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Physics 7440
Spring 2003
Solutions 3
1. Using the Drude model.
In this problem, I asked you to work through an example of the Drude model,
using a ‘Hall bar’ geometry of the type shown in the following figure.
1
2
A
B
3
Hall bar figure. The distance between contacts 1 and 2 is 1 mm. The distance
across the bar between contacts 2 and 3 is 0.1 mm
I gave you the following information: The bar is made of a thin-film
semiconductor material of 100 nm thickness, that a current of 1 mA is passed
from contact A to contact B, that a magnetic field of 1000 Oe is applied in the
direction out of the plane of the figure, and that in this condition, a voltages of
5 mV and 100 V are detected respectively between contacts 1 and 2, and
between contacts 2 and 3.
Your job was to use the Drude model to determine the carrier density and the
scattering rate.
Determining the carrier density
The short route to get at the carrier density is via the Drude prediction for the Hall
coefficient. Recall that the Hall coefficient, RH, is defined as:
RH 
Ey
J x Hz
Within the Drude model, we found that RH is related to the carrier density via:
RH  
1
nec
in cgs units
1
ne
in M KS units
or
RH  
All you have to do is use the measured currents, voltages, and Hall bar geometry
to determine the electric fields and current densities, plug in to solve for RH, and
solve for the carrier density.
Sol 3.1
Physics 7440
Spring 2003
We find:
4
Voltage between 2 and 3 1  10 V
Ey 

 1 V/m
distance between 2 and 3 1 10 4 m
and
current in the bar
1 mA

 1 10 8 A/m 2
cross sectional area of bar width  thickness
Jx 
The field strength is 1000 Oe or 10-1 Tesla so that:
RH 
1 V/m
1

2
1
19
1 10 A/m 10 T n1.6  10 Coulomb 
8
Solving for the carrier density yields:
n  6.3 10 electrons/m
25
3
Determining the scattering rate.
Once we know the electron density, the electronic charge, and the electronic mass,
then a measurement of the electrical conductivity will immediately yield the
scattering rate. Recall that the conductivity is given by:

Jx
Ex
and that within the Drude model, the conductivity takes on the value:

ne 2
m
We already figured out the current density above. The electric field is just:
Ex 
5 mV
 5 V/m
10  10-4 m
Therefore, the conductivity is 2x107 (ohm-m)-1. The scattering rate is then:
 2  10  m 9  10

 6.3  10 m 1.6  10
1
7
25
1
3
31
kg 
19
C
2
 11  1012 sec.
Sol 3.2
Physics 7440
Spring 2003
2. Drift velocity
Calculate the drift velocity, v for an electron in a metal with an applied field of
1000 Volts/cm and relaxation time of   10 14 sec.
OK, the drift velocity is also the terminal velocity or:
vDRIFT 
e E
 177 m / sec
m
3. Use the Boltzmann Eq. to derive a form for the thermal conductivity using:


0 ; F 0 ;
0
t
x
Use the relaxation time approximation and a Fermi-Dirac distribution as the
equilibrium distribution function. Use the electrical conductivity found in class to
generate a value for the Lorenz number i.e., what does our result predict for the
Wiedemann-Franz law.
We have a thermal gradient, so the distribution function is different at different
spatial locations. The relaxation time Boltzmann equation is then:
f  f0
f
v 
r

Again, we look for a first order result
f  f 0  f1
The first order response is given by:
f1  
f0
v
r
In this particular case, the spatial variation is due to the variation of temperature
with position. The equilibrium distribution function will then vary with position
due to its dependence on temperature and indirectly through its dependence on
chemical potential (which depends upon temperature). To be explicit, we have:
 

f1   
f 0 T ,    v
 r

The chain rule then says:
f
 f

f1    0 T  0   v

 T

Sol 3.3
Physics 7440
Spring 2003
Therefore, at this level, we need to understand both the temperature gradient and
the chemical potential gradient before we can proceed with understanding the
final transport physics. The standard approach is to ignore the gradient in
chemical potential. Let's see why we should ignore it. First, notice that the
chemical potential is just some function of temperature. Therefore, it is natural
(and even correct) to relate the chemical potential gradient to the temperature
gradient:
 

T
T
Then, you would apparently have:
f1  
f 0       


 T v
  T
T 
This result is the non-equilibrium distribution that we would get subject only to
the constraint that we have some given temperature gradient and an associated
chemical potential gradient. Because of the chemical potential gradient, we are
Going to have a net particle flow with this particular non-equilibrium distribution.
That's what a chemical potential does: It sets the change in energy due to adding a
particle. If, during the random thermal jiggling, an electron moves one way or
another, it changes the local energy. If you go the wrong way, it costs energy. If
you go the other way, you get overall energy back. Thus, chemical potential
gradients lead to net particle flow. Because the electrons are charged, a net
particle flow will lead to charge separation and the generation of an electric field.
Eventually, the electric field will stop the net particle flow.
Because of this nasty link between chemical potential gradients and electric fields,
we have evolved the idea of the 'electrochemical potential'. Namely, add the
electric potential to the chemical potential to get the electrochemical potential.
Then, gradients of the electrochemical potential lead to net forces on the electrons
and lead to particle flow. In this picture, a gradient in the chemical potential is
thought of as an applied force in exactly the same way as a gradient in the electric
potential leads to an applied force.
The end result is that the initial requirement stated above that the net forces are
zero is usually interpreted as meaning that the gradient in chemical potential is
balanced by some other force, so that the Net Force is zero. I will assume this
condition and will ignore the chemical potential gradient. Then, we have:
f1  
f 0     
T v

T
Next, we want to calculate the associated average current density of heat.
Therefore, we need a definition of the heat current density. Marder's discussion in
the book shows us that the thermal current is:
Sol 3.4
Physics 7440
JQ 
Spring 2003
1
4
3

k  space
d 3 k  f1      v 
 f     2

  3  d k  0
T v  v
T
4 k  space
 

1
3
Let the temperature gradient define the x-direction. Then, we have:
T 1
JQ  
x 4 3
 f     2 
d k 0
vX  v



T


k  space

3
As usual, we can evaluate the angular integrals for various terms. We find that
only the x-component of the velocity vector makes a non-zero contribution:
T 1
JQ  
x 4 3
T 1

x 4 3
 f     2 2 
d k 0
v X  xˆ



T


k  space

3
 f     2 2 2

2
0
k
dkd
cos

d

v
sin

cos


 xˆ





T


k  space

2
Do the angular integrals to find:

 f     2 2 
T 1 1
2
JQ  
4 k dk  0
v  xˆ
x 4 3 3 0
T
 


 f     2 
T 2 1
2

4 k dk  0
  xˆ
x 3m 4 3 0
T
 

In this form, it is clear that you want to convert from an integral over k-space to an
integral over energies. Then, we have:

   
T 2
 f 
J 
d

 D     0 xˆ

x 3m 0
T
  
2
This integral is one of those classic cases where the sharply peaked behavior of
the energy derivative of the Fermi function allows an approximate evaluation of
the integral. However, in this case, you need to go beyond the approximation of a
delta function at the Fermi energy (otherwise, you just get zero). Your best bet is
to evaluate this integral using the Sommerfeld expansion. In Marder's discussion
of the Sommerfeld, you are trying to evaluate something like:

H   d  H   f0  
0


 f 0   
  d    d  H     
0
0
 
Sol 3.5
Physics 7440
Spring 2003
The recipe then tells you to expand the integral. To first order, the expansion says:

H   d  H    
0
2
6
 k BT 
2
H   
Therefore, the solution is:
2
2 

  

T 2   2
2 
  k BT 

JQ  

D

  
x 3m  6
 2  T
    F

Take the derivatives and you find:
JQ  
 2

T 2
 F   kB2T  D  F 
x 3m  3

From Marder, (6.77), we know that the specific heat of the free electron gas is
given by:
 2

2
cv  
kB T D  F 
3



Therefore, we can write the heat current density due to the thermal gradient as:
JQ 
1
T
 vF2 cv
3
x
Finally, the Lorenz number is predicted to be:
1 2
 vF cv
 2 k B2

Watt-Ohm
3
L


 2.4  108
2
2
 T ne 
3e
K2
T
m
4. Pauli Pressure and White Dwarf stars
A star like the Sun is essentially a box of hydrogen, held together by gravity. After
10 billion years or so, the core of the star contains enough helium and other fusion
products that the fusion processes slow down and the star begins to cool and
contract. Eventually, as the star cools, the effects of the Pauli exclusion principle
(namely the Pauli Pressure) become important in stopping further gravitational
collapse. There are two types of fermions inside the star, namely, electrons and
protons. Both make a contribution to the outward pressure:
The Pauli pressure for any particular type of fermion is given by:
PPauli 
N2
N 2 2 k F2
EF 
V 5
V 5 2m
Sol 3.6
Physics 7440
Spring 2003
In this expression, we need to know the number of particles, the volume of the
box they live in, the Fermi wave vector, and the mass. The Fermi wavevector (just
an inverse characteristic length) is given by the density of particles:
13
N

k F   3 2 
V 

In a charge-neutral star, the number of electrons is equal to the number of protons.
Therefore, the number of particles and the density of particles is the same for both
populations. However, the masses of the two particles is quite different. The
electrons are roughly 1800 times lighter. Therefore, the electron Pauli pressure
is 1800 times larger than the proton Pauli pressure.
Therefore, the dominant pressure is given by:
PPauli 
23
N 2 2 k F2 2 2 N 5 3
2 2
N5 3
2 23

3


3 2 


53
53 
V 5 2m
5 2mel V
5 2mel  4 3 
 R 
3

You might imagine that the protons would fall towards the center of the star since
their pressure is so low, but they are attracted to the electrons by the electrostatic
force. This force prevents them from falling into the core.
Let's estimate (this is crude, but does the job) the gravitational energy of a star by
using a result for a uniform sphere of the solar radius and solar mass. Then, the
gravitational energy of the star as something like:
2
GM SUN
EG 
8 RSUN
Rewrite this energy in terms of the solar volume and then take its negative
derivative with respect to solar volume to estimate the inward pressure due to
gravity.
Just take the derivative to find:
PG  
2
2
 1 EG
EG
GM SUN
  GM SUN





4
V
V  8 RSUN  3 V
32 2 RSUN
The mass of the Sun is 2x1030 kg. To very high accuracy, all of the mass is in
protons, each of which is 1.67x10-27 kg. Therefore, the total number of protons in
the Sun is roughly Np=1.2x1057 protons. This number is also the number of
electrons (again, assuming a charge neutral star). To find the radius of a solar
Sol 3.7
Physics 7440
Spring 2003
mass white dwarf, we ask for the radius where the gravitational pressure is
balanced by the electronic Pauli Pressure:
PG  PPauli
or
2
2
GM SUN
2 2 N 5 3  3 

4
5
53
5 2mel RSUN
32 2 RSUN
4 
 
3 
23
Rearrange the terms and solve for the radius to find the radius of the Sun in its
white dwarf stage as:
2
2
N 5 3 32  3 

2
53
mel GM SUN
4 
5  
3 
23
2
RSUN
1.05 x10 

34 2
9 x10
31
1.35 x1095
 6.7 x10
11
3023
 340, 000km
 4 x10 54.4
60
This estimate is comparable to the Earth-Moon radius and is too large by about a
factor of 50 or so from more accurate models for real stars.
For a solar mass of neutrons, the only difference is that we need the neutron Pauli
Pressure, which is roughly 2000 times smaller. Therefore, the neutron
Nuclear stability and neutron stars
Because of the mass difference between the neutron and its decay products:
mn c2  mp c2  me c2
it is energetically favorable for the neutron in free space to decay. The reaction that
actually occurs is:
n  p  e 
However, if the neutron is held inside a box, then the energy is given by the sum of
the rest mass energy, plus the energy level inside the box. If you have a number of
neutrons in the box, then on average, the energy of the typical neutron is:
3
mn c 2  EF ,n
5
Sol 3.8
Physics 7440
Spring 2003
Here, we have listed the Fermi energy of the typical neutron. If the typical neutron
tries to decay and the electron and proton are constrained to remain inside the box,
then the final energy after decay is
mp c2 
3
3
EF , P  me c 2  EF ,el
5
5
The proton and electron energies are again given by their rest mass, plus the energy
associated with being inside the box.
The issue is to determine the condition where the energy of the neutron in the box is
just equal to the proton and electron in the box. Once the neutron energy is less than
the energy of the decay products, it will be stable. Thus, we need:
3
3
3
mn c 2  EF ,n  m p c 2  EF , P  me c 2  EF ,el
5
5
5
Let's assume that the box is a cube of side, L. Then, the volume is L3, and we find that
the neutron is stable if:
m
n
 m p  me  c 2 
3
 EF ,P  EF ,el  EF ,n 
5
or, substituting in the form of the Fermi energy
 mn  m p  me  c 2 
23 1  1
3 2
1
1 
3 2 




2 
5 2
L  m p mel mn 
Solving for the length of the box yields:
12
 3
L 
 10 
 3 
c
2 13
12
 3
 
 10 
 1
1
1 
1




 m p mel mn   mn  m p  me 
 3 
c
2 13
mel
 6 x1013 m
This is a length of order nuclear size scales and explains why the neutrons inside a
nucleus are stable against decay. Another way to say this is that if the density of
neutrons is 1/L3 = 4.6x1036 neutrons/m3, then they will be stable.
Take another look at the balance between gravity and Pauli pressure:
Sol 3.9
Physics 7440
Spring 2003
2
GM S2
2 2 N 5 3  3 

32 2 RS4 5 2mel RS5  4 5 3
 
3 
23
If you rewrite the number of particles in terms of the mass of the star, you find that the
radius is related to the mass of the star by:
2
2
2 2 1 m p  32  3 
RS 
53
5 2mel GM S1 3
4 
 
3 
53
23
or
VS 
1
MS
Therefore, as the star gets more massive, it also gets denser. Eventually, the
density becomes large enough that the protons and electrons are more energetic
than a population of neutrons. At that point, they recombine into neutrons. Then,
because the neutron Pauli pressure is so much lower than the electronic pressure,
the star collapses. The core becomes a neutron star. Usually, the collapse is
violent and blows off the outer layers of the star, yielding a nice nebula and
destroying all habitable planets within a few light years. You get a super nova.
We can estimate the stellar mass where the neutron star is stable by relating the
stellar density to the density for neutron stability. In other words:
MS
N

 4.6  1036 particles/m3
VS VS m p
Plug in the relationship from above for the volume of the star given its mass:
4
mp 

MS
GM S2
4 

   4.6  1036

3
3
2
VS mP  2 2   32 2   3 2   3 


 5 2mel 
4
or
Sol 3.10
Physics 7440
Spring 2003
 32   3 
2 3
M S  4.6  10
36
2 2
4
4
4 
G     mp 
3 
 10 solar masses
2 2 


 5 2mel 
3
Again, this estimate is about 4 times larger than better models yield. Still, it's nice to
know that we're all safe for now.
The radius of the neutron star can be estimated from the white dwarf result by
replacing the electronic mass with the neutron mass. You find a radius that is roughly
2000 times smaller or 170 km. Again, this is about 10x too large, but not bad.
For what it's worth, the major flaw in our calculations is the gravitational model of the
star, NOT the quantum mechanics of the fermions!
Sol 3.11