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BAAO 2019, Теория — Задача 4. Superluminal Jets

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British Astronomy and Astrophysics Olympiad 2018-2019 — Astronomy & Astrophysics Competition Paper — Monday 21st January 2019 · 21 януари 2019 г. · 25 т.

Указания

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

Time: 3 hours plus 15 minutes reading time (no writing permitted). Approx 45 minutes per question.

Questions: All four questions should be attempted.

Marks: The questions carry similar marks.

Solutions: Answers and calculations are to be written on loose paper or in examination booklets. Students should ensure their name and school is clearly written on all answer sheets and pages are numbered. 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 26th 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 paper.

Eligibility: The International Olympiad will be held during August 2019; all sixth form students are eligible to participate, even if they will be attending university from October.

Training Dates and the International Astronomy and Astrophysics Olympiad (IOAA)

The IOAA this year will be held in Zanka, Hungary, from 2nd to 11th August 2019.

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, (Saturday 13th April to Wednesday 17th April 2019). 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 May 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	$m_{\mathrm{e}}$	$9.11 \times 10^{-31}\ \mathrm{kg}$

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.

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 2018/19 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 speed of light is considered to be the speed limit of the Universe, however knots of plasma in the jets from active galactic nuclei (AGN) have been observed to be moving with apparent transverse speeds in excess of this, called superluminal speeds. Some of the more extreme examples can be appearing to move at up to 6 times the speed of light (see Figure 7).

Left: HST image of the galaxy M87 with a bluish jet extending from its bright core towards the lower right. Right: a strip image of the jet with a small box marked, and below it five false-colour frames labelled 1994, 1995, 1996, 1997 and 1998 with slanting lines tracking moving knots, labelled 6.0c, 5.5c, 6.1c and 6.0c. — Figure 7: Left: The jet coming from the elliptical galaxy M87 as viewed by the Hubble Space Telescope (HST). Right: Sequence of HST images showing motion at six times the speed of light. The slanting lines track the moving features, and the speeds are given in units of the velocity of light, $c$ . Credit: NASA / Space Telescope Science Institute / John Biretta.

This can be explained by understanding that the jet is offset by an angle $\theta$ from the sightline to Earth, and that the real speed of the plasma knot, $v$ , is less than $c$ , and from it we can define the scaled speed $\beta \equiv v/c$ .

a. Show, with use of an appropriate diagram, that the apparent value of the scaled transverse speed (for a jet coming towards us) is

\beta_{\mathrm{app}} = \frac{\beta\sin\theta}{1 - \beta\cos\theta}\,.

[5 т.]

b. Determine the relationship between $\beta$ and $\theta$ that maximises $\beta_{\mathrm{app}}$ for a given value of $\beta$ , and hence determine the minimum value of $\beta$ needed to give rise to superluminal apparent speeds (i.e. when $\beta_{\mathrm{app}}^{\mathrm{max}} > 1$ ). [You are given that a graph of $\beta_{\mathrm{app}}$ against $\theta$ has only one turning point in the range $0 < \theta < \pi$ , and that it is a maximum.] [4 т.]

Superluminal jets are not limited just to AGN, as they have also been observed from systems within our own galaxy. A particularly famous one is the ‘microquasar’ GRS 1915+105, which is a low mass X-ray binary consisting of a small star orbiting a black hole. A symmetrical jet with components approaching and receding from us is observed (as expected for jets coming from the poles of the black hole), and the apparent transverse motion of material in those jets has been measured using very high resolution radio imaging. Fender et. al (1999) measure these motions to be $\mu_a = 23.6\ \mathrm{mas\ day^{-1}}$ and $\mu_r = 10.0\ \mathrm{mas\ day^{-1}}$ for the approaching and receding jet respectively (1 mas = 1 milliarcsecond, a unit of angle, and there are 3600 arcseconds in a degree) and the distance to the system as 11 kpc.

c. Calculate $\beta_{\mathrm{app}}$ for both jets, and use your formula from part b. to calculate the minimum value of $\beta$ to explain the apparent superluminal motion. [5 т.]

In practice, for a given $\beta_{\mathrm{app}}$ the values of $\beta$ and $\theta$ are degenerate and it is unlikely that the orientation of the jet is such that $\beta_{\mathrm{app}}$ has been maximised, so the value in part c. is just a lower limit. However, since there are two jets then if we assume that they are from the same event (and so equal in speed but opposite in direction) we can break this degeneracy.

d. Given that $\mu_a$ and $\mu_r$ can be calculated (in radians s $^{-1}$ ) as

\mu_a = \frac{\beta\sin\theta}{1 - \beta\cos\theta}\frac{c}{D} \quad \text{and} \quad \mu_r = \frac{\beta\sin\theta}{1 + \beta\cos\theta}\frac{c}{D}\,,

derive a formula for the distance, $D$ , as a function of $\theta$ , $\mu_a$ and $\mu_r$ (i.e. independent of $\beta$ ), and hence calculate $\theta$ . [4 т.]

e. Show that $\beta\cos\theta$ can be expressed purely as a function of $\mu_a$ and $\mu_r$ , and hence use your value of $\theta$ to calculate the value of $\beta$ . [4 т.]

Since it is a binary system, we can gain information about the masses of the objects by looking at their period and radial velocity. Formally, the relationship is

\frac{(M_{\mathrm{BH}}\sin i)^3}{(M_{\mathrm{BH}} + M_{\star})^2} = \frac{P_{\mathrm{orb}}K_d^3}{2\pi G}\,,

where $M_{\mathrm{BH}}$ is the mass of the black hole, $M_{\star}$ is the mass of the orbiting star, $i$ is the inclination of the orbit, $P_{\mathrm{orb}}$ is the orbital period, and $K_d$ is the amplitude of the radial velocity curve. Normally the inclination can’t be measured, however if we assume that the orbit is perpendicular to the jets then $i = \theta$ and we can measure the mass of the black hole.

f. Greiner et. al (2001) measure $K_d = 140\ \mathrm{km\,s^{-1}}$ , $P_{\mathrm{orb}} = 33.5$ days, and a mass ratio for the two objects of $M_{\mathrm{BH}}/M_{\star} = 12.3$ . Using the assumption that $i = \theta$ , calculate $M_{\mathrm{BH}}$ . Give your answer in $M_{\odot}$ . [3 т.]

Отговори

Решение

Q4 – Superluminal Jets [25 marks]

Attempt at suitable diagram [be flexible] [2]

Derivation diagram: from point A a vertical arrow labelled 1st photon goes down To Earth; a slanted arrow v from A at angle θ to the vertical reaches C; B lies directly above C on a dashed horizontal line from A; from C a second vertical arrow labelled 2nd photon goes down To Earth.

Apparent distance moved in time $\Delta t$ ,

$AB = d_{app} = v\sin\theta\times\Delta t$ [1]

Difference in distances moved by photons,

$BE = c\Delta t - v\cos\theta\times\Delta t$

So apparent time delay,

$\Delta t_{app} = \frac{BE}{c} = \Delta t - \beta\cos\theta\times\Delta t$ [1]

So apparent speed,

$v_{app} = \frac{d_{app}}{\Delta t_{app}} = \frac{\Delta t\times v\sin\theta}{\Delta t - \beta\cos\theta\times\Delta t}$ [1] [5]

$\therefore \beta_{app} = \frac{v_{app}}{c} = \frac{\beta\sin\theta}{1 - \beta\cos\theta}$

[Allow any convincing derivation, given that it was a ‘show that’ question. Any derivation without a diagram scores 3 marks max. Most diagrams with two sightlines, v & θ should score at least 1 mark]

Want to find where $\frac{d\beta_{app}}{d\theta} = 0$ , so need to differentiate our expression [with product / quotient rule]

$\frac{d\beta_{app}}{d\theta} = \frac{\beta\cos\theta(1 - \beta\cos\theta) - \beta\sin\theta(\beta\sin\theta)}{(1 - \beta\cos\theta)^2} = 0$ [1]

$\therefore$ ( $\beta_{app}$ is maximised for a given value of $\beta$ when) $\beta = \cos\theta$ [1] [2]

Putting this into the formula from part a. and evaluating for when $\beta_{app} = 1$ ,

$\beta_{app}^{max} = \frac{\cos\theta\sin\theta}{1 - \cos\theta\cos\theta} = \cot\theta \quad \left(= \frac{\beta}{\sqrt{1 - \beta^2}}\right)$ [1]

If $\cot\theta = 1$ , $\theta = \frac{\pi}{4} \quad \therefore \beta_{min} = \cos\theta = \cos\frac{\pi}{4} = \frac{\sqrt{2}}{2}$ (= 0.707) [1] [2]

We need to convert μ from milliarcseconds a day to radians a second,

$v_a = \mu_ad_{system} = \frac{23.6\times10^{-3}}{3600}\times\frac{2\pi}{360}\times(11\times10^{3}\times3.09\times10^{16})$ [1]

$= 4.50\times10^{8}\ m\ s^{-1} \quad \therefore \beta_{app,a} = 1.50$ [1] [2]

Similarly, for the receding jet, $\beta_{app,r} = 0.636$ [2] [2]

[Values of $\mu_a$ and $\mu_r$ are $1.32\times10^{-12}\ \mathrm{rad\,s^{-1}}$ and $5.61\times10^{-13}\ \mathrm{rad\,s^{-1}}$ respectively]

$\beta_{app}^{max} = \cot\theta = 1.50 \quad \therefore \beta_{min,a} = \cos(\cot^{-1}1.50) = 0.832$ [1] [1]

[No requirement to use the data from the receding jet (which gives $\beta_{min,r} = 0.824$ )]

Rearranging the given formulae,

$D\mu_a(1 - \beta\cos\theta) = \beta c\sin\theta$ and $D\mu_r(1 + \beta\cos\theta) = \beta c\sin\theta$

$D\mu_a\mu_r\left(\frac{1}{\beta} - \cos\theta\right) = \mu_rc\sin\theta$ ① $\quad D\mu_a\mu_r\left(\frac{1}{\beta} + \cos\theta\right) = \mu_ac\sin\theta$ ② [1]

② – ① $\quad D\mu_a\mu_r(2\cos\theta) = (\mu_a - \mu_r)c\sin\theta$

$\therefore D = \frac{c}{2}\left(\frac{\mu_a - \mu_r}{\mu_a\mu_r}\right)\tan\theta$ [1] [2]

[This is just one approach – allow any reasonable attempt that gets the right answer]

$\theta = \tan^{-1}\left(\frac{2D}{c}\left(\frac{\mu_a\mu_r}{\mu_a - \mu_r}\right)\right)$

$= \tan^{-1}\left(\frac{2\times11\times10^{3}\times3.09\times10^{16}}{3.00\times10^{8}}\left(\frac{1.32\times10^{-12}\times5.61\times10^{-13}}{1.32\times10^{-12} - 5.61\times10^{-13}}\right)\right) \quad (= \tan^{-1}2.206)$ [1]

$= 65.6^{\circ} \quad (= 1.15\ \mathrm{rad})$ [1] [2]

Rearranging the given formulae again,

$D\mu_a(1 - \beta\cos\theta) = \beta c\sin\theta$ ① and $D\mu_r(1 + \beta\cos\theta) = \beta c\sin\theta$ ②

② – ① $\quad D\mu_r + D\mu_r\beta\cos\theta + D\mu_a\beta\cos\theta - D\mu_a = 0$ [1]

$\therefore \beta\cos\theta = \frac{\mu_a - \mu_r}{\mu_a + \mu_r}$ [1] [2]

[This is just one approach – allow any reasonable attempt that gets the right answer]

$\beta\cos\theta = \frac{23.6 - 10.0}{23.6 + 10.0} = 0.405$ [1]

$\therefore \beta = \frac{0.405}{\cos 65.6^{\circ}} = 0.981$ [1] [2]

[When they were discovered, these blobs of plasma in the jet from the microquasar were the fastest moving that anyone had ever seen in our galaxy!]

Rewriting the given equation in terms of the ratio $M_*/M_{BH}$ ,

$\frac{M_{BH}^3\sin^3\theta}{M_{BH}^2\left(1 + \frac{M_*}{M_{BH}}\right)^2} = \frac{P_{orb}K_d^3}{2\pi G} \quad \therefore M_{BH} = \frac{P_{orb}K_d^3}{2\pi G}\frac{\left(1 + \frac{M_*}{M_{BH}}\right)^2}{\sin^3\theta}$ [1]

$= \frac{(33.5\times24\times3600)\times(140\times10^{3})^3}{2\pi\times6.67\times10^{-11}}\times\frac{(1 + 12.3^{-1})^2}{\sin^3 65.6^{\circ}}$ [1]

$= 2.93\times10^{31}\ \mathrm{kg} = 14.7\,M_{\odot}$ [1] [3]

[Must be in $M_{\odot}$ for the final mark]

[At the time of discovery, this meant the black hole in the microquasar was the heaviest stellar black hole known. Subsequent measurements have made the distance a little closer than 11 kpc, and so the mass has been revised down, but until the discovery of the black holes behind detections of gravitational waves (with masses about double this) it was still one of heaviest known]

Questions proposed by

END OF PAPER

Questions proposed by:

Dr Alex Calverley (Royal Grammar School, Guildford)

John Hayton (University of Cambridge)

Josh Brown (University of Cambridge)

Оригинал в Архива: BAAO_2019.pdf · официални решения: BAAO_2019_solutions.pdf