1 PHY831 - Midterm 2, Monday October 22nd 2012 Name:

1
PHY831 - Midterm 2, Monday October 22nd 2012
Answer all questions. Time for midterm - 50 minutes
Name:
Formulae that might be helpful
Z
∞
2
e−ax
+bx
−∞
b2
π
dx = ( )1/2 e 4a
a
(1)
Relations between partial derivatives,
∂x
∂y
∂x
∂y
=
z
∂x
∂y
z
∂y
∂x
= 1/
z
∂x
∂w
z
∂y
∂z
x
∂w
∂y
;
Inversion
(2)
z
;
Addition of a variable
(3)
z
∂z
∂x
= −1 ;
Triple Product
(4)
y
If A(x, y), then,
∂A
∂x
=
z
∂A
∂x
+
y
∂A
∂y
x
∂y
∂x
non − natural derivative
;
Response functions are defined as follows;
∂U
∂S
∂Q
∂S
=
=T
,
CV =
∂T V,N
∂S V,N ∂T V,N
∂T V,N
CP =
∂H
∂T
=
P,N
1
κS = −
V
1
κT = −
V
1
V
αP =
∂H
∂S
∂V
∂P
∂V
∂T
P,N
∂V
∂P
∂S
∂T
=−
S,N
=−
T,N
=
P,N
(5)
z
=T
P,N
∂lnV
∂P
∂lnV
∂P
∂lnV
∂T
∂S
∂T
(6)
,
(7)
P,N
,
(8)
,
(9)
S,N
T,N
,
(10)
P,N
For any s we have,
Z
0
∞
xs−1 dx
= Γ(s)ζ(s),
ex − 1
where Γ(s) = (s − 1)! for s a positive integer, and ζ(2) = π 2 /6, ζ(3) = 1.202..., ζ(4) = π 4 /90.
For n even we have,
Z ∞
tn et
dt t
= 2n(1 − 21−n )(n − 1)!ζ(n)
(e + 1)2
−∞
(11)
(12)
while for odd n this integral is zero.
h̄ = 1.055 × 10−34 Js; kB Troom = .025eV ; e = 1.602 × 10−19 C; kB = 1.38 × 10−23 J/K; me = 9.11 × 10−31 kg (13)
2
1. (15 points)
N is the total number of Krypton atoms in a chamber where the surfaces of the chamber are lined with graphite.
Some of the Kr atoms are in the gas phase and others are bound to the graphite surface. If there are a total of
NB sites on the graphite surfaces of the chamber where Kr atoms can bind, with binding energy (i.e. the bound
state is lower in energy), find an expression for the average number of Kr atoms that are bound to the surfaces of
the chamber as a function of temperature. Take the volume of the chamber to be V , ignore the interactions between
Krypton atoms and consider the case where N ≤ NB . Find an explicit solution in the limit where the number of Kr
atoms bound to the surface is small. Under what conditions does this limit hold?
Solution
We assume that the particles can be treated as classical particles. If there are Ns particles bound to the graphite
surface, their Helmholtz free energy is given by,
F = U − T S = −Ns − kB T ln(
NB !
) = −Ns + kB T NB [xln(x) + (1 − x)ln(1 − x)]
Ns !(NB − NS )!
(14)
where x = Ns /NB . The chemical potential of the particles at the surface is given by,
µs =
∂F
= − + kB T ln(x/(1 − x)).
∂Ns
(15)
Particles in the gas phase have chemical potential,
µg = kB T ln(
Ng λ3
).
V
(16)
At equilibrium the two chemical potentials must be the same µs = µg , and using Ng = N − Ns we find,
kB T [ln(N − Ns ) + ln(
λ3
)] = − + kB T ln(x/(1 − x)
V
(17)
Consider the limit where there are not many bound atoms so that Ns /N → 0 and Ns /NB → 0 so that,
Ns
NB λ3 β
≈
e
N
V
(18)
Even at high temperatures atoms are bound to the surface - and the number that are bound depends on the relative
entropy of the surface atoms and gas atoms, as well as the binding energy.
An alternative approach that gives the same final result is to write down the partition function of the whole system,
Z=
N
X
Zsurf ace (Ns )Zgas (N − Ns )
(19)
Ns =0
and hence F = −kB T ln(Z). Then we can find the optimal value of Ns , by solving ∂F/∂Ns = 0. This leads to the
same equation as µs = µg above.
2. (15 points)
Recent observations indicate that the universe is expanding at an increasing rate so there is an intrinsic acceleration
of the size of the universe and the Hubble constant is actually getting larger. This observation led to the 2011 Nobel
prize in physics (Perlmutter, Schmidt, Reiss). Consider the universe to be a container that has size L and volume L3 .
(i) Considering the cosmic microwave background (CMB) radiation to be a ideal Bose gas of photons, find an
expression for the pressure that this gas exerts on the universe.
(ii) Now consider a model for pressure due to dark matter. Treat dark matter as being due to a single type of nonrelativistic spinless fermion of mass m in its ground state. Take the total mass of the universe to be M . Astronomical
measurements suggest that 84% of this mass is dark matter. Find an expression for the pressure due to this fermionic
dark matter.
(iii) Make order of magnitude estimates of the CMB and dark matter pressures, taking the mass of the dark matter
fermions to be that of the electron. Which pressure is likely to be larger? Take the mass of the universe to be 1055 kg,
and the size of the universe to be 1026 m.
3
Solution
(i) The internal energy of the photon gas is given by,
Z
L 3 ∞
e−βh̄kc
U = 2( )
4πk 2 dk(h̄kc)
2π
1 − e−βh̄kc
0
(20)
using x = βh̄kc this reduces to,
∞
x3 dx
ex − 1
(21)
1U
1 π4
= 2 3 3 (kB T )4
3V
π h̄ c 45
(22)
U
1
1
=
2
V
βπ (βh̄c)3
Z
0
The pressure is given by,
P =
which is the pressure exerted by the photon gas on the surface of the universe. The pressure depends only on the
temperature and fundamental constants.
(ii) The internal energy of the degenerate fermi gas is given by,
U =(
L 3
)
2π
kF
Z
4πk 2
0
h̄2 k 2
;
2m
(23)
we also define,
N =(
L 3 4π 3
)
k ;
2π 3 f
F =
h̄2 kF2
2m
(24)
so that,
3
N F
5
(25)
2
2 h̄2
N F =
(6π 2 )2/3 ρ5/3
3
5 2m
(26)
U=
and
P =
where ρ = N/V .
(iii) For the CMB at T = 2.713K, we have,
P =
1
π 2 h̄3 c3
π4
1 34∗3 1 −24 81
(kB T )4 ≈
10
10
(1.38 × 2.713)4 (10−23∗4 ) ≈ 2 × 10−14 N/m2
45
10
27
45
(27)
For dark matter fermions of mass m we have,
P =
2 h̄2
(6π 2 )2/3 ρ5/3 .
5 2m
(28)
We keep m as a free parameter so that N = Mdark /m = 0.84 × 1055 /m and ρ ≈ N/V = 1055 /m × 10−26×3 . we then
find,
P =
1
1
1
(10−68 )(16)(10−40 ) 8/3 = 3 × 10−108 8/3
5
m
m
(29)
where the mass m is in kg. If the dark matter fermions have a mass of order that of the electron, m = 9.11 × 10−31 ,
then the dark matter pressure is roughly 10−28 N/m2 . However since the formula has a factor 1/m8/3 , if the mass
of the dark matter fermions is lower, then it is possible for them to generate a pressure which is large. For example
the current limit on the mass of the electron neutrino is not precise but is around m < 0.5eV /c2 as compared to the
electron mass which is 0.511M eV /c2 . If we use the mass of the electron neutrino in our calculation (Eq. (29)), the
pressure is 1016 times larger, and may be larger than that of the CMB.