Magnetic dipole expansion with an explicit uniform remainder

For each $i$ the field $y\mapsto y_{i}J(y)$ is compactly supported and smooth, and $\mathrm{div}(y_{i}J)=J_i+y_i\mathrm{div}J=J_i$; both $\mathrm{\mid }{y}_{i}J\mathrm{\mid }$ and $J_i$ are integrable, so [F1] gives $\int J_i=0$, that is $\int J=0$; similarly $\mathrm{div}(y_i{y}_{j}J)=y_j{J}_i+y_i{J}_j$ is integrable with compactly supported smooth field $y_i{y}_{j}J$, so [F1] gives $\int (y_j{J}_i+y_i{J}_j)=0$, that is the antisymmetry of $M_{ij}:=\int y_j{J}_i$.

For $t>0$ and $t+h>0$, rationalization gives $\sqrt{t+h}-\sqrt{t}=h\mathrm{/}(\sqrt{t+h}+\sqrt{t})$, so $\mathrm{\mid }\sqrt{t+h}-\sqrt{t}\mathrm{\mid }\le \mathrm{\mid }h\mathrm{\mid }\mathrm{/}\sqrt{t}$ and the difference quotient tends to $1\mathrm{/}(2\sqrt{t})$. Thus $t\mapsto \sqrt{t}$ is differentiable on $(0,\mathrm{\infty })$; repeated product, quotient and chain rules [F5] make it smooth there. With $q(w)={\sum }_i{w}_i^2$ and $f(w)=q(w{)}^{-1\mathrm{/}2}$ on $w\mathrm{\ne }0$, these rules give $Df(w)[u]=-(w\cdot u)\mathrm{\mid }w{\mathrm{\mid }}^{-3}$ and $D^{2}f(w)[u,v]=3(w\cdot u)(w\cdot v)\mathrm{\mid }w{\mathrm{\mid }}^{-5}-(u\cdot v)\mathrm{\mid }w{\mathrm{\mid }}^{-3}$. Differentiating this last expression supplies the third differential used below. Integer powers appearing here are finite products or their reciprocals, so their differentiation uses the product and quotient rules without an additional power-rule assertion.

The moment $m=\frac{1}{2}\int y\times J$ is finite and $M_{ij}=-{\epsilon }_{ijk}{m}_k$: indeed $m_k=-\frac{1}{2}{\epsilon }_{kpq}{M}_{pq}$ by the definition of $M$, and ${\epsilon }_{ijk}{m}_k=-\frac{1}{2}{\epsilon }_{ijk}{\epsilon }_{kpq}{M}_{pq}=-\frac{1}{2}({\delta }_{ip}{\delta }_{jq}-{\delta }_{iq}{\delta }_{jp})M_{pq}=-\frac{1}{2}(M_{ij}-M_{ji})=-M_{ij}$ by the antisymmetry of step 1.1 and the $\epsilon$-identity; consequently $(Mx{)}_i={\sum }_j{M}_{ij}{x}_j={\sum }_j(-{\epsilon }_{ijk}{m}_k)x_j=-(x\times m{)}_i=(m\times x{)}_i$, so $Mx=m\times x$.

Apply the Coulomb dipole expansion [F2] to each component $\rho =J_i$: since $Q_i=\int J_i=0$ and $p_i=\int y{J}_i(y)dy$ satisfies $⟨p_i,x⟩=(Mx{)}_i=(m\times x{)}_i$ by step 2.1, and $\mathrm{\mid }{R}_i(x)\mathrm{\mid }\le 16a^2\mathrm{\parallel }{J}_i{\mathrm{\parallel }}_1\mathrm{/}{r}^3$, summing over the three components gives $A(x)=\frac{{\mu }_0}{4\pi }(\frac{m\times x}{r^3}+R(x))$ with $\mathrm{\parallel }R(x){\mathrm{\parallel }}_2\le 48a^2\mathrm{\parallel }J{\mathrm{\parallel }}_1\mathrm{/}{r}^3$, so $A=\frac{{\mu }_0}{4\pi }\frac{m\times x}{r^3}+O({\mu }_0{a}^2\mathrm{\parallel }J{\mathrm{\parallel }}_1\mathrm{/}{r}^3)$ uniformly for $r\ge 2a$ with the absolute constant $12\mathrm{/}\pi$.

The explicit remainder is $h_2(x,y)=1\mathrm{/}\mathrm{\mid }x-y\mathrm{\mid }-1\mathrm{/}r-⟨x,y⟩\mathrm{/}{r}^3$ with $R(x)=\int J(y)h_2(x,y)dy$; applying [F4] twice to $t\mapsto \mathrm{\mid }x-ty{\mathrm{\mid }}^{-1}$ gives $h_2(x,y)={\int }_0^1(1-s)y_j{y}_k{\mathrm{\partial }}_j{\mathrm{\partial }}_k\mathrm{\mid }x-sy{\mathrm{\mid }}^{-1}ds$ (derivatives in $x$), and since $\mathrm{\mid }x-sy\mathrm{\mid }\ge r-s\mathrm{\mid }y\mathrm{\mid }\ge r\mathrm{/}2$ on the support, the third differential of $f(w)=\mathrm{\mid }w{\mathrm{\mid }}^{-1}$ is $D^{3}f(w)[u,v,z]=3[(u\cdot v)(w\cdot z)+(u\cdot z)(w\cdot v)+(v\cdot z)(w\cdot u)]\mathrm{\mid }w{\mathrm{\mid }}^{-5}-15(w\cdot u)(w\cdot v)(w\cdot z)\mathrm{\mid }w{\mathrm{\mid }}^{-7}$. Cauchy-Schwarz bounds its trilinear operator norm by $C_3\mathrm{\mid }w{\mathrm{\mid }}^{-4}$ with the absolute constant $C_3=24$, so $\mathrm{\mid }{\mathrm{\nabla }}_x{D}^{2}f(x-sy)[y,y]\mathrm{\mid }\le C_3\mathrm{\mid }y{\mathrm{\mid }}^2\mathrm{\mid }x-sy{\mathrm{\mid }}^{-4}\le 16C_3{a}^2\mathrm{/}{r}^4$; hence $\mathrm{\mid }{\mathrm{\nabla }}_x{h}_2\mathrm{\mid }\le 8C_3{a}^2\mathrm{/}{r}^4$ and, because ${\mathrm{curl}}_x(J(y)h_2)={\mathrm{\nabla }}_x{h}_2\times J(y)$, the modulus bound [F5] gives $\mathrm{\parallel }\mathrm{curl}R(x){\mathrm{\parallel }}_2\le 8C_3{a}^2\mathrm{\parallel }J{\mathrm{\parallel }}_1\mathrm{/}{r}^4$.

For the leading term compute with [F3] and the constant field $F=m$, $G=x\mathrm{/}{r}^3$: $\mathrm{div}(x\mathrm{/}{r}^3)=3\mathrm{/}{r}^3-3r^2\mathrm{/}{r}^5=0$ and the Jacobian of $x\mathrm{/}{r}^3$ applied to $m$ is $m\mathrm{/}{r}^3-3x⟨x,m⟩\mathrm{/}{r}^5$ by [F5]; hence $\mathrm{curl}(m\times x\mathrm{/}{r}^3)=\mathrm{curl}(F\times G)=(\mathrm{div}G)m-(DG)m=0-m\mathrm{/}{r}^3+3x⟨x,m⟩\mathrm{/}{r}^5=\frac{3n(m\cdot n)-m}{r^3}$ with $n=x\mathrm{/}r$, because $3x⟨x,m⟩\mathrm{/}{r}^5=3n(m\cdot n)\mathrm{/}{r}^3$; combining with steps 3.1 and 4.1 and differentiating the explicit formula for $A$ under the integral sign [F5] locally on the observation region, where $\mathrm{\mid }x-y\mathrm{\mid }\ge r-a\ge r\mathrm{/}2>0$ uniformly for $y$ in the support gives $B=\mathrm{curl}A=\frac{{\mu }_0}{4\pi }\frac{3n(m\cdot n)-m}{r^3}+O({\mu }_0{a}^2\mathrm{\parallel }J{\mathrm{\parallel }}_1\mathrm{/}{r}^4)$ uniformly in $n=x\mathrm{/}r$ with an absolute constant.

The two $O$ constants depend only on absolute numbers and $\pi$, not on $J$, $a$, $x$ or $r$, and the second was obtained by differentiating the explicit remainder of the first rather than an unspecified $O$ term; Countable Choice is inherited from the Coulomb dipole expansion [F2] and the integrability interfaces of [F1], and no stronger principle is used. This proves both estimates.