Download Physics 4230 Solutions to Problem Set 11 H11sol.0 Problems: Fermi

Physics 4230
Solutions to Problem Set 11
Problems:
Fermi gases! The Pauli Exclusion Principle causes much of the behavior of matter, both
of the type you are quite familiar with e.g, the hardness of solids, and things that are
more extreme, e.g., the behavior of white dwarf stars. Let’s see how all this works:
1. Schroeder Problem 7.19 Think about the electrons around the atoms that form a
solid material. In the case of metals, it’s pretty clear that some of the electrons are
rather free to move around in the material. In fact, you can understand a huge
amount about the behavior of metals by treating these electrons as an ideal noninteracting gas of Fermions. To a surprising degree, you can understand the behavior
of such a system by assuming that the electrons don’t interact! That’s partly because
a solid is charge neutral, so the electron-electron interactions are screened, and
partly good luck! In any case, the first problem has you play with this electron gas
model for copper. If you ever take the Solid State course, you’ll see that the free
electron gas is a hugely important part of our understanding of solids.
OK, so everything traces back to the fact that copper contributes a single valence electron
to the gas, so if you know the atomic density of copper, then you know the electron
density of the gas. The rest follows from the equations of Schroeder’s section 7.3.
So, I looked up the number density of atoms in copper and found it listed in “Solid State
Physics”, by N.David Mermin and N. W. Ashcroft (1976). They have
N/V = 8.47x1022 atoms/cm3
Using that density as the electronic density, we find the following table of information for
the requested quantities:
Dnsty(1/m^3)
kF (1/m)
8.47E+28
1.36E+10
EF
(Joules)
1.13E-18
TF (K)
81927.8
PD
(N/m^2)
3.83E+10
BF
(N/m^2)
6.38E+10
B_measured
(N/m^2)
1.34E+11
The bulk modulus is predicted at 63.5 GPa, and is off by about a factor of two from the
experimentally determined bulk modulus. It’s clear that the Pauli exclusion make a large
contribution to the strength of materials!
H11sol.0
Physics 4230
Problem Set 9
Fall 2009
2. Schroeder Problem 7.22. Relativistic electrons can also be important, say in the
interior of stars. Surprisingly, you’ll find relativistic electrons in some solids like
tungsten!.
Here we are asked to consider highly relativistic electrons. In this picture, the
wavefunctions are still plane waves, the boundary conditions still lead to a digital kspace, but the relationship between the k-vector and the energy is no longer the classical
result. Instead, we have that:
  k   pc  kc
The argument about the ground state would still have you filling a sphere in k-space, so
we still have the requirement that the number of filled states inside the sphere (including
the factor of 2 for spin) is given by the number of particles/2 and therefore by the ksphere volume divided by the volume per k-state:
4 3
 kF
N
3

2  2 3


 L 
Therefore, the radius of the sphere is still given by:
13
N

kF   3 2 
V

So far, the argument is completely the same as for the classical case. However, the
ENERGY associated with this wavevector would now be different:
 F    kF   k F c
At zero temperature, the chemical potential is equal to the Fermi energy, so we have:

13
N
  c  3 2 
V

This result is the desired answer.
b) Find a formula for the total energy of this system in terms of N and .
Really, all that is wanted here is to repeat the calculation of the systems internal energy,
U, as Schroeder did in eqs 7.40 through 7.42. The change from a sum on filled states to
an integral over k-space is completely the same, so we pick up at Eq. 7.42 with:
Physics 4230
n _ MAX

U 
Problem Set 9
Fall 2009
  n  n 2 dn
0
However, the energy relationship is now different. We have:
  n   kc  c
k
2
X
 kY2  kZ2  
hc
n
L
So,
U 
hc
L
n _ MAX

n3 dn 
0
 hc
4 L
4
nMAX
One of the factors of the maximum number can be combined to give you the Fermi
energy, or equivalently, the chemical potential. The other factors combine to give you the
cube of the Fermi wave vector i.e., the density. Here is my approach:
4  2  3

n
1 1 3  L  MAX
3

 N
3
44
16
 2 


3
 L 
3
U
  c 2 nMAX 
4 
L
3
 nMAX

3. Schroeder Problem 7.23. White dwarf stars are charge neutral chucks of matter too,
and you won’t be surprised to see that a model for the electrons in a white dwarf has
a strong overlap with the treatment of electrons in copper. The major difference is
that copper is held together mostly by electrostatic attractive forces between the ionic
cores of the atoms and the electrons. By comparison, a white dwarf is held together
by gravitational forces.
The white dwarf. Here, we basically calculate the radius where a charge neutral
collection of protons and electrons finds a balance between gravitational pressure and the
quantum mechanical Pauli pressure. Since the nuclear energy output has ended in a white
dwarf, the radiation pressure (which is what keeps the Sun rather inflated) has dropped
out of the problem.
Let’s start with the gravitational pressure. Schroeder suggests that you use the
gravitational energy for a uniform density sphere as a good starting point. Then, the
gravitational energy is negative (referenced to infinite separation of all the mass in the
white dwarf) due to the fact that gravity is attractive, and is approximately:
Physics 4230
U  R 
Problem Set 9
Fall 2009
3 GM 2
5 R
Just for fun, notice that this is the same 3/5 that appears in the Fermi energy calculation.
Schroeder asks you to move on with calculation of the electron energies due to the Pauli
principle, but before we do, notice that you can think of the gravitational energy as also
creating an inwards pressure on the mass of the white dwarf. The associated gravitational
pressure due to gravity is easy to calculate:
PG  



R
U  R  
U  R
V
R
V
3 GM 2
20 R 4
b) Next, we assume that the star is composed of equal amounts of protons, electrons,
and neutrons. Basically, it’s hydrogen, plus neutrons that have been created by the
nuclear fusion process. Usually, these neutrons appear in combination with
protons as helium nuclei, but Schroeder is not worrying too much about how the
whole mess is assembled.
In any case, we expect that the star is essentially a box full of electrons and
protons and neutrons, all of which are Fermions. Since all of them are held in the
same box, they all have the same restrictions on k-vector. Further, if they are nonrelativistic, then they all have the same relationship between k-vector and energy,
differing only in the mass:
2
k2

2m
However, since the electrons have the smallest mass, they certainly have the
highest energy and they will have the most important contribution to the energy of
the star (compared to the protons and neutrons). In any case, the average kinetic
energy in the ground state is given by:
U KE
2
3
3
 2 N
 N F  N
 3

5
5 2me 
V
23
To finish the energy, we need to know the number of electrons. While the
electrons certainly dominate the KE, most of the star mass is in the protons and
neutrons. Therefore, you can count the electrons by assuming that half the star
mass is protons, and that the number of electrons is equal to the number of
protons for charge neutrality. Given a stellar mass of M, the number of particles
is then,
Physics 4230
N
Problem Set 9
Fall 2009
M
2m p
And plugging back into the energy gives (assuming a spherical volume):
U KE
3 53 2  2 1 
 N
 3

5
2me 
V
3
23 M
  9  
 2m
5
 p



53
23
2
1
2me R 2
This is the result that Schroeder quotes, complete with all the constants.
c) OK, so we now have a gravitational energy that is attractive and 1/R, along with a
Pauli exclusion kinetic energy that is repulsive, and goes as 1/R^2. The total
energy looks like:
M
3 GM 2
UTOTAL  R   
 0.0088 
m
5 R
 p
53



h2 1
me R2
For small radii, the repulsive part gets large faster. For large radii, the attractive
part dominates, so things are repulsive if the radius is too small, attractive if the
radius is too big, and there is an equilibrium point somewhere. The sketch has an
attractive potential well.
To find the minimum, differentiate:
M

3 GM 2
UTOTAL  R  

2

0.0088

R
5 R2
 mp
53



h2 1
0
me R3
Or
REQ
 1
 2  0.0088 
m
 p
53



h2
5
me 3GM 1 3
This result predicts that as the mass of the star increases, gravitational attraction
pulls it in to a smaller radius, the smaller radius causes the Fermi energy to
increase until you compensate the larger gravitational attraction with a counteracting kinetic energy. It all makes sense.
d) Next we can calculate the properties of a white dwarf with a solar mass. The white
dwarf state is widely considered to be the end point of solar evolution for a star
Physics 4230
Problem Set 9
Fall 2009
like the Sun (after the red giant phase). Let’s see what things might look like in a
few billion years.
Let’s plug and chug with Mathematica:
White Dwarf predictions
Let's put down some of the white dwarf equations from HW 11
uGravity = -3/5 (gNewton*solarMass2)/radius;
solarMass
5
3
uDegeneracy = 0.0088 protonMass
*hPlanck2/electronMass*1/radius2;
howThingsAre = {solarMass2*1030,hPlanck6.6*10-34,protonMass1.67*1027
,electronMass9*10-31,gNewton6.6*10-11};
Sketch of the energy minimum
Here' s the energy function for the solar mass case, so you can see the total energy curve
around the minimum
uTotal = uGravity+uDegeneracy/.howThingsAre
5.75242×1056/radius2-1.584×1050/radius
sketch = Plot[uTotal/.howThingsAre, {radius, 1000000, 20000000}]
1.0
2
10 42
4
10 42
6
10 42
8
10 42
1
10 43
10 7
1.5
10 7
2.0
You can see that there is a minimum around 7 million meters or 7000 km.
The equilibrium radius
In the solutions, we did the differentiation to find that the equilibrium radius is given by :
10 7
Physics 4230
Problem Set 9
eqRadius =
1
2*0.0088 protonMass
solarMass
1
3
Fall 2009
5
3
*hPlanck2/electronMass*5/(3*gNewton*
);
The numerical value for the case of a solar mass is :
eqRadius/.howThingsAre
7.26316×106
or 7.3 kilometers
The equilibrium density
The predicted density is then :
density=solarMass/(4/3 Pi*eqRadius3);
density/.howThingsAre
1.24614×109
So, we have something like a billion kg/m^3, which is HUGE compared with the
density of water (1000 kg/m^3) It's a million times as dense as water. Cool.
e) The Fermi energy is going to come out larger than it would for, say, a material
like copper (see the first problem, above) because the density of the material is
larger. The Fermi energy is related to the Fermi k-vector, which is given by the
cube root of the density. Therefore, the Fermi energy in the non-relativistic region
is going to go as the 2/3 power of the density. If the density is a million times
larger, then the Fermi energy will be 10^4 times larger. Given that copper has a
Fermi energy of something like 10 eV and a Fermi temperature of 100,000K, we
expect that the white dwarf energy will be around 100,000 eV, and the Fermi
temperature will be around 10^9 K.
Given that nuclear fusion temperatures are around 10 million kelvins, we are
really quite close to the ground state.
f) OK, so suppose that we go relativistic! Then, the energy becomes proportional to
the k-vector, rather than the square of the k-vector. That means you go as the
density to the 1/3 instead of 2/3 and THAT is enough to change the dependence
on the stellar radius to 1/R instead of 1/R^2.
Now you’d have a pair of energy terms, one attractive and one repulsive, but
BOTH going as 1/R. There is no stable equilibrium point. The star is going to do
something interesting
g) Mc^2 for the electron is around 500,000 eV, but the one solar mass Fermi energy
is only 100,000 eV, so you’re probably doing OK by predicting a white dwarf end
point for the Sun.
Physics 4230
Problem Set 9
Fall 2009
However, the Fermi energy will continue to increase with density. Again, the
Fermi energy is:
 2 N
F 
 3

2me 
V
2
23
M
 3
 Req



23
M43
Somewhere between 3 and 4 solar masses, you’re going to cross the line between
the non-relativistic electron and the relativistic electron and things are going to
start to lose stability!
4) Schroeder Problem 7.24. Neutron stars!
Schroeder’s argument about the loss of stability at large enough white dwarf mass
assumes that there is SOMETHING that the white dwarf can evolve into. What
that something is, is a neutron star. The electrons and protons eventually can react
to form neutrons and emit a pile of neutrinos…
Neutron stars and the Chandrasekhar mass. We expect that the gravitational
pressure will eventually squeeze the protons and electrons to a high enough
density that the system would be more stable as a collection of neutrons. When the
system realizes that it has reached that point, there is a sudden drop in the Pauli
pressure as the electrons are gobbled up into neutrons. This sudden pressure drop
is a catastrophe for the star as its core suddenly implodes. The resulting shock
wave blows off the outer parts of the star and you get a nova or supernova. Left in
the middle of the disaster is the neutron stellar core, a neutron star.
The mass of the neutron star can vary depending upon the history of the events.
Our previous result, just above, can be used to estimate the radius of a solar mass
neutron star. The only difference in the derivation is in the use of the neutron
mass, rather than the electronic mass. Thus, the radius is just scaled by the neutron
to electron mass ratio of roughly 2000:
RM  neutron  7000km / 2000
 3.5km
Neutron stars are small! Similarly, neutron stars are dense. Since the radius is
down by 2000X, the density is larger by that radius cubed, or 8 Billion times more
dense than the white dwarf.
However, because the Fermi energy is going only as the 2/3 power of density
AND because it also goes as 1/mass of the fermion of interest, the Fermi energy is
only scaling as the mass ratio. Therefore the Fermi energy and Fermi temperature
are both about 2000 times what you get for the white dwarf.
Physics 4230
Problem Set 9
Fall 2009
Finally, the interesting thing about the stability criterion is that the rest mass of a
neutron is… the rest mass of the neutron, and is therefore 2000 times the rest
mass of the electron. That means that the neutron star is very close to the same
instability as the white dwarf star: Somewhere around 3 to 4 solar masses is where
we expect that the neutrons might go relativistic and the star might become
unstable.
For many years, astrophysicists argued about what a neutron star might become.
Yes, the star is predicted to be unstable, but to go into some new form, a new form
must be available. For example, classical mechanics tells you that the electron
around a hydrogen atom is unstable because it should radiate energy. That does
NOT mean that the electrons are unstable. It means that you have a lot to learn
about quantum mechanics. Similarly, people were not certain that there is actually
anything that a neutron star can decay into. We now expect that neutron stars
above a certain size can become black holes.