Physics 4340
Homework 5
Due Wednesday Feb. 18, 2015
We are working with the Sommerfeld Model for the free and independent electron gas.
It’s our first model for the electrons in metals and treats them as essentially a set of free
fermions in a box. The Pauli Exclusion Principle makes the behavior rather different
than the simpler Drude picture as the electrons as a classical charged gas. The model
gives us many successful descriptions of objects composed of dense fermions.
1) The Pauli Pressure (20 points)
One of the major results from the Fermi gas treatment is that the Fermi gas resists
being compressed. The reason is that the Pauli Exclusion Principle fights any effort to
bring Fermions into overlapping volumes. The Pauli Exclusion and Heisenberg
Uncertainty principles together are the first real explanation of the stability of matter.
In this problem, you look at the extent to which this pressure explains the
compressibility or bulk modulus of metals.
In class, we found that the average energy of the free electron gas is give by:
E 
3
NE f
5
If you squeeze on the metal, we expect for the energies of the electrons to generally
increase since the electron density increases and the Fermi energy is an increasing
function of density:
 2N
Ef 
 3

2m 
V 
2
23
Therefore, squeezing on the solid costs energy. The need to put energy into the
system to compress it is interpreted as doing work against an outward pressure from
the electron gas, much as the work of compressing any other type of gas. This
pressure, the Pauli Pressure, is defined by:
PP  
 E
V
Plug and chug to find that:
PP 
2N
Ef
5V
or you can rewrite this result as:
Physics 4340
HW5.1
Spring 2015
Physics 4340
Homework 5
PP 
Due Wednesday Feb. 18, 2015
2 E
3 V
All of the above is a reminder of what we did in class.
a) Bulk Modulus done more correctly. (5 points)
The thermodynamic definition of ‘bulk modulus’, which is the inverse of
compressibility, is:
B  V
P
V
Work out the bulk modulus of the free Fermi gas.
b) Compare your results to the behavior of Sodium metal, where the electron density
9
2
is 2.65x1022 per cc, and the measured bulk modulus is 6.8  10 N / m . (5 points)
c) Look up some data and see how the theory does. (10 points)
Look up a bunch of data for metals. Find out the electron density and bulk
modulus for a variety of metals, and compare your results for the free electron
bulk modulus to the actual measured bulk modulus. Plot the predicted versus
calculated and point out which metals seem to be doing particularly well or
particularly poorly in comparison between the simple theory and reality.
Be sure to see the next page.
Physics 4340
HW5.2
Spring 2015
Physics 4340
Homework 5
Due Wednesday Feb. 18, 2015
Just kidding! Here’s the table of electron density and measured bulk modulus.
It’s from Kittel’s book. All you need to do is the plotting of theory versus
experiment and the identification of the good, the bad, and the ugly.
Element
Li
Na
K
Rb
Cs
Cu
Ag
Au
Be
Mg
Ca
Sr
Ba
Zn
Cd
Al
Ga
In
Pb
Sn
Concentration 10^22 cm^-3
4.7
2.65
1.4
1.15
0.91
8.45
5.85
5.9
24.2
8.6
4.6
3.56
3.2
13.1
9.28
18.06
15.3
11.49
13.2
14.48
B Meas. in 10^11 N/m^2
0.116
0.068
0.032
0.031
0.02
0.73
0.993
0.577
0.997
2.82
6.58
8.62
9.97
1.67
2.14
1.385
1.76
2.43
2.33
0.901
2) Fermi gases in other situations: (30 points)
Liquid 3He
3
He is an atom which is a composite fermion of spin 1/2. Therefore, we expect
that the ground state of a system of many independent 3He atoms should resemble
the spherical k-space ground state we found for the free electron gas. In many
ways, you might expect a real 3He system to be a better example of a Fermi gas
because the interatomic interaction potentials are much weaker than for bare
electrons.
a) Explain why 3He atoms look like spin 1/2 fermions. (5 points)
b) The mass density of liquid 3He is 0.081 g/cm3 at low temperatures. Determine
the Fermi energy and the associated Fermi temperature. (5 points)
Physics 4340
HW5.3
Spring 2015
Physics 4340
Homework 5
Due Wednesday Feb. 18, 2015
Nuclear matter
Free neutrons have somewhat more rest mass than the combination of a free
proton and a free electron i.e.,
mn c 2  mp c 2  mec 2
Therefore, free neutrons are observed to be unstable and eventually decay via:
n  p  e   e
However, you know that if the neutron is confined to a small region such as an
atomic nucleus, that it is stable. Neutrons, protons, and electrons are all spin 1/2
fermions. Treat them as non-interacting particles and calculate the size of the box
necessary to stabilize the neutron against decay. How does this size compare to
nuclear dimensions? (10 points)
Fermions in astrophysics
The white dwarf. Use the solar mass, assume that the entire Sun is a gas of
ionized hydrogen, and determine the number of electrons in the Sun. Then, from
the second problem, you can determine the Pauli pressure of the Sun as a function
of solar volume (why can we ignore the protons here?). Assume that the
gravitational potential energy of the Sun is given by (here, R and M are
respectively the radius and mass of the Sun):
E  R  
GM 2
R
Estimate the gravitational pressure due to this energy. Then estimate the size
white dwarf we expect for a solar mass of material. Compare your result to any
estimate of white dwarf radius you can find. What keeps the Sun at it’s current,
much larger radius? (5 points)
Neutron stars and the Chandrasekhar mass. If you add enough mass to a star,
eventually the gravitational pressure compresses the material to the point where
the electrons and protons are in such a small volume that they become unstable
against recombination into neutrons. Use your white dwarf result along with your
estimate of the neutron-stabilization volume from the nuclear matter section to
determine the mass beyond which the white dwarf is unstable. This mass is
referred to as the Chandrasekhar mass. Then, repeat your calculation of the
balance of gravitational pressure to Pauli pressure (this time for a gas of neutrons)
to determine the radius of the resulting neutron star. (5 points).
Physics 4340
HW5.4
Spring 2015