varvec{a}}). \end{aligned}$$

As there are six faces on $CU_l,$ we also have a total surface of

$$\begin{aligned} |\partial (CU_l)|:=6(2l(\varvec{\delta }+ {\varvec{a}}))^2. \end{aligned}$$

Now, on the surface of a given cube $CU_l,$ each inhomogeneity occupies a total surface of

$$\begin{aligned} S(B({\varvec{z}}_m,{\varvec{a}}+\varvec{\delta })):=|\partial (CU_l)\cap B({\varvec{z}}_m,{\varvec{a}}+\varvec{\delta })|=\pi ({\varvec{a}}+\varvec{\delta })^2 \end{aligned}$$

hence, each cube may contain ${6 (2l(\varvec{\delta }+{\varvec{a}}))^2}/{\pi ({\varvec{a}}+\varvec{\delta })^2}={24l^2}/{\pi }$ inhomogeneities on its surface. Then,

$$\begin{aligned} ({{\aleph }}-1)=\sum _{{\varvec{z}}_j, j\ne m }1=\sum _{l=1}^p \text {cardinal}\{{\varvec{z}}_j\in CU_l\}\ge \frac{24}{\pi }\sum _{l=1}^p l^2\ge p(p + 1)(2p + 1) \end{aligned}$$

which means that p is at most of the order ${{\aleph }}^{1/3}$.

Now, if $(F_l)_{l=1}^p$ stands, respectively, for one of the faces of $(CU_l)_{l=1}^p,$ chosen to have the same orientation (i.e.,  $d(F_{l\pm 1},F_{l})={\varvec{a}}+\varvec{\delta }$ (see Fig. 1), then, for $z_l$ standing for the orthogonal projection of ${\varvec{z}}_m$ on $F_l,$ the distance from a point ${\varvec{z}}_j\in F_l$ to ${\varvec{z}}_m$ is¹⁴

$$\begin{aligned} \begin{aligned} d({\varvec{z}}_m,{\varvec{z}}_j)&=\sqrt{d({\varvec{z}}_m,F_l)^2+d({\varvec{z}}_j,{\varvec{z}}_l)^2},\\&\quad \text { with } d({\varvec{z}}_m,F_l)=d({\varvec{z}}_m, {\varvec{z}}_l)=l(\varvec{\delta }+{\varvec{a}}). \end{aligned} \end{aligned}$$

(A.7)

Analogously, we split each face $F_l$ with concentric squares $(SQ_k)_{k=1}^l$, centered at ${\varvec{z}}_l$ (the orthogonal projection of ${\varvec{z}}_m$ on $F_l$). There is 4 or at most 8 locations that are equidistant from a given square $SQ_k$ to ${\varvec{z}}_l$ which corresponds to the intersections of a circle and a square sharing the same center, see Fig 2. Similarly, for a point $z_p\in SQ_k$, we get, with $z_k$ standing for the orthogonal projection of ${\varvec{z}}_l$ on one side of $SQ_k,$

$$\begin{aligned} \begin{aligned} d({\varvec{z}}_p,{\varvec{z}}_l)&=\sqrt{(d({\varvec{z}}_p,{\varvec{z}}_k)^2+d(SQ_k,{\varvec{z}}_l)^2)}, \text { with } d(SQ_k,{\varvec{z}}_l)=k(\varvec{\delta }+{\varvec{a}}).\quad \end{aligned}\nonumber \\ \end{aligned}$$

(A.8)

As

$$\begin{aligned} d(B_{{\varvec{z}}_m}^{{\varvec{a}}}, B_{{\varvec{z}}_j}^{{\varvec{a}}})=d({\varvec{z}}_m,{\varvec{z}}_j)-{\varvec{a}}, \end{aligned}$$

and for a non-negative function g, with (A.7), we get

$$\begin{aligned} \sum _{j(\ne m)=1}^{{{\aleph }}} g\bigl (d(B_{{\varvec{z}}_m}^{{\varvec{a}}}, B_{{\varvec{z}}_j}^{{\varvec{a}}}) \Bigr )&=\sum _{j(\ne m)=1}^{{{\aleph }}} g\bigl (d({\varvec{z}}_m,{\varvec{z}}_j)-{\varvec{a}}\bigr )\\&= \sum _{l=1}^{{{\aleph }}^\frac{1}{3}} 6\sum _{{\varvec{z}}_j\in F_l}g\bigl (d({\varvec{z}}_j,{\varvec{z}}_m)-{\varvec{a}}\bigr ),\\&=6\sum _{l=1}^{{{\aleph }}^\frac{1}{3}}\sum _{z_j\in F_l}g\Bigl (\bigl (d(F_l,{\varvec{z}}_m)^2+d({\varvec{z}}_l,{\varvec{z}}_j)^2\Bigr )^\frac{1}{2}-{\varvec{a}}\Bigr )\\&\le 6\sum _{l=1}^{{{\aleph }}^\frac{1}{3}}\sum _{k=1}^l8\sum _{z_p\in SQ_k} g\Bigl (\bigl (l^2(\varvec{\delta }+{\varvec{a}})^2+d({\varvec{z}}_l,{\varvec{z}}_p)^2\Bigr )^\frac{1}{2}-{\varvec{a}}\Bigr ), \end{aligned}$$

with (A.8), we get

$$\begin{aligned}&\sum _{j(\ne m)=1}^{{{\aleph }}} g\bigl (d(B_{{\varvec{z}}_m}^{{\varvec{a}}}, B_{{\varvec{z}}_j}^{{\varvec{a}}})\Bigr )\\&\quad \le 6\times 8\sum _{l=1}^{{{\aleph }}^\frac{1}{3}}\sum _{k=1}^l\sum _{p=1}^k g\Bigl (\bigl (l^2(\varvec{\delta }+{\varvec{a}})^2+d({\varvec{z}}_l,SQ_k)^2+d({\varvec{z}}_p,{\varvec{z}}_k)^2\Bigr )^\frac{1}{2}-{\varvec{a}}\Bigr ),\\&\quad \le 6\times 8\sum _{l\ge 1}^{{{\aleph }}^\frac{1}{3}}\sum _{k=1}^l\sum _{p=1}^k g\Bigl (\bigl (l^2(\varvec{\delta }+{\varvec{a}})^2+k^2(\varvec{\delta }+{\varvec{a}})^2+p^2(\varvec{\delta }+{\varvec{a}})^2\Bigr )^\frac{1}{2}-{\varvec{a}}\Bigr ), \end{aligned}$$

which guaranties that

$$\begin{aligned} \sum _{j(\ne m)=1}^{{{\aleph }}} g\bigl (d(B_{{\varvec{z}}_m}^{{\varvec{a}}}, B_{{\varvec{z}}_j}^{{\varvec{a}}})\bigr ) \le 6\times 8\sum _{{\mathop {1\le p\le k\le l}\limits ^{1\le l\le {{\aleph }}^\frac{1}{3}}}} g\Bigl (\bigl [\bigl (l^2+k^2+p^2\Bigr )^\frac{1}{2}({\varvec{c}}_r+1)-1\bigr ]{\varvec{a}}\Bigr ). \end{aligned}$$

(A.9)

To derive (A.6), we start from (A.9), with $g(x)={1/x^q}$, and get

$$\begin{aligned} \sum _{j(\ne m)=1}^{{{\aleph }}} \frac{1}{\bigl (d(B_{{\varvec{z}}_m}^{{\varvec{a}}}, B_{{\varvec{z}}_j}^{{\varvec{a}}})\bigr )^q}&\le 6\times 8\sum _{{\mathop {1\le p\le k\le l}\limits ^{1\le l\le {{\aleph }}^\frac{1}{3}}}} \frac{1}{\Bigl (\bigl [\bigl (l^2+k^2+p^2\Bigr )^\frac{1}{2}({\varvec{c}}_r+1)-1\bigr ]{\varvec{a}}\Bigr )^q},\\&\le 6\times 8\sum _{{1\le l\le {{\aleph }}^\frac{1}{3}}}\sum _{k=1}^l k \frac{1}{\Bigl (\bigl [\bigl (l^2+k^2\Bigr )^\frac{1}{2}({\varvec{c}}_r+1)-1\bigr ]{\varvec{a}}\Bigr )^q},\\&\le \frac{6\times 8}{{\varvec{a}}^q}\sum _{{1\le l\le {{\aleph }}^\frac{1}{3}}} l^2 \frac{1}{\Bigl (\bigl [l({\varvec{c}}_r+1)-1\bigr ]\Bigr )^q}, \end{aligned}$$

hence

$$\begin{aligned} \sum _{j(\ne m)=1}^{{{\aleph }}}\frac{1}{\varvec{\delta }_{mj}^q}\le \sum _{j(\ne m)=1}^{{{\aleph }}} \frac{1}{\bigl (d(B_{{\varvec{z}}_m}^{{\varvec{a}}}, B_{{\varvec{z}}_j}^{{\varvec{a}}})\bigr )^q}\le \frac{c_0}{({\varvec{c}}_r{\varvec{a}})^q}\sum _{{1\le l\le {{\aleph }}^\frac{1}{3}}} l^2 \frac{1}{l^q}. \end{aligned}$$

(A.10)

Using (A.10) and Hölder’s inequality for the inner sum, we obtain

$$\begin{aligned} \sum _{m=1}^{{{\aleph }}} \Bigl (\sum _{{\mathop {j\ne m}\limits ^{j\ge 1}}}^{{{\aleph }}}\frac{\alpha _j}{\varvec{\delta }_{mj}^q}\Bigr )^2&\le \sum _{m=1}^{{{\aleph }}} \Biggl ( \Biggl [\sum _{{\mathop {j\ne m}\limits ^{j\ge 1}}}^{{{\aleph }}} \Bigl (\frac{\alpha _j}{\varvec{\delta }_{mj}^\frac{q}{2}}\Bigr )^2\Biggr ]^\frac{1}{2} \Biggl [\sum _{{\mathop {j\ne m}\limits ^{j\ge 1}}}^{{{\aleph }}}\Bigl (\frac{1}{\varvec{\delta }_{mj}^\frac{q}{2}}\Bigr )^2\Biggr ]^\frac{1}{2}\Biggr )^2,\\&\le \sum _{m=1}^{{{\aleph }}} \Biggl ( \sum _{{\mathop {j\ne m}\limits ^{j\ge 1}}}^{{{\aleph }}}\Bigl (\frac{\alpha ^2_j}{\varvec{\delta }_{mj}^q}\Bigr )~\sum _{{\mathop {j\ne m}\limits ^{j\ge 1}}}^{{{\aleph }}}\frac{1}{\varvec{\delta }_{mj}^q}\Biggr ),\\&\le \frac{c_0}{\varvec{\delta }^q}\sum _{l= 1}^{{{\aleph }}^\frac{1}{3}}l^{(2-q)} \sum _{m=1}^{{{\aleph }}} \sum _{{\mathop {j\ne m}\limits ^{j\ge 1}}}^{{{\aleph }}}\frac{\alpha ^2_j}{\varvec{\delta }_{mj}^q}. \end{aligned}$$

The proof ends with

$$\begin{aligned} \sum _{m=1}^{{{\aleph }}} \sum _{{j\ge 1},{m\ne j}}^{{{\aleph }}}\frac{\alpha ^2_j}{\varvec{\delta }_{mj}^q} =\sum _{j=1}^{{{\aleph }}} \alpha ^2_j\sum _{{\mathop {m\ne j}\limits ^{m\ge 1}} }\frac{1}{\varvec{\delta }_{mj}^q}. \end{aligned}$$

$\square$