Ex2046: Gas in a centrifuge

Ex2046: Gas in a centrifuge
Submitted by: Isaac Weinberg
The problem:
A cylinder of of radius R rotates about its axis with a constant angular velocity Ω. It contains an
ideal classical gas of N particles at temperature T .
Note that the Hamiltonian in the rotating frame is H 0 (r, p; Ω) = H (r, p) − ΩL (r, p) where L (r, p)
is the angular momentum.
It is conceptually useful to realize that formally the Hamiltonian is the same as that of a charged
particle in a magnetic field (”Coriolis force”) plus centrifugal potential V (r). Explain how this
formal equivalence can be used in order to make a shortcut in the above calculation.
(1) Find the density distribution as a function of the radial distance from the axis.
(2) Write what is the pressure on the walls.
The solution:
First, we would like to rearrange the Hamiltonian in order to simplify the partition function calculation:
H=
p2
p2
p2
(p − mΩ × r)2 1
−Ω·L=
−Ω·r×p=
−p·Ω×r =
− m (Ω × r)2
2m
2m
2m
2m
2
(1)
One can see that the term mΩ × r eA(r) and that V (r) = − 21 m (Ω × r)2 which corresponds to
a charged particle in electromagnetic field:
H=
1
(p − eA(r))2 + V (r)
2m
(1) In order to find the density distribution we will find the partition function of a very thin dr
layer of gas distant r from the center of the cylinder:
Z
Z1 (r) =
drdp −βH
e
=
(2π)3
Z
drdp −β [ 1 (p−mΩ×r)2 − 1 m(Ω×r)2 ]
2
e 2m
(2π)3
(2)
We will change the variable p0 = p − mΩ × r and since we are working in a cylindrical coordinates
the integral of equation (2) will become:
Z
Z1 (r) =
dp0 −β 1 p02
e 2m
(2π)3
z=L
Z
dz
z=0
ZN (r) =
Z1N
N!
θ=2π
Z
dθ rdr e
1
βm(Ωr)2
2
=
1
λT
3
1
V e 2 βm(Ωr)
2
(3)
θ=0
N equal the number of atoms in the thin layer
(4)
Because each thin layer is in contact with the layer around it after equilibrium is archived all the
layers have the same chemical potential which we’ll calculate next:
1
∂F
1
1
1
3 N (r) − 12 βm(Ωr)2
µ=
= (− ln(Z1 (r)) + ln(N (r))) = ln λT
e
= ln( λ3T n(r) − m(Ωr)2
∂N
β
β
V
β
2
1
(5)
If we equate µ(r) = µ(0) we can find the distribution of the density:
ln
µ(r) = µ(0)
1
2
= ln λ3T n(0)e− 2 βm(Ω·0)
1
2
λ3T n(r)e− 2 βm(Ωr)
1
2
n(r) = n(0)e 2 βm(Ωr)
(6)
(2) To find the pressure on the walls of the cylinder we will calculate the force on the walls from
the partition function of the entire cylinder, not just a thin layer as we did in the previous section.
Z1 =
z=L
Z
hF iR =
P =
Z1N
N!
r=R
Z
2πL 1 βm(ΩR)2
2
−
1
(7)
e
mΩ2 β
now N equal the number of atoms in the cylinder
(8)
dz
z=0
ZN (r) =
θ=2π
Z
3
dp0 −β 1 p02
e 2m
(2π)3
Z
θ=0
1
2
dr re 2 βm(Ωr) =
dθ
1
λT
r=0
1 ∂ ln(ZN )
1
= N mΩ2 R
1
2
−
β ∂R
1 − e 2 βm(ΩR)
N mΩ2 R
1
F
=
1
A
2πRL 1 − e− 2 βm(ΩR)2
(9)
(10)
We will replace N with nV = nπR2 L and get:
P =
nπR2 LmΩ2 R
mR2 2
1
1
=
n
Ω
1
1
2
2
−
βm(ΩR)
−
2πRL
2
1−e 2
1 − e 2 βm(ΩR)
(11)
If Ω → 0 we can expand the exponent ex = 1 + x... and get:
Ω→0
P ≈ n
mR2 2
1
n
=
Ω
1
2
2
β
1 − 1 − 2 βm(ΩR)
(12)
Which corresponds to the equation of state of an ideal gas.
On the Other hand if Ω → ∞ the force on the walls equals the Centripetal force exerted by the wall
on all the atoms:
Ω→∞
hF iR ≈ N mΩ2 R
(13)
2