Съдържание

Fundamental constants Astronomical data Approximations Gauss’s formulae The Sun The Earth and Moon Условие Решение

IOAA 2023, Теория — Задача 13. LISA

Оригинал на условието (PDF)Съобщи за грешка

Всички материали от IOAA 2023

IOAA 2023 — Theoretical competition · 45 т.

Указания

Do not touch envelopes until the start of the examination.
The theoretical examination lasts for 5 hours and is worth a total of 250 marks.
There are Answer Sheets for carrying out detailed work and Working Sheets for rough work, which are already marked with your student code and question number.
Use only the answer sheets for a particular question for your answer. Please write only on the printed side of the sheet. Do not use the reverse side. If you have written something on any sheet which you do not want to be evaluated, cross it out.
Use as many mathematical expressions as you think may help the evaluator to better understand your solutions. The evaluator may not understand your language. If it is necessary to explain something in words, please use short phrases (if possible in English).
You are not allowed to leave your work desk without permission. If you need any assistance (malfunctioning calculator, need to visit a restroom, etc.), please draw the attention of the supervisor.
The beginning and end of the examination will be indicated by the supervisor. The remaining time will be displayed on a clock.
At the end of the examination you must stop writing immediately. Put everything back in the envelope and leave it on the table.
Once all envelopes are collected, your student guide will escort you out of the examination room.
A list of constants and useful relations are included in the envelope.

Table of Constants

Fundamental constants

Constant	Symbol	Value
Speed of light in vacuum	$c$	$2.998\times10^8\,\mathrm{m\,s^{-1}}$
Planck constant	$h$	$6.626\times10^{-34}\,\mathrm{J\,s}$
Boltzmann constant	$k_B$	$1.381\times10^{-23}\,\mathrm{J\,K^{-1}}$
Stefan-Boltzmann constant	$\sigma$	$5.670\times10^{-8}\,\mathrm{W\,m^{-2}\,K^{-4}}$
Elementary charge	$e$	$1.602\times10^{-19}\,\mathrm C$
Universal gravitational constant	$G$	$6.674\times10^{-11}\,\mathrm{N\,m^2\,kg^{-2}}$
Universal electric constant	$\epsilon_0$	$8.854\times10^{-12}\,\mathrm{m^{-3}\,kg^{-1}\,s^4\,A^2}$
Universal gas constant	$R$	$8.315\,\mathrm{J\,mol^{-1}\,K^{-1}}$
Avogadro constant	$N_A$	$6.022\times10^{23}\,\mathrm{mol^{-1}}$
Wien’s displacement constant	$b=\lambda_mT$	$2.898\times10^{-3}\,\mathrm{m\,K}$
Mass of electron	$m_e$	$9.109\times10^{-31}\,\mathrm{kg}$
Mass of proton	$m_p$	$1.673\times10^{-27}\,\mathrm{kg}$
Mass of neutron	$m_n$	$1.675\times10^{-27}\,\mathrm{kg}$
Mass of Helium nucleus	$m_{\mathrm{He}}$	$6.645\times10^{-27}\,\mathrm{kg}$
Atomic mass unit (a.m.u., Dalton)		$1.661\times10^{-27}\,\mathrm{kg}$

Astronomical data

Data	Symbol	Value
Hubble constant	$H_0$	$70\,\mathrm{km\,s^{-1}\,Mpc^{-1}}$
North Ecliptic Pole ( $J2000.0$ )	$(\alpha_E,\delta_E)$	$(18\mathrm h\,00\mathrm m\,00\mathrm s,+66^\circ33'39'')$
North Galactic Pole ( $J2000.0$ )	$(\alpha_G,\delta_G)$	$(12\mathrm h\,51\mathrm m\,26\mathrm s,+27^\circ07'42'')$
1 jansky	$1\,\mathrm{Jy}$	$10^{-26}\,\mathrm{W\,m^{-2}\,Hz^{-1}}$
1 parsec	$1\,\mathrm{pc}$	$3.086\times10^{16}\,\mathrm m=206\,265\,\mathrm{au}=3.262\,\mathrm{ly}$
1 astronomical unit (au)	$1\,\mathrm{au}$	$1.496\times10^{11}\,\mathrm m$
1 sidereal day	$T_{\mathrm{SD}}$	$23.93444\,\mathrm h=23\mathrm h\,56\mathrm m\,04\mathrm s$
1 tropical year		$365.2422$ solar days
1 sidereal year		$365.2564$ solar days

Approximations and Gauss’s formulae

Approximations

(1+x)^n\approx1+nx

(1+x)(1+y)\approx1+x+y\quad\text{if }x\ll1\text{ and }y\ll1

Gauss’s formulae

Spherical law of cosines: $\cos a=\cos b\cos c+\sin b\sin c\cos A$

Spherical law of sines: $\dfrac{\sin A}{\sin a}=\dfrac{\sin B}{\sin b}=\dfrac{\sin C}{\sin c}$ .

The Sun, Earth and Moon

The Sun

Data	Symbol	Value
Solar luminosity	$L_\odot$	$3.826\times10^{26}\,\mathrm W$
Apparent angular diameter of Sun	$\theta_\odot$	$32'$
Effective temperature of Sun	$T_{\mathrm{eff},\odot}$	$5778\,\mathrm K$
Apparent visual magnitude		$-26.75$
Absolute visual magnitude		$+4.82$
Apparent bolometric magnitude		$-26.83$
Absolute bolometric magnitude		$+4.74$
Distance of the Sun from the Galactic centre		$\approx8\,\mathrm{kpc}$

The Earth and Moon

Data	Symbol	Value
Obliquity of the ecliptic (Earth)	$\epsilon$	$23.5'$
Platonic year (period of precession of Earth’s axis)		$25\,765$ years
Apparent visual magnitude of full Moon		$-12.74$
Apparent angular diameter of Moon	$\theta_L$	$31'$
Inclination of the lunar orbit to the ecliptic		$05^\circ08'43''$
Inclination of the lunar equator to its orbital plane		$6.687^\circ$
Lunar sidereal month	$T_{\mathrm{SL}}$	$27.321661\,\mathrm d=655.71986\,\mathrm h$
Synodic month		$29.530589\,\mathrm d$
Tropical month		$27.321582\,\mathrm d$
Anomalistic month (the average time between two successive Moon’s perigees)		$27.554550\,\mathrm d$
Draconic month (time between two successive transits of the Moon through the same node)		$27.212221\,\mathrm d$

The Solar System

Object	Mean radius [km]	Mass [kg]	Semimajor axis [au]	Eccentricity
Sun	$695\,700$	$1.988\times10^{30}$	—	—
Mercury	$2\,440$	$3.301\times10^{23}$	$0.387$	$0.206$
Venus	$6\,052$	$4.867\times10^{24}$	$0.723$	$0.007$
Earth	$6\,378$	$5.972\times10^{24}$	$1.000$	$0.016\,710$
Moon	$1\,737$	$7.346\times10^{22}$	$3.844\times10^5\,\mathrm{km}$	$0.054\,900$ (range $0.026$ – $0.077$ )
Mars	$3\,390$	$6.417\times10^{23}$	$1.524$	$0.093$
Jupiter	$69\,911$	$1.898\times10^{27}$	$5.203$	$0.048$
Saturn	$58\,232$	$5.683\times10^{26}$	$9.537$	$0.054$
Uranus	$25\,362$	$8.681\times10^{25}$	$19.189$	$0.047$
Neptune	$24\,622$	$1.024\times10^{26}$	$30.070$	$0.009$

General Marking Scheme

Case	Marking
Using incorrect physical concept (despite correct answers)	No points given
Giving correct answer without detailed calculation	Deduct 50% of the marks for that part
Minor mistakes in the calculations, e.g., wrong signs, symbols, substitutions	Deduct 20% of the marks for that part
Units missing from final answers	Deduct 0.5 pts
Too few or too many significant figures in the final answer	Deduct 0.5 pts
Error resulting from another error in an earlier part for which the student already lost marks, if the answer is physically reasonable.	Full points (i.e., no deductions)
Error resulting from another error in an earlier part, where the student should have realised the answer was physically unreasonable.	Deduct 20% of the marks for that part

For example, if due to an error in an earlier part, the student calculates the mass of a star as $2.5\times10^{30}\,\mathrm{kg}$ instead of $2\times10^{30}\,\mathrm{kg}$ , they will only lose marks for the earlier part. However, if, for the same reason, a student calculates the mass as $2\times10^{25}\,\mathrm{kg}$ , they should realize this is wrong (a few times the Earth’s mass) and thus should lose some marks for this part as well.

Условие

The Laser Interferometer Space Antenna (LISA) is a proposed experiment to detect low-frequency gravitational waves. It consists of three spacecraft arranged in an equilateral triangle. A passing gravitational wave changes the distance between the spacecraft, which can be precisely measured (more details are given in the notes below).

One of the sources of low-frequency gravitational waves are compact binary star systems, for example binary white dwarfs. Such a system was recently discovered at a distance of $2.34\,\mathrm{kpc}$ from the Sun. The orbital period of the binary was found to be $414.79\,\mathrm{s}$ and is changing at a rate of $-7.49\times 10^{-4}\,\mathrm{s\,yr^{-1}}$ due to the emission of gravitational waves.

a. Check if this binary system can be detected by LISA. (25 points)

b. Calculate the chirp mass. (5 points)

c. Determine the masses of both components knowing that the ratio between the radius of one of the components to the semi-major axis of the orbit is $0.139$ , and assuming both components follow the mass-radius relation for white dwarfs given in the table below. (15 points)

Notes:

A binary star system with an orbital period $P$ emits gravitational waves with a frequency of $f = 2/P$ .
LISA measures a dimensionless quantity called the characteristic strain amplitude, $S$ , given by
$S = h\sqrt{fT_{\mathrm{obs}}},$
where $T_{\mathrm{obs}} = 4\,\mathrm{yr}$ is the expected duration of the mission. $h$ is the gravitational wave strain, given by:
$h = \frac{2(G\mathcal{M})^{5/3}(\pi f)^{2/3}}{c^4 D},$
where $\mathcal{M}$ is the so-called chirp mass, $f$ is the frequency of the gravitational wave and $D$ is the distance to the system. If we denote the masses of the components of the binary as $M_1$ and $M_2$ , then the chirp mass is given by:
$\mathcal{M} = \frac{(M_1 M_2)^{3/5}}{(M_1+M_2)^{1/5}}.$
The expected sensitivity of LISA as a function of a gravitational wave frequency is presented on the figure below.
The semi-major axis $a$ of the binary system changes due to the emission of gravitational waves at a rate:
$\frac{\Delta a}{\Delta t} = -\frac{64}{5}\frac{G^3}{c^5}\frac{M_1 M_2(M_1+M_2)}{a^3}.$

$M\,(M_\odot)$	$R\,(R_\odot)$
0.48	0.0144
0.50	0.0147
0.52	0.0150
0.54	0.0153
0.56	0.0156
0.58	0.0159
0.60	0.0162
0.62	0.0165
0.64	0.0168

Mass-radius relation for white dwarfs based on theoretical models of Althaus et al. (2013) for white dwarfs of $\log g = 7.7$ .

Complete original LISA sensitivity curve, logarithmic frequency/strain axes and every numeric tick.

The expected sensitivity of LISA as a function of gravitational wave frequency.

Решение

$M$ ( $M_\odot$ )	$R$ ( $R_\odot$ )
0.48	0.0144
0.50	0.0147
0.52	0.0150
0.54	0.0153
0.56	0.0156
0.58	0.0159
0.60	0.0162
0.62	0.0165
0.64	0.0168

Mass–radius relation for white dwarfs based on theoretical models of Althaus et al. (2013) for white dwarfs of $\log_g=7.7$ .

Complete original key LISA sensitivity curve: all logarithmic axes, frequency/strain labels and numeric ticks.

The expected sensitivity of LISA as a function of gravitational wave frequency.

Solution

Part (a)

To determine whether the system can be detected by LISA, we need to determine two quantities: the gravitational-wave frequency and the characteristic strain amplitude.

It is straightforward to calculate the gravitational-wave frequency:

f=\frac{2}{P}=\frac{2}{414.79}=4.8\times10^{-3}\,\mathrm{Hz}.

This frequency is within the LISA band and close to the maximum LISA sensitivity.

Marking	Points
Calculating gravitational wave frequency	2 points

To estimate the characteristic strain amplitude, we need to know the chirp mass $\mathcal M$ . The other required quantities (such as the gravitational wave frequency, distance, and duration of the LISA observations) are already known.

The orbital period change rate $\Delta P/\Delta t$ is given in the problem. We need to link it to $\Delta a/\Delta t$ which we know from Note 3 is linked to the mass function. From Kepler’s third law, we know that:

\frac{a^3}{P^2}=\frac{G(M_1+M_2)}{4\pi^2},

where $M_1$ and $M_2$ are masses of both components of the system. If, due to the emission of gravitational waves, the semi-major axis changes from $a$ to $a+\Delta a$ , then the orbital period changes from $P$ to $P+\Delta P$ , as the masses of both components are constant. Then:

\frac{a^3}{P^2}=\frac{(a+\Delta a)^3}{(P+\Delta P)^2}=\frac{a^3(1+\Delta a/a)^3}{P^2(1+\Delta P/P)^2}=\frac{a^3}{P^2}\left(1+3\frac{\Delta a}{a}-2\frac{\Delta P}{P}\right),

hence $\Delta P/P=\frac32\Delta a/a$ .

Here, we used the fact that $(1+x)^n\approx1+nx$ for $x\ll1$ .

Therefore:

\begin{aligned} \frac{\Delta P}{\Delta t} &=\frac{3P}{2a}\frac{\Delta a}{\Delta t}\\ &=-\frac32\cdot\frac{64}{5}\cdot\frac{P}{a}\frac{G^3}{c^5}\cdot\frac{M_1M_2(M_1+M_2)}{a^3}\\ &=-\frac{96}{5}\frac{G^3}{c^5}\frac{P}{a}\cdot\frac{M_1M_2(M_1+M_2)}{G(M_1+M_2)P^2}\cdot4\pi^2\\ &=-\frac{96}{5}(2\pi)^2\frac{G^2}{c^5}\cdot\frac{M_1M_2}{aP}\\ &=-\frac{96}{5}(2\pi)^2\frac{G^2}{c^5}\cdot\frac{M_1M_2}{P}\cdot\frac{(2\pi)^{2/3}}{G^{1/3}(M_1+M_2)^{1/3}P^{2/3}}\\ &=-\frac{96}{5}(2\pi)^{8/3}\frac{G^{5/3}}{c^5}\cdot\frac{M_1M_2}{(M_1+M_2)^{1/3}}\cdot\frac1{P^{5/3}}\\ &=-\frac{96}{5}(2\pi)^{8/3}\frac{G^{5/3}}{c^5}\cdot\frac{M_1M_2}{(M_1+M_2)^{1/3}}\cdot\frac1{P^{5/3}}\\ &=-\frac{96}{5c^5}(2\pi)^{8/3}\left(\frac{G\mathcal M}{P}\right)^{5/3}\\ &=-\frac{192\pi}{5c^5}(G\mathcal M)^{5/3}\left(\frac{P}{2\pi}\right)^{-5/3}. \end{aligned}

Thus, by knowing the orbital period and its rate of change from observations, we can determine the chirp mass:

\mathcal M=\left(\frac5{192\pi}\right)^{3/5}\frac{c^3}{G}\frac{P}{2\pi}\left(-\frac{\Delta P}{\Delta t}\right)^{3/5}=0.319\,M_\odot,

and find the characteristic strain amplitude $h$ .

Alternatively, if we notice that the characteristic strain amplitude $h$ depends on $(G\mathcal M)^{5/3}$ which we can get directly from the rate of change of the period:

(G\mathcal M)^{5/3}=-\frac{\Delta P}{\Delta t}\left(\frac{P}{2\pi}\right)^{5/3}\frac{5c^5}{192\pi},

we can skip calculating the chirp mass and get the characteristic strain amplitude directly, which is all we need to check if the binary can be detected.

Either way, the gravitational wave strain is:

h=-\frac{5c}{192\pi^2}P\frac{\Delta P}{\Delta t}\frac1D.

If we plug in the numerical values, we get $h=1.1\times10^{-22}$ and $S=8.4\times10^{-20}$ . Checking the plot, this is above the expected sensitivity of LISA at $5\,\mathrm{mHz}$ . Thus, this object should be detected by LISA.

Marking	Points
Calculating the chirp mass or $(G\mathcal M)^{5/3}$ as a function of $P$ and $\Delta P/\Delta t$	15 points
Calculating $h$ and $S$	5 points
Correct conclusion – the system may be detected with LISA	3 points

Part (b)

To determine the masses of both components $M_1$ and $M_2$ we need two simultaneous $M_1$ – $M_2$ relations. The expression for the chirp mass:

\mathcal M=\frac{(M_1M_2)^{3/5}}{(M_1+M_2)^{1/5}},

gives us one relation, and we can calculate the chirp mass of the observed system from the rate of change of the period, if that was not already done in part (a):

\mathcal M=\left(\frac5{192\pi}\right)^{3/5}\frac{c^3}{G}\frac{P}{2\pi}\left(-\frac{\Delta P}{\Delta t}\right)^{3/5}=0.319\,M_\odot.

The second relation can be obtained from the mass–radius relation for white dwarfs given in the table and Kepler’s third law.

We are told that for one of the components (call it ‘1’),

a=\frac{R_1}{0.139}.

From Kepler’s third law we know that:

M_1+M_2=\left(\frac{a}{1\,\mathrm{au}}\right)^3\cdot\left(\frac{P}{1\,\mathrm{yr}}\right)^2,

which will give us the mass of the second component $M_2$ from the mass of the first, $M_1$ .

From here, it is not possible to derive an analytical formula for the masses of the components. Instead, we need to use numerical methods to estimate the result.

Taking the masses and radii listed in the given table as $M_1$ and $R_1$ , we can obtain $a$ and thus $M_1+M_2$ , $M_2$ and finally $\mathcal M$ for each mass:

$M_1$ ( $M_\odot$ )	$R$ ( $R_\odot$ )	$a$ ( $R_\odot$ )	$M_1+M_2$	$M_2$	$\mathcal M$
0.48	0.0144	0.104	0.647	0.167	0.240
0.50	0.0147	0.106	0.689	0.189	0.261
0.52	0.0150	0.108	0.732	0.212	0.283
0.54	0.0153	0.110	0.776	0.236	0.306
0.56	0.0156	0.112	0.823	0.263	0.329
0.58	0.0159	0.114	0.871	0.291	0.354
0.60	0.0162	0.117	0.922	0.322	0.379
0.62	0.0165	0.119	0.974	0.354	0.405
0.64	0.0168	0.121	1.028	0.388	0.431

The actual chirp mass is $\mathcal M=0.319\,M_\odot$ . Therefore, by linear interpolation or graphically, we estimate $M_1=0.55\,M_\odot$ and $M_2=0.25\,M_\odot$ .

(The student does not need to calculate $\mathcal M$ for all values of $M_1$ .)

Graphical solution:

Complete original chirp-mass versus component-mass graph: nine points, connecting curve, axis units and every tick.

Marking	Points
Calculating the chirp mass	5 points
Deriving two $M_1$ – $M_2$ relations	5 points
Determining the masses of both components	10 points

Original PDF comments

The original solution PDF contains two comments attributed to IOAA on this page:

“In previuos erratic version it was "... - 1".”

“In previous erratic version it was L_2/L_1 = ...”

Бележки за изданието

The English question packet includes a Ukrainian answer form for problem 9. The constants table prints Earth’s obliquity as 23.5′ and the Moon’s semimajor axis in km under an au heading. In the eclipse key, the magnitude addition and luminosity-ratio labels differ from its derivation and results; the original PDF correction comments are retained. Aldebaran’s key prints 4412 km/h and then uses 4416 in the angular-speed numerator. The X-ray question and key allocate points to parts (c) and (d) differently. DART’s alternative calculation prints both 0.17622 and 0.17662 m/s. The LISA key labels the component-mass calculation Part (b), repeats one algebra line and prints the period factor with a positive square. These printed discrepancies remain literal.

Оригинал в Архива: theory.pdf · официални решения: th_solutions.pdf