Download headingE2170: Polarization of two-spheres system inside a tube The problem:

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)