Download Exam and Solutions - Stony Brook University

STONY BROOK UNIVERSITY
DEPARTMENT OF PHYSICS AND ASTRONOMY
Comprehensive Examination, August 31, 2011
General Instructions: Twelve problems are given; you should solve four problems.
If you do more than four problems, you must choose which four should be graded,
and only submit those four. You may do two problems from the same field only once,
except for Astronomy, for which you can do up to four problems.
Each problem counts 20 points, and the solution should typically take less than 45
minutes.
Some of the problems cover multiple pages. Make sure you do all the parts of each
problem you choose.
Use one exam book for each problem, and label it carefully with the problem topic
and number and your name.
You may use a one page help sheet, a calculator, and with the proctor’s approval, a
foreign language dictionary. No other materials may be used.
Some potentially useful information:
The atomic mass of hydrogen is 1.00794 amu.
The atomic mass of helium is 4.002602 amu.
1 amu is 1.66×10−27 kg.
𝑐 = 2.998 × 108 ms−1 .
𝐺 = 6.673 × 10−11 m3 kg−1 s−2 .
The solar luminosity is 3.85×1026 W.
The mass of the Sun is 1.989×1030 kg
The radius of the Sun is 7.00 × 108 meters
ℏ = 1.055 × 10−34 J
𝑒 = 1.602 × 10−19 C
𝑘𝐵 = 1.38 × 10−23 J/K
𝑚𝑐2 ≃ 0.5 MeV
ℏ𝑐 ≃ 197 MeV-fm.
1
Astronomy 1
Many astrophysical objects may be approximated as blackbodies, ideal emitters that
absorb all energy incident on them and re-radiate it with a characteristic spectrum.
At the turn of the 20th century, Max Planck synthesized earlier results with the new
quantum theory and found the Planck function,
𝐵𝜆 (𝑇 ) =
2ℎ𝑐2 /𝜆5
,
𝑒ℎ𝑐/𝜆𝑘𝑇 − 1
for power per unit area per unit solid angle per wavelength that gave good agreement
with experimental results. In the expression, 𝑇 is the temperature, 𝜈 is the frequency,
ℎ is Planck’s constant, and 𝑘 is Boltzmann’s constant. There are two important aspects of this expression, one concerning the peak emission and one concerning the
total emission.
a. (3 pts) Derive Wein’s law for the peak emission as a function of wavelength.
Hint: you will have to find a solution to a transcendental equation.
b. (3 pts) By integrating the Planck function, derive an approximate expression
for the total emission of a blackbody and how it depends on the temperature.
c. (5 pts) With the aid of a sketch, combine these to describe the behavior of
blackbody radiation as a function of wavelength and temperature.
d. (3pts) Show that the Rayleigh-Jeans law,
𝐵𝜆 (𝑇 ) ≃
2𝑐𝑘𝑇
,
𝜆4
is an approximation to Planck function at long wavelengths.
e. (3 pts) RXJ 185635-3754 is a bright x-ray source observed by the ROSAT X-ray
observatory that has been identified as an isolated neutron star. The distribution in x-rays is well-fitted by a blackbody with a peak at a wavelength of
4.39 × 10−9 m. Determine the temperature of the corresponding blackbody.
f. (3 pts) Subsequent observations found that the optical flux is about seven times
larger than the Rayleigh-Jeans estimate. How can this be reconciled?
SOLUTION:
(a.) To find the maximum wavelength, maximize the function.
𝑑
𝑑 2ℎ𝑐2 /𝜆5
𝐵𝜆 (𝑇 ) =
=0
𝑑𝜆
𝑑𝜆 𝑒ℎ𝑐/𝜆𝑘𝑇 − 1
2
This gives
0 = 2ℎ𝑐
2
[
−5𝜆
−6
(
ℎ𝑐/𝜆𝑘𝑇
𝑒
−1
)−1
+𝜆
−7
(
ℎ𝑐/𝜆𝑘𝑇
(−1) 𝑒
−1
)−2
(
(
)−1
)−2 ℎ𝑐/𝜆𝑘𝑇 ℎ𝑐
+ 𝜆−7 𝑒ℎ𝑐/𝜆𝑘𝑇 − 1
𝑒
= −5𝜆−6 𝑒ℎ𝑐/𝜆𝑘𝑇 − 1
𝑘𝑇
)−1 ℎ𝑐/𝜆𝑘𝑇 ℎ𝑐
( ℎ𝑐/𝜆𝑘𝑇
−1
𝑒
= 5−𝜆 𝑒
𝑘𝑇
)
( ℎ𝑐/𝜆𝑘𝑇
ℎ𝑐
− 1 − 𝑒ℎ𝑐/𝜆𝑘𝑇
= 5 𝑒
𝜆𝑘𝑇
Let 𝑥 =
ℎ𝑐
𝜆𝑘𝑇
ℎ𝑐/𝜆𝑘𝑇
𝑒
(
−ℎ𝑐
𝜆2 𝑘𝑇
)]
and with some algebra we may write
𝑒−𝑥 = 1 − 𝑥/5
A little trial and error gives 𝑥 ≈ 4.956 in addition to the trivial solution (𝑥 = 0).
Now solve for 𝜆𝑇 to have the form of Wein’s law:
𝜆𝑇 =
ℎ𝑐
𝑘 (4.956)
Apply SI units
ℎ = 6.626 × 10−31 Js
𝑘 = 1.381 × 10−23 J/K
𝑐 = 2.998 × 108 m/s
we have
𝜆𝑇 ≈ 2.9 × 10−3 mK ,
which is Wein’s law.
(b.) To derive an expression for the total power emitted, integrate the Planck function,
here in terms of frequency.
∫ ∞
∫ ∞
2ℎ𝜈 3 /𝑐2
𝑑𝜈
𝐵𝜈 (𝑇 )𝑑𝜈 =
𝑒ℎ𝜈/𝑘𝑇 − 1
0
0
∫ ∞
2
(ℎ𝜈 3 )
𝑑𝜈
= 2 2
ℎ 𝑐 0 𝑒ℎ𝜈/𝑘𝑇 − 1
Substitute 𝑥 =
ℎ𝜈
𝑘𝑇
and 𝑑𝑥 =
ℎ
𝑑𝜈
𝑘𝑇
and
∫
2 (ℎ𝑇 )4 ∞ 𝑥3
=
𝑑𝑥
ℎ3 𝑐2
𝑒𝑥 − 1
0
∫ ∞
𝑥3
2 (ℎ𝑇 )4 𝜋 4
𝜋4
=
where
𝑑𝑥
=
ℎ3 𝑐2 15
𝑒𝑥 − 1
15
0
5 4
𝜎 4
2𝜋 𝑘
=
𝑇 , where 𝜎 =
𝜋
15ℎ3 𝑐2
3
Figure 1: Plot of energy density vs. wavelength that illustrates the behavior of
the Plank function. Two things to note- as the temperature increases, the emission
increases at all wavelengths (the lines do not cross) and the maximum emission occurs
at shorter wavelengths. Image from Wikipedia.
Note that the key is that power ∝ 𝑇 4 . The integral is a constant.
(c.)Sketch and description should resemble this:
(d.) Start with the Planck function
𝐵𝜆 (𝑇 ) =
With 𝑒𝑥 ≈ 1 + 𝑥 for 𝑥 ≪ 1, we have
2ℎ𝑐2 /𝜆5
.
𝑒ℎ𝑐/𝜆𝑘𝑇 − 1
2ℎ𝑐2 /𝜆5
1 + ℎ𝑐/𝜆𝑘𝑇 − 1
2ℎ𝑐2 /𝜆5
=
ℎ𝑐/𝜆𝑘𝑇
2𝑐𝑘𝑇
=
𝜆4
𝐵𝜆 (𝑇 ) ≈
Note that the expression is independent of ℎ and thus classical.
(e.) First, how I came to this: The observation was a “57 eV blackbody.” Accordingly,
57 eV
≈ 6.6 × 105 K.
8.6173 × 10−5 eV/K
Applying Wein’s law (𝜆max 𝑇 ≃ 0.0029 mK),
𝜆max ≃
0.0029 mK
= 4.39 × 10−9 m
5
6.6 × 10 K
4
𝑇 ≃
0.0029 mK
0.0029 mK
= 6.6 × 105 K
=
𝜆max
4.39 × 10−9 m
(f.) The Rayleigh-Jeans law applies to the long wavelength tail in which the emitted
power is proportional to temperature. One obvious solution is that the temperature
is not uniform across the neutron star. Another is an atmosphere on the NS that will
redistribute energy of the radiation, so the assumption of a blackbody is not valid
over the range from X-ray to optical.
5
Astronomy 2
a. (16 points) The Sun shines by nuclear fusion. Estimate the main sequence
lifetime of the Sun. Show all steps and clearly state all assumptions. Use only
the following information:
∙ The solar constant, the radiant flux from the Sun measured at the top of
Earth’s atmosphere, is 1.38 ×106 erg/cm2 /s. This is a direct observation.
∙ The orbital velocity of the Earth is 30 km/s. This can be determined from
the aberration of starlight.
∙ The length of the year is 365.2422 days.
∙ m1 𝐻 = 1.0078 amu
∙ m4 𝐻𝑒 = 4.0026 amu
∙ 1 amu = 1.66 ×10−24 gm
∙ 𝛼 = 1/137
∙ c = 2.99792×105 km/s
∙ G = 6.67×10−8 cm3 sec−2 gm−1
∙ h = 6.62×10−27 erg sec
b. (4 pts) Write down the nuclear reaction(s) involved in the PP I chain.
SOLUTION:
a. Use the Earth’s orbital velocity to determine the distance from the Earth to the
Sun. 2𝜋D = vt, where t is the length of the year
b. Use the inverse-square law to determine the solar luminosity
c. Use Newtonian gravity to determine the mass of the Sun;
m𝐸 v2 /r = Gm⊙ m𝐸 /r2
d. The nuclear reaction is 41 H → 4 He. Use the mass deficit to determine the
energy per reaction, and the number of reactions/sec needed to drive the solar
luminosity.
e. The lifetime of the sun depends on the number of H atoms.
f. For full credit, realize that only about 10% of the solar H is fused into He
For the PP I chain:
6
∙ 1 H + 1 H → 2 D + e+ + 𝜈𝑒
∙ 2 D + 1 H → 3 He + 𝛾
∙ 3 He + 3 He → 4 He + 1 H + 1 H
7
Astronomy 3
Consider a neutron star which is undergoing a burst. The energy source for the burst
is thermonuclear energy release underneath the surface layer of the star. Assume the
configuration during the burst remains spherically symmetric. The energy released
by the burst is in the form of photons. What critical luminosity of photons could
just begin to raise an optically thick but physically thin surface layer? What is the
name associated with this luminosity? Assume the opacity of the material is due to
electron scattering, so that it is 𝜅𝑇 = 0.2(1 + 𝑋) cm2 g−1 , where 𝑋 is the hydrogen
mass fraction.
Observed from Earth, the maximum flux observed from a bursting neutron star is
often taken to be associated with this critical luminosity. During the later stages of
the burst, the observed flux and temperature both steadily decrease. Assume that
the radiation during the later stages can be well-modeled by a blackbody.
Show that with these assumptions, measurements of the maximum flux and the flux
and temperature late in the burst can be used to determine the mass and radius of
the neutron star if the distance to the star is known.
SOLUTION:
The critical luminosity is called the Eddington luminosity or the Eddington limit,
and is given by
4𝑐𝐺𝑀
𝐿𝐸𝑑𝑑 =
.
𝜅𝑇
This can be derived assuming a balance between the outward acceleration due to
radiation pressure and gravitation. The energy flux at a distance 𝐷 from the star’s
center is
𝐿
𝑑𝐸
=
.
𝑑𝐴𝑑𝑡
4𝜋𝐷 2
The momentum of a photon is related to the energy by 𝑝 = 𝐸/𝑐. Then the momentum
transfer rate (the radiation force) to an electron is
𝑑𝑝
𝜎𝑇 𝑑𝑝
𝐿𝜎𝑇
,
=
=
𝑑𝑡
𝑑𝐴 𝑑𝑡
4𝜋𝐷 2
where the Thomson cross section is 𝜎𝑇 . The gravitational force felt by a proton at
the same distance is 𝐺𝑀𝑚𝑝 /𝐷 2 . Equating these,
𝐿
𝐺𝑀𝑚𝑝
=
,
4𝜋𝐷 2
𝐷2
𝐿=
4𝜋𝐺𝑀𝑚𝑝 𝑐
.
𝜎𝑇
Since the opacity is related to the cross section by 𝜅𝑇 = 𝜎𝑇 /𝑚𝑝 we have the desired
result.
8
Observed from the Earth, the radiation is gravitationally redshifted, and the Eddington flux is
√
√
𝐿𝐸𝑑𝑑
2𝐺𝑀
𝑐𝐺𝑀
2𝐺𝑀
1
−
=
1
−
,
𝐹𝐸𝑑𝑑 =
4𝜋𝐷 2
𝑅𝑐2
𝜅𝑇 𝐷 2
𝑅𝑐2
where 𝐷 is the distance and 𝑀 and 𝑅 are the neutron star’s mass and radius, respectively. Credit will be given if the Newtonian expression, without the redshift factor,
is used. Late in the burst, if the radiation is approximately blackbody, the observed
flux 𝐹𝑜𝑏𝑠 and temperature 𝑇𝑜𝑏𝑠 are related by
𝐹𝑜𝑏𝑠 =
4𝜋𝑅2
2𝐺𝑀 −1
4
𝜎𝐵 𝑇𝑜𝑏𝑠
(1 −
) .
2
4𝜋𝐷
𝑅𝑐2
Again, credit is given if the redshift factors are neglected.
Forming the combination
√
4
𝐹𝐸𝑑𝑑 𝜅𝐷 2 𝜎𝐵 𝑇𝑜𝑏𝑠
𝐺𝑀
2𝐺𝑀
=
𝛼
=
(1
−
),
𝐺𝑐3
𝐹𝑜𝑏𝑠 𝐷 2
𝑅𝑐2
𝑅𝑐2
one can solve the quadratic for 𝐺𝑀/𝑅𝑐2 :
𝐺𝑀
1 1√
±
=
1 − 8𝛼.
𝑅𝑐2
4 4
Credit is given for the Newtonian expression 𝐺𝑀/𝑅𝑐2 = 𝛼. Then 𝑀 can be found
from the formula for the Eddington flux:
𝑀=
2𝐺𝑀 −1/2
𝜅𝐷 2 𝐹𝐸𝑑𝑑
(1 −
)
.
𝑐𝐺
𝑅𝑐2
𝑅 can be found from
𝑅=
𝐺𝑀 𝐺𝑀 −1
(
) .
𝑐2 𝑅𝑐2
9
Astronomy 4
This problem derives the local instability condition of a rotating galactic disk, so
called the Toomre instability. The Toomre instability is often characterized by the
parameter 𝑄, which is ∝ 𝜎𝜅/𝐺Σ, where 𝜎, 𝜅, 𝐺, and Σ are the velocity dispersion,
epicyclic frequency, gravitational constant, and surface density, respectively.
We consider a small square region of infinitesimal thickness in the galactic plane with
the size 𝐿 at a galactic distance of 𝑅. The mass within the region is 𝑀 = Σ𝐿2 .
The region is on a circular orbit around the galaxy. The angular rotation speed and
gravitational potential of the galaxy are Ω(𝑅) and Φ(𝑅), respectively.
Four types of forces act on the boundaries of the region: the galactic tidal force,
centrifugal force, self-gravity within the region, and internal pressure. Consider only
forces parallel to the galactic radius vector, and neglect any internal pressure. For
example, the galactic tidal force stretches the region; comparing the gravitational
forces of the galaxy at radii 𝑅 + 𝐿/2 and 𝑅 − 𝐿/2 the galactic tidal force is
( )
( )
𝑑Φ
𝑑Φ
𝑑2 Φ
𝑓G =
−
=
𝐿
(1)
𝑑𝑅 𝑅+𝐿/2
𝑑𝑅 𝑅−𝐿/2 𝑑𝑅2
Here we consider the changes of these forces when the region is perturbed in the
direction of galactic radius by Δ𝐿, where Δ𝐿 << 𝐿. A positive Δ𝐿 corresponds to
a compression of the region.
a. Derive the change of the galactic tidal force (𝑓G ). Does it increase or decrease?
b. Derive the change of the centrifugal force (𝑓c = Ω2 𝑅). Does it increase or
decrease? For this question, use the conservation of angular momentum (𝐽 =
Ω𝑅2 ).
c. Derive the change of the self-gravity of the region (𝑓g ). State whether the
self-gravity increases or decreases.
d. Compare the above three forces, Δ𝑓G , Δ𝑓c , and Δ𝑓g , and derive the threshold
size 𝐿crit for the region to become unstable. Use the epicyclic frequency 𝜅2 =
𝑑2 Φ/𝑑𝑅2 + 3Ω2 .
e. In addition to this condition, the Jeans criterion provides another
√ condition
for instability. It is derived by comparing the free-fall time 𝑡ff = 𝐿/𝐺Σ and
𝑡cross = 𝐿/𝜎. The region is unstable when 𝑡ff < 𝑡cross , which gives the Jeans
length 𝜆J :
𝜎2
𝜆𝐽 =
<𝐿
(2)
𝐺Σ
From this condition and the condition from (d), derive the Toomre Q parameter
and Toomre’s instability condition.
10
SOLUTION:
(a)
𝑑2 Φ
𝑑2 Φ
𝑑2 Φ
(𝐿 − Δ𝐿) −
𝐿 = − 2 Δ𝐿
Δ𝑓G =
𝑑𝑅2
𝑑𝑅2
𝑑𝑅
The tidal force that acts on the region decreases.
(b)
Δ𝐽 = Δ(Ω𝑅2 ) = ΔΩ𝑅2 + 2Ω𝑅Δ𝑅 = 0
Δ𝑓c = Δ(Ω2 𝑅) = 2ΩΔΩ𝑅 + Ω2 Δ𝑅 = −3Ω2 Δ𝑅 = −3Ω2 Δ𝐿
The centrifugal force decreases.
(c)
𝐺𝑀
𝐺𝑀
𝐺𝑀 Δ𝐿
Δ𝐿
+ 2 ≃ −2 2
Δ𝑓g = −
= −2𝐺Σ
2
(𝐿 − Δ𝐿)
𝐿
𝐿 𝐿
𝐿
(3)
(4)
(5)
(6)
The self-gravity decreases.
(d)
∣Δ𝑓G + Δ𝑓c ∣ < ∣Δ𝑓g ∣,
and
𝐿 < 𝐿crit =
2𝐺Σ
𝜅2
(7)
(8)
(e)
and
𝜆J < 𝐿crit
(9)
1 𝜎𝜅
<1
𝑄= √
2 𝐺Σ
(10)
11
AMO 1
The interaction of atoms with a laser field can be described in a simple classical
model (Lorentz model). In each atom, the bound electron subjected to an oscillating
electric field ℰ = ℰ0 𝑒𝑖𝜔𝑡 is assumed to respond like a damped harmonic oscillator with
an undamped resonance frequency 𝜔0 ,
𝑒
(11)
𝑥¨(𝑡) + 𝛽 𝑥(𝑡)
˙ + 𝜔02 𝑥(𝑡) = ℰ(𝑡)
𝑚
a. (8 pts) By considering the atomic dipole moment 𝑒𝑥(𝑡) ≡ 𝛼ℰ(𝑡), show that the
real and imaginary parts of the polarizability 𝛼 are given by
1 𝑒2
𝛽
2
2 2𝑚𝜔𝑜 𝛿 + 𝛽 2 /4
𝑒2
𝛿
𝑅𝑒(𝛼) = −
,
2
2𝑚𝜔𝑜 𝛿 + 𝛽 2 /4
𝐼𝑚(𝛼) = −
(12)
(13)
near resonance with 𝛿 ≪ 𝜔 + 𝜔0 where 𝛿 ≡ 𝜔 − 𝜔0 is the detuning.
b. (5 pts) What physical process(es) give(s) can give rise to the damping coefficient
𝛽 in a gas of atoms? Write an expression for 𝛽 (gas density 𝑛, atomic mass 𝑀,
collisional cross-section 𝜎, temperature 𝑇 ), accounting for all of the terms.
c. (3 pts) Now consider the interaction Δ = 12 𝛼⟨𝑅𝑒(ℰ 2)⟩. What does the real part
of Δ represent? What does it mean that Δ is complex?
d. (4 pts) Derive an expression for the rate at which an atom scatters photons
from the laser beam.
SOLUTION:
a. Using the ansatz 𝑥(𝑡) = 𝑥0 𝑒𝑖𝜔𝑡 yields 𝛼 = (𝑒2 /𝑚)/ [𝜔02 − 𝜔 2 + 𝑖𝛽𝜔]. Sorting into
real and imaginary parts, and making the approximation 𝜔02 − 𝜔 2 ≃ 2𝜔0 𝛿, near
resonance for 𝛿 ≪ 𝜔 + 𝜔0 , lead to the desired results.
b. 1. Radiative damping (i.e. spontaneous emission). Note: The radiative damping rate may be
√ changed by putting the atoms
√in a cavity. 2. Collisions between
atoms. 𝛽 = Γ2 + (𝑛𝜎𝑣𝑡ℎ )2 , where 𝑣𝑡ℎ = 𝑘𝐵 𝑇 /𝑀. – Note: The damping
coefficient also depends on the effects absorption and stimulated emission (saturation; power broadening), which is however not part of the Lorentz model.
c. Real: AC Stark Shift (Optical potential), Imaginary: damping.
d. Calculate scattering rate Γ𝑠𝑐 using 2∣𝐼𝑚(Δ)∣ = ℏΓ𝑠𝑐
12
AMO 2
An optical mug for atoms (“gravito-optical surface trap”; cf. Yu. B. Ovchinnikov, I.
Manek, R. Grimm, PRL 79, 2225 (1997)) can be implemented above a glass surface
as illustrated in fig.1. The surface is covered with an evanescent light field, combined
with a hollow laser beam. The evanescent field, which is formed by total internal reflection at the glass surface, gives rise to a repulsive optical potential 𝑈(𝑧) = 𝑈0 𝑒−2𝑧/𝜉 ,
where 𝑧 is the distance above the surface and 𝜉 is the decay length of the evanescent field. The repulsive optical potential of the hollow laser beam (radius 𝑅), which
passes through the prism vertically, forms the walls of the mug.
a. (5 pts) The combined potential in 𝑧 direction resulting from the combination
of gravity and the repulsive wall is approximately wedge shaped. Derive an
expression for the density distribution 𝑛(𝑧) for an atomic gas of 𝑁 atoms (mass
𝑚) at temperature 𝑇 in this potential. What is the peak density 𝑛0 of the gas?
b. (5 pts) Calculate the corresponding phase space density 𝐷, and give an estimate
for the critical temperature 𝑇𝑐 at which quantum degeneracy sets in.
c. (5 pts) At low temperature, the details of the shape of the potential near the
surface become important, as most atoms reside near the potential minimum.
Calculate the frequency 𝜔𝑧 for harmonic oscillations around the potential minimum. For which range of temperatures is the gas 2D?.
d. (5 pts) How could you conveniently implement evaporative cooling in the setup,
and how would it work?
13
SOLUTION:
−𝑧/𝑧0
, with 𝑧0 = 𝑘𝐵 𝑇 /𝑚𝑔. From
a. Barometric density
0𝑒
∫ 3 distribution 𝑛(𝑧) = 𝑛𝑚𝑔
normalization 𝑑 𝑟 𝑛 = 𝑁 obtain 𝑛0 = 𝜋𝑅2 𝑘𝐵 (𝑁/𝑇 )
√
b. Phase-space density 𝐷 = 𝑛0 Λ3 with thermal de-Broglie wavelength Λ = ℎ/ 2𝜋𝑚𝑘𝐵 𝑇 .
Obtain 𝐷 ∼ 𝑁𝑇 −5/2 , and 𝑇𝑐 from 𝐷 ∼ 1.
2𝑈0
.
c. Total 1D potential 𝑈(𝑧) = 𝑈0 𝑒−2𝑧/𝜉 + 𝑚𝑔𝑧 has minimum at 𝑧 = 𝑧0 = 2𝜉 ln 𝑚𝑔𝜉
1
1 ′′
2
2
2
Harmonic expansion around 𝑧0 , i.e. 2 𝑚𝜔𝐻𝑂 (Δ𝑧) = 2 𝑈 (𝑧0 )(Δ𝑧) , yields
√
𝜔𝐻𝑂 = 2𝑔/𝜉. Gas is 2D for 𝑘𝐵 𝑇 < ℏ𝜔𝐻𝑂 .
a. Evaporative cooling relies on the escape of atoms with more than the average
energy, and subsequent rethermalization of the remaining atoms in the gas.
This can be implemented by reducing the laser power of the evanescent field
such that those atoms can escape toward the surface. It can also be done by
reducing the power of the hollow beam, but the efficiency will be lower.
14
Condensed Matter 1
The lower left plot below is the phonon dispersion relation of neon, measured by
Leake et al. (Phys. Rev. 1969) by inelastic neutron scattering. The crystal has
𝑓 𝑐𝑐 structure. The lattice constant is 𝑎 = 4.43˚
𝐴, and the primitive vectors are
a = 𝑎(0, 1/2, 1/2), b = 𝑎(1/2, 0, 1/2), c = 𝑎(1/2, 1/2, 0).
a. What is the velocity of longitudinal sound waves in the (111) direction?
b. Consider the LA mode at the k = (1/2, 1/2, 1/2) point of the Brillouin zone,
designated ”L” in the picture and at the top of the phonon graph. Its energy is
6.7 meV. What is its group velocity? (Note that the notation is an abbreviation
for k = (2/𝑎)(1/2, 1/2, 1/2).)
c. Consider the same LA mode. What is the normal mode displacement pattern?
(Explain carefully in words, and provide at least a simplified picture.)
d. Explain why three ”branches” are shown to the left of k = (0, 0, 0) and only
two to the right.
e. What are the frequencies of the phonons at k = (1, 1, 1) and at k = (0, 0, 1)
(abbreviated notation; neither is shown explicitly on the graph.)
15
SOLUTION:
(a.) The slope in the limit 𝑘 → 0 is
9.5𝑚𝑒𝑉 /ℏ
Δ𝜔
= 2𝜋 1 1 1 = 1.17 × 103 𝑚/𝑠
Δ𝑞
( , , )
𝑎 2 2 2
(b.) 𝑑𝜔/𝑑𝑘 = 0
(c.) The (111) planes displace perpendicular to the planes parallel to (111) direction
by alternate amounts ±𝑢.
(d.) 120 degree rotational symmetry around the (111) direction (to the “right” of
(000) in the graph) requires that the two transverse branches are degenerate. In the
(110) direction (to the left of (000)) there is only 180 degree rotational symmetry so
transverse branches have different velocities.
(e.) Since (2𝜋/𝑎)(111) is a reciprocal lattice vector, it is “the same” state as 𝑞 = 0
and the frequencies are 0. But 𝑞 = (2𝜋/𝑎)(001) is not a reciprocal lattice vector.
Therefore we can subtract the reciprocal lattice vector (2𝜋/𝑎)(−1, −1, 1) to convert
𝑞 to (2𝜋/𝑎)(110), the ”X point”, where the frequencies are 7 meV (LA branch) and
4.8 meV (doubly degenerate TA)
16
Condensed Matter 2
The free energy density of a superconductor is approximated in the Ginzburg-Landau
theory by
,(
) ,2
2
,
𝑒∗
1 ,, ℏ
2
4
, +ℎ .
∇
−
𝑨
Ψ
𝑓𝑠 (𝒓) − 𝑓𝑛 = 𝑎 ∣Ψ∣ + 𝑏 ∣Ψ∣ +
,
2𝑚∗ , 𝑖
𝑐
8𝜋
Here 𝑒∗ , 𝑚∗ are phenomenological charge and mass of supercurrent carriers, 𝑎(𝑇 ) and
𝑏(𝑇 ) are temperature dependent constants and Ψ(𝒓) is the “wave function” of the
superconducting condensate and 𝑓𝑛 is a free energy of the normal state at the same
temperatures. 𝑨(𝒓) is vector potential of an external magnetic field ℎ.
a. Assuming that close to some temperature 𝑇𝑐 the temperature dependence of
the coefficient 𝑏(𝑇 ) is not significant, 𝑏(𝑇 ) ≈ 𝑏 > 0, and 𝑎(𝑇 ) ≈ 𝛼(𝑇 − 𝑇𝑐 ),
show that 𝑇𝑐 is the critical temperature of the superconducting phase transition in the absence of external fields. Find the temperature dependence of the
“superconducting order parameter”, defined as Δ ≡ ∣Ψ∣, close to the critical
temperature 𝑇𝑐 .
b. A strong external magnetic field ℎ > 𝐻𝑐 (𝑇 ) destroys superconductivity. Under
the assumption that the surface effects can be neglected, and that the magnetic
field is uniform, find the dependence 𝐻𝑐 (𝑇 ) close to 𝑇𝑐 by comparing the condensation energy (the minimum of the above free energy without ℎ) with the
energy of magnetic field 𝐻𝑐2 /8𝜋. 𝐻𝑐 (𝑇 ) is called the critical magnetic field of a
superconductor.
c. Let us look for the solution Ψ(𝒓) under the following conditions: (i) Ψ(𝒓) =
𝑓 (𝑥), ℎ = 0, Ψ(𝑥 < 0) = 0, Ψ(𝑥 → +∞) = Δ with Δ found in a. These
conditions describe the interface between a superconductor and the vacuum
(vacuum is at 𝑥 < 0). Write down an equation for 𝑓 (𝑥). Extract from this
equation the typical length 𝜉 (known as the coherence length) at which order
parameter Ψ can change significantly. Solve the equation or at least sketch the
solution qualitatively. What is the temperature dependence of 𝜉?
d. Using the free energy density 𝑓𝑠 describe qualitatively the reason why weak
magnetic field does not penetrate the superconductor (“Meissner effect”). Apply
this reasoning and the standard properties of the vector potential to derive
“flux quantization” condition for allowed values of magnetic flux threading a
superconducting loop. Take into account that the microscopic BCS theory
shows that the supercurrent is carried by “Cooper pairs”, so that 𝑒∗ = 2𝑒.
17
SOLUTION:
(a.) Let us assume that an external field is absent ℎ = 0 and 𝑨 = 0. We look for a
uniform solution Ψ = 𝑐𝑜𝑛𝑠𝑡. Minimizing the free energy with respect to ∣Ψ∣2 we have
( 𝑎 )1/2 ( 𝛼 )1/2 √
= 2𝑏
𝑇𝑐 − 𝑇 .
𝑎 + 2𝑏∣Ψ∣2 = 0 which gives Δ = − 2𝑏
2
(b.) The minimum of the free energy at the optimal value of Δ derived in a. is − 𝑎4𝑏 .
This is so called “condensation energy”. Together with the energy of diamagnetic
2
ℎ2
currents in the presence of an external field we have 𝑓𝑠 − 𝑓𝑛 ≈ − 𝑎4𝑏 + 8𝜋
. This energy
(
)1/2
√
2
= 2𝜋𝛼/𝑏 (𝑇𝑐 − 𝑇 ). This means that for
is positive for ℎ > 𝐻𝑐 (𝑇 ) = 2𝜋 𝑎𝑏
ℎ > 𝐻𝑐 (𝑇 ) it is favorable to have a normal state and superconductivity is destroyed.
2
ℏ2
′′
(c.) − 2𝑚
+ 𝑎𝑓 + 2𝑏𝑓 3 = 0 or −𝜉 2 𝑔 ′′ + 𝑔 3 = 𝑔, where 𝑔 = 𝑓 /Δ and 𝜉 2 = 2𝑚ℏ∗ ∣𝑎∣ so
∗𝑓
that 𝜉 ∼ (𝑇𝑐 − 𝑇 )−1/2 .
( 𝑎 )1/2
. Let us
(d.) As we saw the condensation energy argument favors ∣Ψ∣ = Δ = − 2𝑏
consider the free energy where the absolute value of the wave function is fixed while
the phase can change in space, i.e., Ψ = Δ𝑒𝑖𝜃 :
(
)2
𝑒∗
ℏ2 Δ2
∇𝜃 − 𝑨 .
𝛿𝑓 =
2𝑚∗
ℏ𝑐
We can see that the presence of magnetic field inside the superconductor the vector
potential can not be compensated by the gradient of the phase (as ∇ × 𝑨 ∕= 0) and
the weak magnetic field is repelled from
∮ the superconductor.
∗ ∮
∗
Considering the loop integral 2𝜋𝑛 = ∇𝜃 ⋅ 𝑑𝒓 = 𝑒ℏ𝑐 𝑨𝜃 ⋅ 𝑑𝒓 = 𝑒ℏ𝑐 Φ we see that the
total magnetic flux Φ through a loop is quantized in units of
Φ0 =
ℎ𝑐
ℎ𝑐
.
=
𝑒∗
2𝑒
18
Nuclear 1
Consider the scattering of a proton with energy 𝐸 = 250 GeV off a Gold nucleus at
rest. Estimate the total and elastic cross sections of proton-nucleus scattering, and
sketch the differential cross section as a function of the scattering angle. Will the
differential cross section have a minimum?
SOLUTION:
1. The mean free path of a proton in the nucleus can be estimated as 𝜆 ≃ (𝜌𝜎)−1 ,
where 𝜌 ≃ 0.18 fm−3 is the nuclear density, and 𝜎 ≃ 3.4 fm2 is the protonnucleon inelastic cross section. Note that a rough estimate for this cross section
can be obtained by simply writing 𝜎 ≃ 𝜋𝑅𝑝2 and using 𝑅𝑝 ≃ 1 fm as a proton
radius. Numerically, we get for the mean free path 𝜆 ≃ 1.6 fm which is much
smaller than the radius of the Gold nucleus, 𝑅𝐴 ≃ 1.1 × (197)1/3 fm ≃ 6.4 fm.
Because the mean free path is short, the nucleus acts as a black disk. The
𝐴
2
total, elastic and inelastic cross sections are thus 𝜎𝑡𝑜𝑡
≃ 2𝜋𝑅𝐴
≃ 2.6 barn,
𝐴
𝐴
2
−13
𝜎𝑒𝑙 ≃ 𝜎𝑖𝑛𝑒𝑙 ≃ 𝜋𝑅𝐴 ≃ 1.3 barn. Units: 1 fm = 10
cm; 1 barn = 10−28 cm2 =
2
100 fm .
2. The angle at which the differential cross section for the scattering off a black
disk has a diffractive minimum is 𝜃𝑑 ≃ 1/(𝑃 𝑅𝐴 ). Using the proton’s momentum
𝑃 ≃ 250 GeV ≃ 1250 fm−1 and the Gold radius 𝑅𝐴 ≃ 6.4 fm we estimate
𝜃𝑑 ≃ 10−4 rad.
19
Nuclear 2
Consider three amplitudes of pion-nucleon scattering:
𝑀 + ≡ 𝑀(𝜋 + 𝑝 → 𝜋 + 𝑝), 𝑀 − ≡ 𝑀(𝜋 − 𝑝 → 𝜋 − 𝑝), and 𝑀 0 ≡ 𝑀(𝜋 − 𝑝 → 𝜋 0 𝑛)
∙ Assuming isosping invariance, find the relation between these scattering amplitudes
∙ When the center-of-mass pion-nucleon energy is close to the mass of the Δ
resonance (isospin 𝐼 = 3/2), it dominates the scattering process. Find the
ratio of the cross sections of the three scattering processing above in this energy
range.
SOLUTION:
a) The isospin of the pion is 𝐼 = 1, with 𝜋∗+, 𝜋 0 and 𝜋 − having the isospin projections
𝐼𝑧 = +1, 0, −1 respectively. The isospin of the nucleon is 𝐼 = 1/2, with proton and
neutron’s isospin projections 𝐼𝑧 = +1/2, −1/2. Therefore the three pion-nucleon
states that enter as the initial and/or final states can be decomposed in the isospin
space as
∣𝜋 + 𝑝 > = ∣3/2, 3/2 >
√
√
1/3∣3/2, −1/2 > − 2/3∣1/2, −1/2 >
∣𝜋 − 𝑝 > =
√
√
∣𝜋 0 𝑛 > =
2/3∣3/2, −1/2 > + 1/3∣1/2, −1/2 >
Isospin invariance implies that the scattering amplitudes depend only on the total
isospin, but not on the isospin projection. We can thus define two isospin amplitudes
𝑀3/2 and 𝑀1/2 , and express the scattering amplitudes of interest as
𝑀+ = 𝑀3/2
𝑀− = (1/3)𝑀3/2 + (2/3)𝑀1/2
√
√
𝑀0 = ( (2)/3)𝑀3/2 − ( 2/3)𝑀1/2
Now it is easy to find the relation that we are looking for
√
2𝑀0 + 𝑀− = 𝑀+
b) The dominance of the Δ(𝐼 = 3/2) resonance allows us to neglect the 𝐼 = 1/2
scattering amplitude. In this case the scattering amplitudes are related as
√
𝑀+ : 𝑀− : 𝑀0 = 1 : 1/3 : 2/3
20
Neglecting the small mass differences (in accord with the assumed isospin invariance)
then yields the ratio of the cross sections:
𝜎+ : 𝜎− : 𝜎0 = 9 : 1 : 2.
This prediction agrees quite well with the experimental data.
21
High Energy 1
Consider Compton scattering, the process of electron-photon scattering,
𝑒− (𝑝) + 𝛾(𝑘) → 𝑒− (𝑝′ ) + 𝛾(𝑘 ′ )
a. (5 pts) Write down the Feynman diagrams which contribute at the tree graph
level, and the corresponding amplitudes in terms of momenta, spinors and polarization vectors 𝜖𝜇 (𝑘) and 𝜖𝜈 (𝑘 ′ ).
b. (5 pts) Prove gauge invariance: show by explicit calculation that the sum of the
amplitudes vanishes if one replaces one of the two polarization vectors of the
photons by its 4-momentum.
c. (5 pts) Now consider quark-gluon scattering. Let the incoming gluon have color
index a, and the outgoing gluon color index b. There is a new Feynman diagram
that contributes. What is it? How are the previous amplitudes modified to take
color into account?
d. (5 pts) Now prove again gauge invariance. First write the amplitude for the
new diagram in terms of the 3-gluon vertex V. Then substitute the expression
for V.
Hint: the 3-gluon vertex is equal to:
𝑎𝑏𝑐
𝑉𝜇𝜈𝜌
(𝑘, 𝑝, 𝑞) = 𝑔𝑓 𝑎𝑏𝑐 [𝜂 𝜇𝜈 (𝑘 − 𝑝)𝜌 + 𝜂 𝜈𝜌 (𝑝 − 𝑞)𝜇 + 𝜂 𝜌𝜇 (𝑞 − 𝑘)𝜈 ]
if all momenta are incoming.
(a.) For diagram I we have
]
−𝑖(/𝑝 + 𝑘/ ) + 𝑚
𝑢¯(𝑝 )/𝜖(𝑘 )
/𝜖(𝑘)𝑢(𝑝)
(𝑝 + 𝑘)2 + 𝑚2
′
′
[
(14)
and for diagram II we have
]
′
/
−𝑖(
𝑝
−
𝑘
)
+
𝑚
/
𝑢¯(𝑝′ )/𝜖(𝑘)
/𝜖(𝑘 ′ )𝑢(𝑝)
′
2
2
(𝑝 − 𝑘 ) + 𝑚
[
(15)
(See figures (2) and (3) for the associated Feynman diagrams.)
(b.) Replacing /𝜖(𝑘) by 𝑘/, and substituting 𝑘/ = (𝑘/ + /𝑝 − 𝑖𝑚) − (/𝑝 − 𝑖𝑚) in (14), and
𝑘/ = (𝑘/ − /𝑝′ + 𝑖𝑚) + (/𝑝′ − 𝑖𝑚) with 𝑘 − 𝑝′ = −𝑝 + 𝑘 ′ in (15), and using
(/𝑝 − 𝑖𝑚)𝑢(𝑝) = 0
𝑢¯(𝑝′ )(/𝑝′ − 𝑖𝑚) = 0
]
[
][
]
[
−𝑖(/𝑝 + 𝑘/) + 𝑚 /𝑝 + 𝑘/ − 𝑖𝑚 = −𝑖 (𝑝 + 𝑘)2 + 𝑚2
22
(16)
(17)
one can see that the propagators cancel, and one finds 𝑢¯(𝑝′ )/𝜖′ (𝑘 ′ )(−𝑖)𝑢(𝑝) for (14)
and 𝑢¯(𝑝′ )(𝑖)/𝜖′ (𝑘 ′ )𝑢(𝑝) for (15). The sum clearly vanishes.
(c.) In (14) and (15) one must insert color matrices 𝑇𝑏 𝑇𝑎 and 𝑇𝑎 𝑇𝑏 , respectively,
giving an extra factor 𝑓𝑎𝑏 𝑐 𝑇𝑐 , so that the sum of diagrams (2) and (3) becomes equal
to
𝑢¯(𝑝′ )(𝑖)/𝜖′ (𝑘 ′ )𝑢(𝑝)𝑓𝑎𝑏 𝑐 𝑇𝑐 .
(18)
(d.) The new Feynman diagram is in figure (4). The new amplitude is
𝛿 𝑑𝑐 (−𝑖) 𝑎𝑏𝑐
𝑉𝜇𝜈𝜌 (𝑘, −𝑘 ′ , 𝑘 ′ − 𝑘)𝜖𝜇 (𝑘)𝜖𝜈 (𝑘 ′ )
𝑢¯(𝑝 )𝛾 𝑇𝑑 𝑢(𝑝)
′
2
(𝑘 − 𝑘 )
(19)
𝑎𝑏𝑐
(𝑘, −𝑘 ′ , 𝑘 ′ − 𝑘) = 𝑔𝑓 𝑎𝑏𝑐 𝜖𝜈 (𝑘 ′ ) [𝑘𝜈 (−𝑘 + 𝑘 ′ )𝜌 − 2𝑘 ⋅ 𝑘 ′ 𝛿𝜈 𝜌 ]
𝑘 𝜇 𝜖𝜈 (𝑘 ′ )𝑉𝜇𝜈𝜌
(20)
′
𝜌
One finds
The term with (−𝑘 +𝑘 ′ )𝜌 = (𝑝−𝑝′ )𝜌 gives no contribution due to current conservation
at the quark-gluon vertex, while the term with 𝑘 ⋅ 𝑘 ′ cancels the contribution of the
first two graphs (again the propagator (𝑘 − 𝑘 ′ )−2 = (−2𝑘 ⋅ 𝑘 ′ )−1 cancels).
𝑘′
𝑘
𝑝+𝑘
𝑝
𝑝′
Figure 2: Diagram I
23
𝑘
𝑘′
𝑝 − 𝑘′
𝑝′
𝑝
Figure 3: Diagram II
𝑘
𝑘′
𝑘 − 𝑘 ′ = 𝑝′ − 𝑝
𝑝
𝑝′
Figure 4: Diagram III
24
High Energy 2
Jets are collimated sprays of hadrons emitted in high-energy collisions. They are the
direct offspring of high-energy quarks and gluons that are ”kicked out” in a highenergy collision. Such collisions occur, for instance, at the Large Hadron Collider
(LHC), a proton synchrotron accelerator/collider of 27 km circumference near Geneva,
Switzerland. At the LHC counter-rotating proton beams of 1014 protons /beam are
accelerated to 3,500 GeV/proton. The beams circulate and are brought into collision
at several interaction regions around the ring where sophisticated particle detectors
are located.
a. Describe how protons are accelerated to energies in this range in a proton synchrotron.
b. A jet is created when a quark or gluon is ejected from a proton during a violent
collision
a. List all types of quarks that have been discovered so far and their charges
and mass estimates
b. Have free quarks or gluons been observed? What mechanism has been
proposed to describe this?
c. Describe the process of jet creation, i.e. the creation of a collimated
spray of hadrons (pions, kaons, protons, neutrons and many other types of
hadrons)
c. The simplest Feynman diagrams that describe the interactions between the
quarks and gluons in two colliding protons are the so-called ”two-to-two” processes.
a. Sketch two Feynman diagrams for processes where two gluons (one from
either proton) produce a pair of gluon jets.
b. Sketch two Feynman diagrams for processes where a quark and antiquark
(one from either proton) produce two gluon jets.
c. Sketch two Feynman diagrams for processes where a gluon and a quark
(one from either proton) produce two jets.
d. Assuming that a quark carries - on average - one-sixth of the proton’s energy,
calculate the maximum transverse energy a quark jet from an average collision
may have.
e. We now discuss measuring the direction and energy of a jet in a collider detector.
25
a. The primary energy loss mechanism for particles in a jet which interact
with a detector is the strong nuclear interaction. Given that a typical
cross section for a hadronic interaction of a pion on a nucleon is around 40
mbarn (1 barn = 10−28 m2 ), estimate the mean free path of a pion in iron
(𝐴𝐹 𝑒 = 55, 𝜌𝐹 𝑒 = 7.87 g/cm3 , 𝑁𝐴 = 6.022 × 1023 )
b. Describe and motivate the layout and structure of a typical collider detector.
SOLUTION:
(a.) The protons are accelerated by radio-frequency electric fields. The beams are
kept at in a circle of constant radius by increasing the strength of magnetic fields in
bending magnets at a rate proportional to the proton momentum.
(b.)
Quark
Charge (units of e)
Mass (GeV/c2 )
up
down
+2/3 -1/3
0.05 0.05
charm
+2/3
1.7
strange top bottom
-1/3
+2/3
-1/3
0.5
175
5
Free quarks have not been observed because of “color confinement”, the idea that the
strength of the strong force increases with separation distance between quarks. The
only free states are bound states of two or three quarks with no net color charge.
Jet creation arises as quarks (or gluons) with significant relative momentum move
apart. The strong interaction force increases to the point it becomes enegetically
favorable (𝐸 = 𝑚𝑐2 ) to create new quarks “in between” the initial quarks. This
process continues until the quarks have low enough relative momentum that they
form colorless bound states.
(c.)
𝑔𝑔 → 𝑔𝑔 processes:
26
𝑞𝑞 → 𝑔𝑔 processes:
𝑔𝑞 → 𝑔𝑞 processes:
(d.) The largest transverse momentum occurs when the jet is at 90𝑜 to the beam,
so
√ that all of the momentum appears in the transverse plane. For this case, with
𝑠 = 𝑥1 𝑥2 𝑝1 𝑝2 with 𝑥𝑖 the fraction of the proton momentum 𝑝1 = 𝑝2 = 𝑝 equal to
𝑥𝑖 = (1/6)𝑝,
𝑝𝑇 = 𝑝/6
(e.) The mean free path ℓ is given by ℓ = (𝑛 × 𝑠𝑖𝑔𝑚𝑎)−1 with 𝑛 the nucleon density.
The nucleon density is related to the given constants by 𝑛 = 𝜌𝑁𝐴 /𝑀 with 𝑀 ≈ 55,
the molar mass. The result is
ℓ ≈ 2 mm.
27