Classical Mechanics Review - Navidromekissel.navidrome.com/.../S07_classical_solutionbank.pdfClassical Mechanics Review (Louisiana State University Qualiﬁer Exam) Jeﬀ Kissel May

Classical Mechanics Review

(Louisiana State University Qualifier Exam)

Jeff Kissel

May 18, 2007

A particle is dropped into a hole drilled straight through the center

of the earth. Neglecting rotational effects, show that the particle’s

motion is simple harmonic. Compute the period and give an estimate

in minutes. Compare your result with the period of a satellite orbiting

near the surface of the earth.

Figure 1: A particle dropped from R⊕.

Only the mass inside the shell of radius r contributes to the gravitationalforce. If we assume the Earth have uniform density, then the mass contributingto Fg is

M

4

3π r3

=M⊕

4

3π R3

⊕

M =M⊕

R3⊕

r3 (1)

Thus, the force experienced by the particle m from mass M is

Fg =GmM

r2

=Gm

r2

(

M⊕

R3⊕

r 3)

Fg =

(

GM⊕m

R3⊕

)

r (2)

= k r

1

Jeff Kissel May 18, 2007 Classical Mechanics

where the last step absorbs all the constants G, M⊕, R⊕, and m into the constant“k,” which shows that Eq. 2 is linearly proportional to the radius. Similarly,Hooke’s force law for simple harmonic motion states that the force of a simpleharmonic oscillator is proportional to the displacement from the equilibrium po-sition. Since the center of the earth would be the particles equilibrium position,r is the displacement, and thus by direct comparison, the particle obeys simpleharmonic motion, with “spring constant” k.

To compute the period τ , we use what we know for a spring with resonantfrequency ω =

√

k/m,

τ =1

f=

2π

ω= 2π

√

m

k

= 2π

√

m

(

R3⊕

GM⊕m

)

τ = 2π

√

R3⊕

GM⊕

τ2 =4π2R3

⊕

GM⊕

(3)

which is Kepler’s 3rd law for planetary motion (for a small particle), againindicating that motion is periodic.

Finally, one can find an estimate of this period by using Earth’s gravitationalacceleration of a small particle at the surface, g = GM⊕

R2

⊕

≈ 9.8 m/s ≈ π2.

τ2 =4π2R3

⊕

GM⊕

= 4π2

(

R⊕

g

)

τ2≈ 4R⊕ (4)

= 4(6.4 × 106 m) [s]

τ2 = 2.57 × 107 s

τ = 5069 s = 84 min (5)

which is approximately equivalent to a satellite in low-Earth orbit. (At 185 km,i.e. when low-Earth orbit becomes stable, the correct period is 88.19 min.)

2



Jeff Kissel

October 11, 2006

Two identical bodies of mass m are attached by identical springs ofspring constant k as shown in the figure.

Figure 1: A system of two masses attached by springs (A) in equilibrium, and(B) after some oscillation.

a) Find the frequencies of free oscillation of this system.

The first goal will be to set up the system’s characteristic equation,

V − ω2T = 0 (1)

where V is the potential energy matrix, T is the kinetic energy matrix, and ω

are the desired eigenfrequencies (frequencies of free oscillation) of the system.The potential on the particles is the sum of each individual spring’s potentials,

V =1

2kd2

1 +1

2kd2

2 +1

2kd2

3 (2)

1

Jeff Kissel October 11, 2006 Classical Mechanics

If we define the displacements between the masses in terms of generalized coor-dinates η,

d1 = (x1 − x(0)1 ) + (0) = η1

d2 = (x2 − x(0)2 ) + (x

(0)1 − x1) = η2 − η1

d3 = (0) + (x(0)2 − x2) = −η2

we can then write the potential as,

V =1

2k(η1)

2 +1

2k(η2 − η1)

2 +1

2k(−η2)

2

=1

2kη2

1 +1

2k(η2

2 + η21 − η1η2 − η2η1) +

1

2kη2

2

=1

2k(

2η22 + 2η2

1 − η1η2 − η2η1

)

(3)

which in the desired matrix form is,

V ≡ 1

2Vijηiηj =

(

2k −k

−k 2k

)

(4)

The kinetic energy matrix is much easier to find because it is neatly diagonal:

T =1

2mv2

1 +1

2mv2

1

T ≡ 1

2Tijηiηj =

(

m 00 m

)

(5)

Once we’ve got these matrices, we want to find the eigenfrequencies, ω ofthe charateristic equation

V − ω2T = 0 (6)

Notice however, that Eq. 4 and Eq. 5 are symmetric, 2 × 2 matrices, so wecan use some trickery and write them as multiples of the identity matrix or theSU(2) spinor matrices:

V = 2k I − k σx

T = m I

Now we can just diagonalize Eq. 6 instead of the usual determinant method,since we know the eigenvalues of I are 1,1, and for any σi are 1,-1. I’ll alsouse the fact that a diagonalized σx is just σz

0 = V − ω2T

= 2k I − k σx − mω2 I

D−1 (mω2 I)D = D−1 (2k I− k σx)D

ω2 (D−1ID) =2k

m(D−1ID) − k

m(D−1σxD)

2


ω2 I =2k

mI − k

mσz (7)

ω2

(

1 00 1

)

=2k

m

(

1 00 1

)

− k

m

(

1 00 −1

)

ω2 =2k ± k

m(8)

So the eigen frequencies for this two mass system are

ω1 =

√

k

mor ω =

√

3k

m(9)

We’ll need the eigenvectors for part (b), so we’ll find them now. A bigadvantage to using this SU(2) group argument is that since Eq. 7 only involesI and σz, we know immediately that the two (normalized) eigenvectors for thesystem are the (normalized) eigenvectors of those two matrices,

for I ⇒ u1 =1√2

(

11

)

(10)

for σz ⇒ u2 =1√2

(

1−1

)

(11)

where u1 corresponds to when the two masses are sloshing back and forth inphase with each other, and u1 is when they oscillate out of phase, i.e. if M1 ismoving to the left, M2 is moving to the right, or vice versa.

b) M1 is displaced from its position by a small distance A1 to theright while M2 is not moved from its position. If the two masses arereleased with zero velocity, what is the subsequent motion of M2?

With the (very safe) assumption that the system is periodic in time, we knowfrom Fourier’s theorem that the positions of each particle as a function of timemay be written as a sum of sines and cosines, whose phase is the eigenfrequen-cies. So,

x1(t) = A sin ω1 t + B cosω1 t + C sin ω2 t + D cosω2 t (12)

x2(t) = E sin ω1 t + F cosω1 t + G sinω2 t + H cosω2 t (13)

However, we can narrow this down at least a little because of the eigenvectors.

ω1 =

√

k

m⇒ u1 = 1

√

2

(

11

)

⇒ A = E, B = F

ω2 =

√

3k

m⇒ u2 = 1

√

2

(

1−1

)

⇒ C = −G, D = −H

which means Eqs. 12 and 13 (and there derivatives) simplify a lil’ bit to

x1(t) = A sin ω1t + B cosω1t + C sin ω2t + D cosω2t

3


→ x1(t) = Aω1 cosω1t − Bω1 sin ω1t + Cω2 cosω2t − Dω2 sinω2t

x2(t) = A sin ω1t + B cosω1t − C sin ω2t − D cosω2t

→ x2(t) = Aω1 cosω1t − Bω1 sin ω1t − Cω2 cosω2t + Dω2 sinω2t

Remember the problem statement said “M1 is diplaced by a small distanceA1 [at t = 0]?” That means the initial conditions (where t = 0, so the sineterms are zero, and cosine terms are one) are

x1(0) = A1 = B + D

x2(0) = 0 = B − D

⇒ B = D =1

2A1

x1(0) = 0 = A + C√

3

x2(0) = 0 = A − C√

3

⇒ A = C = 0

So then holy moly with extra cannoli, the positions of M1 and M2 at a funcionof time are

x1(t) =1

2A1 cos

(

√

k

mt

)

+1

2A1 cos

(

√

3k

mt

)

(14)

x2(t) =1

2A1 cos

(

√

k

mt

)

− 1

2A1 cos

(

√

k

mt

)

(15)

4



Jeff Kissel

May 18, 2007

Exam Iteration Question Number Notes/ChangesSpring 2006 X –Fall 2006 X –

Spring 2007 #3 Added to bank.

A planet is in circular motion about a much more massive star. The

star undergoes an explosion where three percent of its mass is ejected

far away, equally in all directions. Find the eccentricity of the new

orbit for the planet.

Initially the planet is in a circular Keplerian orbit, where we know severalthings. The force and potential are

F =GMm

r2, V = −

GMm

r(1)

For any circular orbit, the radius is given by

F = mac

GMm

r 2c= m

v2t

rc(

For circular orbits, r = 0

⇒ v2

t = r2 + r2θ2 = r2

c θ2

)

GM

rc= θ2r2

c

r3

c =GM

θ2

rc =

(

GM

θ2

)1/3

(2)

Also, from Eq 3.57 on p94 of Goldstein, any mass m orbiting a r−2 central

1


force will have eccentricity,

e =

√

1 +2Eℓ2

mk2(3)

with E as the total energy, ℓ = mr2θ as the angular momentum, and k is thiscase as GMm. Since the orbit is circular, the eccentricity is zero, so the initialtotal energy is

0 =

√

1 +2Eℓ2

mk2

0 = 1 +2Eℓ2

mk2

−1 =2Eℓ2

mk2

E = −mk2

2ℓ2(4)

And finally, the total energy can also be written as the sum of the kinetican potential energies,

E =1

2m

(

r2

c + r2

c θ2

)

−GMm

rc

=1

2m

(

GM

θ

)2/3

θ −GMm

(

GMθ

)1/3

=1

2m

(

GMθ)2/3

− m(

GMθ)2/3

=

(

1

2− 1

)

m(

GMθ)2/3

E = −1

2m

(

GMθ)2/3

(5)

Phew! Now, just after the explosion, the star is at the same radius, rc.However the star has a new mass, M ′ = (1 − λ)M where I’ve denoted λ asthe percentage of mass loss. The only portion of the energy that is effects isthe potential, in which k′ = (1 − λ)k. The angular momentum ℓ = mr2

c θ isindependent of M ′. So, the new energy is

E =1

2mr2

c θ2−

GM ′m

rc

=1

2m

(

GM

θ

)2/3

θ −G ((1 − λ)M)m

(

GMθ

)1/3

=1

2m

(

GMθ)2/3

− (1 − λ) m(

GMθ)2/3

2


=

(

1

2− (1 − λ)

)

m(

GMθ)2/3

E = (−1

2+ λ) m

(

GMθ)2/3

(6)

The ratio of initial energy to final energy is then,

E′

E=

(− 1

2+ λ)

m(

GMθ)2/3

−1

2 m

(

GMθ)2/3

E′

E= −2

(

−1

2+ λ

)

E′ = (1 − 2λ) E (7)

This in turn affects the eccentricity, which means Eq. 3 becomes,

e′ =

√

1 +2E′ℓ2

mk′2

=

√

1 +2(1 − 2λ) Eℓ2

m ((1 − λ) k)2

=

√

1 +(1 − 2λ)

(1 − λ)22ℓ2

mk2E

(

Eq. 4: E = −mk2

2ℓ2

)

=

√

1 −(1 − 2λ)

(1 − λ)22ℓ2

mk2

(

mk2

2ℓ

)

e′ =

√

1 −1 − 2λ

(1 − λ)2(8)

For λ = 0.03, the new eccentricity is

e′ = 0.030928 (9)

3



Jeff Kissel

October 11, 2006

A homogeneous cube each edge of which has a length ℓ, is initiallyin a position of unstable equilibrium with one edge in contact witha horizontal plane. The cube is then given a small displacement andallowed to fall. Find the angular velocity of the cube when one facestrikes the plane if:

a) the edge cannot slip on the plane (friction).b) sliding can occur (no friction).

Figure 1: A cube falling with friction holding one corner stationary, and thesame cube sliding on the corner without friction.

For both scenarios we can use a conservation of energy arguement to solvefor the angular velocity.

a) The initial kinetic energy, Ti is zero, because the the cube is in (unstable)equilibrium. The potential we can say is simply that of a point mass at thecube’s center of mass, a height h above the ground,

Ei = 0

Ti + Vi = mgh

1


Vi =1√

2mgℓ (1)

In this case, the edge that is stationary acts as a pivot around which thecube will rotate. Thus, the kinetic energy of the cube while in motion is onlyrotational. We can use the parallel axis theorem to find the moment of iner-tia rotating about an edge of the cube, noting that the moment of inertia forspinning along the center of mass, ICoM = 1

6mℓ2, and the displacement, d is

h = 1√2

ℓ ;

I|| = ICoM + md2 =1

6mℓ2 + mh2

=1

6mℓ2 + m

(

ℓ√

2

)2

= mℓ2

(

1

6+

1

2

)

=2

3mℓ2 (2)

The final kinetic energy as the cube hits the plane is then

Tf =1

2I|| ω2

f

=1

2

(

2

3mℓ2

)

ω2

f

=1

3mℓ2ω2

f (3)

The final potential is as a point mass at a height, 1

2ℓ

Vf =1

2mgℓ (4)

So conservation of energy dictates that

Vi = Tf + Vf

1√

2mgℓ =

1

3mℓ2ω2

f +1

2mgℓ

1√

2g =

1

3ℓω2

f +1

2g

1

3ℓω2

f = g

(

1√

2−

1

2

)

ωf =

√

3g

2ℓ

(√

2 − 1)

(5)

2


b) In this case, the kinetic energy will be slightly different. Now, because theplane is frictionless, the cube rotates about its center of mass, which is moving(only) in the direction of gravity. The position and velocity of the center ofmass become

~rCoM = yCoM

= h sin(

φ +π

4

)

~vCoM = yCoM

= h cos(

φ +π

4

)

φ

=ωℓ√

2cos(

φ +π

4

)

taking φ to be ω.Though the initial kinetic energy is still zero, the final kinetic energy changes

to

Tf =1

2m~v2

f +1

2ICoMω2

f

=1

2m

(

ωf ℓ√

2

)2

cos2

(

0

φf +π

4

)

+1

2

(

1

6mℓ2

)

ω2

f

=1

4mℓ2ω2

f

(

1

2

)

+1

12mℓ2ω2

f

=5

24mℓ2ω2

f (6)

Notice that the initial and final potential energies will remain the same,because the cube starts in the same position as in case a), and ends in the sameorientation, just displaced by 1

2ℓ.

So, using Eq. 1 and 4, the conservation of energy equation is

Vi = Tf + Vf

1√

2mgℓ =

5

24mℓ2ω2

f +1

2mgℓ

5

24ℓω2

f = g

(

1√

2−

1

2

)

ω2

f =24g

5ℓ

(√

2 − 1

2

)

ωf =

√

12g

5ℓ

(√

2 − 1)

(7)

3



Jeff Kissel

October 11, 2006

A chain of linear density (µ [g/cm]) is hanging vertically above atable. Its lowest point is at a height h above the table. The chain isreleased and allowed to fall. Calculate the force exerted on the tableby the chain when a length x of chain has fallen onto the surface.

Figure 1: A chain suspended a height h above the table at t0, and then x amountfallen onto the table after time t

The force exerted on the table will be equivalent to the sum of the gravita-tional (from whatever part of the chain has already landed on the table) andimpulse (from the part of the chain that has just hit the table) forces. Thegravitational force from the amount of mass m = µx already on the table issimply

Fg = mg = µxg (1)

1


During a time interval dt, the mass of the rope equal to µ(v dt) is hitting thetable. The change in momentum imparted onto the table is then

dp = dm v

= [µ(v dt)] v

= µv2 dt (2)

The impulse force is then

Fimpulse =dp

dt= µv2 (3)

However we would like to know how velocity v is related to x(t). Some onedimensional kinematics should shed light on the matter:

v2− v2

i = 2a∆x

⇒ v2 = 2g(x + h) (4)

So Eq. 3 becomesFimpulse = µv2 = 2µg(x + h) (5)

Finally, the total force of the table from the falling chain is

F = Fg + Fimpulse

= µxg + 2µ(x + h)g

= 3µxg + 2µhg

= µg(3x + 2h) (6)

which is equivalent to the weight of the length 3x of the rope, plus the correctionfor the initial height.

2



Jeff Kissel

October 11, 2006

A thin circular ring of radius R and mass m is constrained to rotateabout a horizontal axis passing through two points on the circumfer-ence. The perpendicular distance from the axis to the center of thering is h.

Figure 1: A thin hoop confined to swing from an off-diameter axis of rotation.

a) Find a Lagrangian for this object.

Looking at the hoop from the side few, it’s apparent that this system is akinto a simple rigid pendulum of length h. Since we know the pendulum’s bob (i.e.the hoop’s center of mass) is confined the plane of the page (or in and out ofthe page for the front view) a convenient generalized coordinate for this systemis the angle φ as seen in Figure 1. The kinetic energy of the hoop is then

T =1

2Iφ2

=1

2

(

mR2 + mh2)

φ2

=1

2m(

R2 + h2)

φ2 (1)

1


where the moment of inertia I was found using the parallel axis thereom,

I|| = ICoM + md2 (2)

noting that d is the distance from the center of mass to the parallel axis, andICoM = 1

2mr2 for a thin hoop.

The potential energy will be similar a simple pendulum:

V = m~g · ~r

= −mgh cosφ (3)

So using Eq. 1 and 3 the Lagrangian for a thin hoop hung from an off diameteraxis of rotation is then

L ≡ T − V

=1

2m(

R2 + h2)

φ2 + mgh cosφ (4)

b) Find the period of small oscillations about this axis.

Retrieving the equation of motion from the usual Lagrangian formalism willyield the frequency of small oscillations ω, from which we can find the period.

d

dt

(

∂L

∂φ

)

−∂L

∂φ≡ 0

d

dt

(

m(

R2 + h2)

φ)

+ mgh sinφ = 0

m(

R2 + h2)

φ = −mgh sinφ

φ = −gh

R2 + h2sin φ (5)

From here, we make the approximation that sinφ ≈ φ so that Eq. 5 becomesa simplified 2nd order differential equation of the form x = −φ2x, whose generalsolution is x(t) = x0 cos (φt) in which φ is the frequency. Hence,

φ ≈ −gh

R2 + h2φ

⇒ φ(t) = φ0 cos

(

√

gh

R2 + h2t

)

(6)

The period of small oscillations is the time it takes Eq. 6 to repeat the initialconditions at t = 0, such that cos (φτ − 2π) = cos (0). From this we arrive atthe familiar expression for the period of a pendulum as expected:

τ =2π

φ(7)

= 2π

(

gh

R2 + h2

)− 1

2

(8)

2


Hoop! There it is!

c) For what value of h is the period minimum?

We wish to minimize τ with respect to h, i.e. set the derivative of Eq. 8equal to zero,

∂τ

∂h= 2π (gh)−

1

2 (R2 + h2)1

2

0 = −1

2(gh)−

3

2 (g) (R2 + h2)1

2 +1

2(gh)

1

2 (R2 + h2)−1

2 (2h)

=−g (R2 + h2)

1

2

2 (gh)3

2

+2h

2 (gh)1

2 (R2 + h2)1

2

=−g (R2 + h2)

1

2 (R2 + h2)1

2 + 2h (gh)

(((((((((2 (gh)

3

2 (R2 + h2)1

2

= −g(R2 + h2) + 2gh2

= −gR2− gh2 + 2gh2

= g (h2− R2)

R2 = h2

⇒ h = ±R (9)

Only the non-trivial, positive value for h is physical, so the shortest periodwould result if the axis of rotation which has a distance

h = R (10)

from the center of mass of the hoop.Sanity check: if one wants to minimize the hoop’s period, Eq. 7 says that

one could maximize the frequency. The axis at which the hoop won’t oscillateat all is zero is through center, and an axis with the maximum frequency is thatwhich has the longest “pendulum arm,” i.e. at the edge of the hoop, or h = R.

3



Jeff Kissel

May 18, 2007

An object is dropped from a tower of height h. The tower is locatedat the equator of the Earth. The rotational speed of the earth is Ω.

Figure 1: Da Erf. Note that +ex points North, and +ey points West, so ~Ω =Ω ex.

(a) If the Earth is treated as perfectly round and uniform and theacceleration due to gravity on a non-rotating earth is g, what is theacceleration of gravity at the tower?

With Earth rotating, we must not only consider the acceleration due togravity, but the centripetal and Coriolis acceleration. At the top of the tower,~agrav = −g ez, the radial vector is ~r = +h ez, and the velocity is ~v = −vr ez.The effective acceleration (in the rotating frame) felt at the tower is then givenas

~aeff = ~agrav + ~acentrifugal + ~aCoriolis

= ~agrav − ~Ω × ~Ω × ~r − 2 Ω × ~v

= (−gez) − (Ωex) × (Ωex) × (hez) − 2(Ωex) × (−vr ez)

= −g ez − hΩ2 (ex × ex × ez) − 2Ωvr (ex ×−ez)

1


= −g ez − hΩ2 (ex ×−ey) − 2Ωvr(−ey)

= −g ez − hΩ2 (−ez) + 2Ωvrey

~aeff = 2Ωvr ey − (g − hΩ2) ez (1)

Yet, at the top of the tower, we assume the object to have no radial velocity,vr = 0. Thus, the effective acceleration is

~geff = −(g − hΩ2) ez (2)

(b) Even though it is released from rest, this object will not landdirectly below the point from which it was dropped. Calculate theamount and direction (N, E, S, W, or elsewhere) of the horizontaldeflection of the object. You may assume the deflection is small.

The Coriolis effect is the only term that will have an effect in the horizontal(ey, East or West) direction, so pulling the first term from Eq. 1 if vr is now~geff t,

FCoriolis = m aCoriolis

= 2mΩvr ey

= 2mΩ(−(g − hΩ2) t) ey

m y = −2mΩ(g − hΩ2) t

y = −2Ω(g − hΩ2) t

⇒ y = −2Ω(g − hΩ2)

(

1

2t2

)

⇒ y = −2Ω(g − hΩ2)

(

1

6t3

)

y = −Ω

3(g − hΩ2) t3 (3)

This quantity y is the horizontal deflect, get we can be more explicit. Thetime spent falling t can be found from the free-fall kinematics of a particlereleased from rest:

∆y = v0 t −

1

2geff t2 ⇒ h =

1

2(g − hΩ2) t2

⇒ t2 =2h

(g − hΩ2)⇒ t =

(

2h

(g − hΩ2)

)1/2

(4)

which means the deflection becomes

y = −Ω

3(g − hΩ2)

(

2h

(g − hΩ2)

)3/2

y = −Ω (2h)3/2

3(g − hΩ2)1/2(5)

The direction of this displacement is in the −ey direction, or to the East.

2



Jeff Kissel

October 11, 2006

Consider a pendulum consisting of a uniform disk of radius, R andmass, m suspended from a massless rod that allows it to swing in theplane of the disk.

Figure 1: A circular disk, rotating in its plane on an axis a distance, d awayfrom the center of mass.

(a) Using the parallel axis theorem, or calculating it directly, findthe moment of inertia I for the pendulum about an axis a distanced(0 < d < R) from the center of the disk

The moment of inertia through the center of a disk of radius R and mass m

is

ICoM =1

2mR2 (1)

Using the parallel axis theorem, which states that the moment of inertia arounda point parallel to the center of mass a distance d away is

Id = ICoM + md2 (2)

So, using Eq. 1 the moment of inertia displaced by |~r| = d is

Id =1

2mR2 + md2

Id =1

2m (R2 + 2d2) (3)

1


(b) Find the gravitational torque on the pendulum when displacedby and angle φ.

From the definition of rotational torque from a given force:

τ = ~r × ~Fg

= |~r| |~F | sinφ

= (d)(−mg) sin φ

= −mgd sinφ (4)

(c) Find the equation of motion for small oscillations and give the fre-quency ω. Further, find the value of d corresponding to the maximumfrequency for fixed R and m.

If we note that τ = Iα, then we know the equation of motion immediatelyfrom Eq. 4(using the small angle approximation that sinφ = φ),

Idα = −mgd φ

α = −mgd

Id

φ

(5)

which is a 2nd order differential equation of the form φ = −ω2φ from which wecan pull out ω. Subbing in Eq. 3,

ω2 =2mgd

m (R2 + 2d2)

ω =

(

2gd

R2 + 2d2

)2

(6)

To find the value of d for which ω is maximized, we shall differentiate withrespect to d and set to zero,

dω

dd=

d

dd

(

(2gd)(R2 + 2d2)−1)

0 =1

2

(

2gd

R2 + 2d2

)

−1

2(

2g (R2 + 2d2)−1 − (2gd)(R2 + 2d2)−2(4d))

0 =1

2

(

R2 + 2d2

2gd

)

1

2(

2g

R2 + 2d2− 8gd2

(R2 + 2d2)2

)

0 =1

2

(

R2 + 2d2

2gd

)1

2

(

2g(R2 + 2d2) − 8gd

(R2 + 2d2)2

)

0 =(

R2 + 2d2)

1

2

(

2gR2 + 4gd2 − 8gd2)

0 =(

R2 + 2d2)

1

2 (2g(R2 − 2d2))

0 =(

R2 + 2d2)

1

2

(

R2 − 2d2)

2


R2 + 2d2 = 0 R2 − 2d2 = 0

d =iR√

2d = ± R√

2(7)

But we know the left set is unphysical and − R√

2is equivalent to R

√

2, so

dmax =R√2

(8)

3



Jeff Kissel

October 11, 2006

A heavy particle of mass m is placed close to the top of a frictionless

vertical hoop with radius R and allowed to slide down the hoop. Find

the angle at which the particle falls off.

We know the point at which the particle falls off the hoop is when the normalforce N acting on the particle goes to zero. So, we need an expression for N .

Figure 1: Ball of mass m on a hoop with radius R.

Before the particle begins to move, it has only potential energy Vi. At theinstant it leaves the hoop, it has both potential Vf and kinetic Tf energies. Theparticle is allowed to slide, so there assumed to be no friction and therefore norolling, thus we need no worry about the moment of inertial of the ball. Theenergy conservation equation is then

Ei = Ef

Vi = Tf + Vf

mgR =1

2mv2 + mgR cos θ (1)

However, Eq. 1 does not contain the normal force N , which we need. It can befound by assuming the hoop is circular, so that the normal force acting on the

1


particle is just its centripetal acceleration subtracted from its weight, or

N = mg cos θ − mac

= mg cos θ − mv2

R(2)

Multiplying this by 1

2R will yield a more useful equation for the kinetic energy,

1

2R → (N = mg cos θ − mac)

1

2RN =

1

2mgR cos θ −

1

2mv2

1

2mv2 =

1

2mgR cos θ −

1

2RN (3)

which we can now plug into the conservation equation (Eq. 1) to arrive atexplicit equation for N ,

mgR =1

2mgR cos θ −

1

2RN + mgR cos θ

1

2RN =

3

2mgR cos θ − mgR

N = 3mg cos θ − 2mg (4)

Since we’ve established that the particle falls off the hoop when N = 0, Eq. 4becomes,

0 = mg(3 cos θ − 2)

= 3 cos θ − 2

cos θ =2

3(5)

So the angle at which a heavy particle of mass M slides off a circular hoop orradius R is

θ ≈ 48.2 (6)

2



Jeff Kissel

October 11, 2006

A solid sphere of radius R and mass M rolls without slipping down a

rough inclined plane of angle θ.

Figure 1: A sphere rolling down a rough ramp.

Take the coefficient of static friction to be µ and calculate the lin-

ear acceleration of the sphere down the plane, assuming that it rolls

without slipping.

In this problem, we can use conservation of energy between the top andbottom of the ramp. At the top of the ramp, the ball begins rolling from restand thus has no initial kinetic energy. This means the initial energy is

Ei = 0

Ti + Vi

= mCoM g x sin θ (1)

1


where the origin is set at the top of the ramp.The final energy has only kinetic terms, and is a sum of the motion of the

center of mass, and the rotation of the sphere,

Ef = Tf + 0

Vf

=1

2mCoM v2 +

1

2ICoM ω2 (2)

Setting Eqs. 1 and 2 equal to each other, plugging in for the moment ofinertia of a sphere, ICoM = 2

5 mCoM R2 and noting that v = x, and ω = xR

,

Ei = Ef

mCoM g x sin θ =5

10 mCoM x2 +1

2

(

2

5 mCoM R2

) (

x

R

)2

x2 =10

7g x sin θ

dx

dt= ξx

1

2 (3)

where I’ve defined a constant ξ =√

107 g sin θ. At this point we integrate both

sides to find x as a function of time, which we can then differentiate twice withrespect to time to get the acceleration, x = aCoM ,

x−

1

2 dxξ

= dt

2x1

2

ξ= t

x =1

4ξ2 t2 ⇒ x = 1

2 ξ2 t ⇒ x =1

2ξ2

x = 12

(√

107 g sin θ

)2

aCoM = 75 g sin θ (4)

Calculate the maximum value of θ for which the sphere will not slip.

The sphere will begin to slip once the x-component of the weight becomesgreater than the maximum frictional force providable by µ. The angle at which

this occurs, θc can be found by setting the normal force, F(x)g equal to the force

of friction Ff = µN ,

mg sin θc = µN

mg sin θc = µ mg cos θc

µ =sin θc

cos θc

= tan θc

⇒ θc = tan−1 (µ) (5)

2



Jeff Kissel

October 11, 2006

A rocket is filled with fuel and is initially at rest. It starts moving by

burning fuel and expelling gases with the velocity u, constant relative

to the rocket. Determine the speed of the rocket at the moment when

its kinetic energy is largest.

Figure 1: A free rocket traveling in an inertial frame, without the force of gravity.

At some time t the rocket, traveling at a velocity v, has some mass m. Aninfinitesimal time dt later, the rocket has lost a mass dm, and gained a speeddv. At this time, an infinitessimal amount of fuel, dm′ is ejected at a velocityu with respect to the rocket. In an inertial frame however, that fuel’s velocityis v − u. Thus, using conservation of momentum, we can get an expression forthe mass of the rocket as a function of time,

procket = procket + pfuel

1


m v = (m − dm)(v + dv) + dm′(v − u)

m v = m v + m dv − v dm′− du dv + v dm′

− u dm′

m dv = u dm′− du dv (1)

From here, we note that to first order, the term du dv is extremely small, so itis dropped, leaving

1

udv =

1

mdm′ (2)

Noting that dm′ = −dm and integrating both sides give us an express forthe mass of the rocket as a function of time:

∫ v

0

1

udv′ = −

∫ m

m0

1

mdm

−

v

u= ln

(

m

m0

)

m = m0e−

v

u (3)

Plugging into the expression of the rocket’s kinetic energy,

T = 1

2mv2 =

1

2m0v

2 e−v

u (4)

which we then maximize with respect to v to find the largest speed vmax,

∂T

∂v=

∂

∂v

(

1

2m0v

2 e−v

u

)

0 = m0v e−v

u−

1

2m0v

21

ue−

v

u

m0v e−v

u =v

2u

m0v e−v

u

vmax = 2u (5)

Thus, the kinetic energy is maximum when the velocity of the rocket is twicethat of the ejected fuel.

2



Jeff Kissel

May 19, 2007

Exam Iteration Question Number Notes/ChangesSpring 2006 #12 –Fall 2006 #12 –

Spring 2007 #12 Added damping.

A small lead ball of mass m is attached to one end of a vertical spring

with the spring constant k. The other end of the spring oscillates up

and down with amplitude A and frequency ω. Determine the motion

of the ball after a long period of time. You may assume that the

ball is subject to a small amount of damping, with the damping force

being given by F = −γv.

Figure 1: A driven harmonic oscillator sitting in a viscous damping fluid.

Hopefully, one can assume that all motion is in the y direction.

1


The kinetic energy of the system is

T =1

2my2 (1)

and the potential, as long as we define the origin to be at the equilibrium pointsuch that the displacement ∆y ≡ y − ye = y, is as follows,

V =1

2ky2 + mgy (2)

This makes the Lagrangian

L ≡ T − V

L =1

2my2

−1

2ky2

− mgy (3)

We can use the form of the Euler-Lagrange equations which has forces notderivable from a potential, Q, to determine the equation of motion. The forcesnot derivable from potential are the driving and damping forces,

Fo = A cos (ωt) (4)

Fd = −γy (5)

So, the equation of motion is

d

dt

(∂L

∂y

)−

∂L

∂y= Qi

d

dt(my) − (−ky − mg) = A cos (ωt) − γy

my + ky + mg = A cos (ωt) − γy

my + γy + ky = A cos (ωt) − mg

y +γ

my +

k

my =

A

mcos (ωt) − g

Let β =γ

m, ω2

0 =k

m, and B =

A

m

y + β y + ω20 y = B cos (ωt) − g

Let x = −g

ω2

0

+ y ⇒ y = x + g

ω2

0

x = y

x = y

x + βx + ω20

(x +

g

ω20

)= B cos (ωt) + g

x + βx + ω20x + g = B cos (ωt) + g

x + βx + ω20x = B cos (ωt) (6)

Yes folks, it’s the driven harmonic oscillator with damping. Your favoritedifferential equation to solve.

2


We’ll start with the homogeneous solution.

0 = xH + βxH + ω20xH

The characteristic equation:

0 = am2± + bm± + c

⇒ 0 = (1)m2± + (β)m± + (ω2

0)

has roots

m = p ± iq

with p = −b

2and q =

√c −

b2

4

⇒ p = −β

2and q =

√ω2

0 −β2

4

Such that the general solution is

ept (C cos (q t) + D sin (q t))

xH(t) = e−1

2βt

(C cos

(√ω2

0 −β2

4t

)+ D cos

(√ω2

0 −β2

4t

))(7)

For the inhomogeneous solution, we can generalize the equation making itcomplex,

¨xI + β ˙xI + ω20xI = Be−iωt

and then take the real part at the end. This way we can assume a complexexponential solution, xI = G ei(ωt), which knocks out the differentiation because

xI = G e−iωt ˙xI = − iω G e−iωt ¨xI = − ω2 G e−iωt (8)

which means all we have to do is find the complex constant G,

Be−iωt = −ω2 G e−iωt + −iωβ G e−iωt + ω20G e−iωt

B = −ω2 G + −iωβ G + ω20G

B = G((ω2

0 − ω2) − iωβ)

G =B

(ω20 − ω2) − iωβ

=B

(ω20 − ω2) − iωβ

((ω2

0 − ω2) + iωβ

(ω20 − ω2) + iωβ

)

G =B(ω2

0 − ω2) + iBωβ

((ω20 − ω2) + ω2β2)

(9)

3


So the inhomogeneous solution is

xI(t) =B(ω2

0 − ω2)

((ω20 − ω2) + ω2β2)

e−iωt +Bωβ

((ω20 − ω2) + ω2β2)

ie−iωt

⇒ xI(t) = ℜe

[B(ω2

0 − ω2)

((ω20 − ω2) + ω2β2)

e−iωt +Bωβ

((ω20 − ω2) + ω2β2)

ie−iωt

]

(ℜe[e−iωt

]= cos (ωt)

ℜe[ie−iωt

]= sin (ωt)

)

xI(t) =B(ω2

0 − ω2)

((ω20 − ω2) + ω2β2)

cos (ωt) +Bωβ

((ω20 − ω2) + ω2β2)

sin (ωt) (10)

Yeah. The complete solution for x(t) is

x(t) = e−1

2βt

(C cos

(√ω2

0 −β2

4t

)+ D cos

(√ω2

0 −β2

4t

))

+B(ω2

0 − ω2)

((ω20 − ω2) + ω2β2)

cos (ωt) +Bωβ

((ω20 − ω2) + ω2β2)

sin (ωt)

(11)

Oh but wait, we have to plug back in for y(t) = x(t) + g

ω2

0

,

y(t) = e−

1

2βt(

C cos

(√ω2

0 −β2

4t

)+ D cos

(√ω2

0 −β2

4t

))

+B(ω2

0 − ω2)

((ω20 − ω2) + ω2β2)

cos (ωt) +Bωβ

((ω20 − ω2) + ω2β2)

sin (ωt) −g

ω20

And remember: β =γ

m, ω2

0 =k

m, and B =

A

m

y(t) = e−

γt

2m

(C cos

(√k

m−

γ2

4m2t

)+ D cos

(√k

m−

γ2

4m2t

))

+A(k − mω2)

(k − mω2)2 + ω2γ2cos (ωt) +

Aωγ

(k − mω2)2 + ω2γ2sin (ωt) −

mg

k

(12)

Which is the position as a function of time, where C and D are constantsdetermined by initial conditions.

OK so what happens after a long time? The exponential damping in thehomogeneous term kills off all resonances that would be found from ω0, and thespring just oscillates according to the inhomogeneous equation. In other words,the spring just sloshes along with the driving frequency, according to

limt→∞

y(t) =A(k − mω2)

(k − mω2)2 + ω2γ2cos (ωt) +

Aωγ

(k − mω2)2 + ω2γ2sin (ωt) −

mg

k

(13)

4



Jeff Kissel

May 17, 2007

Exam Iteration Question Number Notes/ChangesSpring 2006 #13 –Fall 2006 #13 –

Spring 2007 #13 Last sentence changed

In movie cameras and projectors film speed is 24 frames/sec. You

see on the screen a car that moves without skidding, and know that

the real-life diameter of its wheels is 1 m. The wheels on the screen

make 3 turns/sec. What is the speed of the car, assuming it is not

moving in excess of 200 mi/hr?

Figure 1: A wheel that appears to rotate one eighth of its circumference.

In one frame, it appears to the viewer that the wheel has turned and angularvelocity

ω =3 [turns]

[sec]

24 [frames]

[sec]

=1

8

turns

frame(1)

1


However, appearances can be deceiving. To the viewer, “measuring” the po-sition of the wheel discretely, the wheel has turned only 1/8 of its circumference.The car could be moving much faster, such that the wheel makes any integernumber N of extra turns before reaching the final 1/8 (for example, N=1 inFigure 1. The angular velocity should then technically be

ω =

(

N +1

8

)

turns

frame(2)

Finally, to find the (linear) speed v, we must convert Eq. 2 into the correctunits,

ω =

(

N +1

8

)

:[turns]

[frame]× 24 [frames]

[sec]× 2π

[radians]

:[turn]

ω = 48π

(

N +1

8

)

rads/s (3)

and then convert angular to linear speed, noting that the wheel has a 1 mdiameter, and thus a 0.5 m radius.

v = ω r

=

(

48π

(

N +1

8

)

rads/s

)

(0.5 m)

= 24π

(

N +1

8

)

m/s (4)

Thus for N = 0, v = 3π m/s ≈ 6.7 mi/hr, and N = 1, v = 27π m/s ≈

189.7 mi/hr. (1 m/s = 2.23693629 mi/hr).Though this problem may seem pretty lame, it brings up an important con-

cepts in signal processing: digital sampling of analog signals, which may resultin “aliasing.” Check out the Wikipedia article on it!

2



Jeff Kissel

October 11, 2006

A ball of mass m collides with another ball, of mass M = am, initiallyat rest. The collision is elastic and central, i.e. the initial velocity ofthe ball is along the line connecting the centers of the two objects.

Figure 1: Two balls collide with mass ratio a.

(a) Determine how the energy lost by the moving ball depends onthe mass ratio a, and find the value of a for which the energy loss islargest. Describe what happens at that value of the mass ratio.

We can solve this problem using conservation of energy and momentum.The problem does not mention any potential, so we can find the velocity of thesecond ball in terms of the first starting with kinetic energy conservation.

Ei = Ef

1

2mv2

i +

>

01

2Mui

2 =1

2mv2

f +1

2Mu2

f

m v2

i = m v2

f + am u2

f

v2

i = v2

f + au2

f (1)

We can now bring in conservation of momentum to knock out one of theunknowns,

pi = pf

mvi +Mui = mvf + Muf

1


m vi = m vf + amuf

vf = vi − auf (2)

Now, pluggind Eq. 1 into Eq. 2,

v2

i = (vi − auf )2 + au2

f

v2

i = v2

i + a2u2

f − 2aviuf + av2

f

0 = a (a + 1) u 2f − 2 a vi uf

−(a + 1)uf = −2 vi

uf =2vi

(a + 1)(3)

Finally, the kinetic energy lost by the first ball is just that that is gained bythe second,

∆E =1

2Mu2

=1

2am

(

2vi

a + 1

)2

∆E =2amv2

i

(a + 1)2(4)

To find the value for which the energy lost is largest, we can just maximizewith respect to a,

∂∆E

∂a=

∂

∂a

(

2mv2

i a (a + 1)−2)

0 = 2mv2

i (a + 1)−2− 4mv2

i a (a + 1)−3

2mv2

i

1

(a + 1)2=

2a

(a + 1)3 2mv2

i

(a + 1)3 = 2a (a + 1)2

a + 1 = 2a

a = 1 (5)

So, the change in energy is the greatest when the balls have the same exactmass: the first ball’s kinetic energy is split evenly between them during thecollision such that vf and uf are equal, but in opposite direcetions.

(b) Investigate the limiting cases of heavy and light balls and com-ment on your result.

In the limiting case that a ≪ 1, the first ball will be much more massivethan the second. In this case, the first ball will lose virtually none of its energy,but because the second ball is so less massive, it will propel off with uf = 2vi.

lima→0

∆E =2amv2

i

(a + 1)2= 2Mv2

i

2


lima→0

vf = vi −auf = vi

lima→0

uf = 2vi

(a + 1)= 2vi

In the case where a ≫ 1, the energy imparted onto the second ball will beneglegible. In addition, since so little velocity is transferred to the second ball,the first will be forced in the opposite direction with a final velocity equal to itsinitial.

lima→∞

∆E =2amv2

i

(:∞a + 1)2

= 0

lima→∞

vf = vi −2avi

(a + 1) = −vi

lima→∞

uf = −2vi

(:∞a + 1)

= 0

3



Jeff Kissel

October 11, 2006

Two particles of mass m and m+2 move in circular orbit around each

other under the influence of gravity. The period of the motion is T .

They are suddenly stopped and then released, after which they fall

towards each other. If the objects are treated as mass points, find

the time they collide in terms of T .

Figure 1: (Left) Two masses orbiting about each other suddenly stopped, as atwo body problem. (Right) The same system reduced to a one-body problemusing the reduced mass µ.

As noted in the picture, the problem can be simplified to a one body problemusing the reduced mass,

µ =m1m2

m1 + m2

=m2 + 2m

2m + 2

µ =1

2

m + 2

(1 + m−1)(1)

Now a one body problem, we can find the Lagrangian of this more simple system,

V = −k

r

1


T =1

2µr2 +

1

2µ r2θ2

L = T − V

L =1

2µ (r2 + r2θ2) +

k

r(2)

where I’ve defined the gravitational constant of the system k = Gm1m2. Fromhere we crank the usual Lagrangian formalism,

d

dt

(

∂L

∂r

)

−∂L

∂r= 0

d

dt(µr) − (µrθ −

k

r2) = 0

µr − µrθ +k

r2= 0 (3)

d

dt

(

∂L

∂θ

)

−∂L

∂θ= 0

d

dt

(

µr2θ)

− (0) = 0

µr2θ = 0

θ = 0 (4)

It appears as though the sooner will be a little more useful than the latter,so let’s roll with it. We know for circular motion, not only is Eq. 4 true, butr = 0 is also true. Using this fact, we can distill Eq. 3 a bit to get a familiarlaw from my buddy Kepler:

µr − µrθ +k

r2= 0

k

r2= µrθ

r3 =k

µθ2(θ = ω =

2π

T)

r3 =kT 2

4π2µ(5)

Now that we’ve naively re-derived Kepler’s 3rd, let’s stop the masses. TakingEq. 5 as our starting radius, r3

0, we know stop the angular motion of the masses,

i.e. set ˙theta, so the Eq. 3 instead becomes

0 = µr −µrθ +

k

r2

r = −k

µr2

( r =∂r

∂t=

∂r

∂t

∂r

∂r= r

∂r

∂r)

2


r dr = −k

µr2dr

1

2r2 =

k

µr+ C

r2 =2k

µr+ 2C

We can use the initial condition that at r = r0, r = 0, so that C = − kµr0

, andfinally we can solve for the linear merger time,

r =

√

2k

µr−

2k

µr0

dr =

(

2k

µ

(

1

r−

1

r0

))1

2

dt

dt =dr

(

2kµ

(

1

r− 1

r0

))1

2

t =( µ

2k

)1

2

∫

0

r0

(

1

r−

1

r0

)

−

1

2

dr (6)

which is a horribly disgusting integral, that can be solved analytically using afew “u substitutions,” (let u = 1

r, and then u = u0 sec2 θ) but I’ll happily just

leave it as Eq. 6. Numerically evaluated this turns out to be

t ≈T

8√

2(7)

3



Jeff Kissel

October 11, 2006

Three points masses of identical mass are located at (a, 0, 0), (0, a, 2a)and (0, 2a, a). Find the moment of inertia tensor around the origin,the principal moments of inertia, and a set of principal axes.

Figure 1: A system of three identical masses.

The moment of inertia tensor for a discrete mass system is given by

I = Ijk =∑

i

mi

(δjkr2

i − (xjxk)i

)(1)

So for

~r1 = a x r2

1= a2

~r2 = a y + 2a z r2

2 = a2 + 4a2 = 5a2

~r3 = 2a y + a z r2

3= 4a2 + a2 = 5a2

1


the diagonal elements of I are

I11 =∑

i

mi

(r2

i − xixi))

= m(r2

1− x1x1) + m(r2

2− x2x2) + m(r2

3− x3x3)

= m(a2− a2) + m(5a2

− 0) + m(5a2− 0)

= 10ma2 (2)

I22 =∑

i

mi

(r2

i − yiyi))

= m(r2

1− y1y1) + m(r2

2− y2y2) + m(r2

3− y3y3)

= m(a2− 0) + m(5a2

− a2) + m(5a2− 4a2)

= 6ma2 (3)

I33 =∑

i

mi

(r2

i − zizi))

= m(r2

1 − z1z1) + m(r2

2 − z2z2) + m(r2

3 − z3z3)

= m(a2− 0) + m(5a2

− 4a2) + m(5a2− a2)

= 6ma2 (4)

And the off-diagonal elements are

I12 = I21 =∑

i

mi(−xiyi)

= m(−x1y1) + m(−x2y2) + m(−x3y3)

= m(0) + m(0) + m(0)

= 0 (5)

I13 = I31 =∑

i

mi(−xizi)

= m(−x1z1) + m(−x2z2) + m(−x3z3)

= m(0) + m(0) + m(0)

= 0 (6)

I23 = I32 =∑

i

mi(−yizi)

= m(−y1z1) + m(−y2z2) + m(−y3z3)

= m(0) − m(2a2) − m(2a2)

= −4ma2 (7)

So the moment of inertia tensor for this system is

I =

10ma2 0 0

0 6ma2−4ma2

0 −4ma2 6ma2

(8)

2


The principle moments of inertia and principle axes are the eigenvalues andeigenvectors (respectively) of the moment of inertia tensor. So to find eigenval-ues, we find the characteristic equation of I

det[I − λ1

]=

∣∣∣∣∣∣

10ma2− λ 0 0

0 6ma2− λ −4ma2

0 −4ma2 6ma2− λ

∣∣∣∣∣∣

0 = (10ma2− λ)

((6ma2

− λ)(6ma2− λ) − (−4ma2)2

)− 0 + 0

= (10ma2− λ)

(36m2a4

− 12λma2 + λ2− 16m2a4

)

= (10ma2− λ)

(λ2

− 12λma2 + 20m2a4)

= (10ma2− λ)(10ma2

− λ)(2ma2− λ)

⇒ λ1 = 10ma2 λ2 = 10ma2 λ3 = 2ma2 (9)

so there is a two-fold degeneracy in the moments of inertia which means thesystem is akin to a symmetrical top. To find the eigenvectors (principle axes),we demand that the principle moments of inertia obey the eigenvalue equation(I − λ1) · ~x = 0. So for λ1 = λ2 = 10ma2,

0 0 00 −4ma2

−4ma2

0 −4ma2−4ma2

·

x

y

z

=

000

−4ma2 y − 4ma2 z = 0

⇒ y = − z

so, picking the simplest eigenvectors, the principle axes for λ1 & λ2 are

χ1 =

0−11

and χ2 =

100

(10)

And for λ3 = 2ma2,

8ma2 0 0

0 4ma2−4ma2

0 −4ma2 4ma2

·

x

y

z

=

000

8ma2 x = 0

4ma2 y − 4ma2 z = 0

−4ma2 y + 4ma2 z = 0

⇒ x = 0, y = z

and again picking the simplest eigenvector, the principle axis for λ3 is

χ3 =

011

(11)

3



Jeff Kissel

October 11, 2006

Consider a double pendulum consisting of a mass m suspended ona massless rod of length ℓ, to which is attached by a pivot anotheridentical rod with an identical mass m attached at the end, as shownin the Figure 1.

Figure 1: A double pendulum, with identical masses and string lengths.

Using the angles θ1 and θ2 as generalized coordinates,

a) Find a Lagrangian for the system.

Using θ1 and θ2 as generalized coordinates, the position (~r1, ~r2) and velocity(~v1, ~v2) vectors become

~r1 = (ℓ sin θ1) x + (ℓ cos θ1) y (1)

~v1 = ~r1 =(

ℓθ1 cos θ1

)

x −

(

ℓθ1 sin θ1

)

y (2)

~r2 = (ℓ sin θ1 + ℓ sin θ2) x + (ℓ cos θ1 + ℓ cos θ2) y

1


= ℓ (sin θ1 + sin θ2) x + ℓ (cos θ1 + cos θ2) y (3)

~v2 = ~r2 =(

ℓθ1 cos θ1 + ℓθ2 cos θ2

)

x −

(

ℓθ1 sin θ1 + ℓθ2 sin θ2

)

y

= ℓ(

θ1 cos θ1 + θ2 cos θ2

)

x − ℓ(

θ1 sin θ1 + θ2 sin θ2

)

y (4)

In the kinetic energy term of the Lagrangian, we’ll need the squares of thevelocities, which are

v2

1= ~v1 · ~v1 =

(

ℓθ1 cos θ1

)2

+(

ℓθ1 sin θ1

)2

= ℓ2θ2

1

(

cos2 θ1 + sin2 θ1

)

= ℓ2θ2

1(5)

v2

2= ~v2 · ~v2 =

(

ℓθ1 cos θ1 + ℓθ2 cos θ2

)2

+(

ℓθ1 sin θ1 + ℓθ2 sin θ2

)2

= ℓ2θ2

1cos2 θ1 + ℓ2θ2

2cos2 θ2 + 2ℓ2θ1θ2 cos θ1 cos θ2

+ ℓ2θ2

1sin2 θ1 + ℓ2θ2

2sin2 θ2 + 2ℓ2θ1θ2 sin θ1 sin θ2

= ℓ2θ2

1

(

cos2 θ1 + sin2 θ1

)

+ ℓ2θ2

2

(

cos2 θ2 + sin2 θ2

)

+ 2ℓ2θ1θ2 (cos θ1 cos θ2 + sin θ1 sin θ2)

= ℓ2θ2

1+ ℓ2θ2

2+ 2ℓ2θ1θ2 cos (θ2 − θ1) (6)

Finally, noting that ~g = −g y and m1 = m2 = m we can combine the resultsof Eqs. 1, 3, 5, and 6 to find the kinetic and potential energies:

T =1

2m~v2

1+

1

2m~v2

1

=1

2mℓ2θ2

1+

1

2m

(

ℓ2θ2

1+ ℓ2θ2

2+ 2ℓ2θ1θ2 cos (θ2 − θ1)

)

= mℓ2θ2

1+

1

2mℓ2θ2

2+ mℓ2θ1θ2 cos (θ2 − θ1) (7)

V = m~g · ~r1 + m~g · ~r2

= −mgℓ cosθ1 − mgℓ (cos θ1 + cos θ2)

= −2mgℓ cosθ1 − mgℓ cosθ2 (8)

which we combine to form a Lagrangian for a double pendulum with identicalmasses and string lengths:

L ≡ T − V

= mℓ2θ2

1+

1

2mℓ2θ2

2+ mℓ2θ1θ2 cos (θ2 − θ1) + 2mgℓ cosθ1 + mgℓ cos θ2

=1

2mℓ2

(

2θ2

1+ θ2

2+ 2θ1θ2 cos (θ2 − θ1)

)

+ mgℓ (2 cos θ1 + cos θ2) (9)

2


b) Find an approximate Lagrangian that is appropriate for small oscil-lations and obtain from it the equations of motions when θ1 & θ2 ≪ 1.

For small oscillations, we can approximate cosx ≈ 1 − 1

2x2 + O(x4), and

drop those terms which are of higher order than quadratic in θi and θi so thatEq. 9 is

L ≈ mℓ2θ2

1+

1

2mℓ2θ2

2+ mℓ2θ1θ2

(

1 −1

2(θ2 − θ1)2

)

+ 2mgℓ

(

1 −1

2θ2

1

)

+ mgℓ

(

1 −1

2θ2

2

)

where what is canceled above will turn out to be quartic in θi and θi so it isdropped. Simplified, the approximate Lagrangian is then

L ≈ mℓ2θ2

1+

1

2mℓ2θ2

2+ mℓ2θ1θ2 + 2mgℓ − mgℓθ2

1+ mgℓ −

1

2mgℓθ2

2

≈1

2mℓ2

(

2θ2

1+ θ2

2+ 2θ1θ2

)

−1

2mgℓ

(

2θ2

1+ θ2

2

)

+ 3mgℓ (10)

c) Assuming that each angle varies as θ1,2 = A1,2 eiωt, find the frequen-cies for small oscillations.

Performing the usual Lagrange processes using Eq. 10 to find the equationsof motion, we can then use these to solve for the frequencies of small oscillations.In general, Lagrange’s equations of motion are defined by

d

dt

(

∂L

∂qi

)

−∂L

∂qi

= 0 (11)

where qi are the generalized coordinates of the system, and qi are their respectivetime derivatives.

So the equations of motion for θ1 and θ2 are

∂L

∂θ1

= 2mℓ2θ1 + mℓ2θ2

∂L

∂θ1

= −2mgℓ θ1

⇒ 2mℓ2θ1 + mℓ2θ2 + 2mgℓ θ1 = 0

ℓ(

2θ1 + θ2

)

+ 2gθ1 = 0 (12)

∂L

∂θ2

= mℓ2θ2 + mℓ2θ1

∂L

∂θ2

= −mgℓ θ2

⇒ mℓ2θ2 + mℓ2θ1 + mgℓ θ2 = 0

ℓ(

θ1 + θ2

)

+ g θ2 = 0 (13)

3


If we assume θj = Aj eiωt, then θj → −ω2θj , and the equations of motion yieldfour oscillation frequencies:

ℓ(

−2ω2θ1 − ω2θ2

)

+ 2g θ1 = 0

−ω2ℓ(

2A1 eiωt + A2

eiωt)

+ 2gA1 eiωt = 0

ω2ℓ (2A1 + A2) = 2gA1

ω2 =2gA1

ℓ(2A1 + A2)

ω± = ±

√

2gA1

ℓ(2A1 + A2)(14)

ℓ(

−ω′2θ1 − ω′2θ2

)

+ g θ2 = 0

−ω′2ℓ(

A1

eiω′t + A2

eiω′t)

+ gA2

eiω′t = 0

ω′2ℓ (A1 + A2) = gA2

ω′2 =gA2

ℓ(A1 + A2)

ω′

± = ±

√

gA2

ℓ(A1 + A2)(15)

So ’der you go.

4



Jeff Kissel

October 22, 2006

A particle of mass m. at rest initially, slides without friction on awedge of angle θ and and mass M that can move without friction ona smooth horizontal surface.

Figure 1: A mass m slides in the +x direction down a frictionless wedge M ,which can slide across a frictionless plane in the x direction. Note, this picturedoes not represent the actual motion; realistically X is in the −x direction aswill be shown later.

a) What is the Hamiltonian?

First we must find the Lagrangian (L = T − V ), to determine whether it isexplicitly independent of time. If this is so, then the Hamiltonian is simply thetotal energy (H = T + V ).

We can define a coordinate system where the origin is the top of the wedgeat some initial time as in Figure 1, such that the position vector for the topcorner of the wedge at time, t will be

~R = X x + (0) y (1)

and particle’s position vector will be

~r = ~R + (x x + y y)

~r = (X + ℓ cos θ) x + (ℓ sin θ) y (2)

1


where ℓ is the distance the mass has traveled down the ramp. While the particleis on the ramp, it is constrained to move along ℓ, or

tan θ =∆y

∆xsin θ

cos θ=

y − Y

x − X(x − X) sin θ = (y − (0)) cos θ

f : (x − X) sin θ − y cos θ = 0 (3)

which is an equation of constraint related to the normal force.The kinetic energy of the system is that of two free particles, since there is

no friction:

T =1

2m(x2 + y2) +

1

2MX2 (4)

The only potential in the system is gravitational, with ~g = g y. We can ignorethe potential on the wedge because it is constant, so the system’s potential isjust the gravitational potential of the particle,

Vg ≡ − m~g · ~r

V = −mgy (5)

The Lagrangian is then potential subtracted from the kinetic energy,

L ≡ T − V

=1

2m(x2 + y2) +

1

2MX2 + mgy

L =1

2mx2 +

1

2my2 +

1

2MX2 + mgy (6)

which does not explicitly depend on time, so the Hamiltonian is the sum of thekinetic and potential energies,

H =1

2mx2 +

1

2my2 +

1

2MX2

− mgy (7)

b) Derive the equation of motion from the Lagrangian.

With Lagrange multipliers, The Euler-Lagrange equation in general is

d

dt

(

∂L

∂qi

)

−∂L

∂qi

= λ∂f

∂qi

+ µ∂f

∂qi

(8)

yet the second term on the right-hand side is zero because Eq. 3 is not a functionof any qi. From Eq. 8, the equations of motion for each coordinate are,

d

dt(mx) = λ (sin θ)

2


mx = λ sin θ (9)

d

dt(my) − mg = λ (− cos θ)

my − mg = −λ cos θ (10)

d

dt

(

mX)

= λ (− sin θ)

mX = −λ sin θ (11)

Immediately, if we add Eq. 11 to Eq. 9, we see why Figure 1 has the directionof motion for the wedge incorrect,

mx + MX = 0

x = −M

mX (12)

Also, this tells us that the x position of the system’s center of mass stays sta-tionary because the only external force on the system is gravity, which is in they direction. We’ve assumed that the particle’s initial position is at the origin,but if we also assume that the initial velocity of the center of mass is also zerothen from Eq. 12,

∂

∂t

∂x

∂t= −

M

m

∂

∂t

∂X

∂t∂x

∂t∂t = −

M

m

∂X

∂t∂t

∫ x

0

∂x = −M

m

∫ X

0

∂X

x = −M

mX or X = −

m

Mx (13)

giving us an equation relating X in terms of x. We can use the equation ofconstraint (Eq. 3) to get y in terms of x,

(x − X) sin θ = y cos θ

(x +m

Mx) sin θ = y cos θ

y = x(1 +m

M) tan θ (14)

from which only y and x are a function of time, so

y = x(1 +m

M) tan θ

y = x(1 +m

M) tan θ (15)

3


OK, OK, let’s finally get the equations of motion. Starting with Eq. 9,

mx = λ sin θ(

Eq. 10: my − mg = −λ cos θ ⇒ λ = −m(y − g)

cos θ

)

mx = m(g − y) tan θ (16)(

Eq. 15: y = x(1 +m

M) tan θ

)

mx = mg tan θ − mx(1 +m

M) tan2 θ

mx + mx tan2 θ + mxm

Mtan2 θ = mg tan2 θ

m(1 + tan2 θ +m

Mtan2 θ)x = mg tan θ

(cos2 θ

cos2 θ+

sin2 θ

cos2 θ+

m

M

sin2 θ

cos2 θ)x = g

sin θ

cos θ

(

(:1cos2 θ + sin2 θ) + (m/M) sin2 θ

)

x = g sin θ cos θ

x =g sin θ cos θ

1 + (m/M) sin2 θ

x = gM sin θ cos θ

M + m sin2 θ(17)

Since the acceleration in the x direction is a constant in time, ax Eq. 17 hasthe usual 1-D kinematic solution,

x(t) = x0 + v0,xt +1

2axt2

(

set x0 → 0, and v0,x → 0)

x(t) =1

2g

(

M sin θ cos θ

M + m sin2 θ

)

t2 (18)

For the y equation of motion, we plug Eq. 17 back into Eq. 15,

y = x(1 +m

M) tan θ

= g

(

M sin θcos θ

M + m sin2 θ

)

1

M(M + m)

sin θ

cos θ

y = g(M + m) sin2 θ

M + m sin2 θ(19)

which is also constant, so with the same initial conditions (y0 → 0, y0,y → 0)we get a similar free-fall equation,

y =1

2g

(

(M + m) sin2 θ

M + m sin2 θ

)

t2 (20)

4


Finally we get the solution for X plugging Eq. 17 into Eq. 12,

X = −m

Mx

= −m

Mg

M sin θ cos θ

M + m sin2 θ

X = −gm sin θ cos θ

M + m sin2 θ(

X0 → 0, and v0,X → 0)

⇒ X = −1

2g

(

m sin θ cos θ

M + m sin2 θ

)

t2 (21)

One could also solve for the normal force λ via Eq. 9, but the problem doesnot ask for it, so it is left as an exercise to the reader. (Ew. I feel so dirty sayingthat.)

c) What are the constants of motion?

To find the constants of motion, i.e. conserved quantities, we need to expressthe Lagrangian (Eq. 6), in terms of independent coordinates. We can use theconstraint equation (Eq. 3) to solve for X ,

(x − X) sin θ − y cos θ = 0

x − X −y

tan θ= 0

X = x −y

tan θ

⇒ X = x −y

tan θ

⇒ L =1

2mx2 +

1

2my2 +

1

2MX2 + mgy

L =1

2mx2 +

1

2my2 +

1

2M

(

x −y

tan θ

)2

+ mgy (22)

Because this Lagrangian is explicitly independent of x, then the ∂L∂x

term inthe Euler-Lagrange equation will be zero. Thus, the canonical momentum inthe x direction is a conserved quantity.

px =∂L

∂x

= mx + M

(

x −y

tan θ

)

(1)

px = (m + M)x −My

tan θ(23)

General Rule: If the Lagrangian is independent of any generalized coordi-nate, that coordinate is “cyclic,” and therefore it’s respective canonical momen-tum (pi ≡ ∂L/∂qi) is conserved.

5


Also, the Lagrangian does not explicitly depend on time, so as previouslystated, the total energy is conserved (which is why we could write the Hamilto-nian as the sum of kinetic and potential energies).

d) Describe the motion of the system.

Figure 2: An alternate view of the particle-wedge system at times t = 1, 2, 3and 4.

Since friction is not involved in the system, the x coordinate of the center ofmass between the two objects will remain stationary as has been shown in partb. The y coordinate will accelerate as though in slow free-fall (a la Eq. 20), untilthe particle reaches the surface on which the wedge slides. From then on, it willsimply travel according to Eq. 18. Thus, (as in Figure 2) from a reference pointfollowing the center of mass, the wedge and particle will fly apart from eachother. If M ≫ m, it will not move much (X ≈ 0), and the system approximatesthat of a mass sliding down a fixed incline.

6



Jeff Kissel

October 26, 2006

A uniform rod slides with its ends inside a smooth (frictionless) verti-cal circle of radius a. The rod of uniform density and mass m subtendsan angle of 120 at the center of the circle.

Figure 1: A rod, with length L (which is the length of a chord subtending 120)is restricted to slide inside a frictionless circular pipe.

(a) Compute the center of mass moment of inertia of the rod ICoM interms of m and a (not the length of the rod).

In most general terms, the moment of inertia tensor is calculated as

ICoM =

∫

ρ(r)(δijr2 − xixj) d3r (1)

However, we assume the rod is only has one dimension so we need onlyintegrate along dℓ instead of d3r; there are no off-diagonal terms (i.e. xixj = 0);and the rod has constant density, so ρ(r) = m/L, which means Eq. 1 becomesconsiderable less intimidating,

ICoM =m

L

∫ +L/2

−L/2

ℓ2 dℓ = 2m

L

∫ L/2

0

ℓ2 dℓ

1


= 2m

L

ℓ3

3

∣

∣

∣

L/2

0

= 2m

L

1

3

(

L

2

)3

=2

24m

L3

L

ICoM =1

12mL2 (2)

Finally, because we know angle which the rod subtends is 120, it forms anisosceles triangle whose sides are a : a : L, and interior angles are 120 : 30 : 30,which means L/2 = a cos (30), or L =

√3 a. Thus, in terms of m and a, the

center of mass moment of inertia is

ICoM =1

4ma2 (3)

(b) Obtain the potential energy, the kinetic energy, and the La-grangian L(θ(t), θ(t); m, a, g) for this system, where θ(t) the dynamicalvariable, is the instantaneous angular position relative to its equilib-rium position and m, a, and g are constant parameters of the system.

As the question implies later, once we have the center of mass moment ofinertia, we can treat the rod like a pendulum, as long as we displace ICoM tothe center of the circle, a distance d = a sin (30). So, following the parallelaxis theorem,

Id = ICoM + md2

=1

4ma2 + m(a sin (30))2

=1

4ma2 + m

(

1

2a

)2

=1

4ma2 +

1

4ma2

Id =1

2ma2 (4)

Now the kinetic energy T is only rotational, the potential V is only gravita-tional since there is no friction, and the Lagrangian is L = T − V ,

T =1

2Idθ

2

=1

2

(

1

2ma2

)

θ2

T =1

4ma2θ2 (5)

V = mga cos θ (6)

L =1

4ma2θ2 − mga cos θ (7)

2


(c) Compare the dynamics of this system with that of a simple pendu-lum with mass M and length L (exactly, without any approximationssuch as the small oscillation approximation). Compare the parame-ters L and M to a and m.

With out making any approximations (I promise!) we can find the equa-tion of motion, and compare those to that of a simple pendulum. Let’s getLagrangimiphysical!

d

dt

(

∂L

∂θ

)

−∂L

∂θ= 0

d

dt

(

1

2ma2θ

)

+ mga sin θ = 0

1

2ma2θ = −mga sin θ

θ = 2g

asin θ (8)

which is indeed comparable to a simple pendulum’s equation of motion whichis different only by the factor of 2 if L → a and M → m. For a pendulum oflength L, and mass M ,

T =1

2ML2θ2 V = MgL cos θ

L =1

2ML2θ2 − MgL cosθ

d

dt

(

∂L

∂θ

)

−∂L

∂θ= 0

d

dt

(

ML2θ)

+ MgL sin θ = 0

ML 2θ = −MgL sin θ

θ =g

Lsin θ (9)

(c) Find the frequency of small oscillations of the system.

Oh, NOW it’s OK to make approximations, huh? Fine. Gosh.

In using the small angle approximation, sin θ ≈ θ, which means Eq. 8becomes a second-order differential equation of the form x = −ω2x which hasthe general solution x(t) = x0 cos (ωt), where ω is the frequency (small) ofoscillation(s).

θ ≈ −2g

aθ

⇒ ω =

√

2g

a(10)

3



Jeff Kissel

May 18, 2007



A mass m is attached to a spring of spring constant k that can slide

vertically on a pole without friction, and moves along a frictionless

inclined plane as shown in Fig 1. After an initial displacement along

the plane, the mass is released. Find an expression for the x and y

position of the mass as a function of time. The initial displacement

of the mass is x0. You may assume that the object never slides down

the ramp so far that it strikes the floor.

Figure 1: A box of undisclosed materials oscillating on a ramp.

If one sets up the coordinates as in Fig 1, where the equilibrium is at theorigin, then two important simplifications occur:

1) y = ax + b → y = ax (1)

2) ∆x = x − xe → ∆x = x (2)

With these simplifications, the kinetic energy is

T =1

2m(

x2 + y2)

(3)

1


and the potential energy is the sum of that from the spring and that fromgravity,

V =1

2k∆x2 + mgy

(Because of Eq. 2, ∆x = x)

V =1

2kx2 + mgy (4)

Making the Lagrangian for the system in terms of x and y,

L = T − V

=1

2m(

x2 + y2)

−

(

1

2kx2 + mgy

)

L =1

2mx2 +

1

2my2

−1

2kx2

− mgy (5)

However, there is a fixed relationship between x and y, namely the Eq. 1, so

y = ax → y = ax (6)

which makes Eq. 5 a function only of x and x,

L =1

2mx2 +

1

2m(ax)2 −

1

2kx2

− mg(ax)

=1

2mx2 +

1

2ma2x2

−1

2kx2

− mgax

L =1

2m(1 + a2)x2

−1

2kx2

− mgax (7)

Using the Euler-Lagrange equation, we can then find an expression for x(t),

0 =d

dt

(

∂L

∂x

)

−∂L

∂x

=d

dt

(

m(1 + a2)x)

− (−kx − mga)

0 = m(1 + a2)x + kx + mga

x = −k

m(1 + a2)x −

ga

m(1 + a2)

Let ω2 =k

m(1 + a2), and R = −

ga

m(1 + a2)

x = −ω2 + R

Let x′ = −ω2x + R

x′ = −ω2x

x′ = −ω2x

x′

−ω2= −ω2

(

x′− R

−ω2

)

+ R

2


x′

−ω2= x′

x′ = −ω2x′

⇒ x′(t) = A cos (ωt) (8)

Where the last step we know comes from the typical harmonic oscillator solution,with A as some constant. Plugging back in for x, and using the initial conditionsthat x(0) = x0,

−ω2x(t) + R = A cos (ωt)

x(t) = −A

ω2cos (ωt) −

R

ω2

x0 = x(0) = −A

ω2 (1) − R

ω2

x0 + R

ω2 = −A

ω2

A = −ω2x0 − R

= −(−ω2x0 − R)

ω2cos (ωt) −

R

ω2

=

(

x0 +R

ω2

)

cos (ωt) −R

ω2

(

R

ω2= −

ga

m(1 + a2)

(

m(1 + a2)

k

)

= −ga

k

)

(

ω =

√

k

m(1 + a2)

)

x(t) =(

x0 −ga

k

)

cos

(√

k

m(1 + a2)t

)

+ga

k(9)

and using Eq. 1, we can get y(t),

y(t) = ax(t)

y(t) = a(

x0 −ga

k

)

cos

(√

k

m(1 + a2)t

)

+ga

k(10)

3



Jeff Kissel

May 17, 2007


Spring 2007 #22 –

An astronaut is on the surface of a spherical asteroid of radius r andmean density similar to that of the earth. On the earth this astronautcan jump about a height h of 0.5 m. If he jumps on this asteroid, hecan permanently leave the surface.

(a) Taking the radius of the earth as 6.4×106 m, find the largest radiusthe asteroid can have.

With the initial velocity on earth v⊕, the astronaut jumps a height h. If hecan leave the asteroid with this same velocity, we can treat v⊕ as the escapeveloctiy for the asteroid vesc. We can determine his velocity on Earth from 1-Dkinematics (using the usual trick that his final velocity on the way down is equalto his initial velocity on the the way up),

v2

f −7

0

v2

i = 2a∆x

v2

⊕ = 2g⊕h

v⊕ =√

2g⊕h

The escape velocity on the asteroid occurs when the astronaut’s total kineticenergy equals the potential felt from the asteroid, i.e. his total energy is zero,

E =1

2mav2

−GMAma

rA

0 =1

2mav2

esc −GMAma

rA

1

2 mav2

esc =GMAma

rA

1


v2

esc =2GMA

rA

vesc =

√

2GMA

rA

ρA = MA

VA

⇒ MA = ρAVA

MA = 4π3

r3

AρA

=

√

2G

rA

(

4π

3r3

AρA

)

= rA

√

2G4πρA

3(

ρA = ρ⊕ρA = M⊕

4π

3R3

⊕

)

= rA

√

√

√

√2G

4π

3

(

M⊕

4π3

R3⊕

)

vesc = rA

√

2GM⊕

R3⊕

(

g⊕ ≡GM

R2⊕

, vesc = v⊕

)

v⊕ = rA

√

2g⊕R⊕

√

2g⊕h = rA

√

2g⊕R⊕

rA =√

R⊕h (1)

=√

(6.4 × 106 m)(0.5 m)

rA = 1788.85 m (2)

b) How fast could the asteroid rotate and not have the astronautbe flung away from the surface?

Here, the maximum rotation speed would be when the astronaut’s weighton the asteroid exactly equals the asteroid’s centripetal acceleration,

Fa = maac

magA = ma

v2

t,A

rA

(vt = ωr)

gA = ω2

maxrA

2


ωmax =

√

gA

rA

(3)

Yet, we don’t know what gravity is like on this asteroid. We can find out,because its density is equal to Earth’s,

gA ≡GMA

r2

A

=G

r2

A

(

4π

3r3

AρA

)

= GrA

(

4π

3

(

M⊕

4π3

R3

⊕

))

=rA

R⊕

(

GM⊕

R2⊕

)

gA =rA

R⊕

g⊕ (4)

So the maximum rotation the asteroid can have with out flinging the astronautinto space is

ωmax =

√

1

rA

(

rA

R⊕

g⊕

)

ωmax =

√

g⊕R⊕

(5)

=

√

9.8 m/s

6.4 × 106 m

ωmax = 1.23 × 10−3 rads/s (6)

3



Jeff Kissel

May 18, 2007

Exam Iteration Question Number Notes/ChangesSpring 2006 #X –Fall 2006 #24 Added to question bank.

Spring 2007 #24 –

In 1956, G. Kuzmin introduced the simple axisymmetric potential

ΦK(R, z) = − GM√

R2 + (a + |z|)2(1)

as an approximate description of the potential of a disk-like galaxy.

(a) Show that this potential can be though of as the result of twopoint masses placed at R = 0, z = ±a, generating the potential in thehalf-space z > 0 and z < 0 respectively.

In cylindrical polar coordinates, a point mass placed at the origin generatesthe following potential,

ΦPM (R, z) = − GM√R2 + z2

(2)

where R2 = x2 + y2.If the point mass is displaced from the origin a positive distance z0, without

moving radially, then ΦPM (R, z) becomes

ΦPM (R, z + z0) = − GM√

R2 + (z − z0)2= − GM

√

R2 + (z0 − z)2(3)

where I’m able to flip the quantity in parentheses because it is squared.In the region of z < 0, The Kuzmin potential takes the form

ΦK(R, z) = − GM√

R2 + (a − z)2(4)

1


and hence by direct comparison with Eq. 3, for z < 0, Eq. 4 is the potential ofa point mass centered at z = a, and R = 0.

Similarly, the potential of a point mass shifted from the origin a negative z0

(again keeping R′ = R) is

ΦPM (R, z − z0) = − GM√

R2 + (z − (−z0))2= − GM

√

R2 + (z0 + z)2(5)

In the region of z > 0, the Kuzmin potential is

ΦK(R, z) = − GM√

R2 + (a + z)2(6)

so, again by direct comparison with 5, Eq. 6 is the potential of a point masscentered at z = −a, and R = 0.

(b) Show that all the mass must be located on the z = 0 plane anddistributed according to the surface density

Σk(R) =aM

2π(R2 + a2)3/2(7)

If Φ(R, z) below the z = 0 plane acts as a point mass at z = a, then above

the z = 0 plane, Poisson’s Equation (∇2Φ = 4πGρ) contains no mass densityρ. Poisson’s Equation then becomes Laplace’s Equation: ∇2Φ = 0. Similarly,if Φ(R, z) above the z = 0 plane acts as a point mass at z = −a, then thepotential also satisfies Laplace’s Equation in the region above the z = 0 plane.If ∇2Φ(R, z > 0) = ∇2Φ(R, z < 0) = 0, assuming the mass must be somewhere,then all the mass must lie in an infinitely thin disk at z = 0, of mass density Σ.

Taking a Gaussian pill-box surrounding the sheet of mass Σ, we know thepotential at z = 0 must satisfy Gauss’ Law of flux,

∮

pill-box

~∇ΦK(R, 0) · d~a = 4πGΣKA (8)

where the A is the area of the disk. We can simplify the left-hand side a bit,∮

pill-box

~∇ΦK(R, 0) · d~a =

∫

A

∂

∂z

(

ΦK

)

da =∂

∂z

(

ΦK

)

∫

A

da = A∂

∂zΦK

∣

∣

∣

z=0

(9)

However, the partial derivative is discontinuous across z = 0, so we can takethe difference between the partial derivative coming from the top and from thebottom,

4πGΣKA = A

(

∂

∂zΦK(R, +z)

∣

∣

∣

z=0

− ∂

∂zΦK(R,−z)

∣

∣

∣

z=0

)

=∂

∂z

(

− GM√

R2 + (a + z)2

)

∣

∣

∣

z=0

2


− ∂

∂z

(

− GM√

R2 + (a − z)2

)

∣

∣

∣

z=0

= −GM∂

∂z

(

(R2 + (a + z)2)−1/2

) ∣

∣

∣

z=0

+ GM∂

∂z

(

(R2 + (a − z)2)−1/2

) ∣

∣

∣

z=0

= −GM

[

−1

2(R2 + (a + z)2)−3/2(2(a + z))(1)

]

z=0

+ GM

[

−1

2(R2 + (a − z)2)−3/2(2(a − z))(−1)

]

z=0

= GM

(

+(a + (0))

(R2 + (a + (0))2)3/2+

(a − (0))

(R2 + (a − (0))2)3/2

)

∣

∣

∣

z=0

4πGΣK = GM2a

(R2 + a2)3/2

ΣK =Ma

2π (R2 + a2)3/2

√(10)

(c) What is the total mass generating ΦK?

We can integrate the surface density Σ over the disk to find the total mass:

∫

S

Σ da =1

2π

∫

2π

0

∫

∞

0

Ma

(R2 + a2)3/2

R dR dφ

=1

2π

2π

∫

∞

0

Ma

(R2 + a2)3/2

R dR

let u = R2 + a2 ⇒

du = 2R dR1

2du = R dR

0 → a2 , ∞ → ∞

=Ma

2

∫

∞

a2

u−3/2 du =Ma

2

[

−2u−1/2

]

∞

a2

= Ma[

−

01√∞ +

1√a2

]

∫

S

Σ da = M (11)

Thus, the total mass gererating ΦK is M .

3



Jeff Kissel

May 19, 2007



A particle of mass m moves in a force field given by a potential energyV = Krs where both K and s may be positive or negative.

(a) For what values of K and s do stable, circular orbits exist?

(These results are a product of Goldstein Section 3.6 pp89-90.)Circular orbits occur when the effective potential,

Veff = V +ℓ2

2mr2(1)

has an extremum, i.e. when

∂Veff

∂r

∣

∣

∣

r=R=

∂V

∂r

∣

∣

∣

r=R+

∂

∂r

(

ℓ2

2mr2

)

∣

∣

∣

r=R(

|~F (r)| ≡ − |~∇V (r)| = −∂V

∂r

)

0 = (−F (R)) +

(

−ℓ2

mR3

)

F (R) = −ℓ2

mR3(2)

(

ℓ ≡ mr2ω)

F (R) = −mω2R (3)

where I’ve denoted R as the circular radius, of which R = 0.However, these extrema can be either a maxima or minima. Circular orbits

are only stable when the the extrema of the effective potential is a minima, i.e.

1


when the second derivative of the effective potential is positive. This restrictsboth K and s,

∂2Veff

∂r2

∣

∣

∣

r=R=

∂

∂r

(

∂Veff

∂r

)

0 <∂

∂r(−Feff )

∣

∣

∣

r=R

< −∂

∂r

(

F +ℓ2

mr3

)

∣

∣

∣

r=R

< −

(

∂F

∂r

∣

∣

∣

r=R−

3ℓ2

mR4

)

0 < −∂F

∂r

∣

∣

∣

r=R+ 3

ℓ2

mR4

∂F

∂r

∣

∣

∣

r=R< 3

1

R

ℓ2

mR3

F = −∂V

∂r= − (sKrs−1)

∂F

∂r= −s(s − 1)Krs−2

(

Eq. 2: F (R) = −ℓ2

mR3

)

−s(s − 1)KRs−2 < −3F (R)

R(

F = −∂V

∂r= − (sKrs−1)

)

−s(s − 1)KRs−2 < −3

(

−sKRs−1)

R−s(s − 1)

KRs−2 < +3 s

KRs−2

−(s − 1) < +3

s − 1 > −3

s > −2 (4)

where canceling out K demands that it’s positive so

K < 0 (5)

(b) What is the relation between the period P of the orbit and theradius R of the orbit?

We’ve already shown that

F = −∂V

∂r= − (sKrs−1) (6)

We can set this equal to Eq. 3, using r = R and ω = 2π/P because the orbit iscircular,

−sKRs−1 = −mω2R

2


1

ω2=

−mR

−sKRs−1

P 2

(2π)2=

m

sKRs−2

P 2 =4π2m

sKR2−s

P =4π2m

sKR(2−s)/2 (7)

For a quick sanity check, let’s make sure we get Kepler’s 3rd Law whenK = −GMm and s = −1,

PKepler =4π2m

(−1)(−GMm)R(2−(−1))/2

=4π2

m

GMmR(2+1)/2

PKepler =4π2

GMR3/2 (8)

L’chayim!

3



Jeff Kissel

November 1, 2006

All parts of this question call for quick and brief answers.

(a) A bead of mass m slides along a wire bent into a parabolic shape.The wire is pivoted at the origin and is spinning around the vertical.Identify suitable generalized coordinates and constraints to describethe bead’s motion.

Figure 1: A bead confined to a parabolic wire that is spinning about the z axis.

A suitable set of generalized coordinates would be cylindrical coordinates,r, φ, and z, (as shown in Figure 1) where the equation of constraint would bethat z = r2.

(b) A system with generalized coordinates q1, q2 and q3 is described bythe Lagrangian L(q1, q2, q3, q3); that is, the Lagrangian is independentof q1, q2 and time (explicitly). What are the conserved quantities ofthis motion?

Short answer: The conserved quantities are energy, p1, and p2.

Thorough answer: The canonical momentums are defined as

pi ≡∂L

∂qi

(1)

1

Jeff Kissel November 1, 2006 Classical Mechanics

Which means if the Lagrangian is independent of any generalized coordinateqi, then its Euler-Lagrange equation will yield a constant of motion,

d

dt

(

∂L

∂qi

)

−

7

0∂L

∂qi

= 0

d

dt(pi) = 0

⇒ pi = constant (2)

Thus, since the Lagrangian is independent of q1 and q2, p1 and p2 are con-stants of motion.

Also, if the Lagrangian is independent of time then the energy function h isconstant of motion, i.e. energy is conserved, and the Hamiltonian is H = T +V .The proof is as follows,

dL

dt≡

∑

i

∂L

∂qi

dqi +∑

i

∂L

∂qi

dqi +

7

0∂L

∂t

=∑

i

d

dt

(

∂L

∂qi

)

dqi +∑

i

∂L

∂qi

dqi

(

d

dt

(

qi

∂L

∂qi

)

=d

dt

(

∂L

∂qi

)

dqi +∂L

∂qi

dqi

dt

)

dL

dt=

d

dt

(

qi

∂L

∂qi

)

0 =∑

i

d

dt

(

qi

∂L

∂qi

)

−dL

dt

0 =d

dt

(

∑

i

qi

∂L

∂qi

− L

)

⇒ Constant = h =∑

i

qi

∂L

∂qi

− L (3)

(c) If the kinetic energy is a quadratic function of generalized veloci-ties,

∑

k,ℓ akℓqk qℓ, express∑

k qk(∂T/∂qk) in terms of T .

Euler’s theorem states if a function f is a an homogeneous function of degreen in variables xi, then

∑

i

xi

∂f

∂xi

= nf (4)

Since T is quadratic in terms of q, T has degree 2, so

∑

k

qk

∂T

∂qk

= 2T (5)

2


(d) A hollow and a solid sphere, made of different materials so as tohave the same masses M and radii R, roll down an inclined plane,starting from rest at the top. Which one will reach the bottom first?

Short Answer: The solid sphere will reach the bottom faster because it hasa smaller moment of rotational inertia, i.e. less energy is put into rotating thesolid sphere so more energy can by used in linear velocity of the center of mass.

Thorough answer: The moments of inertia for a solid and hollow sphere(abbreviated as SS and HS, respectively) are

ISS =2

5MR2 IHS =

2

3MR2 (6)

Remember v = ωR, so each sphere’s kinetic energy is

TSS =1

2Mv2

CoM,SS +1

2ISSω2

=1

2Mv2

CoM,SS +1

2

(

2

5MR2

)

(vCoM,SS

R

)2

=1

2Mv2

CoM,SS +1

5Mv2

CoM,SS

TSS =7

10Mv2

CoM,SS (7)

THS =1

2Mv2

CoM,HS +1

2IHSω2

=1

2Mv2

CoM,HS +1

2

(

2

3MR2

)

(vCoM,HS

R

)2

=1

2Mv2

CoM,HS +1

3Mv2

CoM,HS

THS =5

6Mv2

CoM,HS (8)

Energy is conserved in this system, so the gravitational potential at the topof the ramp will equal the kinetic energy at the bottom, so

Vi = Tf

Mgh =7

10Mv2

CoM,SS Mgh =5

6Mv2

CoM,HS

v2

CoM,SS =10

7gh v2

CoM,HS =6

5gh (9)

which means the solid sphere will arrive first, since it is traveling at a fasterlinear velocity (10/7 > 6/5) down the ramp.

3


(e) Write down the Hamiltonian for a free particle in three-dimensionalspherical-polar coordinates.

Short Answer: In (physics) spherical-polar coordinates (where θ is the anglefrom the z axis, and φ is the angle from the x axis in the xy plane),

H =p2

r

2m+

p2

θ

2mr2+

p2

φ

2mr2 sin2 θ(10)

Thorough Answer: The Lagrangian in free space (in spherical coordinates)is

L =1

2m(r2 + r2θ2 + r2 sin2 θ φ2) (11)

and Hamiltonian can be found from (and is in fact defined as) a Legendretransformation between qi and pi,

H ≡

∑

i

qipi − L where pi ≡∂L

∂qi

(12)

So, turning the crank on Eq. 11, and plugging back into Eq. 12,

pr ≡∂L

∂r= mr ⇒ r =

pr

m

pθ ≡∂L

∂θ= mr2θ ⇒ θ =

pθ

mr2

pφ ≡∂L

∂φ= mr2 sin2 θφ ⇒ φ =

pφ

mr2 sin2 θ

the Lagrangian as a function of qi and pi is then

L =1

2m

(

(pr

m

)2

+ r2

( pθ

mr2

)2

+ r2 sin2 θ

(

pφ

mr2 sin2 θ

)2)

L =p2

r

2m+

p2

θ

2mr2+

p2

φ

2mr2 sin2 θ(13)

and the Hamiltonian is

H =(pr

m

)

pr +( pθ

mr2

)

pθ +

(

pφ

mr2 sin2 θ

)

pφ

−p2

r

2m−

p2

θ

2mr2−

p2

φ

2mr2 sin2 θ

=p2

r

m+

p2

θ

mr2+

p2

φ

mr2 sin2 θ−

p2

r

2m−

p2

θ

2mr2−

p2

φ

2mr2 sin2 θ

H =p2

r

2m+

p2

φ

2mr2+

p2

φ

2mr2 sin2 θ(14)

4



Jeff Kissel

November 1, 2006


(a) In a bound Keplerian orbit, what is the ratio of the average kineticenergy to the average potential energy?

Short Answer: The only force experienced by a particle in a bound Keplerianorbit is the central, gravitational inverse square force (a power law of ordern = −2). This gravitational force is derivable from a scalar is derivable froma scalar potential V . Any system satisfying these conditions follows the VirialTheorem: the average kinetic energy is twice that of the average potential, or

T = −1

2V (1)

Thorough Answer: An useful scalar quantity known as the “Virial,” or the“Virial of Clausius,” is defined as

G(t) =∑

i

~pi · ~ri (2)

where the summation is over all particles in a given system. Newton’s equationsfor each particle in the system are

Fi =d~pi

dt(3)

which gives us the inspiration to take the full time derivative of the Virial,

dG(t)

dt=

∑

i

d~pi

dt· ~ri +

∑

i

~pi ·d~ri

dt

=∑

i

~Fi · ~ri +∑

i

mi~ri · ~ri

=∑

i

~Fi · ~ri +∑

i

mr2i

dG(t)

dt=

∑

i

~Fi · ~ri + 2T (4)

1


Since we’re interested in the (time) average(d) kinetic energy, T , we’ll inte-grate Eq. 4 over time, from 0 to some time τ and normalize by the same timeτ ,

1

τ

∫ τ

0

dG(t)

dtdt =

∑

i

~Fi · ~ri + 2T

1

τ(G(τ) − G(0)) =

∑

i

~Fi · ~ri + 2T (5)

If we chose τ to be the period and motion is cyclic (as in a Keplerian orbit),the left-hand side vanishes (G(τ) − G(0) = 0) leaving

2T = −∑

i

~Fi · ~ri

T = −1

2

∑

i

~Fi · ~ri (6)

which is the general form of what is known as the Virial Theorem.If the forces are derivable from a potential, (as in a Keplerian orbit) i.e.

~F = −~∇V then the theorem becomes

T = −1

2

∑

i

~∇V · ~ri (7)

and for a single particle moving under a central force (as in a particle in aKeplerian orbit), this reduces to

T = −1

2

∂V

∂rr (8)

If V is a power law (as in a Keplerian orbit), where the exponent is chosenso that the force law goes as rn, then

V = arn+1

∂V

∂rr = (n + 1)V

⇒ T =1

2(n + 1)V (9)

which means for a Keplerian orbit (whose force follows and inverse square law,n = −2), the average kinetic energy and average potential have the followingconstant of proportionality

T = −1

2V (10)

2


(b) Write down the Lagrangian and the Hamiltonian of a Free Particleof mass m in three dimensional polar coordinates.

Short Answer: The Lagrangian and Hamiltonian in polar spherical coordi-nates are

L =1

2m(r2 + r2θ2 + r2 sin2 θ φ2) (11)

H =p2

r

2m+

p2θ

2mr2+

p2φ

2mr2 sin2 θ(12)

Thorough Answer: A free particle is that which is traveling at some velocity~v = ~r, without the the influence of any potentials. Thus, the Lagrangian (L =T − V ) will only contain the kinetic term T .

The Cartesian components of the particle’s position in spherical-polar coor-dinates (where we use the physics definitions of θ and φ: θ is the angle from thez axis, and φ is the angle from the x axis in the xy plane) are,

x = r sin θ cosφ

y = r sin θ sin φ

z = r cos θ

so the components of velocities are,

x = r sin θ cosφ + r cos θ cosφθ − r sin θ sin φφ

y = r sin θ sin φ + r cos θ sinφθ + r sin θ cosφφ

z = r cos θ − r sin θθ

The kinetic energy T of any given particle can always be written as

T =1

2m (x2 + y2 + z2) (13)

So, we can plug in the spherical-polar versions of x, y, and z from above,

2T

m=

[( ˜r2 sin2 θ cos2 φ +−−−−−−−−−−−−→r2 cos2 θ cos2 φθ2 +

r2 sin2 θ sin2 φφ2

+2rr sin θ cos θ cos2 φθ −︷︸︸︷2rr sin2 θ sin φ cosφφ

−((((((((((((2r2 sin θ cos θ sin φ cosφθφ

)

+( ˜r2 sin2 θ sin2 φ +

−−−−−−−−−−−→r2 cos2 θ sin2 φθ2 + r2 sin2 θ cos2 φφ2

+2rr sin θ cos θ sin2 φθ +︷︸︸︷2rr sin2 θ sinφ cos φφ

+((((((((((((2r2 sin θ cos θ sin φ cosφθφ

)

+(r2 cos2 θ + r2 sin2 θθ2 − 2rr sin θ cos θθ

)]

3


= ˜r2 sin2 θ(sin2 φ + cos2 φ) +−−−−−−−−−−−−−−−−−−−→r2 cos2 θ(sin2 φ + cos2 φ)θ2

+

r2 sin2 θ(sin2 φ + cos2 φ)φ2

2rr sin θ cos θ(sin2 φ + cos2 φ)θ + r2 cos2 θ

+ r2 sin2 θθ2 − 2rr sin θ cos θθ(1 = sin2 α + cos2 α

)

= r2 sin2 θ + r2 cos2 θθ2

︸︷︷︸+r2 sin2 θφ2

+ (((((((2rr sin θ cos θθ + r2 cos2 θ + r2 sin2 θθ2

︸︷︷︸−(((((((2rr sin θ cos θθ

= r2(sin2 θ + cos2 θ) + r2(sin2 θ + cos2 θ)θ2

︸︷︷︸+r2 sin2 θφ2

2T

m= r2 + r2θ2 + r2 sin2 θφ2

T =1

2m(r2 + r2θ2 + r2 sin2 θφ2

)(14)

Now as previously mentioned, the potential V for a “free” particle is zero,so the Lagrangian (L = T − V ) as a function of qi and qi is

L =1

2m(r2 + r2θ2 + r2 sin2 θφ2

)(15)

The Hamiltonian can be found from (and is in fact defined as) a Legendretransformation between qi and pi,

H ≡∑

i

qipi − L where pi ≡ ∂L

∂qi

(16)


pr ≡ ∂L

∂r= mr ⇒ r =

pr

m

pθ ≡ ∂L

∂θ= mr2θ ⇒ θ =

pθ

mr2

pφ ≡ ∂L


pφ

mr2 sin2 θ


L =1

2m

((pr

m

)2

+ r2

( pθ

mr2

)2

+ r2 sin2 θ

(pφ

mr2 sin2 θ

)2)

L =p2

r

2m+

p2θ

2mr2+

p2φ

2mr2 sin2 θ(17)

4



H =(pr

m

)pr +

( pθ

mr2

)pθ +

(pφ

mr2 sin2 θ

)pφ

− p2r

2m− p2

θ

2mr2−

p2φ

2mr2 sin2 θ

=p2

r

m+

p2θ

mr2+

p2φ

mr2 sin2 θ− p2

r

2m− p2

θ

2mr2−

p2φ

2mr2 sin2 θ

H =p2

r

2m+

p2φ

2mr2+

p2φ

2mr2 sin2 θ(18)

(c) A system with generalized coordinates q1, q2 and q3 is described bythe Lagrangian L(q1, q2, q3, q3); that is, the Lagrangian is independentof q1, q2 and time (explicitly). What are the conserved quantities ofthis motion?

Short answer: The conserved quantities are energy, p1, and p2.

Thorough answer: The canonical momentums are defined as

pi ≡∂L

∂qi

(19)

Which means if the Lagrangian is independent of any generalized coordinateqi, then its Euler-Lagrange equation will yield a constant of motion,

d

dt

(∂L

∂qi

)−

7

0∂L

∂qi

= 0

d

dt(pi) = 0


Thus, since the Lagrangian is independent of q1 and q2, p1 and p2 are con-stants of motion.

Also, if the Lagrangian is independent of time then the energy function h isconstant of motion, i.e. energy is conserved, and the Hamiltonian is H = T +V .The proof is as follows,

dL

dt≡

∑

i

∂L

∂qi

dqi +∑

i

∂L

∂qi

dqi +

7

0∂L

∂t

=∑

i

d

dt

(∂L

∂qi

)dqi +

∑

i

∂L

∂qi

dqi

(d

dt

(qi

∂L

∂qi

)=

d

dt

(∂L

∂qi

)dqi +

∂L

∂qi

dqi

dt

)

5


dL

dt=

d

dt

(qi

∂L

∂qi

)

0 =∑

i

d

dt

(qi

∂L

∂qi

)− dL

dt

0 =d

dt

(∑

i

qi

∂L

∂qi

− L

)

⇒ Constant = h =∑

i

qi

∂L

∂qi

− L (21)

To recap, we’ve decided that h (energy), p1, and p2 are the conserved quan-tities of this system.

(d) A hollow and a solid sphere, made of different materials so as tohave the same masses M and radii R, roll down an inclined plane,starting from rest at the top. Which one will reach the bottom first?

Short Answer: The solid sphere will reach the bottom faster because it hasa smaller moment of rotational inertia, i.e. less energy is put into rotating thesolid sphere so more energy can by used in linear velocity of the center of mass.


ISS =2

5MR2 IHS =

2

3MR2 (22)


TSS =1

2Mv2

CoM,SS +1

2ISSω2

=1

2Mv2

CoM,SS +1

2

(2

5MR2

)(vCoM,SS

R

)2

=1

2Mv2

CoM,SS +1

5Mv2

CoM,SS

TSS =7

10Mv2

CoM,SS (23)

THS =1

2Mv2

CoM,HS +1

2IHSω2

=1

2Mv2

CoM,HS +1

2

(2

3MR2

)(vCoM,HS

R

)2

=1

2Mv2

CoM,HS +1

3Mv2

CoM,HS

THS =5

6Mv2

CoM,HS (24)

6



Vi = Tf

Mgh =7

10Mv2

CoM,SS Mgh =5

6Mv2

CoM,HS

v2CoM,SS =

10

7gh v2

CoM,HS =6

5gh (25)


(e) A satellite initially at a distance RE (= radius of the Earth) abovethe Earth’s surface is moved out to a distance 2RE. How does itsorbital time period change?

A satellite orbiting around the Earth can be approximated quite accuratelyby a Keplerian orbit. Kepler’s 3rd law says that such an orbit’s period P isproportional to the radius of orbit a taken to the 3/2’s power, so

P 2 ∝ a3

P ′ 2 ∝ (2a)3

P ′ 2 ∝ 8a3

⇒ P ′ 2 = 8P 2 (26)

The satellite’s period increases by a factor of√

8, if the radius of orbit isdoubled.

7



Jeff Kissel

May 18, 2007

A particle of mass m moves under the influence of an attractiveforce F = −k r, where r is the distance from the force center.

What are the conserved quantities of this three dimensional motion?If the Lagrangian of the system is independent of any generalized coordi-

nate qi (i.e. “cyclic” in qi, the its Euler-Lagrange equation demands that therespective canonical momentum pi is a constant of motion,

0 =d

dt

(

∂L

∂qi

)

−

7

0∂L

∂qi(

pi ≡∂L

∂qi

)

0 =d

dt(pi)


In order to determine if this system has any of its canonical momentumsconserved, we must find the Lagrangian, L = T − V . The force of this system(F = −k r) is derivable from a potential V in which we need,

F = −~∇ V

⇒ V = −

∫ r

0

(−k r′) dr′

V =1

2kr2 (2)

The kinetic energy T is that of any moving particle, which spherical polar co-ordinates (where θ is the angle from the z axis and φ is the angle from the x tothe project of the vector onto the xy plane) is

T =1

2m(

r2 + r2θ2 + r2 sin2 θφ2

)

(3)

Combining Eqs. 2 and 3 yields the Lagrangian,

L =1

2m(

r2 + r2θ2 + r2 sin2 θφ2

)

−1

2kr2 (4)

1


Since Eq. 4 is independent of φ, pφ is a constant of motion.Also, if the Lagrangian is independent of time then the energy function h is

a constant of motion. Because of this energy is conserved, and the total energyE of the system is the Hamiltonian E = H = T + V . The proof is as follows,

dL

dt≡

∑

i

∂L

∂qi

dqi

dt+∑

i

∂L

∂qi

dqi

dt+

7

0∂L

∂t

=∑

i

d

dt

(

∂L

∂qi

)

dqi

dt+∑

i

∂L

∂qi

dqi

dt

=∑

i

(

d

dt

(

∂L

∂qi

)

dqi

dt+

∂L

∂qi

dqi

dt

)

(

d

dt

(

qi∂L

∂qi

)

=d

dt

(

∂L

∂qi

)

dqi

dt+

∂L

∂qi

dqi

dt

)

dL

dt=

∑

i

d

dt

(

qi∂L

∂qi

)

0 =∑

i

d

dt

(

qi∂L

∂qi

)

−dL

dt

=d

dt

(

∑

i

qi∂L

∂qi− L

)

(

pi ≡∂L

∂qi

)

0 =d

dt

(

∑

i

qi pi − L

)

⇒ constant =∑

i

qi pi − L ≡ h (5)

Write down an expression for the total energy.

As shown in part (a), the total energy E is conserved, thus it is the Hamil-tonian E = H = T + V . From Eqs. 2 and 3, the total energy as a function ofqi and qi is then

E = H =1

2m(

r2 + r2θ2 + r2 sin2 θφ2

)

+1

2kr2 (6)

Either from (b) or otherwise, derive an orbit equation relating r to θ.

As stated previously, the Lagrangian for the system cyclic of φ, so the systemis spherically symmetric. The total angular momentum vector

~j = ~r × ~p (7)

2


is conserved. The definition of the cross product requires then that ~j is perpen-dicular to ~r (and ~p). This can only be true if ~r lies in a plane whose normalis parallel to ~j. Thus, motion is restricted to a plane, and φ must not only beconstant but zero. If φ = 0, then we can reduce the Lagrangian (Eq. 4) to justa function of r, r, and θ,

E =1

2m(

r2 + r2θ2

)

+1

2kr2 (8)

and pθ is now a constant of motion because Eq. 8 is cyclic in θ. We can findthis constant (denoted as pθ ≡ ℓ), and therefore θ,

pθ ≡∂L

∂θ= mr2θ ≡ ℓ ⇒ θ =

ℓ

mr2(9)

This means the total energy (after killing the φ term) becomes

E =1

2m(

r2 + r2θ2

)

+ V

=1

2m

(

r2 + r2

(

ℓ

mr2

)2)

E =1

2mr2 +

ℓ2

2mr2+ V (10)

From here, we can solve Eq. 10 for r, and then r(θ),

1

2mr2 = E − V −

ℓ2

2mr2

r2 =2E

m−

2V

m−

ℓ2

m2r2

r =

(

2E

m−

2V

m−

ℓ2

m2r2

)1/2

(

r =dr

dθ

dθ

dt= θ

dr

dθ=

ℓ

mr2

dr

dθ

)

mr2

ℓ

dθ

dr=

(

2E

m−

2V

m−

ℓ2

m2r2

)−1/2

dθ =ℓ

mr2

(

2E

m−

2V

m−

ℓ2

m2r2

)−1/2

dr

=1

r2

(

m2

ℓ2

(

2E

m−

kr2

m−

ℓ2

m2r2

))−1/2

dr

=

(

2mE

ℓ2−

mkr2

ℓ2−

1

r2

)−1/21

r2dr

let u = r−1⇒

du = −r−2 dr

r2 = u−2

3


= −

(

2mE

ℓ2−

mk

u2ℓ2− u2

)

−1/2

du

let v = u2⇒

dv = 2u dv

u = v1/2

du = 1

2v−1/2 dv

= −

(

2mE

ℓ2−

mk

vℓ2− v

)

−1/21

2u−1/2 dv

dθ = −1

2

(

2mvE

ℓ2−

mk

ℓ2− v2

)

−1/2

dv

let η =2mE

ℓ2and ǫ = mk

2 dθ = −

(

ηv −ǫ

ℓ2− v2

)

−1/2

dv

ηv −ǫℓ2 − v2 = ηv −

ǫℓ2 − v2 + η2

4−

η2

4

= η2

4−

ǫℓ2 −

(

v2 − ηv + η2

4

)

=(

η2

4−

ǫℓ2

)

−(

v −η2

)2

= −

((

η2

4−

ǫ

ℓ2

)

−

(

v −η

2

)2)−1/2

dv

let y = v −η2

⇒ dy = dv

let a2 = η2

4−

ǫℓ2

2 dθ = −(

a2− y2

)−1/2

dy

2

∫

dθ = −

∫

1√

a2 − y2dy

2 θ = − sin−1

(y

a

)

+ 2 γ

2 θ − 2 γ = − sin−1

v −η2

√

η2

4−

ǫℓ2

v −η2

√

η2

4−

ǫℓ2

= sin (2γ − 2θ)

u2−

mE

ℓ2=

(

m2E2

ℓ4−

mk

ℓ2

)1/2

sin (2(γ − θ))

(sin (A − B) = 2 sinA cosB)

1

r2=

mE

ℓ2+

2

ℓ2

√

m2E2 − mkℓ2 sin (γ − θ) cos (γ − θ) (11)

where γ is an arbitrary integration/phase constant. *Phew*. That’s my finalanswer. The orbit equation for a simple harmonic oscillator potential.

4



Jeff Kissel

November 5, 2006


(a) Positronium is a bound state of an electron and a positron. Whatis the effective mass of positronium?

Short Answer: For classical mechanics purposes, the “effective mass” of asystem of particles is its reduced mass. An electron and a positron differ incharge but have the same mass, so positronium’s effective mass is

m∗

Ps = µPs

=me−me+

me− + me+

=m2

e−

2me−

m∗

Ps =1

2me− (1)

Thorough Answer:

“Effective mass is defined by analogy with Newton’s second law ~F = m~a.Using quantum mechanics it can be shown that for an electron in an externalelectric field E,

a =1

h2·d2ǫ(k)

dk2q E (2)

where a is acceleration, h is reduced Planck’s constant, h = h/2π, k is the wavenumber (often loosely called momentum since k = p/h), ǫ(k) is the energy as afunction of k, or the “dispersion relation” as it is often called. From the externalelectric field alone, the electron would experience a force of ~F = q ~E, where qis the charge. Hence under the model that only the external electric field acts,effective mass m∗ becomes:

m∗ = h2·

(

d2ǫ(k)

dk2

)−1

(3)

For a free particle, the dispersion relation is a quadratic, and so the effectivemass would be constant (and equal to the real mass). In a crystal, the situation is

1


far more complex. The dispersion relation is not even approximately quadratic,in the large scale. However, wherever a minimum occurs in the dispersionrelation, the minimum can be approximated by a quadratic curve in the smallregion around that minimum. Hence, for electrons which have energy close toa minimum, effective mass is a useful concept.

In energy regions far away from a minimum, effective mass can be negative oreven approach infinity. Effective mass, being generally dependent on direction(with respect to the crystal axes), is a tensor. However, for most calculationsthe various directions can be averaged out.

Effective mass should not be confused with reduced mass, which is a con-cept from Newtonian mechanics. Effective mass can only be understood withquantum mechanics.”

http://en.wikipedia.org/wiki/Effective_mass, 2006

Regardless, since this question is asked in the classical mechanics portionof the exam, (and every other classical mechanics reference I looked at referspositronium’s reduced mass) I assume “effective mass” is equivalent to the re-duced mass.

m∗

Ps = µPs (For classical treatment)

=me−me+

me− + me+

=m2

e−

2me−

m∗

Ps =1

2me− (4)

(b) At what point in a bound Kepler orbit is the speed of the satelliteat its minimum?

Figure 1: A particle in bound Kepler Orbit about a more massive object. PointP is the periapsis and point A is the apsis.

Short Answer: Kepler’s Second Law (of Planetary Motion) states that theradius vector of an object caught in an inverse square potential sweeps outequal areas in equal times (as in Figure 1). In order to adhere to this law, theobject must travel fastest at the closest point of orbit, known as periapsis (think

2


perigee, perihelion, or periastron), and slowest at the farthest point, known asapsis (apogee, aphelion, apasteron, etc.). In other words, point A in Figure 1 isthe point at which the satellite’s speed in minimum.

Thorough Answer: The Lagrangian for a particle in a Keplerian orbit (inphysics spherical-polar coordinates; θ is the angle from the z axis, φ is that fromthe +x axis) is

L =1

2m(r2 + r2θ2 + r2 sin θφ2) +

k

r(5)

Because Eq. 5 is explicitly independent of θ and φ we know pθ and pφ areconserved quantities. The later fact restricts motion to a plane, i.e. we canignore the third term in the Lagrangian. The former yields Kepler’s SecondLaw:

L =1

2m(r2 + r2θ2) +

k

r

d

dt

(

∂L

∂θ

)

−

7

0∂L

∂θ= 0

d

dt

(

1

2mr2θ

)

= 0

⇒1

2r2θ = constant (6)

The factor 1

2is inserted because 1

2r2θ is the areal velocity – the triangular area

swept out by the radius vector per unit time.

Figure 2: The area swept out but the radius vector in a time dt.

The interpretation follows from Figure 2 above, the differential area sweptout in a time dt being

dA =1

2r(

r dθ)

(7)

3


⇒dA

dt=

1

2r2 dθ

dt(8)

which from Eq. 6 we know is constant.So, for a particle in Keplerian orbit to sweep out the same area per unit time,

the object must travel fastest at the closest point of orbit, known as periapsis

(think perigee, perihelion, periastron, etc.), and slowest at the farthest point,known as apsis (apogee, aphelion, apasteron, etc.). In other words, point A inFigure 1 is the point at which the satellite’s speed in minimum.

(c) If you have two spheres of the same mass and radius, describe anon-destructive test by which you could distinguish between the onethat is solid and the one that is hollow.

Short Answer: If one rolls both spheres down a ramp, the solid sphere willreach the bottom faster. The solid sphere has a smaller moment of rotationalinertia, i.e. less energy is put into rotating the solid sphere so more energy canby used in linear velocity of the center of mass. Thus, the solid sphere will travelfaster down the ramp.


ISS =2

5MR2 IHS =

2

3MR2 (9)


TSS =1

2Mv2

CoM,SS +1

2ISSω2

=1

2Mv2

CoM,SS +1

2

(

2

5MR2

)

(vCoM,SS

R

)2

=1

2Mv2

CoM,SS +1

5Mv2

CoM,SS

TSS =7

10Mv2

CoM,SS (10)

THS =1

2Mv2

CoM,HS +1

2IHSω2

=1

2Mv2

CoM,HS +1

2

(

2

3MR2

)

(vCoM,HS

R

)2

=1

2Mv2

CoM,HS +1

3Mv2

CoM,HS

THS =5

6Mv2

CoM,HS (11)


Vi = Tf

4


Mgh =7

10Mv2

CoM,SS Mgh =5

6Mv2

CoM,HS

v2CoM,SS =

10

7gh v2

CoM,HS =6

5gh (12)


(d) Which way is a particle moving from east to west in the southernhemisphere deflected by the Coriolis force arising from the Earth’srotation?

Figure 3: A particle moves from East to West in the Southern hemisphere, alongthe vector ~r.

The Coriolis Force is defined as

FCor ≡ −2m(~ω × ~vr) (13)

Thus, as in Figure 3, if ~vr points west, ω is perpendicular to Earth, then−2m(~ω × ~vr) points South.

(There’s not much to this problem, so I won’t bother with a “thorough”answer.If you’re wondering where the definition of the Coriolis force comes from,check out Goldstein pp 174 - 180.)

(e) In a bound Keplerian orbit, what is the ratio of the average kineticenergy to the average potential energy?

Short Answer: The only force experienced by a particle in a bound Keplerianorbit is the central, gravitational inverse square force (a power law of ordern = −2). This gravitational force is derivable from a scalar is derivable froma scalar potential V . Any system satisfying these conditions follows the VirialTheorem: the average kinetic energy is twice that of the average potential, or

T = −1

2V (14)

5


Thorough Answer: An useful scalar quantity known as the “Virial,” or the“Virial of Clausius,” is defined as

G(t) =∑

i

~pi · ~ri (15)

where the summation is over all particles in a given system. Newton’s equationsfor each particle in the system are

Fi =d~pi

dt(16)

which gives us the inspiration to take the full time derivative of the Virial,

dG(t)

dt=

∑

i

d~pi

dt· ~ri +

∑

i

~pi ·d~ri

dt

=∑

i

~Fi · ~ri +∑

i

mi~ri · ~ri

=∑

i

~Fi · ~ri +∑

i

mr2i

dG(t)

dt=

∑

i

~Fi · ~ri + 2T (17)

Since we’re interested in the (time) average(d) kinetic energy, T , we’ll inte-grate Eq. 17 over time, from 0 to some time τ and normalize by the same timeτ ,

1

τ

∫ τ

0

dG(t)

dtdt =

∑

i

~Fi · ~ri + 2T

1

τ(G(τ) − G(0)) =

∑

i

~Fi · ~ri + 2T (18)

If we chose τ to be the period and motion is cyclic (as in a Keplerian orbit),the left-hand side vanishes (G(τ) − G(0) = 0) leaving

2T = −

∑

i

~Fi · ~ri

T = −1

2

∑

i

~Fi · ~ri (19)

which is the general form of what is known as the Virial Theorem.If the forces are derivable from a potential, (as in a Keplerian orbit) i.e.

~F = −~∇V then the theorem becomes

T = −1

2

∑

i

~∇V · ~ri (20)

6


and for a single particle moving under a central force (as in a particle in aKeplerian orbit), this reduces to

T = −1

2

∂V

∂rr (21)

If V is a power law (as in a Keplerian orbit), where the exponent is chosenso that the force law goes as rn, then

V = arn+1

∂V

∂rr = (n + 1)V

⇒ T =1

2(n + 1)V (22)

which means for a Keplerian orbit (whose force follows and inverse square law,n = −2), the average kinetic energy and average potential have the followingconstant of proportionality

T = −1

2V (23)

(f) Write down the Hamiltonian for a free particle in three-dimensionalspherical-polar coordinates.

Short Answer: In (physics) spherical-polar coordinates (where θ is the anglefrom the z axis, and φ is the angle from the x axis in the xy plane),

H =p2

r

2m+

p2θ

2mr2+

p2φ

2mr2 sin2 θ(24)

Thorough Answer: The Lagrangian in free space (in spherical coordinates)is

L =1

2m(r2 + r2θ2 + r2 sin2 θ φ2) (25)

and Hamiltonian can be found from (and is in fact defined as) a Legendretransformation between qi and pi,

H ≡

∑

i

qipi − L where pi ≡∂L

∂qi

(26)


pr ≡∂L

∂r= mr ⇒ r =

pr

m

pθ ≡∂L

∂θ= mr2θ ⇒ θ =

pθ

mr2

pφ ≡∂L


pφ

mr2 sin2 θ

7



L =1

2m

(

(pr

m

)2

+ r2

( pθ

mr2

)2

+ r2 sin2 θ

(

pφ

mr2 sin2 θ

)2)

L =p2

r

2m+

p2θ

2mr2+

p2φ

2mr2 sin2 θ(27)


H =(pr

m

)

pr +( pθ

mr2

)

pθ +

(

pφ

mr2 sin2 θ

)

pφ

−p2

r

2m−

p2θ

2mr2−

p2φ

2mr2 sin2 θ

=p2

r

m+

p2θ

mr2+

p2φ

mr2 sin2 θ−

p2r

2m−

p2θ

2mr2−

p2φ

2mr2 sin2 θ

H =p2

r

2m+

p2φ

2mr2+

p2φ

2mr2 sin2 θ(28)

8



Jeff Kissel

November 15, 2006

A pendulum, free to swing in a vertical plane, has its point of sus-

pension on the rim of a hoop rotating with constant angular speed ω.

Write the Lagrangian and the resulting equations of motion in some

suitable coordinates for this system.

Figure 1: A pendulum suspended from a hoop rotating with frequency ω.

Assuming the hoop to be massless, one can solve this problem in a similarmanner to that of a double pendulum whose first bob has zero mass. If the hoopdoes has some mass (i.e. it is the canonical “thin hoop”), there will simply bea constant kinetic energy term, 1

2Iω2, that is coordinate independent and thus

will not effect the equations of motion. In other words, the center of mass ofthe hoop is motionless, and its mass is balanced out at the (assumed perfect)circular edges, such that it rotation is only apparent because of the attachedpendulum. So we proceed assuming the double pendulum scenario.

The components of position and velocity vectors for each bob are

x1 = a sin θ1 ⇒ x1 = aω cos θ1 (1)

y1 = a cos θ1 ⇒ y1 = − aω sin θ1 (2)

1


x2 = a sin θ1 + ℓ sin θ1 ⇒ x2 = aω cos θ1 + ℓθ2 cos θ2 (3)

y2 = a cos θ1 + ℓ cos θ2 ⇒ y2 = − aω sin θ1 − ℓθ2 sin θ2 (4)

where I’ve used θ1 = ω, which is constant. Using these, we can arrive at thepotential and kinetic energies,

V = m1~g · ~r1︸︷︷︸

ignored for massless hoop

+ m2~g · ~r2

V = −m2g(a cos θ1 + ℓ cos θ2) (5)

T =(((((((((((1

2m1(x

2

1+ y2

1) +

1

2Ihω2

︸︷︷︸

ignored for massless hoop

+1

2m2(x

2

2+ y2

2)

=1

2m2

(

(aω cos θ1 + ℓθ2 cos θ2)2 + (−aω cos θ1 − ℓθ2 cos θ2)

2

)

2T/m2 = (a2ω2 cos2 θ1 + ℓ2θ2

2cos2 θ2 + 2aωℓθ2 cos θ1 cos θ2)

+ (a2ω2 sin2 θ1 + ℓ2θ2

2

sin2 θ2 + 2aωℓθ2 sin θ1 sin θ2)

= a2ω2(sin2 θ1 + cos2 θ1) + ℓ2θ2

2(sin2 θ2 + cos2 θ2)

+ 2aωℓθ2(cos θ1 cos θ2 + sin θ1 sin θ2)(

cosα cosβ + sin α sin β = cos(α − β)sin2α + cos2 β = 1

)

2T/m2 = a2ω2 + ℓ2θ2

2+ 2aωℓθ2 cos(θ1 − θ2)

T =1

2m2

(

a2ω2 + ℓ2θ2

2+ 2aωℓθ2 cos(θ1 − θ2)

)

(6)

Finally, from Eqs. 5 and 6, the Lagrangian for this hoop-pendulum system is

L = T − V

L =1

2m2a

2ω2 +1

2m2ℓ

2θ2

2+ m2aωℓθ2 cos(θ1 − θ2)

+ m2ga cos θ1 + m2gℓ cos θ2 (7)

from which we can arrive at the equations of motion. For θ1, because θ1 = ωand ω = 0,

0 =d

dt

(∂L

∂ω

)

−∂L

∂θ1

=d

dt

(

m2a2ω + m2aℓθ2 cos (θ1 − θ2)

)

−

(

−m2aωℓθ2 sin (θ1 − θ2)(1) − m2ga sin θ1

)

=(

0 +[

m2aℓθ2 cos (θ1 − θ2) − m2aℓθ2 sin (θ1 − θ2)(ω − θ2)])

+ m2aωℓθ2 sin (θ1 − θ2) + m2ga sin θ1

2


= m2aℓθ2 cos (θ1 − θ2) −((((((((((

m2aωℓθ2 sin (θ1 − θ2) − m2aℓθ2

2sin (θ1 − θ2)

+((((((((((

m2aωℓθ2 sin (θ1 − θ2) + m2ga sin θ1

m2a ℓθ2 cos (θ1 − θ2) = m2a ℓθ2

2sin (θ1 − θ2) −m2a g sin θ1

ℓθ2 =ℓθ2

2sin (θ1 − θ2) − g sin θ1

cos (θ1 − θ2)(8)

Equivalently, for θ2,

0 =d

dt

(∂L

∂θ2

)

−∂L

∂θ2

=d

dt

(

m2ℓ2θ2 + m2aωℓ cos (θ1 − θ2)

)

−

(

−m2aωℓθ2 sin (θ1 − θ2)(−1) − m2gℓ sin θ2

)

= m2ℓ2θ2 − m2aωℓ sin (θ1 − θ2)(ω − θ2)

− m2aωℓθ2 sin (θ1 − θ2) + m2gℓ sin θ2

0 = m2ℓ2θ2 − m2aω2ℓ sin (θ1 − θ2) +

((((((((((

m2aωℓθ2 sin (θ1 − θ2)

−((((((((((

m2aωℓθ2 sin (θ1 − θ2) + m2gℓ sin θ2

m2ℓ ℓθ2 =

m2ℓ aω2 sin (θ1 − θ2) −m2ℓ g sin θ2

ℓθ2 = aω2 sin (θ1 − θ2) − g sin θ2 (9)

Eqs. 8 and 9 are alternate versions the same the equation of motion of thesystem. This means we can set Eqs. 8 and 9 to find an expression for θ2 interms of ω, θ1, and θ2. Abbreviating θ1 − θ2 as ∆

aω2 sin ∆ − g sin θ2 =ℓθ2

2sin ∆

cos∆−

g sin θ1

cos∆

ℓθ2

2sin ∆

cos∆=

g sin θ1

cos∆− aω2 sin ∆ − g sin θ2

ℓθ2

2=

g sin θ1

sin∆− aω2 cos∆ −

g sin θ2 cos∆

sin∆

θ2 =

(g sin θ1 − aω2 sin ∆ cos∆ − g sin θ2 cos∆

ℓ sin ∆

) 1

2

θ2 =

(g sin θ1

ℓcsc ∆ −

aω2

ℓcos∆ −

g sin θ2

ℓcot∆

) 1

2

(10)

Remember, thought ω = θ1 is constant, both θ1 and θ2 are still functions oftime, so this does not necessarily mean that θ2 is a constant.

3

Statistical Mechanics Review


Jeff Kissel

October 11, 2006

Consider a system of N non-interacting particles in which the energyof each particle can assume two and only two distinct values: 0 and∆. The total energy of the system is E.

(a) Find the entropy, S of the system as a function of E and N.Determine S in the thermodynamic limit where N ≫ 1 and E ≫ ∆.

In this two level system, the number of particles with energy ∆, n∆ is thethe total amount of energy E, dispersed among each particle. The number ofparticles with no energy, n0 is the number of particles with energy ∆ subtractedfrom the total number N . Summing these statements in equations:

n∆ =E

∆(1)

n0 = N −E

∆(2)

With these numbers we can establish how the system behaves. This two levelenergy system will follow a binomial distribution, such that the number of mi-crostates Ω in the system is

Ω(N, n∆) =

(

Nn∆

)

=N !

n∆! n0!

Ω(E, N) =N !

(

E

∆

)

!(

N −E

∆

)

!(3)

where I’ve substituted in Eqs. 1 and 2. Now that we have the number ofmicrostates, the entropy is

S ≡ kB ln( Ω(E, N) ) (4)

S

kB

= ln(N !) − ln

(

E

∆!

)

− ln

(

(N −E

∆) !

)

1

Jeff Kissel October 11, 2006 Statistical Mechanics

where I’ve used the properties of the natural logarithm to break up the mul-tiplication and division present in Eq. 3. From here, we can use Sterling’sapproximation (ln(x!) = x ln(x) − x) to simplify the expression further,

S

kB

= N ln(N) −N −E

∆ln

(

E

∆

)

+E

∆

−(N −E

∆) ln

(

N −E

∆

)

+(N −

E

∆)

= N ln(N) −E

∆ln

(

E

∆

)

− (N −E

∆) ln

(

N −E

∆

)

S(E, N) = kB

[

N ln(N) −E

∆ln

(

E

∆

)

− (N −E

∆) ln

(

N −E

∆

)]

(5)

This formula is already in the thermodynamic limit that N ≫ 1, becausewe used Sterling’s approximation, and if we take E ≫ ∆, i.e. E/∆ ≫ 1,then as long as E/∆ < N , Eq. 5 holds (note the opposite case is unphysical).If N ≫ E/∆, then one can ignore the second term, and in the third term,(N − E/∆) → N , so the entropy approximately zero.

(b) Find the temperature as function of E.

One can find the temperature from the entropy as it is defined,

T ≡

(

∂S

∂E

)

−1

(6)

1

T= kB

∂

∂E

(

ln(Ω))

1

kBT=

∂

∂E

(

N ln(N) −E

∆ln

(

E

∆

)

− (N −E

∆) ln

(

N −E

∆

))

= −

[

1

∆ln

(

E

∆

)

+E

∆

(

E

∆

)

−1 (

1

∆

)

]

−

[

−1

∆ln

(

N −E

∆

)

+

(

N −E

∆

)(

N −E

∆

)

−1 (

−1

∆

)

]

= −1

∆ln

(

E

∆

)

+1

∆ln

(

N −E

∆

)

−

(

1

∆

)

+

(

1

∆

)

=1

∆

(

ln

(

N −E

∆

)

− ln

(

E

∆

))

=1

∆ln

(

∆

E

(

N −E

∆

))

1

kBT=

1

∆ln

(

N∆

E− 1

)

⇒ T =∆

kB ln(

N∆

E− 1

) (7)

2


(b) For what range of energy values is the temperature negative?

From Eq. 7, the temperature of this system will be negative when

ln

(

E

∆

)

> ln

(

N −E

∆

)

(8)

So, from this expression we can exponentiate both sides and solve for theenergy condition under which the temperatures are negative:

E

∆> N −

E

∆2E

∆> N

E >N∆

2(9)

3



Jeff Kissel

May 18, 2007


Spring 2007 #32 Added to question bank.

A crude estimate of the surface temperature of the Earth is to as-

sume that clouds reflect a fraction η (≈ 0.3) of all sunlight, the rest

being absorbed by the Earth and re-radiated. Treating the Sun as a

blackbody at temp TS = 5800 K, find the surface temperature of the

Earth. You may assume the earth is an ideal absorber and that the

rotation of the earth allows it to emit in all directions. The radius of

the Sun is RS = 6.96 × 105 km and that of the Earth is 6, 400 km.

The flux of sunlight recieved by the Earth Φ(R)S will be the power radiated

by the sun, PS reduced by the reflection factor η, and the distance the Earth isfrom the Sun d,

Φ(r)S =

(1 − η)

4πd2PS (1)

From the Stefan-Boltzman Law for blackbody radiators, we know the sun willradiate a power,

PS = ASσT 4S (2)

where AS is the total surface area of the Sun, σ is the Stefan Boltzmann con-stant, and TS is the temperature of the Sun. So, Eq. 1 becomes

Φ(r)S =

(1 − η)

4πd2PS

=(1 − η)

4πd2PS

(

4πR2SσT 4

S

)

Φ(r)S = (1 − η)

R2S

d2σ T 4

S (3)

1

Jeff Kissel May 18, 2007 Statistical Mechanics

At any given time however, only half of the Earth is facing the sun, so the power

absorbed by the Earth P(a)E will be the flux from the sun Φ

(r)S , gathered by the

cross-section of the Earth ΣE (i.e. the area of a circle with radius RE),

Φ(r)S =

P(a)E

ΣE

P(a)E = ΣE Φ

(r)S

= πR2E (1 − η)

R2S

d2σ T 4

S (4)

Now since we’re treating the Earth as a perfect (blackbody) radiator, we canset the power absorbed equal to the power emitted, and solve for temperature.

P(e)E = P

(a)E

4 πR2EσT 4

E = πR2E (1 − η)

R2S

d2 σ T 4S

T 4E = (1 − η)

R2S

4 d2T 4

S

TE =

(

(1 − η)R2

S

4 d2

)1/4

TS

TE = (1 − η)1/4

√

RS

2 dTS (5)

Plugging in numerical values, (note that the distance from the Sun to Earth d

is not given),

TE = (1 − η)1/4

√

RS

2 dTS

η = 0.3RS = 6.96 × 1010 cm

d = 1.496 × 1013 cm

TS = 5800 K

TE = 255.874 K (6)

2



Jeff Kissel

October 11, 2006

A simplified model of diffusion consists of a one-dimensional lattice,with lattice spacing a, in which an “impurity” makes a random walkfrom one lattice site to an adjacent one, making jumps at time inter-vals t.

Figure 1: An “impurity” taking a random walk either right (r) or left (ℓ) withN = 14 steps, resulting in n = 4 steps to the right.

(a) After N jumps have been made, find the probability that the atomhas moved a distance R = na from its starting point, in the limit oflarge N , and n.

As can be seen in the picture, the particle can either move a step right, r orleft, ℓ (or a distance ±a), and since the total number of steps N = r + ℓ, andthe net number of steps n = |r − ℓ| = R / a, the distance traveled is

R = n a

= |r − ℓ| a

= |r − (N − r)| a

= (2r − N) a (1)

Since the probability that the particle will travel right, PN (r) is just asprobable as it traveling left, PN (ℓ), the total probability of it traveling anywhereis PN (n) = PN (r) + PN(ℓ) = 2PN(r).

1


Note that from Eq. 1, n = 2r−N ⇒ r = N+n2 . This is a “two-level” system,

so it will follow a (not necessarily normalized) binomial distribution, or

PN (n) = 2 PN (r) = 2 ξ

(Nr

)

= 2ξN !

r! ℓ!=

2ξ N !(

N+n2

)!(N − N+n

2

)!

PN (n) =2ξ N !

(N+n

2

)!(

N−n2

)!

(2)

From here, because we’re operating with large N and n, we’ll use the naturallogarithm, and take advantage of Sterling’s approximation (x! = x ln(x) − x)and the binomial expansion (ln(1 + x) = x − 1

2 + O(x3)).

ln(

PN(n))

= ln(2ξ) + ln(N !) − ln

(N + n

2!

)

− ln

(N − n

2!

)

= ln(2ξ) + N ln(N) − N −(

N + n

2ln

(N + n

2

)

− N + n

2

)

−(

N − n

2ln

(N − n

2

)

− N − n

2

)

= ln(2ξ) + N ln(N) − N + n

2ln

(N + n

2

)

−N − n

2ln

(N − n

2

)

− N +N + n

2+

N − n

2

= ln(2) + ln(ξ) + N ln(N) − 1

2(N + n) ln

(1

2N(

1 +n

N

))

− 1

2(N − n) ln

(1

2N(

1 − n

N

))

+1

2(((((((2N + N + N +n − n)

= ln(2) + ln(ξ) + N ln(N)

− 1

2(N + n)

[

−ln(2) + ln(N) + ln(

1 +n

N

)]

−[

−ln(2) + ln(N) + ln(

1 − n

N

)]

=(

1 +:N1

2(N + n) +

1

2(N − n)

)

ln(2) + ln(ξ)

+(((((((((((((((

N − 1

2(N + n) − 1

2(N − n)

)

ln(N)

− 1

2(N + n) ln

(

1 +n

N

)

− 1

2(N − n) ln

(

1 − n

N

)

2


= (1 + N) ln(2) + ln(ξ)

− 1

2(N + n)

(n

N− n2

2N2+

n3

3N3+ O

(n4

4N4

))

+1

2(N − n)

(

− n

N− n2

2N2− n3

3N3− O

(n4

4N4

))

= (1 + N) ln(2) + ln(ξ)

− 1

2

(

n − n2

2N+

n3

3N2+

n2

N− n3

2N2+ O

(n4

4N3

))

− 1

2

(

n − n2

2N− n3

3N2+

n2

N+

n3

2N2+ O

(n4

4N3

))

= (1 + N) ln(2) + ln(ξ) − 1

2(n − n)

− 1

2

(n2

2N+

n2

2N

)

− 1

2

(

− n3

6N2+

n3

6N2

)

−O(

n3

4N4

)

= (1 + N) ln(2) + ln(ξ) − n2

2N+ O

(n3

4N4

)

ln(

PN(n))

= ln(

ξ 2(N+1))

− n2

2N+

O(

n3

4N4

)

where I’ve dropped the terms O(

n3

4N4

)

because N is huge. Phew! We can then

exponentiate both sides to get,

PN (n) = ξ 2(N+1) e−n2

2N

(

R = na ⇒ n2 = R2/a2)

PN(R) = ξ 2(N+1) e−R

2

2Na2 (3)

Normalizing to one will yield the constant ξ,

1 =

∫∞

0

PN (R) dR

= ξ 2(N+1)

∫∞

0

e−R

2

2Na2

(

let x =R

a√

2N⇒ dx =

dR

a√

2N⇒ a

√2N dx = dR

)

= ξ 2(N+1) a√

2N

∫∞

0

e−x2

dx

1 = ξ 2(N+1) a√

2N

√π

2

⇒ ξ =1

a2N√

2πN(4)

3


Thus, for the limit in which N and n are large, the probability of an impurityhas moved a distance R is

PN (R) =1

a2N√

2πN2(N+1) e−

R2

2Na2

=2

a√

2πNe−

R2

2Na2

PN (R) =

(2

Nπa2

) 1

2

e−R

2

2Na2 (5)

(b) The diffusion coefficient is defined by the differential equation

∂n

∂t= D

∂2n

∂x2(6)

where n is the impurity concentration. Find an expression for D inthe model described above.

The number of steps that impurity takes, N should be equivalent to thetotal time traveled, t divided by the time it takes to travel the distance of onelattice structure, τ . Thus the probability (Eq. 5) as a function of distance, xand time, t should be

PN (x, t) =

(2

πa2

τ

t

) 1

2

e−τx

2

2ta2 (7)

The impurity concentration is a function of the probability of each of theM particles in the system. In fact, it would be the sum over all the particle’sprobability that over a time t they’ve moved a distance x (from some startingpoint x0), or

n(x, t) =

M∑

i=1

PN(xi, t) (8)

If this is the case, then we can argue that Eq. 6 must be satisfied forPN(x, t) in place of n(x, t). From this fact we can get a value for D. Beforewe do that, let’s clean up PN (x, t) a little bit to get it ready for some seriousdifferenchumacationating.

PN (x, t) =

(2τ

πa2

) 1

2

t−1

2 e−τ

2a2x2t−1

let Υ ≡(

2τ

πa2

) 1

2

and ≡ τ

2a2

PN (x, t) = Υ t−1

2 e−x2t−1

(9)

4


Now, we can sub this into Eq. 6,

∂PN

∂t= D

∂2PN

∂x2

∂

∂t

(

Υ t−1

2 e−x2 −1)

︸︷︷︸

I

= D∂2

∂x2

(

Υ t−1

2 e−x2t−1)

︸︷︷︸

II

I = −1

2t−

3

2 e−x2t−1

+ t−1

2 e−x2t−1 (x2t−2

)

= − e−x2t−1

2t3

2

+x2e−x2t−1

t5

2

= − (t − 2x2)e−x2t−1

2t5

2

II = D∂

∂x

(

t−1

2 e−x2t−1 (−2xt−1))

= D∂

∂x

(

−2xe−x2t−1

t3

2

)

= D

(

−2e−x2t−1

t3

2

− 2xe−x2t−1

t3

2

(

−2x

t

))

= D

(

−2e−x2t−1

t3

2

− 42x2e−x2t−1

t5

2

)

= D

(

−2te−x2t−1

+ 42x2e−x2t−1

t5

2

)

= −D2(t − 2x2)e−x2t−1

t5

2

I = II

+ (t − 2x2)e−x2t−1

2 t5

2

= + D2(t − 2x2)

e−x2t−1

t5

2

D =1

4

=1

4

2a2

τ

D =a2

2τ(10)

Huh! After all that work, the diffusion coefficient is gratifyingly elegant, de-pending only on the size of the lattice structure!

5



Jeff Kissel

October 11, 2006

Show that

CP = CV +TV α2

κT

where CP and CV are the heat capacities at constant pressure and con-

stant volume, respectively, α is the coefficient of thermal expansion,

and κT is the isothermal compressibility.

The important quantities dealt with in this problem defined as

CV ≡ T

(∂S

∂T

)

V

(1)

CP ≡ T

(∂S

∂T

)

P

(2)

α ≡1

V

(∂V

∂T

)

P

(3)

κT ≡1

V

(∂V

∂P

)

T

(4)

where the subscript on the parenthesis indicates what is being held constant inthe partial derivative. Note also that it is assumed that the number of particlesin the system N is constant, so that dN = 0 and the subscript N is suppressed.

We start by expanding the full differential entropy into partial differentialswith respect to T and V ,

dS =

(∂S

∂T

)

V

dT +

(∂S

∂V

)

T

dV

Plugging in Eq. 1, this becomes

dS =CV

TdT +

(∂S

∂V

)

T

dV

From here we can expand dV in a similar manner to dS,

dS =CV

TdT +

(∂S

∂V

)

T

[(∂V

∂T

)

P

dT +

(∂V

∂P

)

T

dP

]

1


=

[CV

T+

(∂S

∂V

)

T

(∂V

∂T

)

P

]

dT +

(∂S

∂V

)

T

(∂V

∂P

)

T

dP (5)

With the goal to bring CP into the expression we take the pressure to beconstant so the dP = 0 and the second term in Eq. 5 vanishes to get

(∂S)P =

[CV

T+

(∂S

∂V

)

T

(∂V

∂T

)

P

]

∂T

(∂S

∂T

)

P

=CV

T+

(∂S

∂V

)

T

(∂V

∂T

)

P

where using Eq. 2 yields

CP

T=

CV

T+

(∂S

∂V

)

T

(∂V

∂T

)

P

CP = CV + T

(∂S

∂V

)

T

(∂V

∂T

)

P

= CV + T

(∂S

∂V

)

T

(V α) (6)

by applying a rearranged Eq. 3.Now, a useful Maxwell relation is that

(∂S

∂V

)

T

=

(∂P

∂T

)

V

(7)

So Eq. 6 is then

CP = CV + αTV

(∂P

∂T

)

V

(8)

Finally, if we pull apart the partial derivative using the chain rule and mul-tiply by one, we’ll find the desired result:

CP = CV + αTV

[

V

(∂P

∂V

)

T︸︷︷︸

κ−1

T

1

V

(∂V

∂T

)

P︸︷︷︸

α

]

CP = CV +TV α2

κT

√(9)

using Eqs. 3 and 4.

2



Jeff Kissel

November 6, 2006

Lead has a molar mass of M = 207.2 g mole−1. At 25 C and 1 atm ofpressure, it has an isothermal bulk modulus κT of 1.6×1010 Pa, a massdensity ρ = 11.4 g cm−3, and a coefficient of thermal expansion, α of 87×10−6 C−1. Its specific heat at constant pressure CP = 128 J kg−1 C−1.

(a) How big is the difference between CP and its constant volumespecific heat CV ?

One can prove that, (see Classical S’06-#23, F’06-#34 for detailed proof)

CP − CV =TV α2

κT(1)

and the only thing in the problem we’re not given is the volume, for one moleis just V = M/ρ. So

CV − CV =TMα2

ρκT(2)

α = 8.7 × 10−7 K−1 ρ = 1.14 × 104 kg m−3

T = 298 K κT = 1.6 × 1010 Pa

M = 2.072× 105 kg mole−1

CV − CV = 2.56224× 10−19 J kg−1 K−1 (3)

Apparently really small!

(b) The Law of Dulong and Petit states that the heat capacity of anysolid at room temperature arises from the vibrations of the atoms(3N degrees of freedom), which can be calculated by treating thevibrations as a set of 3N classical harmonic oscillators. Does the Lawof Dulong and Petit describe CP or CV ? What would you predict theheat capacity of lead to be if this law is correct?

1

Jeff Kissel November 6, 2006 Statistical Mechanics

The Law of Dulong and Petit (describing the heat capacity at constantvolume, CV ) is the formal name for the following identity,

CV = 3R (4)

for a solid, where R is the gas constant equivalent to 8.314472 J K−1 mole−1.This can be shown as follows:

For classical harmonic oscillators, whose only degrees of freedom, f arequadratic, we can use the equipartition theorem which states that

E =1

2nfRT (5)

where n is the number of moles in the system. Yet CV is defined as

CV ≡(

∂E

∂T

)

V,N

⇒ CV =1

2fnR (6)

So for one mole of a solid, for which there are

(3 dimensions)× (1kinetic f + 1 potential f) = 6

degrees of freedom,CV = 3R

√(7)

Thus, by the power of Dulong and Petit, I predict the heat capacity atconstant volume to be

= 3(8.314472 J K−1 mole−1)

CV = 24.9434 J g−1 K−1 (8)

(c) Find the Debye temperature of lead. How does the specific heatof lead vary with temperature for temperatures well below the Debyetemperature?

For the first part, I’ll just explicitly state the definition of the Debye Tem-perature, TD and everything one needs to know. For the second part, I’ll derivethe Debye Temperature from scratch, and then take the low temperature limit(i.e. take what’s important to the problem from pgs 308-311 of Schroeder, andadded some steps he didn’t do).

So, the Debye Temperature is defined as,

TD ≡ hcs

2kB

(

6N

πV

)1

3

(9)

where cs is the speed of sound in a solid. Converting to what we’ve been givenin the problem,

cs ≡√

κT

ρ

N

V=

ρNA

M

2


⇒ TD =h

2kB

(

κT

ρ

)1

2

(

6ρNA

πM

)1

3

(10)

h = 6.626 × 10−34 J s kB = 1.381× 10−23 J K−1

κT = 1.6 × 1010 Pa ρ = 1.14 × 104 kg m−3

M = 2.072× 105 kg mole−1 NA = 6.02 × 1023 particles mole−1

TD = 113.242 K (11)

Aside from a few subtle differences such as slower speeds, the addition oflongitudinal polarization, and a lower limit on wavelength, sounds waves in asolid are quite similar to light waves. Thus, in a solid, each mode of simpleharmonic oscillation has a set of equally spaced energy levels with unit energyequal to

ǫ = hf =hcs

λ

ǫ =hcsn

2L(12)

where h is Planck’s constant, cs is the speed of sound, L is the length of thecrystal, and n = |~n| is the magnitude of the vector in n-space signifying theshape of the wave. At a temperature, T the number of units of energy the solidcontains follows the Planck Distribution,

nPl =1

eǫβ − 1(13)

(note nPl is entirely different from n from Eq. 12). To find the total thermalenergy, E of the crystal, we add up the energies of all dimensions (or “allowedmodes”),

E = 3∑

nx

∑

ny

∑

nz

ǫ nPl(ǫ) (14)

where the factor of 3 comes from that a phonon has three possible polarizationstates (one longitudinal and two transverse). From here we would normally (i.e.for electromagnetic waves) convert to integrals, but we first must worry aboutthe limits of integration.

As we’ve said, in a crystal the atomic spacing puts a lower limit on thewavelength. Consider the following one-dimensional lattice of atoms (Figure 1).Each mode of oscillation has its own distinct shape, with number of “bumps”equal to n. Because each bump has to contain at least one atom, n cannotexceed the number of atoms in a row.

If a three-dimensional crystal is a perfect cube, then the number of atomsalong any direction in 3

√N , so each sum in Eq. 14 should go from 1 to 3

√N .

If the crystal isn’t a perfect cube, then neither is the corresponding volume in

3


Figure 1: Modes of oscillation of a row of atoms in a crystal lattice.

n-space. However, the sum will still run over a region of n-space whose totalvolume is still n.

Though the sums (integrals) in Eq. 14 depends on nx, ny, and nz in a non-trivial way, the function depends on the magnitude of ~n in a simpler way; itdoesn’t depend on angles at all. Thus, the Debye model approximates the cubewith the first octant of a sphere, with total volume N (preserving the degreesof freedom) and radius

N =1

8

(

4πn3max

3

)

=πn3

max

6

⇒ nmax =

(

6N

π

)1

3

(15)

Figure 2: The sum in in Eq. 14 is technically over a cube in n-space, with aside length of 3

√N . Debye makes the approximation that the cube is a sphere

of radius nmax

4


Remarkably, it is experimentally confirmed that this approximation is exactin both the high- and low- temperature limits, and intermediate temperatures,though not exact are surprisingly good. So, now in Eq. 14 we can convert sumsto integrals, then move to spherical coordinates,

E = 3∑

nx

∑

ny

∑

nz

ǫ nPl(ǫ)

= 3

∫ π2

0

∫ π2

0

∫ nmax

0

n2ǫ

eǫ/kBT − 1n2 sin θ dn dθ dφ

(

∫ π2

0

∫ π2

0

sin θ dθdφ =π

2, ǫ =

hcsn

2L

)

E =3π

2

∫ nmax

0

hcs

2L

n3

ehcsn/2LkBT − 1dn (16)

which looks nasty at first, but if we choose our variables such that

x =hcs

2LkBTn

dx =hcs

2LkBTdn ⇒ kBT dx =

hcs

2Ldn

⇒ xmax =hcsnmax

2LkBT

=1

T

hcs

2LkB

(

6N

πV

)1

3

≡ TD

T(17)

where I’ve finally introduced the Debye temperature,

TD ≡ hcs

2kB

(

6N

πV

)1

3

(18)

then E becomes,

E =

∫ nmax

0

3π

2n3

1

ehcsn/2LkBT − 1

hcs

2Ldn

=

∫ xmax

0

3π

2

(

2LkBT

hcsx

)31

ex − 1kBT dx

(

(

T

TD

)3

=

(

2LkBT

hcs

)3π

6N

)

=

∫ TD/T

0

3π

2

6N

πkBT

(

T

TD

)3x3

ex − 1dx

=9NkBT 4

T 3

D

∫ TD/T

0

x3

ex − 1dx (19)

5


At which point you plug the integral into your favorite computer program tofind E for any temperature if you like. BET, the question ask for limits. First,we can check the Law of Dulong and Petit by taking the limit when T ≫ TD,the upper limit of the integral is much less than 1, so x is always very small andwe can approximate ex ≈ 1 + x in the denominator,

E ≈ 9NkBT 4

T 3

D

∫ TD/T

0

x3

1 + x − 1dx

≈ 9NkBT 4

T 3

D

∫ TD/T

0

x2 dx

≈ 3

9NkBT 4

T 3

D

T 3

D

3T 3

E ≈ 3NkBT (when T ≫ TD) (20)

⇒ CV ≡(

∂E

∂T

)

V,N

≈ 3NkB ≈ 3nR√

(21)

When T ≪ TD, as the question asks, the upper limit on the integral is solarge that by the time we get to it, the integrand is dead (due to the ex inthe denominator). So, we can just replace the upper limit with infinity; theextra modes we’re adding don’t contribute anyways. In this approximation, theintegral evaluates to π4/15, so the total energy is

E ≈ 3π4

5

NkBT 4

T 3

D

(when T ≪ TD) (22)

CV ≈ 12π4NkB

5

(

T

TD

)3

(23)

which is the second part of the question.

6



Jeff Kissel

October 12, 2006

Consider a monatomic ideal gas of mass density, ρ at temperature,T whose atoms have mass m. The number of atoms with velocities~v in the velocity space volume element d3~v is given by the Maxwell-Boltzmann distribution

n(~v) d3~v =ρ

m

(

m2

2πkBT

)3

2

exp

(

−m(

v2x + v2

y + v2z

)

2mkBT

)

(1)

The equation given is wrong. I shall prove otherwise.For the canonical ensemble, partition function for N indistinguishible parti-

cles in an ideal gas is

Z =V N

N !

(

2πmkBT

h2

)3

2

(2)

So for a single (N = 1) free (E = p2/2m) particle in an ideal gas (whereβ = 1/kBT ), the phase space probability distribution is

P(1)(~p, ~q) d3~p d3~q =1

h3

e−βE(~p,~q)

Zd3~p d3~q

=1

h3

e−β

2m~p2

V(

2πmβh2

)3

2

d3~p d3~q

P(1)(~p) d3~p =

(∫

P(~p, ~q) d3~q

)

d3~p

P(1)(~p) d3~p =1

h3

e−β

2m~p2

V(

2πmβ h2

)3

2

V

P(1)(~p) d3~p =

(

β

2πm

)3

2

e−β

2m~p2

d3~p (3)

1


Yet, both this momentum distribution, and some velocity distribution shouldbe normalized to unity, so

1 =

∫

∞

−∞

P(1)(~p)d3~p =

∫

∞

−∞

n(~v) d3~v

⇒ n(~v) d3~v = P(1)(~p) d3~p

=

(

β

2πm

)3

2

e−β

2mm2~v2

m3 d3~v

n(~v) d3~v =

(

mβ

2π

)3

2

e−1

2mβ~v2

d3~v (4)

which differs from the equation given in that there is no ρ/m factor out front,and the argument of the exponential should have either have an extra m in thenumerator, or no m in the denominator.

H’ok, now let’s get onto the problem.

(a) What is the average velocity 〈~v〉?The average velocity is defined as

〈~v〉 =

∫

∞

−∞

v n(v) d3v (5)

However, n(~v) is a gaussian distribution, i.e. even over the interval (−∞,∞),and v, begin linear, is odd. The integration of the product of an even and oddfunction from −∞ to ∞ will equal 0. Thus,

〈~v〉 = 0 (6)

(b) Derive the distribution of speeds P (v) dv.

This can be done by converting the velocity distribution (Eq. 4) into spher-ical coordinates, and integrating over the angular terms.

P (v) dv = n(~v) dvxdvydvz

= n(v) v2 sin θ dvdθdφ

=

(∫ 2π

0

∫ π

0

sin θ dθdφ

)

v2n(v) dv

= 4πv2 n(v)dv

P (v) dv = 4πv2

(

mβ

2π

)3

2

e−1

2mβv2

dv (7)

where I’ve exploited the fact that ~v2 = v · v = v2.

2


(c) What is the most probable speed, v∗?The most probable speed will be the maximum of the speed distribution, so

I’ll denote v∗ as vmax. So, we’ll maximize Eq. 7 with respect to v.

(

∂P (v)

∂v

)

v = vmax

= 4π

(

mβ

2π

)3

2 ∂

∂v

(

v2e−1

2mβv2

)

0 =[

2v e−1

2mβv2

+ v e−1

2mβv2

(−mβv)]

v = vmax

= 2(vmax)e−1

2mβv2 − mβ(vmax)3 e−

1

2mβv2

mβ(vmax)3 e−1

2mβv2

= 2(vmax)e−1

2mβv2

(vmax)2 =2

mβ

v ∗ = vmax =

√

2kBT

m(8)

(d) Obtain expressions for vavg, root mean square speed, vrms, and v∗,then rank them in increasing order.

We can obtain the root mean square velocity, vrms =

√

(v2) by doing theintegral

(v2) =

∫

∞

0

v2 P (v) dv (9)

but we can get it quickly from the Equipartition theorem (E = 12fkBT ) for a

monatomic gas (where the degrees of freedom, f is 3), since the energy is onlykinetic for a ideal (free) particle,

E =

(

1

2mv2

)

3

2kbT =

1

2m (v2)

(v2) =3kBT

m

vrms =

√

3kBT

m(10)

For the average velocity, vavg = v, we’ll actually have to do the integral.

v =

∫

∞

0

v P (v) dv

= 4π

(

mβ

2π

)3

2∫

∞

0

v3 e−1

2mβv2

dv

3


let α =1

2mβ

=4α

3

2

√π

∫

∞

0

v3 e−αv2

dv

=4α

3

2

√π

∂

∂α

(∫

∞

0

v e−αv2

dv

)

let u = v2 du = 2v dv

(0 → 0,∞ → ∞)1

2du = v dv

=4α

3

2

√pi

∂

∂α

(

1

2

∫

∞

0

e−αu du

)

=2α

3

2

√π

∂

∂α

(

e−αu

α

∣

∣

∣

∣

∣

∞

0

)

=2α

3

2

√π

∂

∂α

(

− 1

α

)

=2√π

α3

2

α2

=2√π

(

2

mβ

)1

2

v =

√

8kBT

πm(11)

Thus, since Eq. 8, 10, and 11 all three have the factor√

1/mβ in common,and only differ by 2 < 8/π < 3,

vmax < vavg < vrms (12)

Figure 1: The Maxwell-Boltzmann distribution for T ∼ 100 K.

4



Jeff Kissel

October 11, 2006

A certain material is completely specified by its volume V and tem-perature T . It has an equation of state P = AT 4, where A is a constantindependent of the volume. The heat capacity at fixed volume is mea-sured to be BV T 3.

a) From dimensional analysis, B and A have the same units. Showthat B = 12A.

The heat capacity is defined as

CV ≡ T

(

∂S

∂T

)

V

= BV T3

⇒

(

∂S

∂T

)

V

= BV T2 (1)

Integrating both sides, we’ll get an expression for the entropy,

∫ S

0

dS′ = BV

∫ T

0

T′2

dT′

S =1

3BV T

3 (2)

However, since we’d like to bring A into the equation we have to use the Maxwellrelation

(

∂S

∂V

)

T

=

(

∂P

∂T

)

V

(3)

where the subscript denotes what is being held constant in the derivative. Plug-ging in our equation of state and Eq. 2,

∂

∂V

(

1

3BV T

3

)

T

=∂

∂T

(

AT4)

V

1

3BT

3 = 4AT3

B = 12A√

(4)

1


b) Find the entropy of this material as a function of V and T .

Phh, we dun’ already been up in there. Peep Eq. 2:

S =1

3BV T

3 (2)

c) If this material is cooled adiabatically and reversibly from 20K to10K, by how much does the volume change?

In a reversible adiabatic process, entropy is conserved, and thus is called an“isentropic” process. This means we can set up a conservation equation fromEq. 2,

1

3BViT

3

i =1

3BVfT

3

f

ViT3

i = VfT3

f

Vi

Vf

=

(

Tf

Ti

)3

Vi

Vf

=

(

(10 K)

(20 K)

)3

=1

8

Vf = 8Vi (5)

So the volume increases eight-fold by doubling the temperature.

2



Jeff Kissel

May 18, 2007

Exam Iteration Question Number Notes/ChangesSpring 2006 #27 –Fall 2006 #38 Number Shift

Spring 2007 #38 Part (d) removed

Consider an ideal Fermi gas of spin-1/2 particles in a 3-dimensionalbox. The number of particles per unit volume is n, the mass of theparticles is m, and the energy of the particles is the usual ǫ = p2/2m.Assume that the temperature is quite low.

(a) Find formulas for the Fermi energy ǫF , Fermi wavevector kF , andFermi temperature TF in term of m, n and constants such as h andkB.

To find the Fermi energy (the chemical potential of spin 1/2 particles at zerotemperature), we must find the total number of particles N as T approacheszero.

For any temperature T , one normally finds N via summing over the Fermi-Dirac probability distribution 〈 nj(ǫ) 〉FD (because the particles are of spin1/2),

〈 nj(ǫ) 〉FD =1

eβ(ǫj−µ) + 1(1)

where β = 1kT , ǫj is the kinetic energy of a state j, and µ is the chemical potential

of the particles. However, at the limit of zero temperature, 〈 nj(ǫ) 〉FD decaysto either 0 (when ǫj > µ) or 1 (when ǫj < µ). The energy, at T = 0, at whichthe now-step-function distribution flips from 1 to 0 is the Fermi energy, whosevalue is determined (again) by the number of particles.

To perform the calculation of the number of particles, we first find the densityof states: the Jacobian of transformation between summing over energy statesǫj, and integrating over kinetic energies. First, to account the spin degeneracy

1


of +1/2 and −1/2, we tack on a factor of 2,

∑

j

= 2∑

k

=2

h3

∫ ∞

−∞

∫ ∞

−∞

d3~pd3~q

=2

h3V

∫ ∞

−∞

d3~p

=2V

h3(4π)

∫ ∞

0

p2 dp

let ǫ =p2

2m

p =√

2m ǫ1/2

dp = 12

√2mǫ−1/2 dǫ

p2 =√

2mǫ

=8πV

h3

∫ ∞

0

(√2mǫ

) 1

2

√2mǫ−1/2 dǫ

=8πV

h3

1

2(2m)3/2

∫ ∞

0

ǫ1/2 dǫ

∑

j

⇒∫ ∞

0

4πV

(2m

h2

)3/2

ǫ1/2 dǫ ≡∫ ∞

0

g(ǫ) dǫ (2)

Now we can find the number of particles N , and invert for the Fermi energyǫF .

N(T = 0) =∑

j

〈 nj(ǫ) 〉FD

=

∫ ∞

0

〈 n(ǫ) 〉FD gFD(ǫ) dǫ

=

∫ ǫF

0

(1) g(ǫ) dǫ +

∫ ∞

ǫF

(0) g(ǫ) dǫ

=

∫ ǫF

0

4πV

(2m

h2

)3/2

ǫ1

2 dǫ

= 4πV

(2m

h2

)3/22

3ǫ

3

2

F

N =8π

3V

(2m

h2

)3/2

ǫ3

2

F (3)

ǫF =h2

2m

(1

8

3

π

N

V

)2/3

=h2

8m

(3

π

N

V

)2/3

2


ǫF =h2

2m

(3π2 N

V

)2/3

(4)

In this form it’s nice and quick to pull out the Fermi wave-vector kF , sincethe quantum mechanical kinetic energy is

E =p2

2m=

h2

2mk2 (5)

So by direct comparison between Eqs. 4 and 5,

kF =

(3π2 N

V

) 1

3

(6)

The Fermi terperature TF is similarly quick, as the thermal kinetic energyis,

E = kBT (7)

Inverting for temperature and subbing in 4,

TF =h2

2mkB

(3π2 N

V

) 2

3

(8)

(b) Find the total energy of the gas at zero temperature.

This calculation can be done by summing over possible energy states, weightedby their probabilities (determined by the Fermi-Dirac Distribution),

〈E〉 ≡∑

j

ǫj 〈 nj(ǫj) 〉

=

∫ ∞

0

ǫ 〈 n 〉FD gFD(ǫ) dǫ

〈E〉T=0 =

∫ ǫF

0

ǫ (1) g(ǫ) dǫ +

∫ ∞

ǫF

ǫ (0) g(ǫ) dǫ

=

∫ ǫF

0

ǫ g(ǫ) dǫ

=

∫ ǫF

0

ǫV

2π2

(2m

h2

) 3

2

ǫ1

2 dǫ

=V

2π2

(2m

h2

) 3

2∫ ǫF

0

ǫ3

2 dǫ

=V

2π2

(2m

h2

) 3

2 2

5ǫ

5

2

F

=V

5π2

(2m

h2

) 3

2

ǫ5

2

F

3


(Eq. 3: N =

V

3π2

(2m

h2

) 3

2

ǫ3

2

F

)

〈E〉T=0 =3

5N ǫF (9)

(c) Show that the heat capacity at low temperature is proportionalto T .

The exact proof of this problem unfortunately gets rather lengthly, but isknown as the Sommerfield expansion.

If I denote the constants in Eq. 2 to be CF , and use < n > as shorthand for〈 n(ǫ) 〉FD then the ensemble average energy (as seen in part (b)) is

〈E〉 =

∫ ∞

0

ǫ 〈 n(ǫ) 〉FD gFD(ǫ) dǫ = CF

∫ ∞

0

ǫ3

2 < n > dǫ (10)

However, now we need energies at low temperatures (kT ≪ ǫF ), not justat T = 0. Although the integral in Eq. 10 goes from 0 to ∞, i.e. all positiveenergies, the most interesting region is near ǫ = µ, where 〈 nj(ǫj) 〉 falls offsharply. So, we’ll isolate this region using integration by parts,

let u = 〈 n 〉 v = 25 ǫ5/2

du = − d<n>dǫ dǫ let dv = ǫ3/2

〈E〉kT≪ǫF = CF

[

:02

5ǫ5/2 < n >

∣∣∣∣∞

0

+2

5

∫ ∞

0

ǫ5/2

(− d < n >

dǫ

)dǫ

]

(11)

where the ”surface” term vanishes because at ǫ = 0, ǫ5/2 = 0, and at ǫ = ∞,< n >= (eβ(ǫ−µ) + 1)−1 = 0. This leaves,

〈E〉 =2

5CF

∫ ∞

0

ǫ5/2

(− d < n >

dǫ

)dǫ

(− d < n >

dǫ= − d

dǫ

(1

eβ(ǫj−µ) + 1

)=

β eβ(ǫj−µ)

(eβ(ǫj−µ) + 1)2

)

=2

5CF

∫ ∞

0

ǫ5/2 β eβ(ǫj−µ)

(eβ(ǫj−µ) + 1)2

let x = β(ǫ − µ) ⇒

dx = β dǫ0 → −µβ, ∞ → ∞

〈E〉 =2

5CF

∫ ∞

−µβ

ǫ5/2 ex

(ex + 1)dx (12)

Because, the integrand of this expression falls off exponentially when x =β(ǫ − µ) ≪ 1, we can now make two approximations:

1. Extend the lower limit of the integral (−µβ) to (−∞); this makes theintegral symmetric, and its harmless because the integrand is negligible atnegative ǫ’s because of the shape of the distribution at these temperatures.

4


2. Taylor expand ǫ5/2 around ǫ = µ,

ǫ5/2 ≈ ǫ5/2

∣∣∣∣µ

+ (ǫ − µ)

(d(ǫ5/2)

dǫ

) ∣∣∣∣µ

+1

2!(ǫ − µ)2

(d2(ǫ5/2)

dǫ2

) ∣∣∣∣µ

+ ...

= µ5/2 +5

2(ǫ − µ)µ3/2 +

1

2

15

4(ǫ − µ)2µ1/2 + ...

= µ5/2 +5

2(xkT )µ3/2 +

15

8(xkT )2µ1/2 + ... (13)

So that

〈E〉 ≈ 2

5CF

[µ5/2

∫ ∞

−∞

ex

(ex + 1)2dx

︸︷︷︸(I)

+5

2kTµ3/2

∫ ∞

−∞

xex

(ex + 1)2dx

︸︷︷︸(II)

+15

8(kT )2µ1/2

∫ ∞

−∞

x2 ex

(ex + 1)2dx

︸︷︷︸(III)

+...

]

(I)

∫ ∞

−∞

ex

(ex + 1)dx =

∫ ∞

−∞

(− d < n >

dǫ

)dǫ

=

∫ ∞

−∞

d < n >

=

[

:1< n(−∞) > +:0

< n(∞) >

]

∫ ∞

−∞

ex

(ex + 1)dx = 1 (14)

(II)

∫ ∞

−∞

x ex

(ex + 1)2dx =

∫ ∞

−∞

x ex

(ex + 1)(ex + 1)

e−x

e−xdx

=

∫ ∞

−∞︸︷︷︸even

bounds

x

(ex + 1)(1 + e−x)︸︷︷︸odd

function

dx

∫ ∞

−∞

x ex

(ex + 1)2dx = 0 (15)

5


(III)

∫ ∞


bounds

x2ex

(ex + 1)2︸︷︷︸even

function

dx = 2

∫ ∞

0

x2ex

(ex + 1)2dx

let u = x2

du = 2x dx

v = −1ex+1

let dv = ex

(ex+1)2

= 2

>0

−x2

ex + 1

∣∣∣∣∞

0

+

∫ ∞

0

2x

ex + 1dx

= 4

∫ ∞

0

x

ex + 1dx

(∫ ∞

0

xm

ex + 1dx = (1 − 1

2m) Γ(m + 1) ζ(m + 1)

)

= 4(1 − 1

2) Γ(2) ζ(2)

= 4

(1

2

)(1)

(π2

6

)

=π2

3∫ ∞

−∞

x2ex

(ex + 1)2dx =

π2

3(16)

⇒ 〈E〉 ≈ 2

5CF µ5/2 (1) +

2

5CF

5

2kTµ3/2 (0)

+2

5CF

15

8(kT )2µ1/2

(π2

3

)+ ...

=2

5CF µ5/2 +

π2

4CF (kT )2µ1/2 + ...

Eq. 4: ǫF = h2

2m

(3π2 N

V

) 2

3

2mh2 ǫF =

(3π2 N

V

) 2

3

(2mh2

) 3

2 ǫ3/2F = 3π2 N

VVπ2

(2mh2

) 3

2 = 3 N

ǫ3/2

F

⇒ CF = V2π2

(2mh2

) 3

2 = 32

N

ǫ3/2

F

(17)

=2

5

(3

2

N

ǫ3/2F

)µ5/2 +

π2

4

(3

2

N

ǫ3/2F

)(kT )2µ1/2 + ...

=3

5N

µ5/2

ǫ3/2F

+3π2

8N

µ1/2

ǫ3/2F

+ ...

6


〈E〉 =3

5N ǫF

(µ

ǫF

)5/2

+3π2

8N

(kT )2

ǫF

(µ

ǫF

)1/2

+ ... (18)

Performing a similar Taylor expansion process for N (see Schroeder pp 282-284) yields the expansion that,

µ

ǫF= 1 − π2

12

(kT

ǫF

)2

+ ... (19)

which we can exploit for our purposes,

⇒(

µ

ǫF

)5/2

=

(1 − π2

12

(kT

ǫF

)2

+ ...

)5/2

(Binomial Expansion: (1 + x)p ≈ 1 + px (For x ≪ 1))

≈ 1 − 5

2

π2

12

(kT

ǫF

)2

+ ...

= 1 − 5π2

24

(kT

ǫF

)2

+ ... (20)

⇒(

µ

ǫF

)1/2

=

(1 − π2

12

(kT

ǫF

)2

+ ...

)5/2

≈ 1 − 1

2

π2

12

(kT

ǫF

)2

+ ...

= 1 − π2

24

(kT

ǫF

)2

+ ... (21)

and plug back into the ensemble average energy (Eq. 18),

〈E〉 =3

5N ǫF

(1 − 5π2

24

(kT

ǫF

)2

+ ...

)

+3π2

8N

(kT )2

ǫF

(1 − π2

24

(kT

ǫF

)2

+ ...

)+ ... (22)

and then keeping only terms of order (kT/ǫF )2 or less,

〈E〉 =3

5N ǫF − 3π2

24N

(kT )2

ǫF+

3π2

8N

(kT )2

ǫF+ O

((kT )2

ǫ3F

)

=3

5N ǫF − π2

8N

(kT )2

ǫF+

3π2

8N

(kT )2

ǫF+ O

((kT )2

ǫ3F

)

〈E〉 =3

5N ǫF +

π2

4N

(kT )2

ǫF+ ... (23)

7


Now we can finally get to the question. Remember we are trying to showthat CV ∝ T for low temperatures. This is quick attainable from Eq. 23, andthe definition of the heat capacity at constant volume,

CV ≡(

∂〈E〉∂T

)

V,N

=∂

∂T

(3

5N ǫF +

π2

4N

(kT )2

ǫF+ ...

)

CV =π2

2N

k2

ǫFT ∝ T

√(24)

8


OLD VERSIONS

(d) If the magnetic moment of the particles is µe, show that the para-magnetic susceptibility χ for low fields in the limit of zero temperatureis given by

χ =3

2

nµ2e

ǫF(25)

This part deals with the quantum mechanical property of fermions known asthe Zeeman splitting. When a weak magnetic field is applied to a fermi gas, the”up” (parallel to ~B, |−〉) and ”down” (anti-parallel to ~B, |+〉) energy levels ofthe particles are perturbed such that the energy states of the ”up” particles areincreased, and the ”down” particles are decreased, causing a ”splitting” of whatinitially was a single set of quantized energy states. The Zeeman effect is a 1storder perturbation theory problem in quantum mechanics (Check p245-246 ofGriffiths’, or Wikipedia’s article for more detialed explanations. Merzbacher’s ispretty weak sauce, but it’s on p472-473). Anyways, this is a Stat-Mech problem,so we really only care about the resulting energies, so I’ll only highlight thedetails.

The perturbing hamiltonian is

H = −~µe · ~B (26)

where B is the applied, weak magnetic field, and µe is the magnetic moment ofthe fermions (spin-1/2 particles), given by

~µe = −gJ µB~S

h(27)

where gJ is the Lande g-factor, µB = (eh)/(2m) is the Bohr magneton, and

S is the spin operator. The Lande g-factor is approximately 2 for (spin-1/2)electrons (See Wikipedia’s ”Lande g-factor” for details), so I assume gJ = 2.

Thus, H (Eq. 26, assuming the field | ~B| = B applied in the same direction asµe) becomes,

H = −µeB

= −(− 2 µB S

h

)B

(Si =

h

2σi

)

H = µBσzB (28)

where σz is the Pauli spin matrix, and I’ve assumed that the field was appliedin the z-direction.

9


The total energy of the anti-parallel (”down”, |+〉, or the positive eigenvalueof σz) state is

ǫT = ǫ + µBB

in which, as before, ǫ = p2/(2m) is the kinetic energy of the particles. Theparallel (”up”, |−〉, negative eigenvalue of σz) energies are

ǫT = ǫ − µBB

Which means the Fermi-Dirac distribution (1) becomes

〈 nj(ǫ) 〉FD =1

eβ[(ǫj±µBB)−µ) + 1(29)

i.e. is splits into two distributions,

1

2〈 nj(ǫ − µBB) 〉FD and

1

2〈 nj(ǫ − µBB) 〉FD (30)

where the factors of 1/2 insures the distribution remains normalized. The leftdistribution corresponds to the up spins, i.e. those energy levels displaced bya positive µBB and therefore the probability range over which it extends isfrom µBB < ǫT < ∞. The right distribution are the down spins, displaced bynegative µBB, so the range is µBB < ǫT < ∞.

From here, we can calculate the net magnetization per unit volume M , thederivative of which (with respect to B) is χ. The net magnetization would bethe total number of particles anti-align subtracted from those aligned (in a givenvolume), or

M ≡ µB(N↑ − N↓)

V(31)

We know how to find the number of particles: integrate the probability distribu-tion corresponding the particles over the correct region. Thus, the magnetizationis then

M =µB

V

(∫ ∞

µBB

1

2〈 n(ǫ + µBB) 〉FD gFD(ǫ) dǫ

−∫ ∞

µBB

1

2〈 n(ǫ − µBB) 〉FD gFD(ǫ) dǫ

)

=µB

2V

∫ ∞

0

[〈 nj(ǫ + µBB)〉 − 〈 nj(ǫ − µBB)〉] g(ǫ) dǫ

(∂f(x)

∂x ≡ f(x+h)−f(x−h)2h

⇒ f(x − h) − f(x + h) = −2 h ∂f(x)∂x

)

=µB

2V2 µBB

∫ ∞

0

(− ∂〈 n(ǫ) 〉

∂ǫ

)g(ǫ) dǫ

M =µ2

B B

V

∫ ∞

0

(− ∂〈 n(ǫ) 〉

∂ǫ

)g(ǫ) dǫ (32)

10


At low temperatures (i.e kT ≪ ǫF ), the Fermi-Dirac distribution becomesasymptotically close to a step function with the barrier at the Fermi energy(as discussed in part (a)). Thus, the derivative of the distribution at low tem-peratures is well approximated by a delta function about ǫF , or

limT→0

∫ ∞

0

(− ∂〈 n(ǫ) 〉

∂ǫ

)dǫ =

∫ ∞

0

δ(ǫ − ǫF ) dǫ (33)

which makes Eq. 32 a lot prettier,

M =µ2

B B

V

∫ ∞

0

δ(ǫ − ǫF ) g(ǫ) dǫ

=µ2

B B

Vg(ǫF )

= µ2B CF ǫ

1/2F

B

V(Eq. 17: CF =

3

2

N

ǫ3/2F

)

= µ2B

(3

2

N

ǫ3/2F

)ǫ1/2F

B

V

M =3

2

µ2B

ǫF

N

VB (34)

Finally, if I convert to the notation the proof desires (N/V → n), the mag-netic susceptibility is

χ ≡ ∂M

∂B

=∂

∂B

(3

2

µ2B

ǫFn B

)

χ =3

2

n µ2B

ǫF(35)

which is sometimes referred to as the ”Pauli Susceptibility.”

It’s cool. You’ll remember all of this during the qualifier. I promise.

11



Jeff Kissel

May 18, 2007


Spring 2007 #39 –

A collection of N spin-1/2 atoms are fixed in a solid. The atoms donot interact with each other. The magnetic moment of each atom is±µ0. If a magnetic field H is applied to the solid, each atom has anenergy of µ0H.

a) Find the mean energy 〈E〉 in a magnetic field.

For a single spin-1/2 particle, there are only two possible energy states, asstated in the problem:

E± = ±µ0H (1)

where the − or ”up” state is aligned with the magnetic field H , and the + or”down” state is anti-aligned. The grand canonical partition function for theparticle is then,

Zi ≡∑

s

e−βE(s)

= eβE− + eβE+

= eµ0βH + e−µ0βH

(

cosh θ ≡1

2(eθ + e−θ)

)

Zi = 2 cosh (µ0βH) (2)

where β = 1kBT

as usual.For the entire system of N non-interacting indistinguishable particles, the

partition function becomes

Z ≡1

N !(Zi)

N

1


Z =2N

N !coshN (µ0βH) (3)

where taking the partition function to the Nth power accounts for all the par-ticles, and dividing by N ! corrects for double counting present because of indis-tinguishability.

Once we have the partition function of a system, we can find any otherproperty of the system desired, as long as we remember a few relations. Therelation relevant to the ensemble mean energy 〈E〉 is

〈E〉 ≡1

Z

∂Z

∂β=

∂

∂β(ln (Z)) (4)

From Eq. 4 we see we’ll need the natural log of Eq. 3, so let’s get that outof the way,

ln (Z) = ln

(

2N

N !coshN (µ0βH)

)

= ln

(

2N

N !

)

+ ln(

coshN (µ0βH))

ln (Z) = ln

(

2N

N !

)

+ N ln (cosh (µ0βH)) (5)

Happily, the first term will vanish when plugged into Eq. 4, because it’s inde-pendent of β, so we need not use Sterling’s approximation or anything fancylike that. So, the ensemble mean energy is

〈E〉 = −∂

∂β

(

ln

(

2N

N !

)

+ N ln (cosh (µ0βH))

)

= −N

cosh (µ0βH)sinh (µ0βH) µ0H

〈E〉 = −µ0NH tanh (µ0βH) (6)

b) Find the entropy of this collection.

Again, because we have the partition function of the system, we need onlyto remember the relation between that and the Helmholtz Free Energy F ,

F ≡ −kT ln (Z) (7)

from which we’ll find the entropy using the following Maxwell relation,

S =

(

∂F

∂T

)

V,N

(8)

2


So let’s plug and chug!

F = −kT ln

(

2N

N !coshN (µ0βH)

)

S =∂

∂T

(

−kT ln

(

2N

N !coshN (µ0βH)

))

= −k ln

(

2N

N !coshN (µ0βH)

)

− kT

[

∂

∂T

(

ln

2N

N !+ N ln cosh (µ0βH)

)]

= −k ln (Z) − kT

(

N

cosh (µ0βH)sinh (µ0βH) µ0H

∂β

∂T

)

(

∂β

∂T=

∂

∂T

(

1

kT

)

= −1

k

1

T 2= −

1

kT

1

T

)

= −k ln (Z) +

kT

kT

µ0NH

T

sinh (µ0βH)

cosh (µ0βH)

S = −k ln

(

2N

N !coshN (µ0βH)

)

+µ0NH

Ttanh (µ0βH) (9)

c) The magnetization m of a solid is defined as the net magneticmoment per unit volume. The average magnetic moment is definedvia 〈E〉 = −〈M〉H. For noninteracting moments, the magnetizationtypically obeys a Curie Law where m = χ0 H/T for vanishingly smallH, find the value of the constant χ0 for this problem.

If (as given in the problem statement),

〈E〉 = −〈M〉H (10)

then the average magnetic moment 〈M〉, (which is effectively equivalent to mag-netization m mentioned in the problem for large systems) using Eq. 6 is

〈M〉 = −〈E〉

H

m =µ0NH tanh (µ0βH)

Hm = µ0N tanh (µ0βH) (11)

Now, in the ”Curie Regime”, either for ”vanishingly small H” (i.e. H ≪ 1or for high temperatures (i.e T ≫ 1), the argument of the hyperbolic tangentµ0βH is also very small. However, in the case where |x| ≪ 1,

tanhx ≈ x (12)

So Eq. 11, yields what’s known as the Curie Constant χ0,

limT→∞

or limH→0

m = µ0Nµ0H

kT

3


m(T → ∞; H → 0) =µ2

0N

k

H

T

⇒ χ0 =µ2

0N

k(13)

Doneski.

4



Jeff Kissel

May 18, 2007


Spring 2007 #40 Part (d) removed.

An ideal non-relativistic gas of N spin-1/2 particles of mass m isconfined to move in a two dimensional area of size A = L2. Assumethe energy levels are close enough to be described as a continuousdensity of states.

(a) Find the density of states per unit area g(ǫ).

The ensemble average of any observable O is defined as

〈O〉 =∑

j

OjPj (1)

The density of modes g(ǫ) is the Jacobian of transformation when onechanges this sum to an integral. This conversion for a 2-D system of fermions(where we add a factor of 2 to account for the spin degeneracy) is as follows

∑

j

= 2∑

k

=2

h2

∫

∞

−∞

∫

∞

−∞

d2~pd2~q

=2

h2A

∫

∞

−∞

d2~p

=2A

h3(2π)

∫

∞

0

p dp

let ǫ =p2

2m

p =√

2m ǫ1/2

dp = 12

√2mǫ−1/2 dǫ

1


=4πA

h2

∫

∞

0

(√2m

ǫ1/2) 1

2

√2m

ǫ−1/2 dǫ

=4πA

h2

1

22m

∫

∞

0

dǫ

∑

j

⇒∫

∞

0

4mπA

h2ǫ2 dǫ ≡

∫

∞

0

g(ǫ) dǫ (2)

(b) Find the Fermi temperature in terms of the above quantities.

Here, we evaluate Eq. 1 for the total number of particles N , by summingover the probability distribution of fermions (as per Eq. 1),

nFD =1

eβ(ǫ−µ) + 1(3)

which at the limit of zero temperature splits into two regions (i.e. has values1 or 0). The non-vanishing portion of the distribution is in the energy regime0 < ǫ < ǫF , (the upper limit is the Fermi Energy, or chemical potential atzero temperature), because the Pauli Exclusion Principle dictates that fermions(spin-1/2 particles) cannot occupy the same energy state. Thus instead of allcollapsing to some ground state, they stack up in energy to ǫF when tempera-tures very small.

The result of the integral will yield N(ǫF ), which we can invert to get ǫF (N)

N =∑

j

nFD

NT=0 = limT→0

∫

∞

0

1

eβ(ǫ−µ) + 1g(ǫ) dǫ (4)

=

∫ ǫF

0

(1) g(ǫ) dǫ +

∫

∞

ǫF

(0) g(ǫ) dǫ

=

∫ ǫF

0

4πmA

h2dǫ

N =4πmA

h2ǫF

ǫF =Nh2

4πmA(5)

Finally, we make use of the equipartition theorem which states that every kineticdegree of freedom gets an energy of 1

2kBT . Since there are two degrees offreedom in this 2-D system,

E = kBT

So,

TF =ǫf

kB=

Nh2

4πmkBA(6)

2


(c) Find the total kinetic energy of the gas at zero temperature.

Again, we’ll use Eq. 1 for ensemble average energy, 〈E〉. Because we want thevalue at T = 0, the same limits apply as in part (b). Ready, steady, integrate.

〈E〉T=0 =

∫ ǫF

0

ǫ g(ǫ) dǫ

〈E〉T=0 =

∫ ǫF

0

ǫ4πmA

h2dǫ

=1

2ǫ2F

4πmA

h2

〈E〉T=0 =2πmA

h2ǫ2F (7)

and from Eq. 5,

〈E〉T=0 =2πmA

h2

(

h2

4πmAN

)2

=2πmA

h2

N2h4

16π2m2L4

=N2h2

8πmA

〈E〉T=0 =N2h2

8πmA(8)

(d) Find the chemical potential at all temperatures.

To find the chemical potential at all temperatures (instead of at zero tem-perature, which is the Fermi energy), we’ll find the total number of particles(Eq. 4), as in part (b). However, we can no longer divide the integral into tworegions, so we must solve the integral fully from 0 to ∞. Once we have N(µ),we can invert the expression for µ(N) as we did for the Fermi energy (whichwas the chemical potential at T = 0).

N =

∫

∞

0

nFD(ǫ) g(ǫ) dǫ

=

∫

∞

0

1

eβ(ǫ−µ) + 1

4πmA

h2dǫ

=4πmA

h2

∫

∞

0

1

eβ(ǫ−µ) + 1dǫ

let x = β(ǫ − µ) ⇒

dx = β dǫ

dǫ = kTx

0 → −µβ, ∞ → ∞

=4πmA

h2

∫

∞

−µβ

1

ex + 1kT dx =

4πmkT

h2A

∫

∞

−µβ

1

ex + 1dx

3


(

∫ b

a

1

ex + 1dx = [x − ln (ex + 1)]

ba ; λ ≡

(

h2

2πmkT

)1/2)

=A

λ2[x − ln (ex + 1)]

∞

−µβ(

limx→∞

[

x − ln (ex + 1)]

= x − ln (ex) = x − x = 0)

N =A

λ2

[

(0) −(

(−µβ) − ln (e(−µβ) + 1))]

Nλ2

A=

(

µβ + ln (e−µβ + 1))

eNλ2

A = eµβ eln (e−µβ+1) = eµβ (e−µβ + 1) = (1 + eµβ)

eµβ = eNλ2

A − 1

ln (eµβ) = ln (eNλ2

A − 1)

µβ = ln (eNλ2

A − 1)

µ = kT ln (eNλ2

A − 1) (9)

4


OLD VERSIONS

(d) Find the pressure exerted by the gas at zero temperature. dF =−pdA.

I’m pretty sure that the reason the equation dF = −pdA is to just to es-tablish the sign convention (some texts like to define work, W = dF as positivewhen leaving the system, i.e. W = PdV , and others like W = −P dV , i.e. pos-itive work is done when entering the system. This problem chooses the latter,but since we’re dealing with a 2-D system, dV → dA).

Anyways, because of the convention chosen, the thermodynamic identity is

d〈E〉 = T dS − P dA + µ dN (10)

We assume the number of particles stays constant at 〈N〉, so the third termvanishes, and the question desires the pressure at T = 0, so we’re left with arelation for pressure in terms of what we know from previous parts,

d〈E〉 = − P∂A|N ⇒ P = −(

∂〈E〉∂A

)

N

(11)

So from Eq. 8,

P = −N2h2

8πm

∂

∂A

(

1

A

)

= −N2h2

8πm

(

− 1

A2

)

P =N2h2

8πmA2(12)

5



Jeff Kissel

May 18, 2007

In the early universe, a chemical equilibrium between photons and

e+, e− particles was achieved via the conversion process

e+ + e− ↔ 2γ (1)

The energy of an electron or a positron is given by

E = mc2 +1

2mv2 (2)

Using the fact that photons have zero chemical potential, derive an

equation describing the concentrations n+ and n− of positrons and

electrons as related to each other at a given temperature.

The second law of thermodynamics demands that the entropy S of a givensystem and its surroundings must either remain the same (in equilibrium), orincrease (dStotal ≥ 0). Also, because of this fact the Gibbs free energy G eithermust either remain the same (in equilibrium) or decrease, i.e. dG ≤ 0. For apure system, with only one type of particle, this differential Gibbs free energyis

G ≡ E − TS + PV (3)

dG = dE − TdS − SdT + PdV + V dP

(Thermodynamic ID: dE = TdS − PdV + µdN) (4)

dG = −SdT + V dP + µdN (5)

At constant temperature T and pressure P (a safe assumption for the earlyuniverse), the first two terms vanish, leaving only

dG = µdN (6)

For a system of mixed particles (like in this problem) this generalizes to

dG = µ1dN1 + µ2dN2 + ... =∑

j

µjdNj (7)

1


when there are j types of particles in the system.OK. Now that we’ve got the basics outta the way, let’s get to the problem.

The question established that the positions e+, electrons e−, and photons γ arein equilibrium, so dG = 0. Using the coefficients of Eq. 1 as our dNj ’s, we find

0 =∑

j

µjdNj

= µe+Ne+ + µe−Ne− + µγNγ

= µe+(1) + µe−(1) + µγ(2)

0 = µe+ + µe− + 2µγ (8)

However, the chemical potential for photons is zero (either believe the problemstatement, or check out Schroeder pp 289-290), so

0 = µe+ + µe−

µe+ = −µe− (9)

But wait! The mass of positrons and electrons are the same, which meanstheir energies are the same (via Eq. 2). From Eq. 4, at constant S (we’rein equilibrium, so it’s OK) and at constant V (we’re taking the whole earlyuniverse as our “system;” also OK),

µ =

(

∂E

∂N

)

S,V

(10)

which means the chemical potentials must be equivalent. Hmm... the onlynumber that’s both equivalent and opposite to itself is zero. So,

µe+ = µe− = µγ = 0 (11)

Good to know. Now, the only thing left to do is find the concentrations Ne+

and Ne− .Positrons and electrons are both fermions, which means they obey the Fermi-

Dirac distribution,

〈 nj(Ej) 〉FD =1

eβ(Ej−µ) + 1(12)

where β = 1kT , Ej is the kinetic energy ǫ = 1

2mv2 = p2

2m plus the rest massmc2 (as in Eq. 2) of a state j. In the early universe, where we have to considerrelativistic energies Ej ≫ kT , the exponential will be huge, so we can ignore theadditive factor of 1. Also, we’ve already shown that the chemical potentials ofthe particles we’re working with are zero. This leaves a probability distributionof

〈 nj(Ej) 〉 =1

eβ(Ej−µ) + 1=

1

eβEj= e−βEj (13)

2


The total number of each particle, N defined by summing over the proba-bility distribution of states (i.e. Eq. 13),

N ≡∑

j

〈 nj(E) 〉 (14)

but as we’ve already established, there are lots of energy states so we can safelyconvert this to an integral, as long as we’ve got the conversion factor: the densityof states g(ǫ). ’Scuse me while I grab this ever-so-vital Jacobian,

∑

j

= 2∑

k

= 2

∫ ∞

−∞

∫ ∞

−∞

d3~p d3~q

= 2V

∫ ∞

−∞

d3~p =V

(2π)3

∫ ∞

−∞

d3~k

=2V

(2π)3(4π)

∫ ∞

0

k2 dk

let ǫ =p2

2m=

h2k2

2m⇒

dǫ = h2km dk

k =√

2mh2 ǫ

dǫ = h2

m

√

2mǫh2 dk

=2V

(2π)3(4π)

∫ ∞

0

(

2m

h2 ǫ

)

m

h2

√

h2

2mǫ−1/2 dǫ

=2V

(2π)3(4π)

√2

h3 m3/2

∫ ∞

0

ǫ1/2 dǫ

=V

2π2

2√

2

h3 m3/2

∫ ∞

0

ǫ1/2 dǫ

=

∫ ∞

0

V

2π2

(

2m

h2

)3/2

ǫ1/2 dǫ =

∫ ∞

0

g(ǫ) dǫ (15)

Aaaand back to business. The total number of each particle is

N ≡∑

j

〈 nj(E) 〉

=

∫ ∞

0

〈 n(E) 〉g(ǫ)dǫ

=

∫ ∞

0

(

e−βE)

(

V

2π2

(

2m

h2

)3/2

ǫ1/2

)

dǫ

E = mc2 + 1/2 mv2

= mc2 + p2

2mE = mc2 + ǫ

⇒ e−βE = e−βmc2

e−βǫ

3


=V

2π2

(

2m

h2

)3/2

e−βmc2

∫ ∞

0

e−βǫ ǫ1/2 dǫ

∫∞

0 xne−αx dx = Γ(n+1)αn+1

⇒∫∞

0 x1/2e−αx dx = Γ(3/2)

α3/2

Γ(n + 1) = n!

⇒ Γ(3/2) = (1/2)! =√

π2

=V

2π2

(

2m

h2

)3/2

e−βmc2

( √π

2β3/2

)

=V

4

(

2m

πβh2

)3/2

e−βmc2

=V

4

(

222mπkT

h2

)3/2

e−βmc2

= 2V

(

2mπkT

h2

)3/2

e−βmc2

let λ =

(

h2

2mπkT

)1/2

N = 2V

λ3e−βmc2

(16)

Phew. OK, thankfully we’re done: because energies are the same, and µ’sare same, this calculation for N is valid for both positrons and electrons. If thetotal number of each particle is the same, then the concentrations in the earlyuniverse are equivalent.

The more subtle purpose of this question was to show the reason behindone of the big unsolved questions in physics: we live in a real universe full ofelectrons (as opposed to an “anti-universe” full of positrons), i.e. in the earlyuniverse there must have been slightly more electrons than positrons. But wejust just proved that the concentrations were the same... W T F mate? Whelp,show how they’re different, and I’ll give ya a Nobel Prize.

4



Jeff Kissel

May 18, 2007

A collection of N bosons is contained in a volume V . The spin ofthe particles is 0.

a) Find the temperature at which Bose condensation occurs.First, we’ll find the total number of particles N in terms of the critical

temperature and volume, and then invert for the critical temperature Tc.To find N , we sum the Bose-Einstein distribution,

〈 nj 〉BE =1

e β(ǫj−µ) − 1(1)

over all j possible states. Here, ǫj =p2

j

2mis the kinetic energy of each state

(independent of position, q), µ is the chemical potential, and β = 1kT

.

N ≡∞∑

j=0

〈 nj 〉BE (2)

=

∞∑

j=0

1

e β(ǫj−µ) − 1

N =1

e−βµ − 1+

∞∑

j=1

1

e β(ǫj−µ) − 1(3)

Bose-Einstein condensation occurs when the chemical potential µ is approx-imately zero. Thus, we exclude the first term in Eq. 3 for now, as its limit goesto zero as µ approaches zero. Then we can change the second term to an inte-gral (valid for large N), and proceed finding expression for Tc. The conversioninvolves finding the density of states for bosons, so let’s take a detour for a bit:

∑

j

=

∫ ∞

−∞

∫ ∞

−∞

d3~p d3~q

= V

∫ ∞

−∞

d3~p =V

(2π)3

∫ ∞

−∞

d3~k

1


=V

(2π)3(4π)

∫ ∞

0

k2 dk

let ǫ =p2

2m=

~2k2

2m⇒

dǫ = h2km

dk

k =√

2m~2 ǫ

dǫ = ~2

m

√

2mǫ~2 dk

=V

(2π)3(4π)

∫ ∞

0

(

2m

~2ǫ

)

m

~2

√

~2

2mǫ−

1

2 dǫ

=V

(2π)3(4π)

√2

~3m

3

2

∫ ∞

0

ǫ1

2 dǫ

=V

(2π)3(4π)

(2π)3

h3

√2 m

3

2

∫ ∞

0

ǫ1

2 dǫ

=

∫ ∞

0

2V√π

(

2πm

h2

)3

2 √ǫ dǫ ≡

∫ ∞

0

gBE(ǫ) dǫ (4)

Ok. Back on course, remember we’re finding N at the critical temperature Tc,where µ ≈ 0.

N(Tc) =

∞∑

j=1

〈 nj(µ ≈ 0) 〉BE

=

∫ ∞

0

〈 n(µ ≈ 0) 〉BE gBE(ǫ) dǫ

=2V√

π

(

2πm

h2

)3

2∫ ∞

0

1

e β(ǫ−µ) − 1

√ǫ dǫ

=2V√

π

(

2πm

h2

)3

2∫ ∞

0

ǫ1

2

e βǫ − 1dǫ (5)

let x = βǫ ⇒

dx = βdǫdǫ = kT dx

ǫ1

2 = (kT )1

2 x1

2

=2V√

π

(

2πm

h2

)3

2∫ ∞

0

(kTc)1

2 x1

2

ex − 1kTc dǫ

=2V√

π

(

2πm

h2

)3

2

(kTc)3

2

∫ ∞

0

x1

2

ex − 1

∫∞

0xn−1

ex−1 dx = Γ(n)ζ(n)

Γ(32 ) =

√π

2 ζ(32 ) ≈ 2.612

= V

2√π

(

2πmk

h2

)3

2

T3

2

c

√

π

2ζ(

3

2)

2


N = ζ(3

2) V

(

2πmk

h2

)3

2

T3

2

c

T3

2

c =1

ζ(32 )

N

V

(

h2

2πmk

)3

2

Tc =h2

2πmk

(

1

ζ(32 )

N

V

)2

3

(6)

Hooray! The critical temperature for Bose-Einstein condensation.

b) Find how the number of particles in the lowest energy state varieswith temperature below the condensation temperature.

If the particles are in a 3 dimensional box, the particle’s deBroglie wavelengthmust be half integer multiples of the length L of a side. Hence,

L =1

2nλn ⇒ λn =

2L

n(7)

and therefore the momentum in any given direction is

pi = hλn

=hni

2L(8)

The individual kinetic energies in this 3-D box define a sphere in n-space,

ǫ =|~p|22m

=1

2m

(

(

hnx

2L

)2

+

(

hny

2L

)2

+

(

hnz

2L

)2)

ǫ =h2

8mL2(n2

x + n2y + n2

z) (9)

where n is the radius of the sphere in n-space. The lowest energy state will thenbe

ǫ0 = h2

8mL2 (12 + 12 + 12) =3h2

8mL2(10)

which is a very small energy as long as L is macroscopic. The number of particlesin this ground state is given by Eq. 1, at ǫj = ǫ0,

N0 =1

e β(ǫ0−µ) − 1(11)

When temperatures are below the critical temperature, N0 gets quite large.Now look back at the integrand of Eq. 5: as ǫ goes to zero, the density of

states (the part proportional to√

ǫ ) goes to zero while the Bose-Einstein distri-bution blows up (in proportion to 1/ǫ). Although the product is an integrable

3


function, it’s not clear that this infinite spike at ǫ = 0 correctly correspondsto the sum over all states (Eq. 3). In fact, we’ve already seen from Eq. 11that the number of atoms in the ground state can be enormous when µ ≈ 0,and this number was not included in the calculation of the critical temperaturein part (a). On the other hand, the integral should correctly approximate thenumber of exited particles not in the ground state with energies far away fromthis divergence, i.e. ǫ ≫ ǫ0. If one imagines cutting off the integral at a lowerlimit that is somewhat greater than ǫ0 but much less than kT , one still getsapproximately the same answer,

Nexcited = ζ(3

2) V

(

2πmkT

h2

)3

2

(For T < Tc) (12)

or more simply,

Nexcited =

(

T

Tc

)3

2

N (T < Tc) (13)

If these are all the excited particles, the rest must be in the ground state, andthus it follows that

N = N0 + NexcitedN0 = N − Nexcited

= N −(

T

Tc

)3

2

N

N0 =

(

1 −(

T

Tc

)3

2

)

N (T < Tc) (14)

Props to Schroeder (2000), pgs. 315-318.

4



Jeff Kissel

May 18, 2007

The average energy of a system in thermal equilibrium is 〈E〉.a) Prove that the mean square deviation of the energy 〈 (E − 〈E〉)2 〉is given by

〈 (E − 〈E〉)2 〉 = kBT 2CV (1)

where CV is the heat capacity of the system at constant volume andkB is Boltzmann’s constant.

I’ll proceed by attacking the problem from both sides of Eq. 1.The left-hand side, can by simplified to something a little more reasonable

to decipher, and perhaps more recognizable (note I use some shorthand and callthe mean square deviation of the energy as 〈σ2〉):

〈σ2〉 = 〈 (E − 〈E〉)2 〉= 〈 E2 − 2 E 〈E〉 + 〈E〉2 〉

(

Ensemble average is linear: 〈 A + B 〉 = 〈A〉 + 〈B〉)

= 〈E2〉 − 2〈E〉〈〈E〉〉 + 〈E〉2(

Ensemble average is a constant ⇒ 〈〈E〉〉 = 〈E〉)

= 〈E2〉 − 2〈E〉2 + 〈E〉2

〈σ2〉 = 〈E2〉 − 〈E〉2 (2)

which looks a lot like the definition of the variance that we’re used to fromstatistics.

Now let’s figure out some alternate definitions of each term in Eq. 2. Theensemble average of any operator O is defined as

〈O〉 ≡∑

j

Oj Pj =1

Z

∑

j

Oj e−βEj (3)

where P is the probability distribution of states, Z ≡∑

j e−βEj is the partition

function, β = (kT )−1, and Ej is the energy of the jth state.

1


The ensemble average energy, using Eq. 3, can therefore be written as

〈E〉 =1

Z

∑

j

Ej e−βEj

(

∂

∂β

∑

j

e−βEj =∑

j

(− Ej) e−βEj

)

=1

Z

(

− ∂

∂β

∑

j

e−βEj

)

(

Z ≡∑

j

e−βEj

)

〈E〉 = − 1

Z

∂Z

∂β(4)

which means

〈E〉2 =

(

− 1

Z

∂Z

∂β

)2

=1

Z2

(

∂Z

∂β

)2

(5)

Similarly, the ensemble average of energy squared is

〈E2〉 =1

Z

∑

j

E2

j e−βEj

(

∂∂β

∑

j e−βEj =∑

j (− Ej) e−βEj

∂2

∂β2

∑

j e−βEj =∑

j E2

j e−βEj

)

=1

Z

∂2

∂β2

(

∑

j

e−βEj

)

〈E2〉 =1

Z2

∂2Z

∂β2(6)

So, from Eqs. 5 and 6, Eq. 2 becomes

〈σ2〉 = 〈E2〉 − 〈E〉2 =1

Z2

∂2Z

∂β2− 1

Z2

(

∂Z

∂β

)2

(7)

On to the right-hand side of Eq. 1. The heat capacity CV is defined as

CV ≡(

∂〈E〉∂T

)

V,N

(8)

We can convert the derivative with respect to T to one with respect to β bynoting the following,

∂β

∂T=

∂

∂T

(

1

kT

)

= − (kT )−2(k) = − k

(kT )2= − 1

kT 2

2


CV is then

CV =∂〈E〉∂T

=∂〈E〉∂β

∂β

∂T= − 1

kT 2

∂〈E〉∂β

(

Eq. 4: 〈E〉 = − 1

Z

∂Z

∂β

)

= +1

kT 2

∂

∂β

(

1

Z

∂Z

∂β

)

kT 2CV =

[

1

Z

∂

∂β

(

∂Z

∂β

)

+

(

∂

∂β

1

Z

)

∂Z

∂β

]

=1

Z

∂2Z

∂β2+

(

− 1

Z2

∂Z

∂β

)

∂Z

∂β

kT 2CV =1

Z

∂2Z

∂β2− 1

Z2

(

∂Z

∂β

)2

(9)

which, from Eq. 7 is exactly the MS deviation,

kT 2CV = 〈σ2〉 = 〈E2〉 − 〈E〉2 √(10)

b) Use this result to show that the energy of a macroscopic systemmay ordinarily be considered constant when the system is in thermalequilibrium.

I’ll show this be proving that the standard deviation in energy σrms is verysmall compared with the energy 〈E〉.

We only have a definition for 〈σ2〉, so I’ll just prove the deviation is smallusing that.

σrms

〈E〉 =

( 〈σ2〉〈E〉2

)1/2

=

( 〈E2〉 − 〈E〉2〈E〉2

)1/2

Eq. 1 : 〈E2〉 − 〈E〉2 = kT 2CV

Equipartition Thm. : 〈E〉 = f2

NkT

Eq. 8 : CV = ∂〈E〉∂T = f

2Nk

=

kT 2

(

f2Nk)

(

f2

NkT)2

1/2

=

(

4 kT 2 f2(Nk)

f2(Nk)2T 2

)1/2

=

(

2 kT 2

fN kT 2

)1/2

σrms

〈E〉 =

(

2

fN

)1/2

(11)

For macroscopic systems, in which N is huge, this ratio is really, really, really,really, ridiculously small. For example, for 1 mole (N = NA = 6.02 × 1023) ofmonatomic gas (f = 3),

σrms

〈E〉 = 1.05234× 10−12

3



Jeff Kissel

May 18, 2007

Consider a container of volume 100 cm3 containing a classical idealgas at 1 atm pressure and 350 K.

a) Find the number of particles.

Yo, wassup ideal gas law! (P is pressure, V is volume, N is total number ofparticles, k is Boltzmann’s constant, and T is temperature.)

PV = NkT (1)

N =PV

kT

P = 1 atm = 1.013× 105 PaV = 100 cm3 = 1 × 10−4 m3

k = 1.381 × 10−23 J K−1

=(1.013 × 105 Pa)(1 × 10−4 m3)

(1.381 × 10−23 J K−1)(350 K)

N = 2.0958× 1021 particles (2)

b) Compute the mean kinetic energy of a particle in the gas.

Holler at your Equipartition Theorem! (f is the number of degrees of freedomfor the gas particle.)

〈E〉 =f

2NkT (3)

(Assume the gas is monatomic, i.e. f = 3)

=3

2(2.0958× 1021 particles)(1.381× 10−23 J K−1)(350 K)

〈E〉 = 15.159 J (4)

c) Suppose one counted the number of particles in a small sub-volumeof size 0.1 micron on a side. What is the probability of finding noparticles in this volume?

1


If the small sub-volume v has sides of length L, then the sub-box has volumev = L3. Thus, there must be certain number Nv = V/v sub-volumes in the totalvolume V . Now imagine emptying box V of particles. If we toss one particle inthe box, the probability that it will not land in a given sub-box will be

P

(

one particledoes not land in v

)

=

(

Nv − 1

Nv

)

(5)

The next particle we toss in will have the exact same probability, and the oneafter that, and so on. Thus, if we toss N particles in V , the probability that no

particles will be found in the sub-volume v is

P

(

none of N particlesare in v

)

=

(

Nv − 1

Nv

)N

(6)

=

(

V

v− 1V

v

)N

=

(

v

V

(

V

v− 1

))N

P

(

none of N particlesare in v

)

=(

1 −v

V

)N

(7)

For V = 1 × 10−4 m3 and v = 0.1 µm = 1 × 10−21 m3, the ratio of v/V is1 × 10−17. This means what’s in parentheses is (1− [retarded small]) ≈ 1, andtherefore P = 1N = 1. (Graphing calculators and even Mathematica, aftertaking the natural logarithm, i.e. computing N ln (1 − v/V ), still say that it’sone).

No particles will be found in a 0.1 µm sub volume.

2



Jeff Kissel

May 17, 2007

Exam Iteration Question Number Notes/ChangesSpring 2006 #35 –Fall 2006 #46 Question number shift

Spring 2007 #46 Comment added to part (c)

The pressure in a vacuum system is 10−3 mm Hg. The external pres-sure is 1 atm at 300 K. This is a pinhole in the vacuum system ofarea 10−10 cm2. Assume that any molecule entering the pinhole goesthrough. Use an average molecular weight of air as 29 amu.

(a) How many molecules enter the vacuum system each hour?

The process in which gas at high pressure is moving freely through a pinholeinto a much lower pressure is known as effusion. Effusion problems are usuallydefined such that the flow of particles is from a confined system, at high pressure,outward into an unconfined vacuum. This problem, which has particles leakinginto a confined vacuum, is very similar. The resulting equation of particle flowrate dN/dt will differ only by a negative sign to account for the directionaldifference. Thus, I’ll derive the flow rate for outward leaking gas, and changethe sign at the end.

The pressure P on an area A of the container wall caused by a single airparticle of mass ma over a time period ∆t is

P = ≡ − F

A= −

(

ma∆v

∆t

)

1

A(1)

If we (arbitrarily) chose a reference frame so the particle moves in the x-direction, with average velocity vx, then ∆v as it bounces of the wall in time ∆tshould be 2 vx. This velocity can be approximated by the root-mean-squaredvelocity

√vx = vrms

x defined by the temperature and mass of the particle viathe Equipartition Theorem,

〈E〉Kin = 〈E〉Therm

1


1

2mav2

x =1

2kT

1

2ma v2

x =1

2kT

v2x =

kT

ma√

v2x =

√

kT

ma

≡ vrmsx (2)

This turns Eq. 1 into

P = − ma

∆v

∆t

1

A

= − ma

2 vx

∆t

1

A

≈ − ma

2 vrmsx

∆t

1

A

P = − 2ma

A∆t

√

kT

ma

(3)

For an infinitesimal amount of (non-interacting) air particles dN , the pres-sure of the gas trying to escape becomes

P = − 2(madN)

A dt

√

kT

ma

= − 2ma

A

√

kT

ma

dN

dt(4)

This pressure should be approximately equivalent to the pressure on a wallwith a pinhole poke in it, as the area of the hole (A = 10−10 cm2) is approx-imately the size of the air molecule, allowing barely more than one particlethrough at a time.

We can now re-arrange for the flow rate of air particles dN/dt from a pres-surized contain outward into a vacuum,

dN

dt= − PA

2ma

√

ma

kT(5)

As I’ve mentioned earlier, I’ll now flip the sign, because the direction of flowis opposite from that which we’ve used to derive the flow rate. Thus, the flowrate of particles into a confined vacuum from a pressurized exterior is

dN

dt=

1

2P A

√

1

makT(6)

(Note that I’ve combined the ma’s).Finally, we plug in the numbers given in the problem statement, converting

to mks units. For Pa = 1 atm = 1.013 × 105 Pa, Apinhole = 10−10 cm2 = 1 ×

2


10−14 m2, ma = 29 amu = (29 amu)×(1.661×10−27 kg/amu) = 4.8×10−26 kg,k = 1.381 × 10−23 J K−1, and T = 300 K, the flow rate is

dN

dt=

1

2(1.013 × 105 Pa) (1 × 10−14 m2)

×√

1

(4.8 × 10−26 kg)(1.381 × 10−23 J K−1)(300 K)

= 3.5854× 1013 particles s−1(

3600 s hr−1)

dN

dt= 1.3 × 1017 particles hr−1 (7)

(b) If the volume of the system is 2 liters, by how much does thepressure rise in 1 hour?

Assuming the air obeys the ideal gas law (quite safe at room temperature),we can find the change in pressure by computing

V ∆P = kT∆N

∆P =kT

V∆N (8)

assuming that effusion happens slow enough such that the system remains inthermal equilibrium, i.e. T is constant.

Having computed the flow rate of particles into the vacuum (Eq. 7), overone hour the number of particles in the chamber should increase by

∆N = Nf − Ni =dN

dt× t

= (1.3 × 1017 particles hr−1) × (1 hr)

∆N = 1.3 × 1017 particles (9)

which if volume remains a constant V = 2 L = (2 L)(0.001 L/m3) = 0.002 m2

at room temperature (T = 300 K), the change in pressure after the same hourshould be

∆P =kT

V∆N

=(1.381× 10−23 J K−1)(300 K)

(0.002 m3)(1.3 × 1017 particles)

∆P = 0.267 Pa = 0.002005 mm Hg (10)

(c) How long does it take for the pressure to rise to 750 mm Hg? Note:this is close to the pressure outside the vacuum tank.

We now have a rate for pressure increase from part (b), (assuming the pres-sure increases linearly with time as the ideal gas law dictates),

dP

dt≡ η = 2.005 × 10−3 mm Hg hr−1 (11)

3


which means the it will take

dP = η dt

∆P = η ∆t

(Reset clock such that ti = 0)

∆P = η t

t =∆P

η=

(750 mm Hg)

(2.005× 10−3 mm Hg hr−1)

t = 3.74 × 105 hr ( = 42 yrs !) (12)

to increase the pressure 750 mm Hg.

4



Jeff Kissel

May 18, 2007

Two phases of a pure material coexist along a line Pcoex(T ). Use mA asthe molar mass of the substance. The latent heat of transformation(in J/mole) is L, and the molar volume of each phase are v1 and v2 re-spectively. Assume that phase 1 is the stable phase for temperatureslower than the temperature where they coexist.

a) What is known about the chemical potentials of each phase in thelow temperature region where phase 1 exists, the high temperatureregion where phase 2 exists, and on the coexistence line?

Chemical potential µ is defined as the Gibbs free energy per particle. Theproof is as follows,

G ≡ E − TS + PV (1)

dG = dE − TdS − SdT + PdV + V dP

(Thermodynamic ID: dE = TdS − PdV + µdN) (2)

dG = −SdT + V dP + µdN (3)

Taking the partial derivative of both sides of Eq. 3 with respect to N , holdingT and P constant,

µ =

(

∂G

∂N

)

T,P

(4)

or more simply, since we’ve assumed the number of particles is fixed in additionto temperature and pressure,

G = Nµ (5)

Phase transitions are determined when two phases are in chemical equilib-rium with each other (for example a solid phase in equilibrium with a liquidphase is the melting/freezing phase transition). In order to achieve an overallsystem (system A and the reservoir) increase in entropy (essential for a sponta-neous process), the Gibb free energy is minimized in equilibrium:

dStotal = dSA + dSR ≥ 0 (6)

1


For each part of the system, Eq. 2 applies, (used here in the form dS = dET

+PT

dV −µT

dN) such that for fixed T , P , and N

dStotal = dSA +1

TdER +

P

TdVR −

>0

µ

TdN

(dER = − dER; dVR = − dVA)

= dSA −dEA

T−

P

TdVA

=1

T(TdSA − dEA − PdVA)

= −1

T(dEA − TdS + PdVA)

= −1

T(µdN)T,P

(

(dG)T,P = :0−SdT −:0

V dP + µdN

)

dStotal = −1

TdG (7)

Thus, the temperature and pressure line where the Gibbs free energy, and(via Eq. 5) the chemical potential of each phase is equivalent defines the phasetransition. Therefore, on the coexistence linePcoex(T ), the chemical potentialsare equivalent.

By inspection, one can then infer that at low temperatures where phase 1exists, phase 1 has lower chemical potential than phase 2, i.e. µ1 < µ2 (lowchemical potential corresponds to stronger attraction). At high temperatureswhere phase 2 exists, µ2 < µ1.

b) Derive the Clausius-Clapeyron equation for the slope of the coex-istence curve.

dPcoex

dT=

L

T (v1 − v2)(8)

As established in part (a), the Gibbs free energies along the coexistencecurve for two phases are equivalent. So, for constant N , using Eq. 3,

G1 = G2

−S1dT + V1dP +:0µ1dN = −S2dT + V2dP +:0

µ2dN

(V1 − V2) dP = (S1 − S2) dT

dP

dT=

(S1 − S2)

(V1 − V2)(9)

The molar latent heat L is defined as L ≡ Q/m where Q is the heat requiredto complete a particular phase transition and m is the molar mass. During thisphase transition, temperature is constant, and for such an isothermal process

2


∆S = S1 − S2 = Q/T . Thus Eq. 9 becomes the Clausius-Clapeyron (C-C)relation,

dP

dT=

(S1 − S2)

(V1 − V2)

=Q

T (V1 − V2)

m

m

=Q/m

T (V1/m − V2/m)

dP

dT=

L

T (v1 − v2)(10)

c) If the higher temperature phase 2 can be treated as an ideal gas,and molar volume of phase 1 is so small compared to v2 as to beneglected, find the dependence of the saturated vapor pressure ontemperature.

For high temperatures (as the question dictates), v2 can be neglected, so theC-C relation becomes

dP

dT=

L

Tv1(11)

From the ideal gas law, the molar volume is

PV = NkT or PV = nRT

⇒ Pv = RT

v =RT

P(12)

which means Eq. 11 yields

dP

dT=

LP

RT 2

dP

P=

L

RT 2dT

∫ P

P0

1

PdP =

L

R

∫ T

T0

1

T 2dT

ln

(

P

P0

)

= −L

R

(

1

T−

1

T0

)

P

P0= e

L

R( 1

T0−

1

T)

Psvp(T ) = P0eL

R( 1

T0−

1

T) (13)

which is the saturated vapor pressure as a function of temperature. P0 is theexperimental value of the pressure at temperature T0 (usually room tempera-ture), used as calibration. This is called the “saturated” vapor pressure becausewe ignored v2, i.e. claiming that the system is saturated with phase 1.

3



Jeff Kissel

May 18, 2007

Consider a collection of N particles of spin-1/2 in a volume V . Com-pute the magnetic spin susceptibility of these particles if they can betreated as a non-degenerate gas, but

The order in which we’ll attack both parts of this problem is to find (1)the partition function Z for the system, (2) the ensemble average magneticmoment 〈µ〉, (3) the magnetization per unit volume M , and (4) the magneticsusceptibility χm. In equation form, that’s

Z1 ≡∑

j

eβǫj ⇒ Z =1

N !ZN

1 (1)

⇒ 〈µ〉 ≡∑

j

µjPj → 〈µ〉 = − ∂

∂(βB)ln(Z) (2)

⇒ M =〈µ〉V

⇒ χm =M

B(3)

a) the spin is still treated quantum mechanically.

To “treat the spin quantum mechanically” implies that the magnetic portionof energy states are discrete, i.e. in the presence of an external magnetic field ofstrength B, the fermions (spin-1/2 particles) have only the “spin up” (aligned

with B, and energy ǫ↑ = −µB) and “spin down” (anti-aligned with B, ǫ↓ =+µB). The total energy of a given state is then the usual continuous, free-particle kinetic portion (independent of position) and the discrete, spin-magneticfield interaction portion (independent of position and momentum),

ǫ =p2

2m± µB (4)

This means the partition function for one particle Z1 is

Z1 ≡∑

j

e−βǫj

= eβµB∑

j

e−βp2

2m + e−βµB∑

j

e−βp2

2m

1


=(eβµB + eβµB

) ∑

j

e−βp2

2m

(

coshx =1

2

(ex + e−x

))

= (2 cosh (βµB))

(1

h3

∫ ∞

−∞

∫ ∞

−∞

e−β p2

2m d3~p d3~q

)

= 2 cosh (βµB)V

h3

∫ ∞

−∞

e−β p2

2m d3~p

= 2 cosh (βµB)V

h3

∫ ∞

−∞

e−β p2

2m d3~p

= 2 cosh (βµB)V

h3

[∫ ∞

−∞

e−β p2

2m dp

]3

(∫ ∞

−∞

e−αx2

dx =

√π

α

)

= 2 cosh (βµB)V

h3

(2m

βπ

)3/2

= 2V cosh (βµB)

(2πmkT

h2

)3/2

let λ =h2

2πmkT

Z1 =2V

λ3cosh (βµB) (5)

For a system of N such particles, which are indistinguishable, the partitionfunction (from Eq. 1) is

Z =1

N !

(2V

λ3

)N

coshN (βµB) (6)

Onto the average magnetic moment. Any generic (ensemble) average of anobservable is defined as

〈O〉 =∑

j

O(j)P(j) =1

Z

∑

j

Ojeβǫj (7)

which means the average magnetic moment is

〈µ〉 =1

Z

∑

j

µj e−βǫj

=

∑

j µ e−β p2

2m e±µ βB

∑

j e−β p2

2m e±µ βB

=

∂∂(βB)

(∑

j e−β p2

2m e±µ βB)

∑

j e−β p2

2m e±µ βB

2


=1

Z

∂

∂(βB)(Z)

〈µ〉 =∂

∂(βB)(ln(Z)) (8)

and plugging in Eq. 6

〈µ〉 =∂

∂(βB)

ln

[(1

N !

2V

λ3

)N

coshN (µ βB)

]

=∂

∂(βB)

[

ln

(1

N !

2V

λ3

)N

+ ln(

coshN µ βB)]

=

∂

∂(βB)

[

ln

(1

N !

2V

λ3

)N]

+∂

∂(βB)[N ln (cosh (µ βB))]

(d

ducoshu = sinhu du

)

=N sinh (µβB) (µ)

cosh (µβB)

〈µ〉 = Nµ tanh (µβB) (9)

The magnetization per unit volume is then

M =〈µ〉V

M =Nµ

Vtanh (µβB) (10)

Finally, the magnetic susceptibility for a collection of spin-1/2 particles,whose spin is treated quantum mechanically is

χm =M

B=

Nµ

V Btanh (µβB) (11)

b) their magnetic moments are treated as a classical magnetic momentof µ =

√3µB (recall that s2 = 3/4 for a spin-1/2 systems), but that

the moment can point in any direction, so ǫ = −µB cos(θ), with thecomponent of the magnetization along B being given by µ cos(θ).

Again, we begin by establishing the energies of each state. As before,there remains the continuous free-particle, kinetic energy portion, yet the spin-magnetic field interaction term is now continuous as well and no longer inde-pendent of position, i.e.

ǫ =p2

2m− µB cos θ (12)

3


which means we can’t pull that portion out of the (single particle) partitionfunction as we did in part (a). So,

Z1 =∑

j

e−βǫj

=1

h3

∫ ∞

−∞

∫ ∞

−∞

e−β

(p2

2m−µB cos θ

)

d3~p d3~q

=1

h3

∫ ∞

−∞

∫ ∞

−∞

e−β p2

2m e+βµE cos θ d3~p d3~q

Z1 =

1

h3

∫ ∞

−∞

e−β p2

2m d3~p

︸︷︷︸

Zp

∫ ∞

−∞

e+βµB cos θ d3~q

︸︷︷︸

Zq

(13)

The momentum portion Zp computes very similar to part (a), just with outthe volume,

Zp =1

h3

∫ ∞

−∞

e−β p2

2m d3~p

=1

h3

∫ ∞

−∞

e−β p2

2m d3~p

=1

h3

[∫ ∞

−∞

e−β p2

2m dp

]3

=1

h3

(2m

βπ

)3/2

Zp =

(2πmkT

h2

)3/2

=1

λ3(14)

The position portion of the partition function Zq (independent of momen-tum) is a little trickier, but still doable,

Zq =

∫ ∞

−∞

e+βµB cos θ d3~q

=

∫ 2π

0

∫ π

0

∫ ∞

0

e+βµB cos θ r2 sin θ drdθdφ

if∫

d3~q = Vthen V =

∫ ∫ ∫r2 sin θ drdθdφ

V = 4π∫

r2dr⇒

∫r2dr = V/4π

=V

4π

∫ 2π

0

∫ π

0

e+βµB cos θ sin θ dθdφ

4


=V

4π(2π)

∫ π

0

e+βµB cos θ sin θ dθ

let u = cos θ

du = − sin θ dθ−du = sin θ dθ

0 → 1 π → −1

=V

2

(

−∫ −1

1

eβµBu du

)

=V

2

∫ 1

−1

eβµBu du

=V

2

[1

βµBeβµBu

]1

−1

=V

βµB

1

2

(eβµB − e−βµB

)

(

sinh(x) =1

2

(ex − e−x

))

Zq =V

βµBsinh (βµB) (15)

Combining Eqs. 14 and 15, the partition function for a single particle, andthus for the whole system is

Z1 =V

λ3

sinh (βµB)

βµBsinh (βµB)

Z =

(V

µλ2

)N (sinh (µ βB)

βB

)N

(16)

From here, the rest is just plug and chug, in the same manner as in part (a):find 〈µ〉, then M , and finally χm. ’Ere we go!

〈µ〉 =∂

∂(βB)(ln (Z))

=∂

∂(βB)

ln

[(V

µλ2

)N (sinh (µ βB)

βB

)N]

=

∂

∂(βB)

[

N ln

(V

µλ2

)]

+∂

∂(βB)

[

N ln (sinh (µ βB)) − N lnβB]

(d

dusinh(u) = cosh(u)du

)

=

(N cosh (µβB)(µ)

sinh (µβB)

)

− N

βB

= Nµ coth (µβB) − N1

βB

〈µ〉 = Nµ

(

coth (µβB) − 1

µβB

)

(17)

5


M =〈µ〉V

=Nµ

V

(

coth (µβB) − 1

µβB

)

(18)

χm =M

B

=Nµ

V B

(

coth (µβB) − 1

µβB

)

(19)

and finally, plugging in µ =√

3µB as the problem desires, the magnetic sus-ceptibility for a collection of spin-1/2 particles, whose spin is treated classicallyis

χm =3µBN

V B

(

coth(√

3µBβB)

− 1√3µBβB

)

(20)

6



Jeff Kissel

May 18, 2007

An object of heat capacity C is used as the cold thermal reservoirby a Carnot engine. The hot reservoir has an infinite heat capacity.During the operation of the engine to produce work, the temperatureof the cold reservoir will slowly rise. Assume the starting temperatureof the cold reservoir is Tc and the constant temperature of the hotreservoir is Th.

a) To what temperature will the cold reservoir before the engineceases to produce work?

To cease producing work W , the engine’s efficiency must go to zero. Effi-ciency η is defined as

η ≡benefitcost =

W

Qh

(1)

where Qh is the heat cost, taken from the hot reservoir. For a closed cycle(Carnot) engine, the first law of thermodynamics tells us that energy in thecycle must be conserved. Thus, the heat absorbed must be equal to the heatreleased Qc plus the work produced, or

Qh = Qc + W or W = Qh − Qc (2)

We can plug this into Eq. 1 to get

η =Qh − Qc

Qh

= 1 −Qc

Qh

(3)

The second law of thermodynamics states that the entropy S ≡ Q/T ofany system plus its surroundings can’t decrease. Here, since the engine cycle isclosed, this means the entropy expelled to the cold reservoir must be at least asmuch as that absorbed from the hot reservoir, i.e.

Qc

Tc

≥Qh

Th

⇒Qc

Qh

≥Tc

Th

(4)

Now, or equation for efficient becomes

η ≤ 1 −Tc

Th

(5)

1


The engine’s efficiency will drop to zero once the second term in Eq. 5 goesto unity, in other words when Tc = Th. So, when the cold reservoir (originalyas Tc) reaches that of the hot reservoir Th, the engine will cease to do work.

b) How much heat flows into the cold reservoir until the engine stopsproducing work?

When one asks “how much heat can flow into an object,” we’re concernedabout its heat capacity, C, defined as

C ≡Q

∆T(6)

The engine will stop doing work once the maximum amount heat has flowedinto the cold reservoir, i.e. when the engine has increased the temperature fromits current Tc to Th, and described in part (a). To find the amount of heat thisrequires, we re-arrange Eq. 6, using this temperature change as ∆T ,

Qc→h = C∆Tc→h = C (Th − Tc) (7)

c) What is the total amount of work the engine can produce beforethe process stops?

As the engine does work, it will steadily increase the temperature of thecold reservoir to Th, at which point work will cease. As stated in part (a), theentropy released to the cold reservoir must either remain constant or increaseduring a cycle. We want the maximum amount of work, so we’ll assume thatno entropy “escapes” during a given cycle such that

dQ′

c

T ′

c

=dQ′

h

T ′

h

(8)

Because the temperature of the cold reservoir is increasing every cycle, theamount of heat that can be given decreases over time. Incrementally this heatis dQ′

c= CdT ′

c, from Eq. 6. So, Eq. 8 becomes,

CdT ′

c

T ′

c

=dQ′

h

T ′

h

⇒ dQ′

h = CT ′

h

T ′

c

dT ′

c (9)

The amount of work produced by this incremental cycle in terms of tem-perature, from re-arranging Eq. 1, plugging in Eq. 5 for η, and Eq. 9 fordQ′

h

dW = η dQ′

h

=

(

1 −T ′

c

T ′

h

)(

CT ′

h

T ′

c

dT ′

c

)

dW = C

(

T ′

h

T ′

c

− 1

)

dT ′

c(10)

2


The total work produced by the engine as the temperature of the cold reser-voir increases from some initial T ′

c = Tc to Th when work ceases is

W =

∫ Th

Tc

C

(

Th

T ′

c

− 1

)

dT ′

c

= CTh

∫

Th

Tc

1

T ′

c

dTc − C

∫

Th

Tc

dTc

= C

(

Th

[

ln (T ′

c)]Th

Tc

−[

T ′

c

]Th

Tc

)

= C

(

Th ln

(

Th

Tc

)

− Th + Tc

)

W = C

[

Th

(

ln

(

Th

Tc

)

− 1

)

+ Tc

]

(11)

3



Jeff Kissel

May 18, 2007

Exam Iteration Question Number Notes/ChangesSpring 2006 X –Fall 2006 #52 Added to question bank

Spring 2007 #52 Almost entirely reworded.

A certain gas has an equation of state

P = b +NRT

V(1)

where N is the mole number, V is the volume, T is the absolutetemperature and b is a constant.

(a) Find the value of CP −CV . where CP and CV are the heat capacitiesat fixed pressure and volume.

The equation we need for the difference in heat capacities is

CP − CV = T

(

∂P

∂T

)

V,N

(

∂V

∂T

)

P,N

(2)

which I’ll take a few lines to prove. The variables used will be entropy S, tem-perature T , pressure P , volume V , chemical potential µ, total internal energy E,and total number of particles N . Subscripts on parentheses of partial derivativesindicate what is being held constant.

S = S(T, V, N)

dS =

(

∂S

∂T

)

V,N

dT +

(

∂S

∂V

)

T,N

dV +

(

∂S

∂N

)

T,V

dN

(

∂S

∂T

)

P,N

=

(

∂S

∂T

)

V,N

(1) +

(

∂S

∂V

)

T,N

(

∂V

∂T

)

P,N

+

(

∂S

∂N

)

T,V

(0)

(

∂S

∂T

)

P,N

=

(

∂S

∂T

)

V,N

+

(

∂S

∂V

)

T,N

(

∂V

∂T

)

P,N

1


(

CP ≡ T(

∂S∂T

)

P,N⇒

(

∂S∂T

)

P,N= 1

TCP

CV ≡ T(

∂S∂T

)

V,N⇒

(

∂S∂T

)

V,N= 1

TCV

)

(3)

1

TCP =

1

TCV +

(

∂S

∂V

)

T,N

(

∂V

∂T

)

P,N

CP − CV = T

(

∂S

∂V

)

T,N

(

∂V

∂T

)

P,N(

Maxwell relation:

(

∂S

∂V

)

T,N

=

(

∂P

∂T

)

V,N

)

(4)

CP − CV = T

(

∂P

∂T

)

V,N

(

∂V

∂T

)

P,N

(5)

OK, now that you believe me on that one, we can re-arrange our equationof state (Eq. 1) to find the partial derivatives in Eq. 2.

P = b +NRT

V

(P − b) =NRT

V(P − b)V = NRT

V =NRT

P − bP = b +

NRT

V(

∂V

∂T

)

P,N

=NR

P − b

(

∂P

∂T

)

V,N

=NR

V(6)

which means Eq. 2 becomes

CP − CV = T

(

NR

V

)(

NR

P − b

)

= T

(

NR

V

)(

NR

(NRT/V )

)

= NR

(

T

V

)(

V

T

)

CP − CV = NR (7)

(b) Show that constant volume heat capacity is not dependent on thevolume. Specify how the heat capacity can depend on N and T .

We can kill two birds with one stone by showing that

∂ CV

∂V= 0 (8)

We can do so be using the definition of CV as seen from Eqs. 3, and Eq. 4

2


again,

(

∂ CV

∂V

)

T,N

=∂

∂V

[

(

T∂S

∂T

)

P,N

]

T,N

= T∂

∂V

[

(

∂S

∂T

)

P,N

]

T,N(

∂x∂y f(x, y) = ∂y∂x f(x, y))

= T∂

∂T

[

(

∂S

∂V

)

T,N

]

P,N

= T∂

∂T

[

(

∂P

∂T

)

V,N

]

P,N(

(

∂P

∂T

)

V,N

=∂

∂T

(

b +NRT

V

)

=NR

V

)

= T∂

∂T

(

NR

V

)

∂ CV

∂V= 0

√(9)

(c) This gas undergoes a process at fixed temperature where thevolume is changed from Vi to Vf . Find the change in entropy in termsof the variables N, R, T, Vf and Vi.

The change in entropy for an isothermal process is given by

∆S =Q

T(10)

where Q is the heat leaving the system. The amount heat leaving the systemwill be given by the first law of thermodynamics,

∆U = Q + W

Q = ∆U − W

Equipartion Thm.: U ∝ NRT⇒ Isothermal process: ∆T = 0

⇒ ∆U = 0

Q = −W (11)

The work done to decrease the volume from Vi to Vf can be found by integratingthe pressure,

W = −P∆V

= −

∫ Vf

Vi

P dV

3


(

Eq. 1: P = b +NRT

V

)

= −

∫ Vf

Vi

b dV −

∫ Vf

Vi

NRT

VdV

W = −b(Vf − Vi) − NRT ln

(

Vf

Vi

)

(12)

which means the total heat lost is

Q = −W = b(Vf − Vi) + NRT ln

(

Vf

Vi

)

(13)

and the change in entropy is

∆S =Q

T=

b

T(Vf − Vi) + NR ln

(

Vf

Vi

)

(14)

4



Jeff Kissel

May 18, 2007

Consider a classical gas composed of N particles of mass m that possesa permanent electric dipole moment ℘. The gas is subject to a uniformelectric field of size E in the the x-direction and is enclosed in acontainer of size V .

a) Find the partition function.

The partition function for a single particle is defined by

Z1 ≡∑

j

e−βǫj = Tr[e−βH] (1)

where ǫj is the energy of the jth state, H is the system’s Hamiltonian, and

β = (kT )−1. The trace Tr[O], of some operator O is defined as the sum over allstates, or the integral over all phase space

Tr[O] =1

h3

∫ ∞

−∞

∫ ∞

−∞

∫ ∞

−∞

O d3~p d3~q d~Ω (2)

where h is Planck’s constant, p & q are the generalized momentum and position,and Ω is the solid angle corresponding to internal rotations about θ and φ. Thefactor of h−3 appears to be sure the trace is dimensionless.

Because these particles have a permanent dipole moment ℘, the Hamiltonianis not just the usual free-particle kinetic energy component, but also the electricenergy component generated by the interaction between the electric field E andthe dipole moment. In equation form, that’s

H =~p2

2m− ~℘ · ~E

(~p2

= ~p · ~p = p2

− ~℘ · ~E = −℘E cos θ

)

H =p2

2m− ℘E cos θ (3)

where θ is the angle of the dipole from the z axis (in spherical polar coordinates),

along which the electric field ~E is assumed to point (φ, the angle from the x

1


axis, is free to rotate from 0 to 2π). Using Eqs. 1, 2, and 3 the partition functionfor this system integral form is then

Z1 =1

h3

∫ ∞

−∞

∫ ∞

−∞

∫ ∞

−∞

e−β

(p2

2m−℘E cos θ

)

d3~p d3~q d~Ω

=1

h3

∫ ∞

−∞

∫ ∞

−∞

e−β p2

2m e+β℘E cos θ d3~p d3~q d~Ω

Z1 =

1

h3

∫ ∞

−∞

e−β p2

2m d3~p

︸︷︷︸Zp

∫ ∞

−∞

d3~q

︸︷︷︸Zq

∫ ∞

−∞

e+β℘E cos θ d~Ω

︸︷︷︸ZΩ

(4)

The momentum portion of the partition function Zp (independent of positionand internal angle) can be cleaned up significantly,

Zp =1

h3

∫ ∞

−∞

e−β p2

2m d3~p

=1

h3

[∫ ∞

−∞

e−β p2

2m dp

]3

(∫ ∞

−∞

e−αx2

dx =

√π

α

)

=1

h3

(2m

βπ

)3/2

=

(2πmkT

h2

)3/2

let λ =h2

2πmkT

Zp =1

λ3(5)

where I’ve defined a variable λ which has dimensions of length, typically calledthe “quantum length.”

Because position portion of the partition function Zq (independent of mo-mentum and internal angle) has no integrand, we just evaluate it to be thevolume,

Zq =

∫ ∞

−∞

d3~q = V (6)

Finally, the internal angle portion of the partition function ZΩ (independentof poistion and momentum) is a little trickier, but still doable,

ZΩ =

∫ ∞

−∞

e+β℘E cos θ d~Ω

=

∫ 2π

0

∫ π

0

e+β℘E cos θ sin θ dθ dφ

2


= 2π

∫ π

0

e+β℘E cos θ sin θ dθ

let u = cos θ

du = − sin θ dθ

−du = sin θ dθ

0 → 1 π → −1

= 2π

(−

∫ −1

1

eβ℘Eu du

)

= 2π

∫ 1

−1

eβ℘Eu du

= 2π

[1

β℘Eeβ℘Eu

]1

−1

=2π

β℘E

(eβ℘E − e−β℘E

)

(ex − e−x = 2 sinh(x)

)

Zq =4π

β℘Esinh (β℘E) (7)

Combining Eqs. 5, 6, and 7 yields the partition function for one particle,

Z1 =4πV

λ3

sinh (β℘E)

β℘E(8)

For a system of N such particles, which are indistinguishable, the partitionfunction is

Z = 1N ! ZN

1 =1

N !

(4πV

λ3

)N (sinh (β℘E)

β℘E

)N

(9)

b) Find the net polarization of the sample as a function of the filedstrength, pressure, and temperature. The polarization P is definedas the net dipole moment per unit volume.

As stated in the question, the polarization P is defined as the average dipolemoment 〈℘〉 per unit volume V . In equation form, that’s

P =〈℘〉

V(10)

So, let’s find 〈℘〉. Any generic (ensemble) average of an observable is defined as

〈O〉 =∑

j

O(j)P(j) =1

Z

∑

j

Ojeβǫj (11)

which means the average dipole moment is

〈℘〉 =1

Z

∑

j

℘j e−βǫj

3


=

∑j ℘ e−β p2

2m e+℘·(βE)

∑j e−β p2

2m e+℘·(βE)

=

∂∂(βE)

(∑j e−β p2

2m e+℘·(βE))

∑j e−β p2

2m e+℘·(βE)

=1

Z

∂

∂(βE)(Z)

〈℘〉 =∂

∂(βE)(ln(Z)) (12)

and plugging in Eq. 9

〈℘〉 =∂

∂(βE)

ln

[1

N !

(4πV

λ3

)N (sinh (β℘E)

β℘E

)N]

=∂

∂(βE)

[ln

(1

N !

(4πV

℘λ3

)N)

+ ln(sinhN (℘ βE)

)− ln

((βE)N

)]

=

∂

∂(βE)

[ln

(1

N !

(4πV

℘λ3

)N)]

+∂

∂(βE)

[ln(sinhN (℘ βE)

)− ln

((βE)N

)]

=∂

∂(βE)[N ln (sinh (℘ βE)) − N ln (βE)]

(d

dusinh(u) = cosh(u)du

)

=

(N cosh (℘ βE)(℘)

sinh (℘ βE)

)−

N

βE

〈℘〉 = N℘ coth (℘ βE) − NkT1

E(13)

From which we find the polarization P (using Eq. 10) in terms of electric fieldstrength E, pressure P , and temperature T ,

P =〈℘〉

V

=N

V℘ coth (℘ βE) −

NkT

V

1

E(PV = NkT

⇒ NV = P

kT = Pβ and NkTV = P

)

= Pβ℘ coth (℘ βE) − P1

E

P(E, P, T ) = P℘β

(coth (℘ βE) −

1

℘βE

)(14)

4



Jeff Kissel

May 18, 2007

Exam Iteration Question Number Notes/ChangesSpring 2006 X –Fall 2006 #54 Added to question bank.

Spring 2007 #54 Part (d) changed.

The energy spectrum of neutrinos (spin 1/2 massless particles) isǫ = pc. For a collection of N neutrinos confined to volume V , calculate

I’ll do part (b) first, and then (a).

(b) The Fermi energy ǫF .

I’ll find the Fermi energy by first finding the density of modes g(ǫ) for themassless, spin 1/2 neutrinos, then the total number of particles N(ǫ) at tem-perature T = 0, which I can then invert for ǫ.

The ensemble average of any observable O is defined as

〈O〉 =∑

j

OjPj (1)

The density of modes g(ǫ) is the Jacobian of transformation when onechanges this sum to an integral. This conversion for massless fermions (wherewe add a factor of 2 to account for the spin degeneracy) is as follows

∑

j

= 2∑

k

=2

h3

∫ ∞

−∞

∫ ∞

−∞

d3~pd3~q

=2

h3V

∫∞

−∞

d3~p

=2V

h3(4π)

∫∞

0

p2 dp

1


let ǫ = pc

p = ǫc

dp = 1c dǫ

p2 = ǫ2

c2

=8πV

h3

∫∞

0

(ǫ2

c2

)1

cdǫ

=8πV

h3

∫∞

0

ǫ2

c3dǫ

∑

j

⇒∫

∞

0

8πV

(hc)3ǫ2 dǫ ≡

∫∞

0

g(ǫ) dǫ (2)

On to the total number of particles, which is the sum over the probabilitiesof each state (i.e. Eq. 1 without the observable). The probability distributionfor fermions is the Fermi-Dirac distribution,

nFD =1

eβ(ǫ−µ) + 1(3)

which at the limit of zero temperature splits into two regions (i.e. has values1 or 0). The non-vanishing portion of the distribution is in the energy regime0 < ǫ < ǫF , (the upper limit is the Fermi Energy, or chemical potential atzero temperature), because the Pauli Exclusion Principle dictates that fermions(spin-1/2 particles) cannot occupy the same energy state. Thus instead of allcollapsing to some ground state, they stack up in energy to ǫF when tempera-tures very small.

Thus, at zero temperature, the total number of particles is

N(T = 0) = limT→0

∑

j

PFD

= limT→0

∫ ∞

0

1

eβ(ǫ−µ) + 1g(ǫ) dǫ

=

∫ ǫF

0

(1) g(ǫ) dǫ +

∫∞

ǫF

(0) g(ǫ) dǫ

=

∫ ǫF

0

8πV

(hc)3ǫ2 dǫ =

8πV

(hc)3

∫ ǫF

0

ǫ2 dǫ

=8πV

(hc)3ǫ3F3

N =8

(hc)3πV

3ǫ2F (4)

which we can solve for the Fermi energy of massless, spin 1/2 neutrinos,

ǫ3F =(hc)3

8

3

π

N

V

ǫF =hc

2

(3

π

N

V

)1/3

(5)

2


(a) The Fermi wave vector.

The Fermi wave vector is quickly obtainable from the Fermi energy, becausefor photons,

E = pc

(p = hk)

E =hk

2πc

k =2π

hc E (6)

So,

kF =2π

hc ǫF

=2π

hc

(

hc

2

(3

π

N

V

)1/3)

= πc2

(3

π

N

V

)1/3

kF = c2

(

3π2 N

V

)1/3

(7)

(c) The total energy at T = 0.

The total ensemble averaged energy of the system at T = 0 will be given by

〈E〉T=0 =∑

k

ǫPFD

=

∫ ∞

0

ǫPFDg(ǫ) dǫ

UT=0 =

∫ ǫF

0

ǫ(1)

(8πV

(hc)3ǫ2)

dǫ

=8πV

(hc)3

∫ ǫF

0

ǫ3 dǫ

=8πV

(hc)3ǫ4F4

=2πV

(hc)3

(

hc

2

(3

π

N

V

)1/3)4

=2πV

(hc)3

(

(hc)4

16

(3

π

N

V

)4/3)

=hc

8πV

(3

π

N

V

)(3

π

N

V

)1/3

3


=3

4N

(

hc

2

(3

π

N

V

)1/3)

〈E〉T=0 =3

4N ǫF (8)

(d) The compressibility of the neutrino gas.

We can treat the neutrino gas as an ideal (non-interacting) gas, such that

V =NkBT

P(9)

The (isothermal) compressibility, in terms of the pressure is then

κT ≡ − 1

V

(∂V

∂P

)

= − 1

V

∂

∂P

(NkBT

P

)

=NkBT

V

1

P 2= (P )

1

P 2=

1

P

which is true for any ideal gas. So we need the pressure of the nuetrino gas.The pressure P exerted by neutrinos should in general be

P =1

Z

∑

j

Pj e−βǫj (10)

where Pj is the pressure of any given energy state ǫj .The neutrino’s temperature is a constant T = 0, so there is no heat transfer,

Q = SdT = 0. We can derive an expression for the pressure of a given state Pj

using the thermodynamic identity (assuming the number of particles remainsconstant)

dǫj = *0

dQj + dWj

(Definition of Work: dW = − PdV )

dǫj = −PjdV

⇒ Pj = − ∂ǫj

∂V(11)

For the free neutrino gas, the system is effectively an infinite 3-D infinitesquare well. Tor each dimension, the particle’s deBroglie wavelength must thenbe half integer multiples of the length L of a side. Hence,

L =1

2nλn ⇒ λn =

2L

n(12)

The momentum in any given state is then

pj = hλn

=hnj

2L(13)

making the energy

ǫj = pjc =hcnj

2L∝ 1

L(14)

4


According to Eq. 20, we need the partial derivative of this energy withrespect to V . Acknowledging that L = V 1/3,

Pj = − ∂ǫj

∂V

= − ∂

∂V

(hcnj

2

1

V 1/3

)

= − hcnj

2

(

− 1

3

1

V 4/3

)

= +hcnj

2

1

V 1/3

(1

3

1

V

)

Pj =1

3

ǫj

V(15)

Plugging this result back into Eq. 19,

PT=0 =1

Z

∑

j

Pj e−βǫ

=1

Z

∑

j

(1

3

ǫj

V

)

e−βǫ

=1

3

1

V

( 1

Z

∑

j

ǫj e−βǫ)

(

From Eqs. 1:1

Z

∑

j

ǫj e−βǫ =∑

j

ǫjPj = 〈E〉)

=1

3

〈E〉T=0

V

=1

3

1

V

(3

4NǫF

)

PT=0 =1

4

N

VǫF (16)

Which makes the isothermal compressibility

κT = 4V

NǫF

= 4V

N

(

hc

2

(3

π

N

V

)1/3)

κT = 2hc

(3

π

V 2

N2

)1/3

(17)

5


OLD VERSIONS

b) The pressure exerted by the neutrinos at T = 0.

In general, the probability distribution P in Eq. 1 is

P =1

Z

∑

j

e−βǫj (18)

where Z is the partition function, and ǫj is the energy of the jth state. Thepressure P exerted by neutrinos should then be

P =1

Z

∑

j

Pj e−βǫ (19)

The neutrino’s temperature is a constant T = 0, so there is no heat transfer,Q = SdT = 0. We can derive an expression for the pressure of a given state Pj

using the thermodynamic identity (assuming the number of particles remainsconstant)

dǫj = *0

dQj + dWj

(Definition of Work: dW = − PdV )

dǫj = −PjdV

⇒ Pj = − ∂ǫj

∂V(20)

For the free neutrino gas, the system is effectively an infinite 3-D infinitesquare well. Tor each dimension, the particle’s deBroglie wavelength must thenbe half integer multiples of the length L of a side. Hence,

L =1

2nλn ⇒ λn =

2L

n(21)

The momentum in any given state is then

pj = hλn

=hnj

2L(22)

making the energy

ǫj = pjc =hcnj

2L∝ 1

L(23)

According to Eq. 20, we need the partial derivative of this energy withrespect to V . Acknowledging that L = V 1/3,

Pj = − ∂ǫj

∂V

= − ∂

∂V

(hcnj

2

1

V 1/3

)

6


= − hcnj

2

(

− 1

3

1

V 4/3

)

= +hcnj

2

1

V 1/3

(1

3

1

V

)

Pj =1

3

ǫj

V(24)

Plugging this result back into Eq. 19,

P =1

Z

∑

j

Pj e−βǫ

=1

Z

∑

j

(1

3

ǫj

V

)

e−βǫ

=1

3

1

V

( 1

Z

∑

j

ǫj e−βǫ)

(

From Eqs. 1 and 18:1

Z

∑

j

ǫj e−βǫ =∑

j

ǫjPj = 〈E〉)

PT=0 =1

3

〈E〉T=0

V(25)

c) The heat capacity for 0 < T ≪ ǫF .

Awwww shit. It’s the Sommerfield expansion again. It doesn’t matter thatthe neutrinos are massless, as long as they’re fermions.

If I denote the constants in the density of modes to be CF , and use < n >as shorthand for 〈 n(ǫ) 〉FD then the ensemble average energy (as seen in part(b)) is

〈E〉 =

∫∞

0


∫∞

0

ǫ3

2 < n > dǫ (26)


let u = 〈 n 〉 v = 25 ǫ5/2



[

:02

5ǫ5/2 < n >

∣∣∣∣

∞

0

+2

5

∫ ∞

0

ǫ5/2

(

− d < n >

dǫ

)

dǫ

]

(27)

where the ”surface” term vanishes because at ǫ = 0, ǫ5/2 = 0, and at ǫ = ∞,

7


< n >= (eβ(ǫ−µ) + 1)−1 = 0. This leaves,

〈E〉 =2

5CF

∫∞

0

ǫ5/2

(

− d < n >

dǫ

)

dǫ

(

− d < n >

dǫ= − d

dǫ

(1

eβ(ǫj−µ) + 1

)

=β eβ(ǫj−µ)

(eβ(ǫj−µ) + 1)2

)

=2

5CF

∫ ∞

0


(eβ(ǫj−µ) + 1)2

let x = β(ǫ − µ) ⇒

dx = β dǫ0 → −µβ, ∞ → ∞

〈E〉 =2

5CF

∫∞

−µβ

ǫ5/2 ex

(ex + 1)dx (28)




ǫ5/2 ≈ ǫ5/2

∣∣∣∣µ

+ (ǫ − µ)

(d(ǫ5/2)

dǫ

) ∣∣∣∣µ

+1

2!(ǫ − µ)2

(d2(ǫ5/2)

dǫ2

) ∣∣∣∣µ

+ ...

= µ5/2 +5

2(ǫ − µ)µ3/2 +

1

2

15

4(ǫ − µ)2µ1/2 + ...

= µ5/2 +5

2(xkT )µ3/2 +

15

8(xkT )2µ1/2 + ... (29)

So that

〈E〉 ≈ 2

5CF

[

µ5/2

∫∞

−∞

ex

(ex + 1)2dx

︸︷︷︸

(I)

+5

2kTµ3/2

∫∞

−∞

xex

(ex + 1)2dx

︸︷︷︸

(II)

+15

8(kT )2µ1/2

∫∞

−∞

x2 ex

(ex + 1)2dx

︸︷︷︸

(III)

+...

]

(I)

∫∞

−∞

ex

(ex + 1)dx =

∫∞

−∞

(

− d < n >

dǫ

)

dǫ

8


=

∫ ∞

−∞

d < n >

=

[

:1< n(−∞) > +:0

< n(∞) >

]

∫∞

−∞

ex

(ex + 1)dx = 1 (30)

(II)

∫∞

−∞

x ex

(ex + 1)2dx =

∫∞

−∞

x ex

(ex + 1)(ex + 1)

e−x

e−xdx

=

∫ ∞

−∞︸︷︷︸

even

bounds

x

(ex + 1)(1 + e−x)︸︷︷︸

odd

function

dx

∫∞

−∞

x ex

(ex + 1)2dx = 0 (31)

(III)

∫ ∞

−∞︸︷︷︸

even

bounds

x2ex

(ex + 1)2︸︷︷︸

even

function

dx = 2

∫ ∞

0

x2ex

(ex + 1)2dx

let u = x2

du = 2x dx

v = −1ex+1

let dv = ex

(ex+1)2

= 2

>

0

−x2

ex + 1

∣∣∣∣

∞

0

+

∫ ∞

0

2x

ex + 1dx

= 4

∫ ∞

0

x

ex + 1dx

(∫∞

0

xm

ex + 1dx = (1 − 1

2m) Γ(m + 1) ζ(m + 1)

)

= 4(1 − 1

2) Γ(2) ζ(2)

= 4

(1

2

)

(1)

(π2

6

)

=π2

3∫

∞

−∞

x2ex

(ex + 1)2dx =

π2

3(32)

9


⇒ 〈E〉 ≈ 2

5CF µ5/2 (1) +

2

5CF

5

2kTµ3/2 (0)

+2

5CF

15

8(kT )2µ1/2

(π2

3

)

+ ...

=2

5CF µ5/2 +

π2

4CF (kT )2µ1/2 + ...

Eq. 5: ǫF = h2

2m

(3π2 N

V

) 2

3

2mh2 ǫF =

(3π2 N

V

) 2

3

(2mh2

) 3

2 ǫ3/2F = 3π2 N

VVπ2

(2mh2

) 3

2 = 3 N

ǫ3/2

F

⇒ CF = V2π2

(2mh2

) 3

2 = 32

N

ǫ3/2

F

(33)

=2

5

(

3

2

N

ǫ3/2F

)

µ5/2 +π2

4

(

3

2

N

ǫ3/2F

)

(kT )2µ1/2 + ...

=3

5N

µ5/2

ǫ3/2F

+3π2

8N

µ1/2

ǫ3/2F

+ ...

〈E〉 =3

5N ǫF

(µ

ǫF

)5/2

+3π2

8N

(kT )2

ǫF

(µ

ǫF

)1/2

+ ... (34)


µ

ǫF= 1 − π2

12

(kT

ǫF

)2

+ ... (35)


⇒(

µ

ǫF

)5/2

=

(

1 − π2

12

(kT

ǫF

)2

+ ...

)5/2


≈ 1 − 5

2

π2

12

(kT

ǫF

)2

+ ...

= 1 − 5π2

24

(kT

ǫF

)2

+ ... (36)

⇒(

µ

ǫF

)1/2

=

(

1 − π2

12

(kT

ǫF

)2

+ ...

)5/2

≈ 1 − 1

2

π2

12

(kT

ǫF

)2

+ ...

= 1 − π2

24

(kT

ǫF

)2

+ ... (37)

10



〈E〉 =3

5N ǫF

(

1 − 5π2

24

(kT

ǫF

)2

+ ...

)

+3π2

8N

(kT )2

ǫF

(

1 − π2

24

(kT

ǫF

)2

+ ...

)

+ ... (38)


〈E〉 =3

5N ǫF − 3π2

24N

(kT )2

ǫF+

3π2

8N

(kT )2

ǫF+ O

((kT )2

ǫ3F

)

=3

5N ǫF − π2

8N

(kT )2

ǫF+

3π2

8N

(kT )2

ǫF+ O

((kT )2

ǫ3F

)

〈E〉 =3

5N ǫF +

π2

4N

(kT )2

ǫF+ ... (39)

Now we can finally get to the question. Remember we are trying to showthat CV ∝ T for low temperatures. This is quick attainable from Eq. 39, andthe definition of the heat capacity at constant volume,

CV ≡(

∂〈E〉∂T

)

V,N

=∂

∂T

(3

5N ǫF +

π2

4N

(kT )2

ǫF+ ...

)

CV =π2

2N

k2

ǫFT ∝ T

√(40)

d) The rms fluctuations in the mean energy of the neutrinos of 0 <T ≪ ǫF .

As in part (c) it doesn’t matter that the neutrinos are massless, only thatthey’re fermions.

For any system in equilibrium the mean square fluctuations (variance) is

〈σ2〉 = 〈E2〉 − 〈E〉2 = kT 2CV (41)

(for a detailed proof of this check out statmechS06-32F06-43). From Eq. 40,this becomes

〈σ2〉 = kT 2CV

= kT 2

(π2

2N

k2

ǫFT

)

〈σ2〉 = (kT )3π

2

N

ǫF(42)

11


which means the rms (root mean square) fluctuations are

σrms =

(

(kT )3π

2

N

ǫF

)1/2

(43)

12



Jeff Kissel

May 18, 2007

Exam Iteration Question Number Notes/ChangesSpring 2006 #27 –Fall 2006 #38 Question number shift.

Spring 2007 #55 Part (c) removed.

Consider an ideal Fermi gas of spin-1/2 particles in a 3-dimensionalbox. The number of particles per unit volume is n, the mass of theparticles is m, and the energy of the particles is the usual ǫ = p2/2m.Assume that the temperature is quite low.

a) Find formulas for the Fermi energy ǫF , Fermi wavevector kF , andFermi temperature TF in term of m, n and constants such as h andkB.

To find the Fermi energy (the chemical potential of spin 1/2 particles at zerotemperature), we must find the total number of particles N as T approacheszero.

For any temperature T , one normally finds N via summing over the Fermi-Dirac probability distribution 〈 nj(ǫ) 〉FD (because the particles are of spin1/2),

〈 nj(ǫ) 〉FD =1

eβ(ǫj−µ) + 1(1)

where β = 1kT , ǫj is the kinetic energy of a state j, and µ is the chemical potential

of the particles. However, at the limit of zero temperature, 〈 nj(ǫ) 〉FD decaysto either 0 (when ǫj > µ) or 1 (when ǫj < µ). The energy, at T = 0, at whichthe now-step-function distribution flips from 1 to 0 is the Fermi energy, whosevalue is determined (again) by the number of particles.

To perform the calculation of the number of particles, we first find the densityof states: the Jacobian of transformation between summing over energy statesǫj, and integrating over kinetic energies. First, to account the spin degeneracy

1


of +1/2 and −1/2, we tack on a factor of 2,

∑

j

= 2∑

k

=2

h3

∫ ∞

−∞

∫ ∞

−∞

d3~p d3~q =2

h3V

∫ ∞

−∞

d3~p

=2V

h3(4π)

∫ ∞

0

p2 dp

let ǫ =p2

2m

p =√

2m ǫ1/2

dp = 12

√2mǫ−1/2 dǫ

p2 =√

2mǫ

=8πV

h3

∫ ∞

0

(√2mǫ

) 1

2

√2mǫ−1/2 dǫ

=8πV

h3

1

2(2m)3/2

∫ ∞

0

ǫ1/2 dǫ

∑

j

⇒∫ ∞

0

4πV

(2m

h2

)3/2

ǫ1/2 dǫ ≡∫ ∞

0

g(ǫ) dǫ (2)

Now we can find the number of particles N , and invert for the Fermi energyǫF .

N(T = 0) =∑

j

〈 nj(ǫ) 〉FD

=

∫ ∞

0

〈 n(ǫ) 〉FD gFD(ǫ) dǫ

=

∫ ǫF

0

(1) g(ǫ) dǫ +

∫ ∞

ǫF

(0) g(ǫ) dǫ

=

∫ ǫF

0

4πV

(2m

h2

)3/2

ǫ1

2 dǫ

= 4πV

(2m

h2

)3/22

3ǫ

3

2

F

N =8π

3V

(2m

h2

)3/2

ǫ3

2

F (3)

ǫF =h2

2m

(1

8

3

π

N

V

)2/3

=h2

8m

(3

π

N

V

)2/3

ǫF =h2

2m

(3π2 N

V

)2/3

(4)

In this form it’s nice and quick to pull out the Fermi wave-vector kF , since

2


the quantum mechanical kinetic energy is

E =p2

2m=

h2

2mk2 (5)

So by direct comparison between Eqs. 4 and 5,

kF =

(3π2 N

V

) 1

3

(6)

The Fermi terperature TF is similarly quick, as the thermal kinetic energyis,

E = kBT (7)

Inverting for temperature and subbing in 4,

TF =h2

2mkB

(3π2 N

V

) 2

3

(8)

b) Find the total energy of the gas at zero temperature.

This calculation can be done by summing over possible energy states, weightedby their probabilities (determined by the Fermi-Dirac Distribution),

〈E〉 ≡∑

j

ǫj 〈 nj(ǫj) 〉

=

∫ ∞

0

ǫ 〈 n 〉FD gFD(ǫ) dǫ

〈E〉T=0 =

∫ ǫF

0

ǫ (1) g(ǫ) dǫ +

∫ ∞

ǫF

ǫ (0) g(ǫ) dǫ

=

∫ ǫF

0

ǫ g(ǫ) dǫ

=

∫ ǫF

0

ǫV

2π2

(2m

h2

) 3

2

ǫ1

2 dǫ

=V

2π2

(2m

h2

) 3

2∫ ǫF

0

ǫ3

2 dǫ

=V

2π2

(2m

h2

) 3

2 2

5ǫ

5

2

F

=V

5π2

(2m

h2

) 3

2

ǫ5

2

F

(Eq. 3: N =

V

3π2

(2m

h2

) 3

2

ǫ3

2

F

)

〈E〉T=0 =3

5N ǫF (9)

3


c) If the magnetic moment of the particles is µe, show that the param-agnetic susceptibility χ for low fields in the limit of zero temperatureis given by

χ =3

2

nµ2e

ǫF(10)

This part deals with the quantum mechanical property of fermions known asthe Zeeman splitting. When a weak magnetic field is applied to a fermi gas, the”up” (parallel to ~B, |−〉) and ”down” (anti-parallel to ~B, |+〉) energy levels ofthe particles are perturbed such that the energy states of the ”up” particles areincreased, and the ”down” particles are decreased, causing a ”splitting” of whatinitially was a single set of quantized energy states. The Zeeman effect is a 1storder perturbation theory problem in quantum mechanics (Check p245-246 ofGriffiths’, or Wikipedia’s article for more detialed explanations. Merzbacher’s ispretty weak sauce, but it’s on p472-473). Anyways, this is a Stat-Mech problem,so we really only care about the resulting energies, so I’ll only highlight thedetails.

The perturbing hamiltonian is

H = −~µe · ~B (11)

where B is the applied, weak magnetic field, and µe is the magnetic moment ofthe fermions (spin-1/2 particles), given by

~µe = −gJ µB~S

h(12)

where gJ is the Lande g-factor, µB = (eh)/(2m) is the Bohr magneton, and

S is the spin operator. The Lande g-factor is approximately 2 for (spin-1/2)electrons (See Wikipedia’s ”Lande g-factor” for details), so I assume gJ = 2.

Thus, H (Eq. 11, assuming the field | ~B| = B applied in the same direction asµe) becomes,

H = −µeB

= −(− 2 µB S

h

)B

(Si =

h

2σi

)

H = µBσzB (13)

where σz is the Pauli spin matrix, and I’ve assumed that the field was appliedin the z-direction.

The total energy of the anti-parallel (”down”, |+〉, or the positive eigenvalueof σz) state is

ǫT = ǫ + µBB

4


in which, as before, ǫ = p2/(2m) is the kinetic energy of the particles. Theparallel (”up”, |−〉, negative eigenvalue of σz) energies are

ǫT = ǫ − µBB

Which means the Fermi-Dirac distribution (1) becomes

〈 nj(ǫ) 〉FD =1

eβ[(ǫj±µBB)−µ) + 1(14)

i.e. is splits into two distributions,

1

2〈 nj(ǫ − µBB) 〉FD and

1

2〈 nj(ǫ − µBB) 〉FD (15)

where the factors of 1/2 insures the distribution remains normalized. The leftdistribution corresponds to the up spins, i.e. those energy levels displaced bya positive µBB and therefore the probability range over which it extends isfrom µBB < ǫT < ∞. The right distribution are the down spins, displaced bynegative µBB, so the range is µBB < ǫT < ∞.

From here, we can calculate the net magnetization per unit volume M , thederivative of which (with respect to B) is χ. The net magnetization would bethe total number of particles anti-align subtracted from those aligned (in a givenvolume), or

M ≡ µB(N↑ − N↓)

V(16)

We know how to find the number of particles: integrate the probability distribu-tion corresponding the particles over the correct region. Thus, the magnetizationis then

M =µB

V

(∫ ∞

µBB

1

2〈 n(ǫ + µBB) 〉FD gFD(ǫ) dǫ

−∫ ∞

µBB

1

2〈 n(ǫ − µBB) 〉FD gFD(ǫ) dǫ

)

=µB

2V

∫ ∞

0

[〈 nj(ǫ + µBB)〉 − 〈 nj(ǫ − µBB)〉] g(ǫ) dǫ

(∂f(x)

∂x ≡ f(x+h)−f(x−h)2h

⇒ f(x − h) − f(x + h) = −2 h ∂f(x)∂x

)

=µB

2V2 µBB

∫ ∞

0

(− ∂〈 n(ǫ) 〉

∂ǫ

)g(ǫ) dǫ

M =µ2

B B

V

∫ ∞

0

(− ∂〈 n(ǫ) 〉

∂ǫ

)g(ǫ) dǫ (17)

At low temperatures (i.e kT ≪ ǫF ), the Fermi-Dirac distribution becomesasymptotically close to a step function with the barrier at the Fermi energy

5


(as discussed in part (a)). Thus, the derivative of the distribution at low tem-peratures is well approximated by a delta function about ǫF , or

limT→0

∫ ∞

0

(− ∂〈 n(ǫ) 〉

∂ǫ

)dǫ =

∫ ∞

0

δ(ǫ − ǫF ) dǫ (18)

which makes Eq. 17 a lot prettier,

M =µ2

B B

V

∫ ∞

0

δ(ǫ − ǫF ) g(ǫ) dǫ

=µ2

B B

Vg(ǫF )

= µ2B CF ǫ

1/2F

B

V(Eq. 28: CF =

3

2

N

ǫ3/2F

)

= µ2B

(3

2

N

ǫ3/2F

)ǫ1/2F

B

V

M =3

2

µ2B

ǫF

N

VB (19)

Finally, if I convert to the notation the proof desires (N/V → n), the mag-netic susceptibility is

χ ≡ ∂M

∂B

=∂

∂B

(3

2

µ2B

ǫFn B

)

χ =3

2

n µ2B

ǫF(20)

which is sometimes referred to as the ”Pauli Susceptibility.”

It’s cool. You’ll remember all of this during the qualifier. I promise.

6


OLD VERSIONS

c) Show that the heat capacity at low temperature is proportional toT .

The exact proof of this problem unfortunately gets rather lengthly, but isknown as the Sommerfield expansion.

If I denote the constants in Eq. 2 to be CF , and use < n > as shorthand for〈 n(ǫ) 〉FD then the ensemble average energy (as seen in part (b)) is

〈E〉 =

∫ ∞

0


∫ ∞

0

ǫ3

2 < n > dǫ (21)


let u = 〈 n 〉 v = 25 ǫ5/2



[

:02

5ǫ5/2 < n >

∣∣∣∣∞

0

+2

5

∫ ∞

0

ǫ5/2

(− d < n >

dǫ

)dǫ

]

(22)

where the ”surface” term vanishes because at ǫ = 0, ǫ5/2 = 0, and at ǫ = ∞,< n >= (eβ(ǫ−µ) + 1)−1 = 0. This leaves,

〈E〉 =2

5CF

∫ ∞

0

ǫ5/2

(− d < n >

dǫ

)dǫ

(− d < n >

dǫ= − d

dǫ

(1

eβ(ǫj−µ) + 1

)=

β eβ(ǫj−µ)

(eβ(ǫj−µ) + 1)2

)

=2

5CF

∫ ∞

0


(eβ(ǫj−µ) + 1)2

let x = β(ǫ − µ) ⇒

dx = β dǫ0 → −µβ, ∞ → ∞

〈E〉 =2

5CF

∫ ∞

−µβ

ǫ5/2 ex

(ex + 1)dx (23)




ǫ5/2 ≈ ǫ5/2

∣∣∣∣µ

+ (ǫ − µ)

(d(ǫ5/2)

dǫ

) ∣∣∣∣µ

+1

2!(ǫ − µ)2

(d2(ǫ5/2)

dǫ2

) ∣∣∣∣µ

+ ...

7


= µ5/2 +5

2(ǫ − µ)µ3/2 +

1

2

15

4(ǫ − µ)2µ1/2 + ...

= µ5/2 +5

2(xkT )µ3/2 +

15

8(xkT )2µ1/2 + ... (24)

So that

〈E〉 ≈ 2

5CF

[µ5/2

∫ ∞

−∞

ex

(ex + 1)2dx

︸︷︷︸(I)

+5

2kTµ3/2

∫ ∞

−∞

xex

(ex + 1)2dx

︸︷︷︸(II)

+15

8(kT )2µ1/2

∫ ∞

−∞

x2 ex

(ex + 1)2dx

︸︷︷︸(III)

+...

]

(I)

∫ ∞

−∞

ex

(ex + 1)dx =

∫ ∞

−∞

(− d < n >

dǫ

)dǫ

=

∫ ∞

−∞

d < n >

=

[

:1< n(−∞) > +:0

< n(∞) >

]

∫ ∞

−∞

ex

(ex + 1)dx = 1 (25)

(II)

∫ ∞

−∞

x ex

(ex + 1)2dx =

∫ ∞

−∞

x ex

(ex + 1)(ex + 1)

e−x

e−xdx

=

∫ ∞


bounds

x

(ex + 1)(1 + e−x)︸︷︷︸odd

function

dx

∫ ∞

−∞

x ex

(ex + 1)2dx = 0 (26)

(III)

∫ ∞


bounds

x2ex

(ex + 1)2︸︷︷︸even

function

dx = 2

∫ ∞

0

x2ex

(ex + 1)2dx

8


let u = x2

du = 2x dx

v = −1ex+1

let dv = ex

(ex+1)2

= 2

>0

−x2

ex + 1

∣∣∣∣∞

0

+

∫ ∞

0

2x

ex + 1dx

= 4

∫ ∞

0

x

ex + 1dx

(∫ ∞

0

xm

ex + 1dx = (1 − 1

2m) Γ(m + 1) ζ(m + 1)

)

= 4(1 − 1

2) Γ(2) ζ(2)

= 4

(1

2

)(1)

(π2

6

)

=π2

3∫ ∞

−∞

x2ex

(ex + 1)2dx =

π2

3(27)

⇒ 〈E〉 ≈ 2

5CF µ5/2 (1) +

2

5CF

5

2kTµ3/2 (0)

+2

5CF

15

8(kT )2µ1/2

(π2

3

)+ ...

=2

5CF µ5/2 +

π2

4CF (kT )2µ1/2 + ...

Eq. 4: ǫF = h2

2m

(3π2 N

V

) 2

3

2mh2 ǫF =

(3π2 N

V

) 2

3

(2mh2

) 3

2 ǫ3/2F = 3π2 N

VVπ2

(2mh2

) 3

2 = 3 N

ǫ3/2

F

⇒ CF = V2π2

(2mh2

) 3

2 = 32

N

ǫ3/2

F

(28)

=2

5

(3

2

N

ǫ3/2F

)µ5/2 +

π2

4

(3

2

N

ǫ3/2F

)(kT )2µ1/2 + ...

=3

5N

µ5/2

ǫ3/2F

+3π2

8N

µ1/2

ǫ3/2F

+ ...

〈E〉 =3

5N ǫF

(µ

ǫF

)5/2

+3π2

8N

(kT )2

ǫF

(µ

ǫF

)1/2

+ ... (29)

9



µ

ǫF= 1 − π2

12

(kT

ǫF

)2

+ ... (30)


⇒(

µ

ǫF

)5/2

=

(1 − π2

12

(kT

ǫF

)2

+ ...

)5/2


≈ 1 − 5

2

π2

12

(kT

ǫF

)2

+ ...

= 1 − 5π2

24

(kT

ǫF

)2

+ ... (31)

⇒(

µ

ǫF

)1/2

=

(1 − π2

12

(kT

ǫF

)2

+ ...

)5/2

≈ 1 − 1

2

π2

12

(kT

ǫF

)2

+ ...

= 1 − π2

24

(kT

ǫF

)2

+ ... (32)


〈E〉 =3

5N ǫF

(1 − 5π2

24

(kT

ǫF

)2

+ ...

)

+3π2

8N

(kT )2

ǫF

(1 − π2

24

(kT

ǫF

)2

+ ...

)+ ... (33)


〈E〉 =3

5N ǫF − 3π2

24N

(kT )2

ǫF+

3π2

8N

(kT )2

ǫF+ O

((kT )2

ǫ3F

)

=3

5N ǫF − π2

8N

(kT )2

ǫF+

3π2

8N

(kT )2

ǫF+ O

((kT )2

ǫ3F

)

〈E〉 =3

5N ǫF +

π2

4N

(kT )2

ǫF+ ... (34)

Now we can finally get to the question. Remember we are trying to showthat CV ∝ T for low temperatures. This is quick attainable from Eq. 34, and

10


the definition of the heat capacity at constant volume,

CV ≡(

∂〈E〉∂T

)

V,N

=∂

∂T

(3

5N ǫF +

π2

4N

(kT )2

ǫF+ ...

)

CV =π2

2N

k2

ǫFT ∝ T

√(35)

11



Jeff Kissel

May 18, 2007

Consider a system consisting of two particles, each of which can bein any one of three quantum states of respective energies 0, ǫ, and 3ǫ.The system is in contact with a heat reservoir at temperature T .

a) Write an expression for the partition function Z if the particlesobey classical Maxwell-Boltzmann statistics and are considered dis-tinguishable.

The partition function of any system is defined as

Z ≡

∑

j

e−βǫj (1)

where β = (kT )−1, and ǫj is the energy of the jth state. When particles obeyMaxwell-Boltzmann statistics, it means that the two particles are distinguish-able from each other. Each particle can occupy all three energy states, as shownbelow.

Figure 1: The energy level diagram for a particle in three possible energy states.

The partition function a given particle is

Z1 =∑

j

e−βǫj

= e−β(0) + e−β(ǫ) + e−β(3ǫ)

Z1 = 1 + e−βǫ + e−3βǫ (2)

For N distinguishable particles, each will have the exact same partition function.So, this system of two distinguishable particles will have the partition function

Z = ZN1 =

(

1 + e−βǫ + e−3βǫ)2

(3)

1


b) Write a similar expression if the particles obey Bose-Einsteinstatistics.

If the particles obey Bose-Einstein statistics, it means there are indistin-

guishable from each other, unlike Maxwell-Boltzmann particles. The particles(known as “bosons”) also have integer spin, which means they can occupy thesame energy state. Because of each of these facts, we must draw the system asa whole and count the energies. There are nine possible states,

Figure 2: The energy level diagram of a two-boson,, three-energy-level system.

where we’ve ruled out ǫ4, ǫ7, and ǫ8 because they’re indistinguishable fromǫ2, ǫ3, and ǫ6 respectively. Each state has a total energy ǫj equal to the totalenergy of both particles, so the partition function becomes

Z =∑

j

e−βǫj

= e−βǫ1 + e−βǫ2 + e−βǫ3 + e−βǫ5 + e−βǫ6 + e−βǫ9

= e−β(0+0) + e−β(0+ǫ) + e−β(0+3ǫ) + e−β(ǫ+0) + e−β(ǫ+3ǫ) + e−β(3ǫ+3ǫ)

Z = 1 + e−βǫ + e−3βǫ + e−2βǫ + e−4βǫ + e−6βǫ (4)

c) Write a similar expression if the particles obey Fermi-Dirac statis-tics.

Finally, Fermi-Dirac particles (fermions) are also indistinguishable, but havehalf-integer spins. This has two consequences: 1) they obey the Pauli exclusionprinciple, which means they cannot occupy the same state like bosons can, and2) each energy state will have some degeneracy related to the spin s of theparticles. For example, if s = 1/2, then we would have to tack on an extrafactor of 2 to partition function. In general for spin s, the correction factor goesas 2s + 1. Let’s count the possible states.

2


Figure 3: The energy level diagram of a two-fermion,, three-energy-level system.

Again, energy states ǫ3, ǫ5, and ǫ6 are rules out because of indistinguisha-bility. So, the following the same technique as before, but accounting for the2s + 1 spin degeneracy,

Z = (2s + 1)∑

j

e−βǫj

= (2s + 1)(

e−βǫ1 + e−βǫ2 + e−βǫ4)

= (2s + 1)(

e−β(0+ǫ) + e−β(0+3ǫ) + e−β(ǫ+3ǫ))

Z = (2s + 1)(

1 + e−βǫ + e−3βǫ + e−4βǫ)

(5)

3



Jeff Kissel

May 19, 2007



A box of volume V hold blackbody radiation at a temperature T .

(a) Show that the amount of energy stored in a small range of fre-quencies (f, f + df) is given by

E(f) df =a f3

ebf − 1df (1)

and find expressions for a and b in terms of V, T, kB, h, and c.

The total energy will be given by the energy times the probability of anygiven energy state, summed over all configuration states. Sounds daunting, butyou’ve seen it before.

Photons follow the Planck distribution function,

PPj =

1

eβǫj − 1(2)

so the total energy is

〈E〉 =∑

j

ǫj PPj (3)

where the sum is over j, the number of polarizations (for photons there’s 2:linearly polarized and circularly polarized), and wave number k = 2π

cf .

〈E〉 =∑

j

ǫj PPj

= 2∑

k

ǫk PPk

1


= 21

h3

∫

∞

−∞

∫

∞

−∞

ǫpPPp d3~p d3~q

(

ǫp 6= ǫp(~q) and PPp 6= PP

p (~q)⇒

∫

∞

−∞d3~q = V

)

= 2V

h3

∫

∞

−∞

ǫp PPp d3~p

(

~p = h~k ⇒ d3~p = h3d3~k)

= 2V

h3h3

∫

∞

−∞

ǫkPPk d3~k

= 2V

(2π)3

∫

∞

−∞

ǫkPPk d3~k

= 2V

(2π)3(4π)

∫

∞

0

ǫkPPk k2 dk

ck = ωck = 2πfk = 2π

cf

dk = 2πc

dfǫf = hf

=V

π2

∫

∞

0

hf1

eβhf − 1

(

2π

cf

)2 (

2π

c

)

df

=V

π2

(2π)3

c3

∫

∞

0

hf3

eβhf − 1df

〈E〉 =

∫

∞

0

8πV h

c3

f3

eβhf − 1df (4)

E(f) df =8πV h

c3

f3

eβhf − 1df (5)

So,

a =8πV h

c3(6)

and

b = βh =h

kBT(7)

(b) The Wien displacement law describes the relationship between thepeak intensity in the blackbody energy distribution (Eq. 1) and thetemperature. Derive this relationship. Note: the function x3/(ex − 1)has a maximum at x = 2.82 approximately.

The intensity at a given frequency for a blackbody is given by

I(f) =1

4c u(f) =

1

4c

E(f)

V=

2πh

c2

f3

eβhf − 1(8)

2


Thus if we find a relationship between the maximum frequency fmax and tem-perature T , we find the relationship with the maximum intensity Imax andtemperature T .

As the problem suggests, we can define the dimensionless constant x = βhf ,so that f = x/(βh) and

I(x) =2πh

c2

(

1

βh

)3x3

ex − 1= 2π

(kBT )3

(hc)2x3

ex − 1(9)

such that when we take the maximum with respect to x

∂I(x)

∂x= 2π

(kBT )3

(hc)2∂

∂

(

x3

ex − 1

)

∣

∣

∣

xmax

0 =∂

∂x

(

x3

ex − 1

)

∣

∣

∣

xmax

(10)

we end up with exactly what we need from the hint, that xmax = 2.82. Pluggingback in for x,

xmax = βhfmax

fmax =kBT

hxmax

kB = 1.38 × 10−23 J/Kh = 6.626 × 10−34 J s

xmax = 2.82

fmax =(

5.873 × 1010 (K s)−1)

T (11)

(c) Find the total energy of the blackbody radiation and how it de-pends on V, T, h, c and kB.

The total energy is given by Eq. 4, so we just need to carry out the integral,

〈E〉 =

∫

∞

0

8πV h

c3

f3

eβhf − 1df

Let x = βhf ⇒

f = 1

βhx

df = 1

βhdx

0 → 0 , ∞ → ∞

=8πV h

c3

∫

∞

0

(

1

βh

)3x

ex − 1

(

1

βh

)

dx

=8πV h

c3

(

1

βh

)4 ∫

∞

0

x3

ex − 1dx

(∫

∞

0

yn

ey − 1dy = ζ[n + 1]Γ[n + 1]

)

=8πk4

B

(hc)3V T 4 ζ[4]Γ[4]

3


(

ζ[4]Γ[4] =

(

π4

90

)

(6) =π4

15

)

=

(

8π5k4

B

15(hc)3

)

V T 4

〈E〉 = αV T 4 (12)

where at the last step I’ve grouped all constants together into α, which is knownas the “radiation constant,” related to the Stefan-Boltzmann constant σ in thesame way intensity (which is luminosity for you astronomers) is related to energy,

σ =1

4c α =

2π5k4

B

15h3c2(13)

⇒ I =1

4c〈E〉V

=1

4c

αV T 4

V= σT 4

√(14)

4

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Classical Mechanics Review - Navidromekissel.navidrome.com/.../S07_classical_solutionbank.pdfClassical Mechanics Review (Louisiana State University Qualiﬁer Exam) Jeﬀ Kissel May