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IOAA 2023, Анализ на данни — Задача 2. Isolated black hole

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Всички материали от IOAA 2023

Data Analysis · 75 т.

Table of Constants

Fundamental constants
Speed of light in vacuum	$c$	$=$	$2.998 \times 10^{8}\ \mathrm{m\,s^{-1}}$
Planck constant	$h$	$=$	$6.626 \times 10^{-34}\ \mathrm{J\,s}$
Boltzmann constant	$k_B$	$=$	$1.381 \times 10^{-23}\ \mathrm{J\,K^{-1}}$
Stefan-Boltzmann constant	$\sigma$	$=$	$5.670 \times 10^{-8}\ \mathrm{W\,m^{-2}\,K^{-4}}$
Elementary charge	$e$	$=$	$1.602 \times 10^{-19}\ \mathrm{C}$
Universal gravitational constant	$G$	$=$	$6.674 \times 10^{-11}\ \mathrm{N\,m^{2}\,kg^{-2}}$
Universal electric constant	$\epsilon_0$	$=$	$8.854 \times 10^{-12}\ \mathrm{m^{-3}kg^{-1}s^{4}A^{2}}$
Universal gas constant	$R$	$=$	$8.315\ \mathrm{J\,mol^{-1}\,K^{-1}}$
Avogadro constant	$N_\mathrm{A}$	$=$	$6.022 \times 10^{23}\ \mathrm{mol^{-1}}$
Wien’s displacement constant	$b = \lambda_m T$	$=$	$2.898 \times 10^{-3}\ \mathrm{m\,K}$
Mass of electron	$m_\mathrm{e}$	$=$	$9.109 \times 10^{-31}\ \mathrm{kg}$
Mass of proton	$m_\mathrm{p}$	$=$	$1.673 \times 10^{-27}\ \mathrm{kg}$
Mass of neutron	$m_\mathrm{n}$	$=$	$1.675 \times 10^{-27}\ \mathrm{kg}$
Mass of Helium nucleus	$m_\mathrm{He}$	$=$	$6.645 \times 10^{-27}\ \mathrm{kg}$
Atomic mass unit (a.m.u., Dalton)		$=$	$1.661 \times 10^{-27}\ \mathrm{kg}$

Astronomical data
Hubble constant	$H_0$	$=$	$70\ \mathrm{km\,s^{-1}\,Mpc^{-1}}$
North Ecliptic Pole ( $J2000.0$ )	$(\alpha_\mathrm{E},\ \delta_\mathrm{E})$		$(18^\mathrm{h}00^\mathrm{m}00^\mathrm{s}, +66^{\circ}33'39'')$
North Galactic Pole ( $J2000.0$ )	$(\alpha_\mathrm{G},\ \delta_\mathrm{G})$		$(12^\mathrm{h}51^\mathrm{m}26^\mathrm{s}, +27^{\circ}07'42'')$
1 jansky	1 Jy	$=$	$10^{-26}\ \mathrm{W\,m^{-2}\,Hz^{-1}}$
1 parsec	1 pc	$=$	$3.086 \times 10^{16}\ \mathrm{m}$
			$206\,265$ au
			$3.262$ ly
1 astronomical unit (au)	1 au	$=$	$1.496 \times 10^{11}\ \mathrm{m}$
1 sidereal day	$T_\mathrm{SD}$	$=$	$23.93444$ 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

$(1+x)^n \approx 1 + nx$

$(1+x)(1+y) \approx 1 + x + y$ if $x \ll 1$ and $y \ll 1$

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
Solar luminosity	$L_\odot$	$=$	$3.826 \times 10^{26}\ \mathrm{W}$
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		$\approx$	$8\ kpc$

The Earth and Moon
Obliquity of the ecliptic (Earth)	$\epsilon$	$=$	$23.5^{\circ}$
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	$\theta_\mathrm{L}$	$=$	$31'$
Inclination of the lunar orbit to the ecliptic		$=$	$05^{\circ}08'43''$
Inclination of the lunar equator to its orbital plane		$=$	$6.687^{\circ}$
Lunar sidereal month	$T_\mathrm{SL}$	$=$	$27.321661$ d
			$655.71986$ h
Synodic month		$=$	$29.530589$ d
Tropical month		$=$	$27.321582$ d
Anomalistic month (the average time between two successive Moon’s perigees)		$=$	$27.554550$ d
Draconic month (time between two successive transits of the Moon through the same node)		$=$	$27.212221$ d

The Solar System

Object	Mean radius [km]	Mass [kg]	Semimajor axis [au]	Eccentricity
Sun	$695\,700$	$1.988 \times 10^{30}$	—	—
Mercury	$2\,440$	$3.301 \times 10^{23}$	$0.387$	$0.206$
Venus	$6\,052$	$4.867 \times 10^{24}$	$0.723$	$0.007$
Earth	$6\,378$	$5.972 \times 10^{24}$	$1.000$	$0.016\,710$
Moon	$1\,737$	$7.346 \times 10^{22}$	$3.844 \times 10^{5}$ km	$0.054\,900$ (range $0.026 - 0.077$ )
Mars	$3\,390$	$6.417 \times 10^{23}$	$1.524$	$0.093$
Jupiter	$69\,911$	$1.898 \times 10^{27}$	$5.203$	$0.048$
Saturn	$58\,232$	$5.683 \times 10^{26}$	$9.537$	$0.054$
Uranus	$25\,362$	$8.681 \times 10^{25}$	$19.189$	$0.047$
Neptune	$24\,622$	$1.024 \times 10^{26}$	$30.070$	$0.009$

Data analysis - instructions

Do not touch envelopes until the start of the examination.
The data analysis examination lasts for 3 hours and is worth a total of 125 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.

Условие

In 2022, two independent groups reported the discovery of an isolated black hole based on observations of the gravitational microlensing event OGLE-2011-BLG-0462. In this problem, we will analyze data from the Hubble Space Telescope to reproduce their findings.

Gravitational microlensing occurs when the light of a distant star (the ‘source’) is bent and magnified by the gravitational field of an intervening object (the ‘lens’). The characteristic angular scale of gravitational microlensing events, called the angular Einstein radius $\theta_E$ , depends on the mass $M$ and distance $D_\ell$ from the Earth to the lens:

\theta_\mathrm{E} = \sqrt{\frac{4GM}{c^2}\frac{D_s - D_\ell}{D_s D_\ell}},

where $D_s$ is the distance to the source star. For typical microlensing events observed in the Milky Way, the source stars are in the Galactic bulge, near the Galactic center, so $D_s \approx 8$ kpc.

(a) Calculate the angular Einstein radius in milliarcseconds (mas) for an example lens of $1\ M_\odot$ located at a distance of 1 kpc. [2 т.]

Suppose that at time $t$ the lens and the source are separated by an angle $\theta \equiv u(t)\theta_E$ on the sky. Two images of the source are created on a line through the positions of the source and the lens, at angular distances $\theta_+$ and $\theta_-$ from the lens given by:

\theta_\pm = \frac{1}{2}\left(u \pm \sqrt{u^2+4}\right)\theta_\mathrm{E}.

These two images are magnified, relative to the unlensed brightness of the source. The absolute amplification of the images is:

A_\pm = \frac{1}{2}\left(\frac{u^2+2}{u\sqrt{u^2+4}} \pm 1\right).

The image below shows the geometry of the event. The position of the lens is marked as $L$ , the unlensed position of the source is marked as $S$ , while $A_+$ and $A_-$ mark the positions of the two images of the source. The dashed circle has a radius of one Einstein radius.

Geometry of the microlensing event: a dashed circle of one Einstein radius centred on the lens L; a straight line through L passes the unlensed source position S (small grey circle) inside the circle and reaches the image A+ (elongated ellipse) outside the circle on the left and the image A− (small ellipse) on the circle on the right; the segment A+–L is marked θ+ (blue), S–L is marked θ (orange), L–A− is marked θ− (green). — (A color version of this figure is available at a separate sheet.)

(b) Current telescopes cannot normally resolve this pair of images, but only measure the position of the image centroid, i.e. the brightness-weighted mean of the positions of the two images. Derive an expression for the angular separation $\theta_c$ of the image centroid relative to the lens as a function of $u$ and $\theta_E$ . [8 т.]

(c) Derive an expression for the source deflection $\Delta\theta$ , i.e. the difference between the location of the centroid and the unlensed position of the source, as a function of $u$ and $\theta_\mathrm{E}$ . What is the source deflection when the lens and the source are nearly perfectly aligned ( $u \approx 0$ )? [4 т.]

The source and lens are moving relative to each other in the sky. Thus, both the total magnification of the images and the position of the centroid changes with time, resulting in observable photometric and astrometric microlensing effects. For now, we assume that the source-lens relative motion is rectilinear.

The plot below shows the light curve of the gravitational microlensing event OGLE-2011-BLG-0462, discovered by the OGLE sky survey led by astronomers from the University of Warsaw. The solid line shows the best-fitting light curve model. The Einstein timescale of the event, i.e. the time needed for the source to move by one angular Einstein radius relative to the lens, was $t_\mathrm{E} = 247$ days. The event peaked on 21 July 2011 ( $\mathrm{HJD} = 2455763$ ). The minimal separation between the lens and the source was $u_0 \approx 0$ .

Light curve of OGLE-2011-BLG-0462: upper panel shows I magnitude (13 to 16, brighter upward) against HJD − 2450000 (days) from 5300 to 6300, with data points near 16.5 mag and a sharp peak reaching about 13.3 mag near 5763, and a blue best-fitting model line; lower panel shows residuals from −0.05 to 0.05 scattered around 0.00.

The table below shows the measured positions of the source star against the background objects in the East and North directions based on images from the Hubble Space Telescope.

HJD	E position (mas)	N position (mas)
$2455765.2$	$2.58 \pm 0.13$	$7.29 \pm 0.16$
$2455865.7$	$2.32 \pm 0.12$	$5.44 \pm 0.24$
$2456179.7$	$0.46 \pm 0.14$	$1.62 \pm 0.08$
$2456195.8$	$0.88 \pm 0.36$	$1.56 \pm 0.77$
$2456426.2$	$-1.02 \pm 0.21$	$-0.94 \pm 0.12$
$2456587.7$	$-2.04 \pm 0.07$	$-1.88 \pm 0.40$
$2456956.6$	$-4.54 \pm 0.25$	$-5.16 \pm 0.29$
$2457995.2$	$-11.14 \pm 0.12$	$-15.14 \pm 0.17$

(d) Plot the measured positions of the source star in the East and North directions as a function of time. [10 т.]

(e) The observed motion of the source star is the sum of two effects: rectilinear proper motion of the source and astrometric microlensing effects. Calculate the proper motion (in mas/year) of the source in the East and North directions and its uncertainty. [8 т.]

(f) After subtracting the effects of proper motion from the data, calculate and plot the total resultant astrometric deflection as a function of $u$ . Neglect uncertainty of the proper motion determination. [20 т.]

(g) Analyse the data to determine the angular Einstein radius $\theta_\mathrm{E}$ of the event and its uncertainty. (Hint: it may be helpful to linearise the expression for $\Delta\theta$ ). [16 т.]

(h) For long-timescale events such as OGLE-2011-BLG-0462, the rectilinear approximation of the relative lens-source proper motion is not strictly true and the orbital motion of the Earth has to be taken into account. This allows measurement of a dimensionless quantity called the microlensing parallax, defined as $\pi_\mathrm{E} = (\pi_l - \pi_s)/\theta_\mathrm{E}$ , where $\pi_l$ and $\pi_s$ are parallaxes of the lens and the source, respectively.

For this event $\pi_E = 0.095 \pm 0.009$ . Rearrange the expression for $\theta_\mathrm{E}$ given earlier to calculate the mass of the lens in solar masses and its uncertainty. [7 т.]

Отговори

Решение

Solution

(a)

\theta_\mathrm{E} = \sqrt{\frac{4GM}{c^2}\frac{D_s - D_\ell}{D_s D_\ell}} = \sqrt{\frac{4GM}{\mathrm{au}\,c^2}\left(\frac{\mathrm{au}}{D_\ell} - \frac{\mathrm{au}}{D_s}\right)} = 2.7\,\mathrm{mas}

The angular resolution of modern large ( $D \approx 10\,\mathrm{m}$ ) optical ( $\lambda = 550\,\mathrm{nm}$ ) telescopes is $\theta_0 = 1.22\lambda/D \approx 14\,\mathrm{mas}$ . Thus, $\theta_0 \gg \theta_\mathrm{E}$ , so the images created during microlensing events cannot be resolved by these telescopes.

(2 points)

(b)

\begin{aligned} \theta_c = \frac{\theta_+ A_+ + \theta_- A_-}{A_+ + A_-} &= \frac{\frac{1}{4}\left(u + \sqrt{u^2+4}\right)\left(\frac{u^2+2}{u\sqrt{u^2+4}} + 1\right) + \frac{1}{4}\left(u - \sqrt{u^2+4}\right)\left(\frac{u^2+2}{u\sqrt{u^2+4}} - 1\right)}{\frac{u^2+2}{u\sqrt{u^2+4}}}\theta_\mathrm{E} \\ &= \frac{\left(u + \sqrt{u^2+4}\right)\left(u^2 + 2 + u\sqrt{u^2+4}\right) + \left(u - \sqrt{u^2+4}\right)\left(u^2 + 2 - u\sqrt{u^2+4}\right)}{4(u^2+2)}\theta_\mathrm{E} \\ &= \frac{2u(u^2+2) + 2u(u^2+4)}{4(u^2+2)}\theta_\mathrm{E} = \frac{2u(2u^2+6)}{4(u^2+2)}\theta_\mathrm{E} = \frac{u(u^2+3)}{u^2+2}\theta_\mathrm{E} \end{aligned}

(8 points)

(c)

\Delta\theta = \theta_c - \theta = \frac{u(u^2+3)}{u^2+2}\theta_\mathrm{E} - u\theta_\mathrm{E} = \frac{u(u^2+3) - u(u^2+2)}{u^2+2}\theta_\mathrm{E} = \frac{u}{u^2+2}\theta_\mathrm{E}

(3 points)

\Delta\theta(u = 0) = 0

so there is no deflection when the lens and the source are nearly perfectly aligned.

(1 points)

(d)

Two scatter plots with error bars: left, Position E (mas) from about 2.6 down to −11.1 against Time (HJD-2450000) from about 5700 to 8000; right, Position N (mas) from about 7.3 down to −15.1 against the same time axis; eight data points each, decreasing with time.

(10 points, 5 points for each graph)

(e) I will use the fact that the first epoch of astrometric observations was taken close to the peak of the light curve (that is, $u_1 \approx 0$ , that is, almost no astrometric deflection). Similarly, astrometric deflection is close to zero for the last epoch. Thus,

\begin{aligned} \mu_E &\approx \frac{x_{E,8} - x_{E,1}}{t_8 - t_1} = -2.247 \pm 0.029\,\mathrm{mas/yr} \\ \mu_N &\approx \frac{x_{N,8} - x_{N,1}}{t_8 - t_1} = -3.674 \pm 0.038\,\mathrm{mas/yr} \end{aligned}

(8 points)

No points should be given if a student does not recognize that the deflection is zero during the first epoch, e.g., they try fitting a straight line to all data points.

(f) I fitted a straight line joining the first and last epoch data and then subtracted it from astrometric measurements. This is because the observed path of the source on the sky is the sum of two effects: the rectilinear proper motion of the source and the astrometric deflection:

\begin{aligned} x(E) &= x_1^E + (t - t_1)\mu_\mathrm{E} + \Delta\theta(E) \\ x(N) &= x_1^N + (t - t_1)\mu_\mathrm{N} + \Delta\theta(N), \end{aligned}

where $x_1^E$ is the East position of the source during the first epoch, $x_1^N$ is the North position of the source during the first epoch, $t_1$ is the time of the first observation, and $\Delta\theta(E)$ and $\Delta\theta(N)$ is the astrometric deflection due to microlensing in East and North direction, respectively.

The same two plots of Position E (mas) and Position N (mas) against Time (HJD-2450000), each with a blue straight line joining the first and the last data points; the intermediate points lie off the lines. — Figure 7: The blue lines join the first and the last epoch data points. The source would move along these lines if there wasn’t any black hole in front of it. However, the position of the source that was observed is deflected due to microlensing effects by the black hole.

Thus, the astrometric deflection during $i$ th epoch is:

\begin{aligned} \Delta\theta(E)_i &= x_i^E - x_1^E - (t_i - t_1)\mu_\mathrm{E} \\ \Delta\theta(N)_i &= x_i^N - x_1^N - (t_i - t_1)\mu_\mathrm{N} \end{aligned}

and the total deflection is:

\Delta\theta_i = \sqrt{\Delta\theta(E)_i^2 + \Delta\theta(N)_i^2}.

I will also use the fact that $u_0 \approx 0$ , so $u = (t - t_0)/t_E$ , where $t_0 = 2455763$ is the peak time. Results of my calculations are shown in the table below.

Two plots with error bars against Time (HJD-2450000) from about 5700 to 8000: left, Deflection E (mas) between 0 and about 0.95; right, Deflection N (mas) between 0 and about −1.6; both are zero at the first and the last epoch. — Figure 8: These figures show the astrometric deflection in East and North directions induced by the black hole. (Students are not required to make these plots.)

Plot of Deflection (mas) from 0 to 2.5 against u from 0 to 9 with error bars: zero at u ≈ 0, rising to about 1.6–1.7 mas near u ≈ 1.7–2.7, then falling to 1.0 at u ≈ 3.3, 0.5 at u ≈ 4.8 and 0 at u ≈ 9. — Figure 9: Total astrometric deflection as a function of u. Students are asked to make this plot in part (f).

Epoch	$u$	$\Delta\theta$ (E,mas)	$\Delta\theta$ (N,mas)	$\Delta\theta$ (mas)
1	$0.01$	$0.00 \pm 0.13$	$-0.00 \pm 0.16$	$0.00 \pm 0.21$
2	$0.42$	$0.36 \pm 0.12$	$-0.84 \pm 0.24$	$0.91 \pm 0.27$
3	$1.69$	$0.43 \pm 0.14$	$-1.50 \pm 0.08$	$1.56 \pm 0.16$
4	$1.75$	$0.95 \pm 0.36$	$-1.40 \pm 0.77$	$1.69 \pm 0.85$
5	$2.69$	$0.47 \pm 0.21$	$-1.58 \pm 0.12$	$1.65 \pm 0.24$
6	$3.34$	$0.44 \pm 0.07$	$-0.90 \pm 0.40$	$1.00 \pm 0.41$
7	$4.83$	$0.21 \pm 0.25$	$-0.47 \pm 0.29$	$0.51 \pm 0.38$
8	$9.04$	$0.00 \pm 0.12$	$-0.00 \pm 0.17$	$0.00 \pm 0.21$

(20 points)

(g) We would like to fit the function $\Delta\theta = \frac{u}{u^2+2}\theta_\mathrm{E}$ to the data. Thus, my “new” independent variable would be $x' = u/(u^2+2)$ . Now, I would like to fit the function $y' = \theta_\mathrm{E} x'$ , where $y' = \Delta\theta$ . Thus

\theta_\mathrm{E} = \frac{\sum y'_i x'_i/\sigma_i^2}{\sum x'^2_i/\sigma_i^2} \pm \frac{1}{\sqrt{\sum x'^2_i/\sigma_i^2}} = 4.5 \pm 0.4\,\mathrm{mas}

Plot of Deflection (mas) from 0 to 2.5 against u/(u^2 + 2) from 0 to 0.35 with error bars and a blue straight-line fit through the origin rising to about 1.55 mas at 0.35.

Alternative solution:

The gravitational deflection is described by the formula $\Delta\theta = \frac{u}{u^2+2}\theta_\mathrm{E}$ . It can be demonstrated that this function reaches a local maximum for $u = \sqrt{2}$ with $\Delta\theta_\mathrm{max} = \frac{\sqrt{2}}{4}\theta_\mathrm{E} = 0.354\theta_\mathrm{E}$ . From the data, we can estimate the maximum deflection of $\Delta\theta_\mathrm{max} = 1.59 \pm 0.14\,\mathrm{mas}$ . Thus, $\theta_\mathrm{E} = 4.5 \pm 0.4\,\mathrm{mas}$ .

(16 points)

(h)

Let $\pi_\mathrm{rel} = \pi_l - \pi_s$ be the relative lens–source parallax. From the definition of the angular Einstein radius we have

\theta_\mathrm{E} = \sqrt{\frac{4GM}{c^2}\frac{D_s - D_l}{D_s D_l}} = \sqrt{\frac{4GM}{c^2}\left(\frac{1}{D_l} - \frac{1}{D_s}\right)} = \sqrt{\frac{4GM}{c^2\mathrm{au}}\left(\frac{\mathrm{au}}{D_l} - \frac{\mathrm{au}}{D_s}\right)} = \sqrt{\frac{4GM\pi_\mathrm{rel}}{c^2\mathrm{au}}}.

From the definition of the microlensing parallax $\pi_\mathrm{rel} = \pi_\mathrm{E}\theta_E$ , so this equation becomes:

\theta_\mathrm{E} = \sqrt{\frac{4GM\pi_\mathrm{E}\theta_\mathrm{E}}{c^2\mathrm{au}}},

and hence

M = \frac{\theta_\mathrm{E} c^2 AU}{4G\pi_\mathrm{E}} = 5.8 \pm 0.8 M_\odot.

The uncertainty on $M$ can be determined from the relation:

\frac{\Delta M}{M} = \sqrt{\left(\frac{\Delta\theta_E}{\theta_\mathrm{E}}\right)^2 + \left(\frac{\Delta\pi_\mathrm{E}}{\pi_\mathrm{E}}\right)^2}

(7 points)

Grading

a	Correct value of $\theta_\mathrm{E}$	2
b	Correct formula	8
c	Correct formula	4
d	Correct plots (5 pts for each plot)	10
e	Correct $\mu_E$ within $3\sigma$	3
	Correct $\mu_E$ within $5\sigma$	1
	Incorrect $\mu_E$ ( $> 5\sigma$ )	0
	Uncertainty on $\mu_E$	1
	Correct $\mu_N$ within $3\sigma$	3
	Correct $\mu_N$ within $5\sigma$	1
	Incorrect $\mu_N$ ( $> 5\sigma$ )	0
	Uncertainty on $\mu_N$	1
	No points should be given if the student does not recognize that the deflection is zero during the first epoch
f	Calculation of the impact parameter $u$ for all epochs	4
	Calculation of the total deflection	6
	Calculation of the uncertainties	5
	Graph	5
	Grant full points if results are correct for 7 or 8 epochs
	Grant 60% points if results are correct for 5 or 6 epochs
	Grant 0 points if results are correct for 4 or less than 4 epochs
g	Correct result with an estimate of the uncertainty	16
	Correct results without the estimate of the uncertainty	8
h	Correct result	7
	TOTAL	75

Grading of the graphs:

Students can get 5 pts for each correct graphs
data points with error bars 3 pts
axis labels with units, tick labels - 1 pts
graph is clear, fills the entire area - 1 pts

Grant full points if the graph shows correct data for 7 or 8 epochs. Grant 60% points if the graph shows correct data for 5 or 6 epochs. Grant 0 points if the graph shows 4 or less than 4 epochs.

Оригинал в Архива: DA.pdf · официални решения: DA_sol.pdf