Problems and solutions statistical physics 1

MASSACHUSETTS INSTITUTE OF TECHNOLOGY

Physics Department

8.044 Statistical Physics I Spring Term 2004

Problem Set #1
Due by 1:10 PM, Monday, February 9
Problem 1: A Continuous Random Variable, a Harmonic Oscillator
Take a pencil about 1/3 of the way along its length and insert it between your index
and middle fingers, between the first and second knuckles from the end. By moving
those fingers up and down in opposition you should be able to set the pencil into
rapid oscillation between two extreme angles. Hold your hand at arms length and
observe the visual effect. We will examine this effect.
Consider a particle undergoing simple harmonic motion, x = x0 sin(ωt + φ), where
the phase φ is completely unknown. The amount of time this particle spends between
x and x + dx is inversely proportional to the magnitude of its velocity (its speed) at
x. If one thinks in terms of an ensemble of similarly prepared oscillators, one comes
to the conclusion that the probability density for finding an oscillator at x, p(x), is
proportional to the time a given oscillator spends near x.
a) Find the speed at x as a function of x, ω, and the fixed maximum displacement
x0.
b) Find p(x). [Hint: Use normalization to find the constant of proportionality.]
c) Sketch p(x). What are the most probable values of x? What is the least
probable? What is the mean (no computation!)? Are these results consistent
with the visual effect you saw with the oscillating pencil?
Problem 2: A Discrete Random Variable, Quantized Angular Momentum
In a certain quantum mechanical system the x component of the angular momentum,
Lx, is quantized and can take on only the three values −¯ h. For a given stateh, 0, or ¯
1
of the system it is known that < Lx >= 3
h and < L2
>= 2
h2
. [¯¯ ¯ h is a constant withx 3
units of angular momentum. No knowledge of quantum mechanics is necessary to do
this problem.]
a) Find the probability density for the x component of the angular momentum,
p(Lx). Sketch the result.
b) Draw a carefully labeled sketch of the cumulative function, P(Lx).
1

Problem 3: A Mixed Random Variable, Electron Energy
The probability density p(E) for ﬁnding an electron with energy E in a certain situ-
ation is
p(E) = 0.2 δ(E + E0) E < 0
= 0.8(
1
b
)e−E/b
E > 0
where E0 = 1.5 ev and b = 1.0 ev.
p(E)
E(ev)
a) What is the probability that E is greater than 1.0 ev?
b) What is the mean energy of the electron?
c) Find and sketch the cumulative function P(E).
Problem 4: A Time Dependent Probability, a Quantum Mechanics Example
In quantum mechanics, the probability density for ﬁnding a particle at a position r at
time t is given by the squared magnitude of the time dependent wavefunction Ψ(r, t):
p(r, t) = |Ψ(r, t) 2
= Ψ∗
(r, t)Ψ(r, t).|
Consider a particle moving in one dimension and having the wavefunction given be-
low [yes, it corresponds to an actual system; no, it is not indicative of the simple
wavefunctions you will encounter in 8.04].
iωt i 2
Ψ(r, t) = (2πx2
−
2
−
2x0
)2
0)−1/4
exp
(2x0)2
(2αxx0 sin ωt − α2
x0 sin 2ωt) − (
x − 2αx0 cos ωt
x0 is a characteristic distance and α is a dimensionless constant.
2

a) Find the expression for p(x, t).

b) Find expressions for the mean and the variance [Think; don’t calculate].

c) Explain in a few words the behavior of p(x, t). Sketch p(x, t) at t = 0,
1
4
T,
1
2
T,
3
4
T,
and T where T ≡ 2π/ω.
Problem 5: Bose-Einstein Statistics
You learned in 8.03 that the electro-magnetic ﬁeld in a cavity can be decomposed (a
3 dimensional Fourier series) into a countably inﬁnite number of modes, each with its
own wavevector k and polarization direction . You will learn in quantum mechanics
that the energy in each mode is quantized in units of ¯hω where ω = c k . Each unit| |
of energy is called a photon and one says that there are n photons in a given mode.
Later in the course we will be able to derive the result that, in thermal equilibrium,
the probability that a given mode will have n photons is
p(n) = (1 − a)an
n = 0, 1, 2, · · ·
where a < 1 is a dimensionless constant which depends on ω and the temperature T.
This is called a Bose-Einstein density by physicists; mathematicians, who recognize
that it is applicable to other situations as well, refer to it as a geometric density.
a) Find < n >. [Hint: Take the derivative of the normalization sum, an
, with
respect to a.]
b) Find the variance and express your result in terms of < n >. [Hint: Now take
the derivative of the sum involved in computing the mean, nan
.] For a given
mean, the Bose-Einstein density has a variance which is larger than that of the
Poisson by a factor. What is that factor?
c) Express p(x) as an envelope function times a train of δ functions of unit area
located at the non-negative integers. Show that the envelope decreases expo-
nentially, that is, as e−x/φ
. Express φ in terms of < n > and show that in the
limit of large < n >, φ < n >.→
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Physics Department

Problem Set #2
Due by 1:10 PM: Tuesday, February 17
Problem 1: Two Quantum Particles
x
A possible wavefunction for two particles (1 and 2) moving in one dimension is
1 + x2
2
ψ(x1, x2) =
1
(
x2 − x1
) exp[−
2
2
]
πx2 x0 2x00
where x0 is a characteristic distance.
a) What is the joint probability density p(x1, x2) for ﬁnding particle 1 at x = x1
and particle 2 at x = x2? On a sketch of the x1, x2 plane indicate where the
joint probability is highest and where it is a minimum.
b) Find the probability density for ﬁnding particle 1 at the position x along the line.
Do the same for particle 2. Are the positions of the two particles statistically
independent?
c) Find the conditional probability density p(x1 x2) that particle 1 is at x = x1|
given that particle 2 is known to be at x = x2. Sketch the result.
In a case such as this one says that the motion of the two particles is correlated. One
expects such correlation if there is an interaction between the particles. A surprising
consequence of quantum mechanics is that the particle motion can be correlated even
if there is no physical interaction between the particles.
Problem 2: A Joint Density of Limited Extent
The joint probability density p(x, y) for two random variables x and y is given below.
p(x,y)
6

 0 ≤ x
p(x, y) = 6(1 − x − y) if 0 ≤ y

x + y ≤ 1
y
1
= 0 othewise
1
x
1

a) Find the probability density p(x) for the random variable x alone. Sketch the
result.
b) Find the conditional probability density p(y x). Sketch the result.|
Problem 3: A Discrete Joint Density
A certain system can be completely described by two integers n and l. n = 0, 1, 2, · · ·
but the values of l are constrained by n to the 2n+1 integers −n ≤ l ≤ n. The ﬁgure
below indicates the possibilities up to n = 5.
0 2 4 66 4 2
2
4
n
l
The joint probability density for n and l is
p(n, l) = c e−an
n ≥ 0 and |l| ≤ n
= 0 otherwise
Here a is a parameter with the constraint that e−a
< 1 and c is a normalization
constant with depends on a.
a) Find p(n) and sketch.

b) Find p(l n) and sketch.
|
c) Find p(l) and sketch. [Hint: You will need to sum a geometric series which does
not begin with n = 0; do this by factoring out an appropriate term to reduce
the series to conventional form. Don’t forget that l can be negative.]
d) Find p(n l) and sketch.|
2

Problem 4: Distance to the Nearest Star
Assume that the stars in a certain region of the galaxy are distributed at random
with a mean density ρ stars per (light year)3
. Find the probability density p(r) that
a given star’s nearest neighbor occurs at a radial distance r away. [Hint: This is the
exact analogue of the waiting time problem for radioactive decay, except that the
space is 3 dimensional rather than 1 dimensional.]
Problem 5: Shot Noise
Shot noise is a type of quantization noise. It is the noise that arises when one tries
to represent a continuous variable by one that can take on only discrete values. It
differs from other quantization situations such as an analogue to digital converter. In
the A-D converter the output is uniquely related to the input (but not visa-versa). In
the case of shot noise the output is a random variable; it is the mean of the random
variable that is directly proportional to the variable being represented.
One example of shot noise is the measurement of the intensity of a weak light source
using a photomultiplier tube. The light causes photoelectrons to be ejected from the
photocathode of the tube. Each photoelectron initiates a cascade of electrons down
the dynode chain which results in a well resolved pulse of electrons at the anode of
the tube. The pulse may be 5 nanoseconds wide (successive pulses are easily resolved
except at very high light levels) and contain 106
electrons. For a fixed light intensity
these pulses arrive at random with an average rate rs counts/second.
IA
rs = η
hν
where I is the intensity of the light (power per unit area) at the phototube, A the
photocathode area, ν the frequency of the light, and h Planck’s constant. η is the
quantum efficiency of the photocathode (number of photoelectrons per photon) which
for a good tube can be 20%. From these facts one expects that the the number of
counts detected in a counting interval T will have a Poisson density with mean rsT.
In the phototube example, the intensity of light controls the photocurrent, which is
quantized in units of about 106
electrons. In the scattering of protons from nuclei
the scattering cross-section controls the angular dependent flux of scattered particles,
which is quantized in units of one proton. Two more examples can be found in the
field of electronics. In a vacuum tube, the voltage on the grid controls the plate
current which is quantized in units of one electron. In a field effect transistor, the
gate voltage controls the source to drain current, which is again quantized in units of
one electron.
3

1
0.8
0.6
0.4
0.2
5 10 15 20
The physical consequences of this type of quantization can be illustrated by imagining
that our phototube is used as a detector for an optical spectrometer which is swept
across an atomic line. The intrinsic line shape is shown above. It has a half width at
half height of one unit. The intensity of the light is measured at frequencies separated
by one-tenth of a unit (20 increments per full width of the line). The mean number of
photo-counts in each increment can be increased by increasing the overall intensity of
the light or by increasing the amount of counting time per increment, T. The traces
below represent spectra measured with diﬀering mean counting rates at the peak of
the line.
12 35
10 30
8
25
20
6
15
4
10
2 5
20 5 10 15 205 10 15
300
100
250
80
200
60
150
40
100
20 50
5 10 15 20 0 5 10 15 20
4

Observe the unusual characteristics of this type of ”noise”. The fluctuations are in the
measured signal itself. No signal; no fluctuations. Small signal; small fluctuations.
Large signal; large fluctuations. But the mean of the signal increases faster than the
magnitude of the fluctuations so that the larger the signal, the easier it is to estimate
its true strength. Note that in the absence of a background source of counts, even a
single measured count is sufficient to establish that a signal is indeed present. A low
counting rate only makes the determination of the exact location and width of the
line difficult.
The situation is changed in a fundamental way when there is a source of background
counts. They could arise from dark counts in the detector (in the case of the photo-
tube, thermal emission from the photocathode which takes place even in the absence
of light), or from another contribution to the spectrum of the light which is uniform
across frequency interval being studied. In this case the fluctuations in the measured
background signal can be sufficiently large to mask even the presence of the signal one
is searching for. The following traces demonstrate this effect by keeping the mean
counting rate from the atomic line at its peak at 100, while increasing the mean
counting rate due to the background.
160
140
120
100
80
60
40
20
250
200
150
100
50
0 5 10 15 20 0 5 10 15 20
500
400
300
200
100
1200
1000
800
600
400
200
0 5 10 15 20 0 5 10 15 20
5

Let us now examine shot noise quantitatively using the phototube as a concrete
example. Assume one wants to measure light of constant intensity I in the presence
of a background caused by dark current.
a) Find the mean number of counts to be expected in the time interval T due to
the light signal alone. We shall refer to this as the signal S.
b) If the measurement is repeated in a number of successive time intervals of the
same length, different numbers of counts could be recorded each time. What is
the standard deviation, σ ≡ (Variance)1/2
, of these measured numbers from the
mean. This is referred to as the shot noise N.
c) Find the signal to noise ratio S/N in the absence of background. How does it
depend on the intensity I and on the time T?
r
d) Now consider the case when the background is present. Assume that the dark
counts occur at random with a rate rd. The number of counts recorded in a
time T when I is on is compared with the number in an equal time interval
when I is off; the difference is the signal S which is proportional to I. We will
take the noise N to be the standard deviation in the counts during the interval
when the source is on (in actuality the fluctuations in the recorded counts when
the source is off also add to the uncertainty in determining the amplitude of
the signal). Find the signal to noise ratio for this experiment in the limit when
d rs. Now how does it depend on I and T?
6


Physics Department

Problem Set #3
Due by 1:10 PM: Monday, February 23
Problem 1: Energy in an Ideal Gas
For an ideal gas of classical non-interacting atoms in thermal equilibrium the cartesian
components of the velocity are statistically independent. In three dimensions
2 2 2
+ v + vv
p(vx, vy, vz) = (2πσ2
)−3/2
exp[− x y z
]

2σ2
1
where σ2 2
= kT/m. The energy of a given atom is E |v|= m .2
a) Find the probability density for the energy of an atom in the three dimensional
gas, p(E).
It is possible to create an ideal two dimensional gas. For example, noble gas atoms
might be physically adsorbed on a microscopically smooth substrate; they would be
bound in the perpendicular direction but free to move parallel to the surface.
b) Find p(E) for the two dimensional gas.
Problem 2: Distance to the Center of the Galaxy
A few years ago the radio astronomer Dr. James Moran and his colleagues at the
Harvard-Smithsonian Center for Astrophysics developed a new method for estimating
the distance from the Earth to the center of the galaxy, R0. Their method involves
the observation of water vapor masers, small highly localized clouds of interstellar
water vapor whose molecules emit radiation in a cooperative fashion in an extremely
narrow and stable spectral line (the frequency is 22.3 Gigahertz which corresponds
to a wavelength of 1.3 cm.). A group of such masers is located in a region known to
be very close to the galactic center. The CfA group made extensive measurements
on about 20 of these. From the Doppler shifts of the lines they found the velocity of
each, v , along the line of sight from the earth to the galactic center. By following the
angular position in the sky over a period of 2 years they found the angular velocity,
θ˙⊥, of each object perpendicular to the line of sight.
1

a) The CfA group assumed that the velocities of the radiating clouds were dis-
tributed in a random but isotropic fashion in space, that is, p(v) had spherical
symmetry. Under these conditions the standard deviation of the probability
density for v should be R0 times the standard deviation for θ˙ . The measure-
ments gave σv = 20 kilometers/second and σθ˙ = 9.1 ×10−17
⊥
radians/second.⊥
What is R0 in kilometers and kiloparsecs? [1 parsec = 3.09 × 1013
kilometers
and a rough estimate of the diameter of our entire galaxy is 25 kiloparsecs.]
b) Assume a model in which the observed radiating clouds are randomly dis-
tributed on a spherical shell of radius r. The center of the sphere might be
a newly formed star which spawned the clouds during its birth. Assume also
that their velocities are due only to the expansion of the radius of the shell:
v = ˙rrˆ where ˙r is a constant and ˆr is a unit vector in the radial direction. Use
functions of a random variable to find the probability density for v that would
result.
Problem 3: Acceleration of a Star
The acceleration a of a given star due to the gravitational attraction of its neighbors
is
a
=

∞
i=1
GMi
|ri|3
ri
M
where ri is the vector position of the ith
neighbor relative to the star in question and
i is its mass. The terms in the sum fall off rapidly with increasing distance. For
some purposes the entire sum can be approximated by its first term, that due to the
nearest neighbor. In this approximation
GM
a ≡ |a| = 2r
where r is the radial distance to the nearest neighbor and M is its mass.
a) Find an expression which relates p(a) to p(r). In what region of p(a) would you
expect the greatest error due to the neglect of distant neighbors?
b) Next assume that the stars are distributed at random in space with a uniform
density ρ. Use the results of problem 4 on problem set 2 to find p(a). It is known
that this assumption breaks down at short distances due to the occurrence of
gravitationally bound complexes (binary stars, etc.). In which region of p(a)
would you expect to see deviations from your calculated result due to this effect?
2

c) What other probabilistic aspect of the problem (i.e. what other random vari-
able) would have to be taken into account before our p(a) could be compared
with experimental observations?
Problem 4: Measuring an Atomic Velocity Profile
d
svx
Âtoms emerge from a source in a well collimated beam with velocities v = vxx directed
horizontally. They have fallen a distance s under the influence of gravity by the time
they hit a vertical target located a distance d from the source.
A
s =
v2
x
1
where A = 2
gd2
is a constant. An empirical fit to measurements at the target give
the following probability density px(ζ) for an atom striking the target at position s.
A3/2
1 A
ps(ζ) =
σ2
√
2πσ2 ζ5/2
exp[−
2σ2ζ
] ζ ≥ 0
= 0 ζ < 0
Find the probability density pvx (η) for the velocity at the source. Sketch the result.
3

Image removed for copyright reasons.

Problem 5: Planetary Nebulae
NGC 7293, the Helix Nebula in Aquarius Planetary Nebula Shapley 1
Pictures from the archive at Astronomy Picture of the Day
http://antwrp.gsfc.nasa.gov/apod/
During stellar evolution, a low mass star in the “red supergiant” phase (when fusion
has nearly run its course) may blow oﬀ a sizable fraction of its mass in the form of an
expanding spherical shell of hot gas. The shell continues to be excited by radiation
from the hot star at the center. This glowing shell is referred to by astronomers as a
planetary nebula (a historical misnomer) and often appears as a bright ring. We will
investigate how a shell becomes a ring. One could equally ask “What is the shadow
cast by a semitransparent balloon?”
A photograph of a planetary nebula records the amount of matter having a given
radial distance r⊥ from the line of sight joining the central star and the observer.
Assume that the gas atoms are uniformly distributed over a spherical shell of radius
R centered on the parent star. What is the probability density p(r⊥) that a given
atom will be located a perpendicular distance r⊥ from the line of sight? [Hint: If one
uses a spherical coordinate system (r, θ, φ) where θ = 0 indicates the line of sight,
then r = r sin θ. P(r⊥) corresponds to that fraction of the sphere with R sin θ < r⊥.]⊥
4


Physics Department

Problem Set #4
Due by 1:10 PM: Monday, March 1
Problem 1: Thermal Equilibrium and the Concept of Temperature, I
This problem and the next are taken from Heat and Thermodynamics by Zeman
-
sky.

Systems A, B, and C are gases with coordinates P, V ; P�
, V �
; and P��
, V ��
. When A

and C are in thermal equilibrium, the equation
PV − nbP − P��
V ��
= 0
is found to be satisﬁed. When B and C are in equilibrium, the relation
nB�
P��
V ��
P�
V �
− P��
V ��
+ = 0
V �
holds. The symbols n, b, and B�
represent constants.
a) What are the three functions which are equal to one another at thermal equi-
librium and each of which is equal to t where t is the empiric temperature?
b) What is the relation expressing thermal equilibrium between A and B?
Problem 2: Thermal Equilibrium and the Concept of Temperature, II
Systems A and B are paramagnetic salts with coordinates H, M and H�
, M�
respec-
tively. System C is a gas with coordinates P, V . When A and C are in thermal
equilibrium, the equation
nRCH − MPV = 0
is found to hold. When B and C are in thermal equilibrium, we get
nRΘM�
+ nRC�
H�
− M�
PV = 0
where n, R, C, C�
, and Θ are constants.
a) What are the three functions that are equal to one another at thermal equilib-
rium?
1

b) Set each of these functions equal to the ideal gas temperature T and see if you
recognize any of these equations of state.
Problem 3: Work in a Simple Solid
In the simplest model of an elastic solid
dV = −V KT dP + V αdT
where KT is the isothermal compressibility and α is the thermal expansion coeﬃcient.
Find the work done on the solid as it is taken between state (P1, T1) and (P2, T2) by
each of the three paths indicated in the sketch. Assume that the fractional volume
change is small enough that the function V (P, T) which enters the expression for dV
can be taken to be constant at V = V1 = V (P1, T1) during the process.
Problem 4: Work in a Non-Ideal Gas
An approximation to the equation of state for a real gas is
(P + a/V 2
)(V − b) = NkT
where a, b, and k are constants. Calculate the work necessary to compress the gas
isothermally from V1 to V2 < V1.
2

Problem 5: Work and the Radiation Field

The pressure P due to the thermal equilibrium radiation ﬁeld inside a cavity depends

only on the temperature T of the cavity and not on its volume V ,

1

P = σT4
.
3

In this expression σ is a constant. Find the work done on the radiation ﬁeld as the
cavity is taken between states (V1, T1) and (V2, T2) along the two paths shown in the
diagram.
3


Physics Department

Problem Set #5
Due by 1:10 PM: Monday, March 8
Practice Problem: Exact Differentials
Which of the following is an exact differential of a function S(x, y)? Find S where
possible.
a) 2x(x3
+ y3
)dx + 3y2
(x2
+ y2
)dy
2
S(x, y) = (2x5
+ 5x y3
+ 3y5
)/5 + C
b) ey
dx + x(ey
+ 1)dy
S(x, y) does not exist.
c) (y − x)ex
dx + (1 + ex
)dy
S(x, y) = y + (1 + y − x)ex
+ C
Problem 1: Equation of State for a Ferromagnet
For a ferromagnetic material in the absence of an applied field, H = 0, the spon-
taneous magnetization is a maximum at T = 0, decreases to zero at the critical
temperature T = Tc, and is zero for all T > Tc.
1

For temperatures just below Tc the magnetic susceptibility and the temperature co-
efficient of M might be modeled by the expressions
χ
∂M a
T ≡
∂H T
=
(1 − T/Tc)
+ 3bH2
∂M 1 f(H) 1 M0 1
=
∂T Tc (1 − T/Tc)2
−
2 Tc (1 − T/Tc)1/2
H
where M0, Tc, a, and b are constants and f(H) is a function of H alone with the
property that f(H = 0) = 0.
a) Find f(H) by using the fact that M is a state function.
b) Find M(H, T).
Problem 2: Heat Capacity at Constant Pressure in a Simple Fluid
For a simple fluid show that CP =(∂U/∂T)P +αV P. Since the thermal expansion
coefficient α can be either positive or negative, CP could be either less than or greater
than (∂U/∂T)P . [Hint: Use the first law to find an expression for dQ, then expand
in terms of the variables T and P.]
Problem 3: Heat Supplied to a Gas
5
An ideal gas for which CV = 2
Nk is taken from point a to point b in the figure along
three paths: acb, adb, and ab. Here P2 = 2P1 and V2 = 2V1. Assume that (∂U/∂V )T
= 0.
2

a) Compute the heat supplied to the gas (in terms of N, k, and T1) in each of the
three processes. [Hint: You may wish to find CP first.]
b) What is the “heat capacity” of the gas for the process ab?
Problem 4: Equation of State and Heat Capacity of a Liquid Surface
The surface tension S of a liquid is the work required to increase the free surface area
of the liquid by one unit of area.
For pure water in contact with air at normal pressure, the surface tension has a
constant value S0 at all temperatures for which the water is a liquid.
Certain surfactant molecules, such as pentadecylic acid, can be added to the water.
They remain on the free surface and alter the surface tension. For water of area A
containing N molecules, one can measure this effect. Experiments show:
∂S NkT 2a N 2
=
∂A (A − b)2
−
A AT
∂T A − b
=
∂S A
−
Nk
where k is Boltzmann’s constant and a and b are constants.
a) Find an expression for S(A, T) that reduces to the result for pure water when
N = 0.
Additional experiments determine the heat capacity at constant area CA and the
change in internal energy with area at constant temperature (∂U/∂A)T .
b) Find an expression for the heat capacity at constant surface tension, CS, in
terms of S, CA, (∂S/∂A)T , (∂T/∂S)A , and (∂U/∂A)T .
3

Problem 5: Thermodynamics of a Curie Law Paramagnet
Simple magnetic systems can be described by two independent variables. State vari-
ables of interest include the magnetic field H, the magnetization M, the temperature
T, and the internal energy U. Four quantities that are often measured experimentally
are
∂M
χT , the isothermal magnetic susceptibility,≡
∂H T
∂M
, the temperature coefficient,
∂T H
dQ
CM , the heat capacity at constant M, and≡
dT M
dQ
CH , the heat capacity at constant H.≡
dT H
A particular example of a simple magnetic system is the Curie law paramagnet defined
by an equation of state of the form M = aH/T where a is a constant.
C
For such a system one can show that (∂U/∂M)T = 0 and we shall assume that
M = bT where b is a constant.
a) Use T and M as independent variables and consider an arbitrary simple mag-
netic system (that is, not necessarily the Curie law paramagnet). Express CM
as a derivative of the internal energy. Find an expresson for CH − CM in terms
of a derivative of the internal energy, H, and the temperature coefficient. Write
4

an expression for dU(T, M) where the coefficients of the differentials dT and
dM are expressed in terms of measured quantities and H(T, M).
b) Find explicit expressions for CH(T, M) and U(T, M) for the Curie law param-
agnet. You may assume that U(T = 0, M = 0) = 0.
c) Consider again an arbitrary simple magnetic system, but now use H and M as
the independent variables. Write an experssion for dU(H, M) where the coef-
ficients of the differentials dH and dM are expressed in terms of the measured
quantities and H. Is the coefficient of the dM term the same as in part a)?
d) Find explicit expressons for the coefficients in c) in the case of the Curie law
paramagnet. You will need your result from b) for CH. Convert the coefficients
to functions of H and M, that is, eliminate T. Integrate dU(H, M) to find
U(H, M). Compare your result with that found in b).
e) Using T and M as the independent variables, find the general constraint on an
adiabatic change; that is, find (∂T/∂M)∆Q=0 in terms of a derivative of the
internal energy, H(M, T), and CM (M, T).
f) Evaluate (∂T/∂M)∆Q=0 for the Curie law paramagnet and integrate the result
to find the equation of an adiabatic path in the T, M plane through the point
T0, M0.
g) For the Curie law paramagnet, draw an isothermal path on a plot of M verses
H. Pick a point on that path; show that the slope of an adiabatic path going
through that point is less than the slope of the isothermal path.
5

17

Physics Department


Problem Set #6
Due by 1:10 PM: Monday, March
Problem 1: Classical Magnetic Moments
Consider a system made up of N independent classical magnitic dipole moments
located on fixed lattice sites. Each moment µi has the same length µ, but is free to
rotate in 3 dimensions. When a magnetic field of strength H is applied in the positive
z direction, the energy of the ith
moment is given by i = −miH where mi is the z
component of µi (that is, zˆ = mi).µi ·
The magentization M and the total energy E are given by
N N
M =
i=1
mi E =
i=1
i = −MH
a) What are the physically allowed ranges of values associated with mi, M, and
E?
b) How many microscopic variables are necessary to completely specify the state
of the system?
In a certain limiting case, the accessible volume in phase space for the microcanonical
ensemble is given by
M2
Ω ≈ (2µ)N
exp[−
3
Nµ2
].2
c) Use the microcanonical ensemble to find the equation of state, M as a function
of H and T.
d) Is there some condition under which the solution to c) is unphysical for the
system under consideraton? Explain your answer. For what values of T is the
expression for Ω a good approximation?
e) The probability density p(M) for the z component of a single magnetic moment
can be written as p(m) = Ω /Ω where Ω is given above. What is Ω ?
1
15

f) Find p(m). [Note: For the limit which applies here, an expression for p(m)
including powers of m no higher than the first is adequate.] Sketch p(m) and
check its normalization.
g) Use p(m) to computed < m >. Compare the result with that which one would
expect.
Problem 2: A Strange Chain
F
l
F
A one dimensional chain is made up of N identical elements, each of length l. The
angle between successive elements can be either 00
or 1800
, but there is no difference
in internal energy between these two possibilities. For the sake of counting, one can
think of each element as either pointing to the right (+) or to the left (-). Then one
has
N = n+ + n−
L = l(n+ − n−) = l(2n+ − N)
a) Use the microcanonical ensemble to find the entropy as a function of N and n+,
S(N, n+).
b) Find an expression for the tension in the chain as a function of T, N, and n+,
F(T, N, n+). Notice the strange fact that there is tension in the chain even
though there is no energy required to reorient two neighboring elements! The
“restoring force” in this problem is generated by entropy considerations alone.
This is not simply an academic oddity, however. This system is used as a model
for elastic polymers such as rubber.
c) Rearrange the expression from b) to give the length as a function of N, T, and
F.
2

d) Use the result for the high temperature behavior from c) to ﬁnd an expression
for the thermal expansion coeﬃcient α ≡ L−1
(∂L/∂T)F . Note the sign. Find
a stout rubber band. Hang a weight from it so that its length is extended by
about a factor of two. Now heat the rubber band (a hair drier works well here)
and see if the weight goes up or down.
Problem 3: Classical Harmonic Oscillators
Consider a collection of N indentical harmonic oscillators with negligable (but non-
zero) interactions. In a microcanonical ensemble with energy E, the system is on a
surface in phase space given by
N 2
mω2
q2
pi i
+ = E.
2m 2i=1
a) Find the volume of phase space enclosed, Φ(E), as follows. Transform to new
variables
1
xi = pi 1 ≤ i ≤ N√
2m
mω2
xi = N + 1 ≤ i ≤ 2N
2
qi−N
Note that in terms of these variables the constant energy surface is a 2N dimen-
sional sphere. Find its volume. Find the corresponding volume in p-q space.
b) Find the entropy S in terms of N and E.
c) Find T and express E in terms of N and T.
d) Find the joint probability density for the position coordinate qi and the mo-
mentum coordinate pi of one of the oscillators. Sketch p(pi, qi).
Problem 4: Quantum Harmonic Oscillators
Consider a system of N almost independent harmonic oscillators whose energy in a
microcanonical ensemble is given by
1
E = hωN + ¯¯ hωM.
2
3

a) Find the number of ways, Ω(E), that this energy can be obtained. Note that
N
M = ni
i=1
where ni is the occupation number (0, 1, 2, . . .) of a given harmonic oscillator.
Ω(E) can be looked upon as the number of ways of putting M indistinguishable
balls in N labelled bins. It is also the number of ways of arranging N − 1
indistinguishable partitions and M indistinguishable balls along a line. Looked
at this way, the problem reduces to ﬁnding the number of diﬀerent orderings of
the M + N − 1 items.
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��5+1��)6�45��!��!*1�5
�37)�6)��*)��5
�*)��5 ��8)*�-2)46161��5
b) Find the entropy S in terms of N and M.
c) Find T and express E in terms of N and T.
d) Find the probability that a given oscillator is in its nth
energy eigenstate. Sketch
p(n).
4


Physics Department

Problem Set #7
Due by 1:10 PM: Monday, April 5
Problem 1: Energy of a Film
A surface film has a surface tension S given by S(T, A) = −NkT/(A − b) where b is
a constant with the units of area.
a) Show that the constant area heat capacity CA is independent of area at fixed
temperature.
b) Show that the internal energy of the film is a function of temperature alone and
find an expression for the energy in terms of the heat capacity.
Problem 2: Bose-Einstein Gas
Later in the course we will show that a gas of atoms obeying Bose-Einstein statistics
undergoes a phase transition at a very low temperature to another gas phase with
quite different properties. [This transition was first observed in the laboratory in 1995
and received a great deal of publicity. Wolfgang Ketterle’s group here at MIT are
world leaders in this field. He shared the 2001 Nobel Prize in Physics for his work on
Bose-Einstein Condensation.] The equation of state and the heat capacity of the gas
below the transition temperature are given by the expressions
P(T, V ) = aT5/2
+ bT3
+ cV −2
CV (T, V ) = dT3/2
V + eT2
V + fT1/2
where a through f are constants which are independent of T and V .
a) Find the differential of the internal energy dE(T, V ) in terms of dT and dV .
b) Find the relationships between a through f due to the fact that E(T, V ) is a
state function.
c) Find E(T, V ) as a function of T and V .
d) Find the entropy, S(T, V ), as a function of T and V .
1

Problem 3: Paramagnet
Consider a paramagnetic material where the equation of state relating the magneti-
zation, M, to the applied magnetic ﬁeld, H, is M = AH/(T − T0). T0 and A are
constants and the expression is valid only for T T0.
a) Show that the heat capacity at constant magnetization, CM , is independent of
M at constant T.
C
b) Find an expression for the internal energy, E(T, M), in terms of A, T0 and
M (T).
c) Find the entropy, S(T, M).
Problem 4: Sargent Cycle
C
The diagram above is an approximation to a Sargent cycle run on an ideal gas. A
constant pressure path and a constant volume path are connected by two adiabatic
paths. Assume all processes are quasi-static and that the heat capacities, CP and
V , are constant.
a) Which of the four states (1,2,3 and 4) has the highest temperature and which
has the lowest?
T
b) T2 could be either hotter of colder than T4 depending on the speciﬁc values of P
and V at the four corners of the cycle. Demonstrate graphically one version of
the cycle where T4 is clearly less than T2. Demonstrate another extreme where
4 would necessarily be greater than T2.
2

c) Prove that the eﬃciency of this cycle running as an engine is

η = 1 − γ(T4 − T1)/(T3 − T2) where γ ≡ CP /CV .

d) Find an expression for the total work done, W, in one cycle. Express your
results in terms of N, k, γ, and the T’s.
e) Show that the ﬁrst law, |QH|−|QC| = W , applied to this cycle (together with| |
the assumption that the heat capacities are constants) leads to the requirement
that CP − CV = Nk.
Problem 5: Entropy Change
A mass M of liquid at a temperature T1 is mixed with an equal mass of the same
liquid at a temperature T2. The system is thermally insulated but the liquids are
maintained at some constant (atmospheric perhaps) pressure. Show that the entropy
change of the universe is
(T1 + T2)
2MCP ln ,
2
√
T1T2
and prove that it is necessarily positive.
3


Physics Department

Problem Set #8
Due by 1:10 PM: Monday, April 12
Problem 1: Correct Boltzmann Counting
The calculation we have done so far to obtain the allowed volume in phase space, Ω,
for a classical system is in error. We will demonstrate the results of this error in two
different cases and then propose a remedy.
a) A state variable F is extensive if, after multiplying all the extensive variables in
the expression for F by a scale factor λ and leaving all the intensive variables in
F unchanged, the result is a λ fold increase in F, that is, λF. The expression
we found for the cumulative volume in phase space for an ideal monatomic gas
using the microcanonical ensemble was
4πemE 3N/2
Φ(E, V, N) = V N
.
3N
Use S = k ln Φ and the derived result E = 3
NkT to write S as a function of2
the thermodynamic variables N, V , and T. On physical grounds S should be
extensive. Show that our expression for S(N, V, T) fails the above test for an
extensive variable.
b) Consider a mixing experiment with two ideal gases, 1 and 2. A volume V is
separated into two parts V1 = αV and V2 = (1 − α)V by a movable partition
(0 ≤ α ≤ 1). Let N1 atoms of gas 1 be confined in V1 and N2 atoms of gas 2
occupy V2. Show that if the temperature and pressure are the same on both
sides of the partition, the ideal gas equation of state requires that N1 = αN
and N2 = (1 − α)N where N = N1 + N2. Pulling the partition out allows the
gases to mix irreversibly if the gases are different.
The mixing is irreversible but entropy is a state function so ∆Si ≡ Si(final) −
Si(initial) can be computed for each gas from the expression in a). Show that
∆S1 = αNk ln(1/α) and
∆ST ≡ ∆S1 + ∆S2 = Nk [α ln(1/α) + (1 − α) ln(1/(1 − α))] .
1

This expression for the entropy of mixing is always positive, which is the result
we expect based upon the disorder interpretation of entropy.
Should ST increase as we slide the partition out when the two gases are the
same? This is difficult to answer from an intuitive point of view since the
presence of the partition does restrict the atomic motion.
Macroscopic thermodynamics, however, requires that ∆ST = 0 in this case.
Explain why the internal energy is unchanged in this process, ∆E = 0. Explain
why no work is done, ∆W = 0. The first law of thermodynamics then requires
that ∆Q = 0. But sliding the partition open and closed is certainly a reversible
process (when both gases are the same) so ∆S = ∆Q/T = 0. This, together
with our calculated result that ∆S = 0 is known as Gibbs’ paradox.
c) One could argue that our expression for S is not correct since it is not a quantum
mechanical result. However, in the classical limit of high T and low N/V the
classical calculation should give the correct answer. Evidently, this one does
not.
There is a concept which plays a central role in quantum mechanical calcula-
tions, even though it is not itself a result of quantum mechanics. That concept
is the indistinguishability of identical particles. One argon atom is the same
as all other argon atoms (with the same isotope number). Which particle has
momentum p at a location q does not matter in the overall specification of the
N body system; what matters is that some particle has that particular p and
q. This implies that we have over-counted the number of meaningful states in
phase space by a factor of N! where N is the number of identical particles. To
remedy this situation we should divide the results we have obtained for Ω (and
Φ and ω) by N!. This approach is known as “correct Boltzmann counting”.
Show by direct calculation that this solves the problems raised in parts a) and
b).
2

Problem 2: Torsional Pendulum

In order to measure a certain physical quantity it is necessary to know the equilibrium
angle θ0 of a torsional pendulum. θ0 is determined by the minimum in the potential
energy of the pendulum V (θ) = 1
K(θ − θ0)2
. The pendulum when rotating has a2
1
kinetic energy T = 2
Iθ˙2
where θ˙ ≡ dθ/dt. The precision with which θ0 can be found
is limited because the pendulum is in thermal equilibrium with its environment at a
temperature T.
a) Find the root-mean-square uncertainty (θ − θ0)2
1/2
associated with a mea-
surement of θ due to classical thermal noise.
b) Find θ ˙θ .
Problem 3: Defects in a Solid
A crystalline solid contains N similar, immobile, statistically independent defects.
Each defect has 5 possible states ψ1, ψ2, ψ3, ψ4, and ψ5 with energies 1 = 2 = 0,
3 = 4 = 5 = ∆.
a) Find the partition function for the defects.
b) Find the defect contribution to the entropy of the crystal as a function of ∆
and the temperature T.
c) Without doing a detailed calculation state the contribution to the internal en-
ergy due to the defects in the limit kT ∆. Explain your reasoning.
3

Problem 4: Neutral Atom Trap
A gas of N indistinguishable classical non-interacting atoms is held in a neutral atom
trap by a potential of the form V (r) = ar where r = (x2
+ y2
+ z2
)1/2
. The gas is in
thermal equilibrium at a temperature T.
a) Find the single particle partition function Z1 for a trapped atom. Express your
answer in the form Z1 = ATα
a−η
. Find the prefactor A and the exponents α
and η. [Hint: In spherical coordinates the volume element dx dy dz is replaced
by r2
sin θ dr dθ dφ. A unit sphere subtends a solid angle of 4π steradians.]
b) Find the entropy of the gas in terms of N, k, and Z1(T, a). Do not leave any
derivatives in your answer.
c) The gas can be cooled if the potential is lowered reversibly (by decreasing a)
while no heat is allowed to be exchanged with the surroundings, dQ = 0. Under
these conditions, find T as a function of a and the initial values T0 and a0.
Problem 5: The Hydrogen Atom
The simplest model of hydrogen has bound electronic energy eigenstates n, l, ml
with energies n,l,ml
= −A/n2
where A = 13.6 eV.
|
a) Find the ratio of the number of atoms in the first excited energy level to the
number in the lowest energy level at a temperature T. [Hint: Be careful of the
difference between energy levels and states of the system.] Evaluate this for
T = 300K and T = 1000K.
4

b) To ﬁnd the actual fraction of atoms in any given state one needs the partition
function. Show that the partition function diverges, even when the unbound
states are neglected.
c) Any ideas why statistical mechanics does not seem to work for this, one of
nature’s simplest systems?
5


Physics Department

Problem Set #9
Due by 1:10 PM: Friday, April 16
Problem 1: Why Stars Shine
The two major intellectual advances in physics at the beginning of the twentieth
century were relativity and quantum mechanics. Ordinarily one associates relativity
with high energies and great distances. The realm of quantum mechanics, on the
other hand, is usually thought of as small distances or low temperatures. These
perceptions are inaccurate. At the small end of the distance scale the energy levels of
heavy atoms must be computed using relativity. For example, the electronic energy
bands of lead show relativistic effects. At the large end of the distance scale QM can
play an important role. We will show later in this course that the radius of a white
dwarf or a neutron star depends on Planck’s constant. The purpose of this homework
problem is to show that QM is necessary even earlier in the life history of stars: they
could not shine without it.
Relativity predated quantum mechanics. In the example discussed here, the physical
source of the energy released by stars, this order of events gave rise to uncertainty
(and some acrimony) in the scientific community.
In 1926 Arthur Eddington collected and reviewed all that was known about the in-
terior of stars (Internal Constitution of the Stars, A. S. Eddington, Cambridge Uni-
versity Press, 1926). Relativity, in particular the equivalence of mass and energy,
was well understood at that time. He concluded that the only possible source of
the tremendous energy release in stars was the fusion of hydrogen nuclei into helium
nuclei with the associated conversion of mass to energy.
Some physicists, however, questioned this view. They pointed out that whatever
the chain of reactions leading from hydrogen to helium (we now recognize two, the
proton-proton chain and the carbon cycle), the nuclei would have to surmount the
repulsive coulomb barrier caused by their positive charges before any fusion step could
take place. The temperature in the interior of the sun, 40 million degrees K, was not
hot enough for this to happen.
a) Assume that the proton charge, e = 4.8 × 10−10
esu, is uniformly distributed| |
E
throughout a sphere of radius R = 1.2×10−13
cm. What is the minimum energy
min that another proton (taken to be a point particle) must have if it is to get
within R of the first proton?
1

Problems and solutions statistical physics 1

Problem 2: Two-Dimensional H2 Gas
N molecules of molecular hydrogen H2 adsorbed on a flat surface of area A are in ther-
mal equilibrium at temperature T. On the surface they behave as a non-interacting
two-dimensional gas. In particular, the rotational motion of a molecule is confined
to the plane of the surface. The quantum state of the planar rotation is specified by
a single quantum number m which can take on the values 0, ±1, ±2, ±3, ±4, etc.
There is one quantum state for each allowed value of m. The energies of the rotational
states are given by m = (¯h2
/2I)m2
where I is a moment of inertia.
a) Find an expression for the rotational partition function of a single molecule. Do
not try to reduce it to an analytic function.
b) Find the ratio of the two probabilities p(m = 3)/p(m = 2).
c) Find the probability that m = 3 given that = 9¯h2
/2I. Find the probability
that m = 1 given that ≤ ¯h2
/2I.
d) Find the rotational contribution to the internal energy of the gas in the high
temperature limit where kT ¯h2
/2I.
4

Problem 3: Adsorption on a Realistic Surface
It is unlikely that a clean crystalline face of a real solid would present a flat, featureless
landscape to a particular adsorbed atom similar to the model surface assumed in the
previous problem. In reality it is more likely that the adsorbed atom would see
an energy topography which mirrors the complex atomic structure of the surface.
Under these circumstances one often finds that there are only a finite number of
geometrically different sites to which the adsorbed atom may be bound, each with
a different binding energy. If these binding energies are comparable to the thermal
energy kT, the adsorbed atom may hop from site to site, executing a random walk
across the surface.
For simple materials, or ones of technological importance, the structure of the surfaces
has been determined and the possible binding sites located for particular adsorbed
species. The binding energies of those sites, however, are very hard to determine.
Experimental measurements of these binding energies would be valuable to theorists
who are trying to make ab initio calculations of the electronic structure of the surface
and to other scientists who are trying to model surface processes such as growth and
catalysis.
The first figure on the next page represents a ball-and-stick model of the 111 surface
of silicon based on calculations by Professor John Joannopoulos and his colleagues
here at MIT. The surface atoms have undergone a “reconstruction”, that is, their
geometrical arrangement is different than the ordering one would find along a 111
plane in the bulk material (this is not surprising, since half of the neighboring atoms
have been removed). The pattern is periodic, and the diamond shaped unit cell is
indicated by the dashed lines. Note that only one type of atom, Si, is involved;
balls with the same size and shading indicate atoms in symmetrically equivalent
environments. One can easily imagine, looking at such a picture, that an adsorbed
atom might find several symmetrically different binding sites, each with a different
binding energy.
5

Assume that a certain atom when adsorbed on a given surface of a particular material
can occupy any one of 5 geometrically different sites, labeled 1 through 5. There are
equal numbers of sites 1, 2, and 5. There are twice as many 3 sites as 1 sites, and
twice as many 4 sites as 1 sites. An STM measurement follows one adsorbed atom
and finds the following results for the fraction of time spent on each type of site when
the material is at a temperature of 300K.
Site Type 1 2 3 4 5
Fraction of Time 0.10 0.15 0.15 0.20 0.40
Number of Sites N N 2N 2N N
In the following, assume that the number of adsorbed atoms is much smaller than
the number of available sites. Under these conditions, one can do the statistics by
having only one adsorbed atom on the surface at a time.
a) Which site has the lowest energy; that is, which site has the largest binding
energy?
b) Find the energy of each of the sites relative to that of the most tightly bound
site, E(i) − E(lowest). Express your energies in K.
c) If the experiment were repeated with the sample at liquid nitrogen temperature
(77K), what would be the fraction of time spent at each type of site?
d) While doing the experiment it is possible to make a histogram of the the different
dwell times spent on a given type of site and thereby measure the probability
of the waiting time to escape, p(t). Can you predict the functional form of p(t)
based on our discussion of Poisson processes earlier in the term?
The idea for this problem came from the article Direct Measurement of Surface Dif-
fusion Using Atom-Tracking Scanning Tunneling Microscopy (B. S. Swartzentruber,
Physical Review Letters 76, 459 [January 15, 1996]). Movie images associated with
this paper, in particular a clip of the Si dimer shown on the previous page actually
moving, can be found on the World Wide Web
8


Physics Department

Problem Set #10
Due by 1:10 PM: Wednesday, April 28
Problem 1: The Big Bang
Early in the evolution of the universe, when the universe occupied a much smaller
volume and was very hot, matter and radiation were in thermal equilibrium. How-
ever, when the temperature fell to about 3000K, matter and the cosmic radiation
became decoupled. The temperature of the cosmic (black body) radiation has been
measured to be 3K now. Assuming adiabatic expansion, by what fraction has the
universe increased in volume since the decoupling of cosmic radiation and matter?
Hint: Remember the expression we derived for the Helmholtz free energy of thermal
radiation,
1 π2
F(T, V ) = −
h3 (kT)4
V.
45 c3¯
Note: The cosmic or background radiation is quite diﬀerent from local radiation ﬁelds
associated with a star, nebula or planet which certainly can be in equilibrium with
the accompanying matter.
Problem 2: Comet Hale-Bopp
a) Estimate the surface temperature of comet Hale-Bopp when it is at a distance
of 200 solar radii from the sun (about the same distance from the sun as the
earth). Assume
1) The surface temperature of the sun is 6000K.
2) The absorptivity of both the sun’s and the comet’s surface can be approx-
imated as α(ν, T) = 1.
3) The comet’s temperature changes are only due to its changing distance
from the sun.
b) This temperature is a bit high for a comet which is supposed to be a frozen
slush of various components, including H2O, CO2, and CH4. What might cause
the actual surface temperature to be lower than that calculated in a)?
c) Does the temperature found in a) seem familiar? Do you think this is a coinci-
dence or is it meaningful? How does the temperature found in a) depend on the
radius of the object? What simple evidence do we have that the sun contributes
to the heating of the earth’s surface? What evidence do we have that there is
another heating mechanism as well? What is that other mechanism?
1

Problem 3: Super Insulation

a) Two parallel plates of infinite extent are separated by a vacuum (Fig. a). The
temperature of the hotter plate is TH and that of the colder one is TC. Find an
expression for the heat flux, J, in watts-m−2
across the gap. Assume that both
surfaces act as black bodies and are maintained at temperatures TH and TC.
b) A thin heat conducting sheet (also a black body) is suspended in the vacuum as
shown in Fig. b. Heat can be transferred to the sheet only by radiation through
the vacuum. Find the steady state temperature of the sheet. Hint: What does
the steady state imply about the heat flux from TH to T and from T to TC?
c) The heat flux from TH to TC in b) is a fraction F of that in a). Find F.
d) There is an analytic expression for F in the case of n non-touching suspended
sheets. Find F(n) by using the fact that the heat flux J between any two
adjacent sheets must be the same.
The effect found here, that thermal radiative transfer can be suppressed substan-
tially by interposing free-floating, non-contacting conducting sheets, is used to build
liquid-nitrogen-free liquid helium storage containers. The dewar flask filled with liq-
uid nitrogen, commonly surrounding the liquid helium filled dewar flask (Fig. c), is
replaced with super-insulation, many layers of aluminum coated mylar film separated
by a tenuous mesh of mylar thread (Fig. d).
2

Problem 4: Properties of Blackbody Radiation
For 1 m3
of blackbody radiation at room temperature (300K) find
a) The heat capacity CV ,
b) The number of atoms of a monatomic gas that would give the same CV ,
c) The rms electric field in volts per centimeter.
Problem 5: Radiation Pressure
The radiation field and a monatomic gas exist in thermal equilibrium in a region
of space. If the density of the gas is the STP density of 2.69 × 1019
atoms-cm−3
(Loschmidt’s number), what must the temperature be so that the gas pressure and
the radiation pressure are equal?
In the interior of very high mass stars the radiation pressure dominates the kinetic
pressure.
3


Physics Department

Problem Set #11
Due by 1:10 PM: Wednesday, May 5
Problem 1: Lattice Heat Capacity of Solids
This problem examines the lattice contribution to the heat capacity of solids. Other
contributions may be present such as terms due to mobile electrons in metals or
magnetic moments in magnetic materials.
A crystalline solid is composed of N primitive unit cells, each containing J atoms.
A primitive unit cell is the smallest part of the solid which, through translational
motions alone, could reproduce the entire crystal. The atoms in the unit cell could be
the same or they could differ: diamond has two carbon atoms per primitive unit cell,
sodium chloride has one Na and one Cl. J could be as small as one, as in a crystal
of aluminum, or it could be tens of thousands as in the crystal of a large biological
molecule.
a) The Classical Model Assume each atom in the crystal is statistically independent
of all the others, and that it can vibrate about its equilibrium position as a
harmonic oscillator in each of 3 orthogonal directions. In principle there could
be 3J different frequencies of vibration in such a model; in fact, symmetry
conditions usually introduce degeneracies, reducing the number of frequencies
(but not the number of modes). On the basis of classical mechanics, find the
heat capacity at constant volume (i.e. constant lattice spacing) for this model.
b) The Einstein Model The result of the classical model does not agree with ob-
servation. The heat capacity of the lattice varies with temperature and goes
to zero at T = 0. Again assume that the atoms are statistically independent
and execute harmonic motion about their mean positions. This time find the
heat capacity using quantum mechanics. For simplicity, assume that the 3J
frequencies are identical and equal to ν. What is the limiting behavior of CV
for kT hν and for kT hν?
c) Phonons The result of the Einstein model is in better agreement with measured
heat capacities, but it is still not completely correct. In particular, the lattice
contribution to CV approaches T = 0 as T3
, a more gradual temperature de-
pendence than found in the Einstein model (using 3J different frequencies does
not help). The remaining flaw in the model is that the atomic motions are
not independent. Pluck one atom and the energy introduced will soon spread
throughout the crystal. Rather, the crystal has 3JN normal modes of vibra-
tion, called phonons, each of which involves all of the atoms in the solid. The
1

amplitude of each normal mode behaves as a harmonic oscillator, but the fre-
quencies of the the normal modes span a wide range from almost zero up to
the frequency one might expect when one atom vibrates with respect to ﬁxed
neighbors. A phonon of radian frequency ω is represented by a quantum me-
chanical harmonic oscillator of the same frequency. The density of frequencies
D(ω) is deﬁned such that D(ω0) dω is the number of phonons in the crystal
with frequencies between ω0 and ω0 + dω. Normalization requires that
∞
D(ω) dω = 3JN.
0
The thermodynamic internal energy of the lattice is
∞
E(T) = (ω, T) D(ω) dω
0
where (ω, T) is the mean energy of a quantum oscillator with radian
frequency ω at a temperature T.
i) Write down the full integral expression for E(T). Evaluate the expression
in the limit kT ¯hωmax where ωmax is the highest phonon frequency
in the solid. What is the heat capacity in this limit? You should get the
classical result.
ii) It can be shown that near ω = 0,
3V
D(ω) → ω2
,
2π2 v 3
where V is the volume of the crystal and v is an average sound velocity

in the solid. Find the heat capacity of the lattice for temperatures so low

2

Please see: Ashcroft, Neil W., and N. David Mermin. Solid State Physics.
New York, NY: Holt, Rinehart and Winston, c1976. ISBN: 0030839939.

that only those phonons in the quadratic region of D(ω) are excited. Use
the fact that
x π4∞ 3
dx = .
0 ex 15− 1
Problem 2: Thermal Noise in Circuits I, Mean-Square Voltages and Currents
An arbitrary network of passive electronic components is in thermal equilibrium with
a reservoir at temperature T. It contains no sources.
a) Find the probability density p(v) that a voltage v will exist on a capacitor of
capacitance C. [Hint: consider the capacitor alone as a subsystem.] Find an ex-
pression for the root-mean-square voltage
√
v2 . What is this in microvolts
when T = 300K and C = 100pF?
b) Find p(i) and
√
i2 for the current i through an inductor of inductance
L. What is the root-mean-square current in nanoamps when T = 300K and
L = 1mH?
c) Why does this method not work for the voltage on a resistor?
Problem 3: Thermal Noise in Circuits II, Johnson Noise of a Resistor
When we discussed jointly Gaussian random variables in the first part of this course,
we learned that the noise voltage in a circuit is a random process, a signal which
evolves in time. It will be composed of a variety of different frequency components.
The noise power in a unit frequency interval centered at radian frequency ω, Sv(ω), is
referred to as the power spectrum, or simply the spectrum, of the voltage fluctuations.
The mean square fluctuation on the voltage v2
is obtained by integrating Sv(ω)
over all ω.
The advantage of the approach to circuit noise introduced in Problem 2 is that mean
square voltages and currents in individual lossless components can be found imme-
diately, with out reference to the remainder of the circuit. The disadvantage is that
it does not allow one to find the spectrum of the voltage or current fluctuations in
those components.
There is another approach to determining the noise in circuits which we will introduce
here. It has the advantage of allowing the spectrum of the fluctuations to be found
anywhere in a circuit. The disadvantage is that one must be able to find the AC
3

transmission function from one part of the circuit to another. This method assumes
that the noise power entering the circuit emanates from each of the dissipative com-
ponents (resistors). Thus, one must replace each real resistor with an ideal resistor
plus a noise source. We will determine in this problem what the characteristics of
that noise source must be. In the next problem we will use the method to study the
noise in a simple circuit.
We will find the noise power emanating from a resistor by connecting it to a lossless
transmission line, assuming thermal equilibrium, and using the principle of detailed
balance. A coaxial transmission line which is excited only in the TEM modes behaves
like a one dimensional system. A vacuum filled line of length L, terminated by a short
circuit at each end, supports standing waves of voltage with the dispersion relation
ω = ck, where c is the speed of light and k is the wavevector. We can treat the
transmission line as a one dimensional analog of thermal (black body) radiation.
a) What are the allowed wave vectors kn on the transmission line described above?
b) Assuming that n is a large number, find the density of modes on the line D(ω).
As in 3 dimensions, one need only consider positive ω and notice that there is
only one “polarization” direction for the voltage in this case.
c) Find u1(ω, T), the energy per unit length per unit frequency interval, when the
line is in thermal equilibrium at temperature T.
d) Usually transmission lines operate under conditions where kBT ¯hω. Find
the limiting form of u1(ω, T) under these conditions.
The energy density on the line in part d) was calculated for standing waves, but it
can be regarded as composed of running waves traveling in two directions. If the
line is cut at some point and terminated with a resistor having the characteristic
impedance of the line, waves traveling toward the resistor behave as if the line were
still infinitely long in that direction and they will never return; that is, they are
completely absorbed. If the resistor is at the same temperature as the line, it must
send power to the line equal to the power which flows to it.
e) Find the thermal energy per unit frequency interval flowing out of the resistor,
P(ω). Thus P(ω) dω is the thermal power in the bandwidth dω. Note that it is
“white” noise in that it is independent of frequency (a flat frequency spectrum)
and that it does not depend on the value of the resistance. This power is referred
to as the “Johnson noise” associated with the resistor.
4

f) What is the noise power from a resistor at room temperature in a 10MHz
bandwidth (real frequency as opposed to radian frequency)?
Problem 4: Thermal Noise in Circuits III, Circuit Model for a Real Resistor
Now we will see how the concept of Johnson noise associated with a resistor is used to
ﬁnd real noise voltages in a circuit. The power absorbed by the resistor in Problem
3 depended on the presence of the transmission line. The noise power emitted by
the resistor, however, must depend only on its properties and therefore must be a
universal property of all resistors in thermal equilibrium. (One can imagine making
a transmission line whose characteristic impedance is equal to the resistance of any
resistor one chooses to consider.) For the purposes of circuit analysis, a real resistor
is modeled by an ideal (noiseless) resistor in series with a source of random (noise)
voltage:
a) The situation of a real resistor attached to an impedance-matched transmission
line is then modeled as follows:
Find the power dissipated in the line by the noise source at a frequency ω as a
2
function of vN (ω) .
5

c) Show that equating the results of a) with those of Problem 3e leads to the result

2
(ω) Nthat v = 2RkBT/π.
c) You may recall that a resistor and a capacitor in series form a low pass filter. In

2
(ω) c = v
2 2
(ω) /(1 + (RCω) ).oparticular, for the circuit shown below v

Note that one can now find the power spectrum of the voltage fluctuations

on the capacitor, information that was not available using the simpler method

of Problem 2. Find the power spectrum of the voltage across the capacitor,

2
(ω) cSvc (ω) ≡ v , when it is attached to a real resistor at temperature T.

Sketch your result. Use this result to find the rms voltage on the capacitor
remembering that
=

∞
2 2
(ω) dω
v v
.
0
Compare your answer with that found in problem 2.

6


Physics Department

Problem Set #12
Solutions Available in on the Web: Wednesday, May 12
Problem 1: Two Identical Particles
A system consists of two identical, non-interacting, spinless (no spin variables at all)
particles. The system has only three single-particle states ψ1, ψ2, and ψ3 with energies
1 = 0 2 3 respectively.
a) List in a vertical column all the two-particle states available to the system, along
with their energies, if the particles are Fermions. Use the occupation number
notation (n1, n2, n3) to identify each state. Indicate which state is occupied at
T = 0.
b) Repeat a) for the case of Bose particles.
c) Use the Canonical Ensemble to write the partition function for both Fermi and
Bose cases.
d) Using only the leading two terms in the partition function, ﬁnd the temperature
dependence of the internal energy in each case. Contrast the behavior of the
internal energy near T = 0 in the two cases.
Problem 2: A Number of Two-State Particles
Consider a collection of N non-interacting, spinless Bose particles. There are only two
single-particle energy eigenstates: ψ0 with energy = 0 and ψ1 with energy = ∆.
a) How would you index the possible N-body energy eigenstates in the occupation
number representation? What are their energies? How many N-body states
are there in all?
b) Find a closed form expression for the partition function Z(N, T) using the
Canonical Ensemble.
c) What is the probability p(n) that n particles will be found in the excited state
ψ1?
d) Find the partition function Zd(N, T) that would apply if the N particles were
distinguishable but possessed the same single particle states as above.
1

= ¯
Problem 3: Spin Polarization
Consider a 3-dimensional non-interacting quantum gas of s = 1/2 Fermions. The two
possible spin states are ms = 1/2 and ms = −1/2. In a uniform magnetic field Hzˆ
the single particle energies depend on the direction of the spin relative to the field:
h2
k2
/2m − 2µ0Hms where µ0 is a constant with the units of magnetic moment.
a) Find the density of states as a function of energy separately for the particles
with spin parallel and antiparallel to the external field, D1/2( ) and D−1/2( ).
Make a careful sketch of the total density of states as a function of energy
D( ) = D1/2( ) + D−1/2( ).
b) For N particles in a volume V at absolute zero (T = 0), find an expression for
the minimum magnetic field H0 that will give rise to total polarization of the
spins, that is no particles left with the high energy spin orientation.
c) Evaluate H0 in Tesla (one Tesla = 104
Gauss) for the electrons in copper where
the electronic magnetic moment µ0 = −9.27 × 10−21
ergs-gauss−1
, the particle
mass m = 9.11 × 10−28
grams and the number density of conduction electrons
is n ≡ N/V = 8.45 × 1022
cm−3
.
d) Evaluate H0 in Tesla for liquid 3
He where the magnetic moment is that of the
nucleus with µ0 = 1.075 × 10−23
ergs-gauss−1
, m = 5.01 × 10−24
grams and the
number density of atoms in the liquid is n = 1.64 × 1022
cm−3
.
One can now buy commercial superconducting solenoids for laboratory research that
go up to about 15 Tesla, and at MIT’s Francis Bitter National Magnet Laboratory
one can obtain continuous fields up to 37 Tesla and pulsed fields as high as 68 Tesla.
As you can see, we are a long way from being able to completely polarize either of
these systems by brute force.
Problem 4: Relativistic Electron Gas
Consider a 3 dimensional non-interacting quantum gas of ultra-relativistic electrons.
h k . The density of allowedIn this limit the single particle energies are given by = c¯| |
wavevectors in k space is V/(2π)3
.
a) Find an expression for the Fermi Energy F as a function of c, ¯h, N and V .
b) Find the density of states as a function of energy D( ). Sketch the result.
2

c) Find the total kinetic energy E of the gas at absolute zero as a function of N
and F .
d) Find the pressure exerted by the gas at T = 0. How does it depend on the
particle density N/V ? Is this a stronger or weaker dependence on density than
in the non-relativistic gas?
e) Assume that this gas represents the electrons in a white dwarf star composed
of α particles (a bound state of two neutrons and two protons: a 4
He nucleus)
and electrons. Express the kinetic energy EK in terms of the total mass of the
star M and its radius R (as well as a handful of physical constants including
the α particle mass). Recall that the potential energy of a self gravitating star
3
of radius R with uniform density is given by EP = 5
GM2
/R where G is the−
gravitational constant. Proceed as we did in class to find the equilibrium R as
a function of M by minimizing the total energy, ET = EK + EP . What can you
conclude about the stability of the star in this model?
f) We know from observation that white dwarfs of mass less than a certain critical
mass (of the order of the Sun’s mass) are stable. The results of e) show that
determining this critical mass will require a more sophisticated model of the
star, certainly taking into account the dependence of the mass density on depth
in the star and using a dispersion relation for the electrons, (k), valid for
all energies. However the calculation in e) allows us to find how the critical
mass might depend on the important parameters in the problem: c, ¯h, G, and
a reference mass such as mα. Use your results to find an expression for the
critical mass of a white dwarf, Mc, in terms of these parameters but neglecting
any purely numerical constants of the order of one.
3


Physics Department

Solutions to Problem Set #1
Problem 1: A Continuous Random Variable, a Harmonic Oscillator
a) First find the velocity as a function of time by taking the derivative of the dis-
placement with respect to time.
d
x˙(t) = [x0 sin(ωt + φ)]
dt
= ωx0 cos(ωt + φ)
But we don’t want the velocity as a function of t, we want it as a function of the
position x. And, we don’t actually need the velocity itself, we want the speed (the
magnitude of the velocity). Because of this we do not have to worry about losing the
sign of the velocity when we work with its square.
2 2
x˙ (t) = (ωx0)2
cos (ωt + φ)
= (ωx0)2
[1 − sin2
(ωt + φ)]
= (ωx0)2
[1 − (x(t)/x0)2
]
Finally, the speed is computed as the square root of the square of the velocity.
2 2
x˙(t) = ω(x0 − x (t))1/2
for| | |x(t)| ≤ x0
b) We are told that the probability density for finding an oscillator at x is proportional
to the the time a given oscillator spends near x, and that this time is inversely
proportional to its speed at that point. Expressed mathematically this becomes
p(x) ∝ |x˙(t)|−1
2 2
)−1/2
= C(x0 − x for x x0| |
where C is a proportionality constant which we can find by normalizing p(x).
x0
2 2
∞
p(x)dx = C (x0 − x )−1/2
dx
−x0−∞
x0 dx/x0
= 2C
0 1 − (x/x0)2
let x/x0 ≡ y
1 dy
= 2C
0
√
1 − y2
π/2
= πC
= 1 by normalization
1

The last two lines imply that C = 1/π. We can now write (and plot) the final result.

−1
p(x) = πx0 1 − (x/x0)2 x x0| |
= 0 x x0| |
As a check of the result, note that the area of the shaded rectangle is equal to 2/π. The
area is dimensionless, as it should be, and is a reasonable fraction of the anticipated
total area under p(x), that is 1.
c) The sketch of p(x) is shown above. By inspection the most probable value of x is
±x0 and the least probable accessible value of x is zero. The mean value of x is zero
by symmetry. It is the divergence of p(x) at the turning points that gives rise to the
apparent image of the pencil at these points in your experiment.
COMMENTS If an oscillator oscillates back and forth with some fixed frequency, why
is this p(x) independent of time? The reason is that we did not know the starting
time (or equivalently the phase φ) so we used an approch which effectively averaged
over all possible starting times. This washed out the time dependence and left a time-
independent probability. If we had known the phase, or equivalently the position and
velocity at some given time, then the process would have been deterministic. In that
case p(x) would be a delta function centered at a value of x which oscillated back and
forth between −x0 and +x0.
Those of you who have already had a course in quantum mechanics may want to
compare the classical result you found above with the result for a quantum harmonic
oscillator in an energy eigenstate with a high value of the quantum number n and
the same total energy. Will this probability be time dependent? No. Recall why the
energy eigenstates of a potential are also called “stationary states”.
2

Problem 2: A Discrete Random Variable, Quantized Angular Momentum
a) Using the expression for the normalization of a probability density, along with
expressions for the mean and the mean square, we can write three separate equations
relating the individual probabilities.
p(−¯ h) = 1h) + p(0) + p(¯
1
h p(−¯ h p(¯ ¯h) + 0 × p(0) + ¯ h) = Lx = h−¯
3
2
= h¯2 2 2 2
¯h h) + 0 × p(0) + ¯hp(−¯ p(¯h) = Lx
3
We now have three simple linear equations in three unknowns. The last two can be
simplified and solved for two of our unknowns.

 p(¯h) = 1
2
1
−p(−¯ h) =h) + p(¯ 3 

⇒
2 1
6
p(−¯ h) =h) + p(¯ h) =p(−¯3
Substitute these results into the first equation to find the last unknown.

1 1 1
6
+ p(0) +
2
= 1 ⇒ p(0) =
3
3

b)

Problem 3: A Mixed Random Variable, Electron Energy
a) Let the probabability that E is greater than 1.0 ev be denoted by prob(E 1.0).
∞
prob(E 1.0) = p(E) dE
1
1
= 0.8
∞
e−E/b
dE let E/b ≡ x
1 b
= 0.8
∞
e−x
dx = 0.8
∞
(−e−x
)
b−1 b−1
= 0.8e−1/b
use the fact that b = 1
= 0.8/e = 0.294
b) Now we simply implement the expression for the mean.
∞
E = E p(E) dE
−∞
4

0 E
e−E/b
dE= E × 0.2δ(E + E0) dE +
∞
0.8
−∞ 0 b
= −0.2E0 + 0.8b
∞
xe−x
dx
0
1
= 0.8b − 0.2E0
Using the given values of b = 1 and E0 = 1.5 ev, we find that E = 0.5 ev.
c) To find the cumulative function we simply carry out the defining integral, using
the fact that the integral of a delta function generates a step function at that location
with step height equal to the coefficient of (or the integral under) the delta function.
E
P(E) ≡ p(ζ) dζ
−∞
210-1-2 E(ev)
P(E)
1.0
0.2
Problem 4: A Time Dependent Probability, a Quantum Mechanics Example
Ψ(r, t) = (2πx2 iωt i 2
)2
0)−1/4
exp −
2
−
(2x0)2
(2αxx0 sin ωt − α2
x sin 2ωt) − (
x − 2αx0 cos ωt
0
2x0
a) First note that the given wavefunction has the form Ψ = a exp[ib+c] = a exp[ib] exp[c]
where a, b and c are real. Thus the square of the magnitude of the wavefunction is
simply a2
exp[2c] and finding the probablity density is not algebraicly difficult.
1
exp[−
(x − 2αx0 cos ωt)2
p(x, t) = Ψ(x, t) 2
= 2
]| |
2πx2 2x00
b) By inspection we see that this is a Gaussian with a time dependent mean x =
2αx0 cos ωt and a time independent standard deviation σ = x0.
5

c) p(x, t) involves a time independent pulse shape, a Gaussian, whose center oscillates
harmonically between −2αx0 and 2αx0 with radian frequency ω.
t=1/2 T
t=3/4 T
t=1/4 T
t=0
-2αx0 0 2αx0 x
Those already familiar with quantum mechanics will recognize this as a “coherent
state” of the harmonic oscillator, a state whose behavior is closest to the classical
behavior. It is not an energy eigenstate since p(x) depends on t. It should be
compared with a classical harmonic oscillator with known phase φ and the same
maximum excursion: x = 2αx0 cos ωt. In this deterministic classical case p(x, t) is
given by
p(x, t) = δ(x − 2αx0 cos ωt).
Problem 5: Bose-Einstein Statistics
We are given the discrete probability density
p(n) = (1 − a)an
n = 0, 1, 2, · · ·
a) First we ﬁnd the mean of n.
∞ ∞
n = np(n) = (1 − a) nan
n=0 n=0
S1
The sum S1 can be found by manipulating the normalization sum.
∞
p(n) =
∞
(1 − a)an
= (1 − a)
∞
an
must = 1
n=0 n=0 n=0
Rearranging the last two terms gives the sum of a geometric series:
1∞
n
a = .
n=0 1 − a
6

But note what happens when we take the derivative of this result with respect to the
parameter a.
d ∞∞
n
∞
n−1 1 n 1
a = na = na = S1
da a an=0 n=0 n=0
1d 1
also = =
da 1 − a (1 − a)2
Equating the two results gives the value of the sum we need, S1 = a/(1 − a)2
, and
allows us to finish the computation of the mean of n:
a
n = .
1 − a
c) To find the variance we first need the mean of the square of n.
2 2 n
n =
∞
n p(n) = (1 − a)
∞
n2
a
n=0 n=0
S2
Now try the same trick used above, but on the sum S1.
d
da
∞
n=0
nan
=
∞
n=0
n2
an−1
=
1
a
∞
n=0
n2
an
=
1
a
S2
S1
d a 2a 1
also =
da (1 − a)2
=
(1 − a)3
+
(1 − a)2
Then
n2
= (1 − a)
2a2
(1 − a)3
+
a
(1 − a)2
= 2
a
1 − a
2
+
a
1 − a
= 2 n 2
+ n ,
and
Variance = n2
− n 2
= n 2
+ n
= n (1+ n ).
This is greater than the variance for a Poisson, n , by a factor 1+ n .

7

c)

∞
p(x) = (1 − a)an
δ(x − n)
n=0
∞
= f(x) δ(x − n)
n=0
) = Ce−x/φ
Try f(x ,

then f(x = n) = Ce−n/φ
= C(e−1/φ
)n
= (1 − a)an
.

This tells us that C = 1 − a and exp(−1/φ) = a. We can invert the expression found
above for n to give a as a function of n : a = n /(1+ n ).
n
−1/φ = ln a = ln
1+ n
n +1 1
1/φ = ln
n
= ln 1 +
n
Recall that for small x one has the expansion ln(1 + x) = x − x2
/2 + . . .. Therefore
in the limit n 1, 1/φ → 1/ n which implies φ n .→
8

Physics Department
Problem 1: Two Quantum Particles
a)
p(x1, x2) = |ψ(x1, x2)|2
=
1
πx2
0
(
x1 − x2
x0
)2
exp[−
x2
1 + x2
2
x2
0
]
The ﬁgure on the left shows in a simple way the location of the maxima and minima of
this probability density. On the right is a plot generated by a computer application,
in this case Mathematica.
Max.
Max.
x2
x1
node at x1=x2
implies p(x1,x2) = 0
b)
p(x1) =
∞
−∞
p(x1, x2) dx2
=
1
πx2
0
1
x2
0
exp[−x2
1/x2
0]
∞
−∞
(x2
1 − 2x1x2 + x2
2) exp[−x2
2/x2
0] dx2
=
1
πx2
0
1
x2
0
exp[−x2
1/x2
0] x2
1( πx2
0 ) − 2x1 × 0 +
x2
0
2
( πx2
0 )
=
1
πx2
0
(x2
1/x2
0 + 1/2) exp[−x2
1/x2
0]
1

By symmetry, the result for p(x2) has the same functional form.
1 2 2
p(x2) = (x2/x2
0 + 1/2) exp[−x2/x2
]0
πx2
0
By inspection of these two results one sees that p(x1, x2) = p(x1)p(x2), therefore
x1 and x2 are not statistically independent .
c)
p(x1|x2) = p(x1, x2)/p(x2)
=
πx2
0
πx2
0
(x1 − x2)2
(x2
2 + 1
2
x2
0)
exp[−(x2
1 + x2
2)/x2
0]
exp[−x2
2/x2
0]
=
2
πx2
0
1
(1 + 2(x2/x0)2)
x1 − x2
x0
2
exp[−x2
1/x2
0]
p(x1|x2)
x1 = x2
0 x1
It appears that these particles are anti-social: they avoid each other.
For those who have had some quantum mechanics, the ψ(x1, x2) given here corre-
sponds to two non-interacting spinless1
Fermi particles (particles which obey Fermi-
Dirac statistics) in a harmonic oscillator potential. The ground and ﬁrst excited single
1
Why spinless? If the particles have spin, there is a spin part to the wavefunction. Under these
circumstances the spin part of the wavefunction could carry the antisymmetry (assuming that the
spatial and spin parts factor) and the spatial part of Fermionic wavefunction would have to be
symmetric.
2

particle states are used to construct the two-particle wavefunction. The wavefunc-
tion is antisymmetric in that it changes sign when the two particles are exchanged:
ψ(x2, x1) = −ψ(x1, x2). Note that this antisymmetric property precludes putting
both particles in the same single particle state, for example both in the single particle
ground state.
For spinless particles obeying Bose-Einstein statistics (Bosons) the wavefunction must
be symmetric under interchange of the two particles: ψ(x2, x1) = ψ(x1, x2). We can
make such a wavefunction by replacing the term x1 − x2 in the current wavefunction
by x1 + x2. Under these circumstances p(x1 x2) could be substantial near x1 = x2.|
Problem 2: A Joint Density of Limited Extent
a) The tricky part here is getting correct
1
limits on the integral that must be done to
eliminate y from the probability density.
It is clear that the integral must start at
y = 0, but one must also be careful to get
the correct upper limit. A simple sketch
such as that at the right is helpful.
y
x1
y=1-x
1−x∞
p(x) = p(x, y) dy = 6 (1 − x − y) dy
−∞ 0
1−x
1 1
= 6{ (1 − x)y − y2
} = 6{(1 − x)2
(1 − x)2
2 0
−
2
}
= 3(1 − x)2
0 ≤ x ≤ 1
b)
p(x, y)
p(y|x) = =
6(1 − x − y)
p(x) 3(1 − x)2
2
=
(1 − x)2
(1 − x − y) 0 ≤ y ≤ 1 − x
Note that this is simply a linear function of y.
3

p(x)
2
3
1-x
p(y | x)
0 1 x 0 1-x y
Problem 3: A Discrete Joint Density
n
a) p(n) = p(n, l) = c exp[−an]
l l=−n
does not depend on l
= c (2n + 1) exp[−an]
p(n)
0 5 10 n
b) p(l n) =
p(l, n)
=
c exp[−an]
|
p(n) c(2n + 1) exp[−an]
1
=
2n + 1
|l| ≤ n
= 0 otherwise
p(l | n) for n = 2
1
5
1
5
1
5
1
5
1
5
-2 -1 0 1 2 l
4

c) p(l) = p(n, l) =
∞
c e−an
=
∞
c (e−a
)n
n n=|l| n=|l|
= c(e−a
)|l|
∞
(e−a
)n−|l|
= ce−a|l|
∞
(e−a
)m
n=|l| m=0
=
c
1 − e−a
e−a|l|
p(l)
l-4 -2 0 2 4
d) p(n|l) = p(n, l)/p(l)
a l
e−an
= (1 − e−a
)e | |
n ≥ |l|
= 0 otherwise
0 3 6 9 n
for l = -3
p(n | l )
Problem 4: Distance to the Nearest Star
First consider the quantity p(no stars in a sphere of radius r). Since the stars are
distributed at random with a mean density ρ one can treat the problem as a Poisson
process in three dimensions with the mean number of stars in the volume V given by
n = ρV = 4
πρr3
. Thus3
p(no stars in a sphere of radius r) = p(n = 0)
= exp[−4/3 πρr3
]
5

Next consider the quantity p(at least one star in a shell between r and r +dr). When
the differential volume element involved is so small that the expected number of stars
within it is much less than one, this quantity can be replaced by p(exactly one star
in a shell between r and r + dr). Now the volume element is that of the shell and
n = ρ∆V = 4πρr2
dr. Thus
p(at least one star in a shell between r and r + dr) = p(n = 1)
1
= n (1)
e−n
1! ≈1
=1
≈ n = 4πρr2
dr
Now p(r) is defined as the probability density for the event “the first star occurs
between r and r+dr”. Since the positions of the stars are (in this model) statistically
independent, this can be written as the product of the two separate probabilities found
above.
p(r)dr = p(no star out to r) × p(1 star between r and r + dr)
= 4πρr2
exp[−4/3 πρr3
] dr
Dividing out the differential dr and being careful about the range of applicability
leaves us with
p(r) = 4πρr2
exp[−4/3 πρr3
] r ≥ 0
= 0 r 0
p(r)
0 r

6

0 5 10 15 20 25 30 35
20
60
100
140
Problem 5: Shot Noise
a) This is a Poisson process with a rate equal to rs counts per second and an interval
of T seconds. The probability of getting n counts in the interval is given by the
Poisson expression for p(n) with a mean n = rsT. Since we have defined the
mean as the signal in this case, S = rsT.
b)
40
80
120
T
n
t
n
σ
One tries to use a record such as the one above to determine I through the relation
I = rs(ηA/hν)−1
= n (ηAT/hν)−1
. The problem is to determine n since
all the other factors are known constants. Using a single measurement taken in
an interval T to determine n could be in error by an amount of the order of
σ = (Variance)1/2
=
√
n which we define as the noise, N, for the measurement.
A histogram of the measured results in an interval T would peak near n and
would have a width ∼
√
n .
c) Now consider the signal to noise ratio, S/N.
S/N = n /
√
n =
√
n
ηA 1/2
= rsT = IT
hν
This shows that the signal to noise ratio grows as the square root of both the light
intensity and the counting time.
Note that if one had the entire record shown in the above figure available, about 36T,
one could do better at estimating I by using the number of counts detected in the
expanded interval T = 36T. The signal to noise ratio would be increased by a factor
of T /T =
√
36 = 6.
7

0 20 40 60 80 100 120 140
20
40
60
80
140
d) When I is on, one has a Poisson process with n = (rs + rd)T and σ =
(rs + rd)T. When I is off, one has a Poisson process with n = rdT and
σ =
√
rdT. The signal is now defined as S ≡ n on − n off = rsT. One
tries to estimate this quantity from non − noff for a single measurement. The noise
is associated with the uncertainty in both measurements, but when rd rs the
contribution from rs is small and N ≈
√
rdT. Thus in the small signal limit
rsT ηA
S/N = √
rdT
= I
√
T.
hν
√
rd
Now the signal to noise ratio is linearly proportional to the intensity of the signal but
it still only grows as the square root of the counting interval.
In the example shown below, rd = 100 counts per second and rs = 20 counts per
second. The signal is cycled on for 25 counting intervals, each of length one second,
and off for an equal number of counting intervals.
100
120
S N
T
t
on offoffoff on
n
on
8


Physics Department

Every problem in this set deals with a situation where a variable is a function of one or
more random variables and is therefore itself a random variable. The problems all use
the procedure given in the notes for finding the probability density of the new random
variable when we know the probability densities for the variables that it depends upon.
Problem 1: Energy in an Ideal Gas
In this problem the kinetic energy E is a function of three random variables (vx, vy, vz)
in three dimensions, two random variables (vx, vy) in two dimensions, and just one
random variable, vx, in one dimension.
(a) In three dimensions
2 2
z
2
)E
=
1
2
(v
+ v + vx y
and the probability density for the three velocity random variables is

1 2
x+v2
y+v2
z)/2σ2
e−(v
p(vx, vy, vz) = .
(2πσ2)3/2
Following the procedure, we want first to calculate the cumulative probability
function P(E), which is the probability that the kinetic energy will have a value
which is less than or equal to E. To find P(E) we need to integrate p(vx, vy, vz)
2 2 2
over all regions of (vx, vy, vz) that correspond to z ) ≤ E.
region in (vx, vy, vz) space is just a sphere of radius 2E/m centered at the
origin.
The integral is most easily done in spherical polar coordinates where
v2
= v2
x + v2
y + v2
z
vx = v sin θ cos φ
vy = v sin θ sin φ
vz = v cos θ
dvx dvy dvz = v2
dv sin θ dθ dφ
(v
That
1
2
+ v + vx y
1

This gives
π
2
P(E) =
1
sin θ dθ
2π
dφ
√
2E/m
v e−v2/2σ2
dv
(2πσ2)3/2
0 0 0
√
2E/m
4π 2
e−v2/2σ2
= v dv
(2πσ2)3/2
0
Thus (substituting mσ2
= kT)
2 1 E
p(E) =
dP(E)
= e−E/kT
if E ≥ 0 (0 if E 0)
dE
√
π kT kT
p(E)
3 dimensions
0 E
You can use this to calculate E = 3
kT.2
(b) In two dimensions
E = 1 2
2
(vx + vy
2
)
and the probability density for the two velocity random variables is
1
2
x+v2
y)/2σ2
.e−(v
p(vx, vy) =
(2πσ2)
To ﬁnd P(E) we need to integrate p(vx, vy) over all regions of (vx, vy) that
correspond to 1 2
2
(vx + vy
2
) ≤ E. That region in (vx, vy) space is just a circle of
radius 2E/m centered at the origin.

The integral is most easily done in polar coordinates where

v2
= v2
x + v2
y
vx = v cos θ
vy = v sin θ
dvx dvy = v dv dθ
2

This gives
√
2E/m
e−v2/2σ2
P(E) =
1 2π
dθ
√
2E/m
v e−v2/2σ2
dv =
1
dv
2πσ2
0 σ2
0 0
= kT)
1
p(E) =
dP(E)
= e−E/kt
if E ≥ 0 (and 0 if E 0)
dE kT
p(E)
2 dimensions
0 E
You can use this to calculate E = kT.
(c) In one dimension (not required)
E = 1
v2
2 x
and the probability density for the single velocity random variable is
2
p(vx) = .√
2
1
πσ2
e−vx/2σ2
To ﬁnd P(E) we need to integrate p(vx) over all regions of vx that correspond
2
to 1
vx ≤ E. That region in vx space is just a line from − 2E/m to 2E/m.2
√
2E/m 2v1
P(E) = √
2πσ2
−
√
2E/m
v e−
2σ2
dv
= kT)
1
p(E) =
dP(E)
= e−E/kt
if E ≥ 0 (0 if E 0)
dE
√
πkT E
p(E)
1 dimension
0 E
You can use this to calculate E = 1
kT.2
3

Problem 2: Distance to the Center of the Galaxy
σv 20 km/sec
(a) R0 = = 9.1×10−17 radians/sec
= 2.2 × 1017
km = 7.1 kiloparsecsσθ˙
⊥
(b) Here the random variable v (the velocity along the line from the center of the
galaxy to an observer on the earth) is a function of the three random variables
vx, vy, vz which are the components of the velocities of the water vapor masers.
The first thing to note is that all masers have the same speed ˙r and that the
randomness is only in the direction of their motion. If we choose spherical polar
coordinates
vx = v sin θ cos φ = ˙r sin θ cos φ
vy = v sin θ sin φ = ˙r sin θ sin φ
vz = v cos θ = ˙r cos θ
there will be only two random variables θ and φ because the magnitude v is
fixed.
Then it is convenient to choose the z axis to be the direction from the center
of the galaxy to the earth, because this makes v = vz. Solving the problem
requires us to find p(v ) = p(vz) using the appropriate p(vx, vy, vz) for a fixed
magnitute and random orientation of v. Since v = ˙r cos θ, the possible values
of v will be from −r˙ when θ = π to ˙r when θ = 0. We first want to find P(v ),
the probability that vz ≤ v ; since v = ˙r cos θ this is equal to the probability
that cos θ ≤ v /r˙ or θ ≥ cos−1
(v /r˙).
π
1 2π

P(v ) = dφ sin θ dθ

4π 0 cos−1(v /r˙)
⎧
π ⎪0 if v r˙⎨1 1
= − cos θ = 2
(1 + v /r˙) if − ˙ r,
2 ⎪
r ≤ v ≤ ˙
⎩cos−1(v /r˙) 1 if v r˙
p(v ) =
dP(v)
=
(2 ˙r)−1
if − r˙ ≤ v ≤ ˙r,
dv 0 otherwise.
p(v )P(v )
1
1 / 2 r
v vr r r r
4

Problem 3: Acceleration of a Star
In this problem we want to ﬁnd the probability density function for the acceleration
random variable a when a = GM/r2
, where r is the distance to a nearest neighbor
star of mass M.
a(r)
GM/r2
region of r for which
a a0
a0
0 r
GM/a0
(a)
P(a) =
∞
√
GM/a
pr(r) dr
p(a) =
dP(a)
da
=
1
2a
GM
a
pr
GM
a
Distant neighbors will produce small forces and accelerations, so their eﬀect on
p(a) will be greatest when a is small.
e
(b) We use the pr(r) found in problem 4 of problem set 2:
−4πρ r3/3
pr(r) = 4πρ r2

to calculate

2πρ GM
5/2
e−(4πρ/3)(GM/a)3/2
p(a) =
GM a
p(a)
0 a
5

If there are binary stars and other complex units at close distances, these will
have the greatest effect when r is small or when a is large.
(c) The model considered here assumes all neighbor stars have the same mass M.
To improve the model, one should consider a distribution of M values. One
could even include the binary stars and other complex units, from part (b),
with a suitable distribution.
Problem 4: Atomic Velocity Profile
vx(s)
0 s
A/s
region where
vx η
η
A η2
In this problem we have two random variables related by s = A/v2
. We are givenx
ps(ζ) and are to find p(vx). To find the probability P(v) that vx ≤ v we must calculate
the probability that s ≥ A/v2
. This is (the only possible values of vx are ≥ 0)
A3/2 ∞
1
e−A/(2σ2ζ)
dζ
A
P(vx) =

σ2
√
2πσ2
A/v2 ζ5/2

3/2 2 5/2

σ2
√
2πσ2
2
dP(vx) d A 2vv 2
x/2σ2 2
x/2σ2
e−v
e−vx x
p(vx) =
σ2
√
2πσ2
= − =
v2
xdvx A dvx
)p(vx
0 vx
6

Problem 5: Planetary Nebulae
In this problem we are looking at matter distributed with equal probability over a
spherical shell of radius R. When we look at it from a distance it appears as a ring
because we are looking edgewise through the shell (and see much more matter) near
the outer edge of the shell. If we choose coordinates so that the z axis points from
the shell to us as observers on the earth, then r = x2 + y2 = R sin θ will be the⊥
radius of the ring that we observe. The task for this problem is to find a probability
distribution function p(r⊥) for the amount of matter that seems to us to be in a ring
of radius r⊥. This is a function of the two random variables θ and φ, which are the
angular coordinates locating a point on the shell of the planetary nebula. Since the
nebulae are spherically symmetric shells, we know that p(θ, φ) = (4π)−1
.
We want first to find P(r⊥), the probability that the perpendicular distance from
the line of sight is less than or equal to r⊥. Since r = R sin θ, that is equal to the⊥
probability that sin θ ≤ r⊥/R. This corresponds to the two polar patches on the
nebula: 0 ≤ θ ≤ sin−1
(r⊥/R) in the “top” hemisphere and sin−1
(r⊥/R) ≥ θ ≥ π in
the “bottom” hemisphere.
π
1 2π sin−1(r⊥/R)
P(r⊥) = dφ sin θ dθ + sin θ dθ = 1 − 1 − r⊥/R2
4π 0 0 sin−1(r⊥/R)
p(r⊥) =
dP(r⊥) r
=
⊥
for 0 ≤ r⊥ ≤ R (and 0 otherwise)2dr⊥ R R2 − r
⊥

2
Note: cos sin−1
(r⊥/R) = 1 − r⊥/R2.
p(r )
0 1 r /R
7
2


Physics Department


Problem 1: Thermal Equilibrium and the Concept of Temperature, I
a) Solve each of the two given equations for P .
P =
P(V − nb) P V
P =
V V (1 − nB /V )
Equate these two and factor out 1/V .
P V some constant,
P(V − nb) =
1 − nB /V
=
call it t
Substitution of this result into the first equation gives P V = t, allowing us to find
the functional form for the temperature in each of the three systems.
P V = t for an ideal gas
P V
1 − nB /V
= t for another non-ideal gas
P(V − nb) = t for one non-ideal gas
b) To find the relation expressing thermal equilibrium between system A and system
B we simply equate the two expressions for a common empiric temperature.
P V
P(V − nb) = t =
1 − nB /V
⇒ P(V − nb)(1 − nB /V ) − P V = 0
Problem 2: Thermal Equilibrium and the Concept of Temperature, II
a) Solve each equation for V.
nR cH nR c H
V = ( )( ) V = ( )(Θ + )
P M P M
1

Equate these two and factor out nR/P.
cH c H some constant,
= Θ + =
M M call it h
Substitution into the first equation gives PV/nR = h, so at equilibrium
PV cH c H
= = Θ +
nR M M
b) PV/nR = h looks like the ideal gas law with h → T, so call h ≡ T and thus find
the following equations of state.
PV = nRT for an ideal gas
H
M = c for a Curie Law Paramagnet
T
H
Paramagnet with ordering
M = c phase transition to a
T − Θ
ferromagnet at t = Θ
Problem 3: Work in a Simple Solid
Substitute the given model expression relating volume changes to changes in the
pressure and the temperature, dV = −V KT dP + V α dT, into the differential for
work. As a simplification we are told to replace the actual volume V by its value at
the starting point V1 in the coefficients entering the differential for the work. Of course
the volume itself can not really remain constant, for in that case d/ W = −P dV = 0.
d/ W = −P dV = KT PV1 dP − αPV1 dT
Along path “a”
/ W + dW1 2 = d / W→
where dP=0 where dT=0
2 2
= −αP1V1 dT + KT V1 P dP
1 1
1
= −αP1V1(T2 − T1) +
2
KT (P2
2
− P1
2
)V1
2

Along path “b”
/ W + dW1 2 = d / W→
where dT=0 where dP=0
2 2
= KT V1 P dP − αP2V1 dT
1 1
1
=
2
KT (P2
2
− P1
2
)V1 − αP2V1(T2 − T1)
Along “c” dT and dP are related at every point along the path,
T2 − T1
P
dT = dP,
2 − P1
so the expression for the differential of work can be written as
d
P
/ W = KT PV1 dP − αPV1(
T2 − T1
) dP
2 − P1
P
= KT V1 − αV1
T2 − T1
P dP
2 − P1
Now we can carry out the integral along the path.
W1 2 = KT V1 − αV1
T2 − T1
2
P dP
1
→
P2 − P1
1
2
(P2
2
− P1
2
) =
1
2
(P1 + P2)(P2 − P1)
1 1
=
2
KT (P2
2
− P1
2
)V1 − α(P1 + P2)(T2 − T1)V1
2
Note that the work done along each path is different due to the different contributions
from the α (thermal expansion) term. Path “b” requires the least work; path “a”
requires the most.
3

Problem 4: Work in a Non-Ideal Gas
Often it is best to work with the given differential in the work term if at all possible,
rather than expanding the differential in terms of the other variables as was done in
Problem 3. The form of the differential, d/ W = −P dV , indicates that we should try
to find P as a function of V so that the integration can be carried out simply.
a NkT a
(P +
V 2
)(V − b) = NkT P =
V − b
−
V 2
⇒
Now we can carry out the integration to find the total work done along the path.
2 NkT a
W1→2 =
1
−
V − b
+
V 2
dV
2 2 a
= NkT
1
− ln(V − b) +
1
−
V
a a
= −NkT ln(
V2 − b
) − ( )
V1 − b V2
−
V1
a a
= NkT ln(
V1 − b
) − ( )
V2 − b V2
−
V1
We are given that V2 V1. The equation of state used in the problem is known as
the Van der Waals equation. In it the parameter b is thought of as a small correction
to the total volume V due to the volume taken up by the other atoms and a/V 2
is
a small correction to the pressure due to the weak attraction between the atoms at
large distances (the Van der Waals attraction). Thus in the above expression the
logarithm term is positive and the term after the minus sign is also positive.
Problem 5: Work and the Radiation Field
The differential of work is d/ W = −P dV as in the previous problem and one immedi-
ately thinks about trying to express P in terms of V in order to simplify the integral.
However, along path “a” this is not necessary: along one part dV = 0 and along the
other the temperature, and hence the pressure, is a constant.
W
2 2
1 2 = P dV − P dV→
1 T constant 1 V constant
0
1 2 1
= −
3
σT1
4
dV = σT1
4
(V2 − V1)
1
−
3
1 V1
= σT1
4
V2(
V2
− 1)
3
4

Since the ﬁgure in the problem indicates that V1 V2, the underlined result is positive.
Along path “b” there are no shortcuts and we must prepare to carry out the integral.
Since d/ W is expressed in terms of dV , we convert the T dependence of P into a
function of V .
V T3
= a constant ≡ V1T3

1
T = (
V1
)1/3
T1
V

4
= T1
4
(
V
V1
)4/3
= T1
4
( )−4/3
T
V V1
Now we can carry out the integral over the path to obtain the total work done.
2
1 2 = −
1
1
P dV =
2

T4
(V ) dVW → −
3
σ
1

1
= −
3
2 V

σT1
4
( )−4/3
dV
1 V1
1
= −
3
2 1
σT1
4
V1
1
−
1/3
(
V

V1
)−1/3
V
= σT1
4
V1 (
V2
)−1/3
− 1

1

V
= σT1
4
V1 (
V1
)1/3
− 1

2

Since V1 V2, this quantity is also positive.
5


Physics Department

Problem 1: Equation of State for a Ferromagnet
a) We are looking for the magnetization as a function of the field and the temperature,
M(H, T), so we form the differential of M as follows.
∂M ∂M
dM = dH + dT
∂H ∂TT H
We are given the two coefficients in the expansion, but must make sure that their
cross derivatives are equal as is required for an exact differential.
∂ ∂M 1 a
=
∂T ∂H Tc (1 − T/Tc)2
T
∂ ∂M 1 f (H)
=
∂H ∂T Tc (1 − T/Tc)2
H
The equality of these two expressions requires that f (H) = a. Integration gives
f(H) = aH + c but we are told that f(0) = 0 so we know that c = 0. Thus
f(H) = aH
b) Now we must integrate the exact differential to find the state function M(H, T).
∂M
M(H, T) = dT + g(H)
∂T H
f(H)
=
(1 − T/Tc)
+ M0(1 − T/Tc)1/2
+ g(H)
∂M f (H)
= + g (H) by calculation from above
∂H (1 − T/Tc)T
a
= + 3bH2
as given
1 − T/Tc
Now we set about finding an expression for g(H).
g (H) = 3bH2
g(H) = bH3
+ K
M(H = 0, T = Tc) = 0 K = 0⇒
1

Now putting all the pieces together gives
aH
M(H, T) = M0(1 − T/Tc)1/2
+
(1 − T/Tc)
+ bH3
Problem 2: Heat Capacity at Constant Pressure in a Simple Fluid
Start with the first law of thermodynamics.
d/Q = dU + P dV
The relation for CP we are looking for involves (∂U/∂T)P so it is natural to try to
do our expansion in terms of the variables T and P. We expand both dU and dV in
terms of dT and dP.
∂U ∂U
dU = dT + dP
∂T ∂PP T
∂V ∂V
dV = dT + dP
∂T ∂PP T
∂U ∂V ∂U ∂V
d/Q =
∂T P
+ P
∂T P
dT +
∂P T
+ P
∂P T
dP
Since we need the derivative at constant P the second term in the above expression
will drop out.
d/Q
CP ≡
dT P
∂U ∂V
= + P
∂T ∂TP P
∂U
= + αV P
∂T P
Problem 3: Heat Supplied to a Gas
To find CP we proceed as follows. Rearrange the first law to isolate d/Q.
d/Q = dU + P dV
Expand the differential of the energy in terms of dT and dV .
∂U ∂U
dU = dT + dV
∂T ∂VV T
2

Substitute into the expression for d/Q.
∂U ∂U
d/Q = dT + + P dV
∂T ∂VV T
Form the derivative with respect to T by dividing by dT and specifying the path as
one of constant P.
d/Q ∂U ∂U ∂V
= + + P
dT P
≡ CP
∂T V
∂V ∂TT P
CV
Now use the given information (∂U/∂V )T =0, PV = NkT, and CV = (5/2)Nk in
the equation above to find that
CP = (7/2)Nk.
a) It is easy to find the heat along the two rectangular paths by integration.
c b
∆Q(acb) = CV dT + CP dT
a c
P2 V1
V2 P2
= CV dP + CP dV
P1 Nk V1 Nk
= (19/2)NkT1
d b
∆Q(adb) = CP dT + CV dT
a d
V2 P1
P2 V2
= CP dV + CV dP
V1 Nk P1 Nk
= (17/2)NkT1
Before we compute the heat along the diagonal path, it is useful to find the difference
in internal energy between b and a. Since the internal energy is a state function, it
does not matter what path we use to find it. We already know the heat input ∆Q
along the path adb and the work ∆W is easy to find.
Vd
∆W(adb) = −P dV = P1 dV = −NkT1−
Va
Then the path-independent result for ∆U can be computed along this particular path.
∆U = ∆Q + ∆W = (17/2 − 2/2)NkT1 = (15/2)NkT1
3

The work along the diagonal path ab can be calculated by integration using the V
dependence of P: P = (P1/V1)V .
b V2
∆W(ab) = P dV = −(P1/V1) V dV−
a V1
= −(1/2)(P1/V1)(V 2
− V1
2
) = −(3/2)P1V1 = −(3/2)NkT12
Finally we can compute the heat supplied along the diagonal path.
∆Q = ∆U − ∆W = 9NkT1
b) Examine the constraint along the diagonal path ab.
d/Q = dU + P dV
∂U ∂U
= dT + +P dV
∂T ∂VV T
0CV
d/Q
dT

dV
Cab = CV +≡
dTab ab
Along the path ab
P =
P1
V1
V =
NkT
V
⇒ V 2
=
V1
P1
NkT.
So along ab we can construct an expression relating dV to dT by taking the derivative
of this expression.
V1
2V dV = Nk dT
P1
∂V V1 Nk
=⇒
∂T ab
P1 2V
Finally
V1 Nk P
Cab = CV + = CV + (1/2)Nk = 3Nk.
P1 2 V
P1/V1
4

As a check we can integrate this heat capacity along the path.
b
∆Q(ab) = Cab dT = 3Nk(Tb − Ta) = 9NkT1
a
This is identical to the result we found above in part a).
Comment: CV = (5/2)Nk is an approximation to a diatomic gas where the rotational
degrees of freedom are contributing to CV but the vibrational degrees of freedom are
not (they are frozen out; we will understand why later in the course). If we had
used the monatomic result CV = (3/2)Nk we would have found CP = (5/2)Nk,
∆Q(acb) = (13/2)NkT1, ∆Q(adb) = (11/2)NkT1, ∆U = (9/2)NkT1, ∆Q(ab) =
6NkT1, and Cab = 2Nk.
Problem 4: Equation of State and Heat Capacity of a Liquid Surface
a) Here we follow the standard procedure for integrating an exact differential. Since
the partial derivatives involve A and T and the expressions for them are given in
terms of A and T, it is natural to choose these as the independent variables (even if
the question did not specifically ask for S(A, T)).
dS(A, T) =
∂S
∂T A
dT +
∂S
∂A T
dA
S(A, T) =
∂S
∂T A
dT + f(A)
1
= ∂T
dT + f(A)
∂S A
Nk
= −
A − b
dT + f(A)
NkT
= −
A − b
+ f(A)
Now we go about finding an expression for the unknown function f(A).
NkT∂S
=
(A − b)2
+ f (A) by differnetiation
∂A T
NkT 2aN2
= given
A3(A − b)2
−
5

Equating the last two lines gives an expression for f (A).
2aN2
f (A) = −
A3
aN2
f(A) = + c
A2
lim = c = 0
N 0
S S
→
from above given
Putting it all together gives the final result.
NkT aN2
S(A, T) = −
A − b
+
A2
+ S0
b) Begin by expanding the differential of the internal energy in terms of the chosen
independent variables.
∂U ∂U
dU = dT + dA
∂T A
∂A T
Next write the first law in terms of the differential of work appropriate to a two
dimensional film and then substitute in the above expansion for the energy differential.
/Q = dU − dd /W
= dU − S dA
∂U ∂U
=
∂T A
dT +
∂A T
− S dA
d/Q ∂U ∂U dA
= +
dT ∂T ∂A
− S
dTA T
The constant area heat capacity is easy since constant area means dA = 0.
d/Q
dT

∂U
=
A
≡ CA
∂T A
In finding CS we must be careful to specify the constraint (or path) associated with
the derivative of the area with respect to temperature in the generic expression found
above.

d/Q ∂U ∂U ∂A
= +≡ CS
∂T ∂A
− S
dT ∂TS A T S
CA
6

The last partial derivative in the above line is not among the ones we are told to use
in the ﬁnal expression, so we use the chain rule to convert it to the partial derivatives
requested.
∂U
∂A
CS = CA − T
− S
∂T ∂S
∂S A ∂A T
Problem 5: Thermodynamics of a Curie Law Paramagnet
All the manipulations we perform here for a simple magnetic system mirror those we
carried out in lecture fjor a simple hydrostatic system.
a) Heat capacities for the genereic magnet.
∂U ∂U
dU = dT + dM expansion
∂T ∂MM T
d/Q = dU − d/W = dU − H dM ﬁrst law
∂U ∂U
=
∂T M
dT +
∂M T
− H dM substitution
CM ≡
d/Q
dT M
=
∂U
∂T M
from line above
CH ≡
d/Q
dT H
=
∂U
∂T M
+
∂U
∂M T
− H
∂M
∂T H
from d/Q
∂U ∂M
CH − CM =
∂M T
− H
∂T H
∂U
∂M T
=
CH − CM
∂M
∂T H
+ H by rearrangement
Now substitute into the general expansion of dU to arrive at
 
dU(T, M) = CM (T, M) dT +
CH(T, M) − CM (T, M)
+ H(T, M) dM∂M
∂T H
7

b) Now ﬁnd the results speciﬁc to the Curie law paramagnet.
∂U
CM = bT, = 0 given
∂M T
∂U ∂U
dU = dT + dM expansion
∂T ∂MM T
CM 0
T
U(T, M) = bT dT = (1/2)bT2
+ f(M) integration
0
∂U
= f (M) = 0 (given) ⇒ f(M) = constant = U(T = 0)
∂M T
0 by assumption
U(T) = (1/2)bT2
+ U(T = 0)
∂M ∂ aH aH
= = from eq. of state
∂T H
∂T T H
−
T2
aH aH2
M2
CH − CM = (0 − H)(−
T2
) = = from a)
T2 a
C
M2
H(T, M) = bT +
a
8

c) Here we practice working with an alternative pair of independent variables.

∂U ∂U
dU =
∂H M
dH +
∂M H
dM expansion
∂U ∂U
d/Q = dU − H dM =
∂H M
dH +
∂M H
− H dM first law and substitution
CM ≡
d/Q
dT M
=
∂U
∂H M
∂H
∂T M
dividing the above by dT
∂M
∂H −1 ∂T H
= ∂T ∂M
= −
χT
chain rule
∂T M ∂M H ∂H T
1 ∂U ∂M
CM = −
χT ∂H M
∂T
substitution

H

∂U CM χT
= rearrangement of above∂M∂H
−
M ∂T H
d/Q ∂U ∂M
= from d/Q expressionCH ≡
dT H ∂M H
− H
∂T H
∂U CH
= + H rearrangement of above∂M∂M H ∂T H
Now substitute into the general expansion of dU(H, M) to arrive at
 
CM (H, M) χT (H, M) CH(H, M)
dU(H, M) = − dH +  + H dM∂M ∂M
∂T H ∂T H
Note that the coefficient of the dM term in the expansion of dU(H, M) found here is
different from the coefficient of the dM term in the expansion of dU(T, M) found in
part a).
9

d) Now we use the equation of state associated with a Curie law paramagnet.
∂M ∂ aH a
χT = =≡
∂H T
∂H T T T
∂M aH
=
∂T
−
T2
H
 
H2
bT a
dU(H, M) = − T
dH + 
bT + a T2
+ H dMaH aH
T2 T2− −
bT2/H −bT2/M
bT2
bT2
= dH − dM
H M
ba2
H ba2
H2
=
M2
dH − dM by eliminating T
M3
ba2
H2
U(H, M) = + f(M) integration
2 M2
∂U ba2
H2
= −
M3
+ f (M)
∂M H
Comparison with the coeﬃcient of dM in the diﬀerential form dU(H, M) above shows
that f (M) = 0 which, when integrated, gives f = constant. Thus we can write
ba2
H2
U(H, M) = + constant
2 M2
There is no reason to carry around a constant term in the internal energy which never
responds to any change in the independent variables, so we are free to set the constant
equal to zero.
ba2
H2
U(H, M) =
2 M2
By using the equation of state, M = aH/T, we see that this reduces to the same
result obtained in b), that is U = (b/2)T2
.
10

e) We are looking for the adiabatic constraint on dT and dM.

∂U ∂U
d/Q =
∂T M
dT +
∂M T
− H dM = 0 for an adiabatic path
CM
∂U
CM dT = − H dM by rearrangement−
∂M T
∂U
dT
⇒
dM

T
− H∂M
∆Q=0
= −
CM
This gives the slope of an adiabatic path for any magnetic system.
f) Next we specialize to the case of a Curie law paramagnet.
∂U
= 0, CM = bT Curie law paramagnet
∂M T
dT
dM ∆Q=0
(0 − H) H M
= = = using the general result from e)−
bT bT ab
1
dT =
ab
M dM after rearrangement
(T − T0) =
1
2ab
(M2
− M2
0 ) integration
1
(T − T0) =
2ab
(M − M0)(M + M0)
11

g) An isothermal path in the H M plane is easy to picture from the given equation of
state, M = aH/T. It is a straight line going through the origin with slope (∂M/∂H)T
= a/T. This is shown in the figure accompanying the statement of the problem. In
part f) we found the relation which must hold between dT and dM along an adiabatic
path: dT = (1/ab)M dM. In order to explore an adiabatic path in the H M plane
we must express dT in terms of dM and dH.
aH
T =
M
equation of state
a aH
dT =
M
dH −
M2
dM differential of above
Substitute this general expression for dT into the adiabatic path derivative and sep-
arate the dH and dM terms.
a aH
(1/ab)M dM = dH − dM
M M2
1 H
M2
T T M2
dH =
M2
+ dM = + dM = 1 + dM
a2b M a2b a a abT
This allows us to find the slope of an adiabatic line in the H M plane in terms of the
quantity a/T which is the slope of an isotherm.
dM

dH

a 1 a

=
T 1 + M2
abT
T∆Q=0
Note that the slope of the adiabatic path is less than that of the isothermal path
going through the same point.
12


Physics Department

Problem 1: Classical Magnetic Moments
a)
mi ≤ µ−µ ≤
−µN ≤ M ≤ µN
−N(µH) ≤ E ≤ N(µH)
b) There are 2N microscopic variables necessary to specify the state of the system.
Some possibilities include the x and z component of each spin, the z component and
the angle in the x y plane for each spin, or the polar angles θ and φ for each spin.
c)
H ∂S 1 ∂Ω
= = using S = −k ln Ω−k
T
−
∂M N
Ω ∂M N
1 2M M
= Ω = k−k
Ω
−
(2/3)µ2N (1/3)Nµ2
Nµ2
H
⇒ M(H, T) = the Curie law result
3k T
d) The expression found in c) allows M to grow without bound as T 0. But M is→ | |
bounded by µN. Thus the expression can only be trusted as an approximation when
M(H, T) µN
Nµ2
µN
3kT
⇒ kT (1/3)µH
This result says that the “thermal energy”, kT, must be much greater than the
maximum energy allowable for a single spin.
e) To ﬁnd Ω reduce N by 1 and reduce M by m.
Ω = (2µ)N−1
exp
(M − m)2
2
−
(2/3)(N − 1)µ
1

f)
p(m) =
Ω
Ω
=
(2µ)N−1
(2µ)N
exp − (M−m)2
(2/3)(N−1)µ2
exp − M2
(2/3)Nµ2
=
1
2µ
exp −
M2
− 2Mm + m2
(2/3)(N − 1)µ2
exp
M2
(2/3)Nµ2
1 3mM
≈
2µ
exp
Nµ2
small since MµN
1 3M
≈
2µ
1 + (
Nµ2
) m − µ ≤ m ≤ µ
Now check the normalization.
1µ µ µ 3M
p(m) dm = dm + m dm
−µ −µ 2µ −µ 2Nµ3
= 1 + 0 = 1
g)
m = p(m) m dm
µ µ1 3M
= m dm + m2
dm
−µ 2µ −µ 2Nµ3
=0
µ 13M 3 M
= m =
2Nµ3
−µ 3 N
2

This tells us that the total magnetization M is N times the average moment of an
individual dipole, the result that one would expect on physical grounds.
Problem 2: A Strange Chain
a) The number of ways of choosing n+ elements from a total of N is N!/(N −n+)!n+!.
It follows that
N!
Ω(N, n+) =
(N − n+)!n+!
S(N, n+) = k ln Ω
k{N ln N − (N − n+) ln(N − n+) − n+ ln n+ −N + (N − n+) + n+≈ }
=0
= k{N ln N − (N − n+) ln(N − n+) − n+ ln n+}
b)
∂S

T

F
= −
∂L N,E
∂S ∂n+
= −
∂n+ ∂L
1/2l
k n+
=
N − n+
+ ln(N − n+) − − ln n+−
2l N − n+ n+
2lF
n
−
kT
= ln
N − n+

+

kT
F(N, T, n+) = −
2l
ln
N − n+
n+
c) Now rearrange the last result, take the exponential of both sides and solve for n+.
n
exp[−2lF/kT] =
N − n+
+
1
n+ = N
1 + exp[−2lF/kT]
3

Next, use the expression for n+ to ﬁnd L.
2 1 + exp[]
L = l(2n+ − N) = Nl
1 + exp[]
−
1 + exp[]
1 − exp[−2lF/kT]
= Nl
1 + exp[−2lF/kT]
exp[lF/kT] − exp[−lF/kT]
= Nl
exp[lF/kT] + exp[−lF/kT]
= Nl tanh(lF/kT)
For high temperatures, where kT lF, tanh x x for small x, so→
Nl2
L ≈ F.
kT
The fact that the length L is proportional to the tension F shows that Hooke’s law
applies to this system, at least for high temperatures.
d)
∂L
α ≡ L−1
∂T F
1 1
= (L)
L
−
T
1
= −
T
Problem 3: Classical Harmonic Oscillators
a) In x space
2N
2
E = xi
i=1
is a sphere in 2N dimensions with radius
√
E. Its volume is πN
EN
/N!. The corre-
sponding volume in pq space is
 
π
N
N
2
Ω(E, N) = (
√
2m )N   EN
N! mω2
2π N
1
= EN
ω N!
4

b)
2π N
1
S(E, N) = k ln Ω(E, N) = k ln EN
ω N!
c)
1
T
=
∂S
∂E N
= k
NE−1
{}
{}
=
Nk
E
⇒ E = NkT CN = Nk
d) Let Ω be the volume in a phase space for N−1 oscillators of total energy E− where
2
= (1/2m)p2
+ (mω2
/2)qi . Since the oscillators are all similar, = E/N = kT.i

p(pi, qi) = Ω /Ω

N−1
2π 1
Ω =
Ω
ω (N − 1)!
(E − )N−1

E − N

2π −1
N! 1
=
Ω ω (N − 1)! E E −
ω N N
=
2π E −
1 −
E
≈ −1 ≈exp[− / ]
1
p(pi, qi) =
(2π/ω)
exp[− / ]
1 2 2
= exp[−pi /2mkT] exp[−(mω2
/2kT)q ]
(2π/ω)kT i
 
1 2  = exp[−pi /2mkT]
1
exp[−qi
2
/2(kT/mω2
)]√
2πmkT 2π(kT/mω2)
= p(pi) × p(qi) ⇒ pi and qi are S.I.
5

Problem 4: Quantum Harmonic Oscillators
a)
E = (1/2)¯ hωMhωN + ¯
Ω(E, N) = Ω(M, N) =
(M + N − 1)!
M!(N − 1)!
b)
S(M, N) = k ln Ω(M, N)
k{(M + N − 1) ln(M + N − 1) − M ln M − (N − 1) ln(N − 1)}≈
c)
1 ∂S ∂S ∂M
= =
T ∂E ∂M ∂EN N N
k
=
hω
{1 + ln(M + N − 1) − 1 − ln M}
¯
k M + N − 1
= ln
¯hω M
Neglect 1 compared to N.
M + N
= exp[¯hω/kT]
M
M = N(exp[¯hω/kT] − 1)−1
1 1
E = N¯hω +
2 exp[¯hω/kT] − 1
hω/kT] − 1)−1
→ exp[−¯ hω,(exp[¯ hω/kT] when kT ¯
kT 1
→
hω
−
2
when kT ¯hω.
¯
6

Take careful note of the 1/2 in the last expression. It cancels the other 1/2 appearing
in the expression for E in the high temperature limit.
d) Remove one oscillator in state n with energy
hω(1/2 + n), = ¯= ¯ hω(1/2+ n ), n = M/N.
Then for the reduced phase space N → N − 1 and M .→ M − n
Ω
p(n) =
Ω
Ω(M − n, N − 1)
=
Ω(M, N)
(M + N − n − 2)! M! (N − 1)!
=
(M + N − 1)! (M − n)! (N − 2)!
n terms
=
· · · (M − n − 1)
(M + N − 1)(M + N − 2)

M(M − 1)
· · · (M + N − n − 1)
× (N − 1)

n+1 terms
N M n
≈
M + N M + N
1 n n
=
1+ n 1+ n
Note the the ﬁnal result for p(n) is properly normalized.

0.5 1 1.5 2
0.5
1
1.5
2
7


Physics Department

Problem 1: Energy of a Film
a) The best approach to take here is to find a general expression for CA and then
show that its derivative with respect to A is zero.
d/Q
CA ≡
dT A
∂S
= T by the second law
∂T A
∂CA ∂2
S
= T
∂A ∂A∂TT
∂ ∂S
= T interchanging order of the derivatives
∂T ∂A T A
We use a Maxwell relation to find (∂S/∂A)T . Note that S and S are different
variables. I would normally construct a magic square to fine the equivalent derivatives,
but for clarity I will go through the more fundamental route here.
dE = T dS + S dA
dF = dE − d(TS) = −S dT + S dA
Since F is a state function, the cross derivatives must be equal.
∂S ∂S Nk
= =−
∂A T
∂T
−
A − bA
Substitute this result into the expression for the derivative of the heat capacity.
∂CA
= T
∂ Nk
= 0
∂A T
∂T A − b A
This shows that the heat capacity at constant area does not depend on the area:
CA(T, A) = CA(T).
1

b) Now we ﬁnd the exact diﬀerential for the energy and integrate up.

dE = T dS + S dA
∂S ∂S
= T
∂T A
dT + T
∂A T
+ S dA
CA(T)
−
NkT
A − b
+ S = 0
∂E
= 0⇒
∂A T
E(T, A) = E(T)
T
= CA(T ) dT + E(T = 0)
0
Problem 2: Bose-Einstein Gas
a) In this problem, we just follow the directions.
dE = T dS − P dV
∂S ∂S
= T dT + T − P dV
∂T ∂VV T
CV
dF = dE − d(TS) = −S dT − P dV
∂S ∂S
=⇒ −
∂V T
−
∂T V
∂P ∂S
= (5/2)aT3/2
+ 3bT2
=
∂T ∂VV T
∂S
T − P = (5/2)aT5/2
+ 3bT3
− aT5/2
− bT3
− cV −2
∂V T
= (3/2)aT5/2
+ 2bT3
− cV −2
Collecting this all together gives
dE = (dT3/2
V + eT2
V + fT1/2
)dT + ((3/2)aT5/2
+ 2bT3
− cV −2
)dV
2

b) Use the fact that the energy is a state function which requires that the cross
derivatives must be equal.
∂ ∂E ∂ ∂E
=
∂V ∂T ∂T ∂VV T T V
dT3/2
+ eT2
= (15/4)aT3/2
+ 6bT2
⇒ d = (15/4)a, e = 6b
c) Use the results from b) to simplify the expression for dE in a).
dE = ((15/4)aT3/2
V + 6bT2
V + fT1/2
)dT + ((3/2)aT5/2
+ 2bT3
− cV −2
)dV
Integrate with respect to T ﬁrst.
E = (3/2)aT5/2
V + 2bT3
V + (2/3)fT3/2
+ F(V )
∂E
= (3/2)aT5/2
+ 2bT3
+ F (V ) from above
∂V T
= (3/2)aT5/2
+ 2bT3
− cV −2
from dE
⇒ F = −cV −2
, F = cV −1
+ KE
E = (3/2)aT5/2
V + 2bT3
V + (2/3)fT3/2
+ cV −1
+ KE
d) Proceed just as we did above for E.
∂S ∂S
dS = dT + dV
∂T ∂VV T
CV /T ∂P
from a)
∂T V
= (dT1/2
V + eTV + fT−1/2
)dT + ((5/2)aT3/2
+ 3bT2
)dV
Integrate with respect to T ﬁrst.
S = (2/3)dV T3/2
+ (1/2)eV T2
+2fT1/2
+ G(V )
(5/2)aV T3/2
3bV T2
3

∂S
= (5/2)aT3/2
+ 3bT2
+ G (V ) from above
∂V T
= (5/2)aT3/2
+ 3bT2
from dS
⇒ G (V ) = 0, (V ) = KSG
S(T, V ) = (5/2)aV T3/2
+ 3bV T2
+ 2fT1/2
+ KS
Problem 3: Paramagnet
a) This is virtually identical in approach to problem 1.
d/Q
CM ≡
dT M
∂S
= T
∂T M
by the second law
∂CM
= T
∂2
S
∂M T
∂M∂T
∂ ∂S
= T
∂T ∂M T M
interchanging order of the derivatives
We will need H(T, M) for what follows.
A M
M = H ⇒ H = (T − T0)
AT − T0
We use a Maxwell relation to ﬁnd (∂S/∂M)T .
dE = T dS + H dM
dF = dE − d(TS) = −S dT + H dM
Since F is a state function, the cross derivatives must be equal.
∂S ∂H M
= =−
∂M T
∂T AM
4

Substitute this result into the expression for the derivative of the heat capacity.
∂CM ∂ M
= T
∂M T
∂T
−
A M
= 0
This shows that the heat capacity at constant magnetization does not depend on the
magnetization: CM (T, M) = CM (T).
b)
dE = T dS + H dM
∂S ∂S
= T dT + T + H dM
∂T ∂MM T
CM (T) −MT/A + H = −MT0/A
Do the T integration ﬁrst.
T
E(T, M) = CM (T ) dT + f(M)
0
∂E
= f (M) from above
∂M T
MT0
= −
A
from dE
M2
T0
⇒ f(M) = − + KE
2A
T M2
T0
E(T, M) = CM (T ) dT + KE
0
−
2A
c)
∂S ∂S
dS = dT + dM
∂T M
∂M T
CM (T)/T −M/A from a)
S(T, M) =
T
0
CM (T )
T
dT + g(M)
5

∂S
= g (M) from above
∂M T
M
= −
A
from dS
M2
⇒ g(M) = − + KS
2A
T CM (T ) M2
S(T, M) = dT + KS
0 T
−
2A
Problem 4: Sargent Cycle
a) The equation of state, PV = NkT, shows that if V is held constant, increasing P
increases T. Thus T3 T2. Similarly if P is held constant, increasing V increases T.
Therefore T4 T1.
For an adiabatic process PV γ
= P0V0
γ
, or equivalently TV γ−1
= T0V γ−1
. Since
T
0
γ − 1 0, increasing V adiabatically decreases T. This tells us that T4 T3 and
1 T2. We can put this all together as follows.
T
T1 T2 T3 T3 is the hottest
1 T4 T3
⇒
T1 is the coldest
b) For an isotherm
∂P ∂ NkT NkT P
= = = .
∂V ∂V V
−
V 2
−
VT T
6

For an adiabatic path
∂P
∂V S
=
∂
∂V
P0V γ
0
V γ
T
= −γ
P
V
.
Since γ 1, “adiabats” descend more steeply than isotherms. In the following phase
diagrams, illustrating two extreme versions of the Sargent cycle, isotherms have been
indicated by dashed lines.
T4 T2 T4 T2
c) The definition of an engine’s efficiency together with the first law of thermodynam-
ics leads to η = 1 − ( QC / QH ). So the problem is reduced to finding expressions| | | |
for the heat taken in at high temperature QH and that dumped at low temperature| |
QC . Note that by definition there is no heat transfer along the adiabatic parts of| |
Q
the path. The temperature increases in going from 2 to 3, so QH = CV (T3 − T2) in
a constant V process. The temperature of the gas decreases going from 4 to 1, so
C = CP (T1 − T4) or QC = CP (T4 − T1) in a constant P process. Therefore,| |
QC CP (T4 − T1)
T
η = 1 −
| |
= 1 −
CV (T3 − T2)
= 1 − γ
T4 − T1
.
3 − T2|QH|
d) For a fluid d/W(work done on a substance) = −PdV .
W /W = W1 2 + W2 3 + W3 4 + W4 1≡ d → → → →
W
2 2
1 2 = −
1
P dV = −P1V1
γ
V −γ
dV→
1
7

P1V1
γ 2
V −(γ−1)
=

γ − 1 1

 
 
 1 P1V γ
1
=  
 V γ−1 −P1V1
γ − 1  2 
P2V2
1
=
γ − 1
(NkT2 − NkT1)
W2 3 = 0 since dV ≡ 0→
W
4 4
3 4 = −
3
P dV = −P3V3
γ
V −γ
dV→
3
P3V3
γ 4
V −(γ−1)
=

γ − 1 3

 
 
1 P3V γ
3 = 

V γ−1 −P3V3
γ − 1  4
P4V4
1
=
γ − 1
(NkT4 − NkT3)
W
1 1
4 1 = P dV = −P1 dV→ −
4 4
= −P1(V1 − V4) = P4V4 − P1V1
= Nk(T4 − T1)
1
W = Nk[T4 − T1 +
γ − 1
(T2 − T1 + T4 − T3)]
γ 1
= Nk
γ − 1
(T4 − T1) −
γ − 1
(T3 − T2)
8

e)

∆E = ∆W + ∆Q = 0 for a complete cycle
⇒ Q = −W = |QH| − |QC|
Nk
−W =
γ − 1
[(T3 − T2) − γ(T4 − T1)] from d)
Alternatively we know that
= CV (T3 − T2) − CP (T4 − T1)|QH| − |QC|
= CV [(T3 − T2) − γ(T4 − T1)]
Equating the two expressions gives
Nk
= CV
γ − 1
or
Nk = CV (γ − 1) = CP − CV
Problem 5: Entropy Change
We assume that the mixing is done at constant pressure so we will be using the
constant pressure heat capacity, CP . We will also assume CP is independent of T in
the region of interest. Take each mass of fluid by some reversible path between the
initial and final states to determine ∆E and ∆S. Note that the path taken between
the same initial and final states in the actual mixing is irreversible.
9

No work is done on either fluid in the mixing process. Therefore for each mass of
fluid
dE = d /W/Q + d
= 0
= T dS
∂S ∂S
= T dT + T dP
∂T P
∂P T
= 0
CP
T
C
= CP dT
Simple integration then gives the results
∆E1 = CP (TF − T1), ∆E2 = CP (TF − T2)
The system as a whole is isolated, so
0 = ∆Q1 + ∆Q2
= ∆E1 + ∆E2
= CP (TF − T1) + CP (TF − T2)
P drops out of the final line above, and rearrangement gives us the final temperature:
F = (T1 + T2)/2.
Now compute the ∆Ss for the reversible paths.
∂S ∂S
dS = dT + dP
∂T ∂PP T
= 0
CP /T
TF CP TF
∆S1 = dT = CP ln
T1 T T1
TF CP TF
∆S2 = dT = CP ln
T2 T T2
∆ST = ∆S1 + ∆S2
10

T2
TF
T
= CP (ln + ln
TF
) = CP ln F
1 T2 T1T2
TF T1 + T2
= 2CP ln = 2CP ln√
T1T2 2
√
T1T2
= 2MCP
T1 + T2
ln
2
√
T1T2unit mass

If ∆ST is to be positive for all positive T1 and T2 for which T1 = T2, the argument of
the logarithm must be greater than 1, which requires that T1 + T2 2
√
T1T2 for all
positive T1 = T2. We now prove that this is indeed true.
(T1 − T2)2
0 square of a real number
not = 0 is positive
T2
1 − 2T1T2 + T2
2 0
T2
1 + 2T1T2 + T2
2 4T1T2 add 4T1T2 to both sides
(T1 + T2)2
4T1T2
T1 + T2 2 T1T2
11

�
� �

Physics Department

Problem 1: Correct Boltzmann Counting
a)
4πemE
�3N/2
Φ = V N
3N
3N/2
= V N
[2πemkT] using E = (3/2)NkT
S(N, V, T) = k ln Φ
3N/2
= k ln V N
[2πemkT]
= Nk ln V + (3/2)Nk ln[2πemkT]
Now let N λN, V λV , and T T. Then as a result→ → →
S λNk ln(λV ) +λ(3/2)Nk ln[2πemkT].→
� ��
= λNk ln V
So S → λS because of the failure in the ﬁrst term.
b) The pressure is the same on both sides of the partition, so
P =
N1kT
=
N2kT
.
V1 V2
Now make use of the deﬁnition of α.
N1kT N2kT
=
αV (1 − α)V
We can solve this to put N1 and N2 in terms of α.
N2 =
1 − α
α
N1
N1
N1 + N2
=
N1
(1 + 1−α
α
)N1
= α =
N1
N
N2
N1 + N2
=
N2
( α
1−α
+ 1)N2
= 1 − α =
N2
N
1

� ��
Since the mixing takes place isothermally (because for ideal gases there is no interac-
tion between the molecules), the T term in our expression for S of each gas does not
change.
∆S1 = N1k ln V − N1k ln αV
= N1k ln(1/α) = Nkα ln(1/α)
∆S2 = N2k ln V − N2k ln[(1 − α)V ]
= N2k ln(1/1 − α) = Nk(1 − α) ln(1/1 − α)
∆S1 + ∆S2 = α ln(1/α) + (1 − α) ln(1/1 − α) 0Nk[��
+ + + +
This result is correct if the two gases are different. What should we expect when
the gases are the same? ∆E = 0 since the internal energy of an ideal gas does not
depend on the volume, E(T, V ) = E(T), and the initial and final temperatures are
equal. ∆W = 0 since no work is necessary to slide the partition in and out (there is
no opposing force in the absence of friction). Using these two results in the first law,
∆E = ∆W + ∆Q, tells us that ∆Q = 0. If the process is reversible ∆S = ∆Q/T
and it follows that ∆S = 0. This is not consistent with the detailed calculation above
which indicated a positive ∆S, but which nowhere required that the two gases be
different.
c)
V N

Φ = [2πemkT]3N/2

N!
S(N, V, T) = Nk ln V − Nk ln N +k ln N +(3/2)Nk ln[2πemkT]
neglect compared to
previous term
= Nk ln(V/N) + (3/2)Nk ln[2πemkT]
Now let N λN, V λV , and T T.→ → →
S → λNk ln(V/N) + λ(3/2)Nk ln[2πemkT] = λS
We can summarize the results for the volume-dependent part of the entropies when
2

�
the mixing involves only one gas by constructing a table.

volume-dependent term in the entropy
with partition without partition
old S αNk ln αV + (1 − α)Nk ln(1 − α)V
= Nk[α ln αV + (1 − α) ln(1 − α)V ]
= Nk[α ln α + (1 − α) ln(1 − α)
� ��
+ ln V ] = Nk ln V
+
new S αNk ln αV
αN
+ (1 − α) ln (1−α)V
(1−α)N
= Nk[α ln(V/N) + (1 − α) ln(V/N)]
= Nk ln(V/N) = Nk ln(V/N)
Problem 2: Torsional Pendulum
2) First ﬁnd the Hamiltonian.
H = T + V + 1
Iθ˙2
+ 1
K(θ − θ0)2
2 2
Then use the canonical ensemble expression for the probability density.
p(θ, θ˙) ∝ exp[−
H
]
kT
∝ exp[−
θ˙2
(θ − θ0)2
]
2(kT/I)
] exp[−
2(kT/K)
Notice two important features of this result. First, the probability density factors,
p(θ, θ˙) = p(θ)p(θ˙), so θ and θ˙ are statistically independent. Second, the dependence
on both θ and θ˙ has the Gaussian form. In particular, p(θ) is Gaussian with mean θ0
and variance σ2
= kT/K. Therefore,θ
kT1/2
(θ − θ0)2
= .
K
θ = θ ˙b) Since θ and θ˙ are statistically independent, θ ˙ θ . By inspection,
p(θ˙) is a zero-mean Gaussian, so ˙ θ = 0.θ = 0 which leads to θ ˙
3

�
� �
� � � �
�
Problem 3: Defects in a Solid
a)
Zone = exp[− i/kT] = 2 exp[−0/kT] + 3 exp[−∆/kT]
states
= 2 + 3 exp[−∆/kT]
Ztotal = (Zone)N
= (2 + 3 exp[−∆/kT])N
b)
F = −kT ln Z = −NkT ln(2 + 3 exp[−∆/kT])
∂F
S(T, N) = = kT ln(2 + 3 exp[−∆/kT]) + 3Nk
�
∆
�
exp[−∆/kT]
−
∂T N
kT 2 + 3 exp[−∆/kT]
c) In the high temperature limit where kT ∆, all states are equally probable for a
single defect.
1
pi = 5
i = 1, 2, 3, 4, 5
1 3
= 0 × 2 1
+ ∆ × 3 × = ∆5 5 5
×
E = N = 3
∆N5
Problem 4: Neutral Atom Trap
a) First write down the Hamiltonian for one atom.
2 2 2
px py pz
= + + + arH1
2m 2m 2m
Then compute the partition function
2
y
2
x
2
z1 ∞ ∞ ∞pp p
dpz e− ar
kT r2
sin θ drdθdφe−
e−
e−
Z1 = dpx dpy2mkT 2mkT 2mkT
h3
�
−∞
��
−∞
��
−∞
�� V
��
√
2πmkT
√
2πmkT
√
2πmkT 4π
∞
0
exp[−ar/kT] r2 dr
4

� ��
� �
� �3 �
2πmkT kT
= ( )3/2
4π
∞
y2
e−y
dy
h2 a 0
2
2πmk
)3/2
T9/2
a−3
= 8πk3
(
h2
In order to emphasize the dependence on the important variables, this can be written
in the form Z1 = ATα
a−η
where
2πmk
)3/2
A = 8πk3
( α = 9/2 and η = 3.
h2
b) Remember to include correct Boltzmann counting.
1
Z = Z1
N
N!
F = −kT ln Z = −kT(N ln Z1 − N ln N + N)
= −NkT ln(Z1/N) − NkT
∂F
S = −
∂T N
1 Z1/N
= Nk ln(Z1/N) + Nk + NkT (9/2)
Z1/N T
= Nk ln(Z1/N) + (11/2)Nk
c)
dQ = 0 no heat is exchanged with surroundings
dQ = dS/T process is said to be reversible
dS = 0, S is constant⇒
⇒ Z1 is constant, using the result from b)
5

� �
�
⇒ T9/2
/a3
is constant and = T
9/2
0 /a3
0
�
T
�9/2 �
a
�3
=
T0 a0
�
a
�2/3
T = T0
a0
Problem 5: The Hydrogen Atom
a)
A
H|n, l, m = n, l, m −
n2
|
The lowest energy, −A, corresponds to 1, 0, 0 and is non-degenerate. The next|
lowest energy, −A/4, is four fold degenerate:
2, 0, 0 , 2, 1, 1 , 2, 1, 0 , and 2, 1, −1 .| | | |
The ratio of the number of atoms in the ﬁrst excited energy level to the number in
the ground state depends on both the energies and the degeneracies.
N(−A/4) 4 exp[A/4kT]
= 4 exp[−(3/4)A/kT]=
N(−A) exp[A/kT]
Using the conversion factor 1meV = 11.6K we ﬁnd that 13.6eV = 1.58 × 105
K. Eval-
uating the above ratio gives 4.8 × 10−170
at 300K and 1.6 × 10−51
at 1000K.
b) The degeneracy of the nth
energy level is
1 + 3 + 5 + · · · + (2n − 1) = n2
.
The partition function for a single atom, neglecting the unbound states, is
Z = exp[− i/kT] =
∞
n2
exp[α/n2
],
states i n=1
where α ≡ A/kT. Since α 0, it follows that exp[α/n2
] 1 for all n. Using this we
can set a lower bound for Z, but Z diverges since the lower bound diverges.
Z
∞
n2
which diverges
n=1
6

c) The Coulomb potential is a mathematical oddity in that it produces an inﬁnite
number of bound states with energies less than zero. This situation if modiﬁed in the
real world by the presence of walls (consider the energy levels of a particle in a box)
or by the presence of other atoms. The existence of hydrogen atoms in the interstellar
medium, on the other hand, probably has more to do with the absence of excitation
mechanisms (non-equilibrium) than with the presence of neighboring atoms.
7


Physics Department

Problem 1: Why Stars Shine
a) The electostatic potential outside the charged sphere depends only on r, the mag-
nitude of the distance from the center of the sphere.
e
φ(r) =
| |
r ≥ R
r
The potential energy of another proton, considered to be a point particle, in this field
is
2
e
V (r) = qφ(r) =
r
Then the minimum energy that the second proton must have to get within a radial
distance R of the first is
2
e
Emin = V (R) = =
(4.8 × 10−10
)2
= 1.92 × 10−6
ergs
R 1.2 × 10−13
b) In problem 1 of problem set 3 we found the following expression for the kinetic
energy of a particle in a three dimensional classical gas.
2 1 E
p(E) = exp[−E/kT]√
π kT kT
Now find the probability p+ that a given proton in the stellar plasma has an energy
greater than Emin.
∞
p+ prob(E Emin) = p(E) dE
≡
Emin

22 1 ∞ E ∞
= √
π kT Emin kT
exp[−E/kT] dE = √
π ymin=Emin/kT
√
y exp[−y] dy
2
≈ √
π
√
ymin exp[−ymin]
1

This is going to turn out to be a very small number, probably too small to be repre-
sented on a hand calculator. Therefore, let’s work toward getting its logarithm.
 
2
log10(p+) = log10

Emin + log10 exp[−Emin/kT]√
π kT
Eminlog10 exp[−Emin/kT] =
Emin log10(e) = −(.4343)−
kT kT
Emin =
1.920 × 10−6
kT 1.381 × 10−16 × 4 × 107
= 3.476 × 102
log10(p+) = 1.323 − 1.510 × 102
= −149.6
× 10−149
p+ = 0.2
c)
8kT
v =
πm
=
8 × 1.381 × 10−16
× 4 × 107 1/2
= 9.18 × 107
cm/sec
π × 1.67 × 10−24
2
σ = π(2R)2
= π(2.4 × 10−13
)2
= 1.81 × 10−25
cm
100
n =
ρ
=
1.67 × 10−24
= 5.99 × 1025
protons/cm3
Mproton
L = (nσ)−1
= 9.22 × 10−2
cm
τcollision = L/ v = 1.01 × 10−9
sec
d) The fusion rate per proton is p+ times the collision rate per proton. But in general
a rate equals the reciprocal of the characteristic time between events, so
τ
1.01 × 10−9
× 10140
fusion = τcollision/p+ =
× 10−149
= 5 sec
0.2
The universe is about 15 billion years old, corresponding to a time
Tuniverse = 15 × 109
× 365 × 24 × 60 × 60 = 4.7 × 1017
sec
2

N
If the mass of the sun is 2 × 1033
grams then the number of protons it contains is
given by
2 × 1033
protons =
1.67 × 10−24
= 1.2 × 1057
Then for the entire sun, the total number of fusions per second is found as follows.
number of fusions per second = Nprotons × fusion rate per proton
= Nprotons/τfusion
× 10140
= 2 × 10−84
= 1.2 × 1057
/ 5 sec−1
Problem 2: Two-Dimensional H2 Gas
a)
±3 9¯ m
h2
/2I 2
±2 4¯ m
h2
/2I 2
±1 ¯ m
h2
/2I 2
m
0 0 1
m degeneracy
m = (¯ 2
h2
/2I) m
Zrot,1 = exp[− (state)/kT] =
∞
exp[−(¯ 2
h2
/2IkT) m ]
states m=−∞
∞
= 1 + 2 exp[−(¯h2
/2IkT) j2
]
j=1
3

b)
p(m = 3) Z−1
exp[−9¯h2
/2IkT]
h2
/IkT]=
h2
/2IkT]
= exp[−(5/2)¯
p(m = 2) Z−1 exp[−4¯
c)
h2
/2IkT]
| h2
/2I) =
Z−1
exp[−9¯
p(m = 3 = 9¯
h2
/2IkT])
= 1/2
2(Z−1 exp[−9¯
h2
/2IkT] 1
p(m = 1 ≤ ¯
Z−1
exp[−¯
=| h2
/2I) =
Z−1 + 2(Z−1 exp[−¯ h2
/2IkT])h2
/2IkT]) 2 + exp[¯
d)
2
Zrot,1 =
∞
exp[−(¯ 2
h2
/2IkT) m ] dmh2
/2IkT) m ]
∞
exp[−(¯
m=−∞
→
−∞
∞ 2 1/2
m 2πIkT
= exp[−
h2
)
] dm =
h2 ∝ β−1/2
−∞ 2(IkT/¯ ¯
Gaussian normalization
1 ∂Z1
Erot = N = N −
Z1 ∂β
1 Z1 1
= N = (1/2)N = (1/2)NkT−
Z1
−
2β β
Problem 3: Adsorption on a Realistic Surface
a) E(5) lies lowest since it is occupied most often, even after adjusting for a degeneracy

diﬀerence (at most a factor of two).

b)

p(5)/p(4) =
N exp[−E(5)/kT]
=
1
exp[(E(4) − E(5))/kT] = 2
2N exp[−E(4)/kT] 2

(E(4) − E(5))/k = T ln 4 = 300 ln 4 = 415.9K

4

4
p(5)/p(3) =
N exp[−E(5)/kT]
=
1
exp[(E(3) − E(5))/kT] =
2N exp[−E(3)/kT] 2 1.5
(E(3) − E(5))/k = 300 ln(8/1.5) = 502.2K
4
p(5)/p(2) =
N exp[−E(5)/kT]
= exp[(E(2) − E(5))/kT] =
N exp[−E(2)/kT] 1.5
(E(2) − E(5))/k = 300 ln(4/1.5) = 294.2K
p(5)/p(1) =
N exp[−E(5)/kT]
= exp[(E(1) − E(5))/kT] = 4
N exp[−E(1)/kT]
(E(1) − E(5))/k = 300 ln 4 = 415.9K
p(4)/p(5)

p(3)/p(5)

p(2)/p(5)

p(1)/p(5)

2N
= exp[−415.9/77] = 0.0090
T=77K N
2N
= exp[−502.2/77] = 0.0029
T=77K N
N
= exp[−294.2/77] = 0.0219
T=77K N
N
= exp[−415.9/77] = 0.0045
T=77K N
The four numbers above sum to 0.0383. Now we can use normalization to ﬁnd each
of the probabilities separately.
p(1) + p(2) + p(3) + p(4) + p(5) = 1
p(1) p(2) p(3) p(4)
+ + + + 1 p(5) = 1 = (1.0383)p(5)
p(5) p(5) p(5) p(5)
5
c)

p(5) = 0.9631
p(4) = 0.0087
p(3) = 0.0028
p(2) = 0.0211
p(1) = 0.0043
1.0000
d) The probability of hopping oﬀ a given site (given that the ad-atom is on that site)
in a time increment ∆t beginning at time t is independent of t. This is just like
radioactive decay. Thus the waiting time t1 until the hop occurs has an exponential
probability density. This is just what was found in the STM measurements of Si
dimers on the 001 surface of Si.
6


Physics Department

Problem 1: The Big Bang
If the expansion is adiabatic, ∆S = 0.
∂F
S = −
∂T V
∂ 1 π2
=
h3 (kT)4
V−
∂T
−
45 c3¯
4 π2
= k4
T3
V
h3
From this result we see that the product T3
V remains constant during the expansion.
Therefore
V (3K) 30003
45 c3¯
= = 109
V (3000K) 33
Problem 2: Comet Hale-Bopp
a) The total power radiated by the sun
= surface area of the sun × e(Tsun)
= 4πR2
sun e(Tsun)
The fraction intercepted by the comet
cross-sectional area of the comet
=
4πr2
πR2
comet
=
4π(200Rsun)2
The power absorbed by the comet
πR2
sune(Tsun) × comet
= 4πR2
4π(200Rsun)2
πR2
comet e(Tsun)
=
(200)2
1

The power radiated by the comet
= surface area of the comet × e(Tcomet)
= 4πR2
comet e(Tcomet)
In the steady state, the power absorbed equals the power radiated:
πR2
comet e(Tsun)
= 4πR2
comet e(Tcomet)
(200)2
Note that the radius of the comet cancels out here.
σT4
sun
= 4σT4
comet
(200)2
The Stefan-Boltzmann constant cancels out here. Finally
1 1/4
Tsun 6000
Tcomet = Tsun = = = 300K
16 × 104 20 20
b) The coma surrounding the nucleus could absorb or reflect away part of the incoming
radiation from the sun. Also, the nucleus might not be a “black” absorber. Both
effects would lead to a colder nucleus. On the other hand, the coma would reflect
some radiation emitted from the comet back toward it (a greenhouse effect) and if the
nucleus is a poorer absorber, it is also a poorer emitter. A quantitative estimate of
the temperature of the nucleus requires more information and detailed heat transport
calculations.
c) 300K is close to the earth’s average surface temperature. It is not a coincidence.
The temperature we calculate is independent of the radius of the body and inversely
proportional to the square of the body’s distance from the sun. The sun does heat
the planets:
• Planets get colder as one goes farther from the sun,
• The seasons on the earth depend on the continent’s average inclination toward
the sun.
But, there is also heat coming from the nuclear fission of radioactive elements in the
earth’s interior:
2

• The earth’s surface was once much hotter (molten) when this heat source (as
well as a gravitational component) was much greater than the heat flux from
the sun.
• The earth’s temperature increases with depth below the surface (this is evident
in deep shaft mines) which implies a heat flow toward the surface from the
interior.
Problem 3: Super Insulation
a)
J = σT4
C = σ(T4
C) ≡ J0H − σT4
H − T4
b) Since the sheet is neither cooling nor warming, the flux in from TH must equal the
flux out to TC.
H − T4
) = σ(T4
− T4
H + T4
= 2T4
σ(T4
C) ⇒ T4
C
T4
H + T4 1/4
T = C
2
c)
J = σ(T4
H − T4
)
1
H + T4
= σ[T4
C)]H − 2
(T4
= 1
σ(T4
C) = 1
J0H − T4
2 2
d) The net energy gain for any one sheet is zero, since the temperatures are constant
in the steady state. This implies that the energy flux, J, between any two adjacent
sheets is the same.
3

4
T−1 i−
4
Ti −1 − J/σ4
T= i
4
TiJ/σ
= ⇒
T4
=1 − J/σ
T4
T4
4
J/σ T− =1 H
4
J/σ T− =2 H
4
TH
T4
2 − 2J/σ
− 3J/σ
=
T4
3 =

· · · · · ·

4
TC
4
J/σ T− = H
4
Tn − (n + 1)J/σ
=
Solving the ﬁnal equation for J gives the result

J =

σ

n + 1

=

1

n + 1

4
T− C
4
TH J0
Problem 4: Properties of Blackbody Radiation
a) In class we found the following expression for the internal energy of the thermal
radiation ﬁeld:
U(T, V ) =
1
15
π2
c3¯h3 (kT)4
V ;
therefore
CV =
∂U
=
4
π2 kT
3
kV.
∂T 15 hc¯V
Since (kT/¯hc) has the units of (length)−1
, CV has the units of k, which is correct.
k 1.38 × 10−16
ergs K−1
= = 4.36 K−1
cm−1
¯hc 1.054 × 10−27 erg-s × 300 × 1010 cm s−1
For 1m3
= 106
cm3
and T = 300K,
4
CV = 15
π2
(4.36 × 300)3
× 106
k = 5.89 × 1015
k = 0.813 ergs/K
4

3
which is not much.
b) CV = 2
Nk for a monatomic gas. To get the same CV as above,
3
× 1015
N = 5.892

× 1015

N = 3.93
3
N/V = 3.93 × 1015
/106
= 3.9 × 109
atoms/cm
This is ten orders of magnitude less than the STP density of 2.69 × 1019
atoms/cm3
.
c) The energy stored in an electric field is
1 1
E E dVUE = E E dV UE =
8π
· ⇒
8π
·
For thermal radiation E E is independent of position, so·
1
UE =
1
E E E E dV ⇒ UE /V =
8π
·
8π
·
where UE is in ergs and E is in statvolts/cm (Gaussian CGS units). In a radiation
field the energy stored in the magnetic field UB is equal to that in the electric
field, so
11
UB = UE = 2
U(T, V ) U(T, V )/V = E2
⇒
4π
Now use the result from a)
4
1 kT
U(T, V )/V ≡ u(T) = π2
c¯h
15 c¯h
π2

= (4.36T)4
c¯
h
15
= 7.52 × 10−15
T4
ergs/cm3
E2
= 4π × 7.52 × 10−15
T4
= 7.65 × 10−4
(statvolts/cm)2
1/2
E2
= 2.77 × 10−2
statvolts/cm = 8 volts/cm
5

Problem 5: Radiation Pressure
From class we have the relation P = 1
u(T) for the thermal radiation ﬁeld. Using the3
expression for U(T, V ) = u(T)V from problem 4a we have
1 π2
kT
3
π2
Pradiation = kT = (4.36T)3
kT
3 15 c¯h 45
For a classical gas Pkinetic = nkT. Setting Pradiation = Pkinetic gives
π2
n = (4.36T)3
45
Solving for T and using a STP density of n = 2.69 × 1019
leads to
2.69 × 1019
× 45
1/3
= 1.1 × 106
KT =
π2(4.36)3
6

�
� � �

Physics Department

Problem 1: Lattice Heat Capacity of Solids
a) The heat capacity of a classical harmonic oscillator is k, independent of its fre-
quency.
# of oscillators = (3 degrees of freedom) × (J atoms/unit cell) × (N unit cells) =

3JN.

Therefore, CV = 3JNk.

b) For a quantum harmonic oscillator

1
= hν( n +2
)
n = (exp[hν/kT] − 1)−1
Assuming all oscillators have the same frequency
∂
CV = 3JN
∂T
∂ n
= 3JNhν
∂T
hν exp[hν/kT]
= 3JNhν (
kT2
)
(exp[hν/kT] − 1)2
= 3JNk (
hν
)2 exp[hν/kT]
kT (exp[hν/kT] − 1)2
hν
lim CV = 3JNk ( )2
exp[−hν/kT] energy gap behavior
kT hν kT
lim CV = 3JNk classical result
kT hν
c i)
∞
E(T) = D(ω) (ω, T) dω
0
1 1∞
= D(ω) + dω
0 exp[¯hω/kT] − 1 2
1

�
� ��
� �
� ��
�
For ¯hω kT, → kT independent of ω. Therefore one can move it out from
under the integral.
∞
lim E(T) = kT D(ω) dω = 3JNkT
kT ¯hωmax 0
3JN
lim CV (T) = 3JNk the classical result
kT ¯hωmax
c ii)
¯hωD(ω)
E(T) = D(ω)(¯
∞
hω/2) dω +
∞
dω
0 0 exp[¯hω/kT] − 1
E0, independent of T
For very low T one may use the low ω limiting form of D(ω) since only the oscillators
with low ω will be excited.
¯3V 1 � ∞ hω ω2
E(T) = E0 +
·
dω
2π2 v 3 0 exp[¯hω/kT] − 1
� �3 �
3V kT ∞ (¯hω/kT)3
= E0 + kT d(¯hω/kT)
¯ hω/kT] − 12π2 h v 0 exp[¯
� �3 � ∞ 3
3V kT x
= E0 + kT dx
¯2π2 h v �0 ex
��
− 1 �
π4
15
π2
kT
�3
= E0 + V kT
10 h v ¯
� �3
2 kT
CV (T) = π2
V
¯
k
5 h v
∝ T3
Note that the exponential temperature dependence associated with energy gap be-
havior has been washed out by the distribution of energy gaps associated with the
distribution of harmonic oscillator frequencies, some less than kT at any reasonable
value of the temperature.
2

� �
�
� �
�
Problem 2: Thermal Noise in Circuits I, Mean-Square Voltages and Currents
1
a) The energy stored in the capacitor, E = 2
Cv2
, acts as the Hamiltonian for this
small subsystem. We can then apply the results of the canonical ensemble.
p(v) exp[−1
Cv2
/kT]2
∝
kT
−1/2
= 2π exp[−v2
/2(kT/C)] when normalized
C
Now we do the numbers.
kT = 1.4 × 10−23
× 300 = 4.2 × 10−21
joules
C = 100 pF = 10−10
F
2
v = kT/C = 42 × 10−12
√
v2 = kT/C = 6.5 × 10−6
= 6.5 µV
b) Now the energy stored in the inductor, the eﬀective Hamiltonian for the subsystem,
is E = 1
Li2
.2
p(i) exp[−1
Li2
/kT]2
∝
kT
−1/2
= 2π exp[−i2
/2(kT/L)] when normalized
L
Again we do the numbers.
L = 1 mH = 10−3
H
i2
= kT/L = 4.2 × 10−18
√
i2 = kT/L = 2 × 10−9
A = 2nA
c) This method does not work for a resistor since it does not store energy; rather,

it is completely dissipative. One can not, for example, write a Hamiltonian which

3

� � � �
�
describes only the voltage and current associated with the resistor. Of course, if the
R were in parallel with a C or in series with an L one could make use of the above
results. These are two particularly simple examples of the general result that the
thermal voltage one would measure across a resistor (RMS) depends on the circuit to
which it is connected. Note that this is not the case for a C or an L as shown above.
Problem 3: Thermal Noise in Circuits II, Johnson Noise of a Resistor
a) For a line of length L which is short circuited at each end,
En(x, t) = En sin (nπx/L) sin(ωnt + φn)
� ��
knx
kn = n(π/L) n = 1, 2, 3, · · ·
Then ωn = ckn and φn is some ﬁxed time phase factor. Only one polarization direction
is allowed on a transmission line. For a coaxial cable E is always in the radial direction.
b)
ω0L
#(ω ω0) = #(k ω0/c) = (ω0/c)/(π/L) =
πc
D(ω) =
d#(ω)
dω
=
L
πc
a constant
c) Each mode is an independent harmonic oscillator, so
1
hω (exp[¯ 2
¯ n = ¯ hω/kT] − 1)−1
+ hω
D(ω) n(ω)
u(ω, T) =
L
L 1
¯ hω/kT] − 1)−1
+ 1
hω= hω (exp[¯ 2
¯
πc L
d) When kT ¯hω, n kT. Then→
L
�
1 kT
u(ω, T) → (kT) =
πc L πc
Note that this result for the double energy density (per unit length, per unit frequency
interval) on the transmission line is independent of the frequency in this frequency
region (low frequencies).
4

�
�
�
�
� �
e) Consider the standing wave modes to be made up of two waves propagating in
opposite directions. Then the thermal energy flow in each direction is
c kT
u(ω, T) = .
2 2π
If this flows from the line into the resistor, then by detailed balance in thermal
equilibrium an equal amount must flow out:
kT
Pn(ω) = .
2π
f)
∆ν = 10 MHz = 107
s−1
∆ω = 2π∆ν = 2π × 107
s−1
kT
Pn(ω)∆ω =
2π
× 2π × 107
s−1
= 1.38 × 10−16
ergs/K × 300 K × 107
s−1
= 4.1 × 10−7
ergs/sec = 4.1 × 10−14
watts
Problem 4: Thermal Noise in Circuits III, Circuit Model for a Real Resistor
a)
v
R
line = vN (ω)
R + R
1
= vN (ω)
2
Power
2
v
1
line
v2
N (ω) = =
line R 4R
b)
kT
2π
=
1
4R
v2
N (ω) ⇒ v2
N (ω) = 2RkT/π
5

�
c) We are told that for the circuit shown
1 22
vC(ω) = v0(ω)
1 + (RCω)2
when v0(ω) is a random noise signal with zero mean. Identify R with the ideal resistor
in the model of a real resistor in thermal equilibrium and v0(ω) with the noise source
voltage vN (ω) in the model. Then
1 22
vC(ω) = vN (ω)
1 + (RCω)2
2 R
= kT
π 1 + (RCω)2
v2
C =
∞
v2
C(ω) dω
0
=
2
π
kT
� ∞
0
R
1 + (RCω)2
dω
=
2
π
kT
C
� ∞
0
1
1 + (RCω)2
d(RCω)
=
2 kT � ∞ dx
=
kT
π C 0 1 + x2 C� ��
π/2
This is just the result we found using the canonical ensemble on the capacitor alone.
THE THERMAL NOISE IN A CIRCUIT CAN BE THOUGHT OF (AND
QUANTITATIVELY MODELED) AS ARISING FROM THE DISSIPA-
TIVE ELEMENTS IN THE SYSTEM.
6


Physics Department

Problem 1: Two Identical Particles
a)
Fermions: 1, 1, 0 2 T = 0 state|
1, 0, 1 3|
0, 1, 1 2 + 3|
b)
Bosons: 2, 0, 0 0 T = 0 state|
1, 1, 0 2|
1, 0, 1 3|
0, 2, 0 2 2|
0, 1, 1 2 + 3|
0, 0, 2 2 3|
c) Let β ≡ 1/kT.
+ e− 3β
+ e−( 2+ 3)β
ZF (T) = e− 2β
ZB(T) = 1 + e− 2β
+ e− 3β
+ e−2 2β
+ e−( 2+ 3)β
+ e−2 3β
d)
ZF (T) ≈ e− 2β
+ e− 3β
1 ∂ZF
E F = −
ZF ∂β
2e− 2β
+ 3e− 3β
2 + 3e−( 3− 2)β
=≈
e− 2β + e− 3β 1 + e−( 3− 2)β
1

� � � �
2 + 3e−( 3− 2)β
1 − e−( 3− 2)β
=
1 − (e−( 3− 2)β)
2
2 + ( 2 − 3)e−( 3− 2)β
− 3e−2( 3− 2)β
=
1 − e−2( 3− 2)β
2 + ( 3 − 2)e−( 3− 2)β
≈
This result shows a ﬁnite E at T = 0 and energy gap behavior with ∆ = 3 − 2.
1 ∂ZB
E B = −
ZB ∂β
2e− 2β
≈
1 + e− 2β
ψ
≈ 2e− 2β
This result shows E = 0 at T = 0 and energy gap behavior with ∆ = 2.
Problem 2: A Number of Two-State Particles
The two single-particle states available are indicated below.
1 = ∆
ψ0 = 0
a) Use the number of particles in the upper single particle state as the index for the
many-particle states:
n0, n1 = N − n1, n1 and En1 = n1∆| |
N ⇒ N + 1 many-particle statesn1 = 0, 1, 2, · · ·
b) Let β ≡ 1/kT.
N N
Z(N, T) =
�
e−n1∆β
=
� �
e−∆β
�n1
n1=0 n1=0
2

� ��
�
∞
e−∆β n1
∞
e−∆β n1
=
n1=0
−
n1=N+1
� � � � � � �∞
e−∆β n1
e−∆β
�N+1
∞
e−∆β m
=
n1=0
−
m=0
e−(N+1)∆β
1
=
1 − e−∆β
−
1 − e−∆β
1 − e−(N+1)∆β
=
1 − e−∆β
c)
1
1 − e
p(n1) = e−n1∆/kT
=
1 − e−∆/kT
e−n1∆/kT
−(N+1)∆/kTZ
d) If the particles are distinguishable, similar, and non-interacting, then
� �NN
Zd(N, T) = (Zone particle) = 1 + e−∆/kT
Problem 3: Spin Polarization
a) Recall that in zero ﬁeld the density of states for spin-1
Fermions is2
V 2m
�3/2
D0( ) =
2π2 h2
1/2
0
¯
= 0 0
These are equally divided between spin up and spin down. The application of a ﬁeld
1
shifts all ms = 2
states down in energy by µ0H and all ms = −1
states up by µ0H2
(assuming a positive µ0, that is, a magnetic moment parallel rather than anti-parallel
to S). Thus

1

D1 ( ) = D0( + µ0H)
2 2
1
D 1 ( ) = D0( − µ0H)
2 2

1

−
D( ) = [D0( + µ0H) + D0( − µ0H)]
2
3

� ��
� ��
�
1
b) The ﬁlling of D( ) at T = 0 must stop just short of = µ0H where the ms = 2
−
states would begin to ﬁll.
� µ0H � 2µ0H
1
N = D( ) d = 2
D0( ) d
0−µ0H
a 1/2
a 2
= (2µ0H)3/2
2 3
1 V 2m
�3/2
= (2µ0H)3/2
h2
3 2π2 ¯
6π2
N/V =
4mµ0H
�3/2
h2
¯
(6π2
N/V )2/3
=
4mµ0
H
h2
¯
¯1 h2
(6π2
N/V )2/3
H0 =
µ0 4m
c) Using cgs units
1
H0 = 4.22 × 10−54
(N/V )2/3
H
µ0M
Using M = 9.11 × 10−28
g, µ0 = −9.27 × 10−21
ergs-gauss−1
, and n = 8.45 × 1022
cm−3
gives
0 = 9.6 × 108
gauss = 9.6 × 104
Tesla.
The negative µ0 means that the electron spins are polarized anti-parallel to the di-
rection of H.
4

� �
� � �
�
� �
� �
� ��
d) For 3
He, M = 5.01 × 10−24
g, µ0 = 1.075 × 10−23
ergs-gauss−1
, and n = 1.64 ×
H
1022
cm−3
. So in this system
0 = 5.1 × 107
gauss = 5.1 × 103
Tesla.
Problem 4: Relativistic Electron Gas
D
a)
V 2V
wavevectors(k) =
(2π)3
Dstates(k) =
(2π)3
4 2V
N = πkf
3
kf = (3π2
N/V )1/3
3 (2π)3
⇒
= c¯| | h(3π2
N/V )1/3
h k f = c¯⇒
b)
 
4 1 3
#( ) =  π k3
( )
 2V
= V
1
�3
3 � ��  (2π)3 3π2 c¯h
h)3( /c¯
d# V 1
�3
D( ) = = 2
d π2 c¯h
c)
D( ) = a 2
1
a 3
N = f D( ) d = f a 2
d = 3 f0 0
1
a 4
E = f D( ) d = f a 3
d = 4 f0 0
3
= N f4
d) � � � �
∂E 1
P = =
3 ∂ f = (N/V ) f ∝ (N/V )4/3
−
∂V
−
4
N
∂V N
4N,S
dS=0 at T=0
5

�
This pressure rises less steeply with density, (N/V )4/3
, than is the case for the non-
relativistic gas, (N/V )5/3
.
e) For a white dwarf composed of α particles and electrons,
4
V = πR3
3
1
M ≈ Nαmα =
2
Nemα ⇒ Ne = 2(M/mα)
EK =
3
4
Ne F =
3
4
Nec¯h(3π2
Ne/V )1/3
3 M
�
9π M 1
�1/3
=
2
c¯h(
mα
)
2 mα R3
3
�
9π
�1/3 �
M
�4/3
1
= c¯h
2 2 mα R
E
The R dependence of the two contribu-
tions to the total energy is straight for-
ward: EK = a/R and EP = −b/R where
a and b are known expressions. Then
TOTAL = (a − b)/R which is never sta-
ble. The condition a = b is a special case,
a dividing line between collapse and inﬁ-
nite expansion.
f) The condition a = b probably gives a combination of parameters indicating the
dimensions of a critical mass which would emerge when a more detailed calculation
is done. What mass gives a ∼ b? Neglect numerical factors of the order of one.
M
�4/3
c¯
m
h ∼ GM2
α
c¯h
4/3
Gmα
∼ M2/3
� �3/2
c¯h
M mα∼
Gm2
α
6

The Chandrasekhar limit for the maximum possible mass of a white dwarf is
� � � �3/2
Z ch
MCh = 0.20
A Gm2
mp
p
where Z/A is the average ratio of atomic number to atomic weight of the stellar
constituents. Note that it has the same form as our expression. For Z/A = 0.5 (α
particles) this gives MCh = 1.4MSun.
7

Problems and solutions statistical physics 1

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Problems and solutions statistical physics 1