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 state h, 0, or ¯

1of the system it is known that < Lx >= 3 h and < L2 >= 2 h2. [¯¯ ¯ h is a constant with x 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 2Ψ(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 4T, 1

2T, 3

4T ,

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 ε. 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)a n 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 2ψ(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

y1

= 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. |

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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.

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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

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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 ?

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MASSACHUSETTS INSTITUTE OF TECHNOLOGYPhysics 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 + vvp(vx, vy, vz) = (2πσ2)−3/2 exp[− x y z ]

2σ2

1where 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 = rr where r is a constant and r 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

Mwhere 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?

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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

ˆAtoms 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

1where 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⊥.]⊥

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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 Aand 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

nBP 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 − MP V = 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 dependsonly on the temperature T of the cavity and not on its volume V ,

1P = σT 4 .

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.

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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 2S(x, y) = (2x5 + 5x y3 + 3y5)/5 + C

b) ey dx + x(ey + 1)dy S(x, y) does not exist.

c) (y − x)exdx + (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

5An 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 A

T

∂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 .

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( )

( )

( )

( )

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

dQCM , the heat capacity at constant M , and ≡

dT M

dQCH , 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.

CFor 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.

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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 εi = −miH where mi is the z component of µi (that is, z = mi).µi ·

The magentization M and the total energy E are given by

N N ∑ ∑ M =

i=1

mi E = i=1

εi = −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

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 mω2q2 ∑ 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.

9)5

5+1)645!!*15 37)6)*)5

*)5 8)*-2)461615

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 ) = aT 5/2 + bT 3 + cV −2

CV (T, V ) = dT 3/2V + eT 2V + fT 1/2

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 .

Cb) 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 isη = 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 of 2

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 a 2

1kinetic energy T = 2 Iθ2 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

ε

A crystalline solid contains N similar, immobile, statistically independent defects. Each defect has 5 possible states ψ1, ψ2, ψ3, ψ4, and ψ5 with energies ε1 = ε2 = 0,

3 = ε4 = ε5 = ∆.

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 = AT αa−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 εn,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 | |

Ethroughout 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

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 εm = (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 ε = 9h2/2I. Find the probability that m = 1 given that ε ≤ 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 )4V.

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 T 3, 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 ) = < ε(ω, T ) > D(ω) dω

0

where < ε(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 velocityin 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 kB T >> 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 2function of < vN (ω) >.

5

∫

c) Show that equating the results of a) with those of Problem 3e leads to the result2 (ω) >Nthat < v = 2RkB T/π.

c) You may recall that a resistor and a capacitor in series form a low pass filter. In2(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 fluctuationson the capacitor, information that was not available using the simpler methodof 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.

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Physics Department


Problem Set #12 Solutions Available in on the Web: Wednesday, May 12

Problem 1: Two Identical Particles

ε

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 < ε2 < ε3 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 ε = 0 and ψ1 with energy ε = ∆.

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

ε = ¯

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 Hzthe single particle energies depend on the direction of the spin relative to the field:

h2k2/2m − 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(ε) and D−1/2(ε). Make a careful sketch of the total density of states as a function of energy D(ε) = D1/2(ε) + D−1/2(ε).

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 3He 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 allowed In this limit the single particle energies are given by ε = c¯| |

wavevectors in k space is V/(2π)3 .

a) Find an expression for the Fermi Energy εF as a function of c, h, N and V .

b) Find the density of states as a function of energy D(ε). Sketch the result.

2

c) Find the total kinetic energy E of the gas at absolute zero as a function of N and εF .

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 4He 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

3of radius R with uniform density is given by EP = 5 GM 2/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, ε(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 2x (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 2x(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 )−1/2dx

−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

> = h2 2 2 2h 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)a n n = 0, 1, 2, · · ·

a) First we find the mean of n.

∞ ∞< n >= np(n) = (1 − a) na n

n=0 n=0

S1

The sum S1 can be found by manipulating the normalization sum.

∞p(n) =

∞(1 − a)a n = (1 − a)

∞a n 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 a n=0 n=0 n=0 ( 1 )d 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) ∞

n 2 a n=0 n=0

S2

Now try the same trick used above, but on the sum S1.

d da

∞∑ n=0

na n = ∞∑

n=0

n 2 a n−1 = 1

a

∞∑ n=0

n 2 a n = 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)a nδ(x − n)

n=0

∞= f(x) δ(x − n)

n=0

) = Ce−x/φTry f(x ,

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

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

πx20

(x1 − x2

x0

)2 exp[−x21 + x2

2

x20

]

The figure on the left shows in a simple way the location of the maxima and minima ofthis probability density. On the right is a plot generated by a computer application,in this case Mathematica.

Max.

Max.

x2

x1

node at x1=x2implies p(x1,x2) = 0

b)

p(x1) =∫ ∞

−∞p(x1, x2) dx2

=1

πx20

1

x20

exp[−x21/x

20]

∫ ∞

−∞(x2

1 − 2x1x2 + x22) exp[−x2

2/x20] dx2

=1

πx20

1

x20

exp[−x21/x

20]

[x2

1(√

πx20 ) − 2x1 × 0 +

x20

2(√

πx20 )

]

=1√πx2

0

(x21/x

20 + 1/2) exp[−x2

1/x20]

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/x

2]0πx2

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/x

2 0]

= 2 √ πx2

0

1

(1 + 2(x2/x0)2)

( x1 − x2

x0

)2

exp[−x 2 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

1Why 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−x1 1 = 6 (1 − x)y − y 2 = 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. Thus

3

p(no stars in a sphere of radius r) = p(n = 0)

= exp[−4/3 πρr 3]

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 >= 4πρr 2 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)

= 4πρr 2 exp[−4/3 πρr 3] dr

Dividing out the differential dr and being careful about the range of applicability leaves us with

p(r) = 4πρr 2 exp[−4/3 πρr 3] 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

20406080

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−(vp(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 2over 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

v 2 = v 2 x + v 2 y + v 2 z

vx = v sin θ cos φ

vy = v sin θ sin φ

vz = v cos θ

dvx dvy dvz = v 2 dv sin θ dθ dφ

(v That1 2

+ v + vx y

1

√

√

This gives

∫ π 2P (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−(vp(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

v 2 = v 2 x + v 2 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


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 v 2 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 2to 1 vx ≤ E. That region in vx space is just a line from − 2E/m to 2E/m.

2 ∫ √2E/m 2 v1 P (E) = √

2πσ2 −√

2E/m

v e− 2σ2 dv


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 kiloparsecs σθ⊥

(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).

‖

∫ π1 ∫ 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 r 2

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 given x

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

2dP (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 v x

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⊥

2Note: 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 − αP V1 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 − P12)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 − P12)V1 − αP2V1(T2 − T1)

Along “c” dT and dP are related at every point along the path,

T2 − T1

PdT = dP,

2 − P1

so the expression for the differential of work can be written as

dP

/ W = KT PV1 dP − αP V1( 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 − P12) =

1 2 (P1 + P2)(P2 − P1)

1 1 =

2 KT (P2

2 − P12)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 = σT14 (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 T 3 = a constant ≡ V1T31

T = ( V1

)1/3 T1V

4 = T14 (

VV1 )4/3 = T1

4 ( )−4/3TV V1

Now we can carry out the integral over the path to obtain the total work done.

∫ 2

1 2 = − 1

1 ∫ P dV =

2T 4(V ) dVW → −

3 σ

1

1 = −

3

∫ 2 VσT1

4 ( )−4/3 dV1 V1

1 = −

3

[2 1 σT1

4 V1 1 −

1/3 ( VV1

)−1/3

V= σT1

4 V1 ( V2

)−1/3 − 11( )

V= σT1

4 V1 ( V1

)1/3 − 12

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 ∂V

V 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 − V12) = −(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.

V12V 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).

2aN 2 f ′(A) = −

A3

aN 2 f (A) = + c

A2

lim = c = 0 N 0

S ︸︷︷︸︸︷︷︸S→

from above given

Putting it all together gives the final result.

NkT aN 2 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)bT 2 + f(M) integration

0

∂U = f ′(M) = 0 (given) ⇒ f(M) = constant = U(T = 0)

∂M T

0 by assumption

U(T ) = (1/2)bT 2 + U(T = 0)

∂M ∂ aH aH = = from eq. of state

∂T H ∂T T H

− T 2

aH aH2 M2

CH − CM = (0 − H)(− T 2

) = = from a) T 2 a

CM2

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

substitutionH

∂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 −

T 2 H ( )

H2

bT a

dU (H, M ) = − T dH + bT + a T 2 + H dMaH aH

T 2 T 2 ︸ − ︸ − ︷︷︸ bT 2/H −bT 2/M

bT 2 bT 2 = dH − dM

H M

ba2H ba2H2

= M 2

dH − dM by eliminating T M 3

ba2 H2

U (H, M ) = + f (M ) integration 2 M 2

∂U ba2H2

= − M 3

+ 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 M 2

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 M 2

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)T 2 .

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 (M 2 − M 2 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 −

M 2 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 M 2

1 H M 2 T T M 2 dH = M 2 + 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.

dMdH

a 1 a= <

T 1 + M 2 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

<< µN3kT

⇒ 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

[ − M 2

(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 M <<µN

( ) 1 3M ≈ 2µ

1 + (Nµ2

) m − µ ≤ m ≤ µ

Now check the normalization. ∫ ∫ 1 ∫ µ 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 + m 2 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

kN ln N − (N − n+) ln(N − n+) − n+ ln n+ −N + (N − n+) + n+≈ ︸︷︷︸

=0

= kN ln N − (N − n+) ln(N − n+) − n+ ln n+

b)

∂STF

= − ∂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+.

nexp[−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

2E = 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 √

πN

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−ε where 2ε = (1/2m)p2 + (mω2/2)qi . Since the oscillators are all similar, < ε >= E/N = kT .i

p(pi, qi) = Ω′/Ω

( )N −12π 1 Ω′ =

Ω

ω (N − 1)! (E − ε)N −1

′ ( ) ( E − ε)N2π −1 N ! 1

= Ω ω (N − 1)! E E − ε

ω N ε )N

= 2π ︸ E − ε ︸︸ 1 − ︷︷ E

≈<ε>−1 ≈exp[−ε/<ε>]

1 p(pi, qi) =

(2π/ω) < ε > exp[−ε/ < ε >]

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)¯ hωM hωN + ¯

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 = Nhω +

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), < ε >= ¯ε = ¯ hω(1/2+ < n >), < n >= M/N.

Then for the reduced phase space N → N − 1 and M .→ M − n

Ωp(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 ∂2S = T

∂A ∂A∂T T

∂ ∂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 − b

A

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 ∂V

V T

CV

dF = dE − d(TS) = −S dT − P dV

∂S ∂S =⇒ −

∂V T

− ∂T

V

∂P ∂S = (5/2)aT 3/2 + 3bT 2 =

∂T ∂VV T

∂S T − P = (5/2)aT 5/2 + 3bT 3 − aT 5/2 − bT 3 − cV −2

∂V T

= (3/2)aT 5/2 + 2bT 3 − cV −2

Collecting this all together gives

dE = (dT 3/2V + eT 2V + fT 1/2)dT + ((3/2)aT 5/2 + 2bT 3 − 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

dT 3/2 + eT 2 = (15/4)aT 3/2 + 6bT 2

⇒ d = (15/4)a, e = 6b

c) Use the results from b) to simplify the expression for dE in a).

dE = ((15/4)aT 3/2V + 6bT 2V + fT 1/2)dT + ((3/2)aT 5/2 + 2bT 3 − cV −2)dV

Integrate with respect to T first.

E = (3/2)aT 5/2V + 2bT 3V + (2/3)fT 3/2 + F(V )

∂E = (3/2)aT 5/2 + 2bT 3 + F ′(V ) from above

∂V T

= (3/2)aT 5/2 + 2bT 3 − cV −2 from dE

⇒ F ′ = −cV −2 , F = cV −1 + KE

E = (3/2)aT 5/2V + 2bT 3V + (2/3)fT 3/2 + 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

= (dT 1/2V + eT V + fT −1/2)dT + ((5/2)aT 3/2 + 3bT 2)dV

Integrate with respect to T first.

S = (2/3)dV T 3/2 + (1/2)eV T 2 +2fT 1/2 + G(V ) ︸︷︷︸︸︷︷︸ (5/2)aV T 3/2 3bV T 2

3

( )

∣ ∣

( ) ( )

∂S = (5/2)aT 3/2 + 3bT 2 + G ′(V ) from above

∂V T

= (5/2)aT 3/2 + 3bT 2 from dS

⇒ G ′(V ) = 0, (V ) = KSG

S(T, V ) = (5/2)aV T 3/2 + 3bV T 2 + 2fT 1/2 + 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 ∂2S

∂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 A

M

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 ∂M

M 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

M2T0 ⇒ f(M) = − + KE2A

∫ T M2T0E(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) = − + KS2A

∫ 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)

Tη = 1 − | |

= 1 − CV (T3 − T2)

= 1 − γT4 − 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 −P3V3γ − 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

TC

= 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

∣ ∣ ∣ ∣ ∣

√

T 2TF

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

T 2 1 − 2T1T2 + T 2

2 > 0

T 2 1 + 2T1T2 + T 2

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 Iθ2 + 1 K(θ − θ0)2

2 2

Then use the canonical ensemble expression for the probability density.

p(θ, θ) ∝ exp[− H]

kT

∝ exp[− θ2 (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[−εi/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< ε > = 0 × 2 1 + ∆ × 3 × = ∆5 5 5

×

E = N < ε > = 3 ∆N5

Problem 4: Neutral Atom Trap

a) First write down the Hamiltonian for one atom.

2 2 2px py pz = + + + arH1 2m 2m 2m

Then compute the partition function

2 y

2 x

2 z1 ∞ ∞ ∞pp p

dpz e− ar kT r 2 sin θ drdθdφ e− e− e−Z1 = dpx dpy 2mkT 2mkT 2mkT

h3 −∞ −∞ −∞ V √

2πmkT √

2πmkT √

2πmkT 4π ∞

0exp[−ar/kT ] r2 dr

4

3 2πmkT kT = ( )3/24π

∞ y 2 e−y dy

h2 a 0

2

2πmk )3/2 T 9/2 a−3= 8πk3(

h2

In order to emphasize the dependence on the important variables, this can be written in the form Z1 = AT αa−η where

2πmk )3/2A = 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

⇒ T 9/2/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 × 105K. 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) = n 2 .

The partition function for a single atom, neglecting the unbound states, is

Z = exp[−εi/kT ] = ∞

n 2 exp[α/n 2], 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 > ∞

n 2 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.

eφ(r) =

| | r ≥ R

r

The potential energy of another proton, considered to be a point particle, in this field is

2eV (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

2eEmin = V (R) = =

(4.8 × 10−10)2

= 1.92 × 10−6 ergsR 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

2 ∫2 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/secπ × 1.67 × 10−24

2σ = 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

τ1.01 × 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¯ mh2/2I 2

±2 4¯ mh2/2I 2

±1 ¯ mh2/2I 2 m0 0 1

ε

m ε degeneracy

m = (¯ 2h2/2I) m

Zrot,1 = exp[−ε(state)/kT ] = ∞

exp[−(¯ 2h2/2IkT ) m ] states m=−∞

∞= 1 + 2 exp[−(h2/2IkT ) j2 ]

j=1

3

∑ ∫

∫ ︸︷︷︸

( )

( ) ( )

b) p(m = 3) Z−1 exp[−9h2/2IkT ]

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 ε = 9¯h2/2IkT ])

= 1/2 2(Z−1 exp[−9¯

h2/2IkT ] 1 p(m = 1 ε ≤ ¯

Z−1 exp[−¯=| h2/2I) =

Z−1 + 2(Z−1 exp[−¯ h2/2IkT ])h2/2IkT ]) 2 + exp[¯

d)

2Zrot,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 ∂Z1Erot = N < ε >= 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 degeneracydifference (at most a factor of two).

b)

p(5)/p(4) = N exp[−E(5)/kT ]

= 1

exp[(E(4) − E(5))/kT ] = 22N 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 ] = 4N 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 )4V−

∂T −

45 c3¯

4 π2

= k4T 3V h3

From this result we see that the product T 3V 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.

σT 4 sun = 4σT 4 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 = σT 4 C = σ(T 4 C ) ≡ J0H − σT 4 H − T 4

b) Since the sheet is neither cooling nor warming, the flux in from TH must equal the flux out to TC .

H − T 4) = σ(T 4 − T 4 H + T 4 = 2T 4σ(T 4 C ) ⇒ T 4 C

T 4 H + T 4 )1/4

T = C

2

c)

J = σ(T 4 H − T 4)

1 H + T 4 = σ[T 4 C )]H −

2 (T 4

= 1 σ(T 4 C ) = 1 J0H − T 4 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

[ ]

4T−1 i−4Ti −1 − J/σ 4T= i

4TiJ/σ= ⇒

T 4 = 1 − J/σ

T 4

T 4

4J/σ T− = 1 H 4J/σ T− = 2 H

4TH

T 4 2 − 2J/σ

− 3J/σ

=

T 4 3 =

· · · · · · 4TC

4J/σ T− = H 4Tn − (n + 1)J/σ=

Solving the final equation for J gives the result

J =σ

n + 1 =

1

n + 1 4T− C

4TH 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

c3h3 (kT )4V ;

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,

4CV = 15

π2(4.36 × 300)3 × 106k = 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 × 1015N = 5.892

× 1015N = 3.93 3N/V = 3.93 × 1015/106 = 3.9 × 109atoms/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 > dV UE = 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)

( )41 kT

U(T, V )/V ≡ u(T ) = π2 ch 15 ch

π2

= (4.36T )4 c¯h 15

= 7.52 × 10−15T 4 ergs/cm3

< E2 > = 4π × 7.52 × 10−15T 4 = 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 the 3

expression for U(T, V ) = u(T )V from problem 4a we have

1 π2 kT )3

π2

Pradiation = kT = (4.36T )3kT 3 15 ch 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 × 106KT = 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< ε > = hν(< n > +2 )

< n > = (exp[hν/kT ] − 1)−1

Assuming all oscillators have the same frequency

∂ < ε > CV = 3JN

∂T

∂ < n > = 3JNhν

∂T

hν exp[hν/kT ]= 3JNhν (

kT 2 )

(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) < ε(ω, T ) > dω 0

1 1∞ = D(ω) + dω

0 exp[hω/kT ] − 1 2

1

For hω kT , < ε >→ 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 ∞ 33V 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 > ∝ T 3

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

1a) 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[−v 2/2(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 = 10−3H

< i2 > = kT/L = 4.2 × 10−18

√< i2 > = kT/L = 2 × 10−9A = 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

1hω (exp[¯2

< εn > = ¯ hω/kT ] − 1)−1 + hω

D(ω) < εn(ω) > u(ω, T ) =

L

L 1¯ hω/kT ] − 1)−1 + 1 hω= hω (exp[¯

2 πc L

d) When kT hω, < εn > 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 × 10−16ergs/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)

vR

line = vN (ω)R + R

1 = vN (ω)

2

Power 2v 1line < 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 > ε2 T = 0 state |1, 0, 1 > ε3|0, 1, 1 > ε2 + ε3|

b)

Bosons: 2, 0, 0 > 0 T = 0 state |1, 1, 0 > ε2|1, 0, 1 > ε3|0, 2, 0 > 2ε2|0, 1, 1 > ε2 + ε3|0, 0, 2 > 2ε3|

c) Let β ≡ 1/kT .

+ e−ε3β + e−(ε2+ε3)βZF (T ) = e−ε2β

ZB (T ) = 1 + e−ε2β + e−ε3β + e−2ε2β + e−(ε2+ε3)β + e−2ε3β

d) ZF (T ) ≈ e−ε2β + e−ε3β

1 ∂ZF< E >F = −

ZF ∂β

ε2e−ε2β + ε3e

−ε3β ε2 + ε3e−(ε3−ε2)β

=≈ e−ε2β + e−ε3β 1 + e−(ε3−ε2)β

1

ε2 + ε3e

−(ε3−ε2)β 1 − e−(ε3−ε2)β

= 1 − (e−(ε3−ε2)β )

2

ε2 + (ε2 − ε3)e−(ε3−ε2)β − ε3e

−2(ε3−ε2)β

= 1 − e−2(ε3−ε2)β

ε2 + (ε3 − ε2)e−(ε3−ε2)β≈

This result shows a finite < E > at T = 0 and energy gap behavior with ∆ = ε3 − ε2.

1 ∂ZB< E >B = −

ZB ∂β

ε2e−ε2β

≈ 1 + e−ε2β

ψ

≈ ε2e−ε2β

This result shows < E >= 0 at T = 0 and energy gap behavior with ∆ = ε2.

Problem 2: A Number of Two-State Particles

The two single-particle states available are indicated below.

1 ε = ∆

ψ0 ε = 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 states n1 = 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)∆β1 =

1 − e−∆β−

1 − e−∆β

1 − e−(N +1)∆β

= 1 − e−∆β

c) 1

1 − ep(n1) = e−n1∆/kT =

1 − e−∆/kT

e−n1∆/kT −(N +1)∆/kT Z

d) If the particles are distinguishable, similar, and non-interacting, then NNZd(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 is 2

V 2m 3/2

D0(ε) =2π2 h2 ε1/2 ε > 0

¯

= 0 ε < 0

These are equally divided between spin up and spin down. The application of a field 1shifts 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

1D1 (ε) = D0(ε + µ0H)

2 2

1 D 1 (ε) = D0(ε − µ0H)

2 2

1

−

D(ε) = [D0(ε + µ0H) + D0(ε − µ0H)]2

3

1b) The filling of D(ε) at T = 0 must stop just short of ε = µ0H where the ms = 2

−states would begin to fill.

µ0H 2µ0H 1N = D(ε) dε = 2 D0(ε) dε

0−µ0H aε1/2

a 2 = (2µ0H)3/2

2 3

1 V 2m 3/2

= (2µ0H)3/2

h23 2π2 ¯

6π2N/V =4mµ0H 3/2

h2¯

(6π2N/V )2/3 =4mµ0

H h2¯

¯1 h2(6π2N/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 × 10−28g, µ0 = −9.27 × 10−21ergs-gauss−1, and n = 8.45 × 1022cm−3

gives

0 = 9.6 × 108gauss = 9.6 × 104Tesla.

The negative µ0 means that the electron spins are polarized anti-parallel to the di-rection of H.

4

ε ε

ε ε

d) For 3He, M = 5.01 × 10−24g, µ0 = 1.075 × 10−23ergs-gauss−1, and n = 1.64 ×

H

1022cm−3. So in this system

0 = 5.1 × 107gauss = 5.1 × 103Tesla.

Problem 4: Relativistic Electron Gas

D

a)

V 2V wavevectors(k) =

(2π)3 Dstates(k) =

(2π)3

4 2V N = πkf

3 kf = (3π2N/V )1/3

3 (2π)3 ⇒

ε = c¯| | h(3π2N/V )1/3h k εf = c¯⇒

b) 4 1 ε3#(ε) = π k3(ε)

2V = V

1 3 3 (2π)3 3π2 ch

h)3(ε/c¯

d# V 1 3

D(ε) = = ε2

dε π2 ch

c)

D(ε) = aε2

1 aε3N = f D(ε) dε = f aε2 dε = 3 f0 0

1 aε4E = f εD(ε) dε = f aε3 dε = 4 f0 0

3 = Nεf4

d) ∂E 1

P = =3 ∂εf = (N/V )εf ∝ (N/V )4/3−

∂V −

4 N

∂V N 4

N,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 NeεF =

3

4 Nech(3π2Ne/V )1/3

3 M 9π M 1

1/3

= 2 ch(

mα )

2 mα R3

3

9π 1/3 M 4/3 1

= ch 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 ∼ GM 2 α

ch 4/3

Gmα

∼ M 2/3

3/2 ch

M mα∼ Gm2

α

6

The Chandrasekhar limit for the maximum possible mass of a white dwarf is

3/2Z 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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