Download Common Exam - 2010 Department of Physics University of Utah August 21, 2010

Common Exam - 2010
Department of Physics
University of Utah
August 21, 2010
Examination booklets have been provided for recording your work and your solutions.
Please note that there is a separate booklet for each numbered question (i.e., use
booklet #1 for problem #1, etc.).
To receive full credit, not only should the correct solutions be given, but a sufficient
number of steps should be given so that a faculty grader can follow your reasoning.
Define all algebraic symbols that you introduce. If you are short of time it may be helpful
to give a clear outline of the steps you intended to complete to reach a solution. In some
of the questions with multiple parts you will need the answer to an earlier part in order to
work a later part. If you fail to solve the earlier part you may represent its answer with an
algebraic symbol and proceed to give an algebraic answer to the later part. This is a
closed book exam: No notes, books, or other records should be consulted. YOU MAY
ONLY USE THE CALCULATORS PROVIDED. The total of 250 points is divided
equally among the ten questions of the examination.
All work done on scratch paper should be NEATLY transferred to answer booklets.
SESSION 1
COMMON EXAM DATA SHEET
Physical constants and units
e = −1.602 × 10−19 C = −4.803 × 10−10 esu
1eV = 1.602 × 10−19 J
c = 299792458m · s−1 ≈ 3 × 108 m · s−1
h = 6.626 × 10−34 J · s = 6.626 × 10−27 erg · s = 4.135 × 10−15 eV · s
hc = 1.240 × 104 eV · Å
h̄ = 1.054 × 10−34 J · s = 1.054 × 10−27 erg · s = 6.582 × 10−16 eV · s
k = 1.380 × 10−23 J · K −1 = 1.380 × 10−16 erg · K −1 = 8.617 × 10−5 eV · K −1
G = 6.674 × 10−11 N · m2 · kg −2
g = 9.80 m · s−2
NA = 6.022 × 1023
!0 = 8.854 × 10−12 F · m−1
µ0 = 4π × 10−7 H · m−1
MElectron = 9.109 × 10−31 kg = 5.4858 × 10−4 amu = 511keV /c2
MP roton = 1.673 × 10−27 kg = 1.007276amu = 938.3M eV /c2
MN eutron = 1.675 × 10−27 kg = 1.008665amu = 939.5M eV /c2
1mile = 1609 m
1m = 3.28 f t
Some useful integrals
a, b, n > 0
!
!
√ x
dx
a+bx
√1
dx
x2 a+bx
! (a+bx)n/2
x
=−
√
√
−2(2a−bx) a+bx
2
3b
a+bx
ax
x
!
=
x
!
!
√
a − bx2 −
√ 1
dx
a2 +x2
! √a2 +x2
dx
!
!
=
√
b
2a
√
"
= ln ""x +
!
√1
dx
x a+bx
! (a+bx)(n−2)/2
x
√
"
a2 + x2 ""
"
"
= arcsin xa
1
dx
(x2 ±a2 )3/2
=
ñx
a2 x2 ±a2
1
dx
(a2 −x2 )3/2
=
√x
a2 a2 −x2
√
sin2 xdx = 12 (x − sin x cos x)
2
e−x dx =
! ∞ n−1 −x
x e dx
0
! ∞ n+1 −a2 x2
x e
dx
=
√
π
= Γ(n)
1
2an+2
#
Γ
"
a2 +x2 "
"
x
cos2 xdx = 12 (x + sin x cos x)
!∞
dx
"√ √
"
" a+ a−bx2 "
"
x
a ln "
a2 + x2 − a ln " a+
√ 1
dx
a2 −x2
−∞
0
−
dx = n2 (a + bx)n/2 + a
! √a−bx2
dx
!
=
n+2
2
$
Maxwell’s equations (SI unit system)
∇·D
=
ρ
∇·B
=
0
∇ × E = − ∂B
∂t
∇×H
=
J+
∂D
∂t
Maxwell’s equations (Gaussian unit system)
∇·D
=
4πρ
∇·B
=
0
∇ × E = − 1c ∂B
∂t
∇×H
=
4π
1 ∂D
J+
c
c ∂t
Simplest material equations
D = $·E
B = µ·H
J = σ·E
Problem 1 – General physics
Consider a cylindrical bowl of radius r = 10 cm and depth ! = 5 mm. The bowl sits on a horizontal
surface with its axis vertical, and is filled with black ink at an initial temperature T = 300 K. The dish is
then exposed to sunlight. The sun can be considered an ideal Black Body and at the time of measurement
the Sun is ψ = 70◦ above the horizon. Only η = 75% of the light striking the top of the atmosphere gets
through to the surface of the earth, and the sun has an angular radius θ = 0.26◦ . After ∆t = 60 s seconds,
you observe that the temperature of the ink has increased by ∆T = 1.35 K. The heat capacity of the ink is
Cp = 4185.5 J · kg −1 · K −1 and its density is ρ = 1000 kg · m−3 . The sides and bottom of the bowl are coated
with a perfectly reflecting material and the top surface of the ink radiates as a ideal Black Body and there is
no non-radiative heat transfer. The Stefan-Boltzmann constant, σ = 5.67 × 10−8 W m−2 K −4 .
(a) [8 pts.] Write a differential equation for the ink temperature as a function of time (Hint: θ ≈ RSun /DSun ).
(b) [7 pts.] From the provided data, estimate the surface temperature TSun of the Sun.
(c) [4 pts.] Estimate the ink temperature at which thermal equilibrium is reached. (Note: if you were not
able to answer the previous question, use TSun = 6000K as the Sun temperature). Give two reasons why
such a high temperature cannot be reached in practice with the setup described.
(d) [6 pts.] At night, the same setup is prepared with the ink at an initial temperature T1 = 320 K. How
much time does it take for the water to cool to T2 = 300 K?
Problem 1 – General physics – sample solution
Consider a cylindrical bowl of radius r = 10 cm and depth ! = 5 mm. The bowl sits on a horizontal
surface with its axis vertical, and is filled with black ink at an initial temperature T = 300 K. The dish is
then exposed to sunlight. The sun can be considered an ideal Black Body and at the time of measurement
the Sun is ψ = 70◦ above the horizon. Only η = 75% of the light striking the top of the atmosphere gets
through to the surface of the earth, and the sun has an angular radius θ = 0.26◦ . After ∆t = 60 s seconds,
you observe that the temperature of the ink has increased by ∆T = 1.35 K. The heat capacity of the ink is
Cp = 4185.5 J · kg −1 · K −1 and its density is ρ = 1000 kg · m−3 . The sides and bottom of the bowl are coated
with a perfectly reflecting material and the top surface of the ink radiates as a ideal Black Body and there is
no non-radiative heat transfer. The Stefan-Boltzmann constant, σ = 5.67 × 10−8 W m−2 K −4 .
(a) [8 pts.] Write a differential equation for the ink temperature as a function of time (Hint: θ ≈ RSun /DSun ).
Let Q be the heat contained in the ink. Then
dQ
dt
= πr2 !ρCp dT
= −πr2 σT 4 + η
dt
2
4
4πRSun
σTSun
2
4πDSun
· πr2 sinψ
where we have introduced RSun and DSun the radius and distance of the Sun. After simplification and
identifying θ ≈ RSun /DSun it comes
dT
dt
=
σ
4
(ηθ2 sinψTSun
#ρCp
− T 4)
(b) [7 pts.] From the provided data, estimate the surface temperature TSun of the Sun.
From the data we have, we see that, at first,
dT
dt
=
1.35
60
= 0.0225 Ks−1 . Solving the above equation for
TSun we get
p dT
1
TSun = [ ηθ2 sinψ
( #ρC
+ T 4 )]1/4 . The angle θ = 0.26◦ = 4.538 × 10−3 rad so numerically we get
σ dt
TSun = [ 0.75×(4.538×101−3 rad)2 ×sin70◦ ( 5×10
−3 m×1000kg·m−3 ×4185.5J·kg −1 K −1
5.67×10−8 W ·m−2 ·K −4
× 0.0225K · s−1 + (300K)4 )]1/4 =
5798 K
(c) [4 pts.] Estimate the temperature at which thermal equilibrium is reached. (Note: if you were not able
to answer the previous question, use TSun = 6000K as the Sun temperature). Give two reasons why such
a high temperature cannot be reached in practice with the setup described.
Thermal equilibrium corresponds to
dT
dt
= 0 and using the differential equation, T = (ηθ2 sinψ)1/4 TSun .
Numerically T = (0.75 × (4.538 × 10−3 )2 × sin70◦ )1/4 TSun = 6.172 × 10−2 × TSun . So for TSun = 5798K
we get T = 358K and for TSun = 6000K we get T = 370K.
These temperatures are close to the boiling water temperature under standard atmospheric conditions.
In practice such high temperatures can not be reached in the setup of this exercise because of heat loss
by thermal conduction and because of evaporation, however solar water heater are designed to prevent
water from boiling in the Sun exposed conduits.
(d) [6 pts.] At night, the same setup is prepared with the ink at an initial temperature T1 = 320 K. How
much time does it take for the water to cool to T2 = 300 K?
At night, the heat source disappears from our differential equation which becomes
−A · T 4 with A =
T1
σ
.
"ρCp
−[ 3T1 3 ]T = A∆t so ∆t =
2
Numerically, ∆t =
This differential equation can be integrated as −
"ρCp
(T2−3
3σ
− T1−3 ).
5×10−3 m×103 kg·m−3 ×4185.5J·kg −1 K −1
((300K)−3
3×5.67×10−8 W ·m−2 ·K −4
! T1 dT
T2
T4
dT
dt
σ
= − "ρC
T4 =
p
= −
! ∆t
0
− (320K)−3 ) = 802 s = 13.3 min
A · dt or
Problem 2 – Electrodynamics
(a) [12 pts.] Neglecting edge effects, find the capacitance per unit area of a parallel plate capacitor with
infinite metallic plates in the planes x = 0 and x = d. The capacitor is filled with a media with
permittivity:
!(x) = !0
(x + d)
.
d
(1)
(b) [13 pts.] Neglecting edge effects, find the total capacitance of a parallel plate capacitor with metallic
plates in the planes x = 0 and x = d. The sizes of the plates in z and y directions are Z and Y
respectively. The capacitor is filled with a media with permittivity:
!(y, z) = !0 (1 + sin2 (2πzm/Z) sin2 (2πyn/Y )),
where n and m are integers.
(2)
Problem 2 – Electrodynamics – sample solution
(a) [12 pts.] Neglecting edge effects, find the capacitance per unit area of a parallel plate capacitor with
infinite metallic plates in the planes x = 0 and x = d. The capacitor is filled with a media with
permittivity:
!(x) = !0
(x + d)
.
d
(1)
" and displacement D
" obey the following equations within the capacitor:
Electric field E
" = 0
divD
(2)
" = 0
curlE
(3)
" = !(x)E
"
D
(4)
There is also the boundary condition that may be written in a form
−
!
" · d"s = V,
E
(5)
where V is the potential difference between the two metallic plates and the integral may be taken along
any contour connecting these plates.
" and D
" have only x-components, Dx = D0 ,
This system of equations is satisfied if we assume that both E
where D0 is x-independent, and Ex ≡ E(x) = D0 /!(x).
As follows from the Gauss theorem the surface charge on metallic plates σ = ±D0 and the capacitance
per unit area is
C=
On the other hand,
V =
! d
0
D0
.
V
E(x)dx = D0
(6)
! d
0
dx
!(x)
(7)
Thus,
C=
!0
.
d ln(2)
(8)
(b) [13 pts.] Neglecting edge effects, find the total capacitance of a parallel plate capacitor with metallic
plates in the planes x = 0 and x = d. The sizes of the plates in z and y directions are Z and Y
respectively. The capacitor is filled with a media with permittivity:
!(y, z) = !0 (1 + sin2 (2πzm/Z) sin2 (2πyn/Y )),
(9)
where n and m are integers.
!
Due to the fact that E · dl = 0, the tangential component of the electric field is zero inside the capacitor.
Since ! is independent of x , the normal component of the electric field is constant: E = V /d. On the
other hand, Dn (z, y) = σ(z, y) = E!(z, y). Therefore, the total charge is
Q=
" Y " Z
0
0
#
$
%
$
%
" Y
" Z
V !0
2πzm
2 2πyn
σ dydz =
YZ +
sin
dy
sin2
dz
d
Y
Z
0
0
Since, Q = Ct V , where Ct is the total capacitance we obtain
Ct =
5 !0 Y Z
.
4 d
&
Problem 3 – Classical Mechanics
A uniform cylindrical log of mass M = 5 kg and radius R = 10 cm is initially at rest on the ground when
it is hit by a bullet of mass m = 10 g, moving horizontally (and also perpendicularly to the log’s axis O) with
a velocity v = 1000 m/s as shown in the picture. The bullet lodges very close to the axis O.
(a) [10 pts.] Assuming that no slipping occurs between the log and the ground, find the frequency ν of the
log’s rotation after the impact.
(b) [7 pts.] What fraction of the initial mechanical energy is dissipated into heat during the collision?
(c) [8 pts.] What is the total impulse
!
F dt provided by the friction force between the log and the ground
during the collision (as compared with the initial momentum of the bullet)?
Problem 3 – Classical Mechanics – sample solution
A uniform cylindrical log of mass M = 5 kg and radius R = 10 cm is initially at rest on the ground when
it is hit by a bullet of mass m = 10 g, moving horizontally (and also perpendicularly to the log’s axis O) with
a velocity v = 1000 m/s as shown in the picture. The bullet lodges very close to the axis O.
(a) [10 pts.] Assuming that no slipping occurs between the log and the ground, find the frequency ν of the
log’s rotation after the impact.
The momentum of inertia of a uniform cylinder with respect to its axis is IO = M R2 /2 and with respect
to the point on its rim, IG = IO + M R2 = 3M R2 /2. Angular momentum conservation about point G
gives,
3
mvR = M R2 ω + mRω,
2
which gives the angular frequency,
ω=
mv
.
(3M/2 + m)R
The frequency is
ν=
ω
mv
mv
=
≈
≈ 2.1 Hz.
2π
2π(3M/2 + m)R
3πM R
(b) [7 pts.] What fraction of the initial mechanical energy is dissipated into heat during the collision?
The kinetic energy after the collision is
(IG + mR2 )
ω2
mv 2
m
mv 2 2m
mv 2
=
≈
≈ 0.0013
.
2
2 3M/2 + m
2 3M
2
About 99.9% of the initial kinetic energy is converted into heat.
(c) [8 pts.] What is the total impulse
!
F dt provided by the friction force between the log and the ground
during the collision (as compared with the initial momentum of the bullet)?
One solution is to equate the impulse of the friction force to the decrease of the total momentum.
Alternatively, it is related to the angular momentum increase with respect to point O:
R
"
F dt = IO ω =
M R2
mv
mvR
≈
.
2 (3M/2 + m)R
3
Thus the total impulse provided by the friction force in the collision is close to 1/3 of the initial momentum
mv.
Problem 4 – Quantum Mechanics
A particle of mass m moves in a one-dimensional repulsive delta-function potential U (x) = αδ(x), with
α > 0.
(a) [10 pts.] Derive the boundary conditions for the wave function ψ(x) and its derivative dψ/dx at the
origin. (Hint: integrate the Schrödinger equation over x in the interval (−d, d), and take the limit
d → +0 after that.)
(b) [10 pts.] Find the reflection R and transmission T coefficients for the particle with energy E > 0
incoming on the delta-function barrier from the left.
(c) [5 pts.] Discuss how your results for R and T apply to the attractive potential with α < 0. Also discuss
E → ∞ limit of your solution: is it consistent with classical expectations?
!!""!#
x
FIG. 1: The scattering problem.
Problem 4 – Quantum Mechanics – sample solution
A particle of mass m moves in a one-dimensional repulsive delta-function potential U (x) = αδ(x), with
α > 0.
(a) [10 pts.] Derive the boundary conditions for the wave function ψ(x) and its derivative dψ/dx at the
origin. (Hint: integrate the Schrödinger equation over x in the interval (−d, d), and take the limit
d → +0 after that.)
The Schrödinger equation reads
−
h̄2 d2 ψ(x)
+ αδ(x)ψ(x) = Eψ(x).
2m dx2
(1)
Integrate both sides over x from some −d to d, and take the limit d → +0. Then
dψ(+d)/dx − dψ(−d)/dx
! +d
−d
→
! +d
−d
dx d2 ψ(x)/dx2 =
ψ " (+0) − ψ " (−0), where prime denotes spatial derivative, and
dx δ(x)ψ(x) = ψ(0) irrespective of d, while the right-hand side
as d → 0. Thus we obtain
ψ " (+0) − ψ " (−0) =
! +d
−d
dxEψ(x) → 0 approaches zero
2mα
ψ(0).
h̄2
(2)
The wave function itself is always continuous, ψ(+0) = ψ(−0).
(b) [10 pts.] Find the reflection R and transmission T coefficients for the particle with energy E > 0
incoming on the delta-function barrier from the left.
For x #= 0 Eq.(1) reads, for E > 0,
d2 ψ(x)
2mE
=
−
Eψ(x),
dx2
h̄2
so that its general solution is ψ(x) = a1 eikx + b1 e−ikx , with k =
(3)
"
2mE/h̄2 .
We now set up the scattering problem:
ψ(x) = eikx + Ce−ikx for x < 0 , ψ(x) = Beikx for x > 0.
(4)
Transmission coefficient D = |B|2 while reflection R = |C|2 . Using derived boundary conditions we
obtain two equations for unknown B, C
1 + C = B ; ikB − ik(1 − C) =
2mα
(1 + C).
h̄2
(5)
Solving for C we get
C=
Hence
!
R = |C|2 = 1 +
! ikh̄2
"−1
−1
.
(6)
h̄4 k 2 "−1
1
=
.
2
2
2
mα
1 + 2h̄ E/(mα2 )
(7)
mα
Clearly D = 1 − R.
(c) [5 pts.] Discuss how your results for R and T apply to the attractive potential with α < 0. Also discuss
E → ∞ limit of your solution: is it consistent with classical expectations?
The result, Eq.(7), is not sensitive to the sign of α: hence the well (α < 0) reflect the particle as well as
the barrier (α > 0) does!
In the limit E → ∞: R(E) →
mα2
2h̄2 E
→ 0 while D(E) → 1 −
mα2
2h̄2 E
→ 1. Thus at high energy reflection
disappears and all particles pass through, which is quite reasonable classically.
!!""!#
x
FIG. 1: The scattering problem.
Problem 5 – Thermodynamics
The equilibrium energy of a photon gas in a cavity with a volume V and temperature T is given by the
equation E = αV T 4 . Useful equations for this problem are dE = T dS − P dV and dF = −SdT − pdV .
(a) [9 pts.] Find the entropy of the photon gas.
(b) [9 pts.] Find the pressure of the photon gas under the same conditions.
(c) [7 pts.] The volume of the cavity is expanded adiabatically up to V2 . What is the equilibrium temperature
of the photon gas? Neglect the heat capacity of the cavity.
Problem 5 – Thermodynamics – sample solution
The equilibrium energy of a photon gas in a cavity with a volume V and temperature T is given by the
equation E = αV T 4 . Useful equations for this problem are dE = T dS − P dV and dF = −SdT − pdV .
(a) [9 pts.] Find the entropy of the photon gas.
If the volume of the cavity is kept constant, from the thermodynamic identity dE = T dS − P dV we
have T dS = dE = 4V αT 3 dT . From the third law of thermodynamics, the entropy of a system at zero
temperature is zero. Then, after integration, we have
S=
! S
0
dS =
! T
0
4
4V αT 2 dT = αV T 3 .
3
(1)
(b) [9 pts.] Find the pressure of the photon gas under the same conditions.
The free energy of the photon gas is
1
F = E − T S = − αV T 4 .
3
(2)
Then, the pressure of the photon gas is
"
∂F
P =−
∂V
#
T
1
= αT 4
3
(3)
(c) [7 pts.] The volume of the cavity is expanded adiabatically up to V2 . What is the equilibrium temperature
of the photon gas? Neglect the heat capacity of the cavity.
In an adiabatic expansion of the photon gas the entropy is conserved. That means V1 T13 = V2 T23 or
T2 = T1
$
V1
V2
% 13
.
(4)
Common Exam - 2010
Department of Physics
University of Utah
August 21, 2010
Examination booklets have been provided for recording your work and your solutions.
Please note that there is a separate booklet for each numbered question (i.e., use
booklet #1 for problem #1, etc.).
To receive full credit, not only should the correct solutions be given, but a sufficient
number of steps should be given so that a faculty grader can follow your reasoning.
Define all algebraic symbols that you introduce. If you are short of time it may be helpful
to give a clear outline of the steps you intended to complete to reach a solution. In some
of the questions with multiple parts you will need the answer to an earlier part in order to
work a later part. If you fail to solve the earlier part you may represent its answer with an
algebraic symbol and proceed to give an algebraic answer to the later part. This is a
closed book exam: No notes, books, or other records should be consulted. YOU MAY
ONLY USE THE CALCULATORS PROVIDED. The total of 250 points is divided
equally among the ten questions of the examination.
All work done on scratch paper should be NEATLY transferred to answer booklets.
SESSION 2
COMMON EXAM DATA SHEET
Physical constants and units
e = −1.602 × 10−19 C = −4.803 × 10−10 esu
1eV = 1.602 × 10−19 J
c = 299792458m · s−1 ≈ 3 × 108 m · s−1
h = 6.626 × 10−34 J · s = 6.626 × 10−27 erg · s = 4.135 × 10−15 eV · s
hc = 1.240 × 104 eV · Å
h̄ = 1.054 × 10−34 J · s = 1.054 × 10−27 erg · s = 6.582 × 10−16 eV · s
k = 1.380 × 10−23 J · K −1 = 1.380 × 10−16 erg · K −1 = 8.617 × 10−5 eV · K −1
G = 6.674 × 10−11 N · m2 · kg −2
g = 9.80 m · s−2
NA = 6.022 × 1023
!0 = 8.854 × 10−12 F · m−1
µ0 = 4π × 10−7 H · m−1
MElectron = 9.109 × 10−31 kg = 5.4858 × 10−4 amu = 511keV /c2
MP roton = 1.673 × 10−27 kg = 1.007276amu = 938.3M eV /c2
MN eutron = 1.675 × 10−27 kg = 1.008665amu = 939.5M eV /c2
1mile = 1609 m
1m = 3.28 f t
Some useful integrals
a, b, n > 0
!
!
√ x
dx
a+bx
√1
dx
x2 a+bx
! (a+bx)n/2
x
=−
√
√
−2(2a−bx) a+bx
2
3b
a+bx
ax
x
!
=
x
!
!
√
a − bx2 −
√ 1
dx
a2 +x2
! √a2 +x2
dx
!
!
=
√
b
2a
√
"
= ln ""x +
!
√1
dx
x a+bx
! (a+bx)(n−2)/2
x
√
"
a2 + x2 ""
"
"
= arcsin xa
1
dx
(x2 ±a2 )3/2
=
ñx
a2 x2 ±a2
1
dx
(a2 −x2 )3/2
=
√x
a2 a2 −x2
√
sin2 xdx = 12 (x − sin x cos x)
2
e−x dx =
! ∞ n−1 −x
x e dx
0
! ∞ n+1 −a2 x2
x e
dx
=
√
π
= Γ(n)
1
2an+2
#
Γ
"
a2 +x2 "
"
x
cos2 xdx = 12 (x + sin x cos x)
!∞
dx
"√ √
"
" a+ a−bx2 "
"
x
a ln "
a2 + x2 − a ln " a+
√ 1
dx
a2 −x2
−∞
0
−
dx = n2 (a + bx)n/2 + a
! √a−bx2
dx
!
=
n+2
2
$
Maxwell’s equations (SI unit system)
∇·D
=
ρ
∇·B
=
0
∇ × E = − ∂B
∂t
∇×H
=
J+
∂D
∂t
Maxwell’s equations (Gaussian unit system)
∇·D
=
4πρ
∇·B
=
0
∇ × E = − 1c ∂B
∂t
∇×H
=
4π
1 ∂D
J+
c
c ∂t
Simplest material equations
D = $·E
B = µ·H
J = σ·E
Problem 6 – Classical Mechanics
Two bodies with masses m1 and m2 are connected by a spring with a spring constant k. The equilibrium
distance between the two bodies is L.
(a) [8 pts.] Find the frequency of oscillation of the system.
Body 2 is now charged with charge +Q and body 1 is neutral and non-polarizable. The system is initially
at rest with the distance between the two bodies being L. At time t = 0, an electric field E is turned
on. The direction of the field is along the axis of the system as shown in the figure.
(b) [7 pts.] Write down Newton’s equations of motion for the first and the second bodies.
(c) [10 pts.] Find the distance between two bodies as a function of time.
Problem 6 – Classical Mechanics – sample solution
Two bodies with masses m1 and m2 are connected by a spring with a spring constant k. The equilibrium
distance between the two bodies is L.
(a) [8 pts.] Find the frequency of oscillation of the system.
The position of the center of mass of the system can be found from the equations m1 L1 = m2 L2 and
L = L1 + L2 , where L1 and L2 are the distances from the center of mass to the first and the second body,
respectively. We find that L2 =
m2 +m1
L.
m1
Since the center of mass is immobile we may describe the oscillation of the second body as if it is
1
connected to a spring with a length L2 and a spring constant k2 = k LL2 = k m2m+m
. The frequency of
1
oscillations of the system is
!
m2 + m1
ω= k
m1 · m2
"1
2
.
(1)
Body 2 is now charged with charge +Q and body 1 is neutral and non-polarizable. The system is initially
at rest with the distance between the two bodies being L. At time t = 0, an electric field E is turned
on. The direction of the field is along the axis of the system as shown in the figure.
(b) [7 pts.] Write down Newton’s equations of motion for the first and the second bodies.
m2 x¨2 = eQ − k(x2 − x1 − L),
(2)
m1 x¨1 = k(x2 − x1 − L),
(3)
where x2 and x1 are the coordinates of the second and the first body, respectively.
(c) [10 pts.] Find the distance between two bodies as a function of time.
Let S be the deviation of the distance between bodies 1 and 2 from their equilibrium separation L,
S = x2 − x1 − L. Then from eq. (2) and (3) we find an equation of motion for S.
S̈ = −ω 2 S + β,
where ω is given by eq. (1) and β =
(4)
eQ
.
m2
The equation (4) is an inhomogeneous ordinary differential equation. Its solution can be found as a sum
of a general solution of a homogeneous equation and any solution of an inhomogeneous equation. For
example, it can be chosen as
S = A cos(ωt) + B sin(ωt) +
β
ω2
(5)
From initial conditions S(t = 0) = 0 and Ṡ(t = 0) = 0, we find that B = 0 and A = − ωβ2 .
We have then finally
x2 (t) − x1 (t) = L −
eQ
eQ
cos(ωt) +
.
2
m2 ω
m2 ω 2
(6)
Problem 7 – Electrodynamics
Consider some properties of inhomogeneous plane waves that are also called evanescent waves. These waves
appear in the theory of total internal reflection and in many other examples of near field optics.
Suppose that the electric field of the evanescent wave has only a z component. It has the form
Ez = E0 exp(iky y − κx − iωt),
(1)
where κ, ky are real. The wave propagates in a dielectric medium with electric permittivity # and magnetic
permeability µ.
$ as given by Eq.(1) obeys the wave equation of electrodynamics and find ω(κ, ky ).
(a) [3 pts.] Show that E
(b) [2 pts.] Find the region of (κ, ky ), where the frequency ω is real.
$
(c) [10 pts.] Using the Maxwell equations, find all components of the magnetic field H.
(d) [10 pts.] Prove that if A(t) = A0 exp −iωt and B(t) = B0 exp −iωt, then < ReAReB >= Re(A∗ · B)/2,
where < ... > means time averaging. Using this relation, find all components of the time averaged
$ =< [ReE
$ × ReH]
$ >.
Poynting vector S
Problem 7 – Electrodynamics – sample solution
Consider some properties of inhomogeneous plane waves that are also called evanescent waves. These waves
appear in the theory of total internal reflection and in many other examples of near field optics.
Suppose that the electric field of the evanescent wave has only a z component. It has the form
Ez = E0 exp(iky y − κx − iωt),
(1)
where κ, ky are real. The wave propagates in a dielectric medium with electric permittivity # and magnetic
permeability µ.
$ as given by Eq.(2) obeys the wave equation of electrodynamics and find ω(κ, ky ).
(a) [3 pts.] Show that E
Substituting
Ez = E0 exp(iky y − κx − iωt),
(2)
(∂ 2 /c2 ∂ 2 t − ∇2 )Ez = 0
(3)
into equation
one gets ω 2 = c2 (ky2 − κ2 ).
(b) [2 pts.] Find the region of (κ, ky ), where the frequency ω is real.
The frequency ω is real if ky2 ≥ κ2 .
$
(c) [10 pts.] Using the Maxwell equations, find all components of the magnetic field H.
Using Maxwell equation
$ = −∂ B/∂t
$
curlE
(4)
$ = iω B,
$
i[$k × E]
(5)
one finds
$ = µH.
$ Then
where $k has projections (iκ, ky , 0) and B
ky
Ez
ωµ
iκEz
Hy = −
ωµ
Hx =
(6)
(d) [10 pts.] Prove that if A(t) = A0 exp −iωt and B(t) = B0 exp −iωt, then < ReAReB >= Re(A∗ · B)/2,
where < ... > means time averaging. Using this relation, find all components of the time averaged
" =< [ReE
" × ReH]
" >.
Poynting vector S
To prove the relations one should write ReA = (A + A∗ )/2 and similar for B. Then take into account
that < exp(±2iωt) >= 0
" are
The components of the Poynting vector S
Sx = −Hy Ez
(7)
Sy = Hx Ez
(8)
Sz = 0
(9)
After time averaging one gets
< Sx >= −Re(Hy Ez∗ )/2 =
κ
Re(i|Ez |2 ) = 0.
2ωµ
(10)
and
< Sy >=
ky
ky
|Ez |2 =
|E0 |2 exp (−2κx)
2ωµ
2ωµ
Thus, the energy of the evanescent wave propagates only in the direction with real momentum.
(11)
Problem 8 – Statistical Mechanics
A classical gas of atoms, each of mass m, is maintained at temperature T inside an enclosure. The gas is
described by Maxwell-Boltzmann distribution,
f (!v )d3 v =
!
m "3/2
m!v 2 3
exp[−
]d v.
2πkB T
2kB T
(1)
The atoms emit light which passes through a window of the enclosure and can then be observed as a spectral
line in a spectroscope. A stationary atom would emit light at a sharply defined wavelength λ0 . But, because of
the Doppler effect, the observed wavelength λ of the light from an atom with the component vs of the velocity
along the line of sight is instead approximately given by
λ = λ0 (1 +
vs
),
c
(2)
where c is the speed of light. Note that components of the velocity orthogonal to the line of sight do not affect
the wavelength of the radiation.
As a result, the light arriving at the spectroscope has an intensity in the wavelength range (λ, λ + dλ)
characterized by some intensity distribution I(λ)dλ.
(a) [15 pts.] Derive the distribution function I(λ)dλ of the observed light for a gas of N atoms.
(b) [5 pts.] Observe that Eq.(2) implies that |vs | " c. What restriction does this impose on the temperature
T of the gas?
(c) [5 pts.] Using the result of (b), normalize the distribution by requiring
#∞
0
I(λ)dλ = N I0 . (Here I0 is
the intensity of radiation detected by the spectroscope from a stationary atom.) (Hint: This question
requires making an approximation which relies on the result of question (b).)
Problem 8 – Statistical Mechanics – sample solution
A classical gas of atoms, each of mass m, is maintained at temperature T inside an enclosure. The gas is
described by Maxwell-Boltzmann distribution,
f (!v )d3 v =
!
m "3/2
m!v 2 3
exp[−
]d v.
2πkB T
2kB T
(1)
The atoms emit light which passes through a window of the enclosure and can then be observed as a spectral
line in a spectroscope. A stationary atom would emit light at a sharply defined wavelength λ0 . But, because of
the Doppler effect, the observed wavelength λ of the light from an atom with the component vs of the velocity
along the line of sight is instead approximately given by
λ = λ0 (1 +
vs
),
c
(2)
where c is the speed of light. Note that components of the velocity orthogonal to the line of sight do not affect
the wavelength of the radiation.
As a result, the light arriving at the spectroscope has an intensity in the wavelength range (λ, λ + dλ)
characterized by some intensity distribution I(λ)dλ.
(a) [15 pts.] Derive the distribution function I(λ)dλ of the observed light for a gas of N atoms.
We only need the distribution of the vs component of the velocity as the transverse ones do not affect the
wavelength of the emitted light. Hence, choosing for a moment vs to be along the x-axis, we integrate
over the two orthogonal to it velocity components, vy and vz , in the full range (−∞, ∞) of allowed values.
Note that !v
2
= vs2 + vy2 + vz2 and d3 v = dvs dvy dvz . Then, the number of particles with the s-component
of their velocity in the interval (vs , vs + dvs ) is obtained
dN (vs ) = N
!
m "1/2
mvs2
exp[−
]dvs .
2πkB T
2kB T
(3)
Using (2) we convert this into the number of atoms dN (λ) emitting light in the interval (λ, λ + dλ):
dN (λ) = N
c ! m "1/2
(λ − λ0 )2
exp[−
]dλ,
λ0 2πkB T
δ2
(4)
where δ 2 = 2kB T λ20 /(mc2 ).
The distribution in question follows as
I(λ)dλ = αdN (λ)
(5)
where the unknown normalization constant α is to be fixed by the normalization condition. The distribution is centered around λ0 and its width is given by δ. Observe that the distribution is very sharp:
λ0 /δ =
!
mc2 /(2kB T ) ! 1.
(b) [5 pts.] Observe that Eq.(2) implies that |vs | " c. What restriction does this impose on the temperature
T of the gas?
The average velocity of particles is determined by the temperature alone: v̄ ∼
velocity of the s-component is also given by this expression, v̄s ∼
!
!
kB T /m. Thus average
kB T /m. Hence, vs " c translates
into kB T " mc2 : the thermal energy of the particles in much smaller than their rest energy. The motion
is non-relativistic.
Observe that this condition implies the sharpness of the derived light distribution: its width is much
smaller than the average wavelength, δ/λ0 =
!
2kB T /(mc2 ) " 1.
(c) [5 pts.] Using the result of (b), normalize the distribution by requiring
"∞
0
I(λ)dλ = N I0 . (Here I0 is
the intensity of radiation detected by the spectroscope from a stationary atom.) (Hint: This question
requires making an approximation which relies on the result of question (b).)
Normalization requires α
"∞
0
dN (λ) = N I0 , which leads to
αN # ∞
αN # ∞ −(λ−λ0 )2 /δ2
2
N I0 = √
e
dλ = √
e−x dx.
πδ 0
π −λ0 /δ
(6)
Here we substituted λ = λ0 + δx. Since, for the non-relativistic gas λ0 /δ ! 1, we can replace the
lower limit of the integration by negative infinity, −∞, which results in the standard gaussian integral
√
" ∞ −x2
dx = π. Thus we fix the normalization constant as α = I0 .
−∞ e
Problem 9 – Quantum mechanics
Consider a single 1/2 spin that initially is aligned with the positive direction of the y-axis. At time t = 0, a
magnetic field along the positive direction of the z-axis is suddenly switched on. Interaction of the spin with
this field is described by the Hamiltonian,
Ĥ = −bσ̂z ,
b>0
(a) [9 pts.] Find the wavefunction of the spin at a moment of time t > 0.
(b) [4 pts.] Describe qualitatively the spin dynamics for t > 0.
(c) [5 pts.] Will there be an occasion when the spin will be aligned with the positive direction of the x-axis?
If this is the case, when will it happen?
(d) [4 pts.] What about the spin being aligned with the negative direction of the x-axis? When will it
happen?
(e) [3 pts.] What about the spin being aligned with the the z-axis? When will it happen if it ever does?
For reference, Pauli matrices:


0 1
σ̂x = 
,
1 0


 0 −i 
σ̂y = 
,
i 0


1 0 
σ̂z = 
.
0 −1
Problem 9 – Quantum mechanics – sample solution
Consider a single 1/2 spin that initially is aligned with the positive direction of the y-axis. At time t = 0, a
magnetic field along the positive direction of the z-axis is suddenly switched on. Interaction of the spin with
this field is described by the Hamiltonian,
Ĥ = −bσ̂z ,
b>0
(a) [9 pts.] Find the wavefunction of the spin at a moment of time t > 0.
The eigenstates of the Hamiltonian correspond to two states with sz = ±h̄/2:


1
χsz =+h̄/2 =   e−ibt/h̄ ,
0


0
χsz =−h̄/2 =   eibt/h̄ .
1
(1)
At t < 0 the wave function corresponds to sy = +h̄/2. This eigenfunction of σ̂y is simply


1 1
χsy =+h̄/2 = √  
2 i
(2)
The linear combination of the functions (1) that coincides with (2) at t = 0 is




1 1
i 0
χ(t) = √   e−ibt/h̄ + √   eibt/h̄ .
2 0
2 1
(3)
(b) [4 pts.] Describe qualitatively the spin dynamics for t > 0.
The motion of spin is precession around the z-axis with the angular frequency b/h̄.
(c) [5 pts.] Will there be an occasion when the spin will be aligned with the positive direction of the x-axis?
If this is the case, when will it happen?
Spin will be aligned with the positive direction of the x-axis when its wavefunction becomes


1
χsx =+h̄/2 = c   .
1
From (4) we obtain that it happens when simultaneously
eibt/h̄ = c,
ie−ibt/h̄ = c,
→ e2ibt/h̄ = i,
which happens at
t=
h̄
h̄n
+
,
4b
b
n = 0, 1, 2, ...
(4)
(d) [4 pts.] What about the spin being aligned with the negative direction of the x-axis? When will it
happen?
Similarly, spin is aligned along the negative direction of the x-axis at times shifted by half-period compared to the above,
t=
3h̄ h̄n
+
,
4b
b
n = 0, 1, 2, ...
(e) [3 pts.] What about the spin being aligned with the the z-axis? When will it happen if it ever does?
As long as dissipation of energy is absent, the spin is never going to be along the z-axis. In practice
however, energy dissipation, although it may be small, is always present and at t − ∞ the spin is going
to get aligned with the positive direction of the z-axis, which is its minumum energy (ground) state.
Problem 10 – Modern physics
Consider photons γ with frequency ν = 160 GHz and density n = 2 × 108 m−3 propagating in the negative
x-direction and protons propagating in the opposite direction. The protons interact with the photons as
p + γ → ∆(1232) → p + π 0 with a total cross section σ = 300µb. The mean lifetime of the ∆(1232) resonance
is τ∆ = 5 × 10−25 s. (m∆ = 1232 M eV /c2 , mp = 938 M eV /c2 and mπ0 = 135 M eV /c2 , 1 b = 10−28 m2 ).
(a) [6 pts.] What is the minimum energy protons must have to produce ∆(1232) resonances when interacting
with the photons? You may assume the proton to be ultra-relativistic and verify with your result that
this assumption is well justified.
(b) [6 pts.] A proton of energy E = 1021 eV interacts with a photon, producing a ∆(1232). What is the
mean distance traveled by the ∆(1232) before it decays?
(c) [7 pts.] From the point of view of a proton of energy Ep = 1021 eV , what is the wavelength of the
photons? How does this wavelength compare to the physical size of a proton?
(d) [6 pts.] In the laboratory rest frame, what is the average distance traveled by E = 1021 eV protons
without interacting? Express your result in light years (a light year is the distance travelled at the speed
of light in one year).
Problem 10 – Modern physics – sample solution
Consider photons γ with frequency ν = 160 GHz and density n = 2 × 108 m−3 propagating in the negative
x-direction and protons propagating in the opposite direction. The protons interact with the photons as
p + γ → ∆(1232) → p + π 0 with a total cross section σ = 300µb. The mean lifetime of the ∆(1232) resonance
is τ∆ = 5 × 10−25 s. (m∆ = 1232 M eV /c2 , mp = 938 M eV /c2 and mπ0 = 135 M eV /c2 , 1 b = 10−28 m2 ).
(a) [6 pts.] What is the minimum energy protons must have to produce ∆(1232) resonances when interacting
with the photons? You may assume the proton to be ultra-relativistic and verify with your result that
this assumption is well justified.
For the reaction to be possible, the mass invariant must be (Ep + Eγ )2 − (cPp − cPγ )2 > m2∆ c4 or
Ep2 + Eγ2 + 2Ep Eγ − c2 Pp2 − c2 Pγ2 + 2Pp Pγ c2 > m2∆ c4 which simplifies to
m2p c4 + 2Ep Eγ + 2Pp cEγ > m2∆ c4 and considering the proton to be ultra-relativistic, Ep ≈ cPp so:
Ep >
m2∆ c4 −m2p c4
4hν
and, numerically: Ep >
(1232 M eV )2 −(938 M eV )2
4×6.63×10−34 J·s×160×109 Hz/1.602×10−19 J·eV −1
= 2.4 × 1020 eV
Since we find Ep >> mp c2 for this reaction to be possible, the ultra-relativistic approximation is well
justified.
(b) [6 pts.] A proton of energy E = 1021 eV interacts with a photon, producing a ∆(1232). What is the
mean distance traveled by the ∆(1232) before it decays?
The lifetime of the resonance is going to be affected by time dilation so we need to find the Lorentz
factor of ∆(1232). We can do this by using conservation of energy: Ep + Eγ = γm∆ c2 so γ =
1021 eV
1232×106 eV
Ep +Eγ
m∆ c2
≈
= 8.117 × 1011 . Since γ >> 1, the ∆(1232) can be considered to be going at the speed of light
and the distance it travels is D = γ · τ∆ · c = 8.117 × 1011 × 5−25 s × 3 × 108 m · s−1 = 0.12 mm.
(c) [7 pts.] From the point of view of a proton of energy Ep = 1021 eV , what is the wavelength of the
photons? How does this wavelength compare to the physical size of a proton?
The Lorentz factor of the protons is γ =
Ep
mp c2
=
1021 eV
938×106 eV
= 1.066 × 1012 . The energy of the photons in
the laboratory reference frame is Eγ = h · ν. The energy of the photons in the proton rest frame can be
obtained by a Lorentz transformation: Eγ# = γEγ + βγEγ ≈ 2γEγ as since γ >> 1 we have β ≈ 1. So
hc/λ# = 2γhν and λ# =
Numerically λ# =
c
.
2γν
3×108 m·s−1
1.066×1012 ×160×109 Hz
as the physical size of a proton.
= 8.8 × 10−16 m = 0.88 f m. This wavelength is of the same order
(d) [6 pts.] In the laboratory rest frame, what is the average distance traveled by E = 1021 eV protons
without interacting? Express your result in light years (a light year is the distance travelled at the speed
of light in one year).
The average distance D traveled without interacting is the speed v of the protons multiplying the mean
time before interaction so D =
D ≈
1
.
2σ·n
Numerically, D =
v
.
σ·n(c+v)
We have seen the protons are ultra-relativistic so v ≈ c and
1
2×300×10−6 ×10−28 m2 ×2×108 m−3
= 8.3 × 1022 m. A light year corresponds to
365 × 24 × 3600 s × 3 × 108 m · s−1 = 9.46 × 1015 m and so the mean distance traveled by protons without
interacting is 8.8 × 106 ly.