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BAAO 2020, Теория — Задача 2. Event Horizon Telescope and Super Massive Black Holes

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British Astronomy and Astrophysics Olympiad 2019-2020 — Astronomy & Astrophysics Competition Paper — Monday 20th January 2020 · 20 януари 2020 г. · 35 т.

Указания

This question paper must not be photographed or taken out of the exam room

Time: 3 hours (approximately 60 minutes per question).

Questions: All three questions should be attempted. The last question is worth slightly fewer marks than the rest.

Solutions: Answers and calculations are to be written on loose paper. Students should ensure their name and school is clearly written on all answer sheets and pages are numbered. EACH QUESTION ANSWERED must be started on a new page. A standard formula booklet with standard physical constants should be supplied.

Instructions: To accommodate students sitting the paper at different times, please do not discuss any aspect of the paper on the internet until 8 am Saturday 25th January.

Clarity: Solutions must be written legibly, in black pen (the papers are photocopied), and working down the page. Scribble will not be marked and overall clarity is an important aspect of this exam.

Eligibility: The International Olympiad will be held during September 2020; all sixth form students are eligible to participate, even if they will be attending university from October.

Calculators: Any standard calculator may be used, but calculators cannot be programmable and must not have symbolic algebra capability.

Training Dates and the IOAA (Bogota, Columbia, 13th to 22nd September 2020)

The best students taking this paper that are eligible to represent the UK at the IOAA will be invited to attend the Training Camp to be held in the Physics Department at the University of Oxford, (Friday 27th March to Tuesday 31st March 2020). Astronomy material will be covered; problem solving skills and observational skills (telescope and naked eye observations) will be developed. At the Training Camp a data analysis exam and a short theory paper will be sat. A team of five students (plus one reserve) will be selected for further training. From April there will be mentoring by email to cover some topics and problems, followed by additional training camps in the summer.

Important Constants

Constant	Symbol	Value
Speed of light	$c$	$3.00 \times 10^{8}\ \mathrm{m\,s^{-1}}$
Earth’s rotation period	1 day	24 hours
Earth’s orbital period	1 year	365.25 days
parsec	pc	$3.09 \times 10^{16}\ \mathrm{m}$
Astronomical Unit	au	$1.50 \times 10^{11}\ \mathrm{m}$
Semi-major axis of the Earth’s orbit		1 au
Radius of the Sun	$R_{\odot}$	$6.96 \times 10^{8}\ \mathrm{m}$
Radius of the Earth	$R_{\oplus}$	$6.37 \times 10^{6}\ \mathrm{m}$
Mass of the Sun	$M_{\odot}$	$1.99 \times 10^{30}\ \mathrm{kg}$
Mass of the Earth	$M_{\oplus}$	$5.97 \times 10^{24}\ \mathrm{kg}$
Luminosity of the Sun	$L_{\odot}$	$3.85 \times 10^{26}\ \mathrm{W}$
Stephan-Boltzmann constant	$\sigma$	$5.67 \times 10^{-8}\ \mathrm{J\,m^{-2}\,K^{-4}}$
Gravitational constant	$G$	$6.67 \times 10^{-11}\ \mathrm{m^{3}\,kg^{-1}\,s^{-2}}$
Boltzmann constant	$k_{\mathrm{B}}$	$1.38 \times 10^{-23}\ \mathrm{J\,K^{-1}}$
Permittivity of free space	$\varepsilon_0$	$8.85 \times 10^{-12}\ \mathrm{F\,m^{-1}}$
Permeability of free space	$\mu_0$	$4\pi \times 10^{-7}\ \mathrm{H\,m^{-1}}$
Planck’s constant	$h$	$6.63 \times 10^{-34}\ \mathrm{J\,s}$
Elementary charge	$e$	$1.60 \times 10^{-19}\ \mathrm{C}$
Proton rest mass	$m_{\mathrm{p}}$	$1.67 \times 10^{-27}\ \mathrm{kg}$
Electron rest mass	$m_{\mathrm{e}}$	$9.11 \times 10^{-31}\ \mathrm{kg}$
Wien’s displacement law	$\lambda_{\max} T$	$2.90 \times 10^{-3}\ \mathrm{m\,K}$

Basic calculus formulae:

Chain rule	$\frac{\mathrm{d}}{\mathrm{d}x} f(g(x)) = f'(g(x))g'(x)$
Product rule	$\frac{\mathrm{d}}{\mathrm{d}x}(uv) = \frac{\mathrm{d}u}{\mathrm{d}x}v + u\frac{\mathrm{d}v}{\mathrm{d}x}$
Quotient rule	$\frac{\mathrm{d}}{\mathrm{d}x}\left(\frac{u}{v}\right) = \frac{\frac{\mathrm{d}u}{\mathrm{d}x}v - u\frac{\mathrm{d}v}{\mathrm{d}x}}{v^2}$
Integration by parts	$\int u\frac{\mathrm{d}v}{\mathrm{d}x}\,\mathrm{d}x = uv - \int v\frac{\mathrm{d}u}{\mathrm{d}x}\,\mathrm{d}x$
Standard integral	$\int \frac{1}{x}\,\mathrm{d}x = \ln\lvert x\rvert + C$

Important Formulae

You might find the diagram of an elliptical orbit below useful in solving some of the questions:

Elements of an elliptic orbit:

$a = \mathbf{OA}\ (= \mathbf{OP})$	semi-major axis
$b = \mathbf{OB}\ (= \mathbf{OC})$	semi-minor axis
$e = \sqrt{1 - \frac{b^2}{a^2}}$	eccentricity
F	focus
P	periapsis (point nearest to F)
A	apoapsis (point furthest from F)

Kepler’s Third Law: For an elliptical orbit, the square of the period, $T$ , of an object about the focus is proportional to the cube of the semi-major axis, $a$ (as defined above), such that

T^2 = \frac{4\pi^2}{GM}a^3\,,

where $M$ is the total mass of the system (typically dominated by the central object) and $G$ is the universal gravitational constant.

Vis-Viva Equation: For an elliptical orbit, the speed $v$ of an object at a distance $r$ from the focus is related to the semi-major axis, $a$ , total mass of the system, $M$ , and universal gravitational constant, $G$ , (as defined above), such that

v^2 = GM\left(\frac{2}{r} - \frac{1}{a}\right)\,.

Magnitudes: The apparent magnitudes of two objects, $m_1$ and $m_0$ , are related to their apparent brightnesses, $b_1$ and $b_0$ , via the formula

\frac{b_1}{b_0} = 10^{-0.4(m_1 - m_0)}\,.

The absolute magnitude of an object, $\mathcal{M}$ , is the same as its apparent magnitude when viewed from 10 pc, hence the relationship between apparent and absolute magnitude and distance is

m - \mathcal{M} = 5\log\left(\frac{d}{10}\right),

where $d$ is measured in parsecs.

Note for markers

BAAO 2019/20 Solutions and Marking Guidelines

Note for markers:

Answers to two or three significant figures are generally acceptable. The solution may give more in order to make the calculation clear. Units should be present on final answers when appropriate.
There are multiple ways to solve some of the questions; please accept all good solutions that arrive at the correct answer. If a candidate gets the final (numerical) answer then allow them all the marks for that part of the question (as indicated in red), so long as there are no unphysical / nonsensical steps or assumptions made.

Условие

The Event Horizon Telescope (EHT) is a project to use many widely-spaced radio telescopes as a Very Long Baseline Interferometer (VBLI) to create a virtual telescope as big as the Earth. This extraordinary size allows sufficient angular resolution to be able to image the space close to the event horizon of a super massive black hole (SMBH), and provide an opportunity to test the predictions of Einstein’s theory of General Relativity (GR) in a very strong gravitational field. In April 2017 the EHT collaboration managed to co-ordinate time on all of the telescopes in the array so that they could observe the SMBH (called M87 ) at the centre of the Virgo galaxy, M87, and they plan to also image the SMBH at the centre of our galaxy (called Sgr A).

Left: globe centred on the Americas with yellow telescope sites labelled SMT, PV, SMA, JCMT, LMT, APEX, ALMA and SPT joined by solid and dashed blue baseline lines. Right: simulated black-hole image on a black background: a bright orange crescent ring, brighter on the left, around a dark central shadow. — Figure 3: Left: The locations of all the telescopes used during the April 2017 observing run. The solid lines correspond to baselines used for observing M87, whilst the dashed lines were the baselines used for the calibration source. Credit: EHT Collaboration. Right: A simulated model of what the region near an SMBH could look like, modelled at much higher resolution than the EHT can achieve. The light comes from the accretion disc, but the paths of the photons are bent into a characteristic shape by the extreme gravity, leading to a ‘shadow’ in middle of the disc - this is what the EHT is trying to image. The left side of the image is brighter than the right side as light emitted from a substance moving towards an observer is brighter than that of one moving away. Credit: Hotaka Shiokawa.

Some data about the locations of the eight telescopes in the array are given below in 3-D cartesian geocentric coordinates with $X$ pointing to the Greenwich meridian, $Y$ pointing $90^{\circ}$ away in the equatorial plane (eastern longitudes have positive $Y$ ), and positive $Z$ pointing in the direction of the North Pole. This is a left-handed coordinate system.

Facility	Location	$X$ (m)	$Y$ (m)	$Z$ (m)
ALMA	Chile	2225061.3	−5440061.7	−2481681.2
APEX	Chile	2225039.5	−5441197.6	−2479303.4
JCMT	Hawaii, USA	−5464584.7	−2493001.2	2150654.0
LMT	Mexico	−768715.6	−5988507.1	2063354.9
PV	Spain	5088967.8	−301681.2	3825012.2
SMA	Hawaii, USA	−5464555.5	−2492928.0	2150797.2
SMT	Arizona, USA	−1828796.2	−5054406.8	3427865.2
SPT	Antarctica	809.8	−816.9	−6359568.7

The minimum angle, $\theta_{\min}$ (in radians) that can be resolved by a VLBI array is given by the equation

\theta_{\min} = \frac{\lambda_{\mathrm{obs}}}{d_{\max}}\,,

where $\lambda_{\mathrm{obs}}$ is the observing wavelength and $d_{\max}$ is the longest straight line distance between two telescopes used (called the baseline), assumed perpendicular to the line of sight during the observation.

a. Determine $d_{\max}$ for the observations of M87 (i.e. the solid lines in Fig 3) and hence $\theta_{\min}$ if the EHT uses radio waves of frequency 230 GHz. Give your answer for $d_{\max}$ in km and $\theta_{\min}$ in microarcseconds (1 degree = 3600 arcseconds). [4 т.]

An important length scale when discussing black holes is the gravitational radius, $r_g = \frac{GM}{c^2}$ , where $G$ is the gravitational constant, $M$ is the mass of the black hole and $c$ is the speed of light. The familiar event horizon of a non-rotating black hole is called the Schwartzschild radius, $r_S \equiv 2r_g$ , however this is not what the EHT is able to observe - instead the closest it can see to a black hole is called the photon sphere, where photons orbit in the black hole in unstable circular orbits. On top of this the image of the black hole is gravitationally lensed by the black hole itself magnifying the apparent radius of the photon sphere to be between $(2\sqrt{3 + 2\sqrt{2}})r_g$ and $(3\sqrt{3})r_g$ , determined by spin and inclination; the latter corresponds to a perfectly spherical non-spinning black hole. The area within this lensed image will appear almost black and is the ‘shadow’ the EHT is looking for.

b. (i) Assuming Sgr A* is a non-spinning black hole with mass $4.1 \times 10^{6}\ M_{\odot}$ and a distance from us of 8.34 kpc:

(i) Derive the (unlensed) radius of the photon sphere, $r_{ph}$ , in units of $r_g$ , by considering a balance between the centripetal and (Newtonian) gravitational forces, but with the relativistic correction $v' = v\sqrt{1 - 2r_g/r}$ where $v'$ is the classical velocity and $r = r_{ph}$ when $v = c$ . [3 т.]

b. (ii) Determine the angular diameter (in microarcseconds) of the lensed photon sphere of Sgr A* and hence verify that the EHT can resolve it. [4 т.]

c. The angular diameter of M87* as determined from the images gained by the EHT (shown in Fig 4) is 42 microarcseconds, and the galaxy is 16.8 Mpc away from us. Determine the minimum and maximum possible masses of the SMBH in units of $M_{\odot}$ . [4 т.]

Four orange ring images of M87* titled April 5, April 6, April 10 and April 11, each with a small white beam circle at bottom right; the first has a 50 μas scale bar. A colour bar at right is labelled Brightness Temp. (10^9 K) from 0 to 6. — Figure 4: Four nights of data were taken for M87* during the observing window of the EHT, and whilst the diameter of the disk stayed relatively constant the location of bright spots moved, possibly indicating gas that is orbiting the black hole. Credit: EHT Collaboration.

The EHT observed M87* on four separate occasions during the observing window (see Fig 4), and the team saw that some of the bright spots changed in that time, suggesting they may be associated with orbiting gas close to the black hole. The Innermost Stable Circular Orbit (ISCO) is the equivalent of the photon sphere but for particles with mass (and is also stable). The total conserved energy of a circular orbit close to a non-spinning black hole is given by

E = mc^2\left(\frac{1 - \frac{2r_g}{r}}{\sqrt{1 - \frac{3r_g}{r}}}\right)\,,

and the radius of the ISCO, $r_{\mathrm{ISCO}}$ , is the value of $r$ for which $E$ is minimised.

d. Determine $r_{\mathrm{ISCO}}$ for a non-spinning black hole. Give your answer in units of $r_g$ . [4 т.]

We expect that most black holes are in fact spinning (since most stars are spinning) and the spin of a black hole is quantified with the spin parameter $a \equiv J/J_{\max}$ where $J$ is the angular momentum of the black hole and $J_{\max} = GM^2/c$ is the maximum possible angular momentum it can have. The value of $a$ varies from $-1 \leq a \leq 1$ , where negative spins correspond to the black hole rotating in the opposite direction to its accretion disk, and positive spins in the same direction. If $a = 1$ then $r_{\mathrm{ISCO}} = r_g$ , whilst if $a = -1$ then $r_{\mathrm{ISCO}} = 9r_g$ . The angular velocity of a particle in the ISCO is given by

\omega^2 = \frac{GM}{\left(r_{\mathrm{ISCO}}^{3/2} + ar_g^{3/2}\right)^2}\,.

e. (i) Taking the mass of M87* as $6.5 \times 10^{9}\ M_{\odot}$ :

(i) Determine the period of a particle in the ISCO of M87* for the $a = 1$ , $a = -1$ and $a = 0$ (i.e. non-spinning) cases. Give your answer in days. [4 т.]

e. (ii) One of the bright patches in Fig 4 seemed to move a quarter of the way around the ring between April 5 and 10 (from the left hand side to the bottom). Could it be attributable to gas moving in the ISCO? If so, is the spin likely to be positive or negative? [2 т.]

e. (iii) Determine the minimum and maximum ISCO periods for Sgr A* and hence suggest a possible reason why M87 has been imaged first, even though Sgr A* has a larger angular diameter, given that each ‘exposure’ with the EHT was 7 mins long (with multiple exposures from each observing run added together for the final image from each night). [3 т.]

Funnel-shaped embedding diagram of curved spacetime with dashed and solid arcs labelled ISCO, Lensed radius, Photon Sphere and Event Horizon (shaded ellipse at the bottom). A solid red double arrow runs along the funnel wall from the photon sphere to the ISCO; below, a horizontal r axis with marks r_ph and r_ISCO joined by a dashed blue double arrow. — Figure 5: Due to the curvature of spacetime, the real distance travelled by a particle moving from the ISCO to the photon sphere (indicated with the solid red arrow) is longer than you would get purely from subtracting the radial co-ordinates of those orbits (indicated with the dashed blue arrow), which would be valid for a flat spacetime. Relations between these distances are not to scale in this diagram. Credit: Modified from Bardeen et al. (1972).

The spacetime near a black hole is curved, as described by the equations of GR. This means that the distance between two points can be substantially different to the distance you would expect if spacetime was flat. GR tells us that the proper distance travelled by a particle moving from radius $r_1$ to radius $r_2$ around a black hole of mass $M$ (with $r_1 > r_2$ ) is given by

\Delta l = \int_{r_2}^{r_1}\left(1 - \frac{2r_g}{r}\right)^{-1/2}\mathrm{d}r\,.

f. A particle close to M87* moves directly from $r_{\mathrm{ISCO}}$ to $r_{ph}$ (and subsequently into the black hole). What is the extra distance travelled by it due to the curvature of spacetime, as described in Fig 5? Give your answer in au, and assume M87* is non-spinning. [7 т.]

Отговори

Решение

Q2 – Event Horizon Telescope and Super Massive Black Holes [35 marks]

Longest (solid) baseline from Fig. 3 is JCMT -> PV, so using 3-D Pythagoras,

$d_{\max} = \sqrt{\left(x_{JCMT} - x_{PV}\right)^2 + \left(y_{JCMT} - y_{PV}\right)^2 + \left(z_{JCMT} - z_{PV}\right)^2}$ [1]

$= 10907.93\ \mathrm{km}$ [1] [2]

[Accept using SMA -> PV (giving $d_{\max}$ = 10907.86 km) for full marks, and $d_{\max}$ should be at least 3 s.f. and in km for the final mark. If the student uses PV -> SPT (the longest baseline, but a dashed line so not used for the M87 observation) giving $d_{\max}$ = 11388.83 km then they lose the final mark only]

Wavelength being used,

$\lambda_{\mathrm{obs}} = \frac{c}{f} = \frac{3.00\times10^{8}}{230\times10^{9}} = 1.30\ \mathrm{mm}$ [1]

$\theta_{\min} = \frac{\lambda_{\mathrm{obs}}}{d_{\max}} = \frac{1.30\times10^{-3}}{10907.93\times10^{3}} = 1.20 \times 10^{-10}\ \mathrm{rad} = 24.7\ \mu\mathrm{as}$ [1] [2]

[Must be in microarcseconds (μas) for the final mark]

b. (i)

Balancing centripetal forces and Newtonian gravity,

$\frac{GMm}{r^2} = \frac{mv'^2}{r} \therefore v'^2 = \frac{GM}{r} = v^2\left(1 - \frac{2r_g}{r}\right)$ [1]

Putting in the limit v = c and using the definition of $r_g$ (= GM/c $^2$ ),

$\frac{GM}{r_{ph}} = c^2\left(1 - \frac{2r_g}{r_{ph}}\right) \therefore \frac{GM}{c^2} = r_{ph}\left(1 - \frac{2r_g}{r_{ph}}\right) \quad \therefore r_g = r_{ph}\left(1 - \frac{2r_g}{r_{ph}}\right)$ [1]

$\therefore r_{ph} = 3r_g$ [1] [3]

b. (ii)

$r_g = \frac{GM}{c^2} = \frac{6.67\times10^{-11}\times4.1\times10^{6}\times1.99\times10^{30}}{(3.00\times10^{8})^2} = 6.05 \times 10^{9}\ \mathrm{m}$ [1]

Recognising that the lensed value we want is (3√3) $r_g$ ,

$\left(3\sqrt{3}\right)r_g = 3.14 \times 10^{10}\ \mathrm{m}$ [1]

Determining the angular radius using trigonometry (and the small angle approximation),

$\theta_r = \frac{(3\sqrt{3})r_g}{d} = \frac{3.14\times10^{10}}{8.34\times10^{3}\times3.09\times10^{16}} = 1.22 \times 10^{-10}\ \mathrm{rad}\ \ (= 25.1\ \mu\mathrm{as})$ [1]

Hence the angular diameter,

$\theta_d = 2\theta_r = 2.44 \times 10^{-10}\ \mathrm{rad} = 50.3\ \mu\mathrm{as}\ \ [\theta_d > \theta_{\min} \therefore \text{resolvable}]$ [1] [4]

[Must be in μas and the angular diameter (not radius) for the final mark]

[This is larger than the measured angular diameter of M87* of 42 μas, meaning that Sgr A* is the easiest SMBH to resolve, but as seen later in the question it’s not the easiest to image]

The angular radius is,

$\theta_r = 21\ \mu\mathrm{as} = 1.02 \times 10^{-10}\ \mathrm{rad}$

The lensed physical radius is,

$r_g' = d\theta_r = 16.8 \times 10^{6} \times 3.09 \times 10^{16} \times 1.02 \times 10^{-10} = 5.29 \times 10^{13}\ \mathrm{m}$ [1]

Looking at the minimum mass limit (i.e. the non-spinning case),

$r_g' = \left(3\sqrt{3}\right)r_{g,\min} \therefore r_{g,\min} = \frac{5.29\times10^{13}}{3\sqrt{3}} = 1.02 \times 10^{13}\ \mathrm{m}$ [1]

$\therefore M_{\min} = \frac{r_{g,min}c^2}{G} = 1.37 \times 10^{40}\ \mathrm{kg} = 6.90 \times 10^{9}\ M_{\odot}$ [1] [3]

Using the other lensed $r_g$ limit gives for the maximum mass,

$M_{\max} = 1.48 \times 10^{40}\ \mathrm{kg} = 7.42 \times 10^{9}\ M_{\odot}$ [1] [1]

[The first three marks are for any correct mass (i.e. done in either order), with the fourth for the other one. Final masses must be given in $M_{\odot}$ ]

Want $\frac{\mathrm{d}E}{\mathrm{d}r} = 0$ given $E = mc^2\left(1 - 2r_g r^{-1}\right)\left(1 - 3r_g r^{-1}\right)^{-1/2}$

Applying the product rule, $\frac{\mathrm{d}}{\mathrm{d}r}(uv) = \frac{\mathrm{d}u}{\mathrm{d}r}v + \frac{\mathrm{d}v}{\mathrm{d}r}u$ ,

$u = 1 - 2r_g r^{-1} \qquad v = \left(1 - 3r_g r^{-1}\right)^{-1/2}$

$\frac{\mathrm{d}u}{\mathrm{d}r} = 2r_g r^{-2} \qquad \frac{\mathrm{d}v}{\mathrm{d}r} = -\frac{3}{2}r_g r^{-2}\left(1 - 3r_g r^{-1}\right)^{-3/2}$ [1]

$\therefore \frac{\mathrm{d}E}{\mathrm{d}r} = mc^2\left[2r_g r^{-2}\left(1 - 3r_g r^{-1}\right)^{-1/2} - \frac{3}{2}r_g r^{-2}\left(1 - 3r_g r^{-1}\right)^{-3/2}\left(1 - 2r_g r^{-1}\right)\right]$ [1]

Making the expression in the square bracket equal zero (since m > 0),

$2r_g r^{-2}\left(1 - 3r_g r^{-1}\right)^{-1/2} = \frac{3}{2}r_g r^{-2}\left(1 - 3r_g r^{-1}\right)^{-3/2}\left(1 - 2r_g r^{-1}\right)$ [1]

$2\left(1 - 3r_g r^{-1}\right) = \frac{3}{2}\left(1 - 2r_g r^{-1}\right)$

$\therefore r = 6r_g$ (so $r_{\mathrm{ISCO}}$ = 6 $r_g$ ) [1] [4]

[First mark is for differentiating the expressions for u and v correctly, second mark for having the correct form of dE/dr, third mark for setting the relevant part of the expression to zero, fourth mark for the final answer. Give full credit to use of the quotient rule or other valid approaches]

e. (i)

$r_g = \frac{GM}{c^2} = \frac{6.67\times10^{-11}\times6.5\times10^{9}\times1.99\times10^{30}}{(3.00\times10^{8})^2} = 9.59 \times 10^{12}\ \mathrm{m}$

For the a = 1 case,

$\omega^2 = \frac{GM}{\left(2r_g^{3/2}\right)^2} = \frac{6.67\times10^{-11}\times6.5\times10^{9}\times1.99\times10^{30}}{\left(2(9.59\times10^{12})^{3/2}\right)^2} = 2.45 \times 10^{-10}\ \mathrm{rad^2\,s^{-2}}$ [1]

$T_{a=1} = \frac{2\pi}{\omega} = \frac{2\pi}{\sqrt{2.45\times10^{-10}}} = 4.02 \times 10^{5}\ s = 4.65$ days [1] [2]

Similarly,

$T_{a=0} = 34.15$ days [Allow ecf on this one only] [1] [1]

$T_{a=-1} = 60.42$ days [All periods must be in days] [1] [1]

[First one correctly calculated is worth two marks, the rest are one mark (done in any order)]

e. (ii)

Quarter of a period in 5 days, so orbital period = 20 days

This is in the valid ISCO range (so could be gas moving in the ISCO) [1] [1]

This corresponds to a positive spin [1] [1]

[Allow full ecf on conclusions consistent with their earlier values]

e. (iii)

Minimum ISCO period when a = 1, so using a similar approach to e.(i),

$\omega^2 = 6.15 \times 10^{-4}\ \mathrm{rad^2\,s^{-2}} \qquad \therefore T_{\min} = 253\ \mathrm{s}\ (= 4.22\ \mathrm{mins})$ [1] [1]

Maximum ISCO period when a = -1, $\therefore T_{\max} = 3293\ \mathrm{s}\ (= 54.88\ \mathrm{mins})$ [1] [1]

Possibility that bright objects / gas clouds appear smeared out during exposures [1] [1]

[Allow difficult to determine time evolution of the SMBH on the timescale of the ISCO / any other relevant point from identifying that the period of the ISCO might be similar to the duration of the exposure (unlike with M87*)]

In flat space,

$\Delta l_{\mathrm{flat}} = r_{\mathrm{ISCO}} - r_{ph} = 6r_g - 3r_g = 3r_g$ [Allow ecf] [1]

In curved space,

$\Delta l = \int_{3r_g}^{6r_g}\left(1 - \frac{2r_g}{r}\right)^{-1/2}\mathrm{d}r$

Using the following consecutive substitutions,

$u = \frac{r}{2r_g} \therefore \mathrm{d}u = \frac{1}{2r_g}\mathrm{d}r \qquad \therefore \Delta l = 2r_g\int_{3/2}^{3}\left(1 - \frac{1}{u}\right)^{-1/2}\mathrm{d}u$ [1]

$t^2 = 1 - \frac{1}{u} \therefore 2t\,\mathrm{d}t = \frac{1}{u^2}\,\mathrm{d}u = (1 - t^2)^2\,\mathrm{d}u \qquad \therefore \Delta l = 4r_g\int_{1/\sqrt{3}}^{\sqrt{2}/\sqrt{3}}\frac{1}{(1-t^2)^2}\,\mathrm{d}t$ [1]

Using difference of squares followed by partial fractions,

$\frac{1}{(1-t^2)^2} = \frac{1}{(1-t)^2(1+t)^2} = \frac{2-t}{4(1-t)^2} + \frac{2+t}{4(1+t)^2} = \frac{1}{4}\left[\frac{1}{1-t} + \frac{1}{(1-t)^2} + \frac{1}{1+t} + \frac{1}{(1+t)^2}\right]$ [1]

$\therefore \Delta l = r_g\int_{1/\sqrt{3}}^{\sqrt{2}/\sqrt{3}}\frac{1}{1-t} + \frac{1}{(1-t)^2} + \frac{1}{1+t} + \frac{1}{(1+t)^2}\ \mathrm{d}t$

$\therefore \Delta l = r_g\left[\ln\left(\frac{1+t}{1-t}\right) + \frac{1}{1-t} - \frac{1}{1+t}\right]_{1/\sqrt{3}}^{\sqrt{2}/\sqrt{3}}$ [1]

$= 4.1424r_g$ [1]

So extra distance = $\Delta l - \Delta l_{\mathrm{flat}} = 1.1424r_g$ [so this is a big effect!]

$= 1.1424 \times 9.59 \times 10^{12} = 1.10 \times 10^{13}\ \mathrm{m} = 73\ \mathrm{au}$ [1] [7]

[Accept any alternative method of integrating the function, including using numerical methods. The final answer must be in au for the final mark. To help with ecf for incorrect limits, the integral as a function of the original variable is $\Delta l = \left[r\left(1 - \frac{2r_g}{r}\right)^{1/2} + r_g\ln\left(r\left(1 - \frac{2r_g}{r}\right)^{1/2} - r_g - r\right)\right]_{r_2}^{r_1}$ so the irrational number should be $\left(2\sqrt{6} - \sqrt{3} - \ln\left(2 + \sqrt{3}\right) + \ln\left(5 + 2\sqrt{6}\right)\right) = 4.1424 \ldots$ ]

Questions proposed by

END OF PAPER

Questions proposed by:

Dr Alex Calverley (Royal Grammar School, Guildford)

Alan Sun (University of Cambridge)

Navonil Neogi (University of Cambridge)

Оригинал в Архива: BAAO_2020.pdf · официални решения: BAAO_2020_solutions.pdf