Ex2340: Boltzmann gas confined in a capacitor

Ex2340: Boltzmann gas confined in a capacitor
Submitted by: Tomer Ygael
The problem:
An ideal gas of N spin-less particles of mass m is inserted in between two parallel surfaces. To make
sure that the particles won’t ”escape” a harmonic two dimensional potential is created in such a
way that:
1
2 2
2
z1 < z < z 2
2 mω (x + y )
V (x, y, z) =
∞
else
Let us denote L = z1 + z2
This problem consist of two independent parts. Express your answers using N, m, L, ω, e, E, T
(I) (a) Calculate the classical partition function
Z 3 3
d pd x −βH
e
Z1 =
(2π)3
;
ZN = Z1N
(1)
find the heat capacity C(T ) of the gas.
(b) Calculate the quantum partition function in the limit where L is large. Find what L is
large enough.
Guidance: What are the stationary states |φi of a single particle in a potential V?
Calculate:
X
1 N
Z1 ≡
e−βEr ; ZN ≈
Z
(2)
N! 1
r
To calculate Z1 use factorization of the sum.
(c) Find the heat capacity C(T ) of the gas using the partition function you found in article
(b). Check the behavior of the heat capacity in high temperature - define what is high
temperature and see if you get the classical result from article (a).
(d) Calculate the forces F1 , F2 that the particles apply on the upper and lower walls of the
”box”
~ = E ẑ, assume that the particle are charged with e.
(II) We now add an electrical field E
(e) Write down the one particle Hamiltonian and calculate the classical partition function.
Z1 (β; z1 , z2 , E)
(f) Calculate the forces F1 , F2 that are acting on the upper and lower walls of the ”box”.
What is the resultant force working on the system?
(g) Find the polarization P̃ of the electron gas as a function of the electric field (The polar¯ = P̃dE)
ization P̃ is defined by the formula dW
(h) Write down P(E) = L1 P̃ for a weak E. Define what is a weak field. Bring the expression
you receive to the following form P(E) = χE + O(E 2 ) and find what is the susceptibility
χ.
1
The Solution:
(a) The problem Hamiltonian is as follows:
H=(
p2y
p2x
1
1
p2
+ mω 2 x2 ) + (
+ mω 2 y 2 ) + z
2m 2
2m 2
2m
(3)
All that is left for us to do is input the Hamiltonian into the Boltzmann exponent and solve
a Gaussian integral which has a known solution.
d3 pd3 x −β(( p2x + 1 mω2 x2 )+( p2y + 1 mω2 y2 )+ p2z )
2m
2
2m
2
2m
e
(2π)3
Z∞Z
Z∞Z
Z∞ Zz2
dpy dy −β( p2y + 1 mω2 y2 )
dpx dx −β( p2x + 1 mω2 x2 )
dpz dz −β p2z
e 2m 2
e 2m 2
e 2m
=
2π
2π
2π
Z
Z1 =
−∞
−∞
(4)
(5)
−∞ z1
(6)
Due to symmetry in px , py , pz we can say that the solution over the kinetic parts of the
Hamiltonian is the same as the cube of one of those integrals. The same can be done with the
potential part of the Hamiltonian (excluding the z component), giving us

Z∞
Z1 = 
3 
dp −β  
e 2m
2π
p2
−∞
Z∞
2
dx −β 1 mω2 x2 
e 2
2π
−∞
Zz2
1
dz =
(2π)3
2mπ
β
3
2
2π
(z2 − z1 )
βmω 2
(7)
z1
Giving us
Z1 =
Finally we get ZN =
Z1N
=
m
2πβ
1
2
m
2πβ
1
2
1
L
(βω)2
(8)
N
1
L
(βω)2
We can now calculate the energy and from it the heat capacity.
!
1
m 2
1
log(ZN ) = N log(
) + log(
) + log(L)
2πβ
(βω)2
1
= N − log(β) − 2 log(β) + const
2
2
(9)
(10)
The energy and heat capacity:
∂ log(ZN )
5
= NT
∂β
2
dE
5
= N
dT
2
E = −
C(T ) =
(b) The stationary state are |pz , ny , nx i; where pz =
So now the energy of the system is
2π
L nz
(11)
(12)
; nz = 0, 1, 2....
1
1
p2z
+ ( + ny )ω + ( + nx )ω
2m
2
2
Epz ,ny ,nx =
(13)
Calculating the partition function we get
Z∞ Zz2
Z1 =
p2
dpz dze
z
−β 2m
∞
X
1
1
e−β( 2 +ny )ω+( 2 +nx )ω
−∞ z1
1
1
1
e− 2 βω
e− 2 βω
= L
1 − e−βω 1 − e−βω
!2
1
2mπ 2
1
= L
β
2 sinh 12 βω
(14)
nx ,ny
2mπ
β
2
(15)
(16)
(17)

ZN
= L
2mπ
β
1
1
2 sinh 12 βω
2
!2 N
(18)

We now ask ourselves when is L large enough for us to make the transition from sum on pz
to an integral. Let us write explicitly the sum on pz :
X
e
−β
p2
nz
2m
=
X
β 2π 2 2
− 2m
( L ) nz
e
nz
pz
Z∞
→
β
2π 2 2
nz
dn e− 2m ( L )
(19)
−∞
1
1 2
This transition to integral is only justified if L >> ( mT
)
(c) In the exact same manner as in article (a) we can calculate the energy and the heat capacity
for the quantum partition function.

E =
C(T ) =
=
2mπ
β
1
!2 
1

2 sinh 12 βω
1
1
− N log β − 2N log 2 sinh
βω + const
2
2
∂ log ZN
1
1
−
= N T + N ω coth( βω)
∂β
2
2
1
d coth 2 βω
dE
1
d
1 d
= N + Nω
={
=− 2
}=
dT
2
dT
dT
T dβ
1
N ω d coth 12 βω
1
1
ω
1
N− 2
= N + N ( )2
2
T
dβ
2
2
T sinh2 12 βω
log Zn = N log L
=
3
2
(20)
(21)
(22)
(23)
(24)
Finally giving us
ω
1
ω
eT
C(T ) = N + 2N ( )2 ω
2
T (e T − 1)2
(25)
Taking high temperature - meaning T >> ω - we receive:
1
5
ω
1
C(T ) = N + 2N ( )2
= N
ω
2
2
T (1 + T − 1)
2
(26)
Which is the exact same result as in the classical case.
ZN
(d) The force is giving by F = T ∂ log
∂X , in our case we are looking for the forces on the upper
and lower walls.
We notice that in both classical and quantum cases we have log ZN = N log L + f (T ) finally
giving us the forces on both walls,
∂ log ZN
NT
=−
∂z1
L
∂ log ZN
NT
= T
=
∂z2
L
F1 = T
(27)
F2
(28)
(e) The new Hamiltonian will be constructed by the previous Hamiltonian and an addition of the
electric field
H = Hclass − eEz
(29)
The partition function will be calculated in the same way as article (a) only with the addition
of the field, giving us
d3 pd3 x −β(( p2x + 1 mω2 x2 )+( p2y + 1 mω2 y2 )+ p2z −eEz)
2m
2
2m
2
2m
e
(30)
(2π)3
Z∞Z
Z∞ Zz2
Z∞Z
dpy dy −β( p2y + 1 mω2 y2 )
dpz dz −β( p2z −eEz)
dpx dx −β( p2x + 1 mω2 x2 )
(31)
=
e 2m 2
e 2m 2
e 2m
2π
2π
2π
Z
Z1 =
−∞
−∞
−∞ z1
Due to the same reason we listed in article (a) this integral is equal to
 ∞
3  ∞
2 z
Z
Z2
Z
2
dp −β p  
dx −β 1 mω2 x2 

2m
2
Z1 =
e
e
dzeβeEz
2π
2π
−∞
=
1
(2π)3
−∞
2mπ
β
3
2
2π
βmω 2
(32)
z1
eeβEz2
eeβEz1
−
eβE
(33)
It is easy to see that when E → 0 we can expand the exponents and receive the original
classical partition function.
In the same way as throughout this entire exercise, ZN = Z1N
4
ZN
(f) Calculating the forces on each wall we use the formula F = T ∂ log
resulting in the following
∂X
−eβEeeβEz1
∂ log ZN
= N T eβEz2
∂z1
− eeβEz1 )
(e
∂ log ZN
eβEeeβEz2
= T
= N T eβEz2
∂z2
(e
− eeβEz1 )
F1 = T
(34)
F2
(35)
Using only the allowed parameters we finally get the forces
F1 = −N eE
F2 = T
1
(eeβEL
(36)
− 1)
∂ log ZN
1
= N eE
∂z2
(1 − e−eβEL )
(37)
And the resultant force working the system: Ftot = F1 + F2 = N eE as expected.
(g) The polarization is giving by:
ZN
. Using the properties of log we get
P̃ = T ∂ log
∂E
∂(log e
e
eβEz2 − eβEz1
P̃ = N e
eβE
)
∂E
1
z2 eeβEz2
= Ne −
+
eβE
eeβEz2
∂
(const)
∂E
− z1 eeβEz1
− eeβEz1
+
(38)
(39)
Unless we want an un-physical constant in our equation we will need to choose the middle of
the system as our zero. Thus making z2 = L2 and z1 = − L2 and giving us
1
1
1
P̃ = N e −
+ L coth eβEL
eβE
2
2
(40)
(h) In order to expand P̃ we must first demand
eβEL << 1 → L <<
T
eE
(41)
we can now expand the coth up to first order and receive
P =
=
=
1
1
1
1 x
−
+ L coth eβEL = {coth x ≈ + + O(x3 )} =
eβE
2
2
x 3
!
1
2
Ne
1 N eL
3
4 eβEL N e
−
+
+
+ O(E ) =
eβEL 2 12 eβEL2
3L
Ne
L
N e2 βEL
+ O(E 3 )
12
We can now easily identify χ =
(42)
(43)
(44)
N e2 βL
12
5