Spectral Characteristics of Stochastic Motion in the System of Two Interacting Particles

Abstract

We report on the results of analytic and numerical investigations of spectral characteristics of the stochastic (thermal) motion of two charged particles in the anisotropic electric field of a trap. We propose analytic relations for the spectral density of displacements in such a system with homogeneous and inhomogeneous heat sources, including the spectral densities for each particle, as well as for their total and mutual displacements. The resulting relations have been verified by numerical simulation of the problem.

Appendices

APPENDIX A

To find the correlators of velocities and displacements in the system defined by Eqs. (1), (2), we note that the correlators of random force F_b1(2) satisfy equations

$$\langle {{F}_{{b1}}}\rangle = \langle {{F}_{{b2}}}\rangle \equiv 0,\quad \langle {{F}_{{b1}}}{{F}_{{b2}}}\rangle = 0,$$

$$\langle {{F}_{{b1}}}{{V}_{2}}\rangle = \langle {{F}_{{b2}}}{{V}_{1}}\rangle \equiv 0,\quad \langle {{F}_{{b1}}}{{\xi }_{2}}\rangle = \langle {{F}_{{b2}}}{{\xi }_{1}}\rangle \equiv 0,$$

$$\langle {{F}_{{b1}}}{{\xi }_{1}}\rangle = \langle {{F}_{{b2}}}{{\xi }_{2}}\rangle \equiv 0,\quad \langle {{F}_{{b1}}}{{V}_{2}}\rangle = \langle {{F}_{{b2}}}{{V}_{1}}\rangle \equiv 0,$$

where V₁₍₂₎ = dξ₁₍₂₎/dt are the velocities of particles per degree of freedom. (Here and below in Appendix A, angle brackets 〈 〉 indicate averaging over time for t → ∞.) Considering that 〈ξ₁V₁〉 = 〈ξ₂V₂〉 ≡ 0 and 〈V₁₍₂₎F_b1(2)〉 = ν$T_{{1(2)}}^{0}$ for motion of particles in closed trajectories, where $T_{{1(2)}}^{0}$ is the temperature of heat sources, we can write the equations for particle velocity and displacement correlators in form [9, 10]

$$ - \frac{{\nu \delta {{T}_{{1(2)}}}}}{M} + b\langle {{V}_{{1(2)}}}{{\xi }_{{2(1)}}}\rangle = 0,$$

(A.1)

$$ - a\langle {{({{\xi }_{{1(2)}}})}^{2}}\rangle + b\langle {{\xi }_{1}}{{\xi }_{2}}\rangle + \frac{{{{T}_{{1(2)}}}}}{M} = 0,$$

(A.2)

$$ - \nu \langle {{V}_{{1(2)}}}{{\xi }_{{2(1)}}}\rangle - a\langle {{\xi }_{1}}{{\xi }_{2}}\rangle + b\langle {{({{\xi }_{{2(1)}}})}^{2}}\rangle + \langle {{V}_{1}}{{V}_{2}}\rangle = 0,$$

(A.3)

$$ - 2\nu \langle {{V}_{1}}{{V}_{2}}\rangle - a\langle {{\xi }_{1}}{{V}_{2}}\rangle - a\langle {{\xi }_{2}}{{V}_{1}}\rangle = 0.$$

(A.4)

Here, δT₁₍₂₎ = $T_{{1(2)}}^{0}$ – T₁₍₂₎ and T₁₍₂₎ = M〈$V_{{1(2)}}^{2}$〉 is the doubled kinetic energy of stochastic motion of particles.

Solving Eqs. (A.1)—(A.4), we obtain the energy valance equation in system (1a), (1b); this balance sets in at $T_{1}^{0}$ ≠ $T_{2}^{0}$ and leads to a redistribution of the stochastic kinetic energy between particles [9, 10]:

$$\delta {{T}_{{1(2)}}} - \frac{{{{b}^{2}}(T_{1}^{0} - T_{2}^{0})}}{{2{{b}^{2}} + {{\nu }^{2}}a}},$$

(A.5)

as well as to the following relations for the correlators of velocities and displacements of particles:

$$\langle {{V}_{1}}{{V}_{2}}\rangle = 0,$$

(A.6)

$$\langle \xi _{{1(2)}}^{2}\rangle = \frac{{(2{{a}^{2}} - {{b}^{2}}){{T}_{{1(2)}}} + {{b}^{2}}{{T}_{{2(1)}}}}}{{2a({{a}^{2}} - {{b}^{2}})M}},$$

(A.7)

$$\langle {{\xi }_{1}}{{\xi }_{2}}\rangle = \frac{{ab({{T}_{1}} + {{T}_{2}})}}{{2a({{a}^{2}} - {{b}^{2}})M}},$$

(A.8)

$$\langle {{\xi }_{2}}{{V}_{1}}\rangle \equiv - \langle {{\xi }_{1}}{{V}_{2}}\rangle = - \frac{{b({{T}_{1}} - {{T}_{2}})}}{{2\nu aM}}.$$

(A.9)

APPENDIX B

Any solution F(t) to problem (1), (2) can be written in the form of superposition

$$F(t) = {{C}_{0}} + \sum\limits_{i = 1}^4 {{{C}_{i}}\exp ({{\lambda }_{i}}t).} $$

(B.1)

In this case, quantity C₀ ≡ A₁ = 2〈$\xi _{{1(2)}}^{2}$〉 for function 〈$\Delta _{{1(2)}}^{2}$(t)〉 (see formula (A.7) in Appendix A and [23]). To determine coefficients C_i (i = 1, 2, 3, 4), we can use the initial conditions of the problem: F(0) = 0, dF(0)/dt = 0, d²F(0)/dt² ≡ A₂ = 2T₁₍₂₎/M, d³F(0)/dt³ ≡ A₃ = 2ν$T_{{1(2)}}^{0}$/M [4]. Then we can write the following system of equations for coefficients C_i:

$$\sum\limits_{i = 1}^4 {{{C}_{i}} = - {{A}_{1}},} $$

(B.2)

$$\sum\limits_{i = 1}^4 {{{\lambda }_{i}}{{C}_{i}} = 0,} $$

(B.3)

$$\sum\limits_{i = 1}^4 {\lambda _{i}^{2}{{C}_{i}} = {{A}_{2}},} $$

(B.4)

$$\sum\limits_{i = 1}^4 {\lambda _{i}^{3}{{C}_{i}} = - {{A}_{3}}.} $$

(B.5)

This gives

$${{C}_{1}} = - \frac{{{{A}_{3}} + {{A}_{2}}({{\lambda }_{2}} + {{\lambda }_{3}} + {{\lambda }_{4}}) - {{A}_{1}}{{\lambda }_{2}}{{\lambda }_{3}}{{\lambda }_{4}}}}{{({{\lambda }_{1}} - {{\lambda }_{2}})({{\lambda }_{1}} - {{\lambda }_{3}})({{\lambda }_{1}} - {{\lambda }_{4}})}},$$

(B.6)

$${{C}_{2}} = \frac{{{{A}_{3}} + {{A}_{2}}({{\lambda }_{1}} + {{\lambda }_{3}} + {{\lambda }_{4}}) - {{A}_{1}}{{\lambda }_{1}}{{\lambda }_{3}}{{\lambda }_{4}}}}{{({{\lambda }_{1}} - {{\lambda }_{2}})({{\lambda }_{2}} - {{\lambda }_{3}})({{\lambda }_{2}} - {{\lambda }_{4}})}},$$

(B.7)

$${{C}_{3}} = - \frac{{{{A}_{3}} + {{A}_{2}}({{\lambda }_{1}} + {{\lambda }_{2}} + {{\lambda }_{4}}) - {{A}_{1}}{{\lambda }_{1}}{{\lambda }_{2}}{{\lambda }_{4}}}}{{({{\lambda }_{1}} - {{\lambda }_{3}})({{\lambda }_{2}} - {{\lambda }_{3}})({{\lambda }_{3}} - {{\lambda }_{4}})}},$$

(B.8)

$${{C}_{4}} = \frac{{{{A}_{3}} + {{A}_{2}}({{\lambda }_{1}} + {{\lambda }_{2}} + {{\lambda }_{3}}) - {{A}_{1}}{{\lambda }_{1}}{{\lambda }_{2}}{{\lambda }_{3}}}}{{({{\lambda }_{1}} - {{\lambda }_{4}})({{\lambda }_{2}} - {{\lambda }_{4}})({{\lambda }_{3}} - {{\lambda }_{4}})}}.$$

(B.9)

With account for the roots of the characteristic equation, we can write system (B.6)–(B.9) as

$${{C}_{1}} = - \frac{{{{A}_{3}} - {{A}_{2}}\nu + {{\lambda }_{2}}({{A}_{2}} - {{A}_{1}}\omega _{2}^{2})}}{{2({{\lambda }_{1}} - {{\lambda }_{2}})b}},$$

(B.10)

$${{C}_{2}} = \frac{{{{A}_{3}} - {{A}_{2}}\nu + {{\lambda }_{1}}({{A}_{2}} - {{A}_{1}}\omega _{2}^{2})}}{{2({{\lambda }_{1}} - {{\lambda }_{2}})b}},$$

(B.11)

$${{C}_{3}} = \frac{{{{A}_{3}} - {{A}_{2}}\nu + {{\lambda }_{4}}({{A}_{2}} - {{A}_{1}}\omega _{1}^{2})}}{{2({{\lambda }_{3}} - {{\lambda }_{4}})b}},$$

(B.12)

$${{C}_{4}} = - \frac{{{{A}_{3}} - {{A}_{2}}\nu + {{\lambda }_{3}}({{A}_{2}} - {{A}_{1}}\omega _{1}^{2})}}{{2({{\lambda }_{3}} - {{\lambda }_{4}})b}}.$$

(B.13)

If the particles have the same temperature ($T_{1}^{0}$ = $T_{2}^{0}$ ≡ T), we have T₁₍₂₎ = $T_{{1(2)}}^{0}$. Consequently,

$${{C}_{1}} = - \frac{{2\nu T{{\lambda }_{2}}(1 - a{\text{/}}\omega _{1}^{2})}}{{M({{\lambda }_{1}} - {{\lambda }_{2}})b}},$$

(B.14)

$${{C}_{2}} = \frac{{2\nu T{{\lambda }_{1}}(1 - a{\text{/}}\omega _{1}^{2})}}{{M({{\lambda }_{1}} - {{\lambda }_{2}})b}},$$

(B.15)

$${{C}_{3}} = \frac{{2\nu T{{\lambda }_{4}}(1 - a{\text{/}}\omega _{2}^{2})}}{{M({{\lambda }_{3}} - {{\lambda }_{4}})b}},$$

(B.16)

$${{C}_{4}} = - \frac{{2\nu T{{\lambda }_{3}}(1 - a{\text{/}}\omega _{2}^{2})}}{{M({{\lambda }_{3}} - {{\lambda }_{4}})b}}.$$

(B.17)