Stability of a Linear Plasma Crystal

Abstract—

The stability of a linear chain of dust particles in an external confining force field is studied. It is assumed that the particles are located in a plasma with Maxwellian electrons with a directed flow of cold ions. It is shown that the aperiodic instability of doubling the chain period can develop in a subsonic flow together with the instability of coupled waves. The region of the crystal stability in the space of external parameters is constructed.

Appendices

DERIVATIVES OF THE POTENTIAL OF THE INTERACTION

Since expression (2) determines a function continuous at $z \approx 0$ and $\rho > 0$, to calculate force constants (4), it suffices to restrict oneself to the limit $z \to + 0$. After integration over ${{k}_{z}}$ with allowance for (3) the potential (2) at $z > 0$ and $\nu \to + 0$ is written in the form (see [14])

$$U(\rho ,z) = \int\limits_0^\infty \,dk\frac{{k{{q}_{1}}(k)}}{{\beta (k)}}{{e}^{{ - {{q}_{1}}(k)z}}}{{J}_{0}}(k\rho ),$$

(A.1)

where $\beta (k)$ = $\sqrt {{{{({{k}^{2}} + {{M}^{2}} - 1)}}^{2}} + 4{{k}^{2}}}$ and ${{q}_{1}}{{(k)}^{2}}$ = $({{k}^{2}} + {{M}^{2}} - 1 + \beta (k)){\text{/}}2$. For the numerical calculations, it is convenient to explicitly extract the Cou-lomb potential in expression (A.1) by rewriting it in the form

$$\begin{gathered} U(\rho ,z) = \frac{1}{{\sqrt {{{\rho }^{2}} + {{z}^{2}}} }} \\ \, + \int\limits_0^\infty \,dk\left\{ {\frac{{k{{q}_{1}}(k)}}{{\beta (k)}}{{e}^{{ - {{q}_{1}}(k)z}}} - {{e}^{{ - kz}}}} \right\}{{J}_{0}}(k\rho ). \\ \end{gathered} $$

(A.2)

The integrand in (A.2) for $z = 0$ decreases as $1{\text{/}}{{k}^{2}}$ at $k \to \infty$, which makes it possible to use it for the numerical calculations of the force constants (4).

In a purely ionic flow at $M = 0$, the potential and its derivatives with respect to z can be calculated explicitly,

$$\begin{gathered} {{u}_{{0,0}}} = \frac{1}{a} - \frac{\pi }{2}({{I}_{0}}(a) - {{{\mathbf{L}}}_{0}}(a)), \\ {{u}_{{0,1}}} = {{K}_{0}}(a), \\ {{u}_{{0,2}}} = - \frac{1}{{{{a}^{3}}}} - \frac{1}{a} + \frac{\pi }{2}({{I}_{0}}(a) - {{{\mathbf{L}}}_{0}}(a)), \\ \end{gathered} $$

(A.3)

where ${{I}_{\nu }}(a)$, ${{K}_{\nu }}(a)$, and ${{{\mathbf{L}}}_{\nu }}(a)$ are modified Bessel and Struve functions (e.g., [18]). At $M > 0$ the integrals are calculated by the standard methods.

HAMILTONIAN FORM OF LINEARIZED EQUATIONS

For a linear chain, the Hamiltonian form of the equation of motion (8) is obtained by a simple change of notation. We denote the horizontal displacements as ${{x}_{1}}(k) = x(k)$ and introduce the corresponding momentum ${{p}_{1}}(k) = \dot {x}(k)$, and for vertical displacements, we swap the places of the coordinate and momentum ${{p}_{2}}(k) = z(k)$ and ${{x}_{2}}(k) = \dot {z}(k)$. Then Eqs. (8) are rewritten in the form

$$\begin{gathered} \mathop {\dot {p}}\nolimits_1 (k) = - {{\Omega }_{x}}{{(k)}^{2}}{{x}_{1}}(k) - ig(k){{p}_{2}}(k) = - \frac{{\delta H}}{{\delta {{x}_{1}}(k){\text{*}}}}, \\ \mathop {\dot {p}}\nolimits_2 (k) = {{x}_{2}}(k) = - \frac{{\delta H}}{{\delta {{x}_{2}}(k){\text{*}}}}, \\ \mathop {\dot {x}}\nolimits_1 (k) = {{p}_{1}}(k) = \frac{{\delta H}}{{\delta {{p}_{1}}(k){\text{*}}}}, \\ \mathop {\dot {x}}\nolimits_2 (k) = - {{\Omega }_{z}}{{(k)}^{2}}{{p}_{2}}(k) - ig(k){{x}_{1}}(k) = \frac{{\delta H}}{{\delta {{p}_{2}}(k){\text{*}}}}, \\ \end{gathered} $$

(B.1)

where the Hamiltonian is

$$\begin{gathered} H = \frac{1}{2}\int {dk} \left\{ {{{{\left| {{{p}_{1}}(k)} \right|}}^{2}} - {{\Omega }_{z}}{{{(k)}}^{2}}{{{\left| {{{p}_{2}}(k)} \right|}}^{2}}} \right. \\ \left. {\, + {{\Omega }_{x}}{{{(k)}}^{2}}{{{\left| {{{x}_{1}}(k)} \right|}}^{2}} - {{{\left| {{{x}_{2}}(k)} \right|}}^{2}} + 2ig(k){{p}_{2}}(k){{x}_{1}}(k){\text{*}}} \right\}. \\ \end{gathered} $$

(B.2)

In the stability region, this expression is a positive definite functional. By calculating the work of external forces on the ensemble of particles (see [19]), it can be shown that Hamiltonian (B.2) cannot be identified with energy. The physical meaning of replacing one of the coordinates with the momentum remains a m-ystery.