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IOAA 2016, Теория, английски — Задача 10. Gravitational Lensing Telescope

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Всички материали от IOAA 2016

Theoretical Examination · 20 т.

Внимание: Бележка към темата

Both archived English question editions print T12 AstroSat, but both 27-page English worked keys substitute T12 Exoplanets. No matching AstroSat solution was found in the inspected full year inventory. Summary Answer Sheets and Cover Sheet named by the instructions were not found in this archived source pack. Instructions specify 13 questions totalling 300 points; both question editions contain 12 questions totalling 250 points. Printed numerical/formula discrepancies are retained: T8 uses c/lambda_max for nu_max and prints 1233 pi then 3813; T9 numerical velocity denominator lacks a power of ten and the stated semimajor axis is 20933 km; T11.4 prints a direct rather than inverse mass ratio in the radius-cubed relation; T11.7 prose says 166.67 Hz while its mass calculation uses 142.86 Hz. No inferred corrections are substituted.

Instructions to the Contestants

(1) Total time duration of this examination is five hours.

(2) This examination consists of

Questions	Marks each	Total
Question 1 to 5	10	10 x 5 = 50 marks
Question 6 to 10	20	20 x 5 = 100 marks
Question 11 to 13	50	50 x 3 = 150 marks
	Total	300 marks

(3) Inside the envelope you will find

a. Question paper in English

b. Question paper in your native language (if applicable)

c. Table of Constants in English

d. Table of Constants in your native language (if applicable)

e. A set of Summary Answersheets

f. Cover Sheet

(4) Use only black or blue pen for writing. For figures, you may use pencils.

(5) There is no negative marking.

(6) Some marks will be deducted if the final answer is given without detailed solution.

(7) Some marks will be deducted if the final answers have inappropriate number of significant digits, no units or wrong units.

(8) The necessary values of fundamental and astronomical constants should be taken from the Table of Constants provided to you.

(9) Instructions regarding using the answersheets

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

b. You should also show detailed solution in the blank space provided on each Summary Answersheet. If necessary, you can ask for extra blank sheets from the invigilators.

c. Use separate blank sheets for each question.

d. Write your contestant code, question number and page number on FRONT side of each blank sheet. Write your contestant code and page number on the BACK side of each blank sheet.

e. The page numbers should be continuous, i.e., if you use 20 sheets for the entire examination (including the Summary Answersheets), the page numbers should run from 1 to 40.

f. Write only inside the boxed area.

g. For the work that you do not want to be evaluated, cross that part out.

(10) At the end of the examination

a. On the Cover Sheet, clearly write page numbers for each question.

b. Check that you have written your contestant code on all pages.

c. Put the Cover Sheet, all Summary Answersheets, blank sheets and rough sheets inside the envelope. You may keep question paper and Table of Constants with you.

Table of Constants

Fundamental Constants

Name	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	$e$	$1.602\times10^{-19}\ \mathrm{C}$
Gravitational constant	$G$	$6.674\times10^{-11}\ \mathrm{N\,m^2\,kg^{-2}}$
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	$\lambda_mT$	$2.898\times10^{-3}\ \mathrm{m\,K}$
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	$m_n$	$1.675\times10^{-27}\ \mathrm{kg}$
Atomic mass unit	$u$	$1.661\times10^{-27}\ \mathrm{kg}$

Astronomical Constants

Name	Symbol	Value
Parsec	pc	$3.086\times10^{16}\ \mathrm{m}$
Astronomical unit	$a_\oplus$	$1.496\times10^{11}\ \mathrm{m}$
Mass of the Sun	$M_\odot$	$1.989\times10^{30}\ \mathrm{kg}$
Radius of the Sun	$R_\odot$	$6.955\times10^8\ \mathrm{m}$
Luminosity of the Sun	$L_\odot$	$3.826\times10^{26}\ \mathrm{W}$
Apparent magnitude of the Sun at midday		$-26.72\ \mathrm{mag}$
Solar constant		$1366\ \mathrm{W\,m^{-2}}$
Angular diameter of the Sun		$30'$
Mass of the Earth	$M_\oplus$	$5.972\times10^{24}\ \mathrm{kg}$
Radius of the Earth	$R_\oplus$	$6.371\times10^6\ \mathrm{m}$
Tropical year		$365.242$ solar days $=3.156\times10^7\ \mathrm{s}$

Условие

Einstein's General Theory of Relativity predicts bending of light around massive bodies. For simplicity, we assume that the bending of light happens at a single point for each light ray, as shown in the figure. The angle of bending, $\theta_{\mathrm{b}}$ , is given by

\theta_{\mathrm{b}} = \frac{2R_{\mathrm{sch}}}{r}

where $R_{\mathrm{sch}}$ is the Schwarzschild radius associated with that gravitational body. We call $r$ , the distance of the incoming light ray from the parallel $x$ -axis passing through the centre of the body, as the "impact parameter".

A light ray bends through angle theta_b at impact parameter r from a gravitating body.

A massive body thus behaves somewhat like a focusing lens. The light rays coming from infinite distance beyond a massive body, and having the same impact parameter $r$ , converge at a point along the axis, at a distance $f_r$ from the centre of the massive body. An observer at that point will benefit from huge amplification due to this gravitational focusing. The massive body in this case is being used as a Gravitational Lensing Telescope for amplification of distant signals.

T10.1 Consider the possibility of our Sun as a gravitational lensing telescope. Calculate the shortest distance, $f_{\min}$ , from the centre of the Sun (in A.U.) at which the light rays can get focused. [6 т.]

T10.2 Consider a small circular detector of radius $a$ , kept at a distance $f_{\min}$ centered on the $x$ -axis and perpendicular to it. Note that only the light rays which pass within a certain annulus (ring) of width $h$ (where $h \ll R_{\odot}$ ) around the Sun would encounter the detector. The amplification factor at the detector is defined as the ratio of the intensity of the light incident on the detector in the presence of the Sun and the intensity in the absence of the Sun.

Express the amplification factor, $A_{\mathrm{m}}$ , at the detector in terms of $R_{\odot}$ and $a$ . [8 т.]

T10.3 Consider a spherical mass distribution, such as dark matter in a galaxy cluster, through which light rays can pass while undergoing gravitational bending. Assume for simplicity that for the gravitational bending with impact parameter, $r$ , only the mass $M(r)$ enclosed inside the radius $r$ is relevant.

What should be the mass distribution, $M(r)$ , such that the gravitational lens behaves like an ideal optical convex lens? [6 т.]

Решение

T10.1

Solution:

The rays travelling closer to the gravitational body will bend more. Thus, we get shortest convergence point where the rays just grazing the solar surface will meet each other.

Solar-limb ray and the minimum focus distance.

\theta_{\mathrm{b}} = \frac{2R_{\mathrm{sch}}}{R_{\odot}} \simeq \frac{R_{\odot}}{f_{\min}}

[2.0]

\therefore\ f_{\min} = \frac{R_{\odot}^2}{2R_{\mathrm{sch}}} = \frac{R_{\odot}^2 c^2}{4GM_{\odot}}

[2.0]

= \frac{(6.955 \times 10^8 \times 2.998 \times 10^8)^2}{4 \times 6.674 \times 10^{-11} \times 1.989 \times 10^{30}}\ \mathrm{m}

= 8.188 \times 10^{13}\ \mathrm{m} = \frac{8.188 \times 10^{13}}{1.496 \times 10^{11}}\ \mathrm{AU}

\boxed{f_{\min} = 547.3\ \mathrm{AU}}

[2.0]

T10.2

Solution:

The following figure needs to be drawn. [1.0]

Annular rays meet a detector of radius a, with distances f_min and f_2.

The light bending from the surface of the Sun ( $r = R_{\odot}$ ) will intersect the detector at its centre, as it is kept at $f_{\min}$ .

The boundary of the detector will be intersected by a light ray with $r = R_{\odot} + h$ . This ray will intersect the $x$ -axis at a distance $f_2$ .

f_2 = \frac{(R_{\odot} + h)^2}{2R_{\mathrm{sch}}}

[2.0]

Same argument as in Part 1

For small angles,

a = (f_2 - f_{\min})\theta_2

= \left[\frac{(R_{\odot} + h)^2}{2R_{\mathrm{sch}}} - \frac{R_{\odot}^2}{2R_{\mathrm{sch}}}\right]\frac{2R_{\mathrm{sch}}}{(R_{\odot} + h)} = \frac{2R_{\odot}h + h^2}{R_{\odot} + h}

\simeq 2h

[2.0]

Let original intensity of the incoming radiation be $I_0$ .

The flux at the detector in the presence of Sun is $\Phi_{\odot} = I_0\ 2\pi\ R_{\odot}\ h$ [1.0]

The flux at the detector in the absence of Sun is $\Phi_0 = I_0\ \pi\ a^2$ [1.0]

The amplification is therefore

A_{\mathrm{m}} = \frac{\Phi_{\odot}}{\Phi_0} = \frac{I_0 2\pi R_{\odot} h}{I_0 \pi a^2} = \boxed{\frac{R_{\odot}}{a}}

[1.0]

T10.3

Solution:

Two impact parameters focus at the same point through a transparent spherical mass distribution.

All rays should focus at the same spot. This should be evident from figure drawn on answersheet or otherwise. [2.0]

Let there be two rays with impact parameters $r_1$ and $r_2$ . The corresponding distances of focus will be

f_i = \frac{r_i^2}{2r_{\mathrm{sch}_i}} = \frac{r_i^2 c^2}{4GM(r_i)}

[2.0]

Same argument as in Part 1

The requirement $f_1 = f_2$ implies

\frac{r_1^2}{r_2^2} = \frac{M(r_1)}{M(r_2)}

[1.0]

The required mass distribution is: $\boxed{M(r) \propto r^2}$ [1.0]

Оригинал в Архива: theory-qp.pdf · официални решения: theory_sol.pdf · допълнителни данни: tableofconstants.pdf · допълнителни данни: tableofconstants.pdf · допълнителни данни: Instructions_theory.pdf · допълнителни данни: Instructions_theory.pdf · допълнителни данни: theory.pdf · допълнителни данни: theory.pdf