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headingE2170: Polarization of two-spheres system inside a tube Submitted by: Ido Moskovich The problem: Given two balls in a very long, hollow tube, with length L. The mass of each ball is m, The charge of one ball is −q and the charge of the other one is +q. The ball’s radius is negligible, and the electrostatic attraction between the balls is also negligible. The balls are rigid and can’t pass through each other. The balls are attached to a drop, whose surface tension causes it’s gravity constant γ to work on the balls toward each other (The force does not depend on the distance between the balls). The system is in an external electric uniform field ε̄ = εx̂ and in thermal equilibrium in temperature T. (a) Write the hamiltonian of the system H (p1 , p2 , x1 , x2 ) = Ek + V (x) when Ek is the kinetic energy. Define properly V (x) when x = x2 − x1 an–d write a diagram of V (x). (b) Calculate the partition function Z (β, ε). (c) Find the probability function of x, ρ (x) and the average distance hxi between the balls. Express again ρ (x) by hxi. (d) Find the polarization p as a function of ε. Use the partition function. (e) Develop P (ε) up to first order in the field: P (ε) = P0 + χε + O ε2 . This development is valid in a weak field, Define what is a weak field. Express your answers with L, m, q, γ, T, ε. The solution: 1 (a) Since we assume L hxi and we derive in the continuing that hxi = β(γ−qε) , this derivation is valid in a weak field with respect to the mechanical attraction force γ, ε γ/q (1) The hamiltonian of the system is: H = Ek + V (x) (2) 1 Where the kinetic energy is: Ek = p21 p2 + 2 2m 2m (3) The electrostatic potential is qε (x1 − x2 ). The surface tension potential is γ (x2 − x1 ). Thus, the potential energy is, ( (γ − qε) x V (x) = ∞ 0<x<L else (4) Graph 1: Potential energy vs. distance between the particles. Both axes are normalized with respect to its maximum values. q mT 1 (b) Let us define λ−1 = . Due to the assumption that L β(γ−qε) , it is a good apT 2π~2 proximation to integrate the center of mass, X, from 0 to L while calculating the partition function, 1 Z= (2π~)2 Z −β e 2 p2 1 + p2 2m 2m Z dp1 dp2 L Z dX 0 L e 0 −β(γ−qε)x dx = λ−2 T ·L· 1 − e−β(γ−qε)L β (γ − qε) (5) Thus, the partition function is approximately, Z≈ L 1 2 λT β (γ − qε) (6) 2 (c) The probability density function of x is, ρ(x) = R L 0 e−β(γ−qε)x e−β(γ−qε)x dx = β (γ − qε) · e−β(γ−qε)x (7) The mean value of x is, Z L hxi = x · β (γ − qε) e −β(γ−qε)x dx = −xe 0 e−β(γ−qε)L − 1 −β(γ−qε)L = −Le − β (γ − qε) L −β(γ−qε)x Z + 0 L e−β(γ−qε)x dx (8) 0 (9) Therefore, hxi ≈ 1 β (γ − qε) (10) Using hxi, the probability density function is, ρ(x) ≈ 1 · e−x/hxi hxi (11) (d) The polarization is, hP i = 1 ∂ ln Z 1 ∂ (− ln β (γ − qε)) q = = = qhxi β ∂ε β ∂ε β (γ − qε) (12) (e) The Taylor expansion of the polarization is, q hP i = βγ 1 − ε (γ/q) = q βγ 1+ ε (γ/q) + O(ε2 ) q = T+ γ 2 q T ε + O(ε2 ) γ (13) Thus, q P0 = T γ 2 q χ= T γ 3 (14)