Electrostatic multipoles — Examples

Compute the total charge and the dipole moment by inserting the four points into the finite sums of [F1]: $Q=q+q+(-q)+(-q)=0$, and $p=q(a{e}_1)+q(-a{e}_1)+(-q)(a{e}_2)+(-q)(-a{e}_2)=qa(e_1-e_1)+(-q)a(e_2-e_2)=0$, using linearity and the coordinate expansions of [F2], [F3] and [F5].

Show that every odd Cartesian moment vanishes. The configuration is invariant under the central reflection $x\mapsto -x$: the charge $+q$ at $\pm a{e}_1$ is carried to the charge $+q$ at $\mp a{e}_1$, and the charge $-q$ at $\pm a{e}_2$ to the charge $-q$ at $\mp a{e}_2$, so pairing each point with its image and reindexing the finite sum [F5] gives $M_{i_1\dots i_{2k+1}}={\sum }_{j=1}^4{q}_j{\prod }_{r=1}^{2k+1}{x}_{j,i_r}={\sum }_{j=1}^4{q}_j(-x_{j,i_1})\cdots (-x_{j,i_{2k+1}})=-{\sum }_{j=1}^4{q}_j{\prod }_r{x}_{j,i_r}=-M_{i_1\dots i_{2k+1}}$, hence $M_{i_1\dots i_{2k+1}}=0$; the case $k=0$ is the vanishing dipole $p=0$; total charge $Q=0$ was computed separately in step 1.1.

Compute the degree-two Cartesian moment. By [F1] and the coordinates of [F2], $M_{ij}=q(a{e}_1{)}_i(a{e}_1{)}_j+q(-a{e}_1{)}_i(-a{e}_1{)}_j-q(a{e}_2{)}_i(a{e}_2{)}_j-q(-a{e}_2{)}_i(-a{e}_2{)}_j=q{a}^2{e}_{1i}{e}_{1j}+q{a}^2{e}_{1i}{e}_{1j}-q{a}^2{e}_{2i}{e}_{2j}-q{a}^2{e}_{2i}{e}_{2j}=2q{a}^2({\delta }_{i1}{\delta }_{j1}-{\delta }_{i2}{\delta }_{j2})$, which is $\mathrm{diag}(2q{a}^2,-2q{a}^2,0)$ in the frame basis and has vanishing trace $2q{a}^2(1-1)=0$.

Compute the traceless quadrupole. With [F1], $Q_{ij}={\sum }_k{q}_k(3x_{k,i}{x}_{k,j}-\mathrm{\mid }{x}_k{\mathrm{\mid }}^2{\delta }_{ij})=3M_{ij}-{\delta }_{ij}{\sum }_k{q}_k\mathrm{\mid }{x}_k{\mathrm{\mid }}^2$; each of the four points has $\mathrm{\mid }{x}_k\mathrm{\mid }=a$ because $\mathrm{\mid }\pm a{e}_r{\mathrm{\mid }}^2=a^2\delta (e_r,e_r)=a^2$ by [F2], [F3] and [F6], so ${\sum }_k{q}_k\mathrm{\mid }{x}_k{\mathrm{\mid }}^2=a^2(q+q-q-q)=0$ and $Q_{ij}=3M_{ij}=6q{a}^2({\delta }_{i1}{\delta }_{j1}-{\delta }_{i2}{\delta }_{j2})$, whose trace is $6q{a}^2(1-1)=0$, in agreement with the general tracelessness of [F4].

Record the reference-point behaviour and the scope. Since $Q=p=0$ by steps 1.1 and 2.1, clause (d) of the origin-shift theorem [F4] gives $Q_{ij}^{(b)}=Q_{ij}$ for every reference point $b$, so the value $\mathrm{diag}(6q{a}^2,-6q{a}^2,0)$ is independent of the reference point; no field, force, energy or self-energy of the point charges is computed or claimed, and the computation is a finite source-moment identity for the prescribed stationary charges.

Prove (a). Each summand is the point-charge object of [F1] with $x_0=\pm d\mathrm{/}2$ and charge $\pm q$; each is locally integrable and $C^{\mathrm{\infty }}$ off its pole, so the sum $\phi$ is locally integrable and $C^{\mathrm{\infty }}$ off $\{d\mathrm{/}2,-d\mathrm{/}2\}$, with $E=-\mathrm{\nabla }\phi =E_{+q}+E_{-q}$ there. The regular distribution of a locally integrable function is unaffected by changing it on the two-point null set [F12], and distributional differentiation is linear [F4], so adding the two identities of [F1] in ${\mathcal{D}}^{\mathrm{\prime }}({\mathbb{R}}^3)$ and using [F5] gives $\mathrm{div}E=(q{\delta }_{d\mathrm{/}2}-q{\delta }_{-d\mathrm{/}2})\mathrm{/}{\epsilon }_0$ and $\mathrm{curl}E=0$; the dictionary identifying this spatial equation with the fixed-time reading of the source is [F3], and $E$ is the field object defined in The electric and magnetic fields as fixed-time spatial vector fields. The chart, its origin and the constant ${\epsilon }_0$ are those of [F15], and the distributional and Lebesgue readings used here carry the declared Countable Choice and measure interface of [F16].

Prove the moment and quadrupole clauses of (b). Insert the two points into the finite sums of [F2]: $Q=q+(-q)=0$ and $p=q(d\mathrm{/}2)+(-q)(-d\mathrm{/}2)=qd$; for every $l\ge 1$, $M_{i_1\dots i_l}=q{\prod }_{r=1}^l(d\mathrm{/}2{)}_{i_r}+(-q){\prod }_{r=1}^l(-d\mathrm{/}2{)}_{i_r}=q(1-(-1{)}^l){\prod }_{r=1}^l(d\mathrm{/}2{)}_{i_r}$, which vanishes for even $l$ and equals $2q{\prod }_{r=1}^l(d\mathrm{/}2{)}_{i_r}$ for odd $l$, in particular $M_{i_1\dots i_{2k+1}}=2q(d\mathrm{/}2{)}_{i_1}\cdots (d\mathrm{/}2{)}_{i_{2k+1}}$ and $M_{ij}=0$. Each point has $\mathrm{\mid }{x}_k\mathrm{\mid }=\mid \pm d\mathrm{/}2\mid =\frac{1}{2}\mathrm{\mid }d\mathrm{\mid }=a$ by the norm formula of [F13], and ${\sum }_k{q}_k\mathrm{\mid }{x}_k{\mathrm{\mid }}^2=a^2(q+(-q))=0$, so the second formula of [F2] gives $Q_{ij}=3M_{ij}-{\delta }_{ij}{\sum }_k{q}_k\mathrm{\mid }{x}_k{\mathrm{\mid }}^2=0$; the identities used only finite distributivity [F14].

Prove the reference-point clause of (e). With $Q=0$ the transformation of [F6] reads $Q_{ij}^{(b)}=Q_{ij}-3(b_i{p}_j+b_j{p}_i)+2(b\cdot p){\delta }_{ij}$, and substituting $Q_{ij}=0$ and $p=qd$ from step 2.1 gives $Q_{ij}^{(b)}=-3(b_i{p}_j+b_j{p}_i)+2(b\cdot p){\delta }_{ij}$, which vanishes at $b=0$. If $q\mathrm{\ne }0$, then $p=qd\mathrm{\ne }0$; a vanishing shifted quadrupole contracted with $p_j$ gives $-3\mathrm{\mid }p{\mathrm{\mid }}^2{b}_i-(b\cdot p)p_i=0$, and its inner product with $p$ gives $-4\mathrm{\mid }p{\mathrm{\mid }}^2(b\cdot p)=0$, whence $b=0$. If $q=0$ then all moments and the off-pole potential vanish, and the invariance special case $Q=p=0$ holds.

Establish the far-field identity and bound, proving (c). Fix $y\in \{d\mathrm{/}2,-d\mathrm{/}2\}$ and $x$ with $r=\mathrm{\mid }x\mathrm{\mid }\ge 2a$, and put $h_y(s):=\mathrm{\mid }x-sy{\mathrm{\mid }}^{-1}$ for $s\in [0,1]$. For all such $s$ the reverse triangle inequality gives $\mathrm{\mid }x-sy\mathrm{\mid }\ge r-s\mathrm{\mid }y\mathrm{\mid }\ge r-a\ge r\mathrm{/}2>0$ [F9], so $h_y$ is $C^{\mathrm{\infty }}$ on $[0,1]$ with $h_y(0)=1\mathrm{/}r$; the chain rule and the derivative of the power map give $h_y^{\mathrm{\prime }}(s)=\frac{(x-sy)\cdot y}{\mathrm{\mid }x-sy{\mathrm{\mid }}^3}$ and $h_y^{\mathrm{\prime }\mathrm{\prime }}(s)=-\frac{\mathrm{\mid }y{\mathrm{\mid }}^2}{\mathrm{\mid }x-sy{\mathrm{\mid }}^3}+3\frac{((x-sy)\cdot y{)}^2}{\mathrm{\mid }x-sy{\mathrm{\mid }}^5}$, so $\mathrm{\mid }{h}_y^{\mathrm{\prime }\mathrm{\prime }}(s)\mathrm{\mid }\le \frac{4\mathrm{\mid }y{\mathrm{\mid }}^2}{\mathrm{\mid }x-sy{\mathrm{\mid }}^3}\le \frac{32\mathrm{\mid }y{\mathrm{\mid }}^2}{r^3}$ by [F13] [F8, F10]. The function $s\mapsto (1-s)h_y^{\mathrm{\prime }}(s)+h_y(s)$ has derivative $(1-s)h_y^{\mathrm{\prime }\mathrm{\prime }}(s)$, so [F7] gives $h_y(1)-h_y(0)-h_y^{\mathrm{\prime }}(0)={\int }_0^1(1-t)h_y^{\mathrm{\prime }\mathrm{\prime }}(t)dt$, whose modulus is at most $\frac{32\mathrm{\mid }y{\mathrm{\mid }}^2}{r^3}\cdot \frac{1}{2}=\frac{16\mathrm{\mid }y{\mathrm{\mid }}^2}{r^3}$ by the Riemann monotonicity in [F11] applied to $-C(1-t)\le (1-t)h_y^{\mathrm{\prime }\mathrm{\prime }}(t)\le C(1-t)$ with $C=32\mathrm{\mid }y{\mathrm{\mid }}^2\mathrm{/}{r}^3$, and ${\int }_0^1(1-t)dt=1\mathrm{/}2$ from [F7] applied to $t-t^2\mathrm{/}2$; hence $h_y(1)=\frac{1}{r}+\frac{x\cdot y}{r^3}+R_y(x)$ with $\mathrm{\mid }{R}_y(x)\mathrm{\mid }\le \frac{16\mathrm{\mid }y{\mathrm{\mid }}^2}{r^3}=\frac{16a^2}{r^3}$.

Substitute the two kernels into $\phi$, completing (c). Since $\mathrm{\mid }y\mathrm{\mid }=a$ for both points and $q+(-q)=0$ for the constant terms and $q(d\mathrm{/}2)\cdot x+(-q)(-d\mathrm{/}2)\cdot x=qd\cdot x=p\cdot x$ for the linear terms, step 4.1 gives $\phi (x)=\frac{q}{4\pi {\epsilon }_0}(h_{d\mathrm{/}2}(1)-h_{-d\mathrm{/}2}(1))=\frac{p\cdot x}{4\pi {\epsilon }_0{r}^3}+R(x)$ with $R=\frac{q}{4\pi {\epsilon }_0}(R_{d\mathrm{/}2}(x)-R_{-d\mathrm{/}2}(x))$, so $\mathrm{\mid }R(x)\mathrm{\mid }\le \frac{\mathrm{\mid }q\mathrm{\mid }(16a^2+16a^2)}{4\pi {\epsilon }_0{r}^3}=\frac{32\mathrm{\mid }q\mathrm{\mid }{a}^2}{4\pi {\epsilon }_0{r}^3}=\frac{8\mathrm{\mid }q\mathrm{\mid }{a}^2}{\pi {\epsilon }_0{r}^3}$.

Draw the comparison (d) and the scope (e). The theorem of Uniform Coulomb dipole expansion of a compactly supported charge density with an explicit remainder assumes an $L^1$ density. For $q\mathrm{\ne }0$ the pair cannot be represented by even a locally integrable density: take bumps ${\psi }_j$ of [F17], centred at $d\mathrm{/}2$, with radii decreasing to zero and all supports excluding $-d\mathrm{/}2$. Their source pairings equal $q$, whereas for any $g\in L_{\mathrm{l}\mathrm{o}\mathrm{c}}^1$ the integrals $\int g{\psi }_j$ tend to zero by dominated convergence [F18], using $\mathrm{\mid }g\mathrm{\mid }$ on one fixed compact ball as majorant and the a.e. limit ${\psi }_j\to 0$. This is a contradiction. For $q=0$ the source and off-pole potential are zero and the zero density satisfies the density hypothesis; the identity and bound just proved for the two kernels are the direct twice-fundamental-theorem computation of step 4.1, and substituting the total variation $2\mathrm{\mid }q\mathrm{\mid }$ into that theorem's remainder bound $16a^2\mathrm{\parallel }\rho {\mathrm{\parallel }}_1\mathrm{/}(4\pi {\epsilon }_0{r}^3)$ would give exactly $\frac{8\mathrm{\mid }q\mathrm{\mid }{a}^2}{\pi {\epsilon }_0{r}^3}$, in agreement with step 5.1; no self-energy, self-force, dynamics, bound-state, stability or empirical assertion is made for the singular pair, and for $q\mathrm{\ne }0$ the quadrupole vanishes exactly at the symmetric origin, while for $q=0$ it vanishes at every reference point, by step 3.1, which is (e).