Download 2012 - University of Utah Physics

1. Classical physics
[25 pts.] A particle of mass m is subjected to the potential V = −F x, where F is a constant.
The particle travels from x = 0 to x = L in a time interval T . Find the motion of the
particle x(t) in terms of L, T , F , and m, such that the action is a minimum.
1
1. Solution: Classical physics with Lagrangian
The minimum of the action is obtained when the particle follows the classical trajectory. In
this case, the particle is subjected to a constant force F = −(∂V /∂x) along x, and its classical
motion is a motion with constant acceleration a = F/m. From elementary kinematics,
x(t) = A + Bt + Ct2 , with C = a/2 = F/(2m). Since the particle is at x = 0 when t = 0,
one immediately finds A = 0. The value of B is obtained by imposing that the particle is at
x = L when t = T , namely L = BT +CT 2 , from which B = (L−CT 2 )/T = L/T −F T /(2m).
To summarize,
F 2
L FT
−
t+
t
x(t) =
T
2m
2m
or
x(t) =
Lt F t(t − T )
+
.
T
2m
2
Problem 2: Electricity and Magnetism.
An infinite wire carrying current I lies in the x-axis. Space is field with two different magnetic dielectrics, divided by xy-plane. The magnetic permeability is µ1 at z > 0 and µ2 at
z < 0. Use Maxwell’s equations and Ampere’s law to find vectors B and H in all space,
following steps outlined below
1.[3 points] Write down two Maxwell’s differential equation for H and B in a static case
and integral Ampere’s law.
2.[5 points] Find fields H0 and B0 , created by the given current in a vacuum (µ = µ0 )
everywhere).
3.[5 points] Formulate boundary conditions for tangential and normal components of vectors
B and H at the plane z = 0 using Maxwell’s equations for H and B in the case of given
magnetic medium
4.[12 points] Express field H and B in the medium with given permeability through H0 .
Hints:
1. Differential equation for H does not contain permeability and it is homogenous outside
the current.
2. Pay attention to the boundary conditions for normal components of B and H at z = 0.
1
Solution to Problem 2
1. Maxwell’s equation for vectors H and B have a form
curlH = J,
(1)
where J is a current density. Using the Stokes theorem we get an integral equation
I
Hl dl = I,
(2)
that is called Ampere’s law. Here I is the total current through the wire. The integration is
taken over any close loop with the current inside. Hl is the projection of a field on a loop.
Field B obeys equation
divB = 0
(3)
The two fields are connected by the relation
B = µH
(4)
2. Because of the symmetry it is reasonable to choose a loop as a circle with radius r with
the center in the wire. The loop is in any plane x = const. Eq. (3) means the absence of
a magnetic charge (monopole). Then both fields has only angular components in the polar
coordinates in any plane x = const. Thus H0 = Hl , which is independent of the angle. Then
Eq. (2) gives H0 2πr = I,
I
H0 =
.
(5)
2πr
and B0 = µ0 H0 .
2
3. From the second Maxwell’s equation divB = 0 it follows that the normal component of
B should be continuous at z = 0. From the symmetry of the problem B has only normal
component at z = 0 in the loop we have chosen. Note that B = µH. It follows from Eq.(1)
that tangential component of H is continuous at z = 0, but it is not important for this
problem. Important is that normal component of H might be discontinuous.
4. Eq. (1) does not contain permeability. So H0 obeys this equation both at z > 0 and
z < 0, but because permeability is different in these regions, this solution does not obey the
boundary condition for normal component of B at z = 0. However in both these regions
equation for H is homogeneous (I=0). Thus solution can be multiplied by a constant factor.
Assume that H = a1 H0 at z > 0 and H = a2 H0 at z < 0. From Ampere’s law and continuity
of B we get two equations for coefficients a1 , a2 :
a1 + a2 = 2,
µ 1 a1 = µ 2 a2 .
(6)
(7)
The solution is
a1 =
a2 =
2µ2
,
µ1 +µ2
2µ1
.
µ1 +µ2
(8)
(9)
Finally
2µ2
H0 ,
µ1 + µ2
2µ1
H0 .
=
µ1 + µ2
H1 =
(10)
H2
(11)
(12)
Field B is axillary symmetric
2µ1 µ2
H0 .
µ1 + µ2
At µ1 = µ2 = µ we get H1 = H2 = H0 , B = µH0 .
B=
3
(13)
3 Thermodynamics
Heat capacities are normally positive, but there is an important class of exceptions: systems of
particles held together by gravity, such as stars and star clusters.
a) [6 pts.] For any system of particles held together by mutual gravitational attraction,
U potential = −2U kinetic ,
where each U refers to the total energy (of that type) for the entire system, averaged over some
sufficiently long time period. This result is known as the virial theorem. Consider a system of
two identical particles on circular orbits about their center of mass. Show that the gravitational
potential energy of this system is -2 times the total kinetic energy. Now suppose that you add
some energy to such a system and wait for the system to equilibrate. Does the average total kinetic
energy increase or decrease?
b) [4 pts.] A star can be modeled as a gas of particles that interact with each other only gravitationally. Assuming these particles are monatomic, and that the average temperature over the star
is T , what is the average kinetic energy of a particle in the star in terms of T ?
c) [10 pts.] Express the total energy of the star in terms of the average temperature, T , and
calculate the heat capacity. Note the sign.
d) [5 pts.] The gravitational potential energy of a star of uniform density, total mass, M , and
2
radius, R, is − 3GM
5R . Using this simple model, estimate the average temperature of the sun, whose
mass is 2 × 1030 kg, and whose radius is 7 × 108 m. You may assume that the sun is made entirely
of protons and electrons.
3 Thermodynamics
Heat capacities are normally positive, but there is an important class of exceptions: systems of
particles held together by gravity, such as stars and star clusters.
a) [6 pts.] For any system of particles held together by mutual gravitational attraction,
U potential = −2U kinetic ,
where each U refers to the total energy (of that type) for the entire system, averaged over some
sufficiently long time period. This result is known as the virial theorem. Consider a system of
two identical particles on circular orbits about their center of mass. Show that the gravitational
potential energy of this system is -2 times the total kinetic energy. Now suppose that you add
some energy to such a system and wait for the system to equilibrate. Does the average total kinetic
energy increase or decrease?
b) [4 pts.] A star can be modeled as a gas of particles that interact with each other only gravitationally. Assuming these particles are monatomic, and that the average temperature over the star
is T , what is the average kinetic energy of a particle in the star in terms of T ?
c) [10 pts.] Express the total energy of the star in terms of the average temperature, T , and
calculate the heat capacity. Note the sign.
d) [5 pts.] The gravitational potential energy of a star of uniform density, total mass, M , and
2
radius, R, is − 3GM
5R . Using this simple model, estimate the average temperature of the sun, whose
mass is 2 × 1030 kg, and whose radius is 7 × 108 m. You may assume that the sun is made entirely
of protons and electrons.
Solution
a) Each particle has a kinetic energy of 21 mv 2 , so the total kinetic energy of the system is Ukinetic =
mv 2 . If the distance of each particle from the center of mass is r, then the total potential energy of
2
Gm2
the two-particle system is Upotential = −Gm
2r . The gravitational force on each particle is 4r2 = ma,
where a is just the centripetal acceleration, a = v 2 /r. So
mv 2
Gm2
=
.
r
4r2
Then
Ukinetic = mv 2 =
Gm2
1
= − Upotential ,
4r
2
so
Upotential = −2Ukinetic .
The total energy is
Utotal = Ukinetic + Upotential = Ukinetic − 2Ukinetic
Utotal = −Ukinetic ,
so increasing the total energy decreases the kinetic energy.
b) According to the equipartition theorem, the average kinetic energy of each particle is 32 kT .
c) A star of N particles has a total kinetic energy of Ukinetic = 23 N kT . The total energy is therefore
3
Utotal = − N kT.
2
The heat capacity is C =
dUtotal
dT ,
so
3
C = − N k.
2
d) Using the fact that U kinetic = − 12 U potential ,
−3GM 2
5R
1 GM 2
1 GM M
T =
=
5 N kR
5 kR N
3
1
N kT = −
2
2
M/N is the mass per particle in the star. Since the star should be made of equal numbers of
protons and electrons, and mp me , the mass of each particle is approximately 21 mp .
2
N
m
−11
6.7 × 10
2 × 1030 kg 12 × 1.67 × 10−27 kg
2
kg
1
T =
5
(1.38 × 10−23 J/K) (7 × 108 m)
T ≈ 2 × 106 K
Problem 4: Quantum Mechanics
A semi-infinite square quantum well, shown in the figure is
given by the following potential:
∞
𝑉(𝑥) = �−𝑉0
0
𝑥≤0
0<𝑥<𝑎
𝑥≥𝑎
A particle of mass m is in the ground (bound) state in this well.
It has energy 𝐸, where 𝐸 < 0.
(a) [8 points] Write down the time-independent Schrödinger Equations for region 1 (0 < 𝑥 < 𝑎 ) and
region 2 (𝑥 ≥ 𝑎) shown in the figure, separately. Then write down their general solutions 𝜑1 (𝑥) and
𝜑2 (𝑥). Remember to include separate normalization constants for the two regions. You don’t need to
normalize the functions.
(b) [5 points] State the boundary conditions at (i) 𝑥 = 0 for 𝜑1 (𝑥), and at (ii) 𝑥 → ∞ for 𝜑2 (𝑥),
respectively, and simplify the two functions from their most general form accordingly
(c) [6 points] State the two boundary conditions for 𝜑1 (𝑥) and 𝜑2 (𝑥) at 𝑥 = 𝑎, and write down the
corresponding relationships between the simplified functions from part (b).
(d) [6 points] You are given that 𝑚𝑎2 𝑉0 /ℏ2 = 4. Using this fact, and the relationships between 𝜑1 (𝑥)
and 𝜑2 (𝑥) from part (c), find the ratio 𝜀 = |𝐸|/𝑉0 for the ground state energy E to ONE significant digit.
This is a numerical solution that requires you to use your calculator (hint: Make a table of 𝜀 and the
relevant quantity that is supposed to be zero, and start with 𝜀 = 0.5).
Problem 4 Solution:
(a) The time-independent Schrödinger Equation is:
−
ℏ2 𝑑2
𝜑(𝑥) + 𝑉(𝑥)𝜑(𝑥) = 𝐸𝜑(𝑥)
2𝑚 𝑑𝑥 2
(i) In region 1, 𝑉(𝑥) = −𝑉0, and we have (noting that −𝑉0 < 𝐸 < 0):
−
ℏ2 𝑑2
𝜑 (𝑥) − 𝑉0 𝜑1 (𝑥) = −|𝐸|𝜑1 (𝑥) →
2𝑚 𝑑𝑥 2 1
This gives us a sinusoidal solution of the form
𝑑2
2𝑚(𝑉0 − |𝐸|)
𝜑1 (𝑥) = −
𝜑1 (𝑥)
2
ℏ2
𝑑𝑥
𝜑1 (𝑥) = 𝐴 cos 𝑘𝑥 + 𝐵 sin 𝑘𝑥 = 𝐴′ 𝑒 𝑖𝑘𝑥 + 𝐵′ 𝑒 −𝑖𝑘𝑥 , 𝑘 = �2𝑚(𝑉0 − |𝐸|)/ℏ
(ii) In region 2, 𝑉(𝑥) = 0, and we have (noting that 𝐸 < 0):
−
ℏ2 𝑑2
𝜑 (𝑥) = −|𝐸|𝜑2 (𝑥) →
2𝑚 𝑑𝑥 2 2
And we have an exponential solution of the form
𝑑2
2𝑚|𝐸|
𝜑2 (𝑥) =
𝜑2 (𝑥)
2
ℏ2
𝑑𝑥
𝜑2 (𝑥) = 𝐶𝑒 𝑞𝑥 + 𝐷𝑒 −𝑞𝑥 , 𝑞 = �2𝑚|𝐸|/ℏ
(b) Boundary conditions at 𝑥 = 0 and 𝑥 → ∞:
(i) Because the potential is infinite at 𝑥 = 0, we have 𝜑0 (𝑥) = 𝐴 = 0 , and so
𝜑1 (𝑥) = 𝐵 sin 𝑘𝑥
(ii) Because the particle is bound to the well near 𝑥 = 0, we must have 𝜑2 (𝑥) → 0 as 𝑥 → ∞. Note that
𝐶𝑒 𝑞𝑥 → ∞, and 𝐷𝑒 −𝑞𝑥 → 0 as 𝑥 → ∞. So we must have 𝐶 = 0 and
𝜑2 (𝑥) = 𝐷𝑒 −𝑞𝑥
(c) The wave function and its derivative must both be continuous at 𝑥 = 𝑎.
(i) Continuity of the wave function at 𝑥 = 𝑎: 𝜑1 (𝑎) = 𝜑2 (𝑎) → 𝐵 sin 𝑘𝑎 = 𝐷𝑒 −𝑞𝑎
(ii) Continuity of 𝑑𝜑/𝑑𝑥 at 𝑥 = 𝑎 gives us
𝑑𝜑2
𝑑𝜑1
=
→ 𝑘𝐵 cos 𝑘𝑎 = −𝑞𝐷𝑒 −𝑞𝑎
:
�
�
𝑑𝑥 𝑥=𝑎
𝑑𝑥 𝑥=𝑎
(d) Dividing [2] by [1] yields 𝑘 cot 𝑘𝑎 = − 𝑞
[1]
[2]
Multiplying both sides by the well width a gives us a transcendental equation that is dimensionless:
𝑥 cot 𝑥 + 𝑦 = 0 where
𝑥 = 𝑘𝑎 = �2𝑚𝑎2 (𝑉0 − |𝐸|)/ℏ
and
𝑦 = 𝑞𝑎 = �2𝑚𝑎2 |𝐸|/ℏ
We are given that
𝑚𝑎2 𝑉0 /ℏ2 = 4 → �𝑚𝑎2 𝑉0 /ℏ = 2
so that
𝑥 = 2�2(1 − |𝐸|/𝑉0 ) = 2�2(1 − 𝜀)
and
𝑦 = 2�2|𝐸|/𝑉0 = 2√2𝜀
Trying different values of e starting at 0.5:
e
0.5
0.6
0.4
0.3
0.35
0.37
0.38
0.39
x
2
1.788854
2.19089
2.366432
2.280351
2.244994
2.227106
2.209072
y
2
2.19089
1.788854
1.549193
1.67332
1.720465
1.74356
1.766352
xcot(x)+y
1.084685
1.794514
0.224448
-0.86619
-0.28494
-0.07353
0.028224
0.127514
The answer to one significant digit is 0.4 (0.38 to 2 significant digits).
5. General/Modern Physics
The Sun is powered by nuclear fusion in its core. The reaction is called proton-proton chain (PP
chain), which converts hydrogen nuclei (protons p) into helium nuclei (4 He). The overall reaction
can be thought as that every four protons are converted into one 4 He nucleus, with an energy
release Q. The masses of proton and 4 He nucleus are mp = 1.0076mu and m4 He = 4.0026mu ,
respectively, where mu = 1.66 × 10−27 kg is the atomic mass unit.
The mass and radius of the Sun are M = 1.99 × 1030 kg and R = 6.96 × 108 m, respectively. The
Sun-Earth distance (called astronomical unit) is d = 1.496 × 1011 m.
[N.B. Use the constants given in this problem.]
(a) [5 pts.] What is the energy release Q in units of MeV for each 4p-to-4 He reaction?
(b) [6 pts.] For the reaction to occur, one would naively argue that two protons should reach
a separation of the order of R = 10−15 m for the nuclear force to overcome the Coulomb
repelling force. The core temperature of the Sun is about T = 107 K. For two protons both
moving at the most probable thermal speed, what is the closest separation they can reach?
Your answer should be much larger than R. What effect can help to overcome the Coulomb
barrier, which then can make the nuclear fusion slowly happen in the core of the Sun?
(c) [7 pts.] The majority of the released energy Q is in the form of photons and a small fraction
is in the form of electron neutrinos and kinetic energy of particles. The photon-electron
interaction cross-section is σT = 6.65 × 10−29 m2 . The neutrino-electron interaction crosssection is on the order of σeν = 10−48 m2 . By making the simplification that the Sun is
composed of fully ionized hydrogen gas with uniform density, estimate the mean free paths of
photons and neutrinos in the Sun, in units of the radius of the Sun R . Based on the results,
which can be used to better probe the core of the Sun, photons or neutrinos?
(d) [7 pts.] In the Sun, about 600 million tons of hydrogen are converted into helium per second.
Each 4p-to-4 He reaction produces two electron neutrinos. What is the expected solar neutrino
flux (in units of m−2 s−1 ) at the Earth distance? Neutrino experiments only detect about
one third of the electron neutrino flux predicted by the solar model. To the best of your
knowledge, what is the solution to such a discrepancy?
1
5. General/Modern Physics – Solution
(a) [5 pts.] What is the energy release Q in units of MeV for each 4p-to-4 He reaction?
Q = (4mp − m4 He )c2 = 0.0278mu c2 = 25.9MeV.
(1)
(b) [6 pts.] For the reaction to occur, one would naively argue that two protons should reach
a separation of the order of R = 10−15 m for the nuclear force to overcome the Coulomb
repelling force. The core temperature of the Sun is about T = 107 K. For two protons both
moving at the most probable thermal speed, what is the closest separation they can reach?
Your answer should be much larger than R. What effect can help to overcome the Coulomb
barrier, which then can make the nuclear fusion slowly happen in the core of the Sun?
From energy conservation, the closest separation r of the two protons satisfies
1
1 e2
2 × mp vp2 =
,
2
4π0 r
(2)
where vp2 = 2kT /mp . We then have
r=
e2
= 8.34 × 10−13 m,
8π0 kT
(3)
much larger than R. Quantum tunneling is the main effect to overcome the Coulomb barrier
for the fusion to happen.
(c) [7 pts.] The majority of the released energy Q is in the form of photons and a small fraction
is in the form of electron neutrinos and kinetic energy of particles. The photon-electron
interaction cross-section is σT = 6.65 × 10−29 m2 . The neutrino-electron interaction crosssection is on the order of σeν = 10−48 m2 . By making the simplification that the Sun is
composed of fully ionized hydrogen gas with uniform density, estimate the mean free paths of
photons and neutrinos in the Sun, in units of the radius of the Sun R . Based on the results,
which can be used to better probe the core of the Sun, photons or neutrinos?
The number density of electrons in the Sun is
M
= 8.44 × 1029 m−3 .
3
mp 4πR /3
(4)
1
= 1.78 × 10−2 m = 2.56 × 10−11 R .
ne σT
(5)
ne =
The mean free path of photons is
lγ =
The mean free path of neutrinos is
lν =
1
= 1.19 × 1018 m = 1.70 × 109 R .
ne σeν
(6)
While photons come out of the Sun after many interactions, neutrinos can escape almost
directly from the Sun. Therefore, neutrinos are a better probe of the core of the Sun.
2
(d) [7 pts.] In the Sun, about 600 million tons of hydrogen are converted into helium per second.
Each 4p-to-4 He reaction produces two electron neutrinos. What is the expected solar neutrino
flux (in units of m−2 s−1 ) at the Earth distance? Neutrino experiments only detect about
one third of the electron neutrino flux predicted by the solar model. To the best of your
knowledge, what is the solution to such a discrepancy?
Let ṀH = 600 million tons per second = 6×1011 kg s−1 . The number of 4p-to-4 He reactions per
second is ṀH /(4mp ) and the number of electron neutrinos produced per second is ṀH /(2mp ).
The expected electron neutrino flux at the Earth distance is
F =
ṀH 1
= 6.4 × 1014 m−2 s−1 .
2mp 4πd2
(7)
The discrepancy is explained by a quantum mechanical phenomenon called neutrino oscillation
— electron neutrinos, after produced in the core of the Sun, have a probability of transforming
into other types of neutrinos (i.e., µ and τ neutrinos) during their propagation to the Earth.
3
Problem 6: Electricity and Magnetism:
Pressure of Electromagnetic Radiation: This effect was originally predicted by J. C. Maxwell (1871) and
demonstrated experimentally by P. Lebedev (1900) in Russia.
A monochromatic, linearly polarized (in the x-direction) electromagnetic wave is incident normally on
(i.e. traveling in a direction perpendicular to) a surface at the xy plane.
(a) [5 points] Write down both the electric and magnetic fields as functions of coordinate z and time t, in
terms of the electric field amplitude 𝐸0 , angular frequency 𝜔, and the speed of light c.
�⃗〉, the time-averaged Poynting vectors, in terms
(b) [5 points] Write down both �𝑺⃗, the instantaneous, and 〈𝑺
of the electric field amplitude 𝐸0 , and c, 𝜇0 and/or 𝜖0 .
--------------
�⃗, representing the time-averaged flow (per unit area per unit time) of momentum, is given by
The vector �𝒈
�⃗〉
〈𝑺
��⃗ =
𝒈
(1)
𝑐
(Poynting vector represents flow of energy, and for photons, momentum is given by 𝑝 = 𝐸/𝑐). If the
wave is traveling originally in vacuum and the medium on the other has refractive index 𝑛, then the
coefficient of reflection is given by
(𝑛 − 1)2
𝑅=
(2)
(𝑛 + 1)2
With this assumption:
(c) [5 points] Further assume that the refractive medium does not absorb radiation, find an expression for
the momentum transferred to the medium per unit area over time interval ∆𝑡. Now express the radiation
pressure P on the medium in terms of the radiation intensity I (the intensity is the power carried by the
wave per unit area).
(d) [5 points] Calculate the radiation pressure P (in appropriate SI units) for an incident wave with
intensity of I = 10.0 W/m2, and a receiving medium with refractive index 𝑛 = 2.00.
----------------
(e) [5 points] Alternately, for a non-refractive medium, there is neither reflection nor transmission. All
energy is absorbed. For such a medium, find the radiation pressure P in terms of intensity I.
Problem 6 solution:
(a) Electric field of wave traveling in the ±𝑧 direction (normally incident on the xy plane):
𝜔
�⃗(𝑧, 𝑡) = +𝒙
�𝐸0 cos � 𝑧 ∓ 𝜔𝑡�
(3)
𝑬
𝑐
and the magnetic field:
𝜔
𝐸0
��⃗(𝑧, 𝑡) = ±𝒚
� cos � 𝑧 ∓ 𝜔𝑡�
(4)
𝑩
𝑐
𝑐
The cosine function here can be replaced by sine and an arbitrary, but COMMON phase factor can be
added to the two fields. For the remainder of this problem we will assume the +𝑧 direction for
propagation.
(b) The Poynting Vector, which gives both a direction and magnitude of the flow of energy (power per
��⃗. In vacuum (which is the incident medium for this problem) it
unit area) is defined by �𝑺⃗ = �𝑬⃗ × �𝑯
becomes
1
�𝑺⃗ = �𝑬⃗ × �𝑩
�⃗
(5)
𝜇0
Substituting (3) and (4) into (5), assuming +𝑧 direction of propagation, we get
1
𝜔
𝜔
𝐸02
�⃗ = (𝒙
�×𝒚
�) cos2 � 𝑧 − 𝜔𝑡� = 𝒛�𝜖0 𝐸02 cos2 � 𝑧 − 𝜔𝑡�
𝑺
𝑐
𝜇0
𝑐
𝑐
𝜔
𝑐
1
2
and taking the time average at any location, we have 〈cos2 � 𝑧 − 𝜔𝑡�〉 = , and thus the time averaged
Poynting vector is
1
�⃗〉 = 𝒛�𝜖0 𝐸02
〈𝑺
2
The magnitude of this vector is known as the intensity:
1
𝐼 = 𝜖0 𝐸02
2
(c) By conservation of energy, and assuming no absorption,
an incident wave of intensity I would give a reflected wave
of intensity RI and transmitted (will eventually pass out of
the medium) wave of intensity (1−R)I, as illustrated by
figure 6-1. Thus only fraction R of the energy flow is
reflected.
Figure 6-1
The magnitude of the momentum flow, by the definition
�⃗〉 /c is given by 𝑔 = 𝐼/𝑐.
��⃗ = 〈𝑺
𝒈
The reflected part of the wave then undergoes a change of momentum of ∆𝑝 = −2𝑅𝑔𝐴∆𝑡 over area A and
time interval ∆𝑡. The momentum transferred per unit area over time ∆𝑡 is therefore
�⃗〉�
(𝑛 − 1)2 1
(𝑛 − 1)2 𝜖0 2
∆𝑝
�〈𝑺
2
= 2𝑅𝑔∆𝑡 = 2𝑅
∆𝑡 = 2
𝜖
𝐸
∆𝑡
=
𝐸 ∆𝑡
(𝑛 + 1)2 2𝑐 0 0
(𝑛 + 1)2 𝑐 0
𝑐
𝐴
In other words the radiation pressure on the surface is
(𝑛 − 1)2 𝐼
1 ∆𝑝 (𝑛 − 1)2 𝜖0 2
𝐼
𝑃=
=
𝐸
=
2
= 2𝑅
0
2
2
(𝑛 + 1) 𝑐
𝐴 ∆𝑡 (𝑛 + 1) 𝑐
𝑐
(d) 𝑛 = 2.00 → 𝑅 = (𝑛 − 1)2 /(𝑛 + 1)2 = (1.00)2 /(3.00)2 =1/9. We are given also I = 10.0 W/m2.
kg∙m2 1 1
2 𝐼 2 10.0 s 2 ∙ s ∙ m2
kg∙m 1
N
𝑃=
= ∙
= 7.41 × 10−9 2 ∙ 2 = 7.41 × 10−9 2 = 7.41 × 10−9 Pa
m
9𝑐 9
m
s
m
3.00 × 108
s
where the symbol Pa stands for “pascals”.
(e) With complete absorption, all the energy flow and momentum flow of the wave stops at the surface.
The medium absorbs all the intensity, and all of the momentum flow is transferred to the surface. The
momentum transferred per unit area over time ∆𝑡 is therefore
�⃗〉�
1
∆𝑝
�〈𝑺
= 𝑔∆𝑡 =
∆𝑡 = 𝜖0 𝐸02 ∆𝑡
𝑐
2𝑐
𝐴
In other words the radiation pressure on the surface is now
1 ∆𝑝 𝜖0 2 𝐼
𝑃=
= 𝐸 =
𝐴 ∆𝑡 2𝑐 0 𝑐
This is the same pressure as in the case where 2𝑅 = 1.
Problem 7: Statistical Mechanics
An ensemble of N weakly interacting atoms on a fixed lattice is placed in a uniform external magnetic
field B = 15 T (tesla), directed along the z direction. Each atom (they are distinguishable by their lattice
location) can be either in a Sz = +1/2 (spin-up) or Sz = −1/2 (spin-down) state, and the magnetic potential
energy of atom i is given by Ei = −µB(Sz)i, where µ = 1.0×10-23 J/T (joule/tesla).
The ensemble is immersed in a thermal bath at temperature T. You might find it useful (making the
math less cumbersome) to use the abbreviation 𝑥 = 𝜇𝐵/𝑘𝑇, where k is the Boltzmann Constant.
(a) [8 points] Find general expressions for the fractions f+ and f- of atoms in the spin-up and spin-down
states in terms of x.
(b) [7 points] In one mole (N = NA = 6.0×1023) of atoms, we find a total of N−=1.5×1023 atoms in the spindown state. Find the numerical value of the temperature T in kelvins. Note k = 1.38×10-23 J/K.
(c) [10 points] Find an expression for the entropy of the ensemble, in terms of x, N, and k.
There are a number of different approaches to part (c). In one of the several cases, you might find the
Stirling Approximation useful: ln(𝑛!) ≈ 𝑛 ln 𝑛 − 𝑛 for very large 𝑛.
Solution to Problem 7:
(a) A spin-down atom has energy 𝐸− = +12𝜇𝐵, which is at a higher energy than a spin-up atom at
𝐸+ = −12𝜇𝐵. The Boltzman distribution gives us that
−(𝐸− −𝐸+ )
𝑁−
= 𝑒 𝑘𝑇
= 𝑒 −𝑥
𝑁+
where 𝑥 = 𝜇𝐵/𝑘𝑇
The fraction of spin-down atoms is then given by
𝑓− =
𝑁−
𝑁+ + 𝑁−
Dividing top and bottom by 𝑁+ then gives
𝑁− /𝑁+
𝑁− /𝑁+
𝑒 −𝑥
𝑓− =
=
=
𝑁+ /𝑁+ + 𝑁− /𝑁+ 1 + 𝑁− /𝑁+ 1 + 𝑒 −𝑥
Similarly:
𝑓+ =
𝑁+
𝑁+ /𝑁+
1
1
=
=
=
𝑁+ + 𝑁− 𝑁+ /𝑁+ + 𝑁− /𝑁+ 1 + 𝑁− /𝑁+ 1 + 𝑒 −𝑥
𝜇𝐵
�
𝑘𝑇
𝑓− =
,
𝜇𝐵
1 + exp �− �
𝑘𝑇
exp �−
𝑓+ =
1
1 + exp �−
𝜇𝐵
�
𝑘𝑇
Alternatively we can express the answers in terms of hyperbolic functions:
𝑓− =
𝑓+ =
e−𝑥
e−𝑥/2
cosh(𝑥 ⁄2) − sinh(𝑥 ⁄2)
=
=
−𝑥
+𝑥/2
−𝑥/2
2cosh(𝑥 ⁄2)
1+e
e
+e
1
e+𝑥/2
cosh(𝑥 ⁄2) + sinh(𝑥 ⁄2)
=
=
−𝑥
+𝑥/2
−𝑥/2
2cosh(𝑥 ⁄2)
1+e
e
+e
Yet a different Approach is to use the Partition function:
𝑍 = � exp �−
𝑖
𝐸+
𝐸−
𝜇𝐵
𝜇𝐵
𝐸𝑖
� = exp �− � + exp �− � = exp �+
� + exp �−
� = 𝑒 +𝑥/2 + 𝑒 −𝑥/2
𝑘𝑇
𝑘𝑇
𝑘𝑇
2𝑘𝑇
2𝑘𝑇
The fraction of occupation in each state is equal to the probability of finding an atom in that state:
We have
and
𝑓𝑖 = 𝑝𝑖 =
1
𝐸𝑖
exp �− �
𝑘𝑇
𝑍
+12𝜇𝐵
𝐸−
𝑒 −𝑥/2
𝑒 −𝑥
exp �− � = exp �−
=
� = 𝑒 −𝑥/2 → 𝑓− = 𝑝− = +𝑥/2
𝑘𝑇
𝑘𝑇
𝑒
+ 𝑒 −𝑥/2 1 + 𝑒 −𝑥
−12𝜇𝐵
𝑒 +𝑥/2
1
𝐸+
+𝑥/2
→ 𝑓+ = 𝑝+ = +𝑥/2
=
exp �− � = exp �−
�=𝑒
𝑘𝑇
𝑘𝑇
𝑒
+ 𝑒 −𝑥/2 1 + 𝑒 −𝑥
(b) N = NA = 6.0×1023, N−=1.5×1023  N+= N − N−=4.5×1023  N−/ N+=1/3
But
1
𝜇𝐵
𝑁− 𝑓−
=
= e−𝑥 → e−𝑥 =
→ e𝑥 = 3 → 𝑥 = ln 3 →
= ln 3
𝑁+ 𝑓+
3
𝑘𝑇
and so
𝑇=
𝜇𝐵
1.0 × 10−23 J/T ∙ 15T
=
= 9.9 K
𝑘 ln 3 1.38 × 10−23 J/K ∙ ln 3
(c) In many thermal physics textbooks, the most generalized expression/definition for entropy is (for
quantized energy states of energy 𝐸𝑖 and N total atoms):
𝑆 = −𝑁𝑘 � 𝑝𝑖 ln 𝑝𝑖
𝑖
where 𝑝𝑖 is ther probability of occupation of state i with energy 𝐸𝑖 . Applied to our case, the we have
1
𝑒 −𝑥
1
𝑒 −𝑥
ln
ln �
��
𝑆 = −𝑁𝑘 (𝑝− ln 𝑝− + 𝑝+ ln 𝑝+ ) = −𝑁𝑘 �
�+
�
−𝑥
−𝑥
−𝑥
1+𝑒
1+𝑒
1+𝑒
1 + 𝑒 −𝑥
Expanding, we get:
𝑒 −𝑥
1
𝑆
𝑒 −𝑥
1
−𝑥 )
=−
ln(𝑒
ln(1 + 𝑒 −𝑥 ) −
ln(1) +
ln(1 + 𝑒 −𝑥 )
+
−𝑥
−𝑥
−𝑥
1+𝑒
1+𝑒
1 + 𝑒 −𝑥
𝑁𝑘
1+𝑒
𝑒 −𝑥
1
𝑥𝑒 −𝑥
+
+
=
� ln(1 + 𝑒 −𝑥 )
�
−𝑥
−𝑥
1+𝑒
1+𝑒
1 + 𝑒 −𝑥
And so we have
𝑆 = 𝑁𝐾 �ln(1 + 𝑒
−𝑥 )
𝑥𝑒 −𝑥
+
�
1 + 𝑒 −𝑥
Alternatively: We really should be using the formalism for Canonical Ensemble (system in thermal
bath), which is formulated based on the Partition Function (again for N atoms):
𝜕𝐹
𝜕
𝑇 𝜕𝑍 𝑑𝑥
(𝑁𝑘𝑇 ln 𝑍) = 𝑁𝑘 �ln 𝑍 + � � � ��
𝑆=−
=+
𝜕𝑇
𝜕𝑇
𝑍 𝜕𝑥 𝑑𝑇
𝑥
𝑥
𝑥
𝑥
+
−
+
−𝑥
ln 𝑍 = ln �𝑒 2 + 𝑒 2 � = ln �𝑒 2 (1 + 𝑒 )� = + ln(1 + 𝑒 −𝑥 )
2
𝑥
𝑥
𝑥
𝜕 +𝑥
1
𝑑𝑥
𝑑 𝜇𝐵
𝜇𝐵
𝑥
𝜕𝑍
=
�𝑒 2 + 𝑒 −2 � = �𝑒 +2 − 𝑒 −2 �,
=
� �=− 2=−
2
𝑑𝑇 𝑑𝑇 𝑘𝑇
𝑘𝑇
𝑇
𝜕𝑥 𝜕𝑥
𝑥
so
𝑥
1
𝑇 𝜕𝑍 𝑑𝑥
𝑇 𝑒 +2 − 𝑒 −2 𝑥
𝑥 1 − 𝑒 −𝑥
1
= 𝑥
→
�
�
�
�
=
−
=
−
𝑍 𝜕𝑥 𝑑𝑇
2 𝑒 +𝑥2 + 𝑒 −𝑥2 𝑇
2 1 + 𝑒 −𝑥
𝑍 𝑒 +2 + 𝑒 −𝑥2
𝑥
𝑥 1 + 𝑒 −𝑥 𝑥 1 − 𝑒 −𝑥
𝑥 1 − 𝑒 −𝑥
𝑆 = 𝑁𝑘 � + ln(1 + 𝑒 −𝑥 ) −
=
𝑁𝑘
−
+ ln(1 + 𝑒 −𝑥 )�
�
�
2
2 1 + 𝑒 −𝑥 2 1 + 𝑒 −𝑥
2 1 + 𝑒 −𝑥
𝑥 1 + 𝑒 −𝑥 − 1 + 𝑒 −𝑥
𝑥𝑒 −𝑥
−𝑥 )�
= 𝑁𝑘 �
=
𝑁𝑘
+
ln(1
+
𝑒
+ ln(1 + 𝑒 −𝑥 )�
�
2
1 + 𝑒 −𝑥
1 + 𝑒 −𝑥
which is the same answer as before.
Yet another (perhaps more familiar) approach: For a micro-canonical ensemble, entropy is given by
Boltzmann’s relation: 𝑆 = 𝑘 ln Ω, where Ω is the number of micro-states that corresponds to a given
macro-state specified by energy E instead of temperature T . In our case, we have a canonical ensemble
the macro-state is completely specified by the temperature T (taking µ and B to be constants), which in
turns specifies the fraction of atoms in the spin-down (or spin-up) state. However, the calculation of
average entropy S yields the same results for either approach. But as Prof. Gondolo pointed out, the
fluctuations would be completely wrong.
Counting the number of ways the 𝑁− spin-down atoms can be arranged out of N total atoms (lattice sites)
𝑁 ∙ (𝑁 − 1) ∙ (𝑁 − 2) ⋯ (𝑁 − 𝑁− + 1)
Ω=
𝑁− ∙ (𝑁− − 1) ∙ (𝑁− − 2) ⋯ 1
Here the numerator denotes the choice of 𝑁 possible sites for the first spin-down atom, times (𝑁 − 1) for
the second ….etc. and (𝑁 − 𝑁− + 1) sites for the last. The denominator is the number of different
equivalent re-arrangements (permutations) of the 𝑁− spin-down atoms.
The numerator can be rewritten as
𝑁!
𝑁!
=
(𝑁 − 𝑁− )! 𝑁+ !
and the denominator is obviously 𝑁− ∙ (𝑁− − 1) ∙ (𝑁− − 2) ⋯ 1 = 𝑁− ! and so we have
𝑁!
𝑁!/𝑁+ !
=
Ω=
𝑁+ ! ∙ 𝑁− !
𝑁− !
𝑁 ∙ (𝑁 − 1) ∙ (𝑁 − 2) ⋯ (𝑁 − 𝑁− + 1) =
Alternatively, we can distribute spin-up atoms and get the same expression
𝑁 ∙ (𝑁 − 1) ∙ (𝑁 − 2) ⋯ (𝑁+ + 1) 𝑁!/𝑁− !
𝑁!
Ω=
=
=
𝑁+ ! ∙ 𝑁− !
𝑁+ ∙ (𝑁+ − 1) ∙ (𝑁+ − 2) ⋯ 1
𝑁+ !
Either way the entropy is then given by:
𝑁!
� = 𝑘(ln 𝑁! − ln 𝑁+ ! − ln 𝑁− !)
𝑁+ ! ∙ 𝑁− !
Applying Stirling’s approximation (and noting that 𝑁 = 𝑁+ + 𝑁− ):
𝑆
= (𝑁 ln 𝑁 − 𝑁) − (𝑁+ ln 𝑁+ − 𝑁+ ) − (𝑁− ln 𝑁− − 𝑁− ) = 𝑁 ln 𝑁 − 𝑁+ ln 𝑁+ − 𝑁− ln 𝑁−
𝑘
𝑁
𝑁e−𝑥
= 𝑁 ln 𝑁 − 𝑁+ ln
−
𝑁
ln
−
1 + e−𝑥
1 + e−𝑥
−𝑥 )
= 𝑁 ln 𝑁 − 𝑁+ ln 𝑁 + 𝑁+ ln(1 + e
− 𝑁− ln 𝑁 − 𝑁− ln(e−𝑥 ) + 𝑁− ln(1 + e−𝑥 )
𝑁𝑥e−𝑥
−𝑥
= (𝑁 − 𝑁+ − 𝑁− ) ln 𝑁 + (𝑁+ + 𝑁− ) ln(1 + e ) +
1 + e−𝑥
−𝑥
𝑁𝑥e
= 𝑁 ln(1 + e−𝑥 ) +
1 + e−𝑥
𝑆 = 𝑘 ln Ω = 𝑘 ln �
Again the same answer as before.
𝑆 = 𝑁𝑘 �ln(1 + e−𝑥 ) +
𝑥e−𝑥
�
1 + e−𝑥
8 Special Relativity
Two powerless rockets are heading towards each other on a collision course. As measured by Sarah,
a stationary Earth observer, Rocket A has speed 0.800c, Rocket B has speed 0.600c, both rockets
are 50.0 m in length, and they are initially 2.52 Tm = 2.52 × 1012 m apart. Rocket A is piloted by
Ann, a stationary observer in Rocket A’s frame, and Rocket B is piloted by Bob. It will take Ann
and Bob 50 minutes (in their respective frames) to evacuate their rockets. [h]
a) [5 pts.] What is the proper length of Rocket A?
b) [5 pts.] What is the velocity of Rocket A according to Bob?
c) [5 pts.] What is the length of Rocket A according to Bob?
d) [5 pts.] From Sarah’s perspective, how much time will pass before the rockets collide?
e) [5 pts.] According to Ann, how much time will pass before the collision? Will she escape?
Solution
a) The proper length of Rocket A is L0 = γL, where L is
pthe length perceived by an observer
moving with speed v with respect to Rocket A, and γ = 1/ 1 − (v/c)2 .
γ=√
LA
0 =
5
1
=
3
1 − 0.8002
5
× (50 m) = 83.3 m
3
b) In Sarah’s frame, if we take Rocket A to be moving in the positive x direction, its velocity is
vxA = 0.800c. Then Rocket B is moving in the negative x direction with velocity vxB = −0.600c.
The velocity of Rocket A as observed by Bob is given by
vxA ′ =
vxA − v
,
1 − vxA v/c2
where vxA is Rocket A’s velocity in Sarah’s frame, and v is the velocity of Bob’s frame with respect
to Sarah’s, i.e. v = vxB .
0.800c − (−0.600c)
vxA ′ =
1 − (0.800c)(−0.600c)/c2
vxA ′ = 0.946c.
Note that an observer on Rocket A would measure Rocket B to be approaching at this speed, as
well.
c) Just as in part 1, the length of Rocket A in Bob’s frame is related to the proper length of Rocket
A
A by LA
0 = γL , where now
1
γ=√
= 3.083.
1 − 0.9462
So the length of Rocket A as observed by Bob is
LA =
1
× (83.3 m) = 27.0 m.
3.083
d) Sarah observes that, in the time before the rockets collide, Rocket A travels a distance of
0.800c∆t, while Rocket B travels 0.600c∆t. The total distance traveled by the two rockets is
D = (0.800c + 0.600c)∆t = 2.52 Tm.
∆t =
2.52 × 1012 m
1.4 × (3.0 × 108 m/s)
∆t = 6 × 103 s = 100 mins.
e) The proper time interval, ∆t0 , is related to another observer’s coordinate time interval by
∆t = γ∆t0 . We already calculated the Lorentz factor in part 1: γ = 35 , so the time before the
collision is
3
∆t0 = × 100 mins. = 60 mins.
5
She and her crew will survive.
9. Quantum mechanics
Consider a non-relativistic spinless particle in a potential V (r).
(a) [8 pts.] Using the Schrödinger equation, prove Ehrenfest’s theorem for the time derivative of the expectation value hOi = hψ|O|ψi of an operator O(r, p, t) in a state |ψ(t)i:
i
∂O
d
hOi = h[H, O]i +
.
dt
~
∂t
(b) [10 pts.] Prove that the rate of change of the expectation value of the orbital angular
momentum L is equal to the expectation value of the torque:
d
hLi = hNi,
dt
where
N = r × (−∇V ) .
(c) [7 pts.] Show that dhLi/dt = 0 for any spherically symmetric potential, i.e. for such a
potential angular momentum is conserved.
1
9. Quantum mechanics
Consider a non-relativistic spinless particle in a potential V (r).
(a) Using the Schrödinger equation, prove Ehrenfest’s theorem for the time derivative of
the expectation value hOi = hψ|O|ψi of an operator O(r, p, t) in a state |ψ(t)i:
i
∂O
d
hOi = h[H, O]i +
.
dt
~
∂t
Using Schrödinger’s equation i~(∂Ψ/∂t) = Hψ,
+ * *
+ * +
∂O
∂Ψ
d
∂Ψ hΨ|O|Ψi =
Ψ + ΨO
OΨ + Ψ
∂t
∂t
dt
∂t *
+
* + 1
1
∂O
=−
HΨOΨ +
ΨOHΨ +
i~
i~
∂t
∂O
i
.
= hHO − OHi +
~
∂t
(b) Prove that the rate of change of the expectation value of the orbital angular momentum
L is equal to the expectation value of the torque:
d
hLi = hNi,
dt
where
N = r × (−∇V ) .
Component by component,
d
i
hLx i = h[H, Lx ]i .
dt
~
[H, Lx ] =
1 2
p , Lx + [V, Lx ] .
2m
The first term is zero, because p2 is invariant under rotations; the second term would
be zero too if V were a function of r = |r| only [this is part (b)]. In general, using
[V, px ] = i~
∂V
,
∂x
[V, py ] = i~
2
∂V
,
∂y
[V, pz ] = i~
∂V
,
∂z
we find
[H, Lx ] = [V, ypz − zpy ] = y[V, pz ] − z[V, py ] = yi~
∂V
∂V
− zi~
= i~[r × (∇V )]x .
∂z
∂y
So
dhLx i
= − h[r × (∇V )]x i .
dt
The same goes for the other two components, so
dhLi
= hr × (−∇V )i = hNi.
dt
(b) Show that dhLi/dt = 0 for any spherically symmetric potential, i.e. for such a potential
angular momentum is conserved.
Either proceed as above and notice that [H, L] = 0, or reason as follows. If V (r) =
V (r), then ∇V = (∂V /∂r)r̂, and r × r = 0, so dhLi/dt = 0.
3
10. Classical Physics
A thin, incompressible rod with length l and negligible mass has two point masses A and B (each
with mass m) attached at the two ends. The system is initially put at rest on a smooth horizontal
surface. Another point mass C (with mass m and moving on the surface with initial velocity v0 )
has an elastic collision with mass A at an angle θ (see the figure). After the collision, mass C is
observed to bounce back (i.e., it moves in the opposite direction to its initial motion), with a
velocity v0′ . The motion of the AB system can be decomposed into a center-of-mass velocity Vcm
and a rotation with angular velocity ω around the center of mass.
[N.B. The expressions in (a), (b), and (c) should be in terms of m, l, θ, v0 , v0′ , Vcm , and ω.]
(a) [6 pts.] Write down the expression of momentum conservation of the whole system (A+B+C).
(b) [6 pts.] Write down the expression of energy conservation of the whole system.
(c) [6 pts.] Write down the angular momentum conservation of the whole system about the
position of mass A (its position right before the collision).
(d) [7 pts.] Based on the above three conservation equations, solve Vcm , ω, and v0′ . That is,
express them in terms of m, l, θ, and v0 .
1
10. Classical Physics – Solution
(a) [6 pts.] Write down the expression of momentum conservation of the whole system.
Before the collision, the total momentum is mv0 , along the CA direction. After the collision,
the momentum from C is mv0′ (in the −CA direction) and that from the system AB is 2mVcm
(which should be in the CA direction). We then have
mv0 = −mv0′ + 2mVcm .
(1)
(b) [6 pts.] Write down the expression of energy conservation of the whole system.
In this problem, the total kinetic energy of the whole system is conserved. Before the collision,
the kinetic energy is mv02 /2. After the collision, the kinetic energy from C is mv0′ 2 /2. The
2 /2
kinetic energy from the system AB is the sum of the center-of-mass kinetic energy (2m)Vcm
and the kinetic energy in the center-of-mass reference frame, 2 × [m(ωl/2)2 /2]. We then have
1
1
1
1
2
′2
2
2
mv = mv0 + × (2m)Vcm + 2 × m(ωl/2) .
(2)
2 0
2
2
2
(c) [6 pts.] Write down the angular momentum conservation of the whole system about the
position of mass A (its position right before the collision).
Before the collision, the angular momentum of the whole system about point A is zero.
After the collision, the angular momentum from C is zero. For the system AB, the angular
momentum is the sum of the center-of-mass angular momentum (2m)Vcm (l/2) sin θ and that
in the center-of-mass reference frame, −2 × m(ωl/2)(l/2). We have
0 = (2m)Vcm (l/2) sin θ − 2 × m(ωl/2)(l/2).
(3)
(d) [7 pts.] Based on the above three conservation equations, solve Vcm , ω, and v0′ .
The three equations in (a), (b), and (c) reduce to
v0 = −v0′ + 2Vcm ,
1
2
2
v02 = v0′ + 2Vcm
+ (ωl)2 ,
2
ωl = 2Vcm sin θ.
(4)
(5)
(6)
The solutions for Vcm , ω, and v0′ are
2
v0 ,
3 + sin2 θ
4 sin θ v0
,
3 + sin2 θ l
cos2 θ
v0 .
3 + sin2 θ
Vcm =
ω =
v0′ =
2
(7)
(8)
(9)