Съдържание

Условие 1.1.1.2.1.3.1.4.Summary Answer Sheet — (D1) Dust and Young Stars in Galaxies Graph Paper Решение

IOAA 2018, Анализ на данни — D1. Dust and Young Stars in Star-forming Galaxies

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Всички материали от IOAA 2018

12th International Olympiad on Astronomy and Astrophysics — Data Analysis Competition · 75 т.

Указания

The data analysis competition lasts for 5 hours and is worth a total of 150 points.
Dedicated IOAA Summary Answer Sheets are provided for writing your answers. Enter the final

answers into the appropriate boxes in the corresponding Summary Answer Sheet. On each Answer
Sheet, please fill in

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guide will escort you out of the competition room.

A list of constants is given on the next page.

Table of constants

Group	Quantity	Value
Earth	Mass	$5.98\times10^{24}\ \mathrm{kg}$
Earth	Radius	$6.38\times10^6\ \mathrm{m}$
Earth	Acceleration of gravity	$9.81\ \mathrm{m\,s^{-2}}$
Earth	Obliquity of Ecliptic	23°27′
Earth	Length of Tropical Year	365.2422 mean solar days
Earth	Length of Sidereal Year	365.2564 mean solar days
Earth	Albedo	0.39
Moon	Mass	$7.35\times10^{22}\ \mathrm{kg}$
Moon	Radius	$1.74\times10^6\ \mathrm{m}$
Moon	Mean Earth-Moon distance	$3.84\times10^8\ \mathrm{m}$
Moon	Orbital inclination with the Ecliptic	5.14°
Moon	Albedo	0.14
Moon	Apparent magnitude (mean full moon)	−12.74
Sun	Mass	$1.99\times10^{30}\ \mathrm{kg}$
Sun	Radius	$6.96\times10^8\ \mathrm{m}$
Sun	Luminosity	$3.83\times10^{26}\ \mathrm{W}$
Sun	Absolute Magnitude	4.80 mag
Sun	Surface Temperature	5772 K
Sun	Angular diameter at Earth	30′
Sun	Orbital velocity in Galaxy	$220\ \mathrm{km\,s^{-1}}$
Sun	Distance from Galactic center	8.5 kpc
Physical constants	1 au	$1.50\times10^{11}\ \mathrm{m}$
Physical constants	1 pc	206265 au
Physical constants	Gravitational constant (G)	$6.67\times10^{-11}\ \mathrm{N\,m^2\,kg^{-2}}$
Physical constants	Planck constant (h)	$6.62\times10^{-34}\ \mathrm{J\,s}$
Physical constants	Boltzmann constant	$1.38\times10^{-23}\ \mathrm{J\,K^{-1}}$
Physical constants	Stefan-Boltzmann constant	$5.67\times10^{-8}\ \mathrm{W\,m^{-2}\,K^{-4}}$
Physical constants	Hubble constant	$67.8\ \mathrm{km\,s^{-1}\,Mpc^{-1}}$
Physical constants	Speed of light in vacuum	299792458 m/s
Physical constants	Magnetic Permeability of free space	$4\pi\times10^{-7}\ \mathrm{H\,m^{-1}}$
Physical constants	1 Jansky (Jy)	$10^{-26}\ \mathrm{W\,m^{-2}\,Hz^{-1}}$

\Delta\log_{10}(x)=\frac{\Delta x}{x\ln10}.

Бележки за изданието

The independently archived question and solution PDFs are the source editions used here. D1.4 remains in the questions, answer form and worked key, but the key explicitly marks its 12 marks “discarded”; printed totals of 75 per problem and 150 overall are retained. Section 1.3 calls $F_{1600}$ “unattenuated” although section 1.2 defines it as observed; the source wording is preserved. The answer form labels the mass-function field D2.2.2 and omits a separate orbital-radius field. D2.3.3 calls the $\sin i+\Delta\sin i$ curve an upper limit and the $\sin i-\Delta\sin i$ curve a lower limit; the official figure places them in the opposite vertical order, as the mass function requires. In the key, equation (19) prints a strict inequality at the radius bound, equation (32) leaves the cgs-to-SI conversion implicit, and the D2.3.4 rubric assigns only four of its printed five marks. These source inconsistencies have not been silently corrected.

The solution PDF repeats an older wording of the questions: $B$ replaces $Q$ , $f(\beta)$ replaces $S(\beta)$ , D1.1.3 explicitly requests least-squares fitting, Table 1 is called UV photometry, and the dex explanation is written $\mathrm{dex}(x)=10^x$ . It additionally gives the 21 µJy ALMA limit and explicit redshift-factor hint reproduced with D1.4. Its APOGEE velocities are rounded to 34.85 and 42.57 km/s; its D2.3.2 options are (1) Blue Giant (2) Blue Dwarf (3) Red Giant (4) Red Dwarf (5) Main-Sequence Star (6) White Dwarf. The question edition’s data and five options are retained above; the official answer remains (3) Red Giant.

Условие

As a by-product of the star-forming process in a galaxy, interstellar dust can significantly absorb stellar light in ultraviolet (UV) and optical bands, and then re-emit in far-infrared (FIR), which corresponds to a wavelength range of $10$ – $300\,\mu\mathrm m$ .

1.1.

In the UV spectrum of a galaxy, the major contribution is from the light of the young stellar population generated in recent star-formation processes, thus the UV luminosity can act as a reliable tracer of the star-formation rate (SFR) of a galaxy. Since the observed UV luminosity is strongly affected by dust attenuation, extragalactic astronomers define an index called the UV continuum slope ( $\beta$ ) to quantify the shape of the UV continuum:

f_\lambda=Q\cdot\lambda^\beta

where $f_\lambda$ is the monochromatic flux of the galaxy at a given wavelength $\lambda$ (in the unit of $\mathrm{W\,m^{-3}}$ ) and $Q$ is a scaling constant.

(D1.1.1) AB magnitude is a specific magnitude system. The AB magnitude is defined as:

m_{\mathrm{AB}}=-2.5\log\frac{f_\nu}{3631\ \mathrm{Jy}}

The AB magnitude of a typical galaxy is roughly constant in the UV band. What is the UV continuum slope of this kind of galaxy? (Hint: $f_\nu\Delta\nu=f_\lambda\Delta\lambda$ ) [6 т.]

(D1.1.2) Table 1 presents the observed IR photometry results for a $z=6.60$ galaxy called CR7. Plot the AB magnitude of CR7 versus the logarithm of the rest-frame wavelength on graph paper and labelled as Figure 1. [12 т.]

(D1.1.3) Calculate CR7’s UV slope, plot the best-fit UV continuum on Figure 1 and make a comparison with the results you obtained in (D1.1.1). Is it dustier than the typical galaxy in (D1.1.1)? Please answer with [YES] or [NO]. (Hint: Express $m_{\mathrm{AB}}$ as a function of $\lambda$ and $m_{1600}$ , where $m_{1600}$ is the AB magnitude at $\lambda_0=160\ \mathrm{nm}$ (1600 Å))

Table 1. (Observed Frame) IR Photometry of CR7 at z = 6.60

Band	Y	J	H	K
Central Wavelength (µm)	1.05	1.25	1.65	2.15
AB Magnitude	24.71 ± 0.11	24.63 ± 0.13	25.08 ± 0.14	25.15 ± 0.15

[5 т.]

1.2.

Under the assumption that dust grains in the galaxy absorb the energy of UV photons and re-emit it by blackbody radiation, the relation between the UV continuum slope ( $\beta$ ), UV brightness (at 1600 Å) and FIR brightness could be established:

\mathrm{IRX}\equiv\log\left(\frac{F_{\mathrm{FIR}}}{F_{1600}}\right)=S(\beta)

where $F_{\mathrm{FIR}}$ is the observed far-infrared flux and $F_{1600}$ is the observed flux at rest-frame wavelength 160 nm (1600 Å) (The “flux” $F_\lambda$ is defined as $F_\lambda=\lambda\cdot f_\lambda$ ). Table 2 presents 20 measurements of $\beta$ , $F_{\mathrm{FIR}}$ and $F_{1600}$ in nearby galaxies (Meurer et al. 1999).

Table 2. UV slope, flux and FIR flux of 20 nearby galaxies

Galaxy Name	UV Slope $\beta$	$\log(F_{1600}/(10^{-3}\ \mathrm{W\,m^{-2}}))$	$\log(F_{\mathrm{FIR}}/(10^{-3}\ \mathrm{W\,m^{-2}}))$
NGC4861	-2.46	-9.89	-9.97
Mrk 153	-2.41	-10.37	-10.92
Tol 1924-416	-2.12	-10.05	-10.17
UGC 9560	-2.02	-10.38	-10.41
NGC 3991	-1.91	-10.14	-9.80
Mrk 357	-1.80	-10.58	-10.37
Mrk 36	-1.72	-10.68	-10.94
NGC 4670	-1.65	-10.02	-9.85
NGC 3125	-1.49	-10.19	-9.64
UGC 3838	-1.41	-10.81	-10.55
NGC 7250	-1.33	-10.23	-9.77
NGC 7714	-1.23	-10.16	-9.32
NGC 3049	-1.14	-10.69	-9.84
NGC 3310	-1.05	-9.84	-8.83
NGC 2782	-0.90	-10.50	-9.33
NGC 1614	-0.76	-10.91	-8.84
NGC 6052	-0.72	-10.62	-9.48
NGC 3504	-0.56	-10.41	-8.96
NGC 4194	-0.26	-10.62	-8.99
NGC 3256	0.16	-10.32	-8.44

(D1.2.1) Based on the data given in Table 2, plot the $\mathrm{IRX}-\beta$ diagram on graph paper and labelled as Figure 2 and find a linear fit to the data. Write down your best-fit equation (i.e. $\mathrm{IRX}=a\cdot\beta+b$ ) by the side of your diagram. [14 т.]

(D1.2.2) Quantify the dispersion (in ‘units’ of dex, where for example, $\log(10^9)-\log(10^4)=5\ \mathrm{dex}$ ) between the observed $\mathrm{IRX}_{\mathrm{obs}}$ and predicted $\mathrm{IRX}_{\mathrm{pred}}$ using the following equation:

\sigma=\sqrt{\frac{\sum(\Delta\mathrm{IRX}_i)^2}{N-1}}\quad\text{(unit: dex)},\qquad\Delta\mathrm{IRX}_i=\mathrm{IRX}_{i,\mathrm{obs}}-\mathrm{IRX}_{i,\mathrm{pred}}.

[6 т.]

1.3.

Under the previous assumption of the energy transfer process, the ratio of $F_{\mathrm{FIR}}$ to $F_{1600}$ can be expressed as:

\frac{F_{\mathrm{FIR}}}{F_{1600}}\approx10^{0.4A_{1600}}-1

Where $F_{1600}$ is the unattenuated flux and $A_\lambda$ is the dust absorption (in magnitudes) as a function of wavelength $\lambda$ .

(D1.3.1) Express $A_{1600}$ as a function of IRX. [6 т.]

(D1.3.2) Based on Table 2 data and the function of $A_{1600}(\mathrm{IRX})$ you derived above, plot the $A_{1600}-\beta$ diagram on graph paper and label it as Figure 3 and find a linear fit to the data. Write down your best-fit equation (i.e. $A_{1600}=a^\prime\cdot\beta+b^\prime$ ) by the side of your diagram. [12 т.]

(D1.3.3) If your linear model in (D1.3.2) is correct, what is the expected UV continuum slope $\beta_0$ of a dust-free galaxy? [2 т.]

1.4.

After establishing the local relation between UV continuum slope and IRX, we could probably test this empirical law in the high-redshift universe. In 2016, researchers obtained an Atacama Large Millimeter / submillimeter Array (ALMA) observation of CR7, and the FIR continuum corresponded to a $3\sigma$ upper limit of an FIR flux of $1.5\times10^{-19}\ \mathrm{W\,m^{-2}}$ .

(D1.4.1) Calculate the IRX of CR7. Is it an upper limit or lower limit?

Hint: here $F_{1600}$ should be written in the form of:

F_{1600}=\lambda_0\cdot f_{1600}

where $\lambda_0=160\ \mathrm{nm}$ (1600 Å) and $f_{1600}$ is the observed flux in the rest-frame [6 т.]

(D1.4.2) Is the current observation long enough to show any deviation of CR7 from the $\mathrm{IRX}-\beta$ relationship you just derived in the local universe? Please answer with [YES] or [NO] on the summary answer sheet, give the IRX difference and show the working used to calculate it on the answer sheet. [6 т.]

Summary Answer Sheet — (D1) Dust and Young Stars in Galaxies

Student’s Code: □□□–□□

(D1.1.1) The UV slope of such a typical galaxy is □;

(D1.1.2) See Figure 1;

(D1.1.3) The UV slope of CR7 is □; [YES] or [NO]? □;

(D1.2.1) See Figure 2; The best-fit Equation is: □;

(D1.2.2) The dispersion is □ dex;

(D1.3.1) $A_{1600}$ as a function of IRX: □;

(D1.3.2) See Figure 3; The Best-fit Equation is: □;

(D1.3.3) Expected UV slope of a dust-free galaxy: □;

(D1.4.1) The IRX of CR7 is □; Here the value is a □ limit (please fill with [upper] or [lower]);

(D1.4.2) Is the current observation deep enough ([YES] or [NO])? □.

Graph Paper

Student’s Code: □□□–□□

Blank logarithmic graph paper sheet; the left scale is labelled 10^3.00 to 10^3.50 in steps of 0.05, the right scale 1000 to 3200 in steps of 200. — Graph Paper（log）

Решение

Official solution and grading. Section 1.4 is explicitly marked “discarded” in the official key; its printed questions and worked answers are retained.

1.1. UV continuum slope ( $\beta$ ) [23 marks]

(D1.1.1) The AB magnitude of a typical galaxy is roughly constant, therefore:

f_\nu=\mathrm{const}.\qquad (1)

Conversion between $f_\nu$ and $f_\lambda$ could be expressed as:

f_\nu\Delta\nu=f_\nu\left(\frac{c}{\lambda-\Delta\lambda}-\frac{c}{\lambda}\right)=\frac{f_\nu c}{\lambda^2}\Delta\lambda=f_\lambda\Delta\lambda\Rightarrow f_\lambda\propto\lambda^{-2}\qquad (2)

Therefore, the UV continuum slope should be $\beta=-2$ .

[Equation (2): 3 marks; final result: 3 marks] [6 т.]

(D1.1.2) First of all, we need to convert all the wavelength into rest-frame value, using the definition of redshift:

z=\frac{\lambda_{\mathrm{obs}}}{\lambda_{\mathrm{rest}}}-1\Rightarrow\lambda_0=\frac{\lambda_{\mathrm{obs}}}{1+z}\qquad (3)

Thus, the rest-frame wavelength of four bands are listed in the following table:

Band	Y	J	H	K
Rest-frame Wavelength (Å)	1382	1645	2171	2829

The plot is shown as Figure 1. Detailed standards of grading the plot are in the solution of (D1.1.3).

[Equation (3): 2 marks; Correct wavelengths: 2 marks; Figure: 8 marks] [12 т.]

(D1.1.3) Here we can express the AB magnitude of CR7 (UV continuum only) using $\beta$ and $\lambda_{\mathrm{rest}}$ :

m_{\mathrm{AB}}=-2.5\log f_\nu+C=-2.5\log(\lambda^2 f_\lambda)+C=-2.5(\beta+2)\log\left(\frac{\lambda}{1600\ \text{Å}}\right)+m_{1600}\qquad (4)

Here $m_{1600}$ is the AB magnitude at rest-frame 1600 Å (observed-frame $(1+z)\cdot1600\ \text{Å}$ ). After writing this out, we can fit the data with a linear function $f(\log\lambda)=a\log\lambda+b$ , where the slope $a=-2.5(\beta+2)$ . Least squares fitting result is $a=1.7$ , therefore $\beta=-2.7$ .

(Note: It is not necessary for examinees to express $m_{\mathrm{AB}}$ in the same format as Equation 4. If an examinee derived the correct value of $\beta$ , he or she will get full marks in this part.)

The best-fit UV continuum should be plot as a straight line on Figure 1. [beta: 3 marks]

Grading standard of Figure 1: (1) [1 mark] The figure is large enough on the graph paper (larger than 50% of usable space); (2) [2 marks] horizontal axis is rest-frame wavelength and plotted in log scale; (3) [1 marks] vertical axis is AB magnitude and plotted reversely in linear scale (i.e. brighter magnitude is higher in the panel); (4) [1 mark] Ticks on both axes are properly labeled with suitable spacing; (5) [1 mark] Data points and best-fit continuum are plotted properly; (6) [2 mark] Error bars of data points are presented properly;

Compared to the galaxy in (D1.1), CR7 shows smaller $\beta$ value, indicating its UV continuum is bluer than typical galaxy. Therefore, there should be less dust content in CR7, and the correct answer should be [NO]. [2 marks]

Figure 1. Official fit to the CR7 photometry.

[5 т.]

1.2. UV slope and IRX [20 marks]

(D1.2.1) For the galaxy sample in Table 2, the IRX could be calculated as:

\mathrm{IRX}=\log F_{\mathrm{FIR}}-\log F_{1600}\qquad (5)

[Equation (5): 1 mark]

Therefore, the final dataset we need for plotting is:

Galaxy Name	UV Slope $\beta$	IRX
NGC4861	-2.46	-0.08
Mrk 153	-2.41	-0.55
Tol 1924-416	-2.12	-0.12
UGC 9560	-2.02	-0.03
NGC 3991	-1.91	0.34
Mrk 357	-1.80	0.21
Mrk 36	-1.72	-0.26
NGC 4670	-1.65	0.17
NGC 3125	-1.49	0.55
UGC 3838	-1.41	0.26
NGC 7250	-1.33	0.46
NGC 7714	-1.23	0.84
NGC 3049	-1.14	0.85
NGC 3310	-1.05	1.01
NGC 2782	-0.90	1.17
NGC 1614	-0.76	2.07
NGC 6052	-0.72	1.14
NGC 3504	-0.56	1.45
NGC 4194	-0.26	1.63
NGC 3256	0.16	1.88

The plot is shown as Figure 2. Best-fit function is:

\mathrm{IRX}=0.98\cdot\beta+1.96\qquad (6)

[Equation (6): 3 marks; Figure 2: 10 marks]

which should be plotted as a straight line on Figure 2.

Grading standard of Figure 2: (1) [1 mark] The figure is large enough on the graph paper (larger than 50% of usable space); (2) [2 marks] horizontal axis is $\beta$ , vertical axis is IRX, and both axes are plotted in linear scale; (3) [1 mark] Ticks on both axes are properly labeled with suitable spacing; (4) [5 marks] Data points and best-fit function are plotted properly; (5) [1 mark] Best-fit equation is noted correctly on the same page of Figure 2. [14 т.]

(D1.2.2) With the IRX- $\beta$ relation we derived above, we can calculate the difference between observed IRX and expected value (inferred by $\beta$ ) of each galaxy:

Name	$\mathrm{IRX}_{\mathrm{obs}}$	$\mathrm{IRX}_{\mathrm{exp}}$	$\Delta\mathrm{IRX}$
NGC4861	-0.08	-0.45	0.37
Mrk 153	-0.55	-0.40	-0.15
Tol 1924-416	-0.12	-0.12	0.00
UGC 9560	-0.03	-0.02	-0.01
NGC 3991	0.34	0.09	0.25
Mrk 357	0.21	0.20	0.01
Mrk 36	-0.26	0.27	-0.53
NGC 4670	0.17	0.34	-0.17
NGC 3125	0.55	0.50	0.05
UGC 3838	0.26	0.58	-0.32
NGC 7250	0.46	0.66	-0.20
NGC 7714	0.84	0.75	0.09
NGC 3049	0.85	0.84	0.01
NGC 3310	1.01	0.93	0.08
NGC 2782	1.17	1.08	0.09
NGC 1614	2.07	1.22	0.85
NGC 6052	1.14	1.25	-0.11
NGC 3504	1.45	1.41	0.04
NGC 4194	1.63	1.71	-0.08
NGC 3256	1.88	2.12	-0.24

Therefore, the dispersion of IRX is:

\sigma=\sqrt{\frac{\sum_{i=1}^{N}\Delta\mathrm{IRX}_i^2}{N-1}}=0.28\ \mathrm{dex}\qquad (7)

[Equation (7): 6 marks] [6 т.]

1.3. IRX and $A_{1600}$ : [20 marks]

(D1.3.1) Base on the two assumptions we made in Question (3), we can simplify $F_{\mathrm{FIR}}/F_{1600}$ as:

\frac{F_{\mathrm{FIR}}}{F_{1600}}\approx10^{0.4A_{1600}}-1\qquad (8)

\therefore A_{1600}=2.5\log(1+10^{\mathrm{IRX}})\qquad (9)

[Equation (9): 6 marks] [6 т.]

(D1.3.2) Now we can calculate $A_{1600}$ of all the galaxy in Table 2 with Equation 9:

Name	$\beta$	$A_{1600}$
NGC4861	-2.46	0.66
Mrk 153	-2.41	0.27
Tol 1924-416	-2.12	0.61
UGC 9560	-2.02	0.72
NGC 3991	-1.91	1.26
Mrk 357	-1.80	1.05
Mrk 36	-1.72	0.48
NGC 4670	-1.65	0.99
NGC 3125	-1.49	1.64
UGC 3838	-1.41	1.13
NGC 7250	-1.33	1.47
NGC 7714	-1.23	2.25
NGC 3049	-1.14	2.27
NGC 3310	-1.05	2.63
NGC 2782	-0.90	3.00
NGC 1614	-0.76	5.18
NGC 6052	-0.72	2.93
NGC 3504	-0.56	3.66
NGC 4194	-0.26	4.10
NGC 3256	0.16	4.71

Then we can plot $A_{1600}-\beta$ relation as Figure 3. The best-fit equation is:

A_{1600}=1.92\cdot\beta+4.62\qquad (10)

[Equation (10): 2 marks; Figure 3: 10 marks]

Grading standard of Figure 3: (1) [1 mark] The figure is large enough on the graph paper (larger than 50% of usable space); (2) [2 marks] horizontal axis is $\beta$ , vertical axis is $A_{1600}$ , and both axes are plotted in linear scale; (3) [1 mark] Ticks on both axes are properly labeled with suitable spacing; (4) [5 marks] Data points and best-fit function are plotted properly; (5) [1 mark] Best-fit equation is noted correctly on the same page of Figure 3.

[12 т.]

(D1.3.3) If Equation 10 is correct, then for a dust-free galaxy, the dust attenuation $A_{1600}$ should equal to 0. Therefore, we can derive $\beta_0\approx-2.4$ , which is the expect UV continuum slope of a “dust-free” galaxy.

(Note: The $\beta_0$ result here is conflict with some observations, for example, the UV slope of CR7 which we just derived. In fact, the linear model here is not necessarily correct and it is just a simplification of complex model: we have made two very rough assumption to derive the $A_{1600}-\mathrm{IRX}$ relation, and other physical parameters of galaxy, e.g. metallicity, will also influence the overall $A_{1600}-\beta$ relation.) [2 т.]

1.4. Extend to high-redshift universe: [12 marks] (discarded)

The problem repeated on solution page 3 additionally states that the FIR continuum was undetected at a $3\sigma$ upper limit of $21\,\mu\mathrm{Jy}$ at rest-frame wavelength $160\,\mu\mathrm m$ (Matthee et al. 2017), corresponding to $1.5\times10^{-19}\ \mathrm{J\,s^{-1}\,m^{-2}}$ . Its hint explicitly includes the redshift factor: $F_{1600}=(1+z)\lambda_0 f_{\lambda_0(1+z)}$ .

(D1.4.1) Since we have already known the upper limit of $F_{\mathrm{FIR}}$ of CR7, we only need to derive $F_{1600}$ in order to calculate IRX. First of all, we need to translate the $m_{1600}$ we got in (D 1.3) to $f_{(1+z)1600\text{Å}}$ :

f_{\lambda_0(1+z)}=f_{\nu_0(1+z)}\cdot\frac{c}{[\lambda_0(1+z)]^2}=\frac{c}{[\lambda_0(1+z)]^2}\cdot10^{-0.4m_{1600}}\cdot3631\ \mathrm{Jy}\qquad (11)

and we will derive $f_{\lambda_0(1+z)}=9.3\times10^{-13}\ \mathrm{J\,s^{-1}\,m^{-3}}$ . Therefore:

F_{1600}=(1+z)\cdot\lambda_0\cdot f_{\lambda_0(1+z)}=1.1\times10^{-18}\ \mathrm{J\,s^{-1}\,m^{-2}}\qquad (12)

Thus, the upper limit of CR7’s IRX should be:

\mathrm{IRX}_{\mathrm{CR7}}=\log\frac{F_{\mathrm{FIR}}}{F_{1600}}=-0.87\qquad (13)

[Equation (11), (12), (13): 2 marks respectively] [6 т.]

(D1.4.2) Given the UV continuum slop of CR7 (−2.7), we can infer its expected IRX from the $\mathrm{IRX}-\beta$ relation in local universe:

\widehat{\mathrm{IRX}}_{\mathrm{CR7}}=0.98\cdot\beta_{\mathrm{CR7}}+1.96=-0.69\qquad (14)

The upper limit of observed IRX is 0.18 dex lower than the value predicted by local relation. However, we have already known the local relation has a dispersion of 0.28 dex, which is larger than the 0.18 dex difference. This indicates that the current IRX upper limit is not tight enough (in other words, current observation is not deep enough) to show any difference of CR7 from local galaxies on the diagram of $\mathrm{IRX}-\beta$ . Therefore, the correct answer should be [NO].

[Equation (14): 2 marks; two IRX difference: 2 marks; [NO]: 2 marks] [6 т.]

Теми и трудност

Трудност: неоценена.

Светлина и астрономически измервания · Звездни величини · Междузвездно поглъщане

Предпоставки: Разширена олимпиадна подготовка

Оригинал в Архива: Data Analysis Problems.pdf · официални решения: Data Analysis Solutions.pdf · допълнителни данни: 18_da.pdf · допълнителни данни: 18_da_anssheet.pdf · допълнителни данни: 18_da_graph.pdf · допълнителни данни: 18_da_graph_log.pdf