“2018年IOAA数据分析第1题-产星星系中的尘埃与年轻恒星”的版本间的差异
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Where 𝐹<sub>1600</sub> is the unattenuated flux and 𝐴<sub>λ</sub> is the dust absorption (in magnitudes) as a function of wavelength λ | Where 𝐹<sub>1600</sub> is the unattenuated flux and 𝐴<sub>λ</sub> is the dust absorption (in magnitudes) as a function of wavelength λ | ||
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+ | (D1.3.1) (6 points) Express 𝐴<sub>1600</sub> as a function of IRX. |
2019年9月12日 (四) 12:05的版本
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(D1) Dust and Young Stars in Star-forming Galaxies [75 points]
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 µm.
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 starformation 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 (β) to quantify the shape of the UV continuum:
𝑓λ=𝑄⋅𝜆β
where 𝑓V is the monochromatic flux of the galaxy at a given wavelength 𝜆 (in the unit of W m-3) and 𝑄 is a scaling constant.
(D1.1.1) (6 points) AB magnitude is a specific magnitude system. The AB magnitude is defined as:
$$\mathrm{m}_{\mathrm{AB}}=-2.5 \log \frac{f_{v}}{3631 \mathrm{Jy}}$$
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: 𝑓λ∆𝜈=𝑓𝜆∆𝜆 )
(D1.1.2) (12 points) 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.
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 |
(D1.1.3) (5 points) 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 mABas a function of 𝜆 and 𝑚1600 , where 𝑚1600 is the AB magnitude at 𝜆0=160 nm (1600 Å))
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 (β), UV brightness (at 1600 Å) and FIR brightness could be established:
$$ \operatorname{IRX} \equiv \log \left(\frac{F_{F I R}}{F_{1600}}\right)=S(\beta) $$
where 𝐹FIR is the observed far-infrared flux and 𝐹1600 is the observed flux at rest-frame wavelength 160 nm ( 1600 Å) (The “flux” 𝐹λ is defined as 𝐹λ =𝜆⋅𝑓λ). Table 2 presents 20 measurements of 𝛽, 𝐹FIR and 𝐹1600 in nearby galaxies (Meurer et al. 1999).
Galaxy Name | UV Slope
β |
log(F1600/10-3Wm-2) | log(FFIR/10-3Wm-2) |
---|---|---|---|
NGC 4861 | -2.46 | -9.89 | -9.97 |
Mrk153 | -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) (14 points) Based on the data given in Table 2, plot the IRX−𝛽 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. IRX=𝑎⋅𝛽+𝑏) by the side of your diagram.
(D1.2.2) (6 points) Quantify the dispersion (in ‘units’ of dex, where 𝐟𝐨𝐫 𝐞𝐱𝐚𝐦𝐩𝐥𝐞,𝐥𝐨𝐠(𝟏𝟎𝟗)− 𝐥𝐨𝐠(𝟏𝟎𝟒)=𝟓 𝐝𝐞𝐱) between the observed IRXobs and predicted IRXpred using the following equation:
$$ \sigma=\sqrt{\frac{\sum\left(\Delta \operatorname{lRX} _{i}\right)^{2}}{N-1}}(\text { unit: dex }) \text { where } \Delta \operatorname{IRX} _{i}=\operatorname{IRX}_{\text {i,obs }}-\operatorname{IRX}_{\text {i,pred }} $$
1.3. Under the previous assumption of the energy transfer process, the ratio of 𝐹FIR to 𝐹1600 can be expressed as:
$$ \frac{F_{F I R}}{F_{1600}} \approx 10^{0.4 A_{1600}}-1 $$
Where 𝐹1600 is the unattenuated flux and 𝐴λ is the dust absorption (in magnitudes) as a function of wavelength λ
(D1.3.1) (6 points) Express 𝐴1600 as a function of IRX.