Съдържание

Universal constants Astronomical data Calculus related formulas Условие (T12.1) Stellar mass blackholes (T12.2) Primordial black holes Решение General Marking Principles

IOAA 2025, Теория — Задача 12. Hawking Radiation from Black Holes

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Всички материали от IOAA 2025

IOAA 2025 — Theory Examination · 50 т.

Theory Examination Instructions

You are given an envelope marked B. Please verify the Contestant Code printed on it. Inside you will find:

This Instruction Sheet and/or This Instruction Sheet in your native language (if applicable)
Data Sheet and/or Data Sheet in your native language (if applicable)
A set of Working Sheets
One envelope marked S (please verify the Contestant Code marked on it).

You are allowed to open the envelope S only when the whistle is blown to mark the start of the examination.

Inside the envelope S you will find the following:

Question Paper and/or Question Paper in your native language (if applicable)
A set of Summary Answersheets.

Total duration of this examination is 5 hours.

This examination consists of the following questions:

Questions	Marks each	Subtotal
T01 and T02	10	$2\times10=20$ marks
T03 and T04	15	$2\times15=30$ marks
T05–T08	20	$4\times20=80$ marks
T09 and T10	35	$2\times35=70$ marks
T11 and T12	50	$2\times50=100$ marks
Total		300 marks

Writing your answers:

Summary Answersheet:

For each question a separate Summary Answersheet has been provided. Write your contestant code on each page of the Summary Answersheet.
Final answer(s) for each question / each part of a question must be written in the corresponding box in the Summary Answersheet.

Working Sheets:

Your detailed work must be shown on the Working Sheet.
Write your Contestant Code on both sides of each Working Sheet.
Start a new question on a fresh Working Sheet. Write the question number in the box provided at the top.
Both sides of the Working Sheet should be used. Write only inside the boxed area.
For rough work that you do not want to be evaluated, clearly strike that part out.
For each question write the page number in the box provided at the top of the Working Sheet, starting with 1 (each sheet has 2 pages). For example, if you use only 1 Working Sheet for T01, the question number and page number should be “T01, 1” and “T01, 2” on the FRONT and BACK of that Working Sheet, respectively. Next, if you use 2 Working Sheets for T02, the question number and page number should be “T02, 1” and “T02, 2” for the first Working Sheet, and “T02, 3” and “T02, 4” on the second Working Sheet.
You may ask for extra Working Sheets, if required.
Note that failure to mark the Working Sheets, as instructed above, may result in your answers not being evaluated.

At the end of the examination:

Check that you have correctly written your Contestant Code and page numbers on all used sheets.
Arrange your sheets in the following way: Summary Answersheet of each question (starting with T01) followed by Working Sheet(s) used for that question.
Similarly stack the Summary Answersheet and Working Sheet(s) used for each question, in sequence.
Place the bunch of all your used sheets arranged in the above manner inside the smaller envelope S. All Summary Answersheets must be placed inside the envelope S irrespective of whether you have used them or not.
Only the contents of the smaller envelope S will be evaluated. Any sheet that is not placed inside this envelope will not be evaluated.
Place the smaller envelope S inside the bigger envelope B.
All other sheets, namely, Instruction Sheet(s), Question Paper(s), Data Sheet(s) and unused Working Sheet(s) should be placed directly inside the bigger envelope B.
Tie the thread seal on the envelope B.
Hand over the envelope B to the examiner. Make sure that no sheet is left on the table, except the green and the red card.

General Instructions:

If you need a visit to the toilet during the examination, please raise your hand holding the green card kept on your table. You will be escorted by a volunteer.
If you need any other help during the examination, please raise your hand holding the red card kept on your table.
Use only black or blue pen for writing. For drawing, you may use a pencil.
Some marks may be deducted if the final answer is given without detailed solution.
Some marks may be deducted if the final answers have improper number of significant digits, no units or wrong units.
The necessary values of fundamental and astronomical constants should be taken from the Data Sheet provided to you.

Original IOAA 2025 Data Sheet

Universal constants

Quantity	Symbol	Value
Avogadro constant	$N_A$	$6.022\times10^{23}\ \mathrm{mol^{-1}}$
Boltzmann constant	$k_B$	$1.381\times10^{-23}\ \mathrm{J\,K^{-1}}$
Charge of electron	$e$	$1.602\times10^{-19}\ \mathrm C$
Planck constant	$h$	$6.626\times10^{-34}\ \mathrm{J\,s}$
Speed of light in vacuum	$c$	$2.998\times10^8\ \mathrm{m\,s^{-1}}$
Universal gravitational constant	$G$	$6.674\times10^{-11}\ \mathrm{N\,m^2\,kg^{-2}}$
Universal gas constant	$R$	$8.315\ \mathrm{J\,mol^{-1}\,K^{-1}}$
Stefan-Boltzmann constant	$\sigma$	$5.670\times10^{-8}\ \mathrm{W\,m^{-2}\,K^{-4}}$
Wien’s displacement constant	$b$	$2.898\times10^{-3}\ \mathrm{m\,K}$
Permittivity of free space	$\epsilon_0$	$8.854\times10^{-12}\ \mathrm{m^{-3}\,kg^{-1}\,s^4\,A^2}$
Permeability of free space	$\mu_0$	$1.257\times10^{-6}\ \mathrm{N\,A^{-2}}$
Mass of electron	$m_e$	$9.109\times10^{-31}\ \mathrm{kg}=0.511\ \mathrm{MeV}/c^2$
Mass of proton	$m_p$	$1.673\times10^{-27}\ \mathrm{kg}=938.272\ \mathrm{MeV}/c^2$
Mass of neutron	$m_n$	$1.675\times10^{-27}\ \mathrm{kg}=939.565\ \mathrm{MeV}/c^2$
Mass of deuteron	$m_D$	$3.344\times10^{-27}\ \mathrm{kg}=1875.613\ \mathrm{MeV}/c^2$
Mass of He nucleus	$m_{\mathrm{He}}$	$6.645\times10^{-27}\ \mathrm{kg}=3727.181\ \mathrm{MeV}/c^2$

Astronomical data

Quantity	Symbol	Value
Mass of Sun	$M_\odot$	$1.988\times10^{30}\ \mathrm{kg}$
Radius of Sun	$R_\odot$	$6.957\times10^8\ \mathrm m$
Luminosity of Sun	$L_\odot$	$3.828\times10^{26}\ \mathrm W$
Effective temperature of Sun	$T_{\mathrm{eff},\odot}$	$5772\ \mathrm K$
Apparent magnitude of Sun (in V-band)	$m_{V,\odot}$	$-26.74$
Absolute magnitude of Sun (in V-band)	$M_{V,\odot}$	$+4.82$
Apparent bolometric magnitude of Sun	$m_{\mathrm{bol},\odot}$	$-26.83$
Absolute bolometric magnitude of Sun	$M_{\mathrm{bol},\odot}$	$+4.74$
Solar constant (above atmosphere of Earth)	$S_\odot$	$1361\ \mathrm{W\,m^{-2}}$
Apparent angular diameter of Sun (from Earth)	$\theta_\odot$	$\approx32'$
Mass of Earth	$M_\oplus$	$5.972\times10^{24}\ \mathrm{kg}$
Radius of Earth	$R_\oplus$	$6.378\times10^6\ \mathrm m$
Axial tilt of Earth	$\epsilon$	$23^\circ26'$
Inclination of the lunar orbit w.r.t. the ecliptic		$5^\circ8'43''$
Mass of Jupiter	$M_J$	$1.898\times10^{27}\ \mathrm{kg}$
Radius of Jupiter	$R_J$	$6.991\times10^7\ \mathrm m$
1 Astronomical Unit	1 au	$1.496\times10^{11}\ \mathrm m$
1 parsec	1 pc	$3.086\times10^{16}\ \mathrm m$
1 light-year	1 ly	$9.461\times10^{15}\ \mathrm m$
1 jansky	1 Jy	$10^{-26}\ \mathrm{W\,m^{-2}\,Hz^{-1}}$
1 tropical year		$365.2422$ solar days $=3.156\times10^7\ \mathrm s$ ; $365\ \mathrm d\ 5\ \mathrm h\ 48\ \mathrm{min}\ 46\ \mathrm s$
1 sidereal year		$365.2564$ solar days $=3.156\times10^7\ \mathrm s$ ; $365\ \mathrm d\ 6\ \mathrm h\ 9\ \mathrm{min}\ 13\ \mathrm s$
Rate of precession of Vernal Equinox		$1^\circ$ per $71.6$ years

Calculus related formulas

$\frac{dy}{dx}=\frac{dy}{du}\frac{du}{dx}$ .
$\frac{d}{dx}x^n=nx^{n-1}$ .
$\frac{d}{dx}\sin kx=k\cos kx$ .
$\frac{d}{dx}\cos kx=-k\sin kx$ .
$\frac{d}{dx}\tan kx=k\sec^2kx$ .
$\int x^n\,dx=\frac{x^{n+1}}{n+1}+\text{constant}$ ; for $n\ne-1$ .
$f(x)\simeq f(x_0)+\left.\frac{df}{dx}\right|_{x=x_0}(x-x_0)$ ; for $x\approx x_0$ .

Условие

(T12.1) Stellar mass blackholes

A black hole (BH) typically forms by the gravitational collapse of a massive star at the end of its life cycle to a point called a singularity. Due to the extreme gravity of such an object, nothing that enters the so-called event horizon (a spherical surface with $r=R_\mathrm{SC}$ , where $r$ is the distance from the singularity) is able to escape from it. Here, $R_\mathrm{SC}$ is referred to as the Schwarzschild radius.

(T12.1a) Modelling the origin of Hawking radiation: Consider a pair of particles, each with mass $m$ , produced on either side of the BH horizon. One particle is slightly outside the horizon at $r\approx R_\mathrm{SC}$ , while the other particle is inside the horizon at $r=\kappa R_\mathrm{SC}$ . Assume that the total energy of a particle is the sum of its rest mass energy $mc^2$ and the gravitational potential energy due to the BH. Determine the value of $\kappa$ for which the particle pair has zero total energy. [4 т.]

(T12.1b) Temperature of a black hole: If the particle produced outside the horizon in the above process has enough kinetic energy, it may escape the BH in a process called Hawking radiation. The one inside the horizon, which has negative energy, gets absorbed and decreases the mass of the BH.

Assume that all Hawking radiation is made of photons with a black body spectrum which peaks at the wavelength $\lambda_\mathrm{bb}\approx16R_\mathrm{SC}$ . It is known that for a solar mass BH, $R_{\mathrm{SC},\odot}=2.952\,\mathrm{km}$ .

Obtain an expression for the temperature, $T_\mathrm{bh}$ , of the BH corresponding to this black body radiation, in terms of its mass $M_\mathrm{bh}$ and physical constants. Calculate the Schwarzschild radius, $R_{\mathrm{SC},10\odot}$ , and temperature, $T_{\mathrm{bh},10\odot}$ , for a BH with mass $10M_\odot$ . [4 т.]

(T12.1c) Mass loss of a black hole: Assume that the Hawking radiation is emitted out from the event horizon. Using the mass-energy equivalence, obtain an expression for the rate of mass loss, $dM_\mathrm{bh}(t)/dt$ , in terms of the mass $M_\mathrm{bh}(t)$ of the BH and physical constants.

Hence, obtain an expression for $M_\mathrm{bh}(t)$ for a BH with initial mass $M_0$ . Sketch $M_\mathrm{bh}(t)$ as a function of $t$ from $M_\mathrm{bh}=M_0$ to $M_\mathrm{bh}=0$ . [8 т.]

(T12.1d) Lifetime of a black hole: Obtain an expression for the lifetime $\tau_\mathrm{bh}$ at which a black hole with initial mass $M_0$ completely evaporates due to Hawking radiation, in terms of $M_0$ and physical constants. Calculate the lifetime $\tau_{\mathrm{bh},10\odot}$ (in seconds) for a black hole with $M_0=10M_\odot$ . [3 т.]

(T12.1e) Black hole in a CMB radiation bath: Consider an isolated black hole in space, far away from other bodies, with a current temperature $T_\mathrm{bh}^\mathrm{now}$ , surrounded by the cosmic microwave background (CMB) with a current temperature $T_\mathrm{cmb}^\mathrm{now}=2.7\,\mathrm{K}$ . The black hole can grow in mass by absorbing CMB radiation, and lose its mass by Hawking radiation.

Taking into account the accelerating expansion of the Universe, identify which of the following figures show the long-term time evolution of $T_\mathrm{bh}$ in the following three cases:

(X) $T_\mathrm{bh}^\mathrm{now}>T_\mathrm{cmb}^\mathrm{now}$ , (Y) $T_\mathrm{bh}^\mathrm{now}=T_\mathrm{cmb}^\mathrm{now}$ , (Z) $T_\mathrm{bh}^\mathrm{now}<T_\mathrm{cmb}^\mathrm{now}$ .

All six original black-hole temperature options I–VI with complete curves and native time/temperature labels.

Indicate your answer by ticking the appropriate box (only one) for each case X, Y or Z in the Table given in the Summary Answersheet corresponding to the appropriate figure number. [6 т.]

(T12.2) Primordial black holes

Primordial black holes (PBHs) of much smaller masses can form in the very early Universe. All the following questions are related to PBHs. Here, any processes that increase the mass of the black hole may be neglected.

(T12.2a) PBH evaporating at the current epoch: As you may have noticed from the answers to the previous questions, black holes of solar mass would take a long time to evaporate. However, since PBHs can have a much smaller mass, we may be able to see them evaporating in current times.

Find the initial mass $M_{0,\mathrm{PBH}}$ (in kg), Schwarzschild radius $R_{\mathrm{SC},\mathrm{PBH}}$ (in m), and temperature $T_\mathrm{PBH}$ (in K) of a black hole that may be evaporating away completely at the present epoch, i.e., those with lifetime $\tau_\mathrm{PBH}=14$ billion years. [4 т.]

(T12.2b) Formation of a PBH: In the radiation-dominated early Universe, the scale factor varies as $a(t)\sim t^{1/2}$ . In this era, PBHs form due to the collapse of all energy contained in a region of physical size $ct$ , where $t$ is the age of the Universe at that time.

A PBH with mass of $10^{12}\,\mathrm{kg}$ forms when the age of the Universe is about $10^{-23}\,\mathrm{s}$ . Calculate the age of the Universe, $t_{20}$ , when a PBH of mass $10^{20}\,\mathrm{kg}$ forms. [6 т.]

(T12.2c) Observed spectrum of Hawking radiation from PBH: Consider a PBH of initial mass $10^{10}\,\mathrm{kg}$ which completely evaporates at the end of its lifetime $\tau_\mathrm{PBH}$ . For this part, assume for simplicity that most of the Hawking radiation is emitted at this time, with a temperature corresponding to its initial mass. Also, take the scale factor of the Universe to be evolving as $a(t)\sim t^{2/3}$ .

Calculate the peak wavelength of this Hawking radiation as observed at Earth, $\lambda_\mathrm{earth}$ , at the present epoch (at $t=14$ billion years). [5 т.]

(T12.2d) High energy cosmic radiation from PBH: Now assume that the Hawking radiation emitted at a given time $t$ corresponds to photons emitted with an energy $k_BT_\mathrm{bh}(t)$ . Also, the highest possible temperature for a black hole is the Planck temperature $T_\mathrm{Planck}$ where $k_BT_\mathrm{Planck}=10^{19}\,\mathrm{GeV}$ .

The evolution of the scale factor over relevant time scales is given in the following figure. The scale factor today is set to be unity. $t$ (s) on the time axis represents the age of the universe in seconds.

Complete original scale-factor graph, full logarithmic time/scale axes, ticks, blue curve and orange grid.

If a photon with an energy of $E_\mathrm{det}=3.0\times10^{20}\,\mathrm{eV}$ is observed on Earth, determine the largest and the smallest possible values of the initial mass of the PBH ( $M_0^\mathrm{max}$ and $M_0^\mathrm{min}$ , respectively) which could be responsible for this photon. [10 т.]

Complete original Summary Answersheet T12, SA-23, FRONT: all native blank fields, printed markings, contestant-code boxes, graphs/checkboxes and page edges retained. No field filled.

Complete original Summary Answersheet T12, SA-24, BACK: all native blank fields, printed markings, contestant-code boxes, graphs/checkboxes and page edges retained. No field filled.

Решение

Part 1: Stellar mass black holes

(T12.1a)

Total energy of any particle is $E=E_{\mathrm{grav}}+mc^2$ , so we have

E_1=-\frac{GM_{\mathrm{bh}}m}{R_{\mathrm{SC}}}+mc^2,\qquad E_2=-\frac{GM_{\mathrm{bh}}m}{\kappa R_{\mathrm{SC}}}+mc^2.\qquad[1.0]

We require $E_1+E_2=0$ . Hence,

2mc^2-\frac{GM_{\mathrm{bh}}m}{R_{\mathrm{SC}}}\left(1+\frac1\kappa\right)=0.\qquad[1.0]

Then

\frac1\kappa=\frac{2c^2R_{\mathrm{SC}}}{GM_{\mathrm{bh}}}-1=3,

where we have used $R_{\mathrm{SC}}=2GM_{\mathrm{bh}}/c^2$ . [1.0]

This gives $\boxed{\kappa=\frac13}$ . [1.0]

(T12.1b)

Using Wien’s law, $\lambda_{\mathrm{bb}}=b/T_{\mathrm{bh}}$ , where $b=2.898\times10^{-3}\ \mathrm{m\,K}$ . [1.0]
Hence,

T_{\mathrm{bh}}=\frac{b}{16R_{\mathrm{SC}}}\implies\boxed{T_{\mathrm{bh}}=\frac{bc^2}{32GM_{\mathrm{bh}}}}.\qquad[1.0]

Since $R_{\mathrm{SC}}$ for a solar mass black hole is 2.952 km, for a black hole with mass $10\,M_\odot$ , we get $\boxed{R_{\mathrm{SC},10\odot}\simeq29.52\ \mathrm{km}}$ . [1.0]

Given the significant figures, 29 or 30 km will get full marks.

The temperature of this BH will be

T_{\mathrm{bh},10\odot}=\frac{2.898\times10^{-3}\ \mathrm{m\,K}}{16\times R_{\mathrm{SC},10\odot}},

\boxed{T_{\mathrm{bh},10\odot}=6.1\times10^{-9}\ \mathrm K}.\qquad[1.0]

Half credit lower limit	Full credit range	Half credit upper limit
	$6.0\times10^{-9}$ K to $6.2\times10^{-9}$ K

(T12.1c)

Power emitted by an evaporating black hole $=\sigma T_{\mathrm{bh}}^4\,4\pi R_{\mathrm{SC}}^2$ . [1.0]
This gives

\frac{dM_{\mathrm{bh}}}{dt}=-\frac1{c^2}\sigma T_{\mathrm{bh}}^4\,4\pi R_{\mathrm{SC}}^2.\qquad[1.0]

With $T_{\mathrm{bh}}=b/(16R_{\mathrm{SC}})$ ,

\boxed{\frac{dM_{\mathrm{bh}}}{dt}=-\frac{\pi b^4\sigma c^2}{16^4G^2}\frac1{M_{\mathrm{bh}}^2}}.\qquad[1.0]

This gives

\frac{dM_{\mathrm{bh}}}{M_{\mathrm{bh}}^2}=-\frac{\pi b^4\sigma c^2}{16^4G^2}\,dt,\qquad\text{and hence }\frac{M_{\mathrm{bh}}^3}{3}=-\frac{\pi b^4\sigma c^2}{16^4G^2}t+C.\qquad[1.0]

Since $M_{\mathrm{bh}}=M_0$ at $t=0$ , we get $M_{\mathrm{bh}}^3=M_0^3-3\frac{\pi b^4\sigma c^2}{16^4G^2}t$ . [0.5]

\boxed{M_{\mathrm{bh}}(t)=\left(M_0^3-3\frac{\pi b^4\sigma c^2}{16^4G^2}t\right)^{1/3}}.\qquad[1.0]

The value of $\frac{\pi b^4\sigma c^2}{16^4G^2}=3.868\times10^{15}\ \mathrm{kg^3\,s^{-1}}$ , not necessary here, but gets full credit.

Complete original evaporating-black-hole mass curve, initial mass label, full time-axis lifetime expression and origin.

	Marks
$M_0$ marked on $M_{\mathrm{bh}}(t)$ -axis	0.5
Correct expression for the intercept on the $t$ -axis	0.5
Negative initial slope	0.5
Large negative final slope	0.5
Convex nature of the curve	0.5

(T12.1d)

We know that $M(t)=\left(M_0^3-3\frac{\pi b^4\sigma c^2}{16^4G^2}t\right)^{1/3}$ .
Clearly, $M(t)=0$ when $t=\frac{M_0^3}{3}\frac{16^4G^2}{\pi b^4\sigma c^2}$ . Thus,

\boxed{\tau_{\mathrm{bh}}=\frac{M_0^3}{3}\frac{16^4G^2}{\pi b^4\sigma c^2}}.\qquad[1.0]

With $A\equiv\frac{\pi b^4\sigma c^2}{16^4G^2}\simeq3.868\times10^{15}\ \mathrm{kg^3\,s^{-1}}$ [0.5], this gives

\tau_{\mathrm{bh}}\approx\frac{M_0^3}{3A}=8.617\times10^{-17}\left(\frac{M_0}{\mathrm{kg}}\right)^3\mathrm s.\qquad[0.5]

For $M_0=10\,M_\odot$ ,

\boxed{\tau_{\mathrm{bh},10\odot}\approx6.8\times10^{77}\ \mathrm s}.\qquad[1.0]

(T12.1e)

As the CMB will redshift away eventually, there will be no CMB radiation for the BH to absorb at large $t$ . It will only decrease in mass and increase in temperature. So only the figures which show the temperature of the black hole increasing eventually ( $t\to\infty$ ) are relevant. The derivatives at the origin (now) should distinguish the three cases.

(X) If $T_{\mathrm{bh}}^{\mathrm{now}}>T_{\mathrm{cmb}}^{\mathrm{now}}$ , then the BH is losing more energy now than it is gaining by CMB absorption, and its temperature should increase due to decreasing mass. $(dT/dt)_{\mathrm{now}}>0$ .
(Y) If $T_{\mathrm{bh}}^{\mathrm{now}}=T_{\mathrm{cmb}}^{\mathrm{now}}$ , the temperature will be in equilibrium for some time, i.e. $(dT/dt)_{\mathrm{now}}=0$ , before the CMB temperature redshifts to smaller values and $(dT/dt)_\infty>0$ .
(Z) For $T_{\mathrm{bh}}^{\mathrm{now}}<T_{\mathrm{cmb}}^{\mathrm{now}}$ , initially the BH will gain mass (and hence cool down) by absorbing from CMB compared to what it loses by evaporation. So $(dT/dt)_{\mathrm{now}}<0$ . But eventually, $(dT/dt)_\infty>0$ .

Case	II	III	VI
X		✓
Y	✓
Z			✓

2.0 marks each for one and only one correct tick for the cases X, Y, Z. More than one tick for any case receives no credit for that case. [6.0]

Part 2: Primordial black holes

(T12.2a)

We want $\tau_{\mathrm{PBH}}=14\times10^9\times3.156\times10^7\ \mathrm s=4.418\times10^{17}\ \mathrm s$ . [1.0]
But we have found in (T12.1d) that $\tau_{\mathrm{bh}}=8.617\times10^{-17}(M_0/\mathrm{kg})^3\ \mathrm s$ .

This gives

M_{0,\mathrm{PBH}}^3=\left(\frac{\tau_{\mathrm{PBH}}}{8.617\times10^{-17}}\right)\mathrm{kg^3}\implies\boxed{M_{0,\mathrm{PBH}}\simeq1.7\times10^{11}\ \mathrm{kg}}.\qquad[1.0]

Since $R_{\mathrm{SC}}\propto M_{\mathrm{bh}}$ , and we know $R_{\mathrm{SC},\odot}=2952$ m,

R_{\mathrm{SC},\mathrm{PBH}}\simeq R_{\mathrm{SC},\odot}\left(\frac{M_{0,\mathrm{PBH}}}{M_\odot}\right)\implies\boxed{R_{\mathrm{SC},\mathrm{PBH}}=2.6\times10^{-16}\ \mathrm m}.\qquad[1.0]

Since $T_{\mathrm{bh}}\propto1/M_{\mathrm{bh}}$ and we know $T_{\mathrm{bh},10\odot}$ from (T12.1b),

T_{\mathrm{PBH}}\simeq T_{\mathrm{bh},10\odot}\left(\frac{10M_\odot}{M_{0,\mathrm{PBH}}}\right)\implies\boxed{T_{\mathrm{PBH}}=7.1\times10^{11}\ \mathrm K}.\qquad[1.0]

(T12.2b)

The size of horizon $r_H=ct$ . So the total mass contained within the horizon volume is

M=\frac1{c^2}\rho\left(\frac{4\pi}3\right)\left(\frac{ct}2\right)^3.\qquad[2.0]

With $a(t)\sim t^{1/2}$ and $a(t)\sim1/T$ , one gets $t\sim T^{-2}$ . [2.0]
With $\rho\sim T^4\sim t^{-2}$ , the mass of PBH formed at time $t$ would be $M_{0,\mathrm{PBH}}\propto t$ . [1.0]
The age of the Universe at which PBH of $10^{12}$ kg forms is $t_{12}=10^{-23}$ s.
Hence,

t_{20}=t_{12}\times\left(\frac{10^{20}\ \mathrm{kg}}{10^{12}\ \mathrm{kg}}\right)\implies\boxed{t_{20}=t_{12}\times10^8=10^{-15}\ \mathrm s}.\qquad[1.0]

(T12.2c)

Lifetime of the PBH with $M_{0,\mathrm{PBH}}=10^{10}$ kg will be

\tau_{\mathrm{PBH}}=8.617\times10^{-17}\cdot(10^{10})^3\ \mathrm s=8.617\times10^{13}\ \mathrm s.\qquad[0.5]

As given in (T12.1b), $R_{\mathrm{SC},\odot}=2952$ m. Using $R_{\mathrm{SC}}\propto M$ , we get the peak wavelength of radiation emitted by this $10^{10}$ kg PBH, at the time of emission, as

\lambda_0\simeq16R_{\mathrm{SC}}=16\cdot\frac{10^{10}\ \mathrm{kg}}{M_\odot}\cdot R_{\mathrm{SC},\odot}=2.4\times10^{-16}\ \mathrm m.\qquad[1.5]

Redshifted wavelength received at earth at present time $t_0$ is obtained from

\lambda_{\mathrm{earth}}=\lambda_0\frac{a(t_0)}{a(\tau_{\mathrm{PBH}})}.\qquad[1.0]

With $a(t)\sim t^{2/3}$ , we get

\lambda_{\mathrm{earth}}=\lambda_0\left(\frac{t_{\mathrm{now}}}{\tau_{\mathrm{PBH}}}\right)^{2/3}=2.4\times10^{-16}\ \mathrm m\left(\frac{4.418\times10^{17}\ \mathrm s}{8.617\times10^{13}\ \mathrm s}\right)^{2/3}.\qquad[1.0]

\boxed{\lambda_{\mathrm{earth}}=7.1\times10^{-14}\ \mathrm m}.\qquad[1.0]

(T12.2d)

Using the expression for $T_{\mathrm{bh}}$ found in (T12.1b), the energy of a typical photon emitted at time $t$ for a PBH is

E_{\mathrm{emitted}}(t)=k_BT_{\mathrm{bh}}(t)=\frac{bc^2k_B}{32GM_{\mathrm{bh}}(t)}.\qquad[1.0]

From (T12.1c), we know

M_0^3=3\frac{\pi b^4\sigma c^2}{16^4G^2}t+M_{\mathrm{bh}}(t)^3=3At+\left(\frac{bc^2k_B}{32GE_{\mathrm{emitted}}(t)}\right)^3.\qquad[1.0]

The value of $E_{\mathrm{det}}$ can be different from $E_{\mathrm{emitted}}$ due to redshift. Hence, the smallest possible value of $E_{\mathrm{emitted}}(t)$ is $E_{\mathrm{det}}=3.0\times10^{20}$ eV. This gives the maximum value of the second term as

\left(\frac{bc^2k_B}{32GE_{\mathrm{emitted}}(t)}\right)^3_{\max}=4.3\times10^{-5}\ \mathrm{kg^3}.\qquad[1.0]

Given the value of $A=3.868\times10^{15}\ \mathrm{kg^3\,s^{-1}}$ , the second term can always be neglected for $t\gtrsim10^{-19}$ s. This is true in this problem. So $M_0\approx(3At)^{1/3}$ . [1.0]
The largest possible $M_0$ would be obtained when $t$ is as large as possible, i.e., $t=t_{\mathrm{now}}$ . This will correspond to a BH radiating in its last stages at present in the nearby region. [1.0]
Then $M_0^{\max}=(3At_{\mathrm{now}})^{1/3}=(5.13\times10^{33}\ \mathrm{kg^3})^{1/3}$ . [0.5]
Thus, $\boxed{M_0^{\max}=1.7\times10^{11}\ \mathrm{kg}}$ . [0.5]
Note that this was already found in (T12.2a), so the calculation above is not needed.
The smallest $M_0$ will correspond to a BH that has evaporated as early as possible. Such a BH would have evaporated far away from Earth, and its radiation would be redshifted.
The smallest time of evaporation would correspond to the largest redshift factor, and hence the largest $E_{\mathrm{emitted}}(t)$ , which is $E_{\mathrm{emitted}}(t)=k_BT_{\mathrm{Planck}}$ . [1.0]
The ratio of the scale factors at the time of the above PBH evaporation and now is

\frac{a(t)}{a(t_{\mathrm{now}})}=\frac{E_{\mathrm{det}}}{k_BT_{\mathrm{Planck}}}=\frac{3.0\times10^{20}\ \mathrm{eV}}{10^{19}\ \mathrm{GeV}}=3.0\times10^{-8}.

With $a(t_{\mathrm{now}})=1.0$ , we have $a(t)=3.0\times10^{-8}$ . [1.0]

Using the plot of $a(t)$ vs $t$ from the figure, $t\simeq10^4$ s. Note that for this $t$ , the second term in the original $M_0^3$ expression is still negligible. [1.0]
Therefore, $\boxed{M_0^{\min}\approx(3At)^{1/3}\approx4.9\times10^6\ \mathrm{kg}}$ . [1.0]

Half credit lower limit	Full credit range	Half credit upper limit
	$2.5\times10^6$ kg to $10.0\times10^6$ kg

General Marking Principles

STRICT adherence to given marking scheme.
No double penalty – deduction only at the original place of error; no further penalty in consequent steps, if they are otherwise correct – if final answer is physically unviable or dimensionally incorrect, deduct for final answer also.
Significant figures: one more or one less than appropriate no. of significant figures allowed with full credit.
Units / dimensions:
- In cases where units are not mentioned in the Summary Answersheet, 0.5 deduction for missing units.
- For missing units on the same kind of quantities in a given part, deduct only once.
- For wrong dimensions in final answer because of some error in earlier steps, no credit for final answer, irrespective of no-double penalty principle.
If expression is asked in terms of x, y, …: deduction of final part credit if expression is not in terms of appropriate quantities.
Cumulative part marking: all previous part credits to be added if final expression / number is correct (unless otherwise specified).
Final answer not in Summary Answersheet, but present inside: if final answer is present on Working Sheet, but not transferred on Summary Answersheet, it is to be credited.

Бележки за изданието

The printed question, key and blank answer sheets contain differing expressions and units. The Balmer key uses different rounded colour excesses; the quasar scale changes its angular unit; the neutron-star key prints differing mass-rate units/values. The black-hole-shadow question prints an unsquared radial speed in its comparison equation, while the repeated key uses a square. Atmospheric Seeing asks for a diameter but its key describes a radius. The nucleosynthesis key changes strict inequalities, has a differently grouped decay exponent, and includes an extra time-arrow field also present on its answer sheet. The stellar-profile key omits a factor in one line that appears in the next. The Hawking key prints a separation step differing from its integral and a horizon volume differing from its prose radius. Its answer sheet labels a temperature field with “s”. These native forms are preserved, including every original blank answer page.

Оригинал в Архива: Theory-Questions.pdf · официални решения: Theory-Solution.pdf · допълнителни данни: Theory-Summary-Answersheet.pdf · допълнителни данни: Theory-Instructions.pdf · допълнителни данни: Data_Sheet.pdf · допълнителни данни: IOAA2025_General_Marking_Scheme.pdf