E2173: Polarization of classical polar molecule

E2173: Polarization of classical polar molecule
Submitted by: Arka Prabha Banik
The problem:
Find the polarization P̃ (ξ) and the electric susceptibility χ for gas of N classical molecules with
dipole moment µ, The system’s temperature is T .
The solution:
This problem has been approached in different branches of Physics , let’s discuss it in the way of
Statistical Mechanics .
Suppose, each polar molecule of the cluster having Mass and Moment of Inertia m and I respectively
is feeling an external Electric field E.
we can write down the Lagrangian for each molecule in (r, θ, φ) co-ordinate
1
1
L(x, y, z, θ, φ) = m(ẋ2 + ẏ 2 + ż 2 ) + I(θ̇2 + sin2 θφ̇2 )
2
2
next comes the Energy consideration :
The conjugate momenta as we know are pj =
∂L
∂ q̇j
where j = r, θ, φ
furthar if we notice that we can write the Hamiltonian from Lagrangian with those conjugate
momenta as ,
H0 = Σpj q̇j − L(r, θ, φ)
With furthar simplification it yields
H0 =
p2φ
1
1 2
pr + p2θ +
2m
2I
2I sin2 θ
or more specifically
H0 =
p2φ
1
1 2
(px + p2y + p2z ) + p2θ +
2m
2I
2I sin2 θ
perturbation on the molecule due to the presence of external electric field
Hperturb = −Eµ cos θ
henceforth the modified Hamiltonian reads
H1 = H0 + Hperturb =
p2φ
1 2
1
(px + p2y + p2z ) + p2θ +
− Eµ cos θ
2m
2I
2Isin2 θ
we are now ready to write the single molecule Partition Function :
1
Z
Z1 =
Z
dxdydzdpx dpy dpz
(2π)3
..
Z Z
dθdpθ
2π
Z Z
dφdpφ −βH1
e
2π
now, Writing H1 explicitly , we ’ve
Z
Z1 =
Z
..
β
2
2
(p2
dxdydzdpx dpy dpz − 2m
x +py +pz )
e
(2π)3
Z Z
dθdpθ − β p2 βEµ cos θ
e 2I θ e
2π
Z Z
2
dφdpφ −β pφ 2
e 2I sin θ = I1 .I2 .I3
2π
the first Integral is as usual
Z
I1 =
Z
..
β
2
2
(p2
dxdydzdpx dpy dpz − 2m
m 3/2
x +py +pz )
e
=V(
)
3
(2π)
2πβ
the second integral gives
Z Z
I2 =
dθdpθ − β p2 βEµ cos θ
2πI 1/2
=(
e 2I θ e
)
2π
β
Z
dθ βEµ cos θ
e
2π
and the third integral yields
Z Z
I3 =
2
dφdpφ −β pφ 2
e 2I sin θ =
2π
Z
−β
dpφ e
p2
φ
2I sin2 θ
=(
2πI sin2 θ 1/2
)
β
hence , the single molecule partition function reads
m 3/2 I
)
Z1 = I1 .I2 .I3 = V (
2πβ
β
Z
sin θdθeβEµ cos θ = V (
m 3/2 2I sinh βµE
)
2πβ
β βµE
now Gibbs Prescription is to be followed , i.e, For N molecules the Partition function is
ZN =
1 N
Z
N! 1
So that
ZN =
1 N m 3N/2 2I N sinh βµE N
V (
)
( ) (
)
N!
2πβ
β
βµE
since we’ve learned Polarisation and Electric fields are conjugate variables , so we can write :
1 ∂lnZN
Pe =
β ∂E
hence in this particular problem , it becomes :
Pe = N µ[coth(βµE) −
1
]
βµE
Let’s define a function as , L(u) = coth(u) −
1
u
2
If we expand the function L(u) in the following Taylor expansion series :
L(u) = coth(u) −
1
1
eu + e−u
−
= u
−u
u
e −e
u
=
(1 + u + 12 u2 + 16 u3 + ....) + (1 − u + 12 u2 − 61 u3 + ....) 1
−
(1 + u + 21 u2 + 16 u3 + ....) − (1 − u + 12 u2 − 61 u3 + ....) u
=
1 4
1 6
1 + 21 u2 + 24
u + 720
u + .... 1
−
1 3
1 5
u
u + 6 u + 120 u + ....
=
1 3
3u
1 5
1 7
+ 30
u + 840
u + ....
1 5
1 3
u(u + 6 u + 120 u + ....)
since , u = βµE is usually very small , so in this limit u → 0
L(u) = u/3
so that it yields , Pe → 31 N βµ2 E and hence ...
Susceptibility, χ =
N
3V
βµ2
3