Solutions to Theory Problems

S2‒1
Theory Problem 2: Neutron Stars (10 points)
Part A. Mass and stability of nuclei (2.5 points)
A.1
The given binding energy is often called the Weizsäcker-Bethe mass formula, and the physicalinterpretation of the volume and the surface terms is based on the liquid drop model. The formula
works quite well except for the shell effects. Find 𝐴 to minimize the binding energy per mass number,i.e.,𝑎𝐵= 𝑎𝑉 − 𝑎𝑆 𝐴−1/3 − 𝐶 𝐴2/3 .
𝐴4(S2.1)

As long as 𝐴 is small, the second term is dominantly increasing with increasing 𝐴, and it is eventuallytaken over by the third term which is decreasing. Therefore, the extremal corresponds to the maximumof 𝐵/𝐴. One can explicitly carry out𝑑(𝐵/𝐴)=0𝑑𝐴(S2.2)

to find the following condition,𝑎𝑆 −4/3 𝑎𝐶 −1/3
𝐴−𝐴
= 0.
36(S2.3)

The solution is
𝐴=2𝑎𝑆
.
𝑎𝐶

(S2.4)

From the given numerical values, 𝐴 = 50 (which must be an integer) is concluded.
▷ Note: In reality 𝐵/𝐴 has a maximum for 𝐴 ranging from 56 Fe to 62 Ni. The discrepancy from the answer
in this problem is understood by the approximation of dropping the pairing energy and disregarding amass difference between the proton and the neutron.

A.1

0.9 pt𝐴 = 50

A.2

Take the differentiation of 𝐵(𝑍, 𝐴 − 𝑍)/𝐴 with respect to 𝑍 for a fixed 𝐴, which leads to−2𝑎𝐶𝑍∗
2𝑍 ∗ − 𝐴− 4𝑎sym
= 0.
1/3
𝐴𝐴

(S2.5)

By solving this in terms of 𝑍 ∗ , one finds𝑍∗ =1𝐴⋅ .
𝑎𝐶1+𝐴2/3 24𝑎sym(S2.6)

From this expression one can understand that 𝑍 ∗ ≃ 𝑁 as long as 𝐴 is small enough, while 𝑍 ∗ becomes farsmaller than 𝑁 for large 𝐴. It is obvious from the explicit form that the symmetry energy tends to favor
𝑍 = 𝑁 but the Coulomb interaction tends to favor 𝑍 → 0, and the balance between these competing
effects determines 𝑍 ∗ . Nuclei with too many neutrons (protons) would go through the 𝛽 − decay (the 𝛽 +decay or the electron capture) toward the stable (𝑍, 𝑁 ).
A.2

0.9 pt𝑍 ∗ = 791

Solutions to Theory Problems

S2‒2
A.3 Plugging the binding energy into the given inequality, one sees that the volume terms cancel due
to volume conservation. Then, the condition involves only 𝑎𝑆 and 𝑎𝐶 which are related as
2/3𝑎𝑆 [𝐴𝐴− 2( )22/3] + 𝑎𝐶 [(𝑍/2)2𝑍2−2] > 0.
𝐴1/3
(𝐴/2)1/3(S2.7)

As guided in the problem, the above inequality becomes as simple as21/3 − 1 𝑎𝑆𝑍2>⋅.
𝐴
1 − 2−2/3 𝑎𝐶(S2.8)

Therefore, the numerical coefficient turns out to be 0.7.
▷ Note: The physical interpretation of this result may need some explanations. Using the values of𝑎𝑆 and 𝑎𝐶 , one finds that such a symmetric fission process is possible for 𝑍 2 /𝐴 ≳ 18. For example,lead (Pb) with 𝑍 = 82 and 𝐴 = 208 is a stable element among several isotopes. Now, one can compute822 /208 ≈ 32, which is larger than the threshold 18. The key to resolving this gap is the potential barrier
from the deformation. When a heavy nucleus splits into two fragments, the shape and the surface shouldchange from the stable configuration (which is not necessarily spherical due to interaction) and thus
the surface energy increases. Although some heavy elements are energetically unstable, the lifetime
necessary to overcome the potential barrier can be very large.

A.30.7 pt

𝐶fission = 7.0 × 10−1

Part B. Neutron star as a gigantic nucleus (1.5 points)
B.1

The expression apart from the parametric dependence on 𝐴 can be identified as𝑎grav =3 𝐺𝑚2𝑁,
5 𝑅0(S2.9)

which is re-expressed in terms of 𝑀𝑃 using the given relation to 𝐺, leading to𝑎grav =3 ℏ𝑐 𝑚2𝑁
3 197 fm ⋅ MeV × (939 MeV/𝑐2 )2≃ 6.4 × 10−37 MeV .
=⋅5 𝑅0 𝑀𝑃2
5 1.1 fm × (1.22 × 1022 MeV/𝑐2 )2(S2.10)

Here, 𝑀𝑃 is a quantity often called the Planck mass. The gravitational effect is extremely tiny as comparedto the typical scale in nuclear physics and this scale difference is manifest for this expression of 𝐺 with
𝑀𝑃 in the MeV unit.
The stability is judged from the condition that the binding energy should be positive, i.e.,
𝐵total (𝐴) = 𝑎𝑉 𝐴 − 𝑎sym 𝐴 + 𝑎grav 𝐴5/3 > 0 .

(S2.11)This inequality can be translated into 𝐴 > 𝐴𝑐 with 𝐴𝑐 given by
𝐴𝑐 = (𝑎sym − 𝑎𝑉 3/2) ≃ 4.4 × 1055 .
𝑎grav(S2.12)▷ Note: One may think that one neutron drip is a process with the least change in the surface area
and thus the smallest barrier. This leads to a condition, 𝐵total (𝐴) > 𝐵total (𝐴 − 1) or approximately
𝑑𝐵total (𝐴)/𝑑𝐴 > 0, which is satisfied in a window with 𝐵total < 0. This condition, 𝑑𝐵total /𝑑𝐴 = 0, results insmaller 𝐴𝑐 but it is nontrivial whether such an unstable initial state could be prepared in the nature. Theneutron star is born in the Type-II (core-collapse) supernovae, and a baby star called the proto-neutronstar is an energetic state at high temperature. Neutrinos bring heat out from the proto-neutron star2

Solutions to Theory Problems

S2‒3
within the time scale of 𝒪(10) seconds. What is the possible smallest mass of the neutron star? Thisis not completely understood partly because the computer simulation of the supernovae is a big challenge even today. Although the neutron star mass can become much smaller than 𝑀⊙ theoretically, thesimulation and the observation favor the mass ≳ 1.4𝑀⊙ .

B.1

1.5 pt𝑎grav = 6 × 10−37 MeV
𝐴𝑐 = 4 × 1055

Part C. Neutron star in a binary system (6.0 points)
C.1
From the energy conservation, the free-falling system earns the kinetic energy 12 𝑚𝑣2 from thepotential energy 𝑚𝑔Δℎ, and the velocity takes
𝑣2 = 2𝑔Δℎ = 2Δ𝜙 .

(S2.13)

The time delay can be derived from the standard arguments. Suppose that Clock-II passes two infinitesimally separate points, 𝑧 and 𝑧 + Δ𝑧, in F at time 𝑡 and 𝑡 + Δ𝜏II , then the time interval registered by Clock-IIisΔ𝜏II =𝛾(𝑐Δ𝜏F − 𝛽Δ𝑧) ,𝑐(S2.14)

where the Lorentz transformation is used1 with 𝛽 = 𝑣/𝑐 and 𝛾 = 1/√1 − 𝛽 2 . Because Δ𝑧/Δ𝜏F = 𝑣 andΔ𝜏F = Δ𝜏I , the above expression is written as
Δ𝜏II = 𝛾(1 − 𝛽 2 )Δ𝜏F = √1 − 𝛽 2 Δ𝜏I .

(S2.15)Using the expression of 𝑣2 , one finally arrives atΔ𝜏II = √1 − 2Δ𝜙Δ𝜙Δ𝜏I ≃ (1 − 2 ) Δ𝜏I .
𝑐2𝑐(S2.16)

C.11.0 ptΔ𝜙Δ𝜏II = (1 − 2 )Δ𝜏I𝑐C.2 In terms of the effective speed of light, the total time necessary for the light propagation from N
to E is𝑥𝐸𝑡E-N = ∫𝑥𝑁𝑑𝑥.
𝑐eff (𝑥)(S2.17)The denominator is expanded in terms of the gravitational potential and the leading-order correction isfound to be𝑥𝑡E-N ≃𝐸2𝐺𝑀
𝑥 − 𝑥𝑁1∫ 𝑑𝑥(1 + √ WD ) = 𝐸+ Δ𝑡 ,222𝑐 𝑥𝑁
𝑐𝑐 𝑥 +𝑑(S2.18)

1 Clock‒F is in an inertial frame but Clock‒II is not. Using Clock‒II′ in another free‒falling frame II′ as an inertial reference toClock‒II, the Lorentz transformation is validated for Clock‒II′ seen from Clock‒F.

3

Solutions to Theory Problems

S2‒4
where the time delay Δ𝑡 is identified as√
𝑥=𝑥𝐸𝑥𝐸
2𝐺𝑀WD𝑑𝑥
𝐺𝑀WD𝑥2 + 𝑑2 + 𝑥√√∫)∣=.
Δ𝑡 =log(𝑐3𝑐3𝑥2 + 𝑑2
𝑥2 + 𝑑2 − 𝑥 𝑥=𝑥𝑥𝑁(S2.19)𝑁

Inside the logarithm, the following approximations are made:√𝑥2𝑁 + 𝑑2 + 𝑥𝑁 ≃𝑑2,2|𝑥𝑁 |√𝑥2𝐸 + 𝑑2 − 𝑥𝐸 ≃𝑑2,2𝑥𝐸√𝑥2𝑁 + 𝑑2 − 𝑥𝑁 ≃ 2|𝑥𝑁 | ,(S2.20)

and√𝑥2𝐸 + 𝑑2 + 𝑥𝐸 ≃ 2𝑥𝐸 .

(S2.21)

Then, the simple form of approximated Δ𝑡 isΔ𝑡 ≃𝐺𝑀WD
2𝑥𝐸 ⋅ 2|𝑥𝑁 |2𝐺𝑀WD4|𝑥 |𝑥)=log( 2log( 𝑁2 𝐸 ) .
𝑐3𝑑 /(2𝑥𝐸 ) ⋅ 𝑑2 /(2|𝑥𝑁 |)𝑐3
𝑑C.2(S2.22)

1.8 pt4|𝑥 |𝑥2𝐺𝑀WDlog( 𝑁2 𝐸 )
Δ𝑡 =
3𝑐𝑑

C.3

Because |𝑥𝑁 | = 𝐿 cos 𝜀 ≃ 𝐿 and 𝑑 = 𝐿 sin 𝜀 ≃ 𝐿𝜀 for Δ𝑡max , the answer of C.2 gives
Δ𝑡max =2𝐺𝑀WDlog(4𝑥𝐸 /𝐿𝜀2 )𝑐3(S2.23)

For Δ𝑡min the sign of 𝑥𝑁 is changed. Although the expression of Δ𝑡 is intact, the approximation takes adifferent form as√𝑥2𝑁 + 𝑑2 + 𝑥𝑁 ≃ 2𝑥𝑁 ,√𝑥2𝑁 + 𝑑2 − 𝑥𝑁 ≃𝑑2.
2𝑥𝑁

(S2.24)

Then, the approximated form of Δ𝑡min isΔ𝑡min ≃𝐺𝑀WD
2𝑥 ⋅ 𝑑2 /(2𝑥𝑁 )2𝐺𝑀WDlog( 2 𝐸)=log(𝑥𝐸 /𝐿) ,3𝑐𝑑 /(2𝑥𝐸 ) ⋅ 2𝑥𝑁
𝑐3(S2.25)where 𝑥𝑁 ≃ 𝐿 is used in the last expression. In the difference, Δ𝑡max − Δ𝑡min , one sees that 𝐿 and 𝑥𝐸disappear.
C.3

C.4

1.8 pt

2𝐺𝑀WDΔ𝑡max − Δ𝑡min =log(4/𝜀2 )𝑐3
Using the expansion, cos 𝜀 ≃ 1 − 12 𝜀2 , one can evaluate
𝜀2 ≃ 2 × (1 − cos 𝜀) = 0.00022 .

(S2.26)

From the graph the difference in time delays is roughly read out as(S2.27)

Δ𝑡max − Δ𝑡min ≈ 50 𝜇s
From these numerical values, 𝑀WD is solved as−1𝑀WD = 𝑀⊙ (2𝐺𝑀⊙)
𝑐3

Δ𝑡max − Δ𝑡min
50 𝜇s≃
𝑀 ≃ 0.5 𝑀⊙ .
log(4/𝜀2 )10 𝜇s log(4/0.00022) ⊙4(S2.28)

Solutions to Theory Problems

S2‒5▷ Note: The data in this problem roughly correspond to PSR J1614-2230 [see P.B. Demorest et al., Nature467, 1080-1083 (2010)]. From the Shapiro delay measurement, the White Dwarf mass was estimated as0.500 ± 0.006𝑀⊙ . With the Keplerian orbital parameters in the binary system, the neutron star mass wasconsidered to be 1.97 ± 0.04𝑀⊙ , which was the heaviest neutron star at that time. Since then, several
candidates for heavier neutron stars have been found.

C.4

0.8 pt
𝑀WD /𝑀⊙ = 0.5

C.5

For the circular orbit with the radius 𝑅, the equation of motion is𝑚𝑅𝜔2 = 𝐺𝑚𝑀,𝑅2(S2.29)

if 𝑀 is sufficiently large. From this it is easy to see
𝑅3 𝜔2 = 𝐺𝑀 = (const.)(S2.30)This is nothing but Kepler's third law and the relation holds for more general elliptical orbits around thecenter of mass.
C.5

C.6

0.4 pt3𝑝=−
2
The sum of the kinetic energy and the potential energy is
𝐸=1𝑚𝑀
𝑚𝑅2 𝜔2 − 𝐺.
2𝑅(S2.31)

From the equation of motion this is rewritten as
1 𝑚𝑀
𝐸=− 𝐺.
2𝑅(S2.32)

Therefore, when 𝐸 decreases due to gravitational wave emission, 𝑅 should decrease. Then, 𝜔 shouldincrease. Since the amplitude is proportional to 𝑅2 𝜔2 ∝ 𝑅−1 , it should increase. In summary, both thefrequency and the amplitude should increase as time goes on, as illustrated in (b).
C.6

0.2 pt

The most appropriate profile is (b).

5