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For a chosen reference charge $q$, the separation at which its Coulomb interaction energy equals $k_{B}T$. The charge-$q$ convention is ${\ell }_B=q^2/(4\pi {ϵ}_{\mathrm{abs}}{k}_{B}T)$ in SI, or ${\ell }_B=\beta q^2/ϵ$ in Gaussian units. The commonly tabulated value uses the elementary charge; changing counterion valence requires tracking that reference convention. It sets the strength of counterion condensation.

The positive mobile charge per unit axial length accumulated from cylinder radius $a$ to $r$ is $Q(r)=2\pi {\int }_a^{r}s\rho (s)ds$. For fixed negative line charge $-\lambda$, Gauss's law gives signed radial electric field $E_r=2[-\lambda +Q(r)]/(ϵr)$ in Gaussian units. Its absolute value is the field magnitude. An integral over a local infinite-dilution profile measures condensed charge, rather than every ion in a finite neutral cell.

Electrical resistivity is the inverse of scalar electrical conductivity in an isotropic Ohmic medium. In its rest frame, $\mathbf{E}={\rho }_{\mathrm{el}}\mathbf{j}$, so the electric field does work per unit volume $\mathbf{j}\cdot \mathbf{E}={\rho }_{\mathrm{el}}|\mathbf{j}{|}^2$. In Gaussian units, finite conductivity gives magnetic diffusivity $\eta =c^2/(4\pi \sigma )$; neglecting this diffusion is a distinct approximation from neglecting electron inertia in Hall magnetohydrodynamics.

Rapid circular particle motion perpendicular to a locally uniform magnetic field, superposed on parallel streaming when the parallel velocity is nonzero. In Gaussian units, its signed cyclotron frequency is given in the title and its gyroradius is $v_{\perp }/|{\Omega }_s|$. Slowly varying fields permit a guiding center description.

Hall magnetohydrodynamics retains bulk ion inertia but freezes magnetic flux into the electron velocity ${\mathbf{u}}_e=\mathbf{u}-\mathbf{j}/(en)$. With constant density, negligible electrical resistivity and electron inertia, Gaussian units give $\mathbf{j}=c\nabla \times \mathbf{B}/(4\pi )$ and ${\partial }_t\mathbf{B}=\nabla \times [(\mathbf{u}-c\nabla \times \mathbf{B}/(4\pi en))\times \mathbf{B}]$. The ion velocity still satisfies the magnetohydrodynamic momentum equation. Neglecting that velocity gives electron magnetohydrodynamics, a further approximation rather than a synonym for the complete Hall system.

For a singly charged ion species in Gaussian units, ${\omega }_{pi}=(4\pi n{e}^2/m_i{)}^{1/2}$ is its plasma frequency and $d_i=c\sqrt{m_i/(4\pi n{e}^2)}$. The ion skin depth marks a characteristic scale below which ion and electron motion can separate in a magnetized plasma. Together with the Alfvén speed, it satisfies $v_A{d}_i=c{B}_0/(4\pi en)$, the coefficient of guide-field propagation in electron magnetohydrodynamics.

In Gaussian units, the equation in the title uses the gyrotropic pressure tensor $\mathbf{P}=p_{\perp }\mathbf{I}+(p_{\parallel }-p_{\perp })\mathbf{bb}$. This allows pressure anisotropy. Its magnetic force contains both magnetic pressure and magnetic tension.

The shell-integrated magnetic energy spectrum distributes magnetic energy over wavenumbers. With a solenoidal isotropic spectral tensor normalized to ${\hat{C}}_{ii}=2H(k)$ and the inverse Fourier transform measure $d^{3}k/(2\pi {)}^3$, one has $\mathbb{E}|\mathbf{B}{|}^2={\pi }^{-2}\int k^{2}H(k)dk$. In Gaussian units, $E_B(k)=k^{2}H(k)/(8{\pi }^3)$ integrates to the energy density $\mathbb{E}|\mathbf{B}{|}^2/(8\pi )$. A spectrum per unit mass or normalized to $B^2/2$ has different constant factors. Scale-local fluctuation estimates give $\delta B_l^2$ proportional to $k{E}_B(k)$ at $k\sim l^{-1}$ when the spectrum is normalized consistently.

The curvature vector of a magnetic field line parametrized by arc length. For circular lines, it points towards their center and has magnitude $1/R$. The curvature contribution to magnetic tension is $B^2\mathbf{\kappa }/(4\pi )$ in Gaussian units.

Twice integrating the radial magnetic-field Fokker-Planck equation by parts gives the displayed law when moments and endpoint terms are controlled. In particular $⟨B^2⟩$ grows at rate $10D=2\gamma$, with $\gamma =5{\kappa }_2/4$. The corresponding magnetic energy density is $⟨B^2⟩/(8\pi )$ in Gaussian units.

For $e^{-i\omega t}$, Gaussian units give $\mathbf{ϵ}=\mathbf{I}+4\pi i\mathbf{\sigma }/\omega$, where $\mathbf{\sigma }$ is the summed conductivity tensor. With the equilibrium magnetic field along $z$, write ${ϵ}_{xx}={ϵ}_{yy}=S$, ${ϵ}_{zz}=P$, and ${ϵ}_{xy}=-{ϵ}_{yx}=ig$. For signed cyclotron frequencies ${\Omega }_s=q_{s}B/(m_{s}c)$,Here ${\omega }_{ps}$ is the species plasma frequency. The two circular eigenvalues are $S\mp g$; defining the sign convention before naming their polarizations prevents ambiguity.

Confine the counterions of a uniformly charged infinite cylinder within a coaxial cylinder of radius $R$, with total mobile line charge $\lambda$. The electric field vanishes at $R$, while its surface value at $a$ is fixed by the bare line charge. This gives $a{\varphi }^{\prime }(a)=2\lambda /ϵ$ and $R{\varphi }^{\prime }(R)=0$ in Gaussian units. Sending $R/a\to \infty$ defines the infinite-dilution limit of cylindrical counterions.

Use the source's Gaussian units and neglect the displacement current, as in nonrelativistic magnetohydrodynamics. Write $\Delta p=p_{\perp }-p_{\parallel }$ for the pressure anisotropy. The pressure tensor is $\mathsf{P}=p_{\perp }\mathsf{I}-\Delta p\mathbf{bb}$, so its force density isAmpère's law and the solenoidal magnetic-field constraint give the Lorentz force decompositionAdding these two force densities and using $\mathbf{BB}=B^2\mathbf{bb}$ yieldsThe first term is the gradient of the modified magnetohydrodynamic total pressure; the second is the anisotropic Maxwell stress tensor contribution together with the directional pressure force. Their effective coefficient is the pressure-anisotropy-modified magnetic tension, $\mathcal{T}=B^2/(4\pi )+\Delta p$. At fixed magnetic field, positive pressure anisotropy increases resistance to bending, while negative pressure anisotropy reduces it. When $\mathcal{T}$ becomes negative, bending experiences an amplifying force rather than a restoring force. Gradients of $\Delta p$ can also drive motion, so the curvature interpretation refers specifically to the bending contribution.

Use Gaussian units, plane waves proportional to $e^{i\mathbf{k}\cdot \mathbf{r}-i\omega t}$, and the signed cyclotron frequency ${\Omega }_s=q_s{B}_0/(m_{s}c)$. Write ${\Omega }_e=e{B}_0/(m_{e}c)>0$ for its electron magnitude, so the signed electron value is $-{\Omega }_e$. Define the species plasma frequency by ${\omega }_{ps}^2=4\pi n_{0s}{q}_s^2/m_s$. The linearization of the cold Lorentz force equations givesSolving this linear system and using the species current density ${\mathbf{j}}_s=n_{0s}{q}_s{\mathbf{u}}_s$ gives the conductivity tensorThe magnetized cold-plasma dielectric tensor is $\mathbf{ϵ}=\mathbf{I}+(4\pi i/\omega ){\sum }_s{\mathbf{\sigma }}_s$. ThusAt the stated electron frequency scale, the ion contribution is smaller by $O(Z{m}_e/m_i)$ for $q_i=Ze$ and quasineutrality. To this accuracy,These expressions require negligible collisions and the cold ordering $k{v}_{\mathrm{th},s}\ll |\omega |$; sufficiently close to a cyclotron resonance, thermal broadening or collisions must also be retained.

Use Gaussian units and the kinetic magnetohydrodynamics equationstogether with species continuity equations, quasineutrality and the drift-kinetic equation. Here $p_{\perp },p_{\parallel }$ sum the species pressures. Interpret the thermal speed denominator in the source as the species mass $m_{\alpha }$, so $v_{\mathrm{th}\alpha }^2=2T_{0\alpha }/m_{\alpha }$ and the first adiabatic invariant is $\mu =v_{\perp }^2/(2B)$.

The specified Maxwell-Boltzmann distribution has zero mean motion and isotropic pressure, $p_{\perp 0}=p_{\parallel 0}=p_0$. All equilibrium scalars depend only on $x$, while ${\mathbf{b}}_0\cdot \nabla x=0$ and ${\mathbf{b}}_0\cdot \nabla B_0=0$. Therefore the equilibrium streaming and mirror terms of the drift-kinetic equation vanish. The remaining terms vanish because ${\mathbf{u}}_0=0$ and $E_{\parallel 0}=0$; the parallel electric constraint is also satisfied because the equilibrium parallel pressure gradient is zero. Species continuity equations and the induction equation are stationary, and quasineutrality holds by assumption.

It remains to check the perpendicular force balance. The radial magnetic pressure and magnetic tension giveThis is preciselyThus the Z-pinch is a steady solution of the stated kinetic magnetohydrodynamics system. Its equilibrium electric current is supplied by the small species drifts implicit in this approximation; zero leading common flow does not require zero electric current.

Use Gaussian units, uniform number density $n$, and ion mass density $\rho =n{m}_i$. Here $e>0$ is the magnitude of the elementary charge, as fixed by the printed current relation. Neglect electron inertia and displacement current. ThenFreezing the magnetic flux into this electron velocity, while retaining the ion momentum equation, gives the closed Hall magnetohydrodynamics systemThe pressure enforces the incompressible flow constraint; taking the divergence of momentum determines it up to the usual boundary-dependent constant. The induction equation preserves the solenoidal magnetic-field constraint. One can absorb $B^2/(8\pi )$ into magnetohydrodynamic total pressure, replacing the Lorentz term by magnetic tension. Keeping ion inertia here is important: electron magnetohydrodynamics with fixed ions will be a later limit, not the complete two-field system.

In Gaussian units, the ensemble-averaged magnetic energy density isLet $M_q={\int }_0^{\infty }{B}^{q}FdB$, and assume the relevant moment exists and endpoint terms vanish. Twice applying integration by parts to the conservative magnitude equation givesIn particular, the magnetic stretching moment growth for $q=2$ isThis is a kinematic magnetic dynamo result with prescribed velocity statistics and no Ohmic loss. Once the Lorentz force appreciably changes the flow, this indefinitely exponential kinematic approximation is no longer self-consistent.

Invert the two-dimensional Fourier transform from part (c):In polar coordinates, the angular integral is $2\pi J_0(ks)$, where $J_0$ is a Bessel function of the first kind. The Fourier-Hankel normalization for an isotropic spectrum therefore yieldsThis Hankel inversion of a rotation-measure correlation recovers the scalar spectral tensor coefficient. To obtain the shell-integrated magnetic energy spectrum, take the trace of the three-dimensional tensor:With magnetic energy density $B^2/(8\pi )$ in Gaussian units, the one-dimensional convention ${\int }_0^{\infty }{E}_B(k)dk=\mathbb{E}|\mathbf{B}{|}^2/(8\pi )$ gives $\boxed{E_B(k)=k^{2}H(k)/(8{\pi }^3)}$. A spectrum normalized instead to $\mathbb{E}|\mathbf{B}{|}^2/2$ has coefficient $k^{2}H(k)/(2{\pi }^2)$; the physical normalization must be stated.

Represent a transformation byMultiplication preserves this form and determinant, and the inverse has the same form with parameters $(\overline{u},-v)$. To check the congruence condition, complex conjugation is the identity modulo $2\mathbb{Z}[i]$. Thus $u+v\equiv 1$ says exactly that $M$ fixes the column $(1,1{)}^T$ modulo two. Products and inverses preserve this property. The identity belongs to the set, and changing the common sign of $(u,v)$ does not change either the transformation or its congruence. Hence these transformations form a group.

Because $|u|>|v|$, the denominator has no zero in the disc. Direct calculation givesThe inverse also maps the disc into itself, so $g$ is a disc automorphism, and these identities show that it preserves the hyperbolic metric $2|dz|/(1-|z{|}^2)$.

The action is properly discontinuous. If $C\subset \mathbb{D}$ is compact, put $R={\max }_{z\in C}\rho (0,z)$. If $gC\cap C\ne \varnothing$, choose $z,w\in C$ with $w=gz$. ThenSince $g0=v/\overline{u}$ and $|u{|}^2-|v{|}^2=1$, its distance bound gives $|v/u|\le \tanh R<1$ and therefore $|u{|}^2=1/(1-|v/u{|}^2)\le {\cosh }^{2}R$. There are only finitely many Gaussian integers $u,v$ with these bounds, so only finitely many $g$ satisfy $gC\cap C\ne \varnothing$. This proves the required properly discontinuous group action directly.

If $g0=0$, then $v=0$ and $u$ is one of the Gaussian units $1,-1,i,-i$. The congruence $u-1\in 2\mathbb{Z}[i]$ leaves only $u=\pm 1$. Both give the identity transformation. Thus the point stabilizer of zero is trivial.

The closed Dirichlet region centered at zero is the set of $z$ with $\rho (z,0)\le \rho (z,g0)$ for every $g$. For a nonidentity element, $v\ne 0$. WriteThe distance inequality is equivalent, by pseudohyperbolic distance, to $|z|\le |(z-q)/(1-\overline{q}z)|$. Squaring and subtracting yieldsSince $|z|<1$ and $|q{|}^2=N/(N+1)$, the bisector half-plane is exactlyFor $u=1+i$ or $1-i$ and $v=i$ or $-i$, all the group conditions hold and the products $uv$ are the four numbers $\pm 1\pm i$, with $N=1$. Their four inequalities forceConversely, for any group element $m^2+n^2=|uv{|}^2=N(N+1)<(N+1{)}^2$. As $m,n$ are integers, $|m|,|n|\le N$. Thus every point satisfying those four inequalities satisfies every other one:This proves, with no missing bisectors, the Gaussian-integer disc group with an ideal-square Dirichlet domain formulaEquivalently it is the part of the unit disc outside all four open circles of radius one centered at $1+i,1-i,-1+i,-1-i$. Its sides are the inward arcs of those circles. They meet the unit circle orthogonally, and their ideal vertices are $\boxed{1,i,-1,-i}$; these vertices themselves lie outside the open disc.

The origin has trivial point stabilizer and the group orbit is locally finite by the discontinuity proof, so this closed bisector intersection has translates covering the disc with disjoint interiors, as in the definition of a Dirichlet domain. Side identifications are allowed on its boundary.

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.