Appendix J. Excess field invariance proofs
https://doi.org/10.48550/arXiv.2403.13981
Eq. (48), Eq. (51), Eq. (52) and Eq. (53), which are expressions for the macroscale interfacial excesses of $\Dnu(x)$ and ${x\,\Delta\nu(x)}$, are the most important results of Sec. 9 and among the most important results of the homogenization theory presented in this work.
On first examination these expressions appear to depend on $x_b$ and on how the mesoscale neighbourhood of $x_b$ is partitioned into microscopic intervals. Since any such dependence would make them ill-defined quantities, it is crucial to the importance and generality of these expressions that all choices of $x_b$ and $\Pi(x_b,\ell)$, which satisfy the conditions stated in Sec. 9.6, give the same values for $\mbsx{\Dnu}_0$ and $\mbsx{\Delta\nu}_1$. This section is devoted to proving that this is indeed the case.
J.1 Derivatives of $\bmone$ and $\bmtwo$ with respect to $x_b$
The derivatives of ${\bmany(x_b)}$ with respect to $x_b$. will be used to demonstrate that the surface excesses calculated in Sec. 9.8.1 and Sec. 9.8.2 are independent of $x_b$. This will demonstrate that $x_b$ is a parameter that determines the values of each term on the right hand sides of Eq. (48) and Eq. (51), but not their sums - $\mbs_0(\mxb)$ and $\mbs_1(\mxb)$, respectively. I will assume that ${\amax/\prectheo}$ is sufficiently small that the kernel average ${\expval{\many;\mu}^*_\prectheo}$ can be replaced with a simple average, i.e., \begin{align*}
\bmany(x_b)
&= \frac{1}{\ell}\;\sum_{m} \many(\bar{x}_m,\Delta_m) \Delta_m \tag{192}
\end{align*}
where $
\many(\bar{x}_m,\Delta_m) \equiv
{\Delta_m^{-1}} \int_{x_m^-}^{x_m^+} \left(x - \bar{x}_m\right)^n \Dnu(x) \dd{x}
$. For convenience I have denoted the left-hand and right-hand boundaries of $\interval_m$ by ${x_m^+\equiv\bar{x}_m+\frac{1}{2}\Delta_m\in\Pi(x_b,\ell)}$ and ${x_m^-\equiv \bar{x}_m-\frac{1}{2}\Delta_m\in\Pi(x_b,\ell)}$, respectively.
The simple average, ${\expval{\Dnu}_\intmax(x_b)}$, fluctuates microscopically and continuously within the range ${\interval(0,\precNu)}$ as $x_b$ changes. However, the definition of ${\expval{\;\cdot\;}^*_\ell}$ stipulates that ${x_b\in\Pi(x_b,\ell)}$ and that the average of $\Delta\nu$ is the same on every microinterval, which means that, in general, ${x_b\pm\ell/2\neq x_{\pm M}}$ and that ${0\leq\abs{\expval{\Dnu}_\prectheo(x_b)-\bmzero}<\precNu}$. I will preserve the constraint ${\bmzero=0}$ as $x_b$ changes, which means that every element of ${\Pi(x_b,\ell)}$, including $x_M$ and $x_{-M}$, changes with $x_b$. Therefore, ${\ell=x_M-x_{-M}}$ also changes. I will assume that ${\nu(x_b)\neq\bnu(x_b)}$, leaving the special case ${\nu(x_b)=\bar{\nu}(x_b)}$ to the interested reader. Using primes to denote total derivatives with respect to $x_b$, I can write
\begin{align*}
\dvone{x_b}\bmany(x_b)
& =
\frac{1}{\ell}\sum_{m} \dvone{x_b}\left[\Delta_m\many(\bar{x}_m,\Delta_m)\right] \\
& -\frac{\ell'}{\ell}\bmany(x_b)
\tag{193}
\end{align*}
where \begin{align*}
\dvone{x_b}\bigg[&\Delta_m\many(\bar{x}_m,\Delta_m)\bigg] \\
& =
\left(\frac{\Delta_m}{2}\right)^n
\left[\dxppr\Dnu(\xppr)
+\left(-1\right)^{n+1}
\dxmpr\Dnu(\xmpr)\right] \\
&\qquad\qquad -n\Delta_m\bar{x}'_m\mathcal{M}^{(n-1)}_\Dnu(\bar{x}_m,\Delta_m) \tag{194}
\end{align*}
I will derive expressions for the derivatives of ${\bmone}$ and ${\bmtwo}$ below. Before doing so, I will simplify this task by deducing a relationship between $\dxmpr$ and $\dxppr$ from the constraints ${\expval{\Dnu}_{\Delta_m}(\bar{x}_m)=\bDnu(x_b)=0\implies \bDnu'(x_b)=0}$. \begin{align*}
\bDnu\,'(x_b) &=
\dvone{x_b}
\left(\frac{1}{\Delta_m}\int_{\xmpr}^{\xppr}\Dnu(x)\dd{x}\right)=0 \\
\implies
-\frac{\Delta'_m}{\Delta_m}&\bDnu(x_b)
+ \frac{1}{\Delta_m}\left[\dxppr\Dnu(\xppr) - \dxmpr\Dnu(\xmpr) \right] =0
\end{align*}
${\bDnu=0}$ means that ${\dxppr\Delta\nu(\xppr) = \dxmpr \Delta\nu(\xmpr)}$. Recursively applying this relationship and using the fact that ${x_0\equiv x_b\implies x'_0=1}$ we find that \begin{align*}
x'_m \Delta\nu(x_m) = \Dnu(x_b), \;\;\forall\;x_m\in\Pi(x_b,\ell)
\tag{195}
\end{align*}
Because a microscopic change of $x_b$ cannot change $\ell$ by more than $\amax$, the second term on the right hand side of Eq. (193) is negligible for our purposes. Therefore, by substituting Eq. (195) into Eq. (196) we find that \begin{align*}
\dvone{x_b}\bmany(x_b)
& =
\frac{\Dnu(x_b)}{\ell}\left[1-\left(-1\right)^n\right]
\sum_{m} \left(\frac{\Delta_m}{2}\right)^n \\
& - \frac{n}{\ell}\sum_m
\Delta_m\bar{x}'_m\mathcal{M}^{(n-1)}_\Dnu(\bar{x}_m,\Delta_m)
\tag{196}
\end{align*}
J.1.1 Case I: $\displaystyle \mathrm{d}\bmone/\mathrm{d}x_b$
For $n=1$, Eq. (194) and Eq. (195) mean that \begin{align*}
\dvone{x_b}&\left[\Delta_m\mone(\bar{x}_m,\Delta_m)\right] \\
& = \frac{\Delta_m}{2} \left[\dxppr\Delta\nu(\xppr) + \dxmpr\Delta\nu(\xmpr)\right]
= \Delta_m \Delta\nu(x_b)
\end{align*}
Subsistuting this and ${\bmzero(x_b)=0}$ into Eq. (196) gives \begin{align*}
\dvone{x_b}\bmone(x_b) & = \Dnu(x_b)
\tag{197}
\end{align*}
This result can be derived at greater length without requiring that ${\bDnu\,'(x_b)=0}$ or ${\bDnu(x_b)=0}$.
J.1.2 Case II: $\displaystyle \mathrm{d}\bmtwo/\mathrm{d}x_b$
Inserting $n=2$ in Eq. (196) and using Eq. (194) and Eq. (195) gives \begin{align*}
\dvone{x_b}\bmtwo(x_b)
=
-\frac{2}{\ell}\sum_m\bar{x}'_md_m
\tag{198}
\end{align*}
where ${d_m \equiv \Delta_m \mone(\bar{x}_m,\Delta_m)}$ is the first moment of $\Dnu$ in $\interval_m$. I define $X_d$ to be the first-moment-weighted average of the interval midpoints, $\bar{x}_m$. \begin{align*}
X_d \equiv \frac{\sum_{m}\bar{x}_m d_m}{\sum_m d_m}
\tag{199}
\end{align*}
In the limit $a/l\to 0$ in an infinite macroscopically-uniform material, $X_d$ coincides both with $x_b$ and with ${\frac{1}{2}\left(x_\mm+x_\m\right)}$. Rearranging Eq. (199) and taking the derivative with respect to $x_b$ gives \begin{align*}
\sum_m\bar{x}'_m d_m = X'_d\sum_{m}d_m + X_d\sum_m d'_m - \sum_m \bar{x}_m d'_m \tag{200}
\end{align*}
Using Eq. (195) and ${\int_{\xmpr}^{\xppr}\Dnu(x)\dd{x}=0}$, the derivative of $d_m$ can be expressed as \begin{align*}
d\,'_m & = \left(\frac{\Delta_m}{2}\right)\left[\dxppr\Dnu(\xppr)+\dxmpr\Dnu(\xmpr)\right] \\
& = \Delta_m\Dnu(x_b) \implies
\sum_m d\,'_m = \ell\, \Delta\nu(x_b) \tag{201}
\end{align*}
The last term on the right hand side of Eq. (200) is \begin{align*}
\sum_m \bar{x}_m d\,'_m & = \Delta \nu(x_b) \sum_m \Delta_m \bar{x}_m
\approx
\Delta \nu(x_b) \int_{x_\mm}^{x_\m} \, x \, \dd{x} \\
&= \Delta \nu(x_b) \, \frac{\ell}{2} \, \left(x_\mm+x_\m\right) \tag{202}
\end{align*}
Eq. (198), Eq. (199), Eq. (200), Eq. (201), and Eq. (202) can be combined with ${\sum_m d_m = \ell \bmone(x_b)}$ to show that \begin{align*}
\dvone{x_b}\bmtwo(x_b)
& =
2\,\Dnu(x_b)\left[\left(\frac{x_\mm+x_\m}{2}\right)-X_d\right] \\
& - 2\,X'_d\,\bmone(x_b)
\end{align*}
Now, because ${\frac{1}{2}(x_\mm+x_\m)}$ and ${X_d}$ both get closer $x_b$ as $l$ increases, their difference vanishes and $X'_d$ becomes one in the limit ${\amax/l\to 0}$. Therefore, in this limit, \begin{align*}
\dvone{x_b}\bmtwo(x_b) = - 2\bmone(x_b)
\tag{203}
\end{align*}
To derive Eq. (197) and Eq. (203) we assumed that $a/l$ could be brought to zero without straying into regions having different mesoscale averages $\bar{\nu}$. When this assumption is not valid, some of the terms that were discarded should be considered more carefully.
J.1.3 ${\mathrm{d}\mbs_0/\mathrm{d}\mx}$ and ${\mathrm{d}\mbs_1/\mathrm{d}\mx}$
It is straightforward to use Eq. (48), Eq. (51), Eq. (197), and Eq. (203) to show that
\begin{align*}
\dv{\mbs_0}{x_b} & = 0 \tag{205}\\
\dv{\mbs_1}{x_b} & = \bmone(x_b)\tag{206}
\end{align*}
J.2 Macroscopic moment densities are independent of the choice of microscopic intervals
It is important to demonstrate that, in the limit $a/l\to 0$, our results do not depend on the choice of the set of microintervals, $\Pi(x_b,\ell)$, that partition the space around $x_b$. In this section it is demonstrated that all sets of points which satisfy the requirements explained in Sec. 9.6 and and Sec. 9.7 give the same values of $\nmomone{x_b}$ and $\nmomtwo{x_b}$ and therefore the same values of $\mbs_0$ and $\mbs_1$.
J.2.1 $\nmomone{x_b}$
Let us assume that we are in the limit ${a/l\to 0}$ and that for a particular choice, $\Pi_1(x_b,\ell)$, of the set of microinterval boundary points, we find \begin{align*}
\nmomone{x_b,\Pi_1} = \frac{1}{\ell}\sum_{m} \int_{\xmpr}^{\xppr}\,x\,\Dnu(x)\dd{x}
\end{align*}
where ${\expval{\Dnu}_{\Delta_m}(\bar{x}_m)=0}$ has allowed each integrand ${(x-\bar{x}_m)\Dnu(x)}$ to be simplified to ${x\Dnu(x)}$.
Now suppose that a new set $\Pi_2(x_b,\ell_s)$ of boundary points $s_m\equiv x_m+\delta x_m$ is formed by changing every point $x_m\in\Pi_1(x_b,\ell)$, except $x_0=x_b$, by an amount $\delta x_m$, such that the average of $\Dnu(x)$ on each of the new microscopic intervals remains equal to ${\bDnu(x_b)=0}$ and such that the ordering of the points does not change ($s_{m+1}>s_m,\;\;\forall\; m$). The new set of microintervals partitions the interval ${[s_\mm,s_\m]}$, where ${s_\m-s_\mm = \ell_s}$, ${\abs{s_\m-x_b-\ell_s/2}<\amax}$, and ${\amax/\ell_s\sim \amax/l \to 0}$. I denote the midpoint, width, left-hand boundary, and right-hand boundary of the new $m^{th}$ interval by ${\bar{s}_m}$, ${\Deltas_m}$, ${\smpr}$, and ${\sppr}$, respectively. By construction, the average of $\Dnu(x)$ on each microinterval is zero. Therefore,
\begin{align*}
\int_{\smpr}^{\sppr}\Dnu(s)\dd{s}&=
\int_{\xmpr}^{\xppr} \Dnu(x)\dd{x} =0 \\
\implies
\int_{\xppr}^{\sppr}\Dnu(x)\dd{x} &= \int_{\xmpr}^{\smpr}\Dnu(x)\dd{x}
\tag{207}
\end{align*}
The new average moment density is \begin{align*}
\nmomone{x_b,\Pi_2}&
= \frac{1}{\ell_s}\sum_m \Bigg[\int_{\xmpr}^{\xppr}x\,\Dnu(x)\dd{x} \\
&+ \int_{\xppr}^{\sppr} x \, \Dnu(x) \dd{x}
- \int_{\xmpr}^{\smpr} x \, \Dnu(x) \dd{x}
\Bigg]
\end{align*}
After cancelling terms in the sum, this can be written as \begin{align*}
\nmomone{x_b,\Pi_2}&
= \frac{1}{\ell_s}\sum_{m=-M}^{M-1} \int_{x_m}^{x_{m+1}}x\,\Dnu(x)\dd{x} \\
+\frac{1}{\ell_s}\int_{x_\m}^{s_\m} & x \; \Dnu(x) \dd{x}
+\frac{1}{\ell_s}\int_{s_\mm}^{x_\mm} x \; \Dnu(x) \dd{x}
\tag{208}
\end{align*}
Although the widths of all microintervals defined by ${\Pi_1(x_b,\ell)}$ and ${\Pi_2(x_b,\ell_s)}$ are less than $\amax$, I have not assumed that ${\abs{\bar{s}_m-\bar{x}_m}<\amax}$. If, for example, the width of each new interval was larger than each old interval, i.e., ${0<\Delta_m<\Delta_m^s <\amax}$, this would imply that ${0<s_\m-x_\m\sim Ma \not\ll l}$. Therefore, we cannot immediately dismiss the second and third terms on the right hand side as negligible. However, $\nmomone{x_b,\Pi_1}$ was assumed to be converged with respect to the magnitude of $\ell$. Therefore, without changing its value significantly, I can expand the set $\Pi_1$ to encompass the ranges $[x_\m,s_\m]$ and $[s_\mm,x_\mm]$ by dividing these ranges into microintervals and adding their boundary points to $\Pi_1(x_b,\ell)$ to form a new set ${\Pi_1^{\text{new}}(x_b,\ell^\text{new})\supset\Pi_1(x_b,\ell)}$ containing ${M^\text{new}>M}$ microinterval boundary points on each side of $x_b$, and such that ${0<x_\mm-x_\mmnew<\amax}$, ${0<x_\mnew-x_\m<\amax}$, and ${0<\ell^\text{new}-\ell_s\lesssim\amax}$. Eq. (208) then becomes \begin{align*}
\nmomone{x_b,\Pi_2}&
= \frac{1}{\ell_s}\sum_{m=\mmnew}^{\mnew-1} \int_{x_m}^{x_{m+1}}x\,\Dnu(x)\dd{x} \\
-\frac{1}{\ell_s}&\int_{s_\m}^{x_\mnew} x \; \Dnu(x) \dd{x}
-\frac{1}{\ell_s}\int^{s_\mm}_{x_\mmnew} x \; \Dnu(x) \dd{x} \\
& =
\frac{\ell^\text{new}}{\ell_s}\,
\nmomone{x_b,\Pi_1^\text{new}} + \order{\frac{\amax}{l}}
\end{align*}
Therefore, if $\order{\amax/l}$ terms are neglected, \begin{align*}
\nmomone{x_b,\Pi_2}
&= \nmomone{x_b,\Pi_1}
\tag{209}
\end{align*}
I have assumed that ${s_\m>x_\m}$ and ${s_\mm<x_\mm}$, but a similar procedure can be followed to prove the same result for any other case, such as ${s_\m>x_\m}$ and ${s_\mm>x_\mm}$.
J.2.2 $\nmomtwo{x_b}$
I will now demonstrate that $\nmomtwo{x_b}$ is independent of how the region around $x_b$ is partitioned. I will use the partitions $\Pi_1(x_b,\ell)$, $\Pi_2(x_b,\ell_s)$ and $\Pi_1^{\text{new}}(x_b,\ell_s)$, introduced in the previous section, where $\ell_s>\ell$, and I again assume the limit ${\amax/l\to 0}$, which implies that $\nmomtwo{x_b,\Pi_1}=\nmomtwo{x_b,\Pi_1^{\text{new}}}$. I will show that ${\nmomtwo{x_b,\Pi_1}=\nmomtwo{x_b,\Pi_1^{\text{new}}}=\nmomtwo{x_b,\Pi_1}}$.
\begin{align*}
&\ell_s\nmomtwo{x_b,\Pi_2}
= \sum_{m} \int_{\smpr}^{\sppr}(x-\bar{s}_m)^2 \Dnu(x) \dd{x}
= \sum_{m} \left[\int_{\smpr}^{\sppr}(x-\bar{x}_m)^2 \Dnu(x) \dd{x}
-2\,\left(\bar{s}_m-\bar{x}_m\right)\,\int_{\smpr}^{\sppr}x\,\Dnu(x)\dd{x}\right]
\\
& = \sum_{m} \left[
\int_{\xmpr}^{\xppr}(x-\bar{x}_m)^2 \Dnu(x) \dd{x}
+\int_{\xppr}^{\sppr}x^2 \Dnu(x) \dd{x}
-\int_{\xmpr}^{\smpr}x^2 \Dnu(x) \dd{x}\right.
-2\,\bar{s}_m\,\int_{\smpr}^{\sppr}x\,\Dnu(x)\dd{x}
\left.
+2\,\bar{x}_m\,\int_{\xmpr}^{\xppr}x\,\Dnu(x)\dd{x}
\right] \\
& = \ell\nmomtwo{x_b,\Pi_1}
+ \sum_{m=\m+1}^{\mnew}\int_{\xmpr}^{\xppr}(x-\bar{x}_m)^2 \Dnu(x) \dd{x}
+ \sum_{m=\mmnew}^{\mm-1}\int_{\xmpr}^{\xppr}(x-\bar{x}_m)^2 \Dnu(x) \dd{x}
+2\sum_{m=\m+1}^{\mnew}\bar{x}_m\int_{\xmpr}^{\xppr}x\,\Dnu(x)\dd{x} \\
&+2\sum_{m=\mmnew}^{\mm-1}\bar{x}_m\int_{\xmpr}^{\xppr}x\,\Dnu(x)\dd{x}
-2\,\sum_{m=-M}^M\bar{s}_m\int_{\smpr}^{\sppr}x\,\Dnu(x)\dd{x}
+2\,\sum_{m=-M}^M\bar{x}_m\int_{\xmpr}^{\xppr}x\,\Dnu(x)\dd{x} \\
&=\ell^\text{new}\nmomtwo{x_b,\Pi_1^\text{new}} +2 \sum_{m=\mmnew}^{\mnew}\bar{x}_m \int_{\xmpr}^{\xppr}x\,\Dnu(x)\dd{x}
-2\,\sum_{m=-M}^M\bar{s}_m\int_{\smpr}^{\sppr}x\,\Dnu(x)\dd{x}
\end{align*}
Denoting the dipole weighted mean positions (Eq. (199)) of sets $\Pi_1^{\text{new}}(x_b,\ell_s)$ and $\Pi_2(x_b,\ell_s)$ by $X_d(\Pi_1^{\text{new}})$ and $X_d(\Pi_2)$, respectively, using Eq. (209), and neglecting $\order{\amax/l}$ terms, allows this to be written as
\begin{align*}
& \nmomtwo{x_b,\Pi_2}
= \nmomtwo{x_b,\Pi_1^{\text{new}}} \\
&+ 2 \nmomone{x_b,\Pi_1^{\text{new}}} X_d(\Pi_1^{\text{new}})
- 2 \nmomone{x_b,\Pi_2} X_d(\Pi_2) \\
& = \nmomtwo{x_b,\Pi_1} + 2 \nmomone{x_b,\Pi_1}\left[X_d(\Pi_1)-X_d(\Pi_2)\right]
\end{align*}
In the limit $a/l \to 0$, both $X_d(\Pi_1)$ and $X_d(\Pi_2)$ tend to $x_b$ and so \begin{align*}
\nmomtwo{x_b,\Pi_2} = \nmomtwo{x_b,\Pi_1}
\end{align*}
J.3 Mesoscale averages of $\bmany(x_b)$
In general, both $\bmone(x_b)$ and $\bmtwo(x_b)$ vary microscopically with $x_b$. It can be necessary to know their mesoscale averages over $x_b$, which are denoted by ${\boldmone}$ and ${\boldmtwo}$, respectively. In Sec. 9.8.3 it was argued that idempotency of the mesoscale-averaging operation applied to surface integrals requires both ${\boldmone}$ and ${\boldmtwo}$ to be zero in regions of mesoscale uniformity. In this section I prove that this requirement is satisfied in the ${a/l\to 0}$ limit.
I define ${\ell_1\sim l}$ and ${\ell_2\sim l}$ as the widths of the intervals on which the mesoscale averages of $\many$ and $\bmany$, respectively, are calculated. I consider the average, over all ${u\in\interval(-\ell_2/2,\ell_2/2)}$, of $\bmone(x_b+u)$. As usual, when $\bmone$ is being evaluated at ${x_b+u}$, I partition an interval of width $\ell_2$ centered at a microscopic distance from ${x_b+u}$ into a set of microintervals. I denote the left-hand boundary, right-hand boundary, midpoint, and width of the $m^\text{th}$ microinterval, ${\interval_m(u)}$, by ${\xmpr=\xmpr(u)}$, ${\xppr=\xppr(u)}$, ${\bar{x}_m=\bar{x}_m(u)}$, and ${\Delta_m=\Delta_m(u)}$, respectively. For each value of $u$, the value of ${\ell_1(u)=\ell_1(0)+\Delta\ell_1(u)}$ is chosen such that ${\bDnu(x_b+u)=0}$. It is always possible to choose it such that ${\Delta\ell_1(u)<\amax}$. The values of ${\bar{x}_m}$ and ${\Delta_m}$ are chosen such that ${\expval{\Dnu}_{\Delta_m(u)}(\bar{x}_m(u))=\bDnu(x_b+u)=0}$. To avoid clutter I will only make the dependences of $x_m$, $\xmpr$, $\xppr$, $\Delta_m$, $\bar{x}_m$, and $\ell_1$ on $u$ explicit in my notation when it is necessary for clarity.
The mesoscale average of $\nmomone{x_b}$ is
\begin{align*}
&\expval{\bar{\mathcal{M}}_\nu^{(1)}}_{\ell_2}(x_b)
=\frac{1}{\ell_2}\int_{-\ell_2/2}^{\ell_2/2}\left(\frac{1}{\ell_1}\sum_{m}
\int_{\xmpr}^{\xppr} x \, \Dnu(x) \dd{x}\right) \dd{u} \\
&=\frac{1}{\ell_2}\int_{-\ell_2/2}^{\ell_2/2}\left(\frac{1}{\ell_1}
\int_{-\ell_1/2}^{\ell_1/2} v \, \Dnu(x_b+u+v) \dd{v}\right) \dd{u}
\end{align*}
Because terms of order ${\amax/\ell_1}$ are negligible, we can ignore the dependence of $\ell_1$ on $u$. This allows us to switch the order of the integrations over $u$ and $v$. \begin{align*}
&\expval{\bar{\mathcal{M}}_\nu^{(1)}}_{\ell_2}(x_b) \\
& = \frac{1}{\ell_2\,\ell_1}\int_{-\ell_1/2}^{\ell_1/2}u\;\left(\int_{\ell_2/2}^{-\ell_2/2}
\Delta\nu(x_b+u+v)\dd{v}\right)\dd{u}
\end{align*}
The inner integral vanishes in the limit ${\amax/\ell_2\to 0}$. Therefore, \begin{align*}
\boldmone(\mxb) \equiv \expval{\bmone}_l(x_b) = 0
\end{align*}
Because $\nu_B$ is constant on the microscale, it follows that ${\boldsymbol{\mathcal{M}^{\expval{1}}_{\nu}}}$ is also zero. This is simply a consequence of mesoscale uniformity implying local mesoscale isotropy. By local mesoscale isotropy I mean that, for all $n$,
\begin{align*}
\int_{-\ell/2}^{\ell/2} u^n \,\nu(x+u)\dd{u} =
\int_{-\ell/2}^{\ell/2} u^n \,\nu(x-u)\dd{u} + \order{a/l}
\end{align*}
I now want to prove that the mesoscale average $\boldmtwo$ of $\bmtwo$ is zero, where
\begin{align*}
\nmomtwo{x_b+u}
&= \frac{1}{\ell_1}\sum_{m}
\int_{\xmpr}^{\xppr} \left(x-\bar{x}_m\right)^2 \Dnu(x)\dd{x} \\
&= \frac{1}{\ell_1}\sum_{m} \bigg[\int_{\xmpr}^{\xppr}\left(x-x_b-u\right)^2 \Dnu(x) \dd{x} \\
&-2\left(\bar{x}_m-x_b-u\right)\int_{\xmpr}^{\xppr} x\;\Dnu(x)\dd{x}\bigg] \\
& = \frac{1}{\ell_1} \int_{x_\mm}^{x_\m} \left(x-x_b-u\right)^2 \Delta\nu(x)\dd{x} \\
&-2\left[X_d-(x_b+u)\right]\bmone(x_b+u)
\end{align*}
As defined in Appendix J.1.2, ${X_d=X_d(u)}$ is the first-moment-weighted average of the microinterval midpoints. Its value tends to ${x_b+u}$ in the ${a/\ell_1\to 0}$ limit; therefore, the second term on the right hand side vanishes in this limit. Since $\order{a/l}$ terms can be disregarded, we can assume that ${x_{\pmm}=x_b+u\pm\ell_1/2}$ and express the average over $u$ as \begin{align*}
& \boldmtwo(\mx_b) \\
&=
\frac{1}{\ell_2}\int_{-\ell_2/2}^{\ell_2/2}
\frac{1}{\ell_1}\left(\int_{-\ell_1/2}^{\ell_1/2} v^2 \,
\Delta\nu(x_b+u+v)\dd{v}\right)\dd{u} \\
&=
\frac{1}{\ell_1}\int_{-\ell_1/2}^{\ell_1/2}
v^2 \expval{\Dnu}_{\ell_2}(x_b+v) \dd{v}
\end{align*}
where, to reach the second line from the first, we reverse the order of integration, thereby neglecting the $\order{\amax/\ell_1}$ error made by ignoring the $u-$dependence of $\ell_1$. The value of ${\expval{\Dnu}_{\ell_2}(x_b+v)}$ fluctuates microscopically about zero as $v$ is varied, and its magnitude remains smaller than ${\delta_A^{[\bnu]}/2}$. Therefore, it is possible to choose a value of ${\eta_1<\amax}$ for which ${\int_{-\ell_1/2}^{\ell_1/2+\eta_1}\expval{\Dnu}(x_b+v)\dd{v}=0}$. Therefore, we can add $\eta_1$ to the upper limit of the integral at the expense of a negligible $\order{a/l}$ error. Integrating by parts gives \begin{align*}
\boldmtwo&(\mx_b) \\
& =
-\frac{2}{\ell_1}\int_{-\ell_1/2}^{\ell_1/2+\eta_1}
v \left(\int_{-\ell_1/2}^v \expval{\Dnu}_{\ell_2}(x_b+v') \dd{v'}\right) \dd{v}
\end{align*}
Let ${\eta_2>0}$ be the shortest distance for which ${\int_{-\ell_1/2}^{v-\eta_2}\expval{\Dnu}_{\ell_2}(x_b+v') \dd{v'} = 0}$. Because ${\expval{\Dnu}_{\ell_2}(x_b+v')}$ fluctuates microscopically about zero, ${\eta_2<\amax}$ and ${\abs{\int_{v-\eta_2}^v\expval{\Dnu}_{\ell_2}(x_b+v') \dd{v'}} < \eta_2\,\delta_A^{[\bnu]}/2}$. Therefore, \begin{align*}
\abs{\boldmtwo(\mx_b)}
<
\frac{\amax}{\ell_1}\,\delta_A^{[\bnu]}\,\abs{\int_{-\ell_1/2}^{\ell_1/2+\eta_1}
v \dd{v}}
\sim
\frac{1}{2}\frac{\amax^3}{\ell_2}\delta_B^{[\nu]}
\end{align*}
This vanishes in the ${\amax/\ell_2\to 0}$ limit and so \begin{align*}
\boldmtwo(\mx_b) = 0 + \order{a/l}
\end{align*}
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