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The electric field of this point charge is a translated and scaled version of the radial field in part (ii). If $\mathbf{a}\notin V$ and $\mathbf{a}\notin \partial V$, its divergence vanishes throughout $V$, so the divergence theorem gives zero flux integral.

If $\mathbf{a}$ lies inside $V$, remove a small ball centered at $\mathbf{a}$. On the remaining volume the divergence is zero. Its boundary is $\partial V$ together with the small sphere, whose outward normal for the punctured volume points towards $\mathbf{a}$. That inner sphere contributes $-q/{ϵ}_0$, by the $4\pi$ sphere flux computed in part (ii). Thus the exterior flux must be $q/{ϵ}_0$:This is Gauss's law for a point charge. The assumption that the charge is not on the boundary is essential to these alternatives.

Let the electrolyte permittivity be $\epsilon$ and let ${\kappa }_D=1/{\ell }_D$ be the inverse Debye–Hückel screening length. We reserve $k$ for the imposed lateral wavevector. In the Debye–Hückel approximation, linearization about an electrically neutral bulk electrolyte givesaway from fixed charges. For ionic species of valence $z_i$ and bulk number density $n_i$, ${\kappa }_D^2={\sum }_i{n}_i{z}_i^2{e}^2/(\epsilon k_{B}T)$; the approximation requires $|z_{i}e\varphi |\ll k_{B}T$. Substitution of a lateral cosine into this screened Poisson equation leaves a vertical decay constantThe screened sinusoidal surface-charge mode thus decays faster than a laterally uniform charge mode. A nonzero $k$ retains a finite decay length even in the zero-salt limit.

The exterior media and their boundary conditions are not specified in the paper. Specifying the two surface charge densities alone does not fix the normal derivative on the inside unless the outside response is also fixed. We first take the confined-gap idealization, with the prescribed charge supplying all the displacement flux into the electrolyte between the sheets. The electrostatic interface boundary conditions are thenThis is, for example, the limit of negligible exterior displacement admittance. The electrostatic potential solving these conditions and the Debye–Hückel approximation equation isDifferentiation checks the two surface signs directly. This solution allows a sine as well as a cosine lateral component when the phase is nonzero.

In linear screened electrostatics, the quadratic charging free energy is $\frac{1}{2}\int \sigma \varphi dA$. It is equivalently $\frac{\epsilon }{2}\int [(\nabla \varphi {)}^2+{\kappa }_D^2{\varphi }^2]dV$ with the stated boundary conditions. The second term represents the linearized ionic response; integrating only the bare electric field energy would omit it. If $\mathcal{A}$ denotes membrane area, the wavelength-averaged energy density isUsing $⟨{\cos }^2(kx){⟩}_x=1/2$ and $⟨\cos (kx)\cos (kx+\theta ){⟩}_x=\cos \theta /2$ gives the phase registration of screened charged sheets energyThe positive cosine coefficient shows that, for nonzero charge amplitude and finite separation,At the minimum the electrostatic potential simplifies to $\varphi =\alpha \cos (kx)\sinh (pz)/[\epsilon p\cosh (pd/2)]$. Positive and negative charge patches face opposite signs across the gap. The two charge patterns therefore favor a half-wavelength lateral offset rather than like-charge alignment.

For comparison, if identical electrolyte fills all of space on both sides of infinitesimally thin charge sheets, the electrostatic interface boundary conditions are continuity of $\varphi$, decay at infinity, and the charge-induced derivative jumpsSuperposing the two screened sinusoidal surface-charge modes gives, in particular between the sheets,Its full-space continuation replaces each vertical distance by the corresponding absolute distance. Evaluating the electrostatic potential on both sheets now yieldsThe constant term is the two isolated-sheet self energies, and the cosine term is their screened interaction. The energy depends on the exterior convention, but both stated physical idealizations give the same phase registration of screened charged sheets: unlike charge patches oppose each other. In the full-space convention the optimum interaction energy is $-{\alpha }^2{e}^{-pd}/(4\epsilon p)$, giving an attractive normal force density $-{\alpha }^2{e}^{-pd}/(4\epsilon )$. Phase sensitivity becomes exponentially weak for $pd\gg 1$; if $\alpha =0$ or the separation tends to infinity there is no selected phase.

One can also display how an exterior dielectric response interpolates between these cases. Let $Y_o\ge 0$ be its normal displacement admittance for this lateral Fourier mode: $Y_o=0$ for the confined-gap idealization, $Y_o=\epsilon p$ for identical exterior electrolyte, and $Y_o={\epsilon }_o|k|$ for an ion-free exterior dielectric. DefineThe symmetric and antisymmetric surface-charge combinations giveSince $D_s<D_a$, its cosine coefficient is positive. This makes the phase-minimizing conclusion robust while exposing the boundary information needed for an absolute electrostatic energy.

Use the Gaussian units convention implicit in the $4\pi$ coefficient. For a uniform dielectric and a counterion-only electrolyte, the Boltzmann distribution and Poisson-Boltzmann equation areThe reference electric potential is $\varphi (a)=0$, so $n_0=n(a)$; it is not a prescribed nonzero bulk concentration at infinity. A salt reservoir with additional ion species would require their charge densities as well. With $u=\log (r/a)$, the cylindrical Laplacian is $r^{-2}{d}^2/d{u}^2$. Since $\beta q\varphi =\Phi +2u$ and $r^2=a^2{e}^{2u}$, the factors of $e^{2u}$ cancel:Here primes mean differentiation with respect to $u$, and the surface condition is $\Phi (0)=0$ in this coordinate.

Multiply by ${\Phi }^{\prime }$ to obtain a first integral:The two specified far-field limits set $K=0$. For a solution tending upwards to infinity, take the positive square root. Writing $b=\sqrt{C/2}$ gives $(e^{\Phi /2}{)}^{\prime }=b$, hence the cylindrical counterion-only Poisson-Boltzmann profileFor $b>0$, $\Phi \to \infty$ and ${\Phi }^{\prime }=2b/(1+bu)\to 0$. The corresponding number density is

Apply Gauss's law to a coaxial surface per unit cylinder length. Define the positive accumulated cylindrical screening charge byThe signed outward electric field is $E_r=2[-\lambda +Q(r)]/(ϵr)$, so its magnitude is $2|\lambda -Q(r)|/(ϵr)$. In the branch here, $Q(r)<\lambda$, the field is inward andIn this display, ${\varphi }^{\prime }$ is a radial derivative. The quantities $Q$ are electric charges per unit axial length, not total charges of an infinite cylinder.

At the surface, $a{\varphi }^{\prime }(a)=2\lambda /ϵ$. Define the charge-$q$ Bjerrum length ${\ell }_B=\beta q^2/ϵ$ and the Manning parameter $\eta =\lambda {\ell }_B/q$. Matching the derivative of the solution givesThus the condensed branch exists only for $\eta >1$, and its normalization isThe sign condition $b>0$ is essential: simply squaring $\eta -1$ would produce a spurious subcritical density. Direct integration using $rn(r)dr=n_0{a}^{2}du/(1+bu{)}^2$ yieldsEquivalently, $E(r)=2{\lambda }_c[1+b/(1+b\log (r/a))]/(ϵr)$: the remaining far-field line-charge magnitude is ${\lambda }_c$. The Manning condensed fraction is $1-1/\eta$.

Below threshold, no positive-density profile can satisfy those same far-field conditions. Indeed ${\Phi }^{\prime \prime }\le 0$, so a negative initial slope $2(\eta -1)$ cannot increase to zero. The physical interpretation is the infinite-dilution limit of cylindrical counterions: start with a neutral finite cylindrical cell and send its outer radius to infinity. For $\eta \le 1$, counterions escape to arbitrarily large radii, leaving $n_0\to 0$ and no finite condensed charge. The bare-cylinder Boltzmann distribution $n\propto r^{-2\eta }$ also shows why its radial normalization $\int r^{1-2\eta }dr$ diverges in this regime. In the local zero-density limit, $\Phi =2(\eta -1)u$; this does not obey the condensed branch's imposed limits. At $\eta =1$ the branch has zero amplitude and $\Phi =0$, rather than $\Phi \to \infty$.

Consequently the intended counterion condensation result, interpreted as a local infinite-dilution density, isEvery finite neutral cell still contains enough counterions to neutralize the cylinder; the missing residual charge in the local limiting profile is carried to infinity. Taking the infinite-cell limit before the charge integral differs from integrating the entire finite cell first. In SI units, the same reference-charge Bjerrum length is $q^2/(4\pi {ϵ}_{\mathrm{abs}}{k}_{B}T)$; the electrostatic prefactors must be changed consistently.

Use the Minkowski metric with signature $(-,+,+,+)$. The source point is the intersection of the worldline with the observation event's past light cone: $R^{\mu }{R}_{\mu }=0$ and $R^0>0$. In components this is $c(t-t_r)=|x-y(t_r)|$ with $t_r<t$. For a timelike worldline the retarded solution is unique in the usual complete-source setting.

Writing $R=|x-y(t_r)|$, $n=(x-y)/R$ and $u^{\mu }=\gamma (c,v)$ gives $R\cdot u=-\gamma cR(1-n\cdot v/c)$. Thus the Lienard-Wiechert potentials are $\varphi =q{\mu }_0{c}^2/[4\pi R(1-n\cdot v/c)]$ and $A=q{\mu }_{0}v/[4\pi R(1-n\cdot v/c)]$. At leading order in $1/r$ and first order in $v/c$, set $R\simeq r,n\simeq \hat{r}$ and expand the denominator:

Now use $t_{-}=t-r/c$ and let $a=\dot{v}(t_{-})$. Derivatives of $1/r$ and $\hat{r}$ give near-field terms of order $r^{-2}$; the radiation terms come from ${\partial }_t{t}_{-}=1$ and $\nabla t_{-}=-\hat{r}/c$. The electric field and magnetic field are thereforeThey are transverse to the observation direction. The Poynting vector isIntegrating over a sphere uses $\int {\sin }^2\theta d\Omega =8\pi /3$, yielding the nonrelativistic Larmor formula $P=q^2{\mu }_0|a{|}^2/(6\pi c)$. For simple harmonic motion, $|a{|}^2=A^2{\omega }^4{\cos }^2(\omega t_{-})$, whose period average is $A^2{\omega }^4/2$. Multiplication by the period $2\pi /\omega$ givesThis is the leading radiation energy in the nonrelativistic and far-field approximation.