University of Illinois at Chicago Department of Physics: Classical Mechanics Qualifying Examination

University of Illinois at Chicago
Department of Physics
Classical Mechanics
Qualifying Examination
January 4, 2008
9.00 am – 12:00 pm
Full credit can be achieved from completely correct answers to

4 questions. If the student attempts all 5 questions, all of the
answers will be graded, and the top 4 scores will be counted
toward the exam’s total score.
B
v
C
m
L
M
y
A
x
y
v0
r
!
O O x
m
k
5r0 /4
L 2
A B
L y
L
Spring 1 x Spring 2
m
120 m 120 m
v2 M0 v
3
X X
Decay #2 m2 Decay #1 m3 Decay #3
Classical Mechanics
January 3, 2006
9:00 am-12:00 pm
Full credit can be achieved from completely correct answers

to 4 questions. If the student attempts all 5 questions, all of
the answers will be graded, and the top 4 scores will be
counted toward the exam’s total score
1. A toy consists of two concentric cylinders with inner radius r and outer radius R. A
string is wound around the inner radius and the outer radius can roll without slipping
on a rough floor. The string is pulled at angle with respect to the horizontal.
r

R
a. Draw the free body diagram.

b. Calculate the angular acceleration.
c. Prove that there exists a critical angle c, where if  < c the cylinder rolls away
from the direction it is pulled, and if  > c the cylinder rolls toward the direction
it is pulled.
d. Determine c
2. A positron e+ with energy of 250 GeV/c2 travels along the x axis and collides with a
stationary electron. A single particle V is produced and only V remains after the
collision. Later, V decays into two identical mass (m= 0.1 GeV/c2), unstable muons 
+
and - which have lifetimes of 2 x10-6 s in their rest frame.
a. Calculate the v/c of the positron.

b. What is the mass of particle V?
c. What is the total energy of the particle V in its rest frame?
d. What are the momenta of the electron and positron in the V rest frame?
e. If the muon decays perpendicularly to the x axis in the V rest frame, what
approximate angle does it make with respect to the x axis in the lab frame?
f. How far would the muon travel in one lifetime as measured in the lab frame?
  C
3. A particle of mass m moves in a field F  f (r )r , where f (r ) =  3 and C>0.
r
dl d
a. Calculate , where l  mr 2 .
dt dt
b. Derive the equation of motion for r and show you can write it in form
d 2u m 1 1
 u   2 2 f ( ) , where u  .
d 2
l u u r
d d
Hint. Find the relationship of to for the central force.
d dt

c. Show that a possible solution is spiral orbit of the form r  r0 e . Find all
possible solutions.
d. Show that  varies logarithmically with t for the spiral orbit from part c.
Hint: integrate l to find (t).
4. Two pendulums are coupled by a massless spring with spring constant k. Both
pendulums have massless springs of length L. They are separated by distance D. The
masses are m and 2m. Consider small oscillations.
L L
g
m
2m
X1 X2
a. Solve for the normal modes of the pendulums.

b. Determine the normal coordinates that undergo simple harmonic motion.
5..A bead of mass m moves along a frictionless wire AB. The wire is fixed at point A and
rotates with angular frequency  about the z axis.  is fixed
z

A
 r
B
O
x y
a. Determine the Lagrangian in terms of r,  and azimuthal angle

dr
b. Determine the Lagrange equation as a function of m, , , r and .
dt
c. Solve the equation of motion.
Classical Mechanics
January 4, 2012
9.00 am – 12:00 pm
Full credit can be achieved from completely correct answers to 4

questions. If the student attempts all 5 questions, all of the answers will
be graded, and the top 4 scores will be counted toward the exam’s total
score.
1
Problem 1.
A cylinder of a non-uniform radial density with mass M,
length l and radius R rolls without slipping from rest down
a ramp and onto a circular loop of radius a. The cylinder
is initially at a height h above the bottom of the loop. At h
the bottom of the loop, the normal force on the cylinder is a
twice its weight.
a) Expressing the rotational inertia of the non-uniform

cylinder in the general form (I=βMR2), express the β
in terms of h and a.
2
b) Find numerical value of β if the radial density profile for the cylinder is given by ρ (r ) = ρ 2 r ;
c) If for the cylinder of the same total mass M the radial density profile is given by ρ n ( r ) = ρ n r n ,
where n ∈ 0,1,2,3... , describe qualitatively how do you expect the value of β to change with increasing
n. Explain your reasoning.
Problem 2.
A particle of unit mass is projected with a velocity v0 at right angles to the
radius vector at a distance a from the origin of a center of attractive force, given
⎛ 4 a 2 ⎞
by f (r ) = −k ⎜⎜ 3 + 5 ⎟⎟
⎝ r r ⎠
9k
For initial velocity value given by v02 = , find the polar equation of the resulting orbit.
2a 2
Problem 3.
A simple pendulum of length b and mass m is suspended from a ωt
point on the circumference of a thin massless disc of radius a
that rotates with a constant angular velocity ω about its central
axis. Using Lagrangian formalism, find
a) the equation of motion of the mass m;

b) the solution for the equation of motion for small θ
oscillations. m
2
Problem 4.
A rigid body consists of six particles, each of mass m, fixed to the (0,0,c)
ends of three light rods of length 2a, 2b, and 2c respectively, the
rods being held mutually perpendicular to one another at their
midpoints.
a) Write down the inertia tensor for the system in the

coordinate axes defined by the rods;
b) Find angular momentum and the kinetic energy of the (0,b,0)
system when it is rotating with an angular velocity ω
about an axis passing through the origin and the point (a,0,0)
(a,b,c).
Problem 5.
 
The force of a charged particle in an inertial reference frame in electric field E and magnetic field B
    
( )
is given by F = q E + v × B , where q is the particle charge and v is the velocity of the particle in the
inertial system.
a) Prove that the transformation from a fixed frame to a rotating frame is given by
r r& r& r
& r r r r r
r&= &
r ʹ′ + ω × r ʹ′ + 2ω × vʹ′ + ω × (ω × r ʹ′ )
b) Find the differential equation of motion referred to a non-inertial coordinate system
  
rotating with angular velocity ω = −⎛⎜ q ⎞⎟ B , for small B (neglect B2 and higher order
⎝ 2m ⎠
terms).
3
January 4, 2012
Classical Mechanics Qualifying Exam Solutions

Problem 1.
A cylinder of a non-uniform radial density with mass M,
length l and radius R rolls without slipping from rest down
a ramp and onto a circular loop of radius a. The cylinder
is initially at a height h above the bottom of the loop. At h
the bottom of the loop, the normal force on the cylinder is a
twice its weight.
a) Expressing the rotational inertia of the non-uniform

cylinder in the general form (I=βMR2), express the β
in terms of h and a.
2
b) Find numerical value of β if the radial density profile for the cylinder is given by ρ (r ) = ρ 2 r ;
c) If for the cylinder of the same total mass M the radial density profile is given by ρ n ( r ) = ρ n r n ,
where n ∈ 0,1,2,3... , describe qualitatively how do you expect the value of β to change with increasing
n. Explain your reasoning.
Solution:
a) Centripetal acceleration as the ball rolls around the circular loop at the bottom of the track is ac
= v2/a, and could be expressed from free body diagram equation:
N-W = Mv2/a, where N is the normal force and the W is the weight. We are given that N=2Mg, so
2Mg – Mg – Mv2/a, i. e. v2 = ga
Relating the angular and translational velocities by v=aω, we next use the expression for the total
kinetic energy of rolling object (no slipping)
K = ½ Mv2 + ½ Iω2
1
And apply energy conservation for the ball between its initial position at rest and its position at the
bottom of the loop:
Mgh = ½ Mv2 + ½ Iω2
Substituting for v and for ω (from above) and using I = βMR2 expression we find
h = ½a +½ β a
Rearranging: β = 2h/a -1
b) The moment of inertial about the rotational axis of the cylinder is given by I = ∫ r 2dm . For the
quadratic density profile dm = ρ 2 r 2l 2πrdr
R R
2 2 R6 2 πlR 4 ρ 2
I = ∫ r ρ 2 r l 2πrdr = 2πρ 2l = MR 2 , with M = ∫ ρ (r )dV = ∫ 2πlρ 2 r 3dr = , so
0
6 3 0
2
2
β =
3
c) For an arbitrary value of n:
R R
R n+ 4 2πlρ n R n+2
I = 2πlρ n ∫ r n+3 dr = 2πlρ n , with M = 2πlρ n ∫ r n+1dr = , so
0
n+4 0
n+2
n+2 n+2 1
I= MR 2 β= n = 0, β = ; n → ∞, β → 1
n+4 , n+4, e.g. 2
2
Problem 2.
A particle of unit mass is projected with a velocity v0 at right angles to the radius vector at a distance a
from the origin of a center of attractive force, given by
⎛ 4 a 2 ⎞
f (r ) = −k ⎜⎜ 3 + 5 ⎟⎟
⎝ r r ⎠
9k
For initial velocity value given by v02 = , find the polar equation of the resulting orbit.
2a 2
Solution: Calculating the potential energy

dv ⎛ 4 a 2 ⎞
− = f ( r ) = −k ⎜ 3 + 5 ⎟
dr ⎝ r r ⎠
⎛ 2 a 2 ⎞
Thus, V = −k ⎜ 2 + 4 ⎟
⎝ r 4r ⎠
The total energy is …
1 ⎛ 2 1 ⎞ 1 ⎛ 9k ⎞ 9k
E = To + Vo = vo2 − k ⎜ 2 + 2 ⎟ = ⎜ 2 ⎟ − 2 = 0
2 ⎝ a 4a ⎠ 2 ⎝ 2a ⎠ 4a
Its angular momentum is …
9k
l 2 = a 2vo2 = = constant = r 4θ&2
2
Its KE is …
2 2
1 1 ⎡⎛ dr ⎞ ⎤ 1 ⎡⎛ dr ⎞ ⎤ l 2
T = r&2 + r 2θ&2 = ⎢⎜ ⎟ + r 2 ⎥ θ&2 = ⎢⎜ ⎟ + r 2 ⎥ 4
( )
2 2 ⎣⎢⎝ dθ ⎠ ⎦⎥ 2 ⎣⎢⎝ dθ ⎠ ⎦⎥ r
The energy equation of the orbit is …
2
1 ⎡⎛ dr ⎞ ⎤ l 2 ⎛ 2 a 2 ⎞
T + V = 0 = ⎢⎜ ⎟ + r 2 ⎥ 4 − k ⎜ 2 + 4 ⎟
2 ⎢⎣⎝ dθ ⎠ ⎥⎦ r ⎝ r 4r ⎠
⎡⎛ dr ⎞2 2 ⎤ 9k ⎛ 2 a 2 ⎞
= ⎢⎜ ⎟ + r ⎥ 4 − k ⎜ 2 + 4 ⎟
⎢⎣⎝ dθ ⎠ ⎥⎦ 4r ⎝ r 4r ⎠
2
⎛ dr ⎞ 1 2 2
or ⎜ ⎟ = a − r
⎝ dθ ⎠ 9
( )
dr dφ
Letting r = a cos φ then = −a sin φ
dθ dθ
2
⎛ dφ ⎞ 1 1
So ⎜ ⎟ = ∴φ = θ
⎝ dθ ⎠ 9 3
1
Thus r = a cos θ ( r = a @θ = 0o )
3
3
Problem 3.
A simple pendulum of length b and mass m is suspended from a point on the circumference of a thin
massless disc of radius a that rotates with a constant angular velocity ω about its central axis. Using
Lagrangian formalism, find
a) the equation of motion of the mass m;

b) the solution for the equation of motion for small oscillations.
Solution:
y
a) Coordinates:
ωt
x = a cos ω t + b sin θ
x
y = a sin ω t − b cosθ
x&= −aω sin ω t + bθ&cosθ

y&= aω cos ωt + bθ&sin θ
θ m
1
L = T − V = m ( x&2 + y&2 ) − mgy
2
m
= ⎡⎣ a 2ω 2 + b2θ&2 + 2bθ&aω sin (θ − ω t )⎤⎦ − mg ( a sin ω t − b cosθ )
2
d ∂L
& = mb 2θ&&+ mbaω θ&− ω cos (θ − ω t )
( )
dt ∂θ
∂L
= mbθ&aω cos (θ − ω t ) − mgb sin θ
∂θ
∂L d ∂L
The equation of motion − = 0 is
∂θ dt ∂θ&
ω 2a g
θ − cos(θ − ωt ) +
sin θ = 0
b b
(Note – the equation reduces to equation of simple pendulum if ω → 0 .)
b) For small θ the equation of motion reduces to that of a constant driving force harmonic
oscillator :
g ω 2a
θ + θ = cos(ωt )
b b
4
The general solution for this equation consists of two parts, a complementary function u(t) and
a particular solution v(t). Complementary solution comes from the simple harmonic oscillator
g ⎛ g ⎞
equation: θ + θ = 0 , and could be immediately written as u (t ) = A cos⎜ t − ϕ ⎟ .
b ⎜ b ⎟
⎝ ⎠
Given the form of the driving force, the particular solution could be defined as
v(t ) = B cos(ωt ) .
Taking the derivatives and plugging back in the equation of motion, one finds the expression
ω 2a
for B as B = , and the general solution in form:
g − ω 2b
⎛ g ⎞ ω 2a
θ (t ) = u(t ) + v(t ) = A cos⎜⎜ t − φ ⎟⎟ + 2
cos(ωt )
⎝ b ⎠ g − ω b
5
Problem 4.
A rigid body consists of six particles, each of mass m, fixed to the ends of three light rods of length 2a,
2b, and 2c respectively, the rods being held mutually perpendicular to one another at their midpoints.
a) Write down the inertia tensor for the system in the coordinate axes defined by the rods;
b) Find angular momentum and the kinetic energy of the system when it is rotating with an
angular velocity ω about an axis passing through the origin and the point (a,b,c).
Solution:
a)
I xy = ∑ mi xi yi = 0 since either xi or yi is zero for all six
i
(0,0,c) particles. Similarly, all the other products of inertia are zero.
Therefore the coordinate axes are principle axes.
2 2
( )
I xx = ∑ mi yi2 + zi2 = m ⎡0 + 0 + b2 + ( −b ) + c 2 + ( −c ) ⎤
⎣ ⎦
i
I xx = 2m (b2 + c2 )
I yy = 2m ( a2 + c2 )
(0,b,0)
I zz = 2m ( a 2 + b2 )
(a,0,0) ⎡b2 + c 2 0 0 ⎤
t ⎢ ⎥
I = 2m ⎢ 0 a + c2
2
0 ⎥
⎢ 0
⎣ 0 a 2 + b2 ⎥⎦
⎡ a ⎤
r ω ω ⎢b ⎥
b) ω= 1 ( aeˆ1 + beˆ2 + ceˆ3 ) = 1 ⎢ ⎥
(a 2
+ b2 + c 2 ) 2 ( a2 + b2 + c2 ) 2 ⎢⎣ c ⎥⎦
r tr
Using = I ω ,
L
⎡b2 + c 2 0 0 ⎤ ⎡ a ⎤
r 2mω ⎢ ⎥
L= 1 ⎢ 0 a2 + c2 0 ⎥ ⎢⎢b ⎥⎥
( a 2 + b2 + c 2 ) 2 ⎢ 0
⎣ 0 a 2 + b 2 ⎥⎦ ⎢⎣ c ⎥⎦
⎡ a ( b 2 + c 2 )⎤
r 2mω ⎢ ⎥
L= 1
⎢b ( a 2 + c 2 )⎥
⎢ ⎥
(a 2
+ b2 + c2 ) 2 ⎢c ( a 2 + b 2 )⎥
⎣ ⎦
6
1 r r
Using T = ω ⋅ L
2
⎡ a ( b 2 + c 2 )⎤
1 2mω 2 ⎢ ⎥
T= [ a b c ] (
⎢ b a 2
+ c 2
)⎥⎥
2 ( a 2 + b2 + c2 ) ⎢
⎢c ( a 2 + b 2 )⎥
⎣ ⎦
2
mω
( ) ( ) ( )
T = 2 2 2 ⎡⎣a 2 b2 + c 2 + b2 a 2 + c 2 + c 2 a 2 + b2 ⎤⎦
a +b +c
2mω 2
(
T = 2 2 2 a 2b 2 + a 2 c 2 + b 2 c 2
a +b +c
)
7
Problem 5.
 
The force of a charged particle in an inertial reference frame in electric field E and magnetic field B
    
( )
is given by F = q E + v × B , where q is the particle charge and v is the velocity of the particle in the
inertial system.
a) Prove that the transformation from a fixed frame to a rotating frame is given by
r r& r& r
& r r r r r
r&= &
r ʹ′ + ω × r ʹ′ + 2ω × vʹ′ + ω × (ω × r ʹ′ )
b) Find the differential equation of motion referred to a non-inertial coordinate system
  
rotating with angular velocity ω = −⎛⎜ q ⎞⎟ B , for small B (neglect B2 and higher order
⎝ 2m ⎠
terms).
Solution:
  
a) Let’s consider a coordinate system (i ʹ′, j ʹ′, k ʹ′) , rotating about the axis defined by the unit vector
     
n with respect to the (i , j , k ) system with angular velocity ω = n ω . Position of any point P
in space can be expressed in two systems (in case of common origin) as
       
r = i x + j y + k z = r ʹ′ = i ʹ′xʹ′ + j ʹ′y ʹ′ + k ʹ′z ʹ′
The velocity then can be written

 
  dx  dxʹ′ di ʹ′  di ʹ′   
v =i + ... = i ʹ′ + ... + xʹ′ + ... = vʹ′ + xʹ′ + ... = vʹ′ + ω × r ʹ′
dt dt dt dt
This finding is generally true for any vector, e.g. for derivative of the velocity vectors:
 
⎛ dv ⎞ ⎛ dv ⎞  
⎜ ⎟ = ⎜ ⎟ +ω×v
⎝ dt ⎠ fixed ⎝ dt ⎠ rotating
   
⎛ dv ⎞ ⎛ d (vʹ′ + ω × r ʹ′) ⎞    
⎜ ⎟ = ⎜ ⎟ + ω × (vʹ′ + ω × r ʹ′) =
⎝ dt ⎠ fixed ⎝ dt ⎠ rotating
  
⎛ dvʹ′ ⎞ ⎛ d (ω × r ʹ′) ⎞     
⎜ ⎟ + ⎜ ⎟ + ω × vʹ′ + ω × ω × r ʹ′ =
⎝ dt ⎠ rotating ⎝ dt ⎠ rotating
  
⎛ dvʹ′ ⎞ ⎛ dω ⎞   ⎛ dr ʹ′ ⎞     
⎜ ⎟ + ⎜ ⎟ × r ʹ′ + ω × ⎜ ⎟ + ω × vʹ′ + ω × ω × r ʹ′
⎝ dt ⎠ rotating ⎝ dt ⎠ rotating ⎝ dt ⎠ rotating
Changing notations, one arrives at the transformation equations from a fixed to a rotating frame:
8
r r r r r r& r& r
& r r r r r
v = vʹ′ + ω × r ʹ′ r&= &
r ʹ′ + ω × r ʹ′ + 2ω × vʹ′ + ω × (ω × r ʹ′ )
b) The equation of motion:
r r r r
mr&&= qE + q v × B ( )
Transforming from a fixed frame to a moving rotating frame:
r q r r
ω =− B so ω&= 0
2m
r r r q r r r r r r r r
mr&&ʹ′ − q B × vʹ′ − B × (ω × r ʹ′ ) = qE + q ⎡⎣( vʹ′ + ω × r ʹ′ ) × B ⎤⎦
( ) 2
r r r q r r r r r r r r r
mr&&ʹ′ + q vʹ′ × B + (ω × r ʹ′ ) × B = qE + q vʹ′ × B + q (ω × r ʹ′ ) × B
( ) ( )
2
r&& r q r r r
mr ʹ′ = qE + (ω × r ʹ′ ) × B
2
q r r r q qB
(ω × r ʹ′) × B = ⎛⎜ ⎞⎟ ( r ʹ′)(sin θ )( B ) ∝ B 2
2 2 ⎝ 2m ⎠
r r
Neglecting terms in B 2 , mr&&ʹ′ = qE (Larmor’s theorem)
9
Classical Mechanics
January 4, 2011
9.00 am – 12:00 pm

score.
1
Problem 1.
Two identical rods of mass m and length l are connected to
the ceiling and together vertically by small flexible pieces
of string. The system then forms a physical double
pendulum. Find the frequencies of the normal modes of θ
this system for small oscillations around the equilibrium
position. Describe the motion of each of the normal modes.
Problem 2.
The particle is sliding down from the top of the hemisphere of radius a.
Find: a) normal force exerted by the hemisphere on the particle; b) r
θ
angle with respect to the vertical at which the particle will leave the
a
hemisphere.
Problem 3.
A uniform rectangular plane lamina of mass m and dimensions a
and b (assume b > a) rotates with the constant angular velocity
ω
ω about a diagonal. Ignoring gravity, find: a) principal axes and
moments of inertia; b) angular momentum vector in the body
a
coordinate system; c) external torque necessary to sustain such
rotation.
b
Problem 4.
z
A particle of mass m moves frictionless under the influence of
gravity along the helix z = kθ, r = const, where k is a constant,
and z is vertical. Find: a) the Lagrangian; b) the Hamiltonian.
Determine: c) equations of motion.
2
Problem 5.
A particle of mass m is bound by the linear potential U = kr, where k = const. Find:
a) For what energy and angular momentum will the orbit be a circle of radius r about the origin?
b) What is the frequency of this circular motion?
c) If the particle is slightly disturbed from this circular motion, what will be the frequency of
small oscillations?
3
Classical Mechanics
Tuesday, January 5, 2010

9:00 am – 12:00 pm

score.
Problem 1
A frictionless pulley, constructed from a solid disk of mass M1 and radius
R1, can rotate about its horizontal axis of rotation. A string is wound
around the pulley, with its other end wound around a second pulley of
mass M2 and radius R2 that is falling downwards while maintaining the
horizontal orientation of its axis. Assuming that the string is massless,
does not slip, and remains vertical and taut during the motion, find:
(a) the linear acceleration of the center of mass of the pulley M2
(b) the angular acceleration of the pulley M2
(c) the angular acceleration of the pulley M1
(d) the tension in the string

Problem 2
a) A pion () is moving with a velocity v and decays into a
muon () and an antineutrino ( ). If the antineutrino
moves after the decay perpendicular to the direction of the
, find the energy of the muon and the angle  of the
muon’s direction relative to the  in terms of: the  mass
m, the muon mass m, , and 1/ 1 . (For
simplicity you can assume that the speed of light c=1.)
b) If the  is at rest when it decays into a muon and an antineutrino, find the distance traveled by
the muon before it decays (i.e., during its lifetime) in terms of: m, m, and the muon lifetime
.
Problem 3
A bead of mass m slides under gravity along a
smooth vertical parabolic wire. The shape of the
wire is given by the equation ax2-z=0. The bead
starts from rest at x=x0.
(a) Write the Lagrangian of the system.
(b) Use the Lagrange multiplier method to determine the force that the wire exerts on the bead as
a function of x.

 
Problem 4
A point particle of mass m is moving under the potential
, sin
2
where k is a positive constant.
(a) Write the Lagrangian of the system.
(b) Prove that the origin x=y=0 is a stable equilibrium point and write the Lagrangian
appropriate for small oscillations about this point.
(c) Find the normal frequencies of the system.
(d) Construct the normal coordinates of the system and express the Lagrangian in terms of these
coordinates.
Problem 5
An object of mass m is thrown vertically upward from the earth’s surface with initial speed v0.
There are only two forces acting on the object: its weight and the air resistance which is opposite
to the direction of motion and has a magnitude of kmv2, where k is a positive constant and v is the
object’s speed at time t.
a) Find the maximum height H reached by the mass as a function of: k, v0, and the acceleration of
gravity g.
b) After the mass has reached the maximum height H, it starts falling down. If kv02<g, find the
distance the mass has dropped from its maximum height when it reaches speed v0 as a function
of: k and H.
Classical Mechanics
January 4, 2011
9.00 am – 12:00 pm

score.
1
Problem 1.
Two identical rods of mass m and length l are connected to
the ceiling and together vertically by small flexible pieces
of string. The system then forms a physical double
pendulum. Find the frequencies of the normal modes of θ
this system for small oscillations around the equilibrium
position. Describe the motion of each of the normal modes.
Solution:
ϕ
Let θ (ϕ) be the angle of the top (bottom) rod with vertical.
1 l & 2 1 l 2 1 2 2
T=  m( θ ) + ml θ& + m(lθ& + ϕ& ) + ml ϕ& 
2 2
2 2 12 2 12 
l 3  1  θ 2 θ 2 ϕ 2 
U = mg (1 − cos θ ) + mg  l −  l cos θ + cos ϕ   ≈ mgl  +  +  
2 2  2   4  2 4 
4 2 & 2 ml 2 & 1 mgl

L = T −U = ml θ + θϕ& + ml 2ϕ& 2 − (3θ 2 + ϕ 2 )
6 2 6 4
The Lagrange’s equations are then given by
1  8 && 2  1  8 && 2 1 
 lθ + lϕ&& + gθ  = 0  lθ + lϕ&& + gϕ  = 0
23 3  23 3 2 
1  8 && 2 2  1  && 2 1 2  g
 θ + ϕ&& + ω0 θ  = 0 θ + ϕ&& + ω0 ϕ  = 0 , where ω0 =
2
23 3  2 3 2  l
Assuming small oscillations with θ = A cos ωt and ϕ = B cos ωt gives
3 2 4 2 ω2
 ω0 − ω −  A 
2 3 2   = 0 , which yields normal mode frequencies of
 ω 2
1 2 1 2  B 
 − ω0 − ω 
 2 2 3 
  −2 7 1
 B =  −  A = −2.10 A
6  2 5.27ω 0
2
   3 3
ω = 3 ±
2
ω 0 =  , and 
0.73ω 0  B =  2 7 − 1  A = −1.43 A
2
 7
  3 3 
 
2
Problem 2.
The particle is sliding down from the top of the hemisphere of radius a.
r
Find: a) normal force exerted by the hemisphere on the particle; b) θ
angle with respect to the vertical at which the particle will leave the a
hemisphere.
a) The equation of constraint is f ( r ,θ ) = r − a = 0
T=
2
(
m 2 2 &2
r& + r θ ) V = mgr cos θ
L=
2
(
m 2 2 &2
)
r& + r θ − mgr cos θ
∂L d ∂L ∂f
− +λ =0
∂r dt ∂r& ∂r
∂L d ∂L ∂f ∂f ∂f
− +λ =0 =1 =0
∂θ dt ∂θ& ∂θ ∂r ∂θ
Thus mrθ& 2 − mg cos θ − mr&& + λ = 0
mgr sin θ − mr 2θ&& − 2mrr&θ& = 0
Now r = a , r& = &&
r = 0 so
maθ − mg cos θ + λ = 0
& 2
mga sin θ − ma 2θ&& = 0

g dθ&
θ&& = sin θ and θ&& = θ&
a dθ
g θ& 2
g g
so ∫ θ&dθ& = a ∫ sin θ dθ or 2
= − cos θ +
a a
hence, λ = mg ( 3cosθ − 2 )
b) and when λ → 0 particle falls off hemisphere at

2
θ o = cos −1  
3
3
Problem 3.
A uniform rectangular plane lamina of mass m and dimensions a
and b (assume b > a) rotates with the constant angular velocity
ω about a diagonal. Ignoring gravity, find: a) principal axes and 2
ω
moments of inertia; b) angular momentum vector in the body
coordinate system; c) external torque necessary to sustain such 1
rotation. a
3
b
ma 2 mb 2 m( a 2 + b 2 )
a) I1 = I2 = I 3 = I1 + I 2 = ω
12 12 12 ω1
ω2
ωb ωa
b) ω1 = 1
ω2 = 1
ω3 = 0
(a + b )
2 2 2
(a + b )
2 2 2
r  ma 2  ωb  mb 2  ωa
L = I1ω1eˆ1 + I 2ω2eˆ2 + I 3ω3eˆ3 =   1
ˆ
e1 +   1
eˆ2 + 0eˆ3
 12  (a 2 + b 2 ) 2  12  (a 2 + b 2 ) 2
r mabω
L= 1
( a , b ,0 )
12(a + b )
2 2 2
r
c) In body coordinate system ω = const
r
r dL r r r r
τ = +ω× L = ω× L
dt
i j k
r
τ = ω1 ω2 0 = (ω1L2 − ω2 L1 )eˆ3
L1 L2 0
r mabω 2
τ = 1
(b 2 − a 2 )eˆ3
12(a + b )
2 2 2
4
Problem 4.
z
A particle of mass m moves frictionless under the influence of
gravity along the helix z = kθ, r = const, where k is a constant,
and z is vertical. Find: a) the Lagrangian; b) the Hamiltonian.
Determine: c) equations of motion.
In cylindrical coordinates the kinetic energy and the potential

energy of the spiraling particle are expressed by
& 2  
1
T= m r
& 2 + r2θ& 2 + z
2  
(1)

U = m gz 

Therefore, if we use the relations,
z = kθ & = kθ& 
i.e.,z
 (2)
r = const. 

the Lagrangian becomes
1  r2 2 
L= m & +z
 k2 z & 2  − m gz (3)
2  
Then the canonical momentum is
∂L  r2 
pz = =m  k2 + 1 z
& (4)
∂z
&  
or,
pz
&=
z (5)
 r2 
m  2 + 1
k 
The Hamiltonian is
pz pz2
H = pz z
& − L = pz − + m gz (6)
 r2   r2 
m  2 + 1 2m  2 + 1
k  k 
or,
1 pz2
H = + m gz (7)
2  r2 
m  2 + 1
k 
Now, Hamilton’s equations of motion are
5
∂H ∂H
− =p
&z; =z
& (8)
∂z ∂pz
so that
∂H
− = − m g = p& z (9)
∂z
∂H pz
= =z
& (10)
∂pz  r2 
m  2 + 1
k 
Taking the time derivative of (10) and substituting (9) into that equation, we find the equation of
motion of the particle:
g
z=
&& (11)
r2

 k2 + 1
 
6
Problem 5.
A particle of mass m is bound by the linear potential U = kr, where k = const. Find:
a) For what energy and angular momentum will the orbit be a circle of radius r about the origin?
b) What is the frequency of this circular motion?
c) If the particle is slightly disturbed from this circular motion, what will be the frequency of
small oscillations?
dU
The force acting on the particle is F = − rˆ = −krˆ
dr
k
a) For particle moving on a circular orbit of radius r: mω 2 r = k , i.e. ω 2 =
mr
mv 2 mω 2 r 2 3kr
The energy of the particle is then E = kr + = kr + =
2 2 2
k
Its angular momentum about the orbit is L = mωr 2 = mr 2 = mkr 3
mr
k
b) The angular frequency of circular motion is ω = .
mr
L2
c) The effective potential is U eff = kr + .
2mr 2
The radius r0 of the stationary circular motion is given by
1
 dU eff  L2  L2  3
  = k − = 0 , i.e. r0 =  
 dr  r =r0 2mr03  mk 
4 1
 d 2U eff  3L2 3L2  mk  3  mk  3
As   = =   = 3k  2  , the angular frequency of oscillations about
 dr 2  4
m  L2   L 
  r =r0 2mr r =r0
r0, if it is slightly disturbed from the stationary circular motion, is
1

1  d U eff
2
  mk  3
3k
ωr = = 3k  2  = = 3ω0 , where ω0 is the angular frequency of the

m  dr 2
 r =r0 L  mr0
stationary circular motion.
7
Classical Mechanics
January 6, 2009
9.00 am – 12:00 pm

questions. If the student attempts all 5 questions, all of the
Problem 1
Long chain molecules sometimes have two types of bonds – strong bonds within small
molecular units, and weak bonds between these molecular units. As a model for the normal
mode vibrations in this situation, consider the longitudinal oscillations along the x-axis of the
spring and ball system shown below, where the side walls are rigid. The springs obey
Hooke’s law with spring constants k and αk, and the masses mi are related to each other as
indicated. α and β are constants with α < 1 and β arbitrary.
x-axis
αk k k αk m
m1 = m3 = m m2 =
β
m1 m2 m3
a) Give the Lagrangian that describes the small amplitude vibrations of this system parallel to
the x-axis.
b) Obtain the equations for motion for harmonic oscillations, i.e., xi = ai e iω t , where ai are
m
the amplitudes. Express your final equations in terms of the parameter λ = ω 2 .
k
⎡ ( 1 + α ) −1 0 ⎤
c) Show that the dynamical matrix has the form ⎢ − β 2β −β ⎥ .
⎥
⎢
⎢ 0
⎣ −1 (1 + α )⎥⎦
Find the normal frequencies λο, λ+, λ− by diagonalizing the secular determinant for this
motion.
[Hint: In your answer one of the frequencies, denoted here by λο, should depend only on α.]
d) The eigenvector for λο has a 20 = 0 . What is the relationship between a10 and a30 for this
mode?
e) The eigenvectors for λ+ and λ− do not have a2 = 0, but satisfy a2± = a1± ( 1 − λ± 2β ) .
In this case, use the equations of motion obtained in part (b) to prove that a1± = a3± .
[Hint: To solve this, it is not necessary to use the explicit forms for λ+(α,β) found in part (c).]
For one of these two modes x2 is in phase with x1 and x3, and for the other mode x2 is out of
phase with x1 and x3. Which mode has the higher frequency?
Problem 2
A pointlike mass m is undergoing a three-dimensional projectile motion in a uniform

gravitational field g. Consider that the air resistance is negligible.
a) Write the Hamiltonian function for this problem H = H ( pi , qi ) , choosing as generalized

coordinates the Cartesian coordinates (x,y,z) with the z axis pointing in the vertical direction.
b) Show that these equations lead to the known equations of motion for Projectile motion and
that the Hamiltonian function corresponds to the Total Mechanical energy of the particle.
c) Once the particle reaches the highest point in its trajectory, a retarding force proportional
to the velocity of the particle starts acting. Assume the proportionality constant to be known
and given by k. Calculate the vertical velocity as a function of time for the descending
particle, and find the terminal velocity.
Problem 3
Consider the one-dimensional motion of a rocket in outer space. The rocket is not subject to
the influence of any external force, but rather moves by the reaction of ejecting mass at high
velocities. At some arbitrary time t, the instantaneous total mass of the ship is m(t), and its
instantaneous velocity with respect to an inertial reference system is v(t). During a time
interval dt, a positive mass dm is ejected from the rocket engine with a speed –u with respect
to the rocket. Consider that the rate at which mass is ejected from the rocket is constant and
given by μ. How far will the rocket have traveled once it lost half of its initial mass? Assume
at the initial time the rocket’s total mass is M0, and it started from rest. Suppose the velocities
involved are small enough to allow you to treat the problem with non-relativistic mechanics.
Problem 4
A yo-yo of mass m and rotational inertia I rolls down due to gravity. The end of its string is
attached to a spring of negligible mass and spring constant k. The radius of the axle of the yo-
yo is a, as indicated in the figure. Let x be the extension of the spring measured with respect
to its natural length.
x
gravity
θ
a
(a) Using the generalized coordinates x and θ indicated in the figure, write the Lagrange
equations of motion for the yo-yo.
(b) Find the oscillation frequency of the spring while the yo-yo is rolling down.
(c) Consider a limit of a thin axle (ma2<<I) and solve the differential equation for x you
found in (b). Explain the motion that is described by the solution.
Problem 5
The figure illustrates two disks of radii a and b mounted inside a fixed circular track of
radius c , such that c = a + 2b . The central disc A is mounted to a drive axle at point O .
Disc B is sandwiched between disc A and track C and can roll without slipping when disc
A is driven by an externally applied torque K through its drive axle. Initially, the system is
at rest such that the dashed lines denoting the spatial orientation of discs A and B line up
horizontally in the figure. K is then applied for a time t 0 causing disc A to rotate, such that
at time t 0 the dashed line denoting its spatial orientation makes an angle α with the
horizontal. Disc B rolls between the track and disc A , and its orientation is denoted by the
dashed line making an angle β with the direction towards O . Let the moments of inertia of
1 1
A and B be I A = M A a 2 and I B = M B b 2 , respectively. The angular velocity of A is α&
2 2
& &
while that of B is β − α& + φ .
a) Taking into account that disc B is rolling without slipping, find expressions for the
angles β and φ as a function of a , b , and α . HINT: For rolling without slipping,
the lengths traveled along the perimeters of disks A and B must be equal to the arc
length traveled along the track C.
a
b) Show that the angular velocity of disc B is equal to ω B = α&
2b
c) Calculate the final angular velocities of the two discs at t = t 0 .
Problem 1
a) Let x1, x2, x3 be the displacements of the masses along the x-axis.
1 ⎛ 1 ⎞
T = m ⎜ x&12 + x& 22 + x& 32 ⎟
2 ⎝ β ⎠
U = α kx12 + α kx32 + k ( x1 − x2 ) + k ( x3 − x2 )
1 1 1 2 1 2
2 2 2 2
Then L = T − U
b)
d ∂L ∂L
=
dt ∂x& i ∂xi
i = 1 mx &&1 = −α kx1 − k ( x1 − x2 )
m
i=2 x = k ( x1 − x2 ) + k ( x3 − x2 ) = k ( x1 − 2 x2 + x3 )
&&
β 2
i =3 mx &&3 = −α kx3 − k ( x3 − x2 )
and letting xi = ai e iω t one obtains

αk k
i =1 − ω 2 a1 = − a1 − ( a1 − a2 ) ⇒ (1+α ) a1 − a2 = λ a1
m m
kβ
i=2 − ω 2 a2 =
m 1
( a − 2a2 + a3 ) ⇒ − β a1 + 2 β a2 − β a3 = λ a2
αk
a3 − ( a3 − a2 )
k
i =3 − ω 2 a3 = − ⇒ − a2 + ( 1+α ) a3 = λ a3
m m
c) The last result in (b) is reproduced by the matrix equation
⎡( 1 + α ) −1 0 ⎤ ⎡ a1 ⎤ ⎡ a1 ⎤ (1 + α ) − λ −1 0
⎢ −β ⎥ ⎢ ⎥ ⎢ ⎥
⎢ 2β −β
⎥ ⎢ a2 ⎥ = λ ⎢ a2 ⎥ So, we solve the sec. eq. −β 2β − λ −β =0
⎢⎣ 0 −1 ( 1 + α ) ⎥⎦ ⎢⎣ a3 ⎥⎦ ⎢⎣ a3 ⎥⎦ 0 −1 (1 + α ) − λ
{(1 + α ) − λ }{⎡⎣(1 + α ) − λ ⎤⎦ [ 2 β − λ ] − β } + 1{⎡⎣(1 + α ) − λ ⎤⎦ [ − β ]} = 0
{(1 + α ) − λ }{⎡⎣(1 + α ) − λ ⎤⎦ [ 2 β − λ ] − 2 β } = 0
and expanding {(1 + α ) − λ }{λ 2 − (1 + α + 2 β ) λ + 2αβ } = 0
12
(1 + α + 2 β ) ± ⎡⎢( 1 + α + 2 β )2 − 8αβ ⎤⎥
1 1
Hence: λo = 1 + α λ± =
2 2⎣ ⎦
d) Method 1: We were told in the part (c) hint that λο should depend only on α. Hence, we
can confirm that the 1 + α eigen frequency, indeed, corresponds to the λο solution. From (b):
1
(1+α ) a1 − a2 = λoa1 ⇒
(o)
a2 = 0
(o) (o)
then − β a1 + 2 β a2 − β a3 = λo a2 ⇒ − β a1 − β a3 = 0
(o) (o)
Hence a1 = −a3
(o) (o)
Method 2: By symmetry a1 = −a3 . The ball and spring system is fully symmetric under
reflection about the origin. The only way to satisfy this symmetry when
(o) (o) (o)
a2 = 0 is to have a1 = −a3 . This mode has the form,
e) From part (b):

(1 + α ) a1 − a2 = λ± a1 ⇒ (1 + α − λ± ) a1± = a2±
−a2 + ( 1 + α ) a3 = λ± a3 ⇒ (1 + α − λ± ) a3± = a2±
Hence a1± = a3±
The out of phase mode has the higher frequency. For the in-phase mode all atoms move in the
same direction, so it has the heaviest effective mass and must have the lowest frequency.
2
Problem3
The rocket is an isolated system and to conserve linear momentum, it will have to move in the opposite than the
ejected mass. We assume all motion is in the x direction and eliminate the vector notation. We know that the
dm
rate at which mass is ejected is constant, therefore μ = and use the conservation of linear momentum
dt
before and after the rocket ejected a mass dm.

Initial momentum at time t: m v
NOTE both m and v are functions of time.
Final momentum at time t+dt: (m − dm) (v + dv) + d m(v − u )
NOTE: First term is rocket, second term is expelled gas, where (v‐u) is the velocity of the gas with respect to the
inertial reference system.

Conservation of linear momentum:
p (t ) = p(t + dt )
m v = (m − dm)(v + dv) + dm(v − u )
dv dt uμ dt
m
dt
= uμ → dv = uμ
m(t )
integrating ∫ dv = M ∫ μt

0 (1 − )
M0
⎛ μt ⎞
Gives v(t ) = −u ln⎜⎜1 − ⎟
⎝ M 0 ⎟⎠
∫
Integrating again & using that ln x dx = x ln x − x

⎡⎛ M ⎞ ⎛ μt ⎞ ⎤
x(t ) = u ⎢⎜⎜ 0 − t ⎟⎟ ln⎜⎜1 − ⎟⎟ + 1⎥
⎣ ⎝ μ ⎠ ⎝ M 0 ⎠ ⎦

M0
At t =t 1 , the mass of the rocket has halved μ t 1 = and the distance travelled is
2 2 2
x(t 1 ) = ut 1 (1 − ln 2)
2 2

Electricity & Magnetism

January 7, 2008
9.00 am – 12:00 pm

1
1. Consider a ring of uniform charge density λ and radius R that lies within the xy-plane. The origin of the
coordinate systems is located at the center of the ring.
a) Give the potential at the point P~ = (ρ0 , ϕ, z) in terms of λ, R, ρ0 , ϕ, and z.
b) We next put a conducting plane into the z = d plane. The potential of the conducting plane is fixed at V = 0.
Compute the total potential at a point P~ = (ρ0 , ϕ, z).
c) Give an explicit form of the induced charge density at P~ = (0, 0, d)? Your final answer should contain no integrals
or derivatives.
2
2. a) Using Gauss’ law, compute the capacitance per unit length of two infinitely long cylindrical conductors of radii
a and b that are parallel and separated by a distance d À a, b, as shown in the figure below
a
b
b) Consider next two infinitely long concentric cylinders, as shown in the figure below. The inner cylinder of radius a
is a conductor with linear charge density λ1 > 0. The second cylinder with inner radius b and outer radius c consists
of a material with permitivity ε3 and is uniformly charged with line charge density λ3 < 0 (λ1 > |λ3 |). The space
between the two cylinders (i.e., a < r < b) is filled with a medium of permitivity ε2 . The medium outside the outer
cylinder possesses the permitivity ε0 . Compute the potential difference between a point at |~r| = 2c and the center of
the inner cylinder.
ε0
ε3
b ε2
c
3
3. Two concentric spheres have radii a and b (b > a) and each is divided into two hemispheres by the same horizontal
plane, as shown below. The potential of the upper hemisphere of the inner sphere and the lower hemisphere of the
outer sphere is kept at V . The other hemispheres are at zero potential.
Φ=0
Φ=V
Φ=0
Φ=V
a) Derive an explicit form of the boundary conditions, using a series expansion of the potential in Legendre
Polynomials.
b) Derive an expression for the coefficients (to all order) in the series expansion of the potential in the region
a < r < b. Give the explicit form of the coefficients up to l = 2.
c) Which coefficients in the series vanish for r < a and which vanish for r > b? Why?
4
4. Consider a cylinder of radius R and length L that is uniformly charged with charge density ρ. The cylinder rotates
with a uniform angular velocity ω around the z-axis, which is also the center axis of the cylinder, as shown in the
figure below
ω
h
a) Compute the current density, J, as a function of distance, r, from the center of the cylinder.
~ along the z-axis at ~r = (0, 0, h).

b) Compute the magnetic induction, B,
c) If the charge on the cylinder is kept the same, but redistributed such that the charge density obeys ρ(r) = αrn , do
you expect that the resulting magnetic field is smaller or larger than that you computed in part b)? Explain!
5
5. An electromagnetic plane wave is incident perpendicular to a layered interface, as shown in the figure below. The
indices of refraction of the three media is n1 , n2 and n3 while the permeability of all three regions is the same, µ0 .
The thickness of the intermediate layer is d. Each of the other media is semi-infinite.
d
E
B k1
n1 n2 n3
z=0 z=d
a) State the boundary conditions at both interfaces in terms of the electric fields.
b) Compute the ratio between the incident electric field in medium 1 and the transmitted electric field in medium 3,
2
i..e, compute |Ei /Et | .
2
c) If the thickness d is varied, the ratio |Ei /Et | oscillates. What is the period of the oscillation? Assuming n1 <
2
n2 < n3 , for which values of d is |Ei /Et | the smallest?
6
I. MATHEMATICAL FORMULAE
A. Definitions
Z
1 ρ(~r0 )
Φ(~r) = d3 r 0
4πε0 |~r − ~r0 |
~ (~r) = −∇Φ(~r)
E
Z 0
~ (~r) = µ0
B ~ r0 ) × ~r − ~r
d3 r0 J(~
4π 3
|~r − ~r0 |
Z ~ r0 )
~ (~r) = µ0
A d3 r0
J(~
4π |~r − ~r0 |
Z
∆Φ = − ~ (~r) · d~r
E
Q ∂Φ
C= ; σ = −ε0
∆Φ ∂n
~ = ρ ~ =0
∇E ; ∇B
ε0
~
∂B
~ = −
∇×E ; ~ = µ0 J~
∇×B
∂t
B. Integrals and Series
Z 2π · ¸
dϕ 1 −2b
√ = K where K is the complete elliptic integral
0 a − b cos ϕ a−b a−b
Z b
x3 2a2 + b2
3/2
dx = 1/2
− 2a
0 [a2 + x2 ] [a2 + b2 ]
 
Z c 2 2
·q ¸ hp i
 2 (a + x) + b  2
dx  h i1/2 − 2 (a + x) = (a + c) (a + c) + b2 − (a + c) − a a2 + b2 − a
0 2
(a + x) + b2

Z 
 0 for even l
1
dx Pl (x) = 1 for l = 0
0  l−1
 (−1) 2 l+1 (l+1)(l−1)!
for odd l
2
2 [( l+1
2 )!]
Z 0 Z 1
l
dx Pl (x) = (−1) dx Pl (x)
−1 0
Z 1
2 2
dx [Pl (x)] =
−1 2l + 1
Xh i
Φ(r, θ) = An rn + Bn r−(n+1) Pn (cos θ)
n
1. Consider a ring of uniform charge density  and radius R that lies within the xy-plane. The origin of the
coordinate systems is located at the center of the ring.
   0 , , z in terms of , R,  0 , , and z.
a) Give the potential at the point P
b) We next put a conducting plane into the z  d plane. The potential of the conducting plane is fixed at
V  0. Compute the total potential at a point P    0 , , z.
c) Give an explicit form of the induced charge density at P   0, 0, d? Your final answer should contain no
integrals or derivatives.
Solution:
a) Give the potential at the point P   0 , , z in terms of , R,  0 , , and z.
Let r  Rcos , sin , 0 be a point along the ring, then
2
 
P 1  
dq
 1  dl
 − r
 R 0 d
4 0 P − r 4 0 P 4 0 R cos  −  0  2  R sin  2  z 2
2
 R
4 0
0 d
R2   20  z 2 − 2 0 R cos 
R 1 −2 0 R
 K
4 0 R2   20  z 2 − 2 0 R R   0  z 2 − 2 0 R
2 2
where K is the complete elliptic integral of the first kind.

b) We next put a conducting plane into the z  d plane. The potential of the conducting plane is fixed at
V  0. Compute the total potential at a point P   0 , , z.
Using the method of images, the image charge is a ring of charge density − and radius R that is parallel to
the conducting plane at a distance of 2d from the first ring. The resulting total potential is then
  R 1 −2 0 R
 tot P K 2   2  z 2 − 2 R
4 0 R  2  20  z − 2 0 R
2 R 0 0
−2 0 R
− R 1 K
4 0 R   0  2d − z 2 − 2 0 R
2 2 R  2  20  2d − z 2 − 2 0 R
  0, 0, d? Your final answer should contain no

c) Give an explicit form of the induced charge density at P
integrals or derivatives.
The charge density follows from
  − 0 ∂ tot
∂−z zd
  0, 0, d, we can compute the potential simply via

Note that at P
2
 
P 1  
dq
 1  
dl  R 0 d
 R
4 0 P − r 4 0 P − r 4 0 R2  z2 2 0 R2  z2
and hence
   R
 tot P 1 − 1
2 0 R2  z2 R 2  2d − z 2
and thus
2z − 2d
  − 0 ∂ tot   0 R − 1 2z − − dR
∂−z zd 2 0 2 R  z 2 
2 3/2
R  2d − z 2
2 3/2
R 2  d 2 
3/2
zd
2. a) Using Gauss’ law, compute the capacitance per unit length of two infinitely long cylindrical
conductors of radii a and b that are parallel and separated by a distance d  a, b, as shown in the figure below
a
b
b) Consider next two infinitely long concentric cylinders, as shown in the figure below. The inner cylinder
of radius a is a conductor with linear charge density  1  0. The second cylinder with inner radius b and outer
radius c consists of a material with permitivity  3 and is uniformly charged with line charge density  3  0
( 1  | 3 |). The space between the two cylinders (i.e., a  r  b) is filled with a medium of permitivity  2 .
The medium outside the outer cylinder possesses the permitivity  0 . Compute the potential difference between
a point at |r|  2c and the center of the inner cylinder.
ε0
ε3
b ε2
.
Solution:
a) Using Gauss’ law, compute the capacitance per unit length of two infinitely long cylindrical conductors
of radii a and b that are parallel and separated by a distance d  a, b.
Using Gauss’ law, we have outside the cylinders
 E  da  2rlE  l 1 
 0  E  2 0 r
Moreover, the electric fields of the two cylinders are pointing in the same direction on the line connnecting
them. I thus obtain
d−b
1 dr   d−a 1 dr d − bd − a
Δ  −  E
 tot  dr  −  a r r  −  ln
2 0 b 2 0 ab
and thus
C
Q
 l  C  2 0
|Δ|  d−bd−a l d−bd−a
2 0
ln ab
ln ab
b) Consider next two infinitely long concentric cylinders. The inner cylinder of radius a is a conductor with
linear charge density  1  0. The second cylinder with inner radius b and outer radius c consists of a material
with permitivity  3 and is uniformly charged with line charge density  3  0 ( 1  | 3 |). The space between
the two cylinders (i.e., a  r  b) is filled with a medium of permitivity  2 . The medium outside the outer
cylinder possesses the permitivity  0 . Compute the potential difference between a point at |r|  2c and the
center of the inner cylinder.
We need to compute the electric field in the different regions of the problem.
i) r  a. We have E  0 inside the inner cylinder is zero, since it is a conductor.
ii) a  r  b. We use
 D  da   1 l  2rlD   1 l  D  1 1
2 r
and thus
E  D2  1 r1
2 2
Note that since  1  0 the electric field points radially outwards.
iii) b  r  c. Note that the material in this region is insulating and uniformly charged. The volume charge
density is
 3
c 2− b2 
and thus
r 2 − b 2  r 2 − b 2  r 2 − b 2 
 D  da  1  3
c 2 − b 2 
l  2rlD  1  3
c 2 − b 2 
l  D  1 1r  1   3 2
2 c − b 2 
and thus
r 2 − b 2 
E  D3  1 1r  1   3 2
2 3 c − b 2 
iv) c  r. Here we have
E 1 1  3
2 0 r
We can now compute the potential difference

0 c b a 0
Δ  −  E
 tot  dr  −  E dr −  E dr −  E dr −  E dr
2c 2c c b a
where
c
−  E dr  − 
c
1  1   3 dr   1   3 ln 2
2c 2c 2 0 r 2 0
and
1 1  1   3 r − b 
2 2
b
−  E dr  − 
b
dr  − 
b
1 1 1 − 3 b2 −
b
3 r dr
c c 2 3 r c 2 − b 2  c 2 3 r c − b 2 
2
c 2 3 c 2 − b 2 
 1 1 − 3 b2 ln c   3
2 3 c − b 2 
2 b 4 3
and
a
−  E dr  − 
a
1  1 dr   1 ln b
b b 2 2 r 2 2 a
and
0
−  E dr  0
a
Thus I obtain
Δ   1   3 ln 2  1 1 − 3 b2 ln c   3   1 ln b
a
2 0 2 3 c − b 2 
2 b 2 2 2
3. Two concentric spheres have radii a and b (b  a) and each is divided into two hemispheres by the same
horizontal plane, as shown below. The potential of the upper hemisphere of the inner sphere and the lower
hemisphere of the outer sphere is kept at V. The other hemispheres are at zero potential.
Φ=0
Φ=V
Φ=0
Φ=V
Polynomials.
b) Derive an expression for the coefficients (to all order) in the series expansion of the potential in the
region a  r  b. Give the explicit form
of the coefficients up to l  2.
c) Which coefficients in the series vanish for r  a and which vanish for r  b? Why?
Solution:
Polynomials.
The problem possesses an azimuthal symmetry, hence
r,   ∑A n r n  B n r −n1 P n cos 
n
1 1
  −1 dxr, xP l x  ∑A n r n  B n r −n1   −1 dx P n xP l x
n
 ∑A n r n  B n r −n1  2l2l,n1  2 A a l  B l a −l1 

2l  1 l
n
And thus we have as the boundary conditions

1
 −1 dx a, xP l x  2 A a l  B l a −l1 
2l  1 l
1
 −1 dx b, xP l x  2l 2 1 A l b l  B l b −l1 
b) Derive an explicit expression for the coefficients (to all order) in the series expansion of the potential in
the region a  r  b.
In order to derive the coefficients, I note that
0 for even l
1 1
 −1 dx a, xP l x  V  0 dx P l x  V 1 for l  0 ≡ VI l
l−1 l1l−1!
−1 2
l1 2 for odd l
2 l1 2
!
and similarly
1 0 1
 −1 dx b, xP l x  V  −1 dx P l x  V−1 l  0 dx P l x
hence I obtain
VI l  2 A a l  B l a −l1 
2l  1 l
V−1 l I l  2 A b l  B l b −l1 
2l  1 l
and we can simply solve from the first equation
A l  1l 2l  1 VI l − B l a −l1
a 2
and inserted into the second equation
V−1 l I l  b l 2l  1 VI l  B l b −l1 − b l B l a −l1

2
2l  1
al 2 al
l
−1 l − b l VI l  2 B l a −l1 a l1 − b l
a 2l  1 b a
l 2l1
1− −a VI l  2 B a −l1 1 − a
b 2l  1 l b
l 2l1 −1
B l  2l  1 1 − − a VI l a l1 1 − a
2 b b
and hence
l 2l1 −1
A l  1l 2l  1 VI l − 2l  1 1 − − a VI l a l1 1 − a a −l1
a 2 2 b b
l 2l1 −1
 2l  1 VIl l 1 − 1 − − a 1− a
2 a b b
 2l  1 2l1VI l 2l1 −1 l b l1 − a l1
2 b −a
c) Which coefficients in the series vanish for r  a and which vanish for r  b? Why?
All A l ≡ 0 for r  b and all B l ≡ 0 for r  a, since otherwise the potential diverges.
4. Consider a cylinder of radius R and length L that is uniformly charged with charge density . The
cylinder rotates with a uniform angular velocity  around the z-axis, which is also the center axis of the
cylinder, as shown in the figure below
ω
h
a) Compute the current density, J, as a function of distance, r, from the center of the cylinder.
b) Compute the magnetic induction, B  , along the z-axis at r  0, 0, h.
c) If the charge on the cylinder is kept the same, but redistributed such that the charge density obeys
r  r n , do you expect that the
resulting magnetic field is smaller or larger than that you computed in part b)? Explain!
Solution:
a) Compute the current density, J, as a function of distance, r, from the center of the cylinder.
The current density, J, is given by
Jr   vr   
  r
Jr′   Jr    r
b) Compute the magnetic induction, B , along the z-axis at r  0, 0, h.

The magnetic induction is given using r  0, 0, h by
 r   0
B  d 3 r ′ Jr′   r − r ′
4 |r − r ′ |
3
and hence
 0 L 2 R
r′
B z r  0
4
 d 3 r ′ Jr′  cos′ 2 
4
 0 dz ′  0 d ′  r ′ dr ′   r ′ 3/2
|r − r | 0 h    r ′  2
z′ 2
r ′  2h  z ′  2  R 2
3
0 L R 0 L
    dz ′  dr ′     dz ′ − 2h  z ′ 
2 0 0 h  z ′  2  r ′  2
3/2 2 0 h  z ′  2  R 2
1/2
0
   h  L h  L 2  R 2 − h  L − h h2  R2 − h
2
c) If the charge on the cylinder is kept the same, but redistributed such that the charge density obeys
r  r n , do you expect that the resulting magnetic field is smaller or larger than that you computed in part
b)? Explain!
Since r  r n , more charge is located close to the perimeter of the cylinder, leading to an increase in
Jr (more specifically, the averaged Jr increases, since
R R R
Jr  1
R
 0 rdrJr  1
R
 0 rdr  r  
R
 0 dr r n2   R n3  Q n  2
R n3 2 n  3
where I used
R R
2  rdrr  Q  2  dr r n1  2 R
n2
0 0 n  2
n  2Q

2R n2
On the other hand, the charge is moved further away from r  0, 0, h. This two effects oppose each other, so
in general it depends on h and n whether the magnetic induction is increased or decreased. However, for
h  R, the magnetic induction increases with increasing n.
5. An electromagnetic plane wave is incident perpendicular to a layered interface, as shown in the figure
below. The indices of refraction of the three media is n 1 , n 2 and n 3 while the permeability of all three regions is
the same,  0 . The thickness of the intermediate layer is d. Each of the other media is semi-infinite.
d
E
B k1
n1 n2 n3
z=0 z=d
b) Compute the ratio between the incident electric field in medium 1 and the transmitted electric field in
medium 3, i..e, compute |E i /E t | 2 .
c) If the thickness d is varied, the ratio |E i /E t | 2 oscillates. What is the period of the oscillation? Assuming
n 1  n 2  n 3 , for which values of d is |E i /E t | 2 the smallest?
Solution:
The EM wave contains only components that are perpendicular to the interface. In region 1, there is an
incoming and a reflected wave, in region 2 there is a right-moving and a left-moving wave, and in region 3,
there is only a transmitted wave. Thus the boundary conditions at z  0 are
Ei  Er  E  E−
Ei − Er  E − E−  E − E  n2
c1 c2 i r n 1 E  − E − 
and at z  d we have
E  e ik 2 d  E − e −ik 2 d  E t e ik 3 d
E  e ik 2 d − E − e −ik 2 d  E t e ik 3 d  E e ik 2 d − E e −ik 2 d  n 3 E e ik 3 d
c2 c3  − n2 t
b) Compute the ratio between the incident electric field in medium 1 and the transmitted electric field in
medium 3, i..e, compute |E i /E t | 2 .
From the last two equations, I obtain
E  1 1  n3 ik 3 d −ik 2 d
2 n2 Ete e
E−  1 1 − n3 ik 3 d ik 2 d
2 n2 Ete e
and from the first two equations
2E i  1  nn 21 E   1 − nn 21 E −
 1  nn 21 1 1  nn 32 E t e ik 3 d e −ik 2 d  1 − nn 21 1 1 − n 3 E t e ik 3 d e ik 2 d
n2
2 2
and thus
4 E i  e ik 3 d e −ik 2 d 1  nn 21 1  nn 32  1 − nn 21 1 − nn 32 e i2k 2 d
Et
and hence
2
16 E i  1  nn 21 1  nn 32  1 − nn 21 1 − nn 32 e i2k 2 d
Et
 1  nn 21 1  nn 32  1 − nn 21 1 − nn 32 e −i2k 2 d
 1  nn 21 1  nn 32  1 − nn 21 1 − nn 32
2 2 2 2
 1  nn 21 1  nn 32 1 − nn 21 1 − nn 32 2 cos2k 2 d
 1  nn 21 1  nn 32  1 − nn 21 1 − nn 32
2 2 2 2
 2 1 − nn 21 1 − nn 32
2 2
1 − 2 sin 2 k 2 d
 4 1  nn 21 nn 32 − 4 1 − nn 21 1 − nn 32
2 2 2
sin 2 k 2 d
or alternatively
2
Ei 1  nn 21 nn 32 n2 n3
2 2 2
 1 − 1− n1 1− n2 sin 2 k 2 d
Et 4
c) If the thickness d is varied, the ratio |E i /E t | 2 oscillates. What is the period of the oscillation? Assuming
n 1  n 2  n 3 , for which values of d is |E i /E t | 2 the smallest?
The period of the oscillation is  2 , and |E i /E t | 2 is the smallest for d  2n  1 2 /2.
Mathematical Formulae
Definitions
 ′
r  1
4 0
 d 3 r ′ |rr
− r ′ |
 r  −∇r
E
 r   0  d 3 r ′ Jr′   r − r ′
B
4 |r − r ′ |
3
 ′
 r   0  d 3 r ′ Jr 
A
4 |r − r ′ |
Δ  −  E
 r  dr
  − 0 ∂
Q
C ;
Δ ∂n
   ;
∇E  0
∇B
0
  − ∂B    0J
∇E ; ∇B
∂t
Integrals and Series

2
0 d
 1 K −2b
a−b a−b
where K is the complete elliptic integral
a − b cos 
b
0 dx  2a  b 1/2 − 2a
2 2
x3
2 3/2
a  x 
2
a  b 
2 2
c 2a  x 2  b 2
 0 dx 1/2
− 2a  x  a  c a  c 2  b 2 − a  c − a a2  b2 − a
a  x 2  b 2
0 for even l
1
 0 dx P l x  1 for l  0
l−1 l1l−1!
−1 2
l1 2
for odd l
2 l1 2
!
0 1
 −1 dx P l x  −1 l  0 dx P l x
1
 −1 dx P l x 2  2
2l  1
r,   ∑A n r n  B n r −n1 P n cos 

n
e&m possibilities 2006 (solutions)
An infinitely long cylinder of linear magnetic material of permeability µ is

wrapped with a wire (forming an infinite solenoid of radius R wrapped around
the cylinder). The wire carries a current I and has N loops per unit length.
Ignore the magnetic properties of the wire.

(a) Calculate the field H as a function of s, the radial distance from the
cylinder axis, for s > R and s < R.
⎧ INẑ inside, s < R

First, find H using ∫ H ⋅ dl = I free , so HL=INL ⇒ H = ⎨
⎩ 0 outside, s > R
L
Justification for H=0 outside is that H depends only on I free and Bout = 0 for solenoid without core.

(a) Calculate the magnetic vector potential A as a function of s, the radial distance from the
⎧ µ INẑ inside, s < R

B = µH = ⎨
⎩ 0 outside, s > R
Calculate A from ∫ A ⋅ dl = ∫ B ⋅ da
By symmetry choose integration path shown in figure.
B
s < R : A2π s = Bπ s 2 ⇒ A = sφ̂
2
BR 2
s > R : A2π s = Bπ R ⇒ A =
2
φ̂
2s
(a) Calculate the surface and volume bound currents in the magnetic cylinder. Note that these
currents are sometimes referred to as magnetization currents.
M = χ m H, where χ m is given by µ =µo (1 + χ m ), so M = χ m INẑ

Volume bound current, Jb = ∇ × M = 0, since M is constant.
Surface bound current, Kb = M × n̂, where n̂ = ŝ, so Kb = χ m INφ̂ , collinear with I free

(a) Use the free and bound currents to calculate B as a function of s, the radial distance from the
Find B using ∫ B ⋅ dl = µo Itotal =µo I free + µo Ibound (use path shown in figure)
L
BL=[µo NI + µo χ m NI ]L = [µo NI + µo ( µ / µo − 1)NI ]L = µ NIL, so B = µ NIẑ
as in part (b) for s < R.
Since I bound is collinear with I free , the symmetry is that of an infinitely long solenoid
without a core, which implies that B = 0 for s > R.
A (non-conducting) solid sphere of radius R carries a charge density ρ(r) = k r 2 (where k is a
constant).
Find the electric field at a distance r such that
(a) r ≥ R
∫ E ⋅ da = Qenc / ε o By symmetry of charge distribution E = Er (r) r̂; also da = da r̂

∫ E ⋅ da =Er (r) 4π r
2
Since ρ depends on r, Qenc = ∫ ρ dτ

R
1 kR 5
Er (r) = ∫ kr 4π r dr, or E =
2 2
r̂
4π r 2 ε o 0
5ε o r 2
(b) 0 ≤ r ≤ R
r
1 kr 3
Similar to part (a): Er (r) = ∫ kr 4π r dr, or E = 5ε o r̂
2 2
4π r 2 ε o 0
(c) Show that your answers to parts (a) and (b) are consistent with the boundary conditions on
the electric field.
σ
Eabove − Ebelow = n̂, where "above" is at r = R + and "below" is at r = R − , n̂ = r̂ and σ = 0.
εo
kR 5 kr 3
−
r= R + 5ε r= R −
= 0 for r̂ component (E⊥ ).
5ε o r 2 o
Also Eφ = Eθ = 0, so E is continuous across the surface.
(d) Calculate the electric potential for all r. Let V = 0 at r = ∞.
V = − ∫ E ⋅ dl Take path l to be in r̂ direction and

r
kR 5 kR 5
for r ≥ R V = − ∫ r̂ ⋅ (r̂ dr) =
∞
5ε o r 2 5ε o r
k ⎛ r 4 R4 ⎞
R r
for r ≤ R V = − ∫
kR 5
5ε o r 2
r̂ ⋅ (r̂ dr) − ∫
kr 3
5ε o
r̂ ⋅ (r̂ dr) =
kR 4
− ⎜
5ε o 5ε o ⎝ 4
− ⎟ =
k
4 ⎠ 20ε o
(5R 4 − r 4 )
∞ R
(e) Find the work required to assemble this charge distribution.

Several ways to calculate the energy, including the following that use previous results of this problem.
R
1 k 4π k 2 9
W = ∫ ρVdτ =
40ε o ∫0
2⎡ 4⎤
kr ⎣ 5R − r ⎦ 4π r dr =
4 2
R
2 45ε o
∞ R
εo ε o kR 5 ε o kr 3
OR, W = ∫ E dτ = 2 R∫ 5ε o r 2
τ
2 ∫0 5ε o
dτ which gives the same answer.
2
d +
2
(f) If the non-conducting solid sphere was replaced by a conducting solid sphere with the same
total charge, how does that change your answer to parts (a), (b), and (c)?
(a) E is unchanged
(b) E = 0 inside the conducting sphere
(c) All charges move to the surface. The surface charge density σ is given by
R R
kR 3
4π R σ = ∫ kr dτ = ∫ kr 2 ( 4π r 2 dr) ⇒ σ =
2 2
0 0
5
So, the boundary conditions on E are given by
σ kR 3
E⊥,above − E⊥below = =
ε o 5ε o
kR 5 kR 3
r= R +
−0 = , as expected. As before, Eφ = Eθ = 0, so E is continuous across the surface.
5ε o r 2 5ε o

Consider a sphere of radius R that is uniformly polarized along the z–axis, such that P = P z .
(a) Determine the bound charge distribution (sometimes called the polarization charge).
(a) Derive the electric potential produced by this polarized sphere.
Since the polarization is constant, the volume bound charge is zero. The surface bound charge is given
by σ b = P ⋅ n = P cosθ , where θ is the polar angle in spherical coordinates. There are several ways to
do this problem, one is to use Coulomb's law, the other is to guess the answer (but the problem asks
for a derivation), and the hint suggests solving for the potential using Laplace's equation by using the
∞
⎛ Bl ⎞
following solution: V (r,θ ) = ∑ ⎜ Al r l + l +1 ⎟ Pl (cosθ )
l =0
⎝ r ⎠
∞
For r ≤ R, the lack of charge inside leads to V (r,θ ) = ∑ Al r l Pl (cosθ )
l =0
( )
∞
⎛ Bl ⎞
For r ≥ R, the boundary condition V (∞) = 0 leads to V (r,θ ) = ∑ ⎜ l +1 ⎟⎠ Pl (cosθ )
⎝
l =0 r
∞ ∞
∑ ( Al R ) Pl (cosθ ) = ∑ ⎜⎝ Rl +1l ⎟⎠ Pl (cosθ )
⎛ B ⎞
The potential is continuous on the boundary, r = R : l
l =0 l =0
Bl
Orthogonality of the Legendre polynomials leads to Al Rl = l +1
, or Bl = Al R 2l +1.
R
The boundary condition on the normal derivatives of the potential:

( ∂Vout
∂r
−
∂Vin
∂r
) r= R
=−
σ
εo
∞ ∞
l =0
⎛ Bl ⎞
− ∑ (l + 1) ⎜ l +2
⎝ R l =0
(
⎟⎠ Pl (cosθ ) − ∑ l Al R
l −1
)
Pl (cosθ ) = −
P cosθ
εo
∞
∑ (2l + 1)( Al Rl −1 ) Pl (cosθ ) =
P cosθ
Using Bl = Al R 2l +1, yields
l =0 εo
∞ π π
∑ (2l + 1)( Al R )
l −1 P
Fourier analysis: ∫ Pl (cosθ )Pn (cosθ ) d(cosθ ) = ε o ∫ cosθ Pn (cosθ ) d(cosθ )
l =0 0 0
Noting that cosθ = P1 (cosθ ) and that only the n= l = 1 term is non-zero
(This could also be done using the orthonormality condition, to be given in equation sheet)
⎧ P
⎪ 3ε r cosθ for r ≤ R
P ⎪ o
Only l = 1 term survives: A1 = , so V (r,θ ) = ⎨
3ε o 3
⎪ P R cosθ for r ≥ R
⎪⎩ 3ε o r 2
When an electron is injected at right angles to a steady uniform magnetic field, B, it initially
executes circular motion about the magnetic field lines.
(a) Assume that the electron is non-relativistic with a velocity v << c. Determine the power
radiated in terms of only the electron's velocity, the angular frequency of orbiting, ωc, and any
other necessary constants. To determine the cyclotron frequency, ωc, you may assume that the
velocity is nearly constant for each (nearly) circular revolution.
The radiated power depends upon the square of the acceleration, as given by the Larmor formula:
q2a2
P=
6πε o c 3
The long way to find a : F = qv × B = ma ⇒ qvb = mv 2 / R, so R = mv / qB.
So a = v 2 / R = qvB / m.
Since the orbiting or cyclotron frequency ω c is given by v = ω c R, we have a = ω c v.
The short way to find a : a = v 2 / R and v = ω c R, so a = ω c v.
So, the power radiated per cycle is
q 2 (ω c v)2
Pcycle =
6πε o c 3
(b) Due to the power radiated in part (a), energy is continually lost by the electron and it slows
down. First, derive an expression for the total energy of the electron (kinetic plus potential) by
taking the potential energy to be − µ ⋅ B , where µ is the magnetic moment of a circular loop of
current due to the electron. This expression for the total energy can be written simply in terms of
only the electron mass m and its velocity v.
Next, derive an expression for the time that it takes the energy to fall to 1/e of its initial value.
1 2
mv , and potential, -µ ⋅ B, energies.
First, the total energy is the sum of the sum of the kinetic,
2
If the magnetic moment is taken as the moment of the current loop due to the electron, then
qv qv qvR mv 2
µ = IA, where A = π R and I = 2
, so µ = πR =
2
= ,
2π R 2π R 2 2B
1
so the potential energy is mv 2 (note that µ and B are in opposite directions).
2
dU q 2 (ω c v)2 q 2ω c2 mv 2 q 2ω c2U
Second, P(t) = − = = =
dt 6πε o c 3 6πε o c 3m 6πε o c 3m
−t /τ 6πε o c 3m 6πε o c 3m 3
So, U = U o e , where τ = = , using ω c = qB / m
q 2ω c2 q 4 B2
(c) Describe the direction of polarization of the electromagnetic radiation seen by an observer
within the plane determined by the path of the electron.
The direction of the radiated electric field is opposite to that of the component of the acceleration
perpendicular to observer's line of sight, Erad ∝ −a ⊥ .
v
e–
a
a⊥
Erad ,
observer
Consider a very long solenoid with radius R, n turns per length, and current Io. Coaxial with the
solenoid are two long cylindrical shells of length l. One of the shells is inside the solenoid, of
radius a, and carries a total charge +Q that is uniformly distributed R
over the shell. The other shell is outside the solenoid, of radius b,
and carries a total charge -Q that is uniformly distributed over the
shell. So, a < R < b and l >> a and l >> b (note that the figure does b
a
not have these relative dimensions correct – for the purpose of
calculating electric and magnetic fields the solenoid and cylinders
can be considered infinitely long). The charged cylinders are free
to rotate about their axis. –Q l
+Q
a) The current in the solenoid is gradually reduced to zero.
Calculate the final angular momentum of the two charged
cylinders.
I
When the current is turned off the changing B induces a circumferential E.
dΦ
∫ E ⋅ dl = − dt B . Since B = µonI :
⎧ 1 dI
⎪⎪− 2 µo n dt sφ̂ for s < R
E=⎨ 2
⎪− 1 µ n dI R φ̂ for s > R
⎪⎩ 2 o dt s
This electric field will exert a torque N on the charged cylinders
⎧ 1 2 dI
⎪⎪ r × (QE) = − µ o nQa ẑ for the inner cylinder
2 dt
N=⎨
⎪r × (−QE) = 1 µ nQR 2 dI ẑ for the outer cylinder
⎪⎩ 2
o
dt
0
dI
The angular momentum imparted to the cylinders is given by ∫ N dt = () ∫ dt = −()I o
Io
dt
⎧ 1
⎪⎪ 2 µo nQa I o ẑ for the inner cylinder
2
L=⎨
⎪− 1 µ nQR 2 I ẑ for the outer cylinder
⎪⎩ 2 o o
1
Ltotal = µo nQ(a 2 − R 2 )I o ẑ
2
b) Demonstrate that angular momentum is conserved by considering the angular momentum
before and after the current is turned off. You can assume that the magnetic field is
negligible after the current in the solenoid is turned off.
The final angular momentum of the cylinders comes from the angular momentum in the original
electric and magnetic fields.
The electric field due to the charged cylinders can be derived from Gauss's law in integral
Q
form E = ŝ for a < s < b. The initial magnetic field inside the solenoid is B = µo nI o ẑ.
2πε ols
Qµo nI o Qµo nI o
The angular momentum density is given by r × (ε o E × B) = r̂ × (ŝ × ẑ) = − φ̂
2π l 2π l
The total angular momentum in the fields is given by this density times the volume occupied by
Qµo nI o 2
the region containing both fields, so Lem = − (R − a 2 )φ̂ . This is the same as the total L
2
transferred to the cylinders.
Electricity & Magnetism

January 2, 2007
9:00 AM to 12:00 Noon
Full credit can be achieved from completely correct answers to 4 questions. If the student
attempts all 5 questions, all the answers will be graded, and the top 4 scores will be counted
towards the exam’s total score.
1
Equation Sheet
( )
V (x, y) = ∑ Aekx + Be−kx ( C sin ky + D cos ky ) Note, x and y may be interchanged.
k
⎧0 if m ≠ n
a ⎪
∫0 sin(nπ y / a)sin(mπ y / a) dy = ⎨ a if m = n
⎪⎩ 2
∞
⎛ B ⎞
V (r,θ ) = ∑ ⎜⎝ Al r l + r l+1l ⎟⎠ Pl (cosθ )
l=0
⎧0 if m ≠ l
1 ⎛ d ⎞l 2
( ) π ⎪
l
where Pl (x) = l ⎜ ⎟ x − 1 and ∫ Pl (cosθ )Pm (cosθ ) sin θ dθ = ⎨ 2
2 l! ⎝ dx ⎠ 0
⎪⎩ 2l + 1 if m = l
Po (x) = 1
P1 (x) = x
P2 (x) = (3x 2 − 1) / 2
P3 (x) = (5x 3 − 3x) / 2
P4 (x) = (35x 4 − 30x 2 + 3) / 8
q2a2
P=
6πε oc 3
2
1. An infinitely long cylinder of linear magnetic material of permeability µ is
wrapped with a wire (forming an infinite solenoid of radius R wrapped
around the cylinder). The wire carries a current I and has N loops per unit
length. Ignore the magnetic properties of the wire.
(a) Calculate the field H as a function of s, the radial distance from the
(b) Calculate the magnetic vector potential A as a function of s, the radial distance from the
(c) Calculate the surface and volume bound currents in the magnetic cylinder. Note that
these currents are sometimes referred to as magnetization currents.
(d) Use the free and bound currents to calculate B as a function of s, the radial distance from
the cylinder axis, for s > R and s < R.
2. A (non-conducting) solid sphere of radius R carries a charge density ρ(r) = k r 2 (where k is a

constant).
Find the electric field E at a distance r such that
(a) r ≥ R
(b) 0 ≤ r ≤ R
(c) Show that your answers to parts (a) and (b) are consistent with the boundary conditions
on the electric field.
(d) Calculate the electric potential for all r. Let V = 0 at r = ∞.
(e) Find the work required to assemble this charge distribution.
(f) If the non-conducting solid sphere was replaced by a conducting solid sphere with the
same total charge, how does that change your answer to parts (a), (b), and (c)?
3. Consider a sphere of radius R that is uniformly polarized along the z–axis, such that P = P ẑ .
(a) Determine the bound charge distribution (sometimes called the polarization charge).
(b) State the boundary conditions on the electric potential.
(c) Use the boundary conditions from part (b) to derive the electric potential produced by
this polarized sphere.
(d) Calculate the total flux through a spherical surface, of any radius, that is concentric with
the polarized sphere.
3
4. When an electron is injected at right angles to a steady uniform magnetic field, B, it initially
executes circular motion about the magnetic field lines.
(a) Assume that the electron is non-relativistic with a velocity v << c. Determine the power
radiated in terms of only the electron's velocity, the angular frequency of orbiting, ωc,
and any other necessary constants. To determine the cyclotron frequency, ωc, you may
assume that the velocity is nearly constant for each (nearly) circular revolution.
(b) Due to the power radiated in part (a), energy is continually lost by the electron and it
slows down. First, derive an expression for the total energy of the electron − µ ⋅ B in
terms of only the electron mass m and its velocity v, where µ is the magnetic moment of
a circular loop of current due to the electron.
Next, derive an expression for the time that it takes the energy to fall to 1/e of its initial
value.
(c) Describe the direction of polarization of the electromagnetic radiation seen by an

observer within the plane determined by the path of the electron.
5. Consider a very long solenoid with radius R, n turns per length, R

and current Io. Coaxial with the solenoid are two long
cylindrical shells of length l. One of the shells is inside the
solenoid, of radius a, and carries a total charge +Q that is a b
uniformly distributed over the shell. The other shell is outside
the solenoid, of radius b, and carries a total charge -Q that is
uniformly distributed over the shell. So, a < R < b and l >> a
–Q l
and l >> b (note that the figure does not have these relative +Q
dimensions correct – for the purpose of calculating electric and
magnetic fields the solenoid and cylinders can be considered
infinitely long). The charged cylinders are free to rotate about
their axis, but they are at rest initially.
a) The current in the solenoid is gradually reduced to zero.
Calculate the torque N on the two charged cylinders in I
dI
terms of and other quantities.
dt
b) Calculate the final angular momentum vector (after the current has been reduced to zero)
of the two charged cylinders in terms of I o and other quantities.
c) Calculate the Poynting vector when the current is at its initial value, Io.
d) Demonstrate that angular momentum is conserved by considering the angular momentum
before and after the current is turned off. You can assume that the magnetic field is
negligible after the current in the solenoid is turned off.
4
ELECTROMAGNETISM PRELIMINARY EXAM
January 2004
z 1. The z-axis coincides with the

axis of an infinite conducting
cylinder of inner radius a, outer
radius b. For a < r < b, there is a
current density flowing around
the wire, J = J0 A/m2 .
J
For the three regions
(I) 0 < r < a,
(II) a < r < b

a b
(III) b < r ,
Find:
(a) The B field .
(b) The A field .
And then find
(c) The pressure exerted by the magnetic field on the cylinder.
1
2. A grounded conducting
sphere of radius a is inside a z
θ
sphere of radius b. The region
between radius a and radius b is σ cos θ
vacuum. The outer sphere is 0
non-conducting, and carries a
surface charge density σ0cosθ. a
In the regions: b
(I) a < r < b
(II) b < r
Find the potential Φ.
2
R
q
z
a
3.
A hollow grounded conducting sphere of radius R contains a point

charge q at the point ak .
(a) Find the potential inside the sphere.
(b) Find the vector force on the charge q.
R
q
z
The sphere of the previous problem is now reduced to a conducting

hemisphere, with a conducting flat base. The charge q is still at the
point ak .
(e) Find the potential in the hemisphere.
(f) Find the vector force on the charge q.
3
4. A plane wave of frequency ω, with EI = EI j , is normally incident
on a conducting plane of conductivity σ. The conducting plane fills
the half space z > 0. The conductivity is very high, σ >> εω , so the
displacement current inside the conductor can be neglected. There
is a reflected wave, ER .
(a) Find the B and E fields inside the conductor in terms of their
values at z = 0, as functions of z , σ and ω .
(b) Find the reflected electric field vector in terms of EI , σ and ω .
4
5. An wire stretches along the z
z-axis from z = -a/2 to z = a/2.
An alternating current of angular
frequency ω runs in the wire,
and the radiated EM wave has a a/2
wavelength λ that is much
greater than the wire length a. A
I
good approximation to the
current density in the wire is
J = I0 δ x δ y k a - z sin ωt k , -a/2
2
where k = ω/c .
At large distances r from the wire, where r >> a , and r >> λ ,
find as functions of r, θ, φ , ω and I0 :
(a) The vector potential A .
(b) The magnetic field B.
(c) The electric field E,.
(d) Find the power radiated per unit solid angle as a function of θ, φ,
ω and I0 .
(e) Find the wire's electric dipole moment p, and its magnetic
dipole moment m .
5
Electromagnetism
PhD Qualifying Examination
January 5, 2012
9.00 am – 12:00 pm

questions. If the student attempts all 5 questions, all of the answers
will be graded, and the top 4 scores will be counted toward the
exam’s total score.
Question 1
A hollow metal sphere of radius R and mass M floats on an insulating dielectric liquid of density
ρ and relative dielectric constant εr. When the metal sphere has no charge on it, it floats on the
dielectric liquid as shown in Figure 1(a); i.e., the bottom of the sphere is R/2 below the surface of
the dielectric liquid.
a) Determine the relationship between M and ρ when the sphere is not charged.
b) The hollow metal sphere is now charged with a charge Q. Draw a diagram that shows all
the charges and explain why the sphere sinks further into the dielectric liquid when it is
charged.
c) Find the magnitude of the charge Q to which the sphere must be charged in order for it to
be half submerged as shown in Figure 1(b). Express your answer in terms of ρ, R, εr , the
vacuum permittivity ε0 , the acceleration due to gravity g, and other numerical factors.
Hollow metal
sphere
Q
R
M
R/2
R
Dielectric liquid ρ, εr
Figure 1(a) Figure 1(b)

Question 2
Consider the circuit shown below, in which for times t < 0 the capacitor of capacitance C is
charged to a voltage V0. At t = 0, the switch S is closed, allowing the capacitor to discharge
through a resistor R and an inductor L placed in series.
V0 C
L
a) Using Kirchoff’s voltage law, write down the second-order differential equation
describing the evolution of the charge q on the capacitor for times t > 0.
b) For times t > 0, solve the differential equation obtained in part (a) subject to the boundary
dq
conditions q(t = 0) = q0 and = 0.
dt t =0
c) Explain why the current in the circuit builds up to a maximum value and then decays to
zero. Show that the time t at which the current in the circuit is a maximum is given by
the relation
2Ω
tanh(Ωt ) = ,
α
α2 R 1
where Ω = ω 2 + and α = with ω = .
4 L LC
Question 3
0 d z
E(t)
A cylindrical ‘pill-box’ resonator of radius a and length d is driven at its fundamental TM010
mode for which the oscillating electric component of the RF field may be written as
⎛ 2.405r ⎞
E(t ) = zˆ E0 J 0 ⎜ ⎟ sin (ωt + φ)
⎝ a ⎠
where J0(x) is the Bessel function of zero order whose first zero is at x = 2.405.
a) What is the form of the magnetic component of the oscillating RF field in the cavity?
b) Verify that the average value of the Poynting vector (i.e., S av. ) is zero.
c) What is the stored energy of the oscillating TM010 mode?
Bessel function relations:
∞
(−1) m
J n ( x) = ∑ (12 x)2m+n where Γ(p) = (p – 1)! for p positive integer
m=0 m!Γ(m + n + 1)
1
∂J n ( x) 2 2
∂x
= J n+1 ( x) ∫ xdxJ n (αx) = 12 J n+1 (α)
0
Question 4
Soleniod
(N turns, current I)
I
a a
θ P β P
z α z
Figure 4(a) Figure 4(b)
a) Show that for a single wire loop of radius a carrying current I the axial magnetic field at
point P in Figure 4(a) may be written as
µ0 I
B(θ) = zˆ sin 3 θ
2a ,
where θ is the angle subtended from the axis at point P to the circumference of the loop
and µ0 is the permeability of vacuum.
b) Use the result of part (a) to show that the axial magnetic field at point P for a solenoid of
length d and radius a carrying current I in N turns (Figure 4(b)) is given by
µ 0 NI
B ( α , β) = (cos α − cos β),
2d
where the front and back coils (i.e., ends) of the solenoid subtend angles α and β with its
axis at point P. Verify that your answer reduces to the expected result for the field inside
an infinitely long solenoid (i.e., d >> a).
c) Show that for small distances z << d inside and close to the center of a narrow (d >> a)
finite solenoid that the axial dependence of the magnetic field strength is parabolic in z
and of the form
µ NI ⎡ 2a 2 ⎛ 12 z 2 ⎞⎤
B( z ) = 0 ⎢1 − 2 ⎜1 + ⎟⎥
2 ⎟
d d ⎜ d ⎠⎥⎦ .
⎢⎣ ⎝
Question 5
A material can be anisotropic in either or both its refractive index and absorption. These optical
properties are described by a permittivity tensor, ε , for the x̂ , ŷ , and ẑ directions (i.e., a 3×3
matrix). The wave equation in a non-conducting, non-magnetic medium then reads
∂ 2E
∇2E − µ0 ε =0
∂t 2 .
a) For a electromagnetic wave of frequency ω propagating in direction k with a polarization

unit vector ê and amplitude E0 described by, E(r, t ) = 12 eˆ E 0 exp[i(k.r − ωt )] + c.c. , show
that the refractive index n experienced by the wave in the medium is given by the
expression
1
n2 = [
eˆ * .(ε.eˆ )]
ε0 .
b) For ê = (sin θ,0, cos θ) in the x-z plane, determine the refractive index experienced by the
⎛ no2 0 0 ⎞
⎜ ⎟
wave in a non-absorbing crystalline medium described by ε = ε 0 ⎜ 0 no2 0 ⎟ ,
⎜ 0 0 (no + Δn) 2 ⎟⎠
⎝
where no and ne = no + Δn are the ordinary and extra-ordinary refractive indexes of the
unaxial crystal respectively.
c) What is the walk-off angle φ between the Poynting vector S and the wave vector k of the
wave in the anisotropic medium if its magnetic field amplitude is given by
H = (0, H 0 ,0) = yˆ H 0 ?
d) What is the angle between E and D in the uniaxial crystal?
Electromagnetism
January 6, 2011
9.00 am - 12.00 pm

4 questions. If the student attempts all 5 questions, all of the an-
swers will be graded, and the top 4 scores will be counted towards
the exam’s total score.
Various equations, constants, etc. are provided on the last page of

the exam.
1. Thomson Model of an Atom
One very simplistic way of describing an atom is to use a positive volume charge distribu-
tion ρ(r) (created by the nucleus) which surrounds the electrons and balance their negative
charges.
For the first part of the problem, only consider the positive volume charge distribution ρ(r):



 ρ0 : r ≤ R
ρ(r) = 
  0 : r>R
(a) Find the corresponding electrostatic potential, V , inside and outside the atom. Do not
pay any attention to the contributions of the electrons.
" inside and outside

(b) Using your results from part (a), find the electrostatic vector-field, E,
the atom.
Now consider a single electron inside this atom, moving under the influence of the potential
ρ(r).
(c) Explain why the electron will oscillate inside the atom, described by the charge distri-
bution ρ(r).
(d) Find the frequency of this oscillation.
1
2. Dielectric Sphere
A sphere of homogeneous dielectric material with permittivity ε and radius R is placed in

" 0 . The external electric field far away from the sphere
an otherwise uniform electric field E
" 0 = E0 ẑ.
is given by E
(a) Determine the boundary conditions for the given setup.
(b) Find the potential inside and outside the dielectric sphere. Explain your approach!
(c) Find the electric field E" and polarization P" inside the dielectric sphere.
(d) Sketch the electric field lines for all regions of this setup.
(e) Find the bound volume charge density ρb , and all the bound surface charge densities σb .
2
3. Inductance
Consider an alternating current, I(t), flowing down a straight wire:
I(t) = I0 cos ωt
(a) In the quasistatic approximation, find the induced electric field as a function of distance,
s, from the wire.
(b) Is your solution to part (a) valid for the limit s → ∞? Explain your answer.
The wire runs parallel to the axis of a coil with rectangular cross-section which is connected
to a resistor R. The wire is at a distance d from the coil of height and width h and N turns.
(c) Find the inductance L of the rectangular coil, in terms of N , h, a, and b.
(d) In the quasi-static approximation, what emf is induced in the coil?
(e) Calculate the back emf in the coil, due to the current I(t).
3
4. Momentum of Electromagnetic Fields
Two non-conducting plates, both parallel to the x−y plane, extend over the region 0 ≤ x ≤ a
and 0 ≤ y ≤ b. One plate is located at z = 0 and has a uniform charge density −σ. The
second plate is located at z = h and has a uniform charge density σ. Assume that the
distance h between the plates is much smaller than their length a, and width b, so that edge
" 0 = B0 ŷ.
effects can be ignored. There is a uniform magnetic field B
"
(a) What is the Poynting vector S?
(b) What is the momentum density of the electromagnetic field?
(c) What is the total momentum of the electromagnetic field?

" 0 is turned off in a time ∆t.
Now the magnetic field B
(d) What is the impulse ∆"p in time ∆t experienced by each plate, as derived from the
induced electric field? How does this compare to the field momentum derived in part (c)?
4
5. Dipole Radiation
Consider some time-dependent charge distribution of finite extent, ρ(r, t), whose time de-
pendent dipole moment is given by p"(t) = p0 (t)p̂. In a region of space, the scalar and vector
potentials established by this charge distribution are given by:
% &
1 q r̂·"p(tr ) r̂·p"˙ (tr )
V ("r, t) = 4πε0 r + r2
+ cr
" r, t) = µ0 p"˙ (tr )

A(" 4π r
where p"˙ is the time derivative of the dipole moment, and tr = t − r

c
is the retarded time.
' (
" t) and magnetic field B(r,
(a) Find the electric field E(r, " t) to first order in in terms of 1
r
p̈0 (tr ) [i.e. the second time derivative of p0 (t) evaluated at the retarded time, tr ].
" r, t) = E("r, t)θ̂ and B("

(b) Assume that p"(tr ) = p0 (t)ẑ. Show that E(" " r, t) = B("r, t)φ̂. Find
expressions for E("r, t) and B("r, t).
"
(c) Calculate the Poynting vector S.
(d) Derive explicitly the power radiated to infinity by this time dependent charge distribu-
tion.
5
Equations and Constants
x̂ = sin θ cos φ r̂ + cos θ cos φ θ̂ − sin φ φ̂

ŷ = sin θ sin φ r̂ + cos θ sin φ θ̂ + cos φ φ̂
ẑ = cos θ r̂ − sin θ θ̂
)' Bl
(
V (r) = Al rl + r l+1
Pl (cos θ) ,
l
' (l
1 d
where Pl (x) = 2l l! dx
(x2 − 1) and

*π  0 if m %= l
Pl (cos θ)Pm (cos θ) sin θ dθ =
0  2
if m=l
2l+1
3x2 −1 5x3 −3x

P0 (x) = 1; P1 (x) = x; P2 (x) = 2
; P3 (x) = 2
" · "v = ∂x vx + ∂y vy + ∂z vz (cartesian);

∇
" · "v = 12 ∂r (r2 vr ) + 1 ∂θ (r sinθvθ ) + 1 ∂φ vφ (spherical)
∇ r r sinθ r sinθ
d"a = s dφ dz r̂ + ds dz φ̂ + s ds dφ ẑ (cylindrical);
d"a = r2 sin θ dθ dφ r̂ + r dr dφ θ̂ + r sin θ dr dθ φ̂ (spherical)
*
sin3 (ax) dx = − a1 cos(ax) + 1
3a
cos3 (ax)
6
Electricity and Magnetism
Thursday, January 7, 2010
Full credit can be achieved from completely correct answers to 4 ques-

tions. If the student attempts all 5 questions, all of the answers will
be graded, and the top 4 scores will be counted toward the exams
total score.
Various equations, standard integrals, etc. are provided on the last

page of the exam.
1. Charged sphere
The potential on the surface of a sphere of radius a as a function of polar angle θ is given by
Φ(r, θ)|r=a = V0 cos 2θ,
where V0 is a known constant. Also given is that Φ → 0 as r → ∞.
(a) Find the potential outside the sphere as a function of r (r > a) and θ.
(b) Find the total charge Q on or inside the sphere.
2. Hydrogen atom
The electric charge of electron is distributed in a hydrogen atom according to

e0
ρ(r) = − 3 e−2r/a ,
πa
where e0 and a are fundamental constants.
(a) Find the energy U of the electrostatic interaction of the electron with the nucleus
(proton).
(b) Find the value of the electrostatic potential Φ created by the electron charge distribution
ρ(r) at the center of the atom r = 0.
(c) Find the total electric field E created by the whole atom as a function of r for any r.
Determine the magnitude and the direction (inward or outward).
3. Dielectric bead
A hemispherical bead of radius a made of a dielectric material

with dielectric permittivity ε is placed on an infinite conducting
E0 sheet. The electric field far away from the sheet is perpendicular
to it and equals E0 .
(a) Find electrostatic potential Φ everywhere to the right of the

sheet (both inside and outside of the bead). Hint: look for the
solution in the form Φ = R(r) cos θ.
(b) Find the surface charge density σ at any given point on the
sheet outside the bead r > a.
1
4. Magnetic field of a rotating cylinder
Electric charge is distributed uniformly with constant volume

density α inside a very long cylinder, whose length L is much
greater than its radius a: L ≫ a. The cylinder is rotated around
ω its axis with angular velocity ω.
(a) What is the current density J at any given point inside

a the cylinder? Express the result in cylindrical coordinates, i.e.,
determine Jρ , Jφ , Jz as functions of ρ, φ and z.
L (b) What is the magnetic field at any given point outside the
cylinder, i.e., at distances ρ, such that ρ > a (but ρ ≪ L).
Express the result in cylindrical coordinates, i.e., determine Bρ ,
Bφ , Bz as functions of ρ, φ and z.
(c) What is the magnetic field at any given point inside the
cylinder. Express the result in cylindrical coordinates, i.e., de-
termine Bρ , Bφ , Bz as functions of ρ, φ and z.
5. Radiating ring
A ring of wire of radius a carries current I which varies with time t

z and angle φ along the ring according to I = I0 sin φ cos ωt.
(a) Find the rate of change dp/dt of the electric dipole moment
y
of the ring at time t.
φ
x (b) Find the rate of change dm/dt of the magnetic dipole mo-
ment of the ring at time t.
(c) In dipole approximation, what is the polarization (e.g., circular, linear, none) of radiation
emitted along each of the directions x, y and z? For each linearly polarized case indicate the
orientation of the polarization axis (e.g., x or y or z). Organize your answer in a table like
this:
radiation direction polarization type polarization axis (if linear)
x ... ...
y ... ...
z ... ...
2
Equations
∇ · D = ρ; ∇×E = −dB/dt; ∇×H = J + dD/dt; ∇ · B = 0;
∂ρ/∂t + ∇ · J = 0;
D = εE; B = µH;
E = −∇Φ − ∂A/∂t; B = ∇ × A;
Z
U= d3 r ρ Φ
Z Z
1
p= d3 r ρr; m= d3 r r×J ;
2
X
Φ= (Al r l + Bl /r l+1 )Pl (cos θ);
l
1
P0 = 1; P1 = x; P2 = (3x2 − 1);
2
Z Z
xe dx = −e (1 + x);
−x −x
x2 e−x dx = −e−x (2 + 2x + x2 ).
3
Electricity and Magnetism
Preliminary Exam
January 2009
1
1. Gauss law
A ball of radius R is uniformly charged with volume charge density equal to ρ. Use Gauss
law and superposition principle to answer the following questions.
(a) Find the electric field E(r) outside the sphere as a function of the radius vector r drawn
from the center of the sphere to the observation point.
(b) Find the electric field E(r) inside the sphere as a function of the radius vector r drawn
from the center of the sphere to the observation point.
(c) Consider now the same sphere but now with an empty cavity. The center of the cavity is
at distance a from the center of the sphere. The radius of the cavity is b, such that a+ b < R.
Find the electric field E inside the cavity.
2. Multipole expansion
The surface of a sphere of radius R is charged with surface charge density varying according
to σ = σ0 cos θ. Use the multipole expansion to answer the following questions.
(a) Find the potential Φ inside and outside the sphere as a function of r and θ.
(b) Find the electric field E inside the sphere.
3. Electrostatics of medium
A spherical capacitor consists of two concentric conducting spheres of radii a and b. The
capacitor is filled with dielectric material whose dielectric constant varies according to:
(
ε1 = const, for a < r < c
ε(r) =
ε2 = const, for c < r < b
The charge on the inner conducting sphere is Q.
(a) Find the electric field inside the capacitor as a function of r.

(b) Find the capacitance of this capacitor.
(c) Find the density of the bound charge on the boundary between the dielectric layers at
r = c.
2
4. Magnetostatics of medium
A very long circular solenoid is made out of a wire with n turns per unit length. The radius
of the cylinder is a and is negligible compared to its length l. The interior of the cylinder
is filled with material such that the linear magnetic permeability varies with the distance r
from its axis according to:
(
µ1 = const, for 0 < r < b
µ(r) =
µ2 = const, for b < r < a
The current passing through the wire is equal to I.
(a) Find the magnetic field B inside the solenoid as a function of r.

(b) Find the inductance L of such a solenoid.
5. Radiation
A thin rod of length L is charged uniformly with density λ per unit length. The rod is rotated
with angular velocity ω around an axis passing through one of its ends perpendicular to the
rod.
(a) Find the magnetic dipole moment m of the rod.

(b) Find the electric dipole p moment of the rod.
(c) Find the total radiation energy emitted by the rotating rod per unit time.
Equations
∇D = ρ; ∇×E = −dB/dt; ∇×H = J + dD/dt; ∇B = 0;
D = εε0E; B = µµ0 H;
1
Z Z
p= d3 x ρr; m= d3 x r×J ;
2
dE |pω |2 ω 4 dE µ0 |mω |2 ω 4
= ; = ;
dt 12πε0 c3 dt 12πc3
(Al r l + Bl /r l+1 )Pl (cos θ);

X
Φ=
l
1
P0 = 1; P1 = x; P2 = (3x2 − 1).
2
3
Solutions
1. Plum Pudding Model
(a) Find the corresponding electrostatic potential inside and outside the atom.
For r ≤ R
∇2Vin = − ρε00 .
The solution can be found by integrating twice,
ρ0 2
Vin = − 6ε0
r + ra2 + b,
where a and b are constants. The potential at the center of the atom has to be finite, so a = 0.
Finally,
ρ0 2
Vin = − 6ε0
r + b.
For r > R
∇2Vout = 0.
The solution can be found by integrating twice,
Vout = − cr + d.
But for large r, the potential has to go to zero, therefore d = 0.
Using the boundary condition that the potential has to be continuous at r = R:
ρ0 2 c
6ε0 R + b = R .
1
(b) Find the electrostatic vector-field inside and outside the atom.
For r R:
!Eout = −!∇Vout = c
r2
r̂.
For r ≤ R:
!Ein = −!∇Vin = ρ0!r

3ε0 .
Using the continuity of !E at r = R:
c ρ0 R ρ0 R3
R2
= 3ε0 ⇔c= 3ε0
In summary:
ρ0!r ρ0 R3 r̂ 2 ρ0 ! 2 "
!Ein =
3ε0 ,
!Eout =
3ε0 r2 = 1 Qtotal
4πε0 r2 r̂, b = − ρ6ε
0R
0
, and Vin = − 6ε0
r + R2
(c) Electrons inside the Plum Pudding Model atom will oscillate. Explain why!
The potential energy of an electron inside the atom is
! 2 "
U = eV (r) = − eρ
6ε0
0
r + R 2 = Ar 2 + const
and the force on the electron is:
!F = q!Ein = − eρ0!R
3ε0
(d) Find the frequency of this oscillation.

Use
m∂t2 r = − eρ 0r 2 2ε0
2ε0 ⇔ ∂t r = − eρ0 r
Using
r = r0 eiωt
we can find
#
ρ0 e
ω= 3mε0
2
2. Dielectric Sphere
(a) Determine the boundary conditions for the given setup.

The boundary conditions can be summarized as follows
1. limr→∞ V (!r) = −E0 r cos θ
2. limr→0 V (!r) remains finite.
3. V (R)in = V (R)out
4. εr,in ∂rVin − ∂rVout = 0
(b) Find the potential inside and outside the dielectric sphere. Explain your approach! Use the
method of separation of variables in spherical coordinates to determine the potential inside and
outside the sphere. In combination with the first two boundary conditions, you will find that
Vin = ∑ Al rl Pl (cos θ ) and Vout = −E0 r cos θ ∑ rDl+1l Pl (cos θ )

l l
Using the remaining boundary conditions, Fourier’s Trick and the orthogonality of the Legendre
Polynomials, we can find:
Vin = − ε3E 0
r +2
z and Vout = −E0 r cos θ + εεrr −1 3 cos θ
+2 R E0 r2
(c) Find the electric field !E and polarization !P inside the dielectric sphere.
!Ein = −!∇Vin = −∂zVin ẑ = − 3E0 ẑ

εr +2
and for a linear dielectric:
!P = ε0 χe !E = εr −1
εr +2 ε0 χe E0 ẑ
3
(d) Sketch the electric field lines for all regions of this setup.
(e) Find the bound volume charge density ρb , and all the bound surface charge densities σb .
Since there are no free charge inside the sphere
ρb = 0.
The bound surface charge is
E0 (εr −1)
σb = !P · n̂ = εr +2 ε0 χe cos θ .
4
3. Inductance
(a) In the quasistatic approximation, find the induced electric field as a function of distance s from
the wire. In the quasistatic approximation, the magnetic field of a wire is
µ0 I
B= 2πs .
Using Faraday’s law, we can find for the electric field:
$ % %s
!E d!l = E(s0 )l − E(s)l = − d !B · d!a = − µ0 I dI 1% ds%
dt 2π dt s
& ' s0
µ I ω ( )
⇔ !E(s) = − 02π0 ln s sin(ωt) + K ẑ = 6 × 10−6 sin(ωt) + K ẑ.
(b) Is your answer valid for the limit s → ∞? Explain your answer. No, since ln s diverges for
large s The quasistatic approximation only holds for s ' ct.
(c) Find the self-inductance L of the rectangular coil.

The magnetic field inside a toroid is given by
µ0 NI
B= 2πs .
So, the flux through a single turn is
% b
µ0 NI % 1 µ0 NIh b
Φ = !B · d!a = 2π h s ds = 2π ln a .
a
Which means that the total flux is N times this, and the self-inductance is
µ0 N 2 h b
Φ = LI ⇔ L = 2π ln a .
(d) In the quasi-static approximation, what emf is induced in the coil?

In the quasistatic approximation:
µ0
2πs φ̂ .
!B =
So,
µ0 I %b 1 µ0 Ih b
φ1 = 2π s h ds = 2π ln a .
a
5
This is the flux through only one turn, so the total flux is N times Φ1 :
µ0 Nh b
Φ= 2π ln a I0 cos(ωt).
So,
µ0 Nh
E = − dφ
dt =
b
2π ln a I0 ω sin(ωt) = 2.61 × 10 sin(ωt)
−4 V,
using
ω = 377 1s .
(e) Find the current I(t) in the resistor R.
Ir = E
R = 5.22 × 10−7 sin ωt A.
(f) Calculate the back emf in the coil, due to the current I(t). The back emf Eb is given by:
Eb = −L dI
dt ;
r
Now use the self-inductance of the square coils calculated in part (a):
µ0 N 2 h b
L= 2π ln a = 1.39 × 10−3 H.
Therefore,
Eb = −2.74 × 10−7 cos(ωt) V.
6
4. Momentum of Electromagnetic Fields
(a) What is the Poynting vector !S?

The Poynting vector is given by
!S = 1! !
µ0 E0 × B0
The electric field of a parallel plate capacitor is given by
!E = − σ ẑ
ε0
So
!S = σ B0
c2 µ0 ε0
x̂
(b) What is the momentum density of the electromagnetic field? The momentum density is given
by
1!
!p = !0
E ×H
c2 0
⇒ !p = σ B0 x̂
(c) What is the total momentum of the electromagnetic field? The total momentum is given by
%
!Ptotal = !p dV = QB0 hx̂
(d) What is the impulse ∆!p in time ∆t experienced by each plate, as derived from the induced
electric field? How does this compare to the field momentum derived in part c)?
Now, we consider a closed loop in the x − z-plane of width a and height h. The magnetic flux
through this loop is:
%
!B0 d!a = Bb ahŷ
and for the induced electric field:
$ %
!Eind d!l = −∂t !B0 d!a = − B0 ah .
∆t
Taking the closed loop integral:
7
$
!Eind d!l = Eind 2a = − B0 ah ⇒ |Eind | = − B0 h
∆t 2∆t
The forces on the plates are now:
!F = q!E with !Eind = Eind x̂ at the top plate and !Eind = −Eind x̂ at the bottom plate.
So, the force on both plates with Q on the top plate and −Q on the bottom plate, is given by:
!F = Q B0 h x̂
2∆t
To find the momentum !p:
% QB0 h
!p = !F dt = 2 x̂
So, each plate receives half of the momentum stored in the fields.
8
5. Dipole Radiation
!1"
(a) Find the electric field !E(r,t) and magnetic field !B(r,t) to leading order in powers of r in
terms of p̈0 (tr ) [i.e. the second time derivative of p0 (t) evaluated at the retarded time, tr ].
To find the electric field, use
!E = !∇V − ∂t !A
!1"
Since we only consider terms of powers r in terms of p̈0 (tr ), we only deal with the last term of
V (!r,t).
r̂!p˙ ( ! ")
!∇V = 1
∂ r r̂ = 1 1
∂ r r̂!˙ r ) r̂
p(t
4πε0 cr 4πε0 cr
! "
= 4πε 1 r̂ ∂
r̂!˙ r ) ∂tr
p(t
0 cr ∂t r ∂r
1 r̂!p¨
= − 4πε0 c2 r r̂
Now,
*˙ + ¨ r)
µ0 ∂ !p(tr ) µ0 !p(t
∂t !A = 4π ∂tr r = 4π r
Finally,
¨ ¨
!E = 1 (r̂·!p)r̂−!p(tr )
4πε0 c2 r
The magnetic field!B can be found, as
µ0 ! ˙ r)
!p(t
!B = !∇ × !A =
4π ∇ × r
use the vector identity

* + * +
!∇ × f!v = f !∇ ×!v = !v × !∇ f
* + ! "
!∇ × 1 !p˙ = 1 !∇ × !p˙ −!p × !∇ 1
r r r
!1" 1
!
and ∇ = 2 r̂,which can be neglected.
r r
Now,
!∇ × !p˙ = !∇ × p̂ ṗ = − p̂ × !∇ ṗ
! "
with !∇ ṗ = p̈!∇ (tr ) = p̈ − cr̂
9
So, finally:
!B = µ0 1 µ0 ! ¨
"
4π rc ( p̂ × p̈r̂) = − 4πrc r̂ × p
!
(b) Assume that !p(tr ) = p0 (t)ẑ. Show that !E(!r,t) = E(!r,t)θ̂ and !B(!r,t) = B(!r,t)φ̂ . Find expres-
sions for E(!r,t) and B(!r,t).
!E = 1 p̈((r̂·ẑ)r̂−ẑ) p̈ cos θ r̂−ẑ

4πε0 c2 r = 4πε0 c2 r
using
ẑ = cos θ r̂ − sin θ θ̂ ,
we finally find for the electric field:
!E = p̈ sin θ θ̂
4πε0 c2 r
Similarly for the magnetic field:
!B = − µ0 p̈ (r̂ × ẑ) = µ0
4πcr 4πcr p̈ sin θ φ̂ .
(c) Find the power radiated to infinity by this time dependent charge distribution.
The total power is given by:
%
P = !S d!a
(d) Calculate the Poynting vector !S.

So, we need to find the Poynting vector:
* +
!S = 1 !E × !B
µ0
µ0 p̈20
= 16π 2 c r2
sin2 θ r̂
So, the total power is then:
% µ0 p̈20 % sin2 θ 2
P = !S d!a = 16π 2 c r2
r sin θ dθ dφ
2
µ0 p̈0 % 3
= 8πc sin θ dθ
µ0 p̈20
= 6πc
10
Quantum Problems
1. Consider a quantum system whose state at time t1 is given by |Ψ(t1 )i = √12 (|ψ1 i + |ψ2 i),
where |ψ1 i and |ψ2 i are energy eigenstates with eigenvalues E1 and E2 respectively. (E1 6= E2 .)
(a) Calculate the uncertainty ∆E of the system, as well as the first time t2 > t1 at which
|Ψ(t2 )i becomes orthogonal to |Ψ(t1 )i. Show that (∆E)(∆t) ≥ h̄, where ∆t = t2 − t1 .
(b) Assume that the above system consists of a particle of mass M moving in the Coulomb
potental V (r) = −e2 /r in three spatial dimensions, and that |ψ1 i = |ψ1,0,0 i and |ψ2 i = |ψ2,1,0 i,
where |ψn,`,m i is the energy eigenstate with principal quantum number n, angular momentum
quantum number `, and magnetic quantum number m. Find ∆E and ∆t in (a) as a function
of M , e, and h̄. Also calculate the time-evolving expectation values hL2 i(t) and hLz i(t) for all
t ≥ t1 , where L2 and Lz are the usual angular momentum operators.
2. Two measurements are made in rapid succession on a quantum system originally in the
state |ψi. The first measurement is of an observable B, and the second is of a non-degenerate
observable A. Assume that the first measurement changes the state of the system, and that
immediately after the second measurement the system is again in the state |ψi.
(a) Prove that [A, B] 6= 0̂.
(b) Show, via an explicit example, that the physical situation described above can actually
occur. (Suggestion: Try this for a two-dimensional Hilbert space.) What is the probability of
your particular scenario occuring?
(0)
3. Consider a quantum system with Hamiltonian operator H = H0 + λH1 , and let |ψn i be a
(0)
particular non-degenerate eigenstate of H0 with corresponding eigenvalue En . Assume λ 1.
(0) (0)
(a) Show that if the first-order corrections in λ to both |ψn i and En vanish, then all higher
order corrections to both vanish as well. (Hint: Prove that if the first-order corrections vanish,
(0)
then H1 |ψn i is the zero vector. In this problem, as is standard, we have chosen the first-order
(0) (0)
correction to |ψn i to be orthogonal to |ψn i.)
(b) In the special case of single particle motion in one dimension with H0 = p̂2 /2M + V0 (x̂)
and H1 = V1 (x̂), show that if the first-order corrections in (a) vanish, then H1 = 0̂.
4. Calculate the degree of degeneracy of the indicated energy level for the following multi-
particle systems in three spatial dimensions.
(a) The ground level of 19 identical spin 1/2 fermions moving in an external isotropic harmonic
oscillator potential.
(b) The second excited level of 2 identical spinless bosons confined inside a cubical box.
5. Consider a particle of mass M moving in the three-dimensional spherically symmetric

potential V (r) = −A e−βr , where A, β > 0.
(a) Show that for “small enough” values of β, the system possesses at least one bound state.
(Hint: Use the variational method with trial ground state wavefunction ψ(r) = N e−αr , where
α > 0 and N is a normalization constant. No minimization with respect to α is necessary.)
(b) Do you expect the system to have a bound state for all values of β? Why or why not?
Quantum Solutions
1. (a) Since ∆E is time-independent, we can calculate it at time t1 . We have
E1 + E 2 E12 + E22
hH i = , hH 2 i = ,
2 2
which gives
p |E1 − E2 |
∆E ≡ hH 2 i − hHi2 = .
2
(In the above, we used the fact that |ψ1 i and |ψ2 i are orthogonal since they are eigenstates
corresponding to distinct eigenvalues of the Hamiltonian operator H = H † .) At any later
time t, the state of the system is given by
1
|Ψ(t)i = √ (e−iE1 (t−t1 )/h̄ |ψ1 i + e−iE2 (t−t1 )/h̄ |ψ2 i) .
2
This is orthogonal to |Ψ(t1 )i if and only if ei(E1 −E2 )(t−t1 )/h̄ = −1. The smallest t > t1 at which
this occurs is t2 = t1 + πh̄/|E1 − E2 |. Thus,
πh̄
∆t = and (∆E)(∆t) = πh̄/2 > h̄ .
|E1 − E2 |
(b) The energy levels of the Coulomb potential are
M e4
En,` = − .
2 h̄ n2
Thus,
3M e4 8πh̄3
(∆E) = and (∆t) = .
16 h̄2 3M e4
Since the Coulomb Hamiltonian H is rotationally invariant, we have [H, L2 ] = [H, Lz ] = 0̂,
so that hL2 i and hLz i are time-independent. Moreover, since L2 |ψn,`,m i = `(` + 1)h̄2 and
Lz |ψn,`,m i = mh̄, we have
hL2 i = h̄2 and hLz i = 0 .
(Note that, for all t, |Ψ(t)i is an eigenstate of Lz but not of L2 .)
2. (a) First note that after the second measurement, the system must be in an eigenstate of A
(by the projection postulate). But this state is |ψi by assumption. Thus, |ψi is an eigenstate
of A, say with eigenvalue λ.
Now assume that A and B commute. (We want to show that this leads to a contradiction.)
We then find
AB |ψi = BA |ψi = λB |ψi .
That is, B |ψi is an eigenstate of A, also with eigenvalue λ. But since A is non-degenerate, any
two eigenstates of A with the same eigenvalue must be colinear. Thus, B |ψi = b |ψi for some
(real) number b. In other words, |ψi is an eigenstate of B as well. But if this were true, then
the initial measurement of B would not change the state of the system (again by the projection
postulate) — a contradiction.
(b) Consider a two-dimensional Hilbert space with orthonormal basis { |1i , |2 i }. Let |ψi = |1i,
A = |1i h1|, and B = |1i h2| + |2i h1|. The eigenvalues of A are λ = 1 and λ = 0, with
corresponding eigenvectors |1i and |2i respectively. The eigenvalues of B are λ = ±1, with
corresponding eigenvectors |ψ± i = √12 (|1i ± |2i). If a measurement of B is performed on |ψi
we will obtain one of the two eigenvalues, each with equal probability (=1/2 by Born’s Rule).
Lets say we get λ = +1. The state of the system immediately following the measurement will be
the corresponding eigenstate |ψ+ i. Upon the ensuing measurement of A, we will again obtain
either of the two eigenvalues with equal probability. Lets say we get λ = 1. Then the state of
the system after the second measurement will be |1i = |ψi, as desired. The probability of this
scenario occuring (that is, of obtaining this particular pair of eigenvalues) is 12 × 12 = 14 .
(0) (0) (0) (0)

3. (a) Assume that H0 |ψm i = Em |ψm i, where the |ψm i’s are a complete, orthonormal set
(0) (0)
of eigenstates of H0 . The first-order corrections to En and |ψn i in non-degerate perturbation
theory are given by
(0)
X hψm (0)
|H1 |ψn i
En(1) = hψn(0) |H1 |ψn(0) i and |ψn(1) i = (0) (0)
(0)
|ψm i.
m6=n En − Em
(1) (0)
Since |ψn i is the zero vector, we have by the linear independence of the |ψm i’s that
(0)
hψm |H1 |ψn(0) i = 0 for all m 6= n .
(0) (0) (1) (1)
That is, H1 |ψn i = λ|ψn i for some complex number λ, and thus En = λ. But since En = 0,
(0) (0) (0) (0)
we have that H1 |ψn i is the zero vector. From this we see that H|ψm i = En |ψn i. That is,
(0) (0)
|ψn i is an exact eigenvector of the full Hamiltonian H (with eigenvalue En ), and hence there
are no corrections to all orders in λ.
(0)
(b) We have from (a) that V1 (x̂)|ψn i is the zero vector. In the position representation we have
(using the completeness of position eigenstates)
Z +∞ Z +∞
V1 (x̂)|ψn(0) i = V1 (x̂)|xihx|ψn(0) i dx = ψn(0) (x)V1 (x)|xi dx .
−∞ −∞
(0)
Since this is the zero vector, we have that V1 (x)ψn (x) = 0 for all x. Hence V1 (x) = 0 for all x
(0)
(except possibly at the nodes of ψn (x) which are a set of measure zero), which implies that
V1 (x̂) = 0̂.
4. (a) Here we have Vext (~x) = A (x2 + y 2 + z 2 ), where A > 0. The energy eigenvalues for a
single (spinless) particle of mass M in this potential are given by
Enx ,ny ,nz = (nx + ny + nz + 3/2) h̄ ω
where nx , ny , nz = 0, 1, 2, . . ., and (for simplicity) we have chosen A = 12 M ω 2 . If we define

N = nx + ny + nz , then any two states of the single-particle system with the same N have the
same energy. It is easy to see that the degeneracy of the level corresponding to any fixed N is
simply (N + 1)(N + 2)/2. (It is easy to work out the corresponding ni ’s.)
For our multi-particle system, by the Pauli exclusion principle, each of the above states can
accomodate exactly two of the identical spin 1/2 fermions — one with “spin up”, and one with
“spin down”. To find the degeneracy of the lowest-lying level of the 15 particles, we imagine
filling these single-particle levels from the “bottom up”. The N = 0 level can accomodate 2
fermions, and the N = 1 level can hold 6 fermions. This leaves 11 fermions. The N = 2
level can accomodate 12 fermions. Thus there are 12 distinct ways of filling this level with the
remaining 11 fermions. (We just have to decide which one of the 12 states we don’t occupy.)
Thus, the degeneracy of the ground level of the 15 identical spin 1/2 fermions is 12.
(b) The energy eigenvalues for a single (spinless) particle of mass M in a cubical box of side-
length “a” are given by
h̄2 π 2 2
Enx ,ny ,nz = (n + n2y + n2z ) ,
2M a2 x
where nx , ny , nz = 1, 2, . . .. If we define N = n2x + n2y + n2z , then any two states of the single-
particle system with the same N have the same energy. The ground state of the single-particle
system has N = 3 and is non-degenerate. The first excited state has N = 6 and is 3-fold
degenerate. The second excited state has N = 9 and is also 3-fold degenerate. (It is easy to
work out the corresponding ni ’s.)
For a multi-particle system of identical bosons, each of the above states can accomodate as many
particles as we wish. (There is no exclusion principle.) However, since the bosons are identical,
we must still (as for fermions) be careful not to overcount multi-particle states. A two-particle
state is labelled by an un-ordered pair of single-particle states (which get symmetrized). Lets
work our way up the two-particle spectrum. The ground level of the two bosons has them both
in the N = 3 single-particle level, and is non-degenerate. The first excited level has one particle
with N = 3 and the other with N = 6. It’s degeneracy is 1 × 3 = 3. The second excited
level has either one particle with N = 3 and the other with N = 9, or both with N = 6. It’s
degeneracy is 1 × 3 + 3 × 3 = 12.
5. (a) It is simplest to work with the reduced radial Schrödinger equation

h̄2 d2 `(` + 1)h̄2
Ĥ` un,` (r) ≡ − + + V (r) un,` (r) = En,` un,` (r) ,
2M dr2 2M
where un,` (r) = rRn,` (r) is the reduced radial wavefunction and Rn,` (r) is the full radial
wavefunction. The normalization integral for un,` (r) is
Z ∞
|un,` (r)|2 dr = 1 .
0
Our trial ground state wavefunction in its reduced form now becomes u(r) = Cre−αr , where
C is a normalization constant. (We can actually do this problem without having to normalize
u(r), but I will do the integral anyway below.) Since in our case V (r) → 0 as r → ∞, we have
by the variational principle that if for any fixed β there exists a value of α such that hĤ0 iu < 0,
then there is a bound state of the system for that value of β. Using the integral
Z ∞
n!
rn e−ar dr = 3
0 a
(for n a non-negative integer and a > 0), we easily find that C = 2α3/2 and
h̄2 α2 8Aα3
hĤ0 iu = − .
2M (2α + β)3
As β decreases, the second term approaches the negative constant −A. But the first term
(which is the positive kinetic energy term) can be made as small as we want by decreasing α.
Therefore, for “small enough” β we can always find a “large enough” α so that hĤ0 iu < 0.
(b) It is difficult to answer this question “rigorously”, which is why I only asked the students
what they “expect”. What I am looking for is the following type of response:
“I do not expect there to be a bound state for all β since the expectation value of the potential
energy for any trial wavefunction will go to zero very rapidly as β → ∞. And since the
potential is getting “narrower” in this limit, by the uncertainty principle I don’t expect that
the corresponding expectation value of the kinetic energy can become small enough so that the
sum remains negative.”
(Note that even though for any fixed α in our trial wavefunction in (a) there exists a large
enough β such that hĤ0 iu > 0, we can’t conclude from this that there are no bound states for
large β since hĤ0 iu is in general only an upper bound for the exact ground state energy. That
is, the exact ground state wavefunction may not be of the form from (a).)

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Quantum Mechanics
January 3, 2011
9:00 am - 12:00 noon
attempts all 5 questions, all of the answers will be graded, and the top 4 scores will be
counted toward the exam’s total score.
Useful integration formulas are

Z ∞
xn e−ax dx = n!/an+1 , valid for complex a as long as Re(a) > 0.
0
Z ∞ q
−ax2
e dx = π/a , valid for complex a as long as Re(a) ≥ 0.
−∞
R∞ 2
q R∞ 2
q R∞ 2
q
−γx 2 −γx 1 4 −γx 3
−∞ e dx = π/γ , −∞ x e dx = 2γ
π/γ , −∞ x e dx = 4γ 2
π/γ .
1. For a 1-dimensional simple harmonic quantum oscillator, V (x) = 12 mω 2 x2 , it is more
convenient
q to describe the dynamics by dimensionless position parameter ρ = x/a (a =
h̄
mω
) and dimensionless energy = E/( 21 h̄ω).
(i) Write down the time-independent Schrodinger equation in terms of ρ derivatives on

the eigenfunction u(ρ). Show that it is equivalent to set ω = 2, 2m = 1, h̄ = 1.
1 2
Directly show that u0 (ρ) ∼ e− 2 ρ satisfies your Schrodinger equation. Explain why
it is the ground state and give its energy 0 in the dimensionless unit.
1 2
Directly show that u1 (ρ) ∼ ρe− 2 ρ satisfies the Schrodinger equation as the first
excited state. Also give its energy 1 in the dimensionless unit.
√
The initial wave function u(ρ, 0) is described by u(ρ, 0) = N (2 2ρ− 32 ) exp(− 12 ρ2 ) ,
with N as the normalization constant.
(ii) Find the average energy.
(iii) The momentum operator in the reduced unit is p = −id/dρ. Find its mean value
hpi initially.
(iv) We use a dimensionless time parameter τ = 12 ωt to study how the wave evolves.
Write down the explicit τ dependence of the wave function.
(v) Find hpi as a function of time τ .
2. In the beginning, a non-relativistic particle of mass m = 12 propagates freely as a wave

4
packet with a mean wave number q = 27 . We choose a unit system such that h̄ = 1. Find
the group velocity and the phase velocity.
Let this wave propagate from the remote left toward a repulsive potential V (x) =
1
2
(1 − 9x2 ) in the region [− 13 , 31 ]. The potential vanishes otherwise. We can treat the
potential as a delta-function potential Gδ(x). Give quantitative reasons why. Figure out
the equivalent strength G.
Derive the probabilities of transmission and reflection in the delta potential approxi-
mation, and give the numerical results for the given inputs.
3. For matrices A, B, C, show that [AB, C] = A[B, C] + [A, C]B. Let S1 (or Sx ) be the
x-component of the spin vector operator, etc. Using the algebra of angular momentum,
[Sx , Sy ] = ih̄Sz , and cyclic permutations for a general spin (S = 21 , 1, 32 , · · ·), simplify
[Sx2 , Sz ] and [Sy2 , Sz ] in terms of Sx Sy or Sy Sx . Then calculate [S 2 , Sz ]. The trace of a
matrix A is defined as Tr A = m hm|A|mi summing over the basis vectors m. Show that
P
Tr (AB) = Tr (BA) by working out the component sum. Based on some earlier steps,
show that Tr (Sx Sy ) = 0.
An unknown particle (X) of spin S and mass MX couples to the fixed target nucleus
of spin I by a feeble spin-dependent contact interaction
V = gδ 3 (r)S · I
.
What are possible numbers of S · I in general? Show all these possibilities for the
special case S = 12 , I = 92 .
Justify the following trace relations,
Tr (Si Sj ) = CS δij , or similarly Tr (Ii Ij ) = CI δij ,
when the corresponding mS , or mI states are summed respectively. Determine coefficients

CS and CI in terms of general S and I respectively.
Calculate the transition probability
|hkk, m0S , m0I |V|k, mS , mI i|2

X
0
from an incoming X plane wave described by eik·r to the outgoing X wave eik ·r . The sum
adds up all spin states mS and m0S of the initial and final spin states of the X particle,
as well as mI and m0I of nucleus.
After averaging the initial spins of the X particle and the nucleus, find the total cross
section in the Born approximation. (Hints: The usual potential scattering in the Born
approximation is
M 2 Z 3
2
dσ

i(kf −ki )·r

= 2 4 d rV (r)e .
dΩ 4π h̄
The above formula has to be generalized to incorporate the spins of S and I.)
4. A quantum particle is in a two dimensional potential V (r) = −V0 exp(−r/a) (with

V0 > 0). Let the trial wave function be ϕ(r; β) = Ce−βr . Determine the normalization C
in terms of the attenuation parameter β.
Find the average position and average momentum x̄ = hxi, ȳ = hyi, p̄x = hpx i,
p̄y = hpy i.
Determine hr 2 i and hp2 i. (Hints: d2 rψ ∗ (r)∇2 ψ(r) = − d2 r|∇ψ(r)|2 .)
R R
What are hx2 i and hp2x i?

Calculate the product h(x − x̄)2 ih(px − p̄x )2 i, and simplify the result in comparison
with the Heisenberg uncertainty.
Find the average kinetic energy and average potential energy.
h̄
Now we set 2m = 1, V0 = 8, a = 12 . Verify that β = 1 minimizes the energy expectation
value.. Using the variational principle, estimate the ground state energy.
5. The dynamic of a three-state system of configurations |1i,|2i,|3i, is governed by the

Hamiltonian H0 = − i,j |iihj|, which is
P
−|1ih1| − |1ih2| − |1ih3|
−|2ih1| − |2ih2| − |2ih3|

−|3ih1| − |3ih2| − |3ih3| .
Work out the Hamiltonian matrix elements hi|H0 |ji, and present the corresponding 3 × 3
matrix.
Find eigen energies as well as the corresponding eigenstates.
A small perturbation gV = g|1ih1| is applied (g 1). Find the 1st order and 2nd
order corrections of the ground state energy .
Quantum Mechanics
January 4, 2010.
9:00 am – 12:00 pm

Various equations, standard integrals, etc. are provided on the last

page of the exam.
Question 1
In heavy (large Z) hydrogen-like atoms, where to a very good approximation the reduced mass µ
is equal to the electron mass me, relativistic corrections to the electron’s kinetic energy need to be
taken into account due to its large orbital velocity.
a) Show that the first-order correction to the Hamiltonian due to the electron’s
2
1  p2 
relativistic kinetic energy is given by Hˆ 1 = −   .
2  2m 
2me c  e 
1 Z
b) Verify that = for the 1s state of the hydrogen-like atom.
r a0
1
c) Similarly, evaluate for the 1s state of the hydrogen-like atom.
r2
ˆ p2 Ze 2
d) Use the fact that the unperturbed Hamiltonian is H 0 = − , and the results
2me 4πε 0 r
from parts (a), (b) and (c), to evaluate the first-order perturbation to the electron’s
kinetic energy due to the relativistic correction.
Question 2
Consider an attractive delta-function

potential − Λδ(x) , where the
V(x)
parameter Λ characterizes the
strength of the delta-function,
positioned at the center of an infinite
square well of width 2a; that is, a
potential given by
V(x ≤ –a) = ∞
‒a 0 a x
V(–a < x < 0) = 0
V(x = 0) = –Λδ(x)
‒Λδ(x)
V(0 < x < a) = 0
V(x ≥ a) = ∞
a) Show that the equation determining the energy E of the eigenstate bound to the delta-
function is given by
h2κ
tanh( κa ) = ,
mΛ
where hκ = 2m E and m is the mass of the particle.
b) What is the minimum strength of the delta-function for which a state with E < 0
exists?
Question 3
Quantum dots are important material systems in nanoscience.

Consider an electron of charge e and mass me confined in an idealized quantum dot with V(r) = 0
for r < a and V(r) = ∞ for r > a; that is, an infinite spherical well of radius a. The eigenstates of
this potential are given by
s r
r ψ nlm = C nl jl  nl Ylm (θ, φ) ,
 a 
where Ylm(θ,φ) is the spherical harmonic associated with the usual angular momentum quantum
numbers l and m, and Cnl is the normalization constant for the lth spherical Bessel function jl with
roots snl. The first three spherical Bessel functions with ρ = s nl r a are
sin ρ sin ρ cos ρ  3 1 3

j 0 (ρ) = j1 (ρ) = − j 2 (ρ) =  3 −  sin ρ − 2 cos ρ
ρ ρ2 ρ ρ ρ ρ
and the first few roots are given below:
snl l=0 l=1 l=2 l=3 l=4

n=1 π 4.49 5.76 6.99 9.36
n=2 2π 7.73 9.10 10.42
n=3 3π
The spherical Bessel functions also satisfy the relation

a
 s r   s r  a3
∫
0
r 2 dr jl  nl  jl  n′l  =
 a   a  2
[ jl −1 (s nl )]2 δ nn′ .
a) What are the energies of the lowest four eigenstates?

b) Evaluate the normalization constants C10 and C11.
 sin ρ cos ρ  
c) Consider the state r ψ = A 2 −  sin θ cos φ + sin ρ  , where A is a constant.
 ρ  ρ 
 ρ
With what probability is the ground state energy measured? And what is the probability
of measuring a z-component of the angular momentum with a value +ħ?
d) In the presence of a strong homogeneous magnetic field B, the energy states split – the
Zeeman effect. Neglecting spin (the normal Zeeman effect), the additional contribution
e
to the Hamiltonian can be written as H 1 = L • B . Amongst the four lowest states
2 me
identified in part (a), at what magnetic field strength will states originating from one
energy level overlap energetically with those from another?
Question 4
Field emission, a quantum tunneling phenomenon, occurs when a DC electric field EDC applied
perpendicular to a material surface becomes sufficiently strong to allow electrons in the material
to tunnel out into the vacuum.
V(x)
εF + Φ
‒eEDCx
Tunneled
εF electron
Electrons
0
In materials like metals, the zero-point energy E of the electrons in the metal is usually defined in
terms of the Fermi energy εF (the energy of the last filled electronic state at zero Kelvin), which
is located at an energy Φ (the material work function) below the vacuum energy.
a) Using the approximation for the one-dimensional tunneling probability through a ‘thick’
barrier,
 b 
∫
2m
T ≈ exp − 2 dx 2 (V ( x) − E )  ,

2
 h 
 a 
of thickness b ‒ a, determine the transmission probability for an electron at an arbitrary
energy E; that is, at an energy E ′ = ε F + Φ − E below the top of the barrier.
b) In a real material at non-zero temperature T, the energy distribution of electrons is given
by the Fermi function
1
f (E) = ,
 ε −E
1 + exp − F 
 k B T 
where kB is Boltzmann’s constant. For exp[(Φ − E ′) k B T ] >> 1 , determine the energy

below the top of the barrier for which electron field emission is most probable.
Question 5
Consider the following spatial-coordinate-independent two-particle Hamiltonian for a spin

system consisting of two identical spin ½ particles:
Hˆ = A + BS1 • S 2
where A and B are constants.
a) Find the 4×4 matrix representation of Ĥ in the uncoupled basis set
1  0  0  0
       
 0 1  0  0
↑↑ =   , ↑↓ =   , ↓↑ =   , and ↓↓ =   ,
0 0 1 0
       
 0  0  0 1
       
where ↑ represents a spin-up particle with S z = 12 h and a spin-down particle with

S z = − 12 h is ↓ .
b) Calculate the eigenvalues of Ĥ .
c) What are the eigenstates of the diagonalized Hamiltonian?
d) Verify your results by finding the eigenvalues of the four states in the coupled basis
S, M representing the total spin S and its z-component M.
Equation Sheet
1 3 3 ± iφ
Y00 (θ, φ) = Y10 (θ, φ) = cos θ Y1±1 (θ, φ) = m e sin θ
4π 4π 8π
Y20 (θ, φ) =
5
16π
(
3 cos 2 θ − 1 ) Y2±1 (θ, φ) = m
15 ±iφ
8π
e sin θ cos θ
15 ± 2iφ
Y2± 2 (θ, φ) = e sin 2 θ
32π
  Ze 2 
2
µ  1 = −13.6 Z eV 4πε 0 h 2 1 1 1
En = − 2   a0 = = +
 2h  4πε   n2 n2 µe 2 µ me M nucleus
  0

3 3
Z 2
 Zr   Z 2  Zr   Zr 
R10 (r ) = 2  exp −  R20 (r ) = 2  1 −  exp − 
 a0   a0   2a 0   2 a 0   2a 0 
3
1  Z  2 Zr  Zr 
R21 (r ) =   exp − 
3  2a 0  a 0  2a 0 
∞
( ) Γ (m2+1 ) (12 ) =
∫
0
dx x m exp − ax 2 =
2a ( m +1) / 2
; Γ(n + 1) = nΓ(n) , Γ(n + 1) = n! , Γ π
∞
so that ∫ dx x
2n
(
exp − λ2 x 2 =) 1.3.5 ... (2n + 1) π
2 n λ2 n+1
0
∫
n!
dx x n e − λx =
λ n+1
0
2 2(2 A − 3Bx)( A + Bx )3 2
( A + Bx ) 2
3
∫ dx A + Bx =
3B ∫ dx x A + Bx = −
15 B 2
Quantum Problems
1. Consider a particle of mass M moving in the spherically symmetric potential V (r) =

−A α/`n(1 + α r) in three spatial dimensions (A, α > 0).
(a) Prove that the ground state energy of the system is strictly less than the corresponding
ground state energy for the potential Vc (r) = −A/r (for any A, α > 0).
(b) Assuming α to be “small”, expand V (r) in a power series in α and calculate the ground
state energy complete to order α2 .
2. Consider a particle moving in the potential V (x) in one spatial dimension, and let Π be
the parity operator.
(a) Assume that the time-evolving state of the system is such that h Π i(t1 ) 6= h Π i(t2 ) for
some times t1 and t2 . Show that V (x) is not a symmetric potential.
(b) Assume instead that V (x) is symmetric and that h Π i(t) = 1. If a measurement of the
energy is performed at time t, show that the probability of finding the system in the first
excited state is equal to zero.
3. Consider a quantum system with a two-dimensional Hilbert space H. Let {|1i , |2i} be an
orthonormal basis for H, and let the time-dependent Hamiltonian operator for the system
be given by (λ > 0)

H(t) = λ cos(θ(t)) |1i h1| + sin(θ(t)) |1i h2| + sin(θ(t)) |2i h1| − cos(θ(t)) |2i h2| .
(a) Assume that θ(t < 0) = 0 and θ(t > 0) = π/2. If the system is in the ground state at
time t = 0− , what is the probability that it will be found in the ground state upon an energy
measurement at time t = 0+ ?
(b) Assume instead that θ(t) is “slowly-varying”, and that θ(t1 ) = 0 and θ(t2 ) = π/2. If the
system is in the state |2i at time t1 , what is the state at time t2 ?
4. Let A and B be two Hermitian operators on a finite-dimensional Hilbert space H.

(a) Assume that A and B do not commute. Prove that the operator C ≡ i [A, B] has (at
least) one eigenvalue which is real and negative, and (at least) one eigenvalue which is real
and positive. Is this result still true if H is infinite-dimensional?
(b) Assume instead that A and B commute. Prove that the operator AB has a zero eigenvalue
if and only if either A or B has a zero eigenvalue. Is this result still true if H is infinite-
dimensional?
5. The total angular momentum operator J~ = S ~1 + S~2 + L

~ for a hydrogen atom has con-
~1 ), the spin of the proton (S
tributions from the spin of the electron (S ~2 ), and the relative
~
orbital angular between the electron and proton (L).
(a) How many linearly independent states of the system are there with total angular mo-
mentum quantum number j ? (Ignore radial excitations.)
(b) Write down a maximal set of linearly independent states with j = 0, each expressed as a
linear combination of the states |s1 , ms1 i ⊗ |s2 , ms2 i ⊗ |`, m` i.
Quantum Solutions
1. (a) First we show that V (r) < Vc (r) for all r > 0. This amounts to showing that 1/r <
α/`n(1 + αr), or equivalently
`n(1 + αr) < αr
which clearly holds. Next, we will use the variational theorem: The expectation value of the Hamilto-
nian operator in any state |ψi is greater than or equal to the ground state energy of the system (with
equality holding if and only if |ψi is the exact ground state). Specifically, we take the expectation
value of the Hamiltonian for our system
−h̄2 2
H= ∇ + V (r)
2M
in the Coulombic ground state |ψc i. Since V (r) < Vc (r) for all r > 0, we see that hHi|ψc i is less
than the Coulomb ground state energy. But by the variational theorem hHi|ψc i is greater than the
ground state energy for V (r).
(b) We have that (after some algebra)
A A Ar
V (r) = − − α − α2 + O(α3 ) ≡ Vc (r) + α V1 (r) + α2 V2 (r) + O(α3 ) .
r 2 12
Thus, in order to obtain the ground state energy correct to order α2 , we must do perturbation
theory about Vc (r) to second-order in V1 (r) and to first-order in V2 (r). Since V1 (r) is a constant, the
associated first-order correction is just α V1 (r), and the associated second-order correction vanishes.
The first-order correction from V2 (r) gives −α2 A hri/12, where the expectation value is taken in the
Coulomb ground state. Putting this all together we obtain
2
M A2 A 2 h̄
Eground = − − α − α + O(α3 ) ,
2h̄2 2 8M
where the first term is the Coulombic ground state energy, and we have used the fact that hri =
3h̄2 /2M A.
2. (a) The “equation of motion” for h Π i(t) is
d h Π i(t) i i
= h [ H, Π ] i = h [ V (x), Π ] i .
dt h̄ h̄
Here H = p2 /2M + V (x) is the Hamiltonian operator of the system, and in the last equality we have
used that [ p2 , Π ] = 0. Since by assumption h Π i(t) changes with time, we know that [ V (x), Π ] 6= 0.
But this commutator is zero if V (x) is symmetric.
(b) The eigenvalues of Π are ±1, with symmetric (+1) and anti-symmetric (−1) wavefunctions
associated with the corresponding eigenvectors. Thus, h Π i(t) = 1 means that the state |ψ(t)i has
a symmetric wavefunction. A measurement of the energy will project this state onto the different
energy eigenstates |ψn i (n = 0, 1, 2, . . .) with associated probabilities |hψ(t)|ψn i|2 . (Here n = 0
corresponds to the ground state, n = 1 to the first excited state, etc.) But these probabilities will
be zero for odd n since V (x) symmetric implies that wavefunction of |ψn i is anti-symmetric in these
cases. In particular, |hψ(t)|ψ1 i|2 = 0.
3. First note that the eigenvalues of H are ±λ independent of θ(t).
(a) Here we use the sudden approximation: The state of the system immediately after the change
(t = 0+ ) will be the same as the state of the system immediately before the change (t = 0− ). We
have that
H(t = 0− ) = λ |1i h1| − |2i h2| ,

so that the ground state at t = 0− is |ψ0 i = |2i (remember that λ > 0). Thus, |ψ(t = 0− )i =
|ψ(t = 0+ )i = |2i. But
H(t = 0+ ) = λ |1i h2| + |2i h1| ,

so that the ground state at t = 0+ is given by |φ0 i = √12 (|1i − |2i). The probability that the
system will be found in the ground state upon an energy measurement at time t = 0+ is thus
|hφ0 |ψ0 i|2 = 1/2.
(b) Here we use the adiabatic approximation: If at time t1 the system is in a non-degenerate eigenstate
of a “slowly-varying” Hamiltonian H(t), and if this energy level does not cross any other level as a
function of t, then the state of the system at any later time t2 is the eigenstate of H(t2 ) which is
related by continuity to the relevant eigenstate of H(t1 ). Note that since the eigenvalues of H(t)
are ±λ for all t, we have that the energy levels are nondegenerate and do not cross. Moreover, the
state of the system at time t1 is the ground state of H(t1 ). Thus, in the adiabatic approximation
the state of the system at time t2 will be the ground state of H(t2 ), which is given by √12 (|1i − |2i).
4. (a) Since H is finite-dimensional,

we are assured that Tr(C) exists and is given by Tr(C) =
i Tr([ A, B ]) = i Tr(AB) − Tr(BA) = 0. But Tr(C) is simply the sum of the eigenvalues of C.
Moreover, C is Hermitian (note the factor of i) and hence has only real eigenvalues. Since we know
that [ A, B ] 6= 0, all of the eigenvalues of C cannot be zero. Hence the sum of the eigenvalues being
zero and real implies that there is at least one positive and at least one negative eigenvalue. If the
dimension of H is infinite, then Tr(C) may not exist and thus the above proof does not go through.
An explicit counterexample to the claim is provided by letting H be the Hilbert space associated
with a single particle in one spatial dimension, and setting A = p̂ and B = x̂ so that C = h̄I (where
I is the identity operator on H). In this case we see that all of the eigenvalues of C are positive.
(b) (i) Assume that A (say) has a zero eigenvalue with corresponding eigenvector |ψi. We then have
AB |ψi = BA |ψi = 0. Hence |ψi is an eigenvector of AB with zero eigenvalue. (ii) Conversely,
assume that AB has a zero eigenvalue with corresponding eigenvector |φi, that is, AB |φi = 0.
There are two possibilities here. First, we can have B |φi = 0, in which case B has a zero eigenvalue.
Second, it may be that B |φi = 6 0. But then B |φi is an eigenvector of A with eigenvalue zero.
Note that this proof did not use the finite-dimensionality of H in any way, so that it is valid in the
infinite-dimensional case.
5. (a) Both the electron and the proton have spin 1/2. That is, s1 = s2 = 1/2. These may combine
~≡S
to either give either s = 0 or s = 1 (where S ~1 + S
~2 ). In the former case, the only way to obtain
total angular momentum j is to have ` = j. This gives (2j + 1) linearly independent states. When
s = 1 we can obtain total angular momentum j with ` = j − 1, j, j + 1 if j ≥ 1, while for j = 0 we
can only use ` = j + 1 = 1. Thus, the s = 1 case gives 3(2j + 1) linearly independent states for
j ≥ 1 and only (2j + 1) = 1 state for j = 0. Putting this all together, the total number of linearly
independent states is 4(2j + 1) for j ≥ 1, and 2(2j + 1) = 2 for j = 0.
(b) The two linearly independent states with j = 0 are given by
1
√ |1/2, 1/2i ⊗ |1/2, −1/2i − |1/2, −1/2i ⊗ |1/2, 1/2i ⊗ |0, 0i
2
and
1
√ |1/2, 1/2i ⊗ |1/2, 1/2i ⊗ |1, −1i + |1/2, −1/2i ⊗ |1/2, −1/2i ⊗ |1, 1i
3
1
− √ |1/2, 1/2i ⊗ |1/2, −1/2i + |1/2, −1/2i ⊗ |1/2, 1/2i ⊗ |1, 0i .
2
These are straightforwardly obtained using the standard techniques for calculating Clebsch-Gordon
coefficients.
Quantum Mechanics (Draft 2010 Nov.)
1. For a 1-dimensional simple harmonic quantum oscillator, V (x) = 12 mω 2 x2 , it is more
convenient to describe the dynamics by dimensionless position parameter ρ = x/a (a =
!
h̄
mω
) and dimensionless energy # = E/( 12 h̄ω).
Write down the time-independent Schrodinger equation in terms of ρ derivatives on the
1 2
eigenfunction u(ρ). Directly show that u0 (ρ) ∼ e− 2 ρ satisfies your Schrodinger equation.
Explain why it is the ground state and give its energy #0 in the dimensionless unit.
1 2
Directly show that u1 (ρ) ∼ ρe− 2 ρ satisfies the Schrodinger equation as the first excited
state. Also give its energy #1 . √
The initial wave function u(ρ, 0) is described by u(ρ, 0) = N(3 2ρ − 4) exp(− 21 ρ2 ) ,
with N as the normalization constant.
(i) Find the average energy.
(ii) The momentum operator in the reduced unit is ℘ = −id/dρ. Find its mean value
$℘% initially.
(iii) We use a dimensionless time parameter τ = 21 ωt to study how the wave evolves.
Write down the explicit τ dependence of the wave function.
(iv) Find $℘% as a function of time τ .
d2 2
− dρ2 u(ρ) + ρ u(ρ) = #u(ρ) .
d d 2
2
We can pretend that ω = 2, 2m = 1, h̄ = 1. dρ u0 (ρ) = −ρu(ρ), dρ2 u0 (ρ) = (ρ − 1)u(ρ),
so u0 is a solution with #0 = 1. u0 has no node, thus it is a ground state. Similarly u1 ,

with one node, is the first excited solution for #1 = 3.
2ρ ρ2 1 √ 1
u(ρ, 0) = N(3 √ − 4)e− 2 = Nπ 4 (3u1 (ρ) − 4u0 (ρ)) , 1 = N 2 π(32 + 42 ) , N= 1
2 5π 4
u(ρ, 0) = 15 (3u1 (ρ) − 4u0(ρ)) , u(ρ, τ ) = 51 (3u1(ρ)e−3iτ − 4u0 (ρ)e−iτ )
9 16 43
The average energy is $#% = 25
×3+ 25
×1 = 25
.
√ ρ2 √ ρ2
u(ρ, τ ) = 1
1 (3 2ρe−3iτ − 4e−iτ )e− 2 , u∗ (ρ, τ ) = 1
1 (3 2ρe3iτ − 4eiτ )e− 2
5π 4 5π 4
√ ρ2
−i(d/dρ)u(ρ, τ ) = − i
1 (3 2(1 − ρ2 )e−3iτ + 4ρe−iτ )e− 2
5π 4
" ∞ √ # $ 2
√
$℘%τ = − i
1 12 2 ρ2 e+2iτ − e−2iτ + ρ2 e−2iτ e−ρ dρ = 12 2
25
sin 2τ
25π 2 −∞
The initial $℘%τ is zero, but it changes as a sine function.
2. In the beginning, a non-relativistic particle of mass m = 12 propagates freely as a wave

4
packet with a mean wave number q = 27 . We choose a unit system such that h̄ = 1. Find
the group velocity and the phase velocity.
1
Let this wave propagates from the remote left toward a repulsive potential V (x) =
9(x − 31 )2 in the region [− 13 , 31 ]. The potential vanishes otherwise. We can treat the
potential as a delta-function potential. Give quantitative reasons why.
Calculate the probabilities of transmission and reflection in the delta potential ap-
proximation.
4 p 8
The momentum is 27
, the particle velocity (i.e. the group velocity) vg = m
= 27
.
p2 16
K = 2m = 729 . ω = Kh̄ = 729
16
, the phase velocity is vphase = ωk = 27
4
.
The wavelength of the particle is 2π/q = 54π/4, which is much greater than the
potential width of a size only 2/3. The height of the potential is also much greater than
K. Therefore we can treat the potential as delta function V (x) = Gδ(x). The strength
G is given by the integral
" 1
3
G= 9(x − 13 )dx = 2
9
− 13
We integrate both sides of Schrodinger equation over a small interval around the origin.
Aeikx + B −ikx ,


 for x < 0 ,
ϕk (x) =
Ceikx + D −ikx , for x > 0 ,


The B component represents the reflection and the C piece is the transmission. Although
for an incoming wave from the left, we do not need the D component, we keep it for other
future purpose temporarily. Continuity of the wave function and its kink in derivatives
around the origin gives
A+B =C +D
h̄2 .
2m
ik[(A − B) − (C − D)] + G(C + D) = 0
We can solve A and B in terms of C and D,
1 + miG + miG
) * ) *) *
A h̄2 k h̄2 k
C
= miG
B − h̄2 k 1 − miG
h̄2 k
D
Now, we require D = 0 for the property of an incoming particle from the left. So
T (k) ≡ C/A = h̄2 k/(h̄2 k + miG) , R(k) ≡ B/A = −miG/(h̄2 k + miG) .

+ 4
+2
+ + 16
|T |2 = +
4
27
3
+
= = 64%
+ i 27 25
+ +
+ +
27
Transmission probability is 64%, and the rest 36% is reflected.
3. For matrices A, B, C, show that [AB, C] = A[B, C] + [A, C]B. Let S1 (or Sx ) be the
x-component of the spin vector operator, etc. Using the algebra of angular momentum,
[Sx , Sy ] = ih̄Sz , and cyclic permutations for a general spin (S = 12 , 1, 32 , · · ·), simplify
[Sx2 , Sz ] and [Sy2 , Sz ] in terms of Sx Sy or Sy Sx . Then calculate [S 2 , Sz ]. The trace of a
2
matrix A is defined as Tr A = m $m|A|m% summing over the basis vectors m. Show that
,
Tr (AB) = Tr (BA). Based of some earlier steps, show that Tr (Sx Sy ) = 0

A dark matter (DM) particle of spin S and mass MDM couples to the fixed target
nucleus of spin I by a weak spin-dependent contact interaction
V = gδ 3 (r)S · I
.
What are possible numbers of S · I in general? Show all these possibilities for the
special case S = 12 , I = 92 .
Justify the following trace relations,
Tr (Si Sj ) = CS δij , similarly Tr (Ii Ij ) = CI δij ,

Determine coefficients CS and CI in terms of general S and I respectively.
Calculate the transition probability
|$kk, m%S , m%I |V|k, mS , mI %|2
-
!
from an incoming DM plane wave described by eik·r to the outgoing DM wave eik ·r . The
sum adds up all spin states mS and m%S of the initial and final spin states of the DM
particle, as well as mI and m%I of nucleus.
After average the initial spins of the DM particle and the nucleus, find the total cross
section in the Born approximation. (Hints: The usual potential scattering in the Born
approximation is
M 2 ++
+2
dσ
+"
3 i(kf −ki )·r +
+
= 2 4 + d rV (r)e + .
dΩ 4π h̄
The above formula has to be generalized to incorporate the spins of S and I.)
[AB, C] = ABC − CAB = ABC − ACB + ACB − CAB = A[B, C] + [A, C]B.
[Sy2 , Sz ] = Sy [Sy , Sz ] + [Sy , Sz ]Sy = iSy Sx + iSx Sy .
[Sx2 , Sz ] = Sx [Sx , Sz ] + [Sx , Sz ]Sy = −iSx Sy − iSx Sy .
As [Sz2 , Sz ] = 0, we have [Sx2 + Sy2 + Sz2 , Sz ] = 0, so [S 2 , Sz ] = 0.
Tr (AB) = i,j Aij Bji = i,j Bji Aij = Tr (BA), and Tr[A, B] = 0. So Tr(Sx Sy ) = 0.
, ,
Let J denote the magnitude number of the sum J = S + I.

h̄2
S · I = 21 [(S + I)2 − S 2 − I 2 ] = 2
[J(J + 1) − S(S + 1) − I(I + 1)] ,
for discrete choices of J = S + J, S + J − 1, · · · , |J − S|.
For the special case S = 21 and I = 92 , we have two possibilities,
(a) J = 5 , S · I = 49 ,
(b) J = 4 , S · I = − 11 4
.
The overall tracelessness is checked.
About Tr (Si Sj ) = CS δij , we have verified the case i = x, j = y. Other unequal i (= j
cases are also true if we permute indices.
Symmetry also implies Tr (Sx Sx ) = Tr (Sy Sy ) = Tr (Sz Sz ) = CS .
h̄2
Tr (Sx Sx ) = 1
3
Tr S 2 = 3
(2S + 1)S(S + 1).
3
h̄4
We note that Tr (S·I)2 = = Tr [(Si Ii )(Sj Ij )] = (2S +1)S(S +1)(2I +1)I(I +1).
,
i,j 9
g 2 h̄4
|$kk, m%S , m%I |V|k, mS , mI %|2 =
-
9
(2S + 1)S(S + 1)(2I + 1)I(I + 1)
1 1
Dividing it by 2S+1 2I+1
for the average of initial spins, we obtain the unpolarized cross
section as
dσ M2
= DM2 g 2 S(S + 1)I(I + 1) .
dΩ 36π
4. Variation principle.
A quantum particle in a two dimensional potential V (r) = −V0 exp(−r/a). Let the
trial wave function be ϕ(r; β) = Ce−βr . Determine the normalization C in terms of the
attenuation parameter β.
Find the average position and average momentum x̄ = $x%, ȳ = $y%, p̄x = $px %,
p̄y = $py %.
Determine $r2 % and $p2 %. (Hints: d2 rψ ∗ (r)∇2 ψ(r) = − d2 r|∇ψ(r)|2.)
. .
What are $x2 % and $p2x %?

Calculate the product $(x − x̄)2 %$(px − p̄x )2 %, and simplify the result in comparison
with the Heisenberg uncertainty.
Find the average kinetic energy and average potential energy.
h̄
Now we set 2m = 1, V0 = 8, a = 12 . Using the variation principle, estimate the ground
state energy. The optimized β turns out to be a simple number.
Normalization is given by
1! πC 2
"
2 −2βr 2
C (2πr)e dr = 2πC = =1
(2β)2 2β 2
!
2
C=β π
. By symmetry x̄ = 0, ȳ = 0, p̄x = 0, p̄y = 0.
3! 3
"
$r 2 % = C 2 (2πr 3 )e−2βr dr = 2πC 2 =
(2β)4 2β 2
"
2
2
$p % = h̄ C 2
(2πr)β 2 e−2βr dr = h̄2 β 2 .
By symmetry, $p2x % = $p2y % = 12 $p2 % = 21 h̄2 β 2 . Similarly, $x2 % = 12 $r2 % = 3

4β 2
. Therefore
2 2 3 2 h̄2
$(x − x̄) %$(px − p̄x ) % = 8
h̄ ,
well above the lowest bound 4
by Heisenberg’s uncertainty
principle.
2 2
Therefore, the average kinetic energy $K% = h̄2m
β
, and
"
$V % = −C 2 (2πr)V0 e−(2β+1/a)r dr = −V0 β 2 /(β + 1 2
2a
) .
For the given parameters,

$E% = β 2 − 8β 2 /(β 2 + 1)2 ,
which is minimized at β = 1 with ($E%)min = −1, as a good estimate of the ground state
energy.
4
5. Matrix diagonalization.
The dynamic of a three-state system of configurations |1%,|2%,|3%, is governed by the Hamil-
tonian H0 = − i,j |i%$j|, which is
,
−|1%$1| − |1%$2| − |1%$3|
−|2%$1| − |2%$2| − |2%$3|

−|3%$1| − |3%$2| − |3%$3| .
Work out the Hamiltonian matrix elements $i|H0 |j%, and present the corresponding 3 × 3
matrix.
Find eigen energies as well as the corresponding eigenstates.
A small perturbation gV = g|1%$1| is applied (g * 1). Find the 1st order and 2nd
order corrections of the ground state energy .
The matrix of Hamiltonian of H0 is

 
1 1 1
− 1 1 1 
 
1 1 1
. It is easy to see the eigen-column-vectors and eigen-values are

     
1 −2 0
Na  1  , Ea = −3 ; Nb  1  , Eb = 0 ; Nc  1  , Ec = 0 .
     
1 1 −1
The normalizations are Na = √13 ,Nb = √16 , Nc = √12 . As b and c are degenerate in energy,
their other superpositions are also acceptable. The ground state and the two excited
states are
|a%0 = √13 (|1% + |2% + |3%)
|b%0 = √1 (−2|1% + |2% + |3%)
6
|c%0 = √1 (+|2% − |3%)

3
We find the following matrix elements of the perturbed interaction,
0 $a|gV |a%0 = 13 g
0 $b|gV |a%0 = √1 (−2g)

18
0 $c|gV |a%0 = 0g
The ground energy is corrected by perturbation as
2 2
Ea = −3 + 31 g − 27
g +··· .
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Thermodynamics & Statistical Mechanics

January 8, 2008
9.00 am – 12:00 pm

1
Problem 1: A cylindrical container of length L is separated into two compartments by a thin
piston, originally clamped at a position L/3 from the left end. The left compartment is filled with
1 mole of helium gas at 5 atm of pressure; the right
compartment is filled with argon gas at 1 atm of pressure.
These gases may be considered ideal. The cylinder is L
submerged in 1 liter of water, and the entire system is
initially at the uniform temperature of 25°C, and thermally He Ar
5 atm 1 atm
isolated from the surroundings. The heat capacities of the water
cylinder and the piston may be neglected. When the piston
is unclamped, the system ultimately reaches a new L/3
equilibrium situation.
(a) What is the change in the temperature of the water?
(b) How far from the left end of the cylinder will the piston come to rest?
⎛ ∂S ⎞ ⎛ ∂S ⎞
(c) Starting from dS = ⎜ ⎟ dV + ⎜ ⎟ dT , find the total increase in the entropy of the
⎝ ∂V ⎠T ⎝ ∂T ⎠V
system.
(d) Now consider a slightly different situation, in which the left side of the cylinder contains 5
moles of real (not ideal) gas, with attractive intermolecular interactions. The right side still
contains 1 mole of an ideal gas. As before, the piston is initially clamped at a position L/3 from
the left end. When the piston is unclamped and released, does the temperature of the water
increase, decrease, or stay the same? Does the internal energy of the gas increase, decrease, or
remain the same? Explain your reasoning.
Problem 2: A copper block is cooled from TB to TA using a Carnot engine operating in reverse
between a reservoir at TC and the copper block. The copper block is then heated back up to TB by
placing it in thermal contact with another reservoir at TB. ( TC > TB > TA )
(a) What is the limiting value of the heat capacity per mole for the copper block at high
temperatures?
(b) Find the total entropy change of the universe in the cyclic process B → A → B and show that
it is greater than zero?
(c) How much work is done on the system, consisting of the copper block and the Carnot engine?
(d) For the cyclic path B → A → B , does the system absorb heat from the reservoirs or reject
heat?
2
Problem 3: Two magnetic spin systems with N1 = 500 spins and N2 = 1000 spins are placed in
an external magnetic field H. They are initially thermally isolated from one another and are
prepared with spin excess values of m1 = +20 and m2 = +250, where m = N ↑ − N ↓ . The magnetic
moment of each spin is denoted by µ m . Assume that one can use the Gaussian approximation for
the number of states of each system with energies between E and E + dE:
2N ⎛ E2 ⎞
Ω(N, E ) = exp⎜⎜ − ⎟dE
⎟
⎝ 2Nµ m H
2 2
2πNµ m2 H 2 ⎠
(a) What is the total energy in each system (in units of µ m H ) and the initial temperature of each
system (in units of µ m H / k )?
(b) The two systems are brought into thermal contact. What are the spin excess values m1eq and
m2eq and the final temperature, after they reach equilibrium?
(c) What is ∆S , the total change in the entropy of the combined system (in units of k), and what
is the probability of finding the system in the initial configuration relative to the probability of
finding the system in the equilibrium configuration?
Problem 4: Consider a 2-dimensional polymer,

consisting of N links, each of length l. Each link has
four allowed orientations, pointing in the +x or −x
direction, or in the +y or −y direction.
(a) Show that the average end-to-end

r r distance squared
for this polymer is given by < L • L >=< L2 >= Nl 2
(b) Now consider a force f that extends the length of the

polymer in the x-direction. In this simplified two-
dimensional picture, the three allowed energy levels of
each link are (i) ε = −fl , if the link is oriented in the direction of the applied force, ε = +fl , if the
link is oriented opposite to the direction of the applied force, and (iii) ε = 0 , if the link is
oriented perpendicular to the direction of the applied force. Find the average extension < Lx > in
the x-direction as a function of the applied force, and show that, at low forces (fl << kBT), the
polymer behaves like a Hookean spring.
(c) The polymer is stretched quasistatically and isothermally such that the average extension in
the x-direction is 5% of its unperturbed size L = N l ? Assume that the polymer still behaves as
a Hookean spring. What is the change in the Helmholtz free energy of the polymer?
(d) Now consider that the polymer is stretched quasistatically and adiabatically. Will the
temperature of the polymer increase or decrease in this process?
3
Problem 5: Consider a system consisting of impurity atoms in a semiconductor. Suppose that
the impurity atom has one "extra" electron (with two degenerate spin states), compared to the
neighboring atoms (e.g. a phosphorus atom occupying a lattice site in a silicon crystal). The extra
electron is easily removed, contributing to conduction electrons, and leaving behind a positively
charged ion.
(a) What is the probability that a single donor atom is ionized? Express your result in terms of
the ionization energy I, and the chemical potential of the "gas" of ionized electrons.
(b) If every conduction electron comes from an ionized donor, and the conduction electrons
behave like an ideal gas, write down an expression that relates the number of conduction
electrons N c and the number of donor atoms N d , in terms of the volume V of the sample, the
temperature T and fundamental constants.
(c) Show that in the limit of low (kT << I ) and high (kT >> I ) temperatures, the ratio N c / N d has
the expected values.
(d) Write down an expression for the Gibbs free energy of the conduction electrons in terms of
Nc, T and V and show that it is an extensive quantity.
Equations and constants:
kB =1.381×10−23 J/K; NA = 6.022×1023; R = 8.315 J/mol/K; 1 atm = 1.013×105 N/m2
Hyperbolic functions
e x − e −x e x + e −x sinh x
sinh x = cosh x = tanh x =
2 2 cosh x
Inequality
ln( y ) ≤ y − 1
Maxwell's relations
⎛ ∂T ⎞ ⎛ ∂P ⎞ ⎛ ∂T ⎞ ⎛ ∂P ⎞ ⎛ ∂T ⎞ ⎛ ∂V ⎞ ⎛ ∂T ⎞ ⎛ ∂V ⎞
⎜ ⎟ = −⎜ ⎟ ; ⎜ ⎟ = −⎜ ⎟ ; ⎜ ⎟ =⎜ ⎟ ; ⎜ ⎟ =⎜ ⎟
⎝ ∂V ⎠S ⎝ ∂S ⎠V ⎝ ∂V ⎠P ⎝ ∂S ⎠T ⎝ ∂P ⎠ S ⎝ ∂S ⎠ P ⎝ ∂P ⎠V ⎝ ∂S ⎠T
Ideal gas
⎛ VZ ⎞
3/2
⎛ h2 ⎞
µ = −k BT ln⎜⎜ int ⎟;
⎟
v q = ⎜⎜ ⎟⎟
⎝ Nv q ⎠ ⎝ 2πmk BT ⎠
4

January 8, 2008
9.00 am – 12:00 pm

1
Problem 1
A cylindrical container of length L is separated into two compartments by a thin piston,

originally clamped at a position L/3 from the left end. The left compartment is filled with 1 mole
of helium gas at 5 atm of pressure; the right compartment
is filled with argon gas at 1 atm of pressure. These gases
may be considered ideal. The cylinder is submerged in 1 L
liter of water, and the entire system is initially at the
uniform temperature of 25°C, and thermally isolated from He Ar
5 atm 1 atm
the surroundings. The heat capacities of the cylinder and water
the piston may be neglected. When the piston is
unclamped, the system ultimately reaches a new L/3
equilibrium situation.
(a) What is the change in the temperature of the water?
For the system consisting of the cylinder, the piston and the gases,
∆U = Q + W = Q
The expanding gas on the left side of the cylinder does work on the right side of the cylinder,
with net work done on the system W = 0.
If ∆U > 0 ⇒ Q > 0 , i.e. the system will absorb heat from the water. The temperature of the gases
in the cylinder will increase, while the temperature of the water will decrease, in violation of the
second law of thermodynamics. Similarly, if ∆U < 0 ⇒ Q < 0 , while would result in a decrease
in the temperature of the gases and an increase in the temperature of water, again in violation of
the second law. The only situation possible is ∆U = Q = 0 . No heat is absorbed or rejected by the
system, and the temperature of the water, and the gases, remains unchanged.
(b) How far from the left end of the cylinder will the piston come to rest?
The initial situation, before the piston is unclamped, gives
PLVL n 5V / 3
PLVL = nL RT and PRVR = nR RT ⇒ = L = = 5/2
PRVR nR 2V / 3
2
The number of moles of gas on the right side nR = nL = 2 / 5
5
When the piston is unclamped, the pressure on both sides is equal
nL RT nR RT V n
Pf = = ⇒ Lf = L = 5 / 2
VLf VRf VRf nR
The piston will move to a distance 5L/7 from the left side of the cylinder.
2
⎛ ∂S ⎞ ⎛ ∂S ⎞
(c) Starting from dS = ⎜ ⎟ dV + ⎜ ⎟ dT , find the total increase in the entropy of the
⎝ ∂V ⎠T ⎝ ∂T ⎠V
system.
⎛ ∂S ⎞ ⎛ ∂S ⎞ ⎛ ∂S ⎞
dS = ⎜ ⎟ dV + ⎜ ⎟ dT = ⎜ ⎟ dV since dT = 0
⎝ ∂V ⎠T ⎝ ∂T ⎠V ⎝ ∂V ⎠T
⎛ ∂S ⎞ ⎛ ∂P ⎞ ⎛ ∂P ⎞ dV
Use the Maxwell relation ⎜ ⎟ =⎜ ⎟ to get dS = ⎜ ⎟ dV = nR for an ideal gas.
⎝ ∂V ⎠T ⎝ ∂T ⎠V ⎝ ∂T ⎠V V
The total entropy change of the system is then given by
⎛V ⎞ ⎛V ⎞ ⎛ 5V / 7 ⎞ 2 ⎛ 2V / 7 ⎞ 2
∆S = ∆SL + ∆SR = nL R ln⎜⎜ Lf ⎟⎟ + nR R ln⎜⎜ Rf ⎟⎟ = R ln⎜ ⎟ + R ln⎜ ⎟ = R ln(15 / 7) + R ln(3 / 7)
⎝ VLi ⎠ ⎝ VRi ⎠ ⎝ V /3 ⎠ 5 ⎝ 2V / 3 ⎠ 5
∆S = 3.52 Joules/K
(d) Now consider a slightly different situation, in which the left side of the cylinder contains 5
moles of real (not ideal) gas, with attractive intermolecular interactions. The right side still
contains 1 mole of an ideal gas. As before, the piston is initially clamped at a position L/3 from
the left end. When the piston is unclamped and released, does the temperature of the water
increase, decrease, or stay the same? Does the internal energy of the gas increase, decrease, or
remain the same? Explain your reasoning.
We still have ∆U = Q , as in part (a), since W = 0 .
If the system was thermally isolated from the surroundings, we would have had ∆U = Q = 0 .
For a real gas, U is a function of volume and temperature. An increase in volume increases the
intermolecular distance between the gas molecules, thus increasing the potential energy
contribution to the internal energy. Therefore, if the system was thermally isolated, an increase
in the potential energy of the gas would be compensated by a decrease in the kinetic energy, to
maintain ∆U = 0 , and the gas would cool. Since our system is in thermal contact with water, the
cooling gas will absorb heat from the water, and the water will cool. The water and the gas will
equilibrate to a temperature lower than the initial temperature of 25°C. From ∆U = Q we can see
that the total internal energy of the gas will increase.
3
Problem 2
A copper block is cooled from TB to TA using a Carnot engine operating in reverse between a
reservoir at TC and the copper block. The copper block is then heated back up to TB by placing it
in thermal contact with another reservoir at TB. ( TC > TB > T A )
(a) What is the limiting value of the heat capacity per mole for the copper block at high
temperatures?
From the equipartition theorem U = f / 2kT where f ≈ 6N are the quadratic degrees of freedom
corresponding to 3N vibrational modes in the solid. Therefore U = 3NkT = 3nRT and the heat
capacity C = 3nR , which gives heat capacity per mole of ≈25 J/K.
(b) Find the total entropy change of the universe in the cyclic process B → A → B and show that
it is greater than zero?
For the reversible path B → A

TA
C ⎛T ⎞
∫ dT = C ln⎜⎜ A ⎟⎟ < 0
B→A
∆S system =
T T
B ⎝ TB ⎠
⎛T ⎞
B→A
∆Sreservoir B→A
= − ∆Ssystem = C ln⎜⎜ B ⎟⎟ > 0
⎝ TA ⎠
For the irreversible path A → B

⎛T ⎞
A →B
∆Ssystem B→A
= −∆Ssystem = C ln⎜⎜ B ⎟⎟ > 0
⎝ TA ⎠
TB
− ∫ CdT
A →B QR − Qsystem TA − TB
∆Sreservoir = = = =C <0
TA
TR TB TB TB
⎛T ⎞ ⎛ T ⎞
∆Scycle = C ln⎜⎜ B ⎟⎟ − C ⎜⎜1 − A ⎟⎟
⎝ TA ⎠ ⎝ TB ⎠
⎛ 1⎞ ⎛ TB ⎞ T
Use the inequality ln( y ) ≤ y − 1or ln⎜⎜ ⎟⎟ ≥ 1 − y to get ln⎜⎜ ⎟⎟ ≥ 1 − A ⇒ ∆Scycle ≥ 0
⎝y⎠ ⎝ TA ⎠ TB
(c) How much work is done on the system, consisting of the copper block and the Carnot engine?
The work is done along the path B → A when the block is cooled. In this process, work is done
on the Carnot engine, which is run in reverse, as a refrigerator. In each Carnot cycle, an amount
of heat dQ2 is extracted from the copper block and a larger amount of heat dQ1 = dQ2 + dW is
transferred to the reservoir at Tc, where dW is the work done on the refrigerator.
4
dQ1 dQ2
Also, from the thermodynamic definition of temperature =
TC T Reservoir at temperature Tc
where T is the instantaneous temperature of the copper block.
dQ1
The infinitesimal work done in each cycle is given by dW
⎛T ⎞ ⎛T ⎞
dW = dQ1 − dQ2 = dQ2 ⎜ c − 1⎟ = −CdT ⎜ c − 1⎟ dQ2
⎝T ⎠ ⎝T ⎠
where we have written dQ2 = −CdT is the heat rejected by the Copper block
at temperature T
system in each cycle.
TA
⎛ TC ⎞ ⎛T ⎞
Net work done is W = −C ∫ ⎜ − 1⎟dT = CTC ln⎜⎜ B ⎟⎟ − C (TB − T A )
TB ⎝ T ⎠ ⎝ TA ⎠
(d) For the cyclic path B → A → B , does the system absorb heat from the reservoirs or reject
heat?
In the cyclic path, there is no change in the state functions. Therefore, ∆U = 0
From the first law, ∆U = Q + W

From the result of part (c), you can show that W > 0 , using the same inequality as in part (b) and
since TC > TB
Therefore W > 0 ⇒ Q < 0 ⇒ net heat is rejected by the system to the reservoirs.
Problem 3
Two magnetic spin systems with N1 = 500 spins and N2 = 1000 spins are placed in an external
magnetic field H. They are initially thermally isolated from one another and are prepared with
spin excess values of m1 = +20 and m2 = +250, where m = N ↑ − N ↓ . The magnetic moment of
each spin is denoted by µ m . Assume that one can use the Gaussian approximation for the
number of states of each system with energies between E and E + dE:
2N ⎛ E2 ⎞
Ω(N, E ) = exp⎜⎜ − ⎟dE
⎟
⎝ 2Nµ m H
2 2
2πNµ m2 H 2 ⎠
(a) What is the total energy in each system (in units of µ m H ) and the initial temperature of each
system (in units of µ m H / k )?
1 ⎛ ∂S ⎞ ⎛ ∂ ln Ω ⎞ E Nµ m2 H 2
=⎜ ⎟ = k⎜ ⎟ = −k ⇒ T = −
T ⎝ ∂E ⎠ N ,H ⎝ ∂E ⎠ Nµ m2 H 2 Ek
5
N µmH
But E = −mµ m H ⇒ T =
m k
µmH µmH
Therefore, T1initial = 25 and T2initial = 4
k k
E1initial = −20 µ m H and E 2initial = −250 µ m H
(b) The two systems are brought into thermal contact. What are the spin excess values m1eq and
m2eq and the final temperature, after they reach equilibrium?
N1 N2 m2eq N 2
At equilibrium T1eq = T2eq ⇒ = ⇒ = =2
m1eq m2eq m1eq N1
Also, the total energy of the combined system is a constant
E1initial + E 2initial = E1eq + E 2eq ⇒ m1initial + m2initial = m1eq + m2eq = 270

⇒ m1eq + 2m1eq = 270 ⇒ m1eq = 90 and m2eq = 180
µmH
T1eq = T2eq = 5.55
k
(c) What is ∆S , the total change in the entropy of the combined system (in units of k), and what
is the probability of finding the system in the initial configuration relative to the probability of
finding the system in the equilibrium configuration?
⎛ ⎡ ⎞
1 ⎛⎜ E1eq E 2eq ⎞⎟⎤
2 2
⎜ ⎟
⎜ exp ⎢ − 2 ⎜
+ ⎥ ⎟
⎛ Ω1final Ω final ⎞ ⎢ 2 µ m H ⎝ N1
2
N 2 ⎠⎦⎥ ⎟
⎛Ω ⎞ ⎜ ⎣ ⎟
∆S = Sf − Si = k ln⎜⎜ f ⎟⎟ = k ln⎜⎜ initial 2
⎟ = k ln
⎟
⎝ Ωi Ω Ω ⎜ ⎡ ⎛ ( ) ( )
initial 2 ⎞ ⎤ ⎟
initial
⎠ ⎝ 1 ⎠ ⎜ exp⎢− 1 ⎜ E1 initial 2
E
2
+ 2 ⎟⎥ ⎟
⎜ 2 ⎜
⎢⎣ 2µ m H ⎝ N1
2
N 2 ⎟⎥ ⎟
⎝ ⎠⎦ ⎠
Simplify to get
∆S =
k ⎡ E eq 2 E eq 2
⎢ − 1
− 2
+
(
E1initial ) + (E
2
2 )
initial 2
⎥= ⎢ 1
( )
⎤ k ⎡ m initial 2 − m eq 2
1 ( )
2
m initial − m2eq
+ 2
2
⎤
⎥
2µ m2 H 2 ⎢⎣ N1 N2 N1 N2 ⎥⎦ 2 ⎢⎣ N1 N2 ⎥⎦
∆S = 7.35k which is greater than zero, as expected for an irreversible process.
p initial Ω initial
= = exp( − ∆S / k ) = exp( −7.35 ) ≈ 6.4 × 10 −4
p eq Ω eq
6
Problem 4
Consider a 2-dimensional polymer, consisting of N links,

each of length l. Each link has four allowed orientations,
pointing in the +x or −x direction, or in the +y or −y
direction.
(a) Show that the average end-to-end

r r distance squared for
this polymer is given by < L • L >=< L2 >= Nl 2
< L2 >=< L2x > + < L2y >= N < l x2 > +N < l y2 >
Since each link has four allowed orientations, with equal energy, which we can set equal to zero,
the statistical weight of each orientation is 1 and the partition function for a single link is Z1 = 4
l 2 (1) + l 2 (1) + 0 + 0
We can write < l x2 >= = l2 /2
4
l 2
(1) + l 2
(1) + 0+0
Similarly, < l y2 >= = l2 /2
4
Therefore, < L2 >= Nl 2 / 2 + Nl 2 / 2 = Nl 2
(b) Now consider a force f that extends the length of the polymer in the x-direction. In this
simplified two-dimensional picture, the three allowed energy levels of each link are (i) ε = −fl , if
the link is oriented in the direction of the applied force, ε = +fl , if the link is oriented opposite to
the direction of the applied force, and (iii) ε = 0 , if the link is oriented perpendicular to the
direction of the applied force. Find the average extension < Lx > in the x-direction as a function
of the applied force, and show that, at low forces (fl << kBT), the polymer behaves like a
Hookean spring.
The average projection of each link in the x-direction is given by
− l exp( −ε − / kT ) + l exp( −ε + / kT ) + 0(exp( −ε 0 / kT ) + 0(exp( −ε 0 / kT )

< l x >=
exp( −ε − / kT ) + exp( −ε + / kT ) + (exp( −ε 0 / kT ) + (exp( −ε 0 / kT )
− l exp( −fl / kT ) + l exp(fl / kT ) sinh(fl / kT )

< l x >= =l
exp( −fl / kT ) + exp(fl / kT ) + 2 cosh(fl / kT ) + 1
Average extension in the x-direction is
sinh(fl / kT )
< Lx >= N < l x >= Nl
cosh(fl / kT ) + 1
7
In the limit fl << kT , sinh( x ) ≈ x and cosh( x ) ≈ 1, where x = fl / kT
(fl / kT ) Nfl 2
< Lx >= Nl =
2 2kT
2kT
Rewrite as f = < Lx >
Nl 2
2kT
The polymer behaves like an entropic spring, with a spring constant k sp =
Nl 2
(c) The polymer is stretched quasistatically and isothermally such that the average extension in
the x-direction is 5% of its unperturbed size L = N l ? Assume that the polymer still behaves as
a Hookean spring. What is the change in the Helmholtz free energy of the polymer?
For a quasistatic process at constant temperature, ∆F = W

The work done to stretch the polymer is W =
1
2
(
1 ⎛ 2kT ⎞
k sp ( ∆L )2 = ⎜ 2 ⎟ 0.05 N l
2 ⎝ Nl ⎠
)
2
= 0.05kT
Therefore ∆F = 0.05kT
(d) Now consider that the polymer is stretched quasistatically and adiabatically. Will the
temperature of the polymer increase or decrease in this process?
When the polymer is stretched, the configuration of the polymer goes from more random to less
random, which tends to decrease the configurational entropy. However, in a reversible, adiabatic
process, the total entropy of the system does not change (isoentropic process). Therefore, the
temperature of the polymer must increase to compensate for the loss of configurational entropy.
More formally, consider S(T , L )
⎛ ∂S ⎞ ⎛ ∂S ⎞
dS = ⎜ ⎟ dT + ⎜ ⎟ dL = 0
⎝ ∂T ⎠L ⎝ ∂L ⎠T
⎛ ∂T ⎞ ⎛ ∂S ⎞ ⎛ ∂T ⎞
or ⎜ ⎟ = −⎜ ⎟ ⎜ ⎟
⎝ ∂L ⎠ S ⎝ ∂L ⎠T ⎝ ∂S ⎠ L
The first term on the right is negative, since the entropy of the chain decreases as the length
increases at constant temperature. The second term on the right is related to the inverse of the
heat capacity, and is positive.
⎛ ∂T ⎞
Therefore ⎜ ⎟ > 0.
⎝ ∂L ⎠ s
8
Problem 5
Consider a system consisting of impurity atoms in a semiconductor. Suppose that the impurity
atom has one "extra" electron (with two degenerate spin states), compared to the neighboring
atoms (e.g. a phosphorus atom occupying a lattice site in a silicon crystal). The extra electron is
easily removed, contributing to conduction electrons, and leaving behind a positively charged
ion.
(a) What is the probability that a single donor atom is ionized? Express your result in terms of
the ionization energy I, and the chemical potential of the "gas" of ionized electrons.
Taking the system to be a single donor atom, there are three possible states: one ionized state
with no electron, and two un-ionized states (with one electron present, either spin-up or spin-
down).
The energies and Gibbs factors of these states are:
ionized: ε = 0, n = 0 , Gibbs factor = exp(0) = 1

unionized: ε = −I, n = 1, Gibbs factor = exp(− ( −I − µ ) / kT ) = exp((I + µ ) / kT )
The grand partition function can be written as
Ξ = 1 + 2 exp((I + µ ) / kT )
and the probability that the donor atom is ionized is
1
P (ionized ) =
1 + 2 exp((I + µ ) / kT )
(b) If every conduction electron comes from an ionized donor, and the conduction electrons
behave like an ideal gas, write down an expression that relates the number of conduction
electrons N c and the number of donor atoms N d , in terms of the volume V of the sample, the
temperature T and fundamental constants.
Nc
P (ionized ) =
Nd
Also, for an ideal gas of conduction electrons with two spin states per particle, the chemical
⎛ VZ int ⎞ ⎛N v ⎞
potential is µ = −kT ln⎜⎜ ⎟ = kT ln⎜ c q
⎟ ⎜ 2V
⎟
⎟
⎝ Ncv q ⎠ ⎝ ⎠
3/2
⎛ h2 ⎞
where Z int = 2 for the two spin orientations, and v q = ⎜⎜ ⎟⎟ is the quantum volume.
⎝ 2πmkT ⎠
Therefore, we can write
Nc 1 1
= =
Nd ⎛ Ncv q ⎞ 1 + (N c / V )v q exp(I / kT )
1 + 2⎜⎜ ⎟ exp(I / kT )
⎟
⎝ 2V ⎠
9
To solve for N c rewrite the equation in terms of dimensionless quantities:
where x = N c / N d and y = (N d / V )v q exp(I / kT )

1
x=
1 + xy
− 1 ± 1 + 4y
or x 2 y + x − 1 = 0 ⇒ x =
2y
Since x is a positive quantity, retain the solution with the plus sign to get:
Nc =
V
2v q exp(I / kT )
( 1 + 4(N d / V )v q exp(I / kT ) − 1 )
(c) Show that in the limit of low (kT << I ) and high (kT >> I ) temperatures, the ratio N c / N d has
the expected values.
Nc 1
You can start with =
N d 1 + (N c / V )v q exp(I / kT )
to show that, in the limit T → 0 , exp(I / kT ) >> 1 and v q >> V / N c
⇒ N c2 = d exp(− I / kT ) ⇒ N c = exp(− I / 2kT )

Nc V NV N dV
⇒ =
N d N c v q exp(I / kT ) vq vq
In the limit T → 0 , N c / N d → 0
In the high T limit, exp(I / kT ) → 1 and v q → 0 ⇒ N c / N d → 1 as expected, with all the donor
atoms ionized.
(d) Write down an expression for the Gibbs free energy of the conduction electrons in terms of
Nc, T and V and show that it is an extensive quantity.
⎛ Ncv q ⎞
Gibbs free energy G = µN c = kTN c ln⎜⎜ ⎟⎟
⎝ 2V ⎠
If the size of the system doubles, V → 2V and N c → 2N c while N c / V , v q and T remain

unchanged.
The Gibbs free energy doubles, as expected for an extensive quantity.
10
2 2 cosh x
Inequality
ln( y ) ≤ y − 1
Maxwell's relations
⎛ ∂T ⎞ ⎛ ∂P ⎞ ⎛ ∂T ⎞ ⎛ ∂P ⎞ ⎛ ∂T ⎞ ⎛ ∂V ⎞ ⎛ ∂T ⎞ ⎛ ∂V ⎞
⎜ ⎟ = −⎜ ⎟ ; ⎜ ⎟ = −⎜ ⎟ ; ⎜ ⎟ =⎜ ⎟ ; ⎜ ⎟ =⎜ ⎟
⎝ ∂V ⎠ S ⎝ ∂S ⎠V ⎝ ∂V ⎠ P ⎝ ∂S ⎠T ⎝ ∂P ⎠ S ⎝ ∂S ⎠ P ⎝ ∂P ⎠V ⎝ ∂S ⎠T
Ideal gas
⎛ VZ ⎞
3/2
⎛ h2 ⎞
µ = −k BT ln⎜⎜ int ⎟;
⎟
v q = ⎜⎜ ⎟⎟
⎝ Nv q ⎠ ⎝ 2πmk BT ⎠
11
1

Thermodynamics and Statistical Physics

Qualifying Exam
January 5, 2007
9:00am-12:00pm

4 questions. If the student attempts all 5 questions, all of the answers
will be graded, and the top 4 scores will be counted towards the exam’s
total score.
2
Notation:
1
β=
kB T
Z ∞
2 ¡ ¢
erfc (z) = √ dx exp −x2 erfc is known as the complimentary error function
π z
Integrals:
Z
dx ln x = x ln x − x
Z
dx
= ln x
x
Z ∞ r
¡ ¢ 1 π ¡√ ¢
dx exp −ax2 = erfc ab
b 2 a
Z a ¡ ¢
1 − exp −a2
dxerfc (x) = √ + aerfc(a)
0 π
Expansions:
1
= 1 + x + x2 + x3 + . . . for x < 1
1−x
x2 x3
exp(x) = 1 + x + + + ...
2!· 3! ¸
¡ ¢ 1
erfc(x) = exp −x2 √ + . . . for x → ∞
πx
sinh(x) = x + ... for x → 0
cosh(x) = 1 + ... for x → 0
3
1. Consider a system consisting of N non-interacting particles each with spin S = 1 in an external magnetic field,
H. For H = 0, all of a single particle’s spin projections, Sz , are degenerate with energy E = 0.
a) Plot the energies of ALL spin projections, Sz , of a single particle as a function of H.

b) Calculate the partition function of the system as a function of temperature T and H.
c) Compute the average energy, hEi of the system. What is the form of hEi in the limit T → 0? Briefly explain
what this result implies for the spin states of the spins (1-2 sentences maximum). What is the form of hEi in
the limit βgµB H ¿ 1, where β = 1/kB T , g is the gyromagnetic ratio, and µB is the Bohr magneton?
d) Compute the specific heat of the system in the high temperature limit for constant H.
2. Consider a three-dimensional box with sides of length L, as shown below
L
x
It contains an ideal gas of non-interacting spin-less particles each with kinetic energy
m→−
ε= v2
2
The temperature of the gas is T , and the particles are uniformly distributed throughout the box.
a) What is the normalized velocity distribution of the gas?

b) We now open the front side of the box (the shaded side facing the +x-direction, as shown in the figure) for a
given time ∆t. Using the result from (a), compute the number of particles that escape from the box in time ∆t.
To this end, consider these two steps: (i) Divide the box into slices of width dx and compute first the number
of particles in a given slice at a distance x from the opening that have escape through the opening in time ∆t.
(ii) In order to find the total number of escaped particles, integrate the result you obtained in (i).
c) How does the total number of escaped particles depend on ∆t in the limit ∆t → 0?
4
3. Suppose one mole of an ideal gas is subjected to the cyclic process shown below (with temperature T1, T2 and
T3 in states 1, 2 and 3, respectively)
p
1
p1
p2=p3
3 2
V1=V3 V2 V
1 ⇒ 2 is a free adiabatic expansion, i.e. an expansion against zero applied pressure (like expanding into a
vacuum).
2 ⇒ 3 is a constant pressure compression step
3 ⇒ 1 is a constant volume heating step
Step 1 ⇒ 2 is irreversible, but steps 2 ⇒ 3 and 3 ⇒ 1 are reversible
a) What is the change in internal energy, ∆U , for the entire cyclic process 1 ⇒ 2 ⇒ 3 ⇒ 1.
b) Use the First Law of Thermodynamics to calculate ∆U for the process 1 ⇒ 2.
c) Use the First Law of Thermodynamics to calculate ∆U for the process 2 ⇒ 3.
d) Use the First Law of Thermodynamics to calculate ∆U for the process, 3 ⇒ 1.
e) Using your answers to parts (a) – (d), show that the following result is obtained for 1 mole of an ideal gas:
Cp − CV = R
where CV . is the specific heat for constant volume, Cp. is the specific heat for constant pressure, and R is the
ideal gas constant.
4. Consider a system consisting of M non-interacting molecules at temperature T . Each of these molecules exhibits
vibrations with energies
µ ¶
1
En = ~ω0 n + where n = 0, 1, 2, 3, ....
2
a) Show that the single particle partition function is given by

1
Z1 = h i
β~ω0
2 sinh 2
and compute the partition function, ZM , for the M molecule system.

b) Compute the free energy F .and the average energy hEi the entire system.What is the form of hEi at high
temperatures?
c) Compute the entropy S and the specific heat CV of the system.
5
5. Consider a monoatomic ideal gas.

a) What is the internal energy and the equation of state of an ideal gas?
b) Compute the entropy of an ideal gas as a function of T and V for constant particle number N starting from
dU = T dS − pdV
c) Compute the chemical potential of the ideal gas as a function of p and T starting from the Gibbs-Duhem
relation
SdT − V dp + N dµ = 0

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¶
1. Ideal gas of two-state atoms
Consider an ideal monoatomic gas made of N atoms each of which has only 2 internal
states: a ground state and an excited state with energy gap equal to ∆. The gas is in a
sealed container with no energy exchange with outside world. Initially, the gas is prepaired
in such a way that all the atoms are in their ground state internally, but the gas is in thermal
equilibrium with respect to kinetic motion of the atoms, characterized by temperature T1 .
After some time, however, due to collisions, the internal degree of freedom of the atoms is
also excited and thermalized.
a) Find the temperature of the gas, T2 , after the internal degree of freedom thermalizes.
Assume ∆ kT1 and calculate the difference T2 − T1 up to order ∆. Does the temperature
increase or decrease?
b) Find the change in the entropy of the gas, S2 − S1 , after the complete thermalization.
Assume ∆ kT1 and work up to order ∆. Does the entropy increase or decrease?
Hint: The entropy of the same gas without the internal degree of freedom is:
3
Skin = kN ln T + (T independent terms).
2
2. van der Waals equation of state
Consider van der Waals equation of state:

nkT
p= − an2 ;
1 − bn
where p is the pressure, n is the density and T is the temperature. a and b are positive
coefficients.
a) There are 2 different types of isotherms p(n), depending on the value of T . For some
values of T the pressure is a monotonous function of the density, for other values it is not.
Sketch these two types of the isotherms on the p vs n plot.
b) Now, on the same plot, sketch the curve where the derivative (∂p/∂n)T vanishes.
c) Shade the region of p and n where the system is thermodynamically unstable towards
phase separation (not even metastable). Write the stability condition (inequality) you use.
d) Express the coordinates pc , nc and Tc of the critical point in terms of a and b.
1
3. Maxwell relations
Consider a rubber band of length L which is being stretched by external force f .
a) Write down the thermodynamic identity (1st law of thermodynamics) relating change in
the internal energy dU to infinitesimal change in length dL, and to supplied heat T dS.
b) In one experiment the length of the band is fixed to L = 1m and the temperature of
the band T = 300K is raised by a small amount ∆T = 3K. This causes the force needed
to maintain the length of the band to increase by the amount ∆f = 1.2N. In another
experiment, the band is stretched from L to L + ∆L at constant temperature T . As a
result the band exchanges heat with the environment. What is the amount of this heat for
∆L = 2cm? Is the heat released or absorbed by the band?
4. Cosmic microwave background
Cosmic microwave background (CMB, or relic) radiation is an isotropic radiation with a

black body spectrum at temperature T = 2.7 K.
a) Find the density n of the CMB photons. How many relic photons are there on average
inside a volume of space V = 1 cm3 ?
b) Find the rate at which a ball of radius R = 1 cm is struck by relic photons.
You may find useful the following combination of constants:

k
= 436.7 K−1 m−1
h̄c
as well as this integral
Z ∞ x2
dx = 2ζ(3) = 2.404 . . .
0 ex − 1
2
5. Ultrarelativistic Fermi gas at T = 0
Matter inside a star can be compressed to such an extent that the Fermi energy of the
electrons becomes much larger than their rest energy. Consider electron gas at T = 0 and
given chemical potential µ, such that µ me c2 . In this regime Coulomb interaction is
negligible.
a) What is the maximum momentum of an electron in such a gas?
b) What is the density of the electrons at this value of µ.
c) What is the total energy E of such a gas in a volume V containing N electrons. The
answer should not contain µ.
d) What is the pressure P of the gas in terms of µ?
e) A nucleus of the substance, called A, can capture an electron and undergo a transforma-
tion:
A + e− → B + ν (1)
The mass of nucleus B is larger than the mass of A, therefore the reaction is energetically
forbidden under normal conditions. However, at sufficiently high pressure P > Pmin the
reaction is allowed. Explain why and calculate Pmin , given the masses mA and mB . Neglect
the masses of electron and neutrino (mB − mA me ).
3
1. Ideal gas of two-state atoms
Set temperature units to make k = 1.
a) The fraction of atoms in excited state after thermalization:

1
p= (1)
e∆/T2 +1
Energy conservation:
3 3 ∆
T1 = T2 + ∆/T2 (2)
2 2 e +1
Expanding to order ∆:
∆
T2 − T1 = − <0 (3)
3
Temperature decreases.
b) The entropy of the two-state system:
∆2

∂ ∆/T
N T ln 1 + e−∆/T = N ln 1 + e−∆/T + N ∆/T

S∆ = = N ln 2 + O (4)
∂T e +1 T2
Alternatively, using p from (1):
∆2

S∆ /N = −p ln p − (1 − p) ln(1 − p) = ln 2 + O , (5)
T2
The total entropy of the gas
∆2

3
S = Skin + S∆ = N ln T + N ln 2 + O (6)
2 T2
The change in the entropy to order ∆/T :

3 3 3 ∆ ∆
S2 −S1 = N ln T2 + ln 2 − ln T1 =N ln 1 − + ln 2 = N ln 2 − >0
2 2 2 3T1 2T1
(7)
Entropy increases.
1
2. van der Waals
a)
2
1.75
1.5
1.25
0.75
0.5
0.25
0.5 1 1.5 2
b) Set k = 1. From
∂p T
= − 2an = 0 (8)
∂n T (1 − bn)2
we find
T = 2an(1 − bn)2 (9)
and substituting (9) into the equation of state p(n, T ) we find
p = an2 − 2abn3 . (10)
See Figure.
c) Stability condition: (∂p/∂n)T > 0.
d) The critical point is at the maximum of the curve given by (8). Two conditions defining
the critical point: (∂p/∂n)T = 0 and (∂ 2 p/∂n2 )T = 0. The second condition gives:
2
∂ p 2bT
2
= − 2a = 0. (11)
∂n T (1 − bn)3
Substituting into (11) the value of T from (9) and solving for n we find:
1
nc = . (12)
3b
Substituting this into (10) and into (9) we find, respectively:
a 8a
pc = ; Tc = (13)
27b2 27b
2
3. Maxwell relations
a) dU = f dL + T dS
b) Using the relation for the free energy: d(U − T S) = f dL − SdT , we find the Maxwell
relation:
∂S ∂f
=− (14)
∂L T ∂T L
Thus

∂f ∆f 1.2 N
∆Q = T ∆S = −T ∆L = −T ∆L = −300 K 2 · 10−2 m = −2.4 J. (15)
∂T L ∆T 3K
Heat is released since ∆Q < 0.
4. Cosmic microwave background
a)
3
d3 pV

1 2ζ(3) kT
Z
n=2 = = 4.0 · 108 m−3 ; N = nV = 400. (16)
(2π~)3 e|p|/kT − 1 π2 ~c
b) The density of photons with velocities v inside an element of solid angle dΩv is
dΩv
dn = n . (17)
4π
The flux of those photons is dJ = dnv. The rate at which the photons with the given
velocity are hitting a small surface element of area σ is
σ · dJ = σ · vdn (18)
The total rate from the photons with all values of v is (integrate over semisphere, use |v| = c):
dΩv 2π 1 1
Z Z
σ · vn = σcn cos θ d(cos θ) = σcn (19)
4π 4π 0 4
Therefore rate per unit area is cn/4. The total rate for the area of the ball is
3
cn 2 2ζ(3) kT
Γ = 4πR ×2
= πR c 2 = 3.76 × 1013 s−1 (20)
4 π ~c
Another derivation: For the photons with velocity v the ball presents a cross section equal
to πR2 . The rate at which these photons strike the ball is πR2 cdn. Integrating over all v
we find Γ = πR2 cn, as before.
3
5. Ultrarelativistic Fermi gas at T = 0
a) F = µ and pF = F /c = µ/c.
b)
d3 p 1 µ 3
Z
n=2 = . (21)
|p|<pF (2π~)3 3π 2 ~c
c)
d3 p 1 µ4
Z
~c ~c
E = V ·2 3
|p|c = V · 2 3
= V · 2 (3π 2 n)4/3 = 2 (3π 2 N )4/3 V −1/3 (22)
|p|<pF (2π~) 4π (~c) 4π 4π
where we used Eq. (21) and n = N/V .
d)
4/3
µ4

∂E ~c 2N 1
P =− = 3π = (23)
∂V N 12π 2 V 12π 2 (~c)3
e) The condition for the reaction to be energetically allowed is
m A c 2 + µ > mB c 2 (24)
Thus µmin = (mB − mA )c2 . Therefore, from (23):
1 ((mB − mA )c2 )4
Pmin = (25)
12π 2 (~c)3
4
1


Qualifying Exam
January 6, 2012
9:00am-12:00pm

4 questions. If the student attempts all 5 questions, all of the answers
will be graded, and the top 4 scores will be counted towards the exam’s
total score.
2
Notation:
1
β=
kB T
∫ z
2 ( )
erf (z) = √ dx exp −x2 erf is known as the error function
π 0
∫ ∞
2 ( )
erfc (z) = √ dx exp −x2 erfc is known as the complimentary error function
π z
Integrals:
∫
dx ln x = x ln x − x
∫
dx
= ln x
x
∫ ∞ √
( ) 1 π (√ )
dx exp −ax = 2
erfc ab
2 a
∫b a
1 − exp (−a2 )
dx erfc (x) = √ + a erfc(a)
π
∫ n
0
√
π (√ ) √
1/2
dx x exp(−x) = erf n − n exp (−n)
2
∫ n
0
3 √ 1√
dx x3/2 exp(−x) = erf( n) − n exp(−n)(3 + 2n)
0 4 2
Expansions:
1
= 1 + x + x2 + x3 + . . . for x < 1
1−x
x2 x3
exp(x) = 1 + x + + + ...
2! [ 3! ]
( 2) 1
erfc(x) = exp −x √ + ... for x → ∞
πx
sinh(x) = x + ... for x → 0
cosh(x) = 1 + ... for x → 0
3
1. Consider a system consisting of N non-interacting particles each with isospin I = 3/2. The
energies of the states with different Iz are given by
E(Iz = −3/2) = E1 ; E(Iz = −1/2) = E2

E(Iz = 1/2) = E3 ; E(Iz = 3/2) = E3
with E1 < E2 < E3 and ∆12 = E2 − E1 ≪ ∆23 = E3 − E2 .
a) Without using the partition function, give the value of the total energy, ⟨E⟩, at temperatures
T = 0, ∆12 ≪ T ≪ ∆23 , and ∆23 ≪ T . Provide a justification for your results. Sketch ⟨E⟩ as
a function of temperature.
b) What is the occupation of the Iz -states for temperature T → ∞ Without using the partition
function, give a value of the specific heat for temperature T → ∞. Provide a justification for
your results.
c) Without using the partition function, give the value of the average isospin per particle, ⟨Iz ⟩,
at temperatures T = 0, ∆12 ≪ T ≪ ∆23 , and ∆23 ≪ T . Provide a justification for your results.
Sketch ⟨Iz ⟩ as a function of temperature.
d) Using the partition function, compute the average isospin per particle, ⟨Iz ⟩, in the limit
T → ∞. How does you result related to those in part c)?
2. Consider an ideal gas of N0 non-interacting spin-less particles each with kinetic energy
m−→
ε= v2
2
that is contained in a box. The temperature of the gas is T0 , and the particles are uniformly
distributed throughout the box.
Compute the total energy E0 = ⟨E⟩ of the N0 particles in the box. Next, one instantaneously
removes all particles from the gas that possess a kinetic energy larger than nkB T (n is an
arbitrary real, positive number). How many particles remain in terms of N0 ? What is the new
total energy, Enew in terms of E0 ? After the remaining particles have returned to equilibrium,
what is the new temperature, Tnew of the gas in terms of T0 ?
4
3. Suppose one mole of an ideal gas is subjected to the cyclic process shown below (with temper-
ature V1 , V2 and V3 in states 1, 2 and 3, respectively)
p
1
p1
p2=p3
3 2
V1=V3 V2 V
1 ⇒ 2 is an isothermal expansion.
2 ⇒ 3 is an adiabatic expansion.
3 ⇒ 1 is an isochoric heating step.
All steps are reversible
a) What is the change in internal energy, ∆U , for the entire cyclic process 1 ⇒ 2 ⇒ 3 ⇒ 1.
b) Use the First Law of Thermodynamics to calculate ∆U , δQ, and δW for the process 1 ⇒ 2.
c) Use the First Law of Thermodynamics to calculate ∆U , δQ, and δW for the process 2 ⇒ 3.
d) Use the First Law of Thermodynamics to calculate ∆U , δQ, and δW for the process 3 ⇒ 1.
e) Is the total work done in a cycle positive or negative? What is the efficiency, η, of this cycle?
In which limit does one obtain η = 1.
4. Consider a system consisting of M non-interacting molecules at temperature T . Each of these

molecules possesses vibrations with energies
( )
1
E n = ~ ω0 n + where n = 0, 1, 2, 3, ..., N0
2
Let us first consider the case N0 = ∞

a) Using the partition function, compute the total energy, ⟨E⟩, of the system for temperature
T → 0 and T → ∞. Explain your results. At what temperature occurs the crossover from the
T → 0 to the T → ∞ behavior of ⟨E⟩?
b) Compute ⟨n⟩ for T → ∞. What is the physical interpretation of ⟨n⟩? What is the relation
of ⟨n⟩ to the partition function and to ⟨E⟩?
c) Consider next the case where N0 is a finite, integer number (i.e., N0 < ∞). What is now the
form of ⟨E⟩ for temperature T → ∞
d) Compute the specific heat, CV , of the system in the limit T → ∞ for the two cases N0 = ∞
and N0 < ∞. Explain the difference in CV between these two cases.
5
5. Consider a monoatomic ideal gas.

a) Compute the entropy of an ideal gas as a function of T and V for constant particle number
N starting from
dU = T dS − pdV
b) Compute the chemical potential of the ideal gas as a function of p and T starting from the
Gibbs-Duhem relation
SdT − V dp + N dµ = 0

Qualifying Exam
January ?, 2012
9:00am-12:00pm

Notation:
 1
kBT
z
erfz  2

 0 dx exp−x 2  erf is known as the error function

erfcz  2

 z dx exp−x 2  erfc is known as the complimentary error function
Integrals:
 dx ln x  x ln x − x
 dx  ln x
x

 b dx exp−ax 2   1  erfc a b
2 a
 0 dx erfcx 1 − exp−a 
a 2
 a erfca

n 
 0 dx x 1/2 exp−x  2
erf n − n exp−n
n
 0 x 3/2 exp−x  3 erf n  − 1 n exp−n3  2n
4 2
Expansions:
1  1  x  x 2  x 3 … for x  1
1−x
2 3
expx  1  x  x  x …
2! 3!
erfcx  exp−x 2  1  … for x → 
x
sinhx  x  . . . for x → 0
coshx  1  . . . for x → 0
Problem 1
Consider a system consisting of N non-interacting particles each with isospin I  3/2. The
energies of the states with different I z are given by
EI z  −3/2  E 1 ; EI z  −1/2  E 2
EI z  1/2  E 3 ; EI z  3/2  E 3
with E 1  E 2  E 3 and Δ 12  E 2 − E 1  Δ 23  E 3 − E 2 .
a) Without using the partition function, give the value of the total energy, 〈E, at
temperatures T  0, Δ 12  T  Δ 23 , and Δ 23  T. Provide a justification for your results.
Sketch 〈E as a function of temperature.
b) What is the occupation of the I z -states for temperature T →  Without using the
partition function, give a value of the specific heat for temperature T → ? Provide a
justification for your results.
c) Without using the partition function, give the value of the average isospin per particle,
〈I z , at temperatures T  0, Δ 12  T  Δ 23 , and Δ 23  T. Provide a justification for your
results. Sketch 〈I z  as a function of temperature.
d) Using the partition function, compute the average isospin per particle, 〈I z , in the limit
T → . How does you result relate to those in part c)?
Solutions:
a) At T  0, all particles are in the ground state, and hence 〈E  NE 1 . For
Δ 12  T  Δ 23 , the two lowest states are equally populated, while the two higher energy
states are empty, and thus 〈E  NE 1  E 2 /2. For Δ 23  T, all states are equally populated,
and hence 〈E  NE 1  E 2  2E 3 /4.
b) For T → , all I z -states are equally populated with N/4 particles being in each of these
four states. Hence, by increasing temperature, no more energy can be stored in the system, and
hence C  0 as T → .
c) At T  0, all particles are in the ground state, and hence 〈I z   −3N/2. For
Δ 12  T  Δ 23 , the two lowest states are equally populated, while the two higher energy
states are empty, and thus 〈I z   N/2−3/2 − 1/2  −N. For Δ 23  T, all states are equally
populated, and hence 〈I z   0.
d) The partition function is given by

Z  e −E 1  e −E 2  2e −E 3
and hence
〈I z   N − 3 e −E 1  − 1 e −E 2  1 e −E 3  3 e −E 3
Z 2 2 2 2
−E 2 −E 1 
−2  −2 e
3 1
 2 e −E 3 −E 1  
1 3
e −E 3 −E 1 
N 2
1  e −E 2 −E 1   2e −E 3 −E 1 
0
since   0.This agrees with the result in c).
Problem 2
Consider an ideal gas of N 0 non-interacting spin-less particles each with kinetic energy
 mv
2
2
that is contained in a box. The temperature of the gas is T, and the particles are uniformly
distributed throughout the box.
a) What is the normalized velocity distribution of the gas?
b) What is the total energy, E 0  〈E of all particles in the box?
c) Instantaneously remove all particles from the gas that possess a kinetic energy larger
than nk B T (n is an arbitrary real, positive number). How many particles remain in terms of N 0 ?
What is the new total energy, E new in terms of E 0 ? After the remaining particles have returned
to equilibrium, what is the new temperature, T new of the gas in terms of T? For which n does
one obtain T new  T/2?
Solutions:
a)
3/2 2
Pv  m exp − m v
2k B T 2k B T
b) Due to the equipartition theorem
〈   m v  3 k B T
2
2 2
and hence
E 0  〈E  3 Nk B T
2
c)We instantaneously removed all particles with a kinetic energy

 kin  1 mv 2 ≥ nk B T
2
The number of remaining particle, N new , is given by

3/2 ′

2
N new  N 0 m d 3 v exp − mv
2k B T 2k B T
3/2 vc
4  dv v 2 exp − mv
2
 N0 m
2k B T 0 2k B T
where
vc  2nk B T
m
I next perform the variable transformation
2 2k B Tx dv 
x  mv  dx  mv dv  m m
2mx dv
2k B T kBT kBT kBT
and thus
3/2 n
N new  N 0 m
2k B T
4 k B T
2m
 0 dx 2kmB T x 1/2 exp−x
n
 N0 2  dx x 1/2 exp−x
 0

 N0 2 erf n − n exp−n
 2
Next, we compute the remaining energy that is contained in the system after the particles
are removed.
3/2 ′

2
E new  N 0 m d 3 v 1 mv 2 exp − mv
2k B T 2 2k B T
3/2
 N0 m m 4  v c dv v 4 exp − mv 2
2k B T 2 0 2k B T
I next perform the variable transformation
2 2k B Tx dv 
x  mv  dx  mv dv  m m
2mx dv
2k B T kBT kBT kBT
and thus
3/2 2
E new  N 0 m m 4 k B T  n dx 2k B T x 3/2 exp−x
2k B T 2 2m 0 m
n
 N 0 2k B T  dx x 3/2 exp−x
 0
 N 0 k B T 3  erf n − 6 n exp−n − 4n 3/2 exp−n

2 
After equilibration, the new temperature is given by

E new  3 N new k B T new
2
or
kBT
N0 3  erf n  − 6 n exp−n − 4n 3/2 exp−n
T new n  2 E new  2
2 
3 k B N new 3 
kBN0 2
 2
erf n  − n exp−n
3  erf n  − 6 n exp−n − 4n 3/2 exp−n
 1 
T
6 erf n  − n exp−n
2
and thus
T new  T/2
requires
1 3  erf n  − 6 n exp−n − 4n exp−n  1
3/2
6  2
2
erf n  − n exp−n
and thus
n  1. 527
Problem 3
Suppose one mole of an ideal gas is subjected to the cyclic process shown below (with
temperature V 1 , V 2 and V 3 in states 1, 2 and 3, respectively)
p
1
p1 •
p2 = p3 • • 2
3
V1 = V 3 V2 V
1  2 is an isothermal expansion.
2  3 is an isobaric expansion.
3  1 is an isochoric heating step.
All steps are reversible
a) What is the change in internal energy, ΔU, for the entire cyclic process 1  2  3  1.
b) Use the First Law of Thermodynamics to calculate ΔU, Q, and W for the process
1  2.
c) Use the First Law of Thermodynamics to calculate ΔU, Q, and W for the process
2  3.
d) Use the First Law of Thermodynamics to calculate ΔU, Q, and W for the process
3  1.
e) Is the total work done in a cycle positive or negative? What is the efficiency, , of this
cycle? In which limit does one obtain   1.
Solutions:
a) Since we have carried out a cyclic process, namely 1  2  3  1 we must have

ΔU  0, or
Q 12  W 12  Q 23  W 23  Q 31  W 31  0
b) Since this is an isothermal expansion, we have

ΔU 12  Q 12  W 12  0
and thus
Q 12  −W 12  Nk B T 1 ln V 2
V1
c)
T3
Q 23  T C p dT  C p T 3 − T 2   0
2
V3
W 23  −  p 2 dV  p 2 V 2 − V 3   Nk B T 2 V 2 − V 1   Nk B T 1 1 − V 1
V2 V2 V2
ΔU 23  Q 23  W 23
d) We have W 31  0 since there is no change in volume, but

T1
Q 31  T C V dT  C V T 1 − T 3 
3
ΔU 31  Q 31
Let us check that

ΔU tot  0
I have
ΔU 12  0
ΔU 23  C p T 3 − T 2   Nk B T 1 1 − V 1
V2
ΔU 31  C V T 1 − T 3 
and thus
ΔU tot  C p T 3 − T 2   Nk B T 1 1 − V 1  C V T 1 − T 3 
V2
 5 Nk B T 3 − T 1   Nk B T 1 1 − V 1  3 Nk B T 1 − T 3 
2 V2 2
 Nk B T 3 − Nk B T 1 V 1  Nk B T 3 − Nk B T 2 V 3
V2 V2
 Nk B T 3 − T 2 V 3  p 3 − p 2 V 3  0
V3 V2
e) The total work done is
W tot  W 12  W 23  W 31  −Nk B T 1 ln V 2  Nk B T 1 1 − V 1
V1 V2
 Nk B T 1 1 − V 1 − ln V 2 0
V2 V1
Thus the system is performing work.
The efficiency is defined via
|W tot | Nk B T 1 ln VV 21 − 1 − V1
V2
1− V1
V2
   1−
Q in Nk B T 1 ln VV 21 ln V2
V1
V1
One obtains   1 for V2
 0.
Problem 4
Consider a system consisting of M non-interacting molecules at temperature T. Each of
these molecules possesses vibrations with energies
En   0 n  1 where n  0, 1, 2, 3, . . . , N 0
2
Let us first consider the case N 0  
a) Using the partition function, compute the total energy, 〈E, of the system for temperature
T → 0 and T → . Explain your results. At what temperature occurs the crossover from the
T → 0 to the T →  behavior of 〈E?
b) Compute 〈n for T → . What is the physical interpretation of 〈n? What is the relation
of 〈n to the partition function and to 〈E?
c) Consider next the case where N 0 is a finite, integer number (i.e., N 0  ). What is now
the form of 〈E for temperature T → ?
d) Compute the specific heat, C V , of the system in the limit T →  for the two cases
N 0   and N 0  . Explain the difference in C V between these two cases.
Solutions:
a) Since the molecules do not interact with each other, the partition function of a single
molecule is
 
 0
Z1  ∑ exp − 0 n 1
2
 exp −
2 ∑exp− 0  n
n0 n0
 0 1 1
 exp − 
2 1 − exp− 0  exp
 0 
− exp − 2 0
2
 1
 0
2 sinh 2
The partition function of the entire system is then
ZM  ZM −M 1
1  2 M  0
sinh 2
The average energy is given by
 0
cosh
〈E   − ∂ ln Z M  − M ∂ 1  M  0 2
∂ Z 1 ∂ 2 sinh  0 2Z 1 2 sinh 2  0
2 2
 0  0
 M 2 sinh
 0  0 cosh 2
 M  0
cosh 2
2 2 2 sinh 2  0 2 sinh  0
2 2
 M  0 1
2 tanh  0
2
For T → 0, we expand 〈E  in the limit  →  which yields
 1 ≈ M  0
〈E   M 0
2 tanh  0 2
2
Hence each molecule has the energy 2 0 associated with the zero-point fluctuations.
For T → , we expand 〈E  in the limit  → 0 which yields
  
 1  cosh 2 0  exp 2 0  exp − 2 0
〈E   M 0
M 0
M 0
2 tanh  0 2 sinh  0 2 exp  0 − exp −  0
2 2 2 2
≈ M  0 
2  M  Mk B T
2 2 0 
2
This is the classical result expected from the equipartition theorem. The crossover between

these two limits occurs at 2 0 ≈ 1.
b) We have
〈E  M  0 〈n   1  M kBT
2
and thus
k T
〈n   1  B
2  0
kBT
and hence for  0
 1, I obtain
k T
〈n  ≈ B
 0
The excitation of each quantum n of oscillation requires the energy   0 .
c)
In this case, we have
∑ n0
N0
 0 n  12 exp − 0 n  1
∑ n0
N0
 0 n  12 exp− 0 n
〈E   2

∑ n0
N0
exp − 0 n  12 ∑ n0
N0
exp− 0 n
∑ n0
N0
 0 n  1
 0 1 N 0  1  N 0 N 0  1
 2

∑ n0
N0
1 N0  1 2 2
  0 1  N 0 
2
Or Alternatively,
N0 N0
 0
Z1  ∑ exp − 0 n 1
2
 exp −
2
∑exp− 0  n
n0 n0
 
 0
 exp −
2
∑exp− 0  − n
∑ exp− 0  n
n0 nN 0 1

 0
 exp − 1 − ∑exp− 0  nN 0 1
2 1 − exp− 0  n0
 0 1 − exp− 0  N 0 1 1 − exp−N 0  1 0 
 exp − 
2 1 − exp− 0  2 sinh
 0
2
The partition function of the entire system is then
ZM  ZM 1
The average energy is given by
1 − exp−N 0  1 0 
〈E   − ∂ ln Z M  − M ∂
∂ Z 1 ∂ 
2 sinh 2 0
  0  0
M N 0  1 0 exp−N 0  1 0  sinh 2 0 − 1 − exp−N 0  1 0  cosh
− 
2 2
2Z 1 sinh 2 2 0
 0
M N 0  1 0 exp−N 0  1 0  sinh − 1 − exp−N 0  1 0  cosh
− 1−exp−N 0 1 0 
2
2  0
2  0
sinh 2
2 sinh 2
 0
M 0 N 0  1 exp−N 0  1 0  sinh − 1 − exp−N 0  1 0  cos
− 2
1 − exp−N 0  1 0  sinh
 0
2
 0 N  1 exp−N 0  1 0 
 M  0 coth −2 0
2 2 1 − exp−N 0  1 0 
 0 N 0  1
 M 0 1 coth 
2 2 1 − expN 0  1 0 
and for   0 I obtain
1  N 0  1
〈E   M 0
 0 1 − 1  N 0  1 0  12 N 0  1 2  0  2
1  N 0  1
 M 0
 0 1 − 1 − N 0  1 0 − 12 N 0  1 2  0  2
 M 0 1 1− 1
 0 1 1
N 0  1 0
2
 M 0 1 N 0  1
2
d) N 0  
∂〈E 
CV   ∂ Mk B T  Mk B
∂T ∂T
N0  
∂〈E 
CV  0
∂T
For N 0   the number of quanta one can excite is not bounded, and one can therefore
increase the energy of the system with increasing temperature. For N 0   this is not possible,
hence there is an upper bound for the energy stored in the system, and hence C V has to go to
zero for T → 
Problem 5
Consider a monoatomic ideal gas.

b) Compute the entropy of an ideal gas as a function of T and V for constant particle
number N starting from
dU  TdS − pdV
c) Compute the chemical potential of the ideal gas as a function of p and T starting from
the Gibbs-Duhem relation
SdT − Vdp  Nd  0
Solutions:
U  3 Nk B T; pV  Nk B T
2
b) Compute the entropy of an ideal gas as a function of T and V for constant particle
number N starting from
dU  TdS − pdV
We start from
p
dS  dU  dV  3 Nk B dT  Nk B dV
T T 2 T V
We can now integrate the above expression from a state with entropy S 0 at T 0 and V 0 to obtain
3/2
ST, V − S 0 T 0 , V 0   3 Nk B ln T  Nk B ln V  Nk B ln T V
2 T0 V0 T0 V0
c) Compute the chemical potential of the ideal gas as a function of p and T starting from
the Gibbs-Duhem relation
SdT − Vdp  Nd  0
Thus
d  − S dT  V dp
N N
Using next
ST, V − S 0 T 0 , V 0   3 Nk B ln T  Nk B ln V
2 T0 V0
5/2 3/2
T V T p0
 Nk B ln  Nk B ln p
T0 V0 T0
and
V  NkpB T
I obtain
5/2
1 S T , V   k B ln T p0 dp
d  − dT  k B T p
N 0 0 0 T0 p
and by integrating
T p dp
T, p − T 0 , p 0   −  dT 1 S 0 T 0 , V 0   k B 5 ln T p0  k B T 
T0 N 2 T0 p p0
p
 − T − T 0 S 0 T 0 , V 0   5 k B T ln T − T − T 0 ln T 0 − T 0  − T − T 0  ln T 0   k B T ln p 0
p
N 2
 − T − T 0 S 0 T 0 , V 0  − 5 k B T ln T  5 k B T − T 0   k B T ln p 0
p
N 2 T0 2
1. Ideal Gas in a Gravitational Field: Consider a cubic container of edge L (volume
V = L3 ) containing an ideal gas of N particles, initially at temperature T .
(a) Near the surface of the earth, the gravitational field acting on all the particles of the
gas is g (acting in the height direction). Find the density of the gas as a function of vertical
position in the container, ρ(z).
(b) Calculate the entropy of the gas as a function of N , the volume V and g, in the small-g
limit mgL/kB T << 1 (find the limiting behavior, don’t just set g = 0!).
(c) Now, suppose the container is launched into deep space, so that now no gravitational
field acts on the gas. Under the condition that the temperature of the container is held fixed
during the flight, find the change in entropy relative to (b), again in the limit mgL/kB T <<
1. Explain your answer.
(d) Suppose you carry out the same experiment, except that now the container is heavily
insulated so no heat can be transferred to or from the gas; it starts on earth at temperature
T . Find the final temperature Tf of the gas when it reaches deep space where g = 0.
Explain your answer.
2. Extensible molecule: Consider a long molecule which is composed of N chemical

units (‘monomers’), each of which can be in one of two states, of different lengths a and
b, with b > a. The whole molecule therefore can be between N a and N b in length. The
energy of a monomer in the longer state is larger than a monomer in the shorter state.
You may consider the thermodynamic limit N >> 1 to simplify the calculations.
(a) Calculate the equilibrium length of the entire molecule as a function of temperature T .
(b) Calculate the root-mean-square fluctuation in the length of the entire molecule as a
function of temperature T .
(c) Now, suppose that the molecule is forced to be a fixed length L(N a < L < N b), so that
(L − N a)/(b − a) of its monomers are in the stretched (length b) state. Find the internal
energy E(N, L) and the entropy S(N, T, L).
(d) From (c) calculate the Helmholtz free energy F (N, T, L), and finally the force needed
to extend the molecule to to length L at fixed temperature T .
1
3. Neutron Star: Consider a neutron star, a macroscopic body composed of neutrons,
at a density of 1014 g/cm3 . The temperature of the star’s interior is approximately 107 K.
For this problem you should consider the star to be a noninteracting Fermi gas of neutrons.
(a) Determine whether the neutrons are relativistic or nonrelativistic, by estimating their
kinetic energy.
(b) Determine whether or not the neutrons are reasonably well considered to be a zero-
temperature Fermi gas.
(c) Estimate the pressure in the neutron star.
(d) Use (c) to estimate the mass of this neutron star.
4. Free (Joule) expansion of a gas: Consider a gas initially at temperature Ti , in a

thermally isolated vessel of volume Vi . Suppose that a door is opened, allowing the gas
to abruptly and freely expand into a final total volume Vf , without any work being done,
and without any heat transfer to the gas. At the end of this process, the gas reaches a
final temperature Tf .
(a) In general, what is the change in internal energy E as a result of this expansion?
(b) Suppose the gas is an ideal gas: what is the change in entropy of the gas as a result of
the expansion? Explain your result in simple terms.
(c) Now consider a non-ideal (interacting) gas. Show that the temperature change can be
expressed as
Z Vf
T 2 ∂(p/T )

∆T = − dV
Vi CV ∂T V
where the integration is done using the equilibrium equation of state of the gas. Make sure
you justify the use of the equilibrium equation of state.
(d) Explain with justification whether in the general (non-ideal gas) case you expect the
temperature change ∆T to be positive or negative.
2
5. Low-temperature paramagnet: Consider a paramagnetic crystal of spin-1/2 spins.
You may neglect all interactions between the spins.
(a) At low temperature (kB T << µB), find the entropy per spin as a function of external
magnetic field B (the magnetic moment per spin is µ, so that each spin has energies ±µB).
(b) Does the result of (a) violate any of the laws of thermodynamics? Explain your answer.
(c) Suppose the crystal is cooled in the presence of a magnetic field Bi to a temperature
Ti , and then is thermally isolated. Now the magnetic field is slowly reduced; since the
system is thermally isolated this is a reversible process. If the final value of the field is
Bf < Bi , find the final temperature Tf of the spins.
3
Constants
kB = 1.381 × 10−23 J/K h = 6.626 × 10−34 J·sec c = 2.998 × 108 m/sec

mp = 1.673 × 10−27 kg mn = 1.675 × 10−27 kg me = 9.109 × 10−31 kg
e = 1.602 × 10−19 C G = 6.67 × 10−11 N m2 /kg2
At room temperature T = 300 K and kB T = 4.1 × 10−21 J
Integrals √
x2
Z ∞
dx exp − 2 = 2πσ 2
−∞ 2σ
x2

1 ∞
Z
n
√ dx x exp − 2 = 1 · 3 · 5 · · · (n − 1)σ n
2πσ 2 −∞ 2σ
for n = 2, 4, 6 · · ·
Series
N (N + 1)
1+ 2+ 3+ ···+ N =
2
1 − xP
1 + x + x2 + · · · xP −1 = (geometric series)
1−x
xn xn
X∞ X∞
= − ln(1 − x) = ex
n=1
n n=0
n!
∞ ∞ ∞
X 1 X 1 π2 X 1 π4
s
= ζ(s) Zeta function ζ(2) = = ζ(4) = =
n=1
n n=1
n2 6 n=1
n4 90
∞ ∞
X 1 X 1
ζ(1) = =∞ ζ(3) = = 1.202056 · · ·
n=1
n n=1
n3
ex − e−x ex + e−x sinh x
2 2 cosh x
d d
sinh x = cosh x cosh x = sinh x cosh2 x − sinh2 x = 1
dx dx
4
Stirling’s approximation for log of factorial
ln n! = n ln n − n + O(1)
Combinations
N!
CnN =
n!(N − n)!
Classical ideal gas partition function

3N
p2

1 1
Z Z
3 3
Z(N, T, V ) = d r d p exp −
N! h3 V 2mkB T
3N/2
VN

2πmkB T
=
N! h2
5
1. (a) Use Boltzmann distribution
ρ(z) = Ae−βmgz
normalization determined by
L
AL2
Z Z
3 2
d rρ(z) = N = AL dze−βmgz = (1 − e−βmgL )
0 βmg
N βmgL
rho(z) = e−βmgz
V (1 − e−βmgL )
(b) Probably easiest to use canonical partition func

3/2 !N
L
1 2πmkB T dz −βmgz
Z
Z= L3 e
N! h2 0 L
Take limit mgL/kB T << 1 right at the start, then integral is

L L
dz −βmgz dz 1 1
Z Z
2
e = 1 − βmgz + (βmgz) + · · ·
0 L 0 L 2! 3!
1
= 1 − βmgL + (βmgL)2 − (βmgL)3 + · · ·
2
so
3/2 !N
2πmkB T 1 1
Z = e−βF = V 1 − βmgL + (βmgL)2 + · · ·
h2 2 3!
so

3 2πmkB T 1 1 2
−F = N kB T ln V − ln N + 1 + ln + ln 1 − βmgL + (βmgL) + · · ·
2 h2 2 3!
need to expand log to 2nd order in βmgL,

V 3 2πmkB T 1 1 2
−F = N kB T ln + 1 + ln + βmgL + (βmgL) + · · ·
N 2 h2 2 12
finally use S = −(∂F/∂T )N,V

" 2 #
V 5 3 2πmkB T 1 mgL
S(N.T, V, g) = N kB ln + + ln −
N 2 2 h2 12 kB T
note sign of g doesn’t matter, and S goes down as g goes up, since fewer positional states
are occupied; the gravitational field orders the gas
6
(c) easy to see that if temp held constant, sending gas into space raises its disorder
2
N kB mgL
∆S = S(N, T, V, 0) − S(N, T, V, g) = +···
12 kB T
so entropy goes up
(d) now send gas up without heat transfer; δS = 0, or S(N, T, V, g) = S(N, Tf , V, 0), giving
us 2
3 1 mgL 3
ln T − = ln Tf
2 12 kB T 2
or 3/2 " 2 #
Tf 1 mgL
= exp −
T 12 kB T
or " 2 # " 2 #
1 mgL 1 mgL
Tf = T exp − + ··· = T 1− + ···
18 kB T 18 kB T
so temperature goes down when gas is launched into space without heat transfer.
7
2. (a) Since elements are independent, we can compute the partition function for one
element, Z1 = 1 + e−β , and then compute ZN = (1 + e−β )N . Since the energy is just
proportional to the number of elements in state b, we can compute
∂ ln Z N
hNB i = − = β
∂β e +1
Note that hNA i = N − hNB i, so
N (b − a)
hLi = a hNA i + b hNB i = N a +
eβ + 1

D 2
E D
2
E
(b) L2 = (aNA + bNB ) = (aN + (b − a)NB )
= a2 N 2 + 2a(b − a)N hNB i + (b − a)2 NB2

2 2 2
hLi = [aN + (b − a) hNB i] = a2 N 2 + (b − a)2 hNB i + 2a(b − a)N hNB i
so h
i
2 2
L2 − hLi = (b − a)2 NB2 − hNB i

and
2 ∂2 N eβ
NB2 − hNB i =

ln Z =
∂(β)2 (eβ + 1)2
giving RMS fluctuation √
N (b − a)
q
2
hL2 i − hLi =
eβ/2 + e−β/2
(c) If we force length to be L we force NB = (L − N a)/(b − a). The energy is E = NB ,

and the entropy is just the log
of the number of ways to choose NB of the monomers to
N
be in state b, S = kB ln NB . So,
N!
S = kB ln
NB !(N − NB )!
which can be written in the large N and NB limit as

N N
S = kB NB ln + (N − NB ) ln
NB N − NB
so
NB N − NB
F = E − ST = NB + kB T NB ln + (N − NB ) ln
N N
where NB /N = (L − N a)/[N (b − a)] and where 1 − NB /N = (N b − L)/[N (b − a)].
8
(d) Force is given by L derivative of free energy at fixed temperature

∂F ∂NB ∂F
f= =
∂L T ∂L ∂NB
f = ( + kB T [ln NB /N − ln(1 − NB /N )]) /(b − a)

or
+ kB T ln L−N a
N b−L
f=
b−a
Note force diverges to infinite tension if L → N b; also force diverges to infinite compression
if L → N a.
9
3. (a) At 107 K we estimate KE per neutron
kB T = 1.38 × 10−23 J/K ×107 K ≈ 10−16 J
compare to neutron rest energy
mc2 = 1.6 × 10−27 × 9 × 1016 kg m/sec2 ≈ 10−10 J >> kB T
these neutrons are nonrelativistic (by a factor of 106 or so)
(b) compute Fermi energy for nonrelativistic gas

2 · 4πV p3F
N=
h3 3
so 2/3
p2 h2

3 N
F = F =
2m 2m 8π V
plug in numbers, compute density ρ = 1014 g/cm3 = 1017 kg/m3

N/V = ρ/m = (1017 kg/m3 )/(1.6 × 10−27 kg) ≈ 1044 /m3 so
F ≈ 10−67 J2 sec2 × (1043 m−3 )2/3 )/(10−27 kg) ≈ 10−11 J >> kB T
so neutron star can be considered a zero-temperature Fermi gas
(c) ground state energy of Fermi gas

3
E= N F ≈ N F
5
pressure is volume derivative

∂E 2E 2
p=− = = N F /V
∂V 3V 5
estimate
p ≈ F N/V ≈ 10−11 J×1044 m−3 = 1033 Pa
(d) pressure is generated gravitationally, so if the star has radius R we can estimate
p ≈ GM 2 /R4 = GM 2/3 (M/R3 )4/3
so p 3/2 1
M=
G ρ2
plug in to find
M ≈ (1033 N/m2 /10−10 N m2 /kg2 )3/2 /(1034 kg2 /m6 ) ≈ 1029 kg
is the mass needed to generate this pressure.
Note that this mass corresponds to a neutron star size of R ≈ 104 m, and is roughly the
mass of our sun (which is 2 × 1030 kg).
More careful analysis shows that the critical mass needed to create a neutron star (to force
electrons to combine with protons) is about 3 × 1030 kg.
10
4. (a) During the free expansion, no work is done on the gas; also no heat is transferred to
the gas. Then, by the first law of thermodynamics (∆E = ∆W + ∆Q) we conclude that
the internal energy E does not change.
(b) For an ideal gas, E = 3N kB T /2, so the temperature does not change during a free
expansion. Since the ideal gas entropy is

3
S(N, T, V ) = N kB ln V + ln T + const
2
we have an entropy change of N kB ln Vf /Vi . The sign reflects the fact that the process is
irreversible; the amount reflects the fact that the number of positional states has increased
by a factor Vf /Vi per gas molecule.
(c) The free expansion is an irreversible process (the entropy change is positive) but it is
subject to the constraint that E be fixed. Writing the equilibrium equation of state for the
gas T (E, V ), we see that since E is fixed, the final value of temperature is uniquely deter-
mined, Tf = T (E, Vf ). Therefore we can compute Tf by considering a quasi-equilibrium
and reversible path: we imagine keeping E fixed while slowly increasing V .
We therefore consider the change of temperature with volume, at fixed internal energy E.

∂T ∂E ∂T
=−
∂V E ∂V T ∂E)V
We can determine the unknown derivatives using the first law in the form
dE = T dS − pdV
which holds even for an irreversible process. This tells us

∂E ∂S ∂E ∂S
=T −p and =T = CV
∂V T ∂V T ∂T V ∂T V
Plug together to obtain

∂T 1 ∂S
=− T −p
∂V E CV ∂V T
and use the Maxwell relation (∂S/∂V )T = (∂p/∂T )V to find
T 2 1 ∂p

∂T 1 ∂p p
=− T −p =− − 2
∂V E CV ∂T V CV T ∂T V T
or finally
T2

∂T ∂p/T
=−
∂V E CV ∂T V
11
(d) Go back to
∂T ∂T ∂E 1 ∂E
=− =−
∂V E ∂E)V ∂V T CV ∂V T
The specific heat is always positive (it is mean-square fluctuation of energy); on the other
hand, when a gas is expanded at constant temperature, its internal energy will always
go down, since the molecules are on average farther apart, and the repulsive interactions
∂E
(keeping it in the gas phase) become weaker. Therefore we must have ∂V T
> 0. The
free expansion of a gas therefore always reduces its temperature.
12
5. (a) For one spin, Z1 = eβµB + e−βµB = 2 cosh βµB
For N spins, ZN = (Z1 )N = (2 cosh βµB)N , so free energy is
−F = kB T ln ZN = N kB T [ln cosh βµB + ln 2]
Calculate S = −(∂F/∂T )B

S µB µB µB
= kB ln cosh + ln 2 − tanh
N kB T kB T kB T
(b) If B 6= 0, then we have S → 0 as T → 0 in accord with third law. But, for the case
B = 0, we have S → kB ln 2 as T → 0, violating the third law. The origin of this is the
highly degenerate ground state for B = 0; all states actually have the same (zero) energy
in this case.
(c) When we cool the crystal in a field Bi to temperature Ti , the entropy per spin will
reach
S(Bi , Ti ) µBi µB µBi
= kB ln 2 cosh − tanh
N k B Ti k B Ti k B Ti
which will be near to zero if µBi >> kB Ti .
If the spins are thermally isolated and then the field is slowly reduced, the entropy of the
spins must stay fixed. However, this means that when the field reaches B2 , we must have
S(B1 , T1 )/N = S(B2 , T2 )/N
Since the entropy is a function of B/T we must have Bi /Ti = Bf /Tf , or Tf /Ti = Bf /Bi .
Thus, if Bf < Bi , Tf < Ti . This effect is often used to cool solids to very low temperature
(< 0.1K).
13

January 9, 2009
9.00 am – 12:00 pm
Full credit can be achieved from completely correct answers

to 4 questions. If the student attempts all 5 questions, all of
the answers will be graded, and the top 4 scores will be
counted toward the exam’s total score.
Problem 1
A vertical cylinder contains n moles of an ideal gas, and is closed off by a piston of
mass M and area A. The acceleration due to gravity is g. The molar specific heat Cv (at
constant volume) of the gas is a constant independent of temperature. The heat
capacities of the piston and the cylinder are negligibly small, and any frictional forces
between the piston and the cylinder walls can be neglected. The whole system is
thermally isolated. Initially the piston is clamped in position so that the gas has a
volume Vo and a temperature To. The piston is now released and, after some
oscillations, comes to rest in a final equilibrium state corresponding to a larger volume
of the gas. Assume that the external air pressure is negligible.
(a) Does the temperature of the gas increase, decrease, or stay the same. Explain your
answer.
(b) Does the entropy of the gas increase, decrease, or remain the same? Explain your
answer.
(c) Express the ratio Tf/Vf in terms of M, g, A, n and the gas constant R, where Tf and Vf
are the final temperature and volume, respectively, of the gas.
(d) What is the net work done by the gas in this process? Express your answer in
terms of Vo, Vf, M, g, and A.
(e) What is the final temperature Tf? Express your answer in terms of To, Vo, M, g,
A, Cv, n and R.
Problem 2
The diagram shows the energy flow in a heat engine. Hot Reservoir, T h
Qh
(a) State the definition of the efficiency of this heat
engine in terms of the heat Qh absorbed from the hot Engine
reservoir, and the heat Qc rejected to the cold reservoir. W
Do not assume that the engine is reversible.
Qc
(b) Based upon entropy considerations, derive an
inequality between the efficiency of the engine, and the Cold Reservoir, T c
temperatures of the reservoirs.
For parts (c)-(e), replace the reservoirs by two identical but finite bodies, each
characterized by a heat capacity at constant pressure C, which is independent of
temperature. The bodies remain at constant pressure and undergo no change of phase.
Initially, their temperatures are T1 and T2, respectively; finally, as a result of the
operation of the heat engine, the bodies will attain a common final temperature Tf.
2
(c) What is the total amount of work W done by the engine? Do not assume that the
engine is operated reversibly. Express your answer in terms of C, T1, T2, and Tf.
(d) As in part (b), use entropy considerations to derive an inequality relating Tf to the
initial temperatures T1 and T2.
(e) For given initial temperatures T1 and T2, what is the maximum amount of work
obtainable by operating an engine between these two bodies?
Problem 3
Consider a paramagnetic material consisting of N non-interacting spin-1 particles (with

r µ r
magnetic dipole moment µ = S , S z = mh , m = 0,±1). Suppose that these spins reside
h
in (and are in thermal equilibrium with) a crystal lattice, which is in good thermal contact
with a reservoir at temperature T. We apply a magnetic field of magnitude B in the z-
direction.
(a) Write the partition function of the paramagnetic system in terms of kBT and µB.
(b) How does the average energy of the N spins vary with temperature T?
(c) Calculate the spin contribution to the entropy of the crystal and show that it is only a
function of µB / k BT .
(d) What are the limiting values of the entropy as T → 0 and as T → ∞ ? How would
your results change if the system consisted of spin-1/2 particles?
(e) The crystal is initially in contact with a reservoir at T = T1 and the magnetic field has
a magnitude B1. The crystal is now thermally isolated, and the magnetic field strength is
slowly turned down to a value B2. Calculate the final temperature T2 of the spins (and
therefore the temperature of the crystal lattice).
Problem 4
(a) Consider a spinless quantum mechanical particle of mass m in a simple harmonic

oscillator potential in one dimension of frequency ν and at temperature T. The energy
levels for this particle are given by E n = (n + 1/ 2)hν . Write down the partition function,
Z1, of a single such particle in the oscillator. Express your result in terms of the
dimensionless quantity α = hν / k BT
(b) Consider N spinless, distinguishable particles placed in this oscillator. Write down
the partition function ZN for N such particles.
3
(c) Now consider two identical spinless particles placed in this oscillator. Write down
the partition function Z2 for the two particles at temperature T as an expansion in
ξ = e −α up to and including order ξ 4 .
(d) Repeat part (c) for two identical Fermi particles, of spin 1/2, put in such an oscillator,
one with spin up and the other with spin down.
(e) How does the partition function from part (d) change if the two fermions are both in
spin-up states in the oscillator.
Problem 5
The internal energy U and entropy S of an ideal monatomic gas at temperature T are,
respectively:
U = 3Nk BT / 2 ; S = Nk B [ln( no / n ) + 5 / 2]
Here kB is the Boltzmann constant, N is the number of atoms, n = N/V is the average
concentration, and no = (Mk BT / 2πh 2 ) , with M the atomic mass.
3/2
(a) Show that the chemical potential µ of the gas is related to the pressure P of the gas
by µ = −k BT ln(n o k BT / P ) .
Two such gases, X and Y, are in equilibrium with surface sites at which the gases bind
to a metal surface. In the presence of X and Y simultaneously there are just three
possible configurations of each surface site: (i) the surface site is empty (denoted by E
below); (ii) the surface site is occupied by one X atom, with energy εx relative to the
empty site; (iii) the surface site is occupied by one Y atom, with energy εy relative to the
empty site. Excited configurations at the site are not bound, and multiple occupancy is
forbidden.
E X Y
(i) (ii) (iii)
(b) The metal surface is at equilibrium with gases X and Y simultaneously at

temperature T. In terms of the parameters defined above, and chemical potentials µx
and µy, write down the grand partition function for the configurations of one site.
4
(c) Calculate the probability that the site is (i) empty, (ii) occupied by X, (iii) occupied by
Y.
(d) At room temperature, and fixed partial pressures of Px = Py = 0.5 atm, the sites are
occupied in the ratio nE:nX:nY = 1:1:2. What is the energy difference ∆ε = ε X − ε Y ?
Ignore any differences in the atomic masses Mx and MY. Express your answer in units
of kBT.
(e) The partial pressures are now doubled to 1 atm each. What is the new value of the
ratio nE:nX:nY?
2 2 cosh x
Maxwell's relations
⎛ ∂T ⎞ ⎛ ∂P ⎞ ⎛ ∂T ⎞ ⎛ ∂P ⎞ ⎛ ∂T ⎞ ⎛ ∂V ⎞ ⎛ ∂T ⎞ ⎛ ∂V ⎞
⎜ ⎟ = −⎜ ⎟ ; ⎜ ⎟ = −⎜ ⎟ ; ⎜ ⎟ =⎜ ⎟ ; ⎜ ⎟ =⎜ ⎟
⎝ ∂V ⎠ S ⎝ ∂S ⎠V ⎝ ∂V ⎠ P ⎝ ∂S ⎠T ⎝ ∂P ⎠ S ⎝ ∂S ⎠ P ⎝ ∂P ⎠V ⎝ ∂S ⎠T
5

January 9, 2009
9.00 am – 12:00 pm
Full credit can be achieved from completely correct

answers to 4 questions. If the student attempts all 5
questions, all of the answers will be graded, and the top 4
scores will be counted toward the exam’s total score.
Problem 1
A vertical cylinder contains n moles of an ideal gas, and is closed off by a piston of
mass M and area A. The acceleration due to gravity is g. The molar specific heat
Cv (at constant volume) of the gas is a constant independent of temperature. The
heat capacities of the piston and the cylinder are negligibly small, and any frictional
forces between the piston and the cylinder walls can be neglected. The whole
system is thermally isolated. Initially the piston is clamped in position so that the
gas has a volume Vo and a temperature To. The piston is now released and, after
some oscillations, comes to rest in a final equilibrium state corresponding to a
larger volume of the gas. Assume that the external air pressure is negligible.
(a) Does the temperature of the gas increase, decrease, or stay the same. Explain
your answer.
First law: ∆U = Q + W
System is thermally isolated ⇒ Q = 0
The expanding gas does work on the surroundings ⇒ W < 0
Therefore, from first law ∆U < 0
For an ideal gas, U is a function of T only, and hence the gas cools.
(b) Does the entropy of the gas increase, decrease, or remain the same? Explain
your answer.
This process is an irreversible expansion of the gas. For a thermally

isolated system undergoing an irreversible process, the entropy of the
system must increase.
(c) Express the ratio Tf/Vf in terms of M, g, A, n and the gas constant R, where Tf
and Vf are the final temperature and volume, respectively, of the gas.
Mg
The final pressure, at equilibrium, is given by Pf =
A
Mg nRTf T Mg
⇒ = ⇒ f = .
A Vf Vf AnR
(d) What is the net work done by the gas in this process? Express your
answer in terms of Vo, Vf, M, g, and A.
The net work done by the gas in pushing the piston up by a height h is
Mg Mg
Mgh = ( Ah ) = (Vf − Vo )
A A
2
(e) What is the final temperature Tf? Express your answer in terms of To, Vo,
M, g, A, Cv, n and R.
The change in the internal energy of the gas can be expressed as
∆U = nCV (Tf − To )
From first law ∆U = W where W is the work done on the gas.
Mg
nCV (Tf − To ) = − (Vf − Vo ) = Mg Vo − Mg Vf
A A A
Use the result from part(c) to eliminate Vf:
Mg Mg ⎛ nRTf A ⎞ Mg
nCV (Tf − To ) = Vo − ⎜ ⎟= Vo − nRTf
A A ⎜⎝ Mg ⎟⎠ A
Rearrange and solve for Tf
Mg ⎛ MgVo ⎞⎛ 1 ⎞
(nCV + nR )Tf − nCV To = Vo ⇒ Tf = ⎜ CV To + ⎟⎜ ⎟
A ⎝ nA ⎠⎜⎝ CV + R ⎟⎠
Problem 2
The diagram shows the energy flow in a heat engine. Hot Reservoir, T h
Qh
(a) State the definition of the efficiency of this heat
engine in terms of the heat Qh absorbed from the hot Engine
reservoir, and the heat Qc rejected to the cold reservoir. W
Do not assume that the engine is reversible.
Qc
The efficiency of the engine is given by
W Qh − QC Q Cold Reservoir, T c
η= = = 1− C
Qh Qh Qh
(b) Based upon entropy considerations, derive an inequality between the efficiency
of the engine, and the temperatures of the reservoirs.
The combined entropy of the two reservoirs and the engine cannot decrease
in this process, i.e. ∆Sh + ∆Sc + ∆Sengine ≥ 0
3
Qh Qc
where ∆S h = − ; ∆Sc = − ; and ∆Sengine = 0
Th Tc
Q h Qc Q Q Q T
which gives − + ≥0⇒ c ≥ h ⇒ c ≥ c .
T h Tc Tc T h Qh Th
Q T
Therefore, η = 1 − C ≤ 1 − C , where the equality holds for a reversible engine.
Qh Th
For parts (c)-(e), replace the reservoirs by two identical but finite bodies, each
characterized by a heat capacity at constant pressure C, which is independent of
temperature. The bodies remain at constant pressure and undergo no change of
phase. Initially, their temperatures are T1 and T2, respectively; finally, as a result of
the operation of the heat engine, the bodies will attain a common final temperature
Tf.
(c) What is the total amount of work W done by the engine? Do not assume that
the engine is operated reversibly. Express your answer in terms of C, T1, T2, and
Tf.
Apply first law to the combined system consisting of the two bodies
and the engine ∆U = Q + W ⇒ ∆U = W , since no heat is absorbed from
the surroundings (all heat exchange is internal to the system). Note
that, in the first law as stated, W is the work done on the system.
The total change in the internal energy of the combined system is

∆U = ∆U h + ∆U c + ∆U engine = C∆Th + C∆Tc + 0
= C(Tf − T1 ) + C(Tf − T2 ) = C (2Tf − T1 − T2 )
Therefore, work done by the engine becomes − ∆U = C (T1 + T2 − 2Tf )
(d) As in part (b), use entropy considerations to derive an inequality relating Tf to

the initial temperatures T1 and T2.
Once again, ∆Sh + ∆Sc + ∆Sengine ≥ 0
However, unlike in part (b), the temperature of the two bodies is now
changing.
Tf T
− dQh f CdTh ⎛ Tf ⎞
Therefore, ∆Sh = ∫T Th T∫ Th =C ln⎜⎜⎝ T1 ⎟⎟⎠ , where − dQh is the heat
=
1 1
absorbed by the hot body in each infinitesimal step of a reversible path

that takes the hot body from an initial temperature T1 to a final
temperature Tf.
4
T Tf
dQc f CdTc ⎛T ⎞
Similarly, ∆Sc = ∫ =∫ =C ln⎜⎜ f ⎟⎟ .
T2 Tc T2 Tc ⎝ T2 ⎠
As before, ∆Sengine = 0
Note that, since entropy is a function of state, we can pick any

reversible path to calculate the entropy change between the initial and
the final states of each of the bodies, even if the engine operates
irreversibly between them.
The condition ∆Sh + ∆Sc + ∆Sengine ≥ 0 leads to:

⎛T ⎞ ⎛T ⎞ ⎛ T2 ⎞
C ln⎜⎜ f ⎟⎟ + C ln⎜⎜ f ⎟⎟ ≥ 0 ⇒ C ln⎜ f ⎟ ≥ 0 ⇒ Tf ≥ T1T2
⎜T T ⎟
⎝ T1 ⎠ ⎝ T2 ⎠ ⎝ 1 2 ⎠
(e) For given initial temperatures T1 and T2, what is the maximum amount of work
obtainable by operating an engine between these two bodies?
The maximum amount of work is done by a reversible engine, for which

Tf = T1T2 , which, from part (c) is: W = C T1 + T2 − T1T2 . ( )
Problem 3
Consider a paramagnetic material consisting of N non-interacting spin-1 particles

r µ r
(with magnetic dipole moment µ = S , S z = mh , m = 0,±1). Suppose that these
h
spins reside in (and are in thermal equilibrium with) a crystal lattice, which is in
good thermal contact with a reservoir at temperature T. We apply a magnetic field
of magnitude B in the z-direction.
(a) Write the partition function of the paramagnetic system in terms of kBT and µB.
In the presence of a magnetic field, the allowed energy states of a single

spin are: ε = − µB, 0, + µB
The partition function of a single spin becomes:
Z1 = exp( µB / k BT ) + 1 + exp( − µB / k BT ) = 1 + 2 cosh( βµB )
(b) How does the average energy of the N spins vary with temperature T?
The average energy of a single spin is given by
5
1 ⎛ ∂Z ⎞ 2µB sinh( βµB )
〈E 〉 = − ⎜⎜ ⎟⎟ = −
Z ⎝ ∂β ⎠ 1 + 2 cosh( βµB )
(c) Calculate the spin contribution to the entropy of the crystal and show that it is
only a function of µB / k BT .
F = U − TS , where U = N 〈E 〉 and F = −Nk BT ln(Z1 )
Therefore, S =
U F
− = −2Nk B
(βµB )sinh( βµB ) + Nk ln(1 + 2 cosh( βµB )
B
T T 1 + 2 cosh( βµB )
(d) What are the limiting values of the entropy as T → 0 and as T → ∞ ? How
would your results change if it was a spin-1/2 particle?
In the limit T → 0 (or B → ∞ ), only the lowest (ground) state is occupied.

The multiplicity of the ground state is Ω = 1 . Therefore, S = k B ln Ω → 0
In the limit T → ∞ (or B → 0 ), all accessible states of the system are

equally likely. Each spin has three allowed orientations. The total
number of accessible states in the high temperature limit is Ω = 3 N , and
hence S = k B ln Ω → Nk B ln 3
For a spin-1/2 particle, the entropy as T → 0 would still be 0. As T → ∞ ,

the total number of accessible states in the high temperature limit would
be Ω = 2 N , and hence S = k B ln Ω → Nk B ln 2
(e) The crystal is initially in contact with a reservoir at T = T1 and the magnetic field
has a magnitude B1. The crystal is now thermally isolated, and the magnetic field
strength is slowly turned down to a value B2. Calculate the final temperature T2 of
the spins (and therefore the temperature of the crystal lattice).
In a quasistatic, adiabatic process, the entropy of the system remains

unchanged. Since the entropy of the spin system is a function of the
dimensionless parameter µB / k BT , constant entropy in this adiabatic
B1 B2 ⎛B ⎞
demagnetization process implies = ⇒ T2 = T1 ⎜⎜ 2 ⎟⎟ .
T1 T2 ⎝ B1 ⎠
6
Problem 4
(a) Consider a spinless quantum mechanical particle of mass m in a simple

harmonic oscillator potential in one dimension of frequency ν and at temperature T.
The energy levels for this particle are given by E n = (n + 1/ 2)hν . Write down the
partition function, Z1, of a single such particle in the oscillator. Express your result
in terms of the dimensionless quantity α = hν / k BT
Z1 = ∑ exp( −E n / k BT ) = exp( −hν / 2k BT )∑ exp( −nhν / k BT )

n n
exp( −α / 2) 1
Z1 = =
1 − exp( −α ) exp(α / 2) − exp( −α / 2)
(b) Consider N spinless, distinguishable particles placed in this oscillator. Write

down the partition function ZN for N such particles.
1
For N distinguishable particles, Z N = Z1N =
(exp(α / 2) − exp(−α / 2)N
(c) Now consider two identical spinless particles placed in this oscillator. Write
down the partition function Z2 for the two particles at temperature T as an
expansion in ξ = e −α up to and including order ξ 4 .
The energy levels are given by E n = (n1 + n 2 + 1)hν ,

where n1 = 0,1,2..... and n2 = 0,1,2.......
The lowest five energy levels are E n = hν , 2hν , 3hν , 4hν , 5hν ,......... with
degeneracy g n = 1, 1, 2, 2, 3........ , respectively, for identical bosons.
The partition function becomes

Z 2 = e −α + e −2α + 2e −3α + 2e −4α + 3e −5α + ........
(d) Repeat part (c) for two identical Fermi particles, of spin 1/2, put in such an
oscillator, one with spin up and the other with spin down.
For two distinguishable fermions, the energy levels are the same as in
part (c), but now the degeneracy of each level becomes
g n = 1, 2, 3, 4, 5........
Thus, the partition function becomes

Z 2 = e −α + 2e −2α + 3e −3α + 4e −4α + 5e −5α + ........
7
(e) How does the partition function from part (d) change if the two fermions are
both in spin-up states in the oscillator.
For two indistinguishable fermions, all double-occupancy states are

forbidden as a result of the Pauli-exclusion principle, so the lowest five
energy levels are given by E n = 2hν , 3hν , 4hν , 5hν , 6hν ...... , with
degeneracy g n = 1, 1, 2, 2, 3.... , and the partition function becomes
Z 2 = e −2α + e −3α + 2e −4α + 2e −5α + 3e −6α + ........
Problem 5
The internal energy U and entropy S of an ideal monatomic gas at temperature T

are, respectively:
U = 3Nk BT / 2 ; S = Nk B [ln( no / n ) + 5 / 2]
Here kB is the Boltzmann constant, N is the number of atoms, n = N/V is the

average concentration, and no = (Mk BT / 2πh 2 ) , with M the atomic mass.
3/2
(a) Show that the chemical potential µ of the gas is related to the pressure P of the
gas by µ = −k BT ln(no k BT / P ) .
⎛ ∂F ⎞
Since F = U − TS and µ = ⎜ ⎟ ,
⎝ ∂N ⎠T ,V
⎛ ∂U ⎞ ⎛ ∂S ⎞ ⎡ ⎛V ⎞ ⎤
we have µ = ⎜ ⎟ −T ⎜ ⎟ , where S = Nk B ⎢ln⎜ n o ⎟ + 5 / 2⎥
⎝ ∂N ⎠T ,V ⎝ ∂N ⎠T ,V ⎣ ⎝N ⎠ ⎦
Therefore
3k BT ⎡ ⎛V ⎞ 5⎤ ⎛ N ⎞⎛ V ⎞ ⎛V ⎞ ⎛n k T ⎞
µ= − k BT ⎢ln⎜ n 0 ⎟ + ⎥ − Nk BT ⎜⎜ ⎟⎟⎜ − 2 n o ⎟ = −k BT ln⎜ n o ⎟ = −k BT ln⎜ o B ⎟
2 ⎣ ⎝ N ⎠ 2⎦ ⎝ Vn o ⎠⎝ N ⎠ ⎝N ⎠ ⎝ P ⎠
Two such gases, X and Y, are in equilibrium with surface sites at which the gases
bind to a metal surface. In the presence of X and Y simultaneously there are just
three possible configurations of each surface site: (i) the surface site is empty
(denoted by E below); (ii) the surface site is occupied by one X atom, with energy
εx relative to the empty site; (iii) the surface site is occupied by one Y atom, with
energy εy relative to the empty site. Excited configurations at the site are not
bound, and multiple occupancy is forbidden.
8
E Y
X
E
(i) (ii) (iii)
(b) The metal surface is at equilibrium with gases X and Y simultaneously at

temperature T. In terms of the parameters defined above, and chemical potentials
µx and µy, write down the grand partition function for the configurations of one site.
For the empty site ε = 0, n = 0, and the Gibbs factor is exp(0) = 1;
For the site occupied by an X atom, ε = εx, n = 1, and the Gibbs factor is
exp(− (ε x − µ x ) / k BT ) ;
Similarly, for the site occupied by a Y atom, ε = εY, n = 1, and the Gibbs
factor is exp(− (ε Y − µY ) / k BT ) ;
Therefore, the grand partition function can be written as:
Ξ = 1 + exp(− (ε X − µ X ) / k BT ) + exp(− (ε Y − µY ) / k BT )
= 1 + λ x exp( −ε x / k BT ) + λY exp( −ε Y / k BT
where λ = exp( µ / k BT )
(c) Calculate the probability that the site is (i) empty, (ii) occupied by X, (iii)
occupied by Y.
The probabilities for (i), (ii), and (iii), are:
1
℘E =
Ξ
λ x exp( −ε x / k BT )
℘x =
Ξ
λ exp( −ε Y / k BT
℘Y = Y
Ξ
9
(d) At room temperature, and fixed partial pressures of Px = Py = 0.5 atm, the sites
are occupied in the ratio nE:nX:nY = 1:1:2. What is the energy difference
∆ε = ε X − ε Y ? Ignore any differences in the atomic masses Mx and MY. Express
your answer in units of kBT.
n x ℘x λ X
= = exp( − ∆ε / k BT )
nY ℘Y λY
From part (a), λ can be expressed in terms of the partial pressure P and
P
temperature T as: λ = exp( µ / k BT ) = .
n 0 k BT
Assuming that the atomic masses Mx and MY are approximately equal,
λ P n 1
we have x = x = 1 , which gives exp( −∆ε / k BT ) = x = ⇒ ∆ε = k BT ln 2 .
λY PY nY 2
(e) The partial pressures are now doubled to 1 atm each. What is the new value
of the ratio nE:nX:nY?
The probability that a state is occupied is proportional to its Gibbs factor.

Under conditions of 0.5 atm partial pressure for X and Y, the Gibbs factors for
the 3 states are in the ratio 1:1:2. When the partial pressure of each gas is
doubled, the Gibbs factor for the state with the empty site is unchanged,
whereas the Gibbs factors for the states with the site occupied by X or Y are
doubled. Therefore, the ratio nE:nX:nY is now 1:2:4.
10
kB =1.381×10−23 J/K; NA = 6.022×1023; R = 8.315 J/mol/K; 1 atm = 1.013×105

N/m2
2 2 cosh x
Maxwell's relations
⎛ ∂T ⎞ ⎛ ∂P ⎞ ⎛ ∂T ⎞ ⎛ ∂P ⎞ ⎛ ∂T ⎞ ⎛ ∂V ⎞
⎜ ⎟ = −⎜ ⎟ ; ⎜ ⎟ = −⎜ ⎟ ; ⎜ ⎟ =⎜ ⎟ ;
⎝ ∂V ⎠ S ⎝ ∂S ⎠V ⎝ ∂V ⎠ P ⎝ ∂S ⎠T ⎝ ∂P ⎠ S ⎝ ∂S ⎠ P
⎛ ∂T ⎞ ⎛ ∂V ⎞
⎜ ⎟ =⎜ ⎟
⎝ ∂P ⎠V ⎝ ∂S ⎠T
11
Thermodynamics and Statistical Mechanics

January 7, 2011
9:00 AM to 12:00 Noon
1
Equation Sheet
"
$
# exp[!bx
2
] dx =
!"
b
"
1 $
#x
2
exp[!bx 2 ] dx =
0
4 b3
"
#x
2
exp[!x] dx = 2
0
"
x $2
# dx ex ! 1
=
6
0
!
x #2
" dx ex + 1
=
12
0
"
x2
# dx x
e !1
= 2 $ (3), where $ (3) can be considered to be just a number.
0
!
x2
" dx x
e +1
= 3 # (3)
2
0
1
n=
e(! " µ )/ kT ± 1
sinh(x) = 1 (e x ! e! x )
2
cosh(x) = 1 (e x + e! x )
2
d
!sinh(x) #$ = cosh(x)
dx "
d
!cosh(x) #$ = sinh(x)
dx "
!
1
" xm = 1 # x
m=0
n
1 " x n+1
! x = 1" xm
m=0
2
Problem 1
A gas of N identical classical non-interacting atoms is held in a potential V (r) = ar , where
( )
1/ 2
r = x2 + y2 + z2 . The gas is in thermal equilibrium at temperature T.
(a) Find the single particle partition function Z1 of an atom in the gas. Express your answer
in the form Z1 = AT ! a "# and provide an expression for the prefactor A and the
exponents ! and ! . [Hint: convert the integral in r to spherical coordinates.]
(b) Find an expression for the entropy S of this classical gas.
Problem 2
A classical ideal gas is taken from state a to state b in

the figure using three different paths: acb, adb, and
ab. The pressure p2 = 2 p1 and the volume V2 = 2 V1.
(a) The heat capacity CV = 52 Nk . Starting from the First Law of Thermodynamics derive a
value for C p . No credit will be given for this part if you just state the answer.
(b) Compute the heat supplied to the gas along each of the three paths, acb, adb, and ab, in
terms of N, k, and T1.
(c) What is the heat capacity Cab of the gas for the process ab?
3
Problem 3
Consider a one-dimensional stretched elastic string that is fixed at its two ends and vibrates
only in a direction perpendicular to its length. The string consists of a very large number N
of atoms arranged in a single row. Let the energies of vibration be quantized in units of hf,
where f is the vibration frequency. This string is in thermal equilibrium with a heat bath at
temperature T.
(a) Determine an expression for the thermal energy of this string in terms of an integral over
the variable x = ! / kT .
(b) Identify a characteristic temperature that separates low T and high T behavior.
Determine an expression for the thermal energy of this string in the limit of low and high
T. Comment on these results in the context of the equipartition theorem.
Problem 4
Consider a spherical drop of liquid water containing Nl molecules surrounded by N - Nl

molecules of water vapor. The drop and its vapor may be out of equilibrium.
(a) Neglecting surface effects write an expression for the Gibbs free energy of this system if
the chemical potential of liquid water in the drop is µl and the chemical potential of water
in the vapor is µv. Rewrite Nl in terms of the (constant) volume per molecule in the
liquid, vl, and the radius r of the drop.
(b) The effect of the surface of the drop can be included by adding a piece Gsurface = !A to the
free energy, where ! is the surface tension (! > 0) and A is the surface area of the drop.
Write Gtotal with the surface piece expressed in terms of r. Make two qualitative hand-
sketches of Gtotal: one sketch with ( µl ! µv ) > 0 and one sketch with ( µl ! µv ) < 0.
Describe the behavior of the drop in these two cases.
(c) Under appropriate conditions, there is a critical radius, rc, that separates drops which
grow in size from those that shrink. Determine this critical radius.
(d) Assume that the vapor behaves as an ideal gas and recall that the chemical potential of an
ideal gas is given by µv = µvo + kT ln( p / p o ) . Write the chemical potential difference
( µv ! µl ) in terms of the vapor pressure and a reference pressure p o , where p o is taken
to be the pressure of a vapor in equilibrium with a large flat surface of water. Then,
derive and comment on the dependence of the relative humidity p / p o on rc .
4
Problem 5
Consider a paramagnetic material whose magnetic particles have angular momentum J,

which is a multiple of !. The projections of the angular momentum along the z-axis can take
2J – 1 values ( J z = ! J ,! J + 1,! J + 2,..., J ), which leads to 2J – 1 allowed values of the z –
component of a particle’s magnetic moment ( µ z = ! J " µ ,!(J + 1)" µ ,..., J " µ ). The energy of
the magnetic moment in a magnetic field pointing in the +z direction is ! µ z B .
(a) Derive an expression for the partition function Z1 of a single magnetic particle in a
magnetic field B pointing in the +z direction. Write your answer in terms of hyperbolic
sin functions, where sinh(x) = 1 (e x ! e! x ) . You may find it convenient to use the
2
variable b = ! µ B" , where ! = 1 .
kT
(b) Derive an expression for the average energy of the particle in part (a). Write your
cosh(x)
answer in terms of the hyperbolic cotangent function coth(x) = .
sinh(x)
(c) Use the expression for the average energy in part (b) to determine the magnetization M
(the average z-component of the total magnetic moment) of a system of N identical,
independent magnetic particles. Comment on its behavior as T ! 0 .
5

January 8, 2010
9.00 am - 12.00 pm

4 questions. If the student attempts all 5 questions, all of the an-
swers will be graded, and the top 4 scores will be counted towards
the exam’s total score.
Various equations, constants, etc. are provided on the last page of

the exam.
1. Van der Waals Fluid
A mole of fluid obeys the van der Waals equation of state:
RT
p= V −b − Va2 ,
where R, a and b are positive constants, and V > b.
(a) Show that the specific heat at constant volume (CV ) is a function of T only.
(b) Is the results still true if the fluid obeys the Dieterici equation of state

RT 1
p= V −b exp − V RT ?
Here, R, a and b are positive constants, and V > b.
(c) Find the entropy S(T, V ) of the van der Waals fluid in the case when CV is independent
of T .
1
2. Simple paramagnet
A simple paramagnet consists of N spin- 21 particles immersed in an external magnetic field

Hext pointing in the z-direction. Each spin- 12 particle can only be in either an “up” or
“down” state, and the energy of a particle with spin pointing down is +µB, while the energy
of particles with spin up is −µB.
(a) Show that the thermodynamic potential is

G(T, Hext) = −aT log 2 cosh b HText ,
where T is the temperature. Determine the constants a and b.
(b) Find the expression of the magnetization M and the magnetic enthalpy H = G + T S
in terms of Hext and T. Verify that H = −M Hext .
(c) Show that the expression for the entropy S obeys the Planck formulation of the third
law (i.e. CV → 0 for T → 0).
(d) Is the expression of the isotheral susceptibility χT in terms of Hext and T consistent
with the postulates of thermodynamics? [Hint: Show that χT → 0 for T → 0.]
2
3. Degenerate Fermi Gas
The crystal structure of metallic copper (Cu) is face-center cubic (fcc) with a unit-cell length
of a = 3.615 Å. This structure results in a density of atoms of ρCu = 8.46 × 1028 m−3 . Each
Cu atom in the crystal donates 1e− to the conduction band, which leads to a density of
states (g (ε)) for the 3-dimensional Fermi gas:
3 1
1 2m 2
g(ε) = 2π 2 h̄2
ε2 ,
where m is the effective mass of the conduction electrons. For this problem, you can assume
m to be the free electron mass me .
(a) In the low temperature limit (i.e. T = 0 K), find the Fermi energy EF and the total
energy U0 of all electrons in the conduction band (per unit volume). Express your answer in
eV.
(b) Electrons that can participate in a current have energies that correspond to an occu-
pancy n(ε) which is neither too close to 1 (no empty final states available for the moving
electrons) nor too small (no electrons left in the conduction band). Calculate the energy
interval that is occupied by the electrons that participate in the current at room temper-
ature (i.e. T = 300 K), assuming that the occupancy for these electrons varies between
n1 (ε) = 0.1 and n2 (ε) = 0.9.
(c) Using your answers to part (b), find the fraction of “current carrying” electrons N1
to the total number of electrons in the conduction band Ntotal (i.e. NN 1
total
). Assume that
within the range of allowed occupancies, the density of states is energy-independent, and
the occupancy will vary linearly with energy. [Hint: Use the Taylor series expansion of the
occupancy around the Fermi-energy εf up to the linear term.]
3
4. The Ideal Gas
One mole of ideal gas with a constant heat capacity CV is placed at a constant pressure p1
inside a cylinder, which is thermally insulated from the environment. The volume of the
cylinder can be changed using a piston, which moves without friction along the vertical axis.
At the beginning of the experiment, the pressure of the gas is abruptly changed from p1 to p2 ,
resulting in a decrease in the volume from V1 at pressure p1 to V2 at pressure p2 . Similarly,
the temperature T1 at pressure p1 will increase to T2 at pressure p2 after thermodynamic
equilibrium has been reached.
(a) Determine the type of compression that took place here. Explain!
(b) Find the temperature T2 and the volume V2 in terms of V1 , p2 and T1 after the ther-
modynamic equilibrium has been reached. [Hint: Use the equation of state for an ideal gas
and the relationship between CV and CP ].
After thermodynamic equilibrium has been established in part (a), the pressure is abruptly
reset to its original value p1 , resulting in Vf and Tf at pressure p1 .
(c) Find the final values of the temperature Tf and volume Vf in terms of p1 , V2 , and T2
after thermodynamic equilibrium is reached again. Use the first law of thermodynamics and
the adiabatic equation to compute the difference in the temperatures (Tf − T1 ).
Comment on both the sign and the relative magnitude of the temperature difference. What
happens in the limit of very small changes in pressure?
4
5. Einstein Solid
Consider two interacting Einstein solids that are in a box with adiabatic walls. The number
of oscillators in system A is NA = 4 and the number in system B is NB = 6. The total
number of units of heat energy is q = 20.
(a) For the most probable macrostate, determine the number of units of heat energy in each
of the two systems. Explain your approach.
(b) Determine the probability of finding the system in the most probable macrostate.
5
Equations and Constants
kB = 1.381 × 10−23 J
K; me = 9.109 × 10−31 kg; mp = 1.673 × 10−27 kg;
e = 1.602 × 10−19 C; h = 6.626 × 10−34 J s; c = 2.998 × 108 m
s;
NA = 6.023 × 1023 ; R = 8.315 molJ K
Hyperbolic Functions:
sinh x = 12 (exp[x] − exp[−x]); cosh x = 12 (exp[x] + exp[−x]);
sinh x
tanh x = cosh x
Maxwell’s

relations:

∂T
∂V = − ∂P ∂S ; ∂T
∂V P =− ∂P
∂S T ;
∂T
∂P S = ∂V
∂S P ;
S V
∂T ∂V
∂P V = ∂S T
Ideal gas:
pV = N kB T ; CV + N kB T = Cp
Fermi-Dirac distribution:

ε−εf
−1
n̄F D = exp kB T +1
Bose-Einstein distribution:

ε−εf
−1
n̄BE = exp kB T −1
Boltzmann distribution:
ε−ε

n̄Boltzmann = exp − k Tf
B
6
Solutions
1. Van der Waals Fluid
(a) Show that the specific heat at constant volume (CV ) is a function of T only.
Remember the

∂S
CV = T · ∂T V
.
To show that CV is a function of T only, we need to show that

∂CV
∂V
= 0.
T
So:

∂CV ∂ ∂S ∂2S ∂ ∂S
∂V
= ∂V
T· ∂T V
=T· ∂V ∂T
=T· ∂T ∂V
T T V T
Using

∂F ∂F ∂2F ∂2F
F (T, V ) = −S dT − p dV = dT + dV and =
∂T V ∂V T ∂V ∂T ∂T ∂V
∂S ∂p
⇔ ∂V T
= ∂T V
Plugging this into the equation above, we can find:
∂2p

∂CV
∂V
=T· (∂T )2 V
T
Using
RT a
p= V −b
− V2

∂CV
⇔ ∂V
=0
T
1
(b) Is the results still true if the fluid obeys the Dieterici equation of state?
No, for the Dieterici equation of state, we find that:
∂2p
1
(∂T )2 V
= 1
R2 V 2 T 3 (V −b)
e− RT V 6= 0.
(c) Find the entropy S(T, V ) for the van der Waals fluid.
Remember that we can write:

∂S ∂S 1 ∂p
dS = ∂T V
dT + ∂V
dV = C
T V
dT + ∂T
dV
T V
Using:
∂p R
∂T
= V −b
we can find for the entropy S:

1 R
dS = C
T V
dT + V −b
dV
⇔S= CV T1 dT + R ln(V − b)
R
If CV is independent of T , this can be simplified to:
S = CV ln T + R ln(V − b) + constants
2
2. Simple paramagnet
(a) Show that the thermodynamic potential

Using the partition function Z for the spin- 12 system:
Z = eβµHext + e−βµHext = 2 cosh (βµHext ), for one particle.
For N particles:
ZN = [2 cosh (βµHext )]N
Using G = −kB T ln ZN :
G = −KB T ln [2 cosh (βµHext )]N = −N kB T ln [2 cosh (βµHext )]
µ
This means that a = N kB and b = kB
.
(b) Find the expression of the magnetization M and the magnetic enthalpy H = G + T S
in terms of Hext and T. Verify that H = −M Hext .
The magnetization M is given by
M = µ̄z · N .
To find µ̄z , we use:
µHext µHext
e kB T e kB T
µz (s) · P (s) = µP↑ − µP↓ , where P↑ =
P
µ̄z = Z
= µH
2 cosh k ext
and
s B T
µHext µHext
− −
e kB T e kB T
P↓ = Z
= µH
2 cosh k ext
.
T B
µHext µHext

µ −
⇒ µ̄z = µH
2 cosh k ext
e kB T
−e kB T
= µ tanh µH ext
kB T
.
B T
So,
M = N µ tanh tanh µH ext

kB T
.
3
The entropy S is given by:

∂G
S=− ∂T V
Using the results from part (a), we find that:

∂G N µHext
S=− ∂T V
= N k ln 2 cosh muH ext
kB T
− T
tanh muH ext
kB T
Using H = G + T S:

H = −N KB T ln 2 cosh muH ext
kB T
+ N KB T ln 2 cosh muH ext
kB T
− T N µH
T
ext
tanh muH ext
kB T
⇔ H = −M · Hext
(c) Show that the expression of the entropy S obeys the Planck formulation of the third
principle.
We have shown in part (b) that the entropy S is given by:

∂G N µHext
S=− ∂T V
= N k ln 2 cosh muH ext
kB T
− T
tanh muH ext
kB T
The third principle states that the specific heat will go to 0 for T approaching 0K.
we can use:
∂S
CV = T ∂T
∂S N µ2 Hext
2
⇔ T ∂T = 1
kB T muHext
cosh2 kB T
1
Since cosh x → ∞ for large x, and CV is proportional to coshx
, the third principle is fulfilled.
4
(d) Is the expression of the isotheral susceptibility χT in terms of Hext and T consistent
with the postulates of thermodynamics?
The isothermal susceptibility is give by:
N µ2

∂M 1
χT = ∂Hext T
= kB T µHext

cosh2 kB T

µHext µHext
For kB T
1, cosh kB T
→ 1, which means that in the limit of high temperatures:
N µ2
χT = kB T
Curie’s Law.

µHext µHext
In the limit of low temperature, kB T
→ ∞, cosh kB T
→ ∞, so that:
lim χT = 0
T →0
5
3. Degenerate Fermi Gas
(a) For the low temperature limit (i.e. T = 0K), find the Fermi energy EF and the total
energy U0 of all electrons in the conduction band (per unit volume). Express your answer in
eV.
The number of electrons per unit volume N is given by:
R∞
N= g(ε)f (ε) dε.
0
At T = 0, all the states are completely filled up to the Fermi-energy (εf ):
εRf εRf εRf 3/2 3/2

1 2m 1 2m 2 3/2
N= g(ε)f (ε) dε = g(ε) dε = 2π 2 h̄2
ε1/2 dε = 2π 2 h̄2
ε
3 f
0 0 0
h̄2 3/2
⇔ εf = 2m
(3π 2 N )
Assuming h̄ = 1.05 × 10−34 Js, me = 9.1 × 10−31 kg, and N = 8.46 × 1028 m−3 :
εf = 1.1 × 10−18 J ≈ 7 eV
The total energy U0 is given by:
εRf 3/2 εRf 3/2

1 2m 2 1 2m 5/2
U0 = g(ε)ε dε = 2π 2 h̄2
ε3/2 /, dε = 5 2π 2 h̄2
εf
0 0
2
U0 = 5
N εf
(b) Calculate the energy interval that is occupied by the electrons that participate in the
current flow at room temperature (i.e. T = 300K).The occupancy at non-zero temperature
is given by:
1
n̄(ε) = ε−εf
+1
e kB T
So, the lower limit is given by:
1
0.9 = ε1 −εf
+1
e kB T
ε1 = εf − kB T ln 9
and the upper limit:
6
1
0.1 = ε2 −εf
+1
e kB T
ε2 = εf + kB T ln 9
This will result in a total energy interval at room temperature:
∆ε = 0.11 eV
(c) Using your answers to part (b), find the fraction of “current carrying” electrons N1 over
N1
the total number of electrons in the conduction band Ntotal (i.e. Ntotal
).
∆ε
First, find the number of “current carrying” electrons N1 in the energy window εf ± 2
,
using ∆ε = 0.11 eV and εf = 7 eV :
εf + ∆ε
2
R
N1 = n̄(ε)g(ε) dε
εf − ∆ε
2
For a small energy window, we can assume that g(ε) is independent of energy ε (i.e. g(εf ±
∆ε
2
) = 32 εNf , and n̄ varies linearly with energy, i.e. n̄(εf ± ∆ε
2
) = 12 − 4kεB T (using only the
linear term in the series expansion of the Fermi-Dirac distribution).
Using these approximations, we can find for N1 :
εf + ∆ε
2
∆ε
2
3N R 1 ε

3N 1
n̄(ε)g(ε) dε ≈ −
R
N1 = 2εf 2 4kB T
dε = 2εf 2
∆ε
εf − ∆ε
2
− ∆ε
2
N1
So, the fraction of charges contributing to the current N
is now given by:
N1 3N 3∆ε
N
= 3N εf
∆ε = 4εf
N1
N
= 0.012
This means that only 1.2% of all the electrons in the conduction band contribute to the
current in Cu at 300K.
7
4. The Ideal Gas
(a) Determine the kind of compression that took place here. Explain!
This is an adiabatic compression with ∆Q = 0.
(b) Find the temperature T2 and the volume V2 in terms of V1 , p2 and T1 after the thermo-
dynamic equilibrium has been reached.
For adiabatic processes
∆U = ∆Q − pdV
Combining these two equations, we will get:
CV (T2 − T1 ) = −p2 (V2 − V1 )
We also know that CV = ∂U

∂T
or ∆U = CV ∆T . The equation of state for an ideal gas is given
by pV = N kB T and CP − CV = N kB .
This will lead to:
(T2 − T1 ) = − CpV2 (V2 − V1 ) ⇔ T2 = − CpV2 (V2 − V1 ) + T1
N kB T2
Now use p2 V2 = N kB T2 ⇔ V2 = p2
:

T2 = − CpV2 N kB T2
p2
− V1 + T1
This can now be rearranged to:
T2 = − p2 V1 +C
Cp
V T1
8
(c) Find the final values of the temperature Tf and volume Vf in terms of p1 , V2 , and T2
after the thermodynamic equilibrium is reached again.
Rearranging the solution to part (a) will lead to:
N kB Tf
Tf = − CV T2C+p
p
1 V2
and Vf = p1
Using CV ∆T + p∆V = 0 for the two parts of the experiments, we will find:
CV (T2 − T1 ) + p2 (V2 − V1 ) = 0
CV (Tf − T2 ) + p1 (V1 − V2 ) = 0
⇔ CV (Tf − T1 ) + p2 (V2 − V1 ) + p1 (Vf − V2 ) = 0
using pV = N kB T
∆(pV ) = p∆V + V ∆p = N kB ∆T
Using CV ∆T + p∆V = 0:

p∆V
p∆V + V ∆p = N kB CV
Using CV + N kB T = CP , we can find:
CP p∆V + CV V ∆p = 0
Using this expression for the two experiments:
CP p2 (V2 − V1 ) + CV V2 (p2 − p1 ) = 0
CP p1 (Vf − V2 ) + CV Vf (pf − p2 ) = 0
⇔ p2 (V2 − V1 ) + p1 (Vf − V2 ) = CCVP [V2 (p1 − p2 ) + Vf (p2 − pf )]
Using pf = p1 and Vf = V1 , we can plug this into the equation above and find:
CV (Tf − T1 ) = CCVP [V2 (p1 − p2 ) − V1 (p1 − p2 )]

Tf − T1 = C1P (V2 − V1 ) (p1 − p2 )
Eliminating (V2 − V1 ), we can finally find:
Tf − T1 = CV
(CP )2
(p1 − p2 )2 V2
P2
9
This result shows that the temperature difference Tf − T1 will always be positive, as required
by the thermodynamic principles. Also, it appears that the temperature difference is only a
2nd -order correction term, which will become very small for ∆P → 0.
10
5. Einstein Solid
(a) For the most probable macrostate, determine the number of units of heat energy in each
of the two systems. Explain your approach.
The most probable state will have the highest multiplicity Ω:
!
q+N −1 (q+N −1)!
Ω= = q!(N −1)!
q
and
Ωtotal = ΩA · ΩB
For NA = 4 and NB = 6, we can assemble the following table:
qA ΩA qB ΩB Ωtotal
8 165 12 6188 1.021 × 106
7 120 13 8568 1.028 × 106
6 84 14 11628 0.98 × 106
So, qA = 7 and QB = 13 will be the most probable state.
(b) Determine the probability of finding the system in the most probable macrostate.
To determine the probability of a macrostate, we need to calculate:
ΩA ΩB
P = Ωtotal
using
! !
q + Ntotal − 1 20 + 10 − 1
Ωtotal = = = 10.02 × 107
q 20
ΩA ΩB
⇔P = Ωtotal
= 0.1
11
SOLUTIONS
Thermodynamics and Statistical Mechanics

January 7, 2011
9:00 AM to 12:00 Noon
1
Equation Sheet
"
$
# exp[!bx
2
] dx =
!"
b
"
1 $
#x
2
exp[!bx 2 ] dx =
0
4 b3
"
#x
2
exp[!x] dx = 2
0
"
x $2
# dx ex ! 1
=
6
0
!
x #2
" dx ex + 1
=
12
0
"
x2
# dx x
e !1
= 2 $ (3), where $ (3) can be considered to be just a number.
0
!
x2
" dx x
e +1
= 3 # (3)
2
0
1
n=
e(! " µ )/ kT ± 1
sinh(x) = 1 (e x ! e! x )
2
cosh(x) = 1 (e x + e! x )
2
d
!sinh(x) #$ = cosh(x)
dx "
d
!cosh(x) #$ = sinh(x)
dx "
!
1
" xm = 1 # x
m=0
n
1 " x n+1
! x = 1" xm
m=0
2
Problem 1
A gas of N identical classical non-interacting atoms is held in a potential V (r) = ar , where
( )
1/ 2
r = x2 + y2 + z2 . The gas is in thermal equilibrium at temperature T.
(a) Find the single particle partition function Z1 of an atom in the gas. Express your answer
in the form Z1 = AT ! a "# and provide an expression for the prefactor A and the
exponents ! and ! . [Hint: convert the integral in r to spherical coordinates.]
Solution:
Z1 = " exp[! E / kT ] where E = p 2 / 2m + V (r).
1$ '$ # '$ # '

#
" exp[! px2 / 2mkT ]dpx ) & " exp[! p y / 2mkT ]dp y ) & " exp[! pz2 / 2mkT ]dpz )
2
Z1 = &
h3 &% !#
)( &%
!#
)( &%
!#
)(
, 2, #
* " d+ " d- " dr exp[!ar / kT ]r
2
sin +
0 0 0
Using equation sheet values for integrals leads to
3/ 2 3 3/ 2
$ 2, mkT ' $ kT ' $ 2, mk '
Z1 = &
% h2 )(
( )
4, 2 & ) = 8, k 3 &
% a ( % h2 )(
T 9/ 2 a !3 = AT . a !/
(b) Find an expression for the entropy S of this classical gas.
Solution:
1 N # "F &
Use F = !kT ln Z with Z = Z1 , then S = ! %
N! $ "T (' N ,V
F = !kT )* ! ln N ! + N ln Z1 +, - !kT )* ! N ln N + N + N ln Z1 +,
# " ln Z1 & # " ln Z1 & 9
S = k )* ! N ln N + N + N ln Z1 +, + kTN % ( , where % ( =
$ "T ' N ,V $ "T ' N ,V 2T
) 11 Z +
S = kN . + ln 1 /
*2 N,
3
Problem 2
A classical ideal gas is taken from state a to state b in

the figure using three different paths: acb, adb, and
ab. The pressure p2 = 2 p1 and the volume V2 = 2
V1.
(a) The heat capacity CV = 52 Nk . Starting from the First Law of Thermodynamics derive a
value for C p . No credit will be given for this part if you just state the answer.
Solution:
" !Q %
Cp = $ . From the first Law dU = dQ ( pdV, let dQ = dU + pdV.
# !T '& p
" !U % " !U %
Write dU in terms of V and T as dU = $ ' dT + $
# !V '& T
dV.
# !T & V
" !U %
The internal energy of an ideal gas depends only on temperature, so $ = 0.
# !V '& T
" !U %
Combining these results leads to dQ = $ dT + pdV = CV dT + pdV.
# !T '& V
" !Q % " !V %
Therefore, C p = $ ' = CV + p $ = CV + Nk, the latter is obtained from pV = NkT .
# !T & p # !T '& p
So, C p = 27 Nk.
(b) Compute the heat supplied to the gas along each of the three paths, acb, adb, and ab, in
terms of N, k, and T1.
Solution:
Paths acb, and adb consist of constant pressure and constant volume processes.
c b p2 V 2
V1 p
Q(acb) = ! CV dT + ! C p dT = ! 5
2
Nk dp + ! 27 Nk 2 dV
a c p1
Nk V
Nk
1
V1 p V 2p
= 52 Nk ( p2 " p1 ) + 27 Nk 2 (V2 " V1 ) = 52 Nk 1 ( p1 ) + 27 Nk 1 (V1 ) = 192 NkT1
Nk Nk Nk Nk
4
d b V2 p2
p V
Q(adb) = ! C p dT + ! CV dT + = ! 7
2
Nk 1 dV + ! 52 Nk Nk2 dp
a d V1
Nk p1
p1 V p 2V
= 27 Nk (V2 " V1 ) + 52 Nk 2 ( p2 " p1 ) = 27 Nk 1 (V1 ) + 52 Nk 1 ( p1 ) = 172 NkT1
Nk Nk Nk Nk
The heat along path ab can be calculated by taking the difference between !U and W .
!U between states a and b can be calculated along any path because it is a state function.
W along path ab is also easy to calculate.
First, calculate !U = W + Q for any path. We already have Q(adb), so
d b V2
W (adb) = " # p dV " # p dV = " # p1 dV = " p1V1 = "NkT1
a d V1
!U = W (adb) + Q(adb) = 17
2
NkT1 " NkT1 = 15
2
NkT1
b V2
Now, for W (ab) = " # p dV = " # p1
V
V1
p
( )
3p V
dV = " 1 V22 " V12 = " 1 1 = " 23 NkT1
2V1 2
a V1
Q(ab) = !U " W (ab) = 15

2 ( )
NkT1 " " 23 NkT1 = 9NkT1
(c) What is the heat capacity Cab of the gas for the process ab?
Solution:
! dQ $ ! 'U $ ! 'U $
Cab = # & , so consider dQ = dU + pdV = # & dT + # dV + pdV = CV dT + pdV
" dT % ab " 'T % V " 'V &% T
! dV $
(Cab = CV + p #
" dT &% ab
! dV $ p NkT NkTV1
Derive an expression for # & using p = 1 V and p = )V2 =
" dT % ab V1 V p1
NkV1 ! dV $ NkV1
differentiate: 2VdV = dT or # &% =
p1 " dT ab 2Vp1
NkV1 Nk
(Cab = CV + p = CV + = 3Nk
2Vp1 2
5
Problem 3
Consider a one-dimensional stretched elastic string that is fixed at its two ends and vibrates
only in a direction perpendicular to its length. The string consists of a very large number N
of atoms arranged in a single row. Let the energies of vibration be quantized in units of hf,
where f is the vibration frequency. This string is in thermal equilibrium with a heat bath at
temperature T.
(a) Determine an expression for the thermal energy of this string in terms of an integral over
the variable x = ! / kT .
Solution:
1
The Planck distribution n =
e! / kT " 1
1
(note that equation sheet has general expression n = ).
( ! " µ )/ kT
e ±1
!
U = 2# , where the 2 is from the two polarizations of vibrations and the energy levels
! / kT
n e "1
hcs nhcs
are given by ! = hf = = , L is the length of the string and cs is the speed of sound of
" 2L
the vibrations.
N
!
For N atoms in this string, we can write U = 2 # dn .
0 e! / kT " 1
hcs N
Let’s rewrite this integral in terms of x = ! / kT with the limit xmax = .
2LkT
x
! 2L $ max
x
U = 2# & ( )
kT
2
( dx
" hcs % 0 ex ' 1
(b) Identify a characteristic temperature that separates low T and high T behavior.
Determine an expression for the thermal energy of this string in the limit of low and high
T. Comment on these results in the context of the equipartition theorem.
Solution:
T hcs N
Rewrite xmax in terms of a characteristic temperature as xmax = = D , where TD is a
2LkT T
characteristic (Debye) temperature for 1-d vibrations. Now, we can rewrite the energy integral
6
TD /T
2NkT 2 x
as U =
TD " dx
ex ! 1
.
0
At high T, we approximate e x ! 1 + x , so that

TD /T
2NkT 2 2NkT 2 # TD &
U!
TD " dx =
TD %$ T ('
= 2NkT . This agrees with the equipartition result that
0
assigns an average thermal energy of 1 kT to each quadratic degree of freedom. The number of
2
vibrational modes for each polarization of a 1-dimensional string of atoms approaches N for
large N. Each mode has 2 degrees of freedom (one for the kinetic energy and one for the
potential energy), plus there are two polarizations.
"
$2
x
At low T, we let TD / T ! " . The integral # dx = (from equation sheet), so that
x 6
0 e ! 1
" 2 NkT 2
U! . Equipartition theorem does not say anything about this low T result.
3TD
7
Problem 4
Consider a spherical drop of liquid water containing Nl molecules surrounded by N - Nl

molecules of water vapor. The drop and its vapor may be out of equilibrium.
(a) Neglecting surface effects write an expression for the Gibbs free energy of this system if
the chemical potential of liquid water in the drop is µl and the chemical potential of water
in the vapor is µv. Rewrite Nl in terms of the (constant) volume per molecule in the
liquid, vl, and the radius r of the drop.
4" r 3
Solution: G = µl N l + µv (N ! N l ) = N µv + N l ( µl ! µv ) = N µv + ( µl ! µ v )
3vl
(b) The effect of the surface of the drop can be included by adding a piece Gsurface = !A to the
free energy, where ! is the surface tension (! > 0) and A is the surface area of the drop.
Write Gtotal with the surface piece expressed in terms of r. Make two qualitative hand-
sketches of Gtotal: one sketch with ( µl ! µv ) > 0 and one sketch with ( µl ! µv ) < 0.
Describe the behavior of the drop in these two cases.
4! r 3
Solution: Gtotal = N µv + ( µl " µv ) + 4!# r 2
3vl
G
For ( µl ! µv ) > 0, the drop radius is
always zero, so the drop just evaporates.
400
300
200
100
r
0 1 2 3 4 5 6
G
For ( µl ! µv ) < 0, the drop radius is
either zero or the drop increases in size 20
depending upon whether or not the
radius is larger than a critical radius. 10
r
1 2 3 4 5 6
- 10
- 20
8
(c) Under appropriate conditions, there is a critical radius, rc, that separates drops which
grow in size from those that shrink. Determine this critical radius.
Solution:
We saw in part (b) that for ( µl ! µv ) < 0 the drop either evaporates completely or grows.
The critical radius separates these two types of behaviors and can be determined from the
maximum in G.
dG 4" rc2
=! ( µv ! µl ) + 8"# rc = 0
dr rc vl
2# vl
rc =
µ v ! µl
(d) Assume that the vapor behaves as an ideal gas and recall that the chemical potential of an
ideal gas is given by µv = µvo + kT ln( p / p o ) . Write the chemical potential difference
( µv ! µl ) in terms of the vapor pressure and a reference pressure p o , where p o is taken
to be the pressure of a vapor in equilibrium with a large flat surface of water. Then,
derive and comment on the dependence of the relative humidity p / p o on rc .
Solution:
If po is the pressure of a vapor in equilibrium with a large flat surface of water, then the
reference chemical potential is just the chemical potential of the bulk liquid, µvo = µl .
2! vl
Therefore, µv ! µl = kT ln( p / p o ) and rc = .
kT ln( p / p o )
So, p / p o = exp(2! vl / kTrc ) . Smaller critical radii require higher relative humidity, so
drops grow more easily in a more humid environment.
9
Problem 5
Consider a paramagnetic material whose magnetic particles have angular momentum J,

which is a multiple of !. The projections of the angular momentum along the z-axis can take
2J – 1 values ( J z = ! J ,! J + 1,! J + 2,..., J ), which leads to 2J – 1 allowed values of the z –
component of a particle’s magnetic moment ( µ z = ! J " µ ,!(J + 1)" µ ,..., J " µ ). The energy of
the magnetic moment in a magnetic field pointing in the +z direction is ! µ z B .
(a) Derive an expression for the partition function Z1 of a single magnetic particle in a
magnetic field B pointing in the +z direction. Write your answer in terms of hyperbolic
sin functions, where sinh(x) = 1 (e x ! e! x ) . You may find it convenient to use the
2
variable b = ! µ B" , where ! = 1 .
kT
Solution:
The allowed energies are E = J ! µ B,(J + 1)! µ B,...," J ! µ B .

The partition function is
! J " µ B# !( J +1)" µ B# !( J +2)" µ B# J " µ B#
Z1 = e +e +e + ... + e , where # = 1
kT
Let b = " µ B# ,
Z1 = e! Jb + e!( J +1)b + e!( J +2)b + ... + e Jb
= e! Jb $&1 + eb + e2b + ... + e2Jb ')

% (
$
( ) ( )
2 2J '
= e! Jb &1 + eb + eb + ... + eb )
&% )(
$
( )
2J +1 '
b
&1 ! e ) e! Jb ! e( J +1)b
! Jb & )=
=e
& 1 ! eb ) 1 ! eb
&% )(
To write this in terms of sinh functions,
e! Jb ! e( J +1)b " e!b/2 % e!( J +1/2)b ! e( J +1/2)b sinh ()(J + 1 / 2)b *+
Z1 = $ !b/2 ' = =
1! e b $
#e '
& e !b/2
!e b/2 sinh(b / 2)
10
(b) Derive an expression for the average energy of the particle in part (a). Write your
cosh(x)
answer in terms of the hyperbolic cotangent function coth(x) = .
sinh(x)
Solution:
1 dZ 1 db dZ
E=! =!
Z d" Z d " db
( ) ( ) ( )
J + 1 sinh b cosh #( b J + 1 %) ! 1 sinh #( b J + 1 %) cosh b
=!
sinh(b / 2)
( )
'µ B
2 2 $ 2 & 2 $ 2 & 2
sinh #$(J + 1 / 2)b %&
( )
2
sinh b
2
(
($ 2 ) ($ ( )
= !' µ B # J + 1 coth # b J + 1 % ! 1 coth b %
2 )& 2 2 )&
(c) Use the expression for the average energy in part (b) to determine the magnetization M
(the average z-component of the total magnetic moment) of a system of N identical,
independent magnetic particles. Comment on its behavior as T ! 0 .
Solution:
The average z-component of the magnetic moment of a single particle is just the average
energy times -1/B. The total M will be the average energy times –N/B.
(
$# 2 ) $# ( 2 &' 2)
M = N ! µ " J + 1 coth " b J + 1 % ( 1 coth b %
2 &'
As T ! 0 , b ! " , and the coth functions go to 1, so M = NJ ! µ . This is equivalent to

having all N particles in their lowest energy magnetic state with µ z = J ! µ .
11

University of Illinois at Chicago Department of Physics: Classical Mechanics Qualifying Examination

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University of Illinois at Chicago Department of Physics: Classical Mechanics Qualifying Examination

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University of Illinois at Chicago

Full credit can be achieved from completely correct answers to

Full credit can be achieved from completely correct answers

a. Draw the free body diagram.

a. Calculate the v/c of the positron.

a. Solve for the normal modes of the pendulums.

a. Determine the Lagrangian in terms of r,  and azimuthal angle

Full credit can be achieved from completely correct answers to 4

a) Expressing the rotational inertia of the non-uniform

a) the equation of motion of the mass m;

a) Write down the inertia tensor for the system in the

Classical Mechanics Qualifying Exam Solutions

a) Expressing the rotational inertia of the non-uniform

Mgh = ½ Mv2 + ½ Iω2

c) For an arbitrary value of n:

Solution: Calculating the potential energy

a) the equation of motion of the mass m;

x&= −aω sin ω t + bθ&cosθ

The velocity then can be written

b) The equation of motion:

Full credit can be achieved from completely correct answers to 4

Tuesday, January 5, 2010

Full credit can be achieved from completely correct answers to 4

(a) the linear acceleration of the center of mass of the pulley M2

(b) the angular acceleration of the pulley M2

(c) the angular acceleration of the pulley M1

(d) the tension in the string

(a) Write the Lagrangian of the system.

(a) Write the Lagrangian of the system.

(c) Find the normal frequencies of the system.

Full credit can be achieved from completely correct answers to 4

4 2 & 2 ml 2 & 1 mgl

The Lagrange’s equations are then given by

Assuming small oscillations with θ = A cos ωt and ϕ = B cos ωt gives

a) The equation of constraint is f ( r ,θ ) = r − a = 0

mga sin θ − ma 2θ&& = 0

b) and when λ → 0 particle falls off hemisphere at

In cylindrical coordinates the kinetic energy and the potential

The radius r0 of the stationary circular motion is given by

Full credit can be achieved from completely correct answers to 4

A pointlike mass m is undergoing a three-dimensional projectile motion in a uniform

a) Write the Hamiltonian function for this problem H = H ( pi , qi ) , choosing as generalized

and letting xi = ai e iω t one obtains

e) From part (b):

Electricity & Magnetism

Full credit can be achieved from completely correct answers to

a) Give the potential at the point P~ = (ρ0 , ϕ, z) in terms of λ, R, ρ0 , ϕ, and z.

~ along the z-axis at ~r = (0, 0, h).

B. Integrals and Series

where K is the complete elliptic integral of the first kind.

  0, 0, d? Your final answer should contain no

  0, 0, d, we can compute the potential simply via

We can now compute the potential difference

 ∑A n r n  B n r −n1  2l2l,n1  2 A a l  B l a −l1 

And thus we have as the boundary conditions

V−1 l I l  b l 2l  1 VI l  B l b −l1 − b l B l a −l1

b) Compute the magnetic induction, B , along the z-axis at r  0, 0, h.

Integrals and Series

r,   ∑A n r n  B n r −n1 P n cos 