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IPhO 2023, Теория — Задача 2. Neutron Stars

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Theory — IPhO 2023 Tokyo Japan — Q2 — English (Official) — Neutron Stars (10 points) · 10 т.

Условие

We discuss the stability of large nuclei and estimate the mass of neutron stars theoretically and experimentally.

Part A. Mass and stability of nuclei (2.5 points)

The rest-energy of a nucleus $m(Z,N)c^2$ consisting of $Z$ protons and $N$ neutrons is smaller than the sum of rest-energies of protons and neutrons, hereafter called nucleons, by the binding energy $B(Z,N)$ , where $c$ is the speed of light in vacuum. Ignoring minor corrections, we can approximate the binding energy consisting of the volume term with $a_V$ , the surface term with $a_S$ , the Coulomb energy term with $a_C$ , and the symmetry energy term with $a_{\mathrm{sym}}$ in the following way.

m(Z,N)c^2 = Am_N c^2 - B(Z,N), \qquad B(Z,N) = a_V A - a_S A^{2/3} - a_C \frac{Z^2}{A^{1/3}} - a_{\mathrm{sym}} \frac{(N-Z)^2}{A}, \qquad (1)

where $A = Z + N$ is the mass number and $m_N$ is the nucleon mass. In the calculation, use $a_V \approx 15.8\ \mathrm{MeV}$ , $a_S \approx 17.8\ \mathrm{MeV}$ , $a_C \approx 0.711\ \mathrm{MeV}$ , and $a_{\mathrm{sym}} \approx 23.7\ \mathrm{MeV}$ ( $\mathrm{MeV} = 10^6$ electron volts).

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

For large nuclei with a large enough mass number $A > A_c$ with a threshold $A_c$ , these nuclei stay stable against nuclear fission because of the sufficiently large binding energy due to gravity.

Part C. Neutron star in a binary system (6.0 points)

Some neutron stars are pulsars regularly emitting electromagnetic waves, which we call "light" for simplicity here, at a constant period. Neutron stars often make binary systems with a White Dwarf. Let us consider the star configuration shown in Fig. 1, where a light pulse from a neutron star N to the Earth E passes near a White Dwarf W of the binary system. Measuring these pulses influenced by the star's gravity leads to an accurate estimation of the mass of W as explained below, resulting in the estimation of the mass of N.

Two configurations of a neutron star N, a White Dwarf W and the Earth E on the x-axis. In (a) N is at x_N < 0 (left of x=0) and E at x_E > 0, with W a perpendicular distance d below the point x=0. In (b) x=0 is at the left, W is a distance d below it, N is at x_N > 0 and E at x_E > 0. — Fig. 1: Configurations with the $x$ -axis along the line connecting N and E. (a) for $x_N < 0$ and (b) for $x_N > 0$ .

A.1 Under the condition of $Z = N$ , determine $A$ for maximizing the binding energy per nucleon, $B/A$ . [0,9 т.]

A.2 Under the condition of fixed $A$ , the atomic number of the most stable nucleus $Z^*$ is determined by maximizing $B(Z, A-Z)$ . For $A = 197$ , calculate $Z^*$ using Eq. (1). [0,9 т.]

A.3 A nucleus having large $A$ breaks up into lighter nuclei through fission in order to minimize the total rest-mass energy. For simplicity, we consider one of multiple ways to break a nucleus with $(Z, N)$ into two equal nuclei, which occurs when the following energy relation holds,

m(Z,N)c^2 > 2m(Z/2, N/2)c^2.

When this relation is written as

Z^2/A > C_{\mathrm{fission}} \frac{a_S}{a_C},

obtain $C_{\mathrm{fission}}$ up to two significant digits. [0,7 т.]

B.1 We assume that $N = A$ and $Z = 0$ is realized for sufficiently large $A$ and Eq. (1) continues to hold with the addition of the gravitational binding energy. The binding energy due to gravity is

B_{\mathrm{grav}} = \frac{3}{5} \frac{GM^2}{R},

where $M = m_N A$ and $R = R_0 A^{1/3}$ with $R_0 \simeq 1.1 \times 10^{-15}\ \mathrm{m} = 1.1\ \mathrm{fm}$ are the mass and the radius of the nucleus, respectively.
For $B_{\mathrm{grav}} = a_{\mathrm{grav}} A^{5/3}$ , obtain $a_{\mathrm{grav}}$ in the MeV unit up to the first significant digit. Then, ignoring the surface term, estimate $A_c$ up to the first significant digit. In the calculation, use $m_N c^2 \simeq 939\ \mathrm{MeV}$ and $G = \hbar c / M_P^2$ where $M_P c^2 \simeq 1.22 \times 10^{22}\ \mathrm{MeV}$ and $\hbar c \simeq 197\ \mathrm{MeV} \cdot \mathrm{fm}$ . [1,5 т.]

C.1 As shown in the figure below, under the constant gravitational acceleration $g$ we place two levels I and II with the height difference $\Delta h (> 0)$ . Set the identical clocks at I, II, and $F$ , the free-falling system, denoted by clock-I, clock-II, and clock-F, respectively.

Levels I and II separated by height difference Δh; a ball F falls freely under gravity g from level I toward level II with velocity v; the ground is hatched below level II. — Set-up of the thought experiment.

We assume that an observer sits with clock-F, and initially $F$ is placed at the same height as that of clock-I and its velocity is zero. Since the clocks are identical, they register equal time intervals, $\Delta \tau_F = \Delta \tau_{\mathrm{I}}$ . Then, we let $F$ fall freely, and work in the frame of $F$ , which is considered to be inertial. In this frame, clock-II passes by clock-F with velocity $v$ , so that the time dilation of clock-II can be determined by the Lorentz-transformation. When time $\Delta \tau_{\mathrm{I}}$ elapses on clock-F, time $\Delta \tau_{\mathrm{II}}$ elapses on clock-II.
Determine $\Delta \tau_{\mathrm{II}}$ in terms of $\Delta \tau_{\mathrm{I}}$ up to the first order in $\Delta \phi / c^2$ , where $\Delta \phi = g \Delta h$ is a difference of the gravitational potential, i.e., the gravitational potential energy per unit mass. [1 т.]

C.2 Under the gravitational potential $\phi$ , time delays change the effective speed of light, $c_{\mathrm{eff}}$ , observed at the infinity, though the local speed of light is $c$ . When $\phi(r=\infty) = 0$ , $c_{\mathrm{eff}}$ can be given up to the first order in $\phi/c^2$ as

c_{\mathrm{eff}} \approx \left(1 + \frac{2\phi}{c^2}\right) c

including the effect of space distortion, which was not featured in C.1. We note that the light path can be approximated as a straight line.
As shown in Fig. 1 (a), we take the $x$ -axis along the light path from the neutron star N to the Earth E and place $x = 0$ at the point where the White Dwarf W is the closest to the light path. Let $x_N\ (< 0)$ be the $x$ -coordinate of N, $x_E\ (> 0)$ be that of E, and $d$ be the distance between W and the light path.
Estimate the changes of the arrival time $\Delta t$ of the light from N to E caused by the White Dwarf with mass $M_{\mathrm{WD}}$ and evaluate the answer in a simple form disregarding higher order terms of the following small quantities: $d/|x_N| \ll 1$ , $d/x_E \ll 1$ , and $GM_{\mathrm{WD}}/(c^2 d) \ll 1$ . If necessary, use the following formula.

\int \frac{dx}{\sqrt{x^2+d^2}} = \frac{1}{2} \log \left( \frac{\sqrt{x^2+d^2}+x}{\sqrt{x^2+d^2}-x} \right) + C. \qquad (\log\ \mathrm{is\ the\ natural\ logarithm})

[1,8 т.]

C.3 As shown below, in a binary star system N and W are assumed to be moving in circular orbits with zero eccentricity around the center of mass $G$ on the orbit plane. Let $\varepsilon$ be the orbital inclination angle measured from the orbit plane to the line directed toward E from $G$ , and let $L$ be the length between N and W and $M_{\mathrm{WD}}$ be the mass of the White Dwarf. In the following, we assume $\varepsilon \ll 1$ .

Two circular orbits around the center of mass G; the White Dwarf W on the larger orbit with orbital phase φ and the neutron star N on the smaller orbit; the inclination angle ε is measured from the orbit plane to the line toward E. — Binary star system.

We observe light pulses from N on E far away from N. The light path to E varies with time depending on the configuration of N and W. The delay in the time interval of arriving pulses on E has the maximum value $\Delta t_{\mathrm{max}}$ for $x_N \simeq -L$ and the minimum value $\Delta t_{\mathrm{min}}$ for $x_N \simeq L$ (see Fig. 1 (b) for the configuration). Calculate $\Delta t_{\mathrm{max}} - \Delta t_{\mathrm{min}}$ in a simple form disregarding higher order terms of small quantities as done in C.2. We note that the delays due to gravity from stellar objects other than W are assumed to cancel out in $\Delta t_{\mathrm{max}} - \Delta t_{\mathrm{min}}$ . [1,8 т.]

C.4 The below figure shows the observed time delays as a function of the orbital phase $\varphi$ for the binary star system with $L \approx 6 \times 10^6\ \mathrm{km}$ and $\cos \varepsilon \approx 0.99989$ . Estimate $M_{\mathrm{WD}}$ in terms of the solar mass $M_{\odot}$ and show the results for $M_{\mathrm{WD}}/M_{\odot}$ up to the first significant digit. Here the approximate relation, $GM_{\odot}/c^3 \approx 5\ \mu\mathrm{s}$ , can be used. [0,8 т.]

Graph of Δt [μs] versus orbital phase φ; vertical axis ticks -20, -10, 0, 10, 20; a sharp positive peak reaching about 21 μs and a broad negative dip reaching about -23 μs; the horizontal zero line is dotted. — Observed time delays $\Delta t$ as a function of the orbital phase $\varphi$ (see the figure in **C.3**) to locate N and W on the orbits.

C.5 In the binary system of neutron stars, two stars release energy and angular momentum by emitting gravitational waves and eventually collide to merge. For simplicity, let us consider only a circular motion with the radius $R$ and the angular velocity $\omega$ and then $\omega = \chi R^p$ holds with the constant $\chi$ depending on neither $\omega$ nor $R$ if relativistic effects are ignored. Determine the value for $p$ . [0,4 т.]

C.6 The amplitude of the emitted gravitational wave from the binary system in C.5 is proportional to $R^2 \omega^2$ . Figure below qualitatively shows four different temporal profiles of the observed gravitational waves before the two-star collision. Select the most appropriate profile from (a) to (d). [0,2 т.]

Four qualitative wave profiles labelled (a)–(d) plotted against t: (a) decreasing amplitude with increasing frequency, (b) increasing amplitude with increasing frequency, (c) increasing amplitude with decreasing frequency, (d) decreasing amplitude with decreasing frequency. — Observed data profiles of gravitational waves.

Решение

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 physical interpretation 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.,

\frac{B}{A} = a_V - a_S A^{-1/3} - \frac{a_C}{4}A^{2/3}. \qquad (S2.1)

As long as $A$ is small, the second term is dominantly increasing with increasing $A$ , and it is eventually taken over by the third term which is decreasing. Therefore, the extremal corresponds to the maximum of $B/A$ . One can explicitly carry out

\frac{d(B/A)}{dA} = 0 \qquad (S2.2)

to find the following condition,

\frac{a_S}{3}A^{-4/3} - \frac{a_C}{6}A^{-1/3} = 0. \qquad (S2.3)

The solution is

A = \frac{2a_S}{a_C}. \qquad (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 $^{56}\mathrm{Fe}$ to $^{62}\mathrm{Ni}$ . The discrepancy from the answer in this problem is understood by the approximation of dropping the pairing energy and disregarding a mass difference between the proton and the neutron.

A.1 — 0.9 pt: $A = 50$

A.2 Take the differentiation of $B(Z, A-Z)/A$ with respect to $Z$ for a fixed $A$ , which leads to

-2a_C\frac{Z^*}{A^{1/3}} - 4a_{\mathrm{sym}}\frac{2Z^* - A}{A} = 0. \qquad (S2.5)

By solving this in terms of $Z^*$ , one finds

Z^* = \frac{1}{1 + \dfrac{a_C}{4a_{\mathrm{sym}}}A^{2/3}} \cdot \frac{A}{2}. \qquad (S2.6)

From this expression one can understand that $Z^* \simeq N$ as long as $A$ is small enough, while $Z^*$ becomes far smaller 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 $Z \to 0$ , and the balance between these competing effects determines $Z^*$ . Nuclei with too many neutrons (protons) would go through the $\beta^-$ decay (the $\beta^+$ decay or the electron capture) toward the stable $(Z, N)$ .

A.2 — 0.9 pt: $Z^* = 79$

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 $a_S$ and $a_C$ which are related as

a_S\left[A^{2/3} - 2\left(\frac{A}{2}\right)^{2/3}\right] + a_C\left[\frac{Z^2}{A^{1/3}} - 2\frac{(Z/2)^2}{(A/2)^{1/3}}\right] > 0. \qquad (S2.7)

As guided in the problem, the above inequality becomes as simple as

\frac{Z^2}{A} > \frac{2^{1/3} - 1}{1 - 2^{-2/3}} \cdot \frac{a_S}{a_C}. \qquad (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 of $a_S$ and $a_C$ , one finds that such a symmetric fission process is possible for $Z^2/A \gtrsim 18$ . For example, lead (Pb) with $Z = 82$ and $A = 208$ is a stable element among several isotopes. Now, one can compute $82^2/208 \approx 32$ , 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 should change 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.3 — 0.7 pt: $C_{\mathrm{fission}} = 7.0 \times 10^{-1}$

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 as

a_{\mathrm{grav}} = \frac{3}{5}\frac{Gm_N^2}{R_0}, \qquad (S2.9)

which is re-expressed in terms of $M_P$ using the given relation to $G$ , leading to

a_{\mathrm{grav}} = \frac{3}{5}\frac{\hbar c\, m_N^2}{R_0 M_P^2} = \frac{3}{5}\cdot\frac{197\ \mathrm{fm} \cdot \mathrm{MeV} \times (939\ \mathrm{MeV}/c^2)^2}{1.1\ \mathrm{fm} \times (1.22 \times 10^{22}\ \mathrm{MeV}/c^2)^2} \simeq 6.4 \times 10^{-37}\ \mathrm{MeV}. \qquad (S2.10)

Here, $M_P$ is a quantity often called the Planck mass. The gravitational effect is extremely tiny as compared to the typical scale in nuclear physics and this scale difference is manifest for this expression of $G$ with $M_P$ in the MeV unit.

The stability is judged from the condition that the binding energy should be positive, i.e.,

B_{\mathrm{total}}(A) = a_V A - a_{\mathrm{sym}}A + a_{\mathrm{grav}}A^{5/3} > 0. \qquad (S2.11)

This inequality can be translated into $A > A_c$ with $A_c$ given by

A_c = \left(\frac{a_{\mathrm{sym}} - a_V}{a_{\mathrm{grav}}}\right)^{3/2} \simeq 4.4 \times 10^{55}. \qquad (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, $B_{\mathrm{total}}(A) > B_{\mathrm{total}}(A-1)$ or approximately $dB_{\mathrm{total}}(A)/dA > 0$ , which is satisfied in a window with $B_{\mathrm{total}} < 0$ . This condition, $dB_{\mathrm{total}}/dA = 0$ , results in smaller $A_c$ but it is nontrivial whether such an unstable initial state could be prepared in the nature. The neutron star is born in the Type-II (core-collapse) supernovae, and a baby star called the proto-neutron star is an energetic state at high temperature. Neutrinos bring heat out from the proto-neutron star within the time scale of $\mathcal{O}(10)$ seconds. What is the possible smallest mass of the neutron star? This is 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 $M_\odot$ theoretically, the simulation and the observation favor the mass $\gtrsim 1.4M_\odot$ .

B.1 — 1.5 pt: $a_{\mathrm{grav}} = 6 \times 10^{-37}\ \mathrm{MeV}$ ; $A_c = 4 \times 10^{55}$

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 $\frac{1}{2}mv^2$ from the potential energy $mg\Delta h$ , and the velocity takes

v^2 = 2g\Delta h = 2\Delta\phi. \qquad (S2.13)

The time delay can be derived from the standard arguments. Suppose that Clock-II passes two infinitesimally separate points, $z$ and $z+\Delta z$ , in F at time $t$ and $t+\Delta\tau_{\mathrm{II}}$ , then the time interval registered by Clock-II is

\Delta\tau_{\mathrm{II}} = \frac{\gamma}{c}(c\Delta\tau_{\mathrm{F}} - \beta\Delta z), \qquad (S2.14)

where the Lorentz transformation is used $^{1}$ with $\beta = v/c$ and $\gamma = 1/\sqrt{1-\beta^2}$ . Because $\Delta z/\Delta\tau_{\mathrm{F}} = v$ and $\Delta\tau_{\mathrm{F}} = \Delta\tau_{\mathrm{I}}$ , the above expression is written as

\Delta\tau_{\mathrm{II}} = \gamma(1-\beta^2)\Delta\tau_{\mathrm{F}} = \sqrt{1-\beta^2}\,\Delta\tau_{\mathrm{I}}. \qquad (S2.15)

Using the expression of $v^2$ , one finally arrives at

\Delta\tau_{\mathrm{II}} = \sqrt{1 - 2\frac{\Delta\phi}{c^2}}\,\Delta\tau_{\mathrm{I}} \simeq \left(1 - \frac{\Delta\phi}{c^2}\right)\Delta\tau_{\mathrm{I}}. \qquad (S2.16)

C.1 — 1.0 pt: $\Delta\tau_{\mathrm{II}} = \left(1 - \frac{\Delta\phi}{c^2}\right)\Delta\tau_{\mathrm{I}}$

$^{1}$ 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 to Clock-II, the Lorentz transformation is validated for Clock-II' seen from Clock-F.

C.2 In terms of the effective speed of light, the total time necessary for the light propagation from N to E is

t_{\mathrm{E\text{-}N}} = \int_{x_N}^{x_E}\frac{dx}{c_{\mathrm{eff}}(x)}. \qquad (S2.17)

The denominator is expanded in terms of the gravitational potential and the leading-order correction is found to be

t_{\mathrm{E\text{-}N}} \simeq \frac{1}{c}\int_{x_N}^{x_E}dx\left(1 + \frac{2GM_{\mathrm{WD}}}{c^2\sqrt{x^2+d^2}}\right) = \frac{x_E - x_N}{c} + \Delta t, \qquad (S2.18)

where the time delay $\Delta t$ is identified as

\Delta t = \frac{2GM_{\mathrm{WD}}}{c^3}\int_{x_N}^{x_E}\frac{dx}{\sqrt{x^2+d^2}} = \frac{GM_{\mathrm{WD}}}{c^3}\log\left(\frac{\sqrt{x^2+d^2}+x}{\sqrt{x^2+d^2}-x}\right)\Bigg|_{x=x_N}^{x=x_E}. \qquad (S2.19)

Inside the logarithm, the following approximations are made:

\sqrt{x_N^2+d^2}+x_N \simeq \frac{d^2}{2|x_N|}, \qquad \sqrt{x_N^2+d^2}-x_N \simeq 2|x_N|, \qquad (S2.20)

and

\sqrt{x_E^2+d^2}-x_E \simeq \frac{d^2}{2x_E}, \qquad \sqrt{x_E^2+d^2}+x_E \simeq 2x_E. \qquad (S2.21)

Then, the simple form of approximated $\Delta t$ is

\Delta t \simeq \frac{GM_{\mathrm{WD}}}{c^3}\log\left(\frac{2x_E \cdot 2|x_N|}{d^2/(2x_E) \cdot d^2/(2|x_N|)}\right) = \frac{2GM_{\mathrm{WD}}}{c^3}\log\left(\frac{4|x_N|x_E}{d^2}\right). \qquad (S2.22)

C.2 — 1.8 pt: $\Delta t = \frac{2GM_{\mathrm{WD}}}{c^3}\log\left(\frac{4|x_N|x_E}{d^2}\right)$

C.3 Because $|x_N| = L\cos\varepsilon \simeq L$ and $d = L\sin\varepsilon \simeq L\varepsilon$ for $\Delta t_{\max}$ , the answer of C.2 gives

\Delta t_{\max} = \frac{2GM_{\mathrm{WD}}}{c^3}\log(4x_E/L\varepsilon^2) \qquad (S2.23)

For $\Delta t_{\min}$ the sign of $x_N$ is changed. Although the expression of $\Delta t$ is intact, the approximation takes a different form as

\sqrt{x_N^2+d^2}+x_N \simeq 2x_N, \qquad \sqrt{x_N^2+d^2}-x_N \simeq \frac{d^2}{2x_N}. \qquad (S2.24)

Then, the approximated form of $\Delta t_{\min}$ is

\Delta t_{\min} \simeq \frac{GM_{\mathrm{WD}}}{c^3}\log\left(\frac{2x_E \cdot d^2/(2x_N)}{d^2/(2x_E) \cdot 2x_N}\right) = \frac{2GM_{\mathrm{WD}}}{c^3}\log(x_E/L), \qquad (S2.25)

where $x_N \simeq L$ is used in the last expression. In the difference, $\Delta t_{\max} - \Delta t_{\min}$ , one sees that $L$ and $x_E$ disappear.

C.3 — 1.8 pt: $\Delta t_{\max} - \Delta t_{\min} = \frac{2GM_{\mathrm{WD}}}{c^3}\log(4/\varepsilon^2)$

C.4 Using the expansion, $\cos\varepsilon \simeq 1 - \frac{1}{2}\varepsilon^2$ , one can evaluate

\varepsilon^2 \simeq 2 \times (1 - \cos\varepsilon) = 0.00022. \qquad (S2.26)

From the graph the difference in time delays is roughly read out as

\Delta t_{\max} - \Delta t_{\min} \approx 50\,\mu\mathrm{s} \qquad (S2.27)

From these numerical values, $M_{\mathrm{WD}}$ is solved as

M_{\mathrm{WD}} = M_\odot\left(\frac{2GM_\odot}{c^3}\right)^{-1}\frac{\Delta t_{\max} - \Delta t_{\min}}{\log(4/\varepsilon^2)} \simeq \frac{50\,\mu\mathrm{s}}{10\,\mu\mathrm{s}\,\log(4/0.00022)}M_\odot \simeq 0.5\ M_\odot. \qquad (S2.28)

▷ Note: The data in this problem roughly correspond to PSR J1614-2230 [see P.B. Demorest et al., Nature 467, 1080-1083 (2010)]. From the Shapiro delay measurement, the White Dwarf mass was estimated as $0.500 \pm 0.006\,M_\odot$ . With the Keplerian orbital parameters in the binary system, the neutron star mass was considered to be $1.97 \pm 0.04\,M_\odot$ , 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: $M_{\mathrm{WD}}/M_\odot = 0.5$

C.5 For the circular orbit with the radius $R$ , the equation of motion is

mR\omega^2 = G\frac{mM}{R^2}, \qquad (S2.29)

if $M$ is sufficiently large. From this it is easy to see

R^3\omega^2 = GM = (\mathrm{const.}) \qquad (S2.30)

This is nothing but Kepler's third law and the relation holds for more general elliptical orbits around the center of mass.

C.5 — 0.4 pt: $p = -\frac{3}{2}$

C.6 The sum of the kinetic energy and the potential energy is

E = \frac{1}{2}mR^2\omega^2 - G\frac{mM}{R}. \qquad (S2.31)

From the equation of motion this is rewritten as

E = -\frac{1}{2}G\frac{mM}{R}. \qquad (S2.32)

Therefore, when $E$ decreases due to gravitational wave emission, $R$ should decrease. Then, $\omega$ should increase. Since the amplitude is proportional to $R^2\omega^2 \propto R^{-1}$ , it should increase. In summary, both the frequency and the amplitude should increase as time goes on, as illustrated in (b).

C.6 — 0.2 pt: The most appropriate profile is (b).

Оригинал в Архива: IPhO_2023_Q2.pdf · официални решения: IPhO_2023_S2.pdf