(a) Generating population

From the previous part in the problem, in the field of view, we have G stars at region A,B,C = 80,100,120 pc = 6,10,14 stars. Absolute magnitude of a G star is given as \[ M_{V} \sim N(M_{V,sun},\sigma^2=0.3) \] where \(M_{V,sun}=4.83\).

  1. Draw \(M_{V}\), 30 stars (= 6+10+14) in total.
  2. Shuffle.
  3. From index \(i=1,...,30\), assign \(i=1,...,6\) in region A; assign \(i=7,...,16\) in region B; assign \(i=17,...,30\) in region C.
## [1] "Abs mag of population, by region A,B,C"
## [[1]]
## [1] 3.545208 5.065041 5.107178 4.515202 4.530597 4.520837
## 
## [[2]]
##  [1] 5.355537 4.769594 4.550109 4.330918 4.371462 6.153207 4.903443
##  [8] 4.561240 4.588702 5.081728
## 
## [[3]]
##  [1] 4.569507 4.441424 4.555450 3.937709 4.190469 3.635943 4.095508
##  [8] 4.668809 4.574817 5.623922 4.244680 4.361497 4.676296 4.285378

(b) Obtaining sample

Sample is obtained from observation. The limitation is that we can observe only stars with \(m_{V} \leq 10\) in any region.

  1. Find absolute magnitude of \(m_{V} = 10\) for each region \(= M_{V}^{cut,region}\).
  2. From population, given the region, select only stars with \(M_{V} \leq M_{V}^{cut,region}\) as our sample.
## [1] "Abs mag cutoff, by region A,B,C"
## [1] 5.484550 5.000000 4.604094
## [1] "Abs mag of sample, by region A,B,C"
## [[1]]
## [1] 3.545208 5.065041 5.107178 4.515202 4.530597 4.520837
## 
## [[2]]
## [1] 4.769594 4.550109 4.330918 4.371462 4.903443 4.561240 4.588702
## 
## [[3]]
##  [1] 4.569507 4.441424 4.555450 3.937709 4.190469 3.635943 4.095508
##  [8] 4.574817 4.244680 4.361497 4.285378

b.1 Mean absolute magnitude: Population VS Sample

Calculate mean absolute magnitude of population, and sample. Then, compare.

## [1] "Mean Abs mag of population and sample"
## [1] 4.593714
## [1] 4.427163

Malmquist bias: average absolute magnitude of sample is brighter than of population, because fainter stars are unobservable.

b.2 Mean distance: Population VS Sample

  1. Calculate mean distance of sample, given known region.
## [1] "Mean distance of sample, given known region"
## [1] 104.1667
  1. Find average luminosity for all the stars in the field of view. Since all stars in the field of view are G stars drawn from the distribution in part (a), the average luminosity \(= (4\pi(10pc)^2)*10^\frac{M_{V,sun}-constant}{-2.5}\), where \(M_{V,sun}=4.83\). Hence, this average luminosity is equivalent to the absolute magnitude \(M_{V,sun}=4.83\).
  2. Transform absolute magnitude of sample to its apparent magnitude by \[ M-m=-5\log_{10}[\frac{d}{10pc}] \] where \(d=80,100,120 pc\) for region A,B,C.
## [1] "Apparent magnitude of sample"
## [[1]]
## [1] 8.060658 9.580491 9.622628 9.030652 9.046047 9.036287
## 
## [[2]]
## [1] 9.769594 9.550109 9.330918 9.371462 9.903443 9.561240 9.588702
## 
## [[3]]
##  [1] 9.965413 9.837330 9.951356 9.333615 9.586375 9.031849 9.491414
##  [8] 9.970724 9.640587 9.757404 9.681284
  1. Find distance of sample by comparing the absolute magnitude \(M_{V,sun}=4.83\) with apparent magnitude of sample by \[ M_{V,sun}-m=-5 \times \log_{10} [\frac{d}{10pc}] \] .
## [[1]]
## [1] 44.27226 89.14526 90.89199 69.20387 69.69626 69.38369
## 
## [[2]]
## [1]  97.25654  87.90665  79.46640  80.96409 103.44003  88.35844  89.48297
## 
## [[3]]
##  [1] 106.43455 100.33813 105.74776  79.56517  89.38712  69.24204  85.56237
##  [8] 106.69516  91.64680  96.71208  93.38062
##  [1] 4.569507 4.441424 4.555450 3.937709 4.190469 3.635943 4.095508
##  [8] 4.574817 4.244680 4.361497 4.285378
  1. Find mean distance of sample.
## [1] "Mean distance of sample, given known G star, but unknown region"
## [1] 86.84084

Malmquist bias: given the field of view and star type, more number of stars is at closer range, and less number is at further range. Hence, the mean distance would be biasly close.

(c) Metallicity

  1. Draw 30 numbers (= 6+10+14) of metallicity as \[ \frac{Z}{Z_{sun}}=\frac{N_{2}+0.5}{6} \] where \(N_2 \sim U\{1,2,3,4,5,6\}\).
  2. Correct absolute magnitude of population by the drawn metallicity as \[ M=M_{initial} + -0.87 \times \log_{10}[\frac{Z}{Z_{sun}}] \] .
## [1] "Metallicity / Metallicity of Sun"
## [[1]]
## [1] 0.06368793 0.06368793 0.06368793 0.06368793 0.06368793 0.06754736
## 
## [[2]]
##  [1] 0.07371645 0.06368793 0.07371645 0.06544124 0.06368793 0.06754736
##  [7] 0.07371645 0.07371645 0.06368793 0.07371645
## 
## [[3]]
##  [1] 0.06368793 0.06368793 0.07018502 0.06368793 0.06544124 0.06368793
##  [7] 0.06368793 0.06368793 0.07371645 0.06368793 0.06368793 0.07371645
## [13] 0.06368793 0.06544124
## [1] "Absoluate magnitude of population with metallicity"
## [[1]]
## [1] 3.427336 4.947169 4.989306 4.397330 4.412725 4.426121
## 
## [[2]]
##  [1] 5.297835 4.651722 4.492407 4.223565 4.253590 6.058492 4.845742
##  [8] 4.503539 4.470830 5.024026
## 
## [[3]]
##  [1] 4.451635 4.323551 4.476560 3.819836 4.083116 3.518071 3.977635
##  [8] 4.550936 4.517116 5.506049 4.126808 4.303796 4.558424 4.178025
  1. Re-obtain sample.
## [1] "Absoluate magnitude of sample with metallicity"
## [[1]]
## [1] 3.427336 4.947169 4.989306 4.397330 4.412725 4.426121
## 
## [[2]]
## [1] 4.651722 4.492407 4.223565 4.253590 4.845742 4.503539 4.470830
## 
## [[3]]
##  [1] 4.451635 4.323551 4.476560 3.819836 4.083116 3.518071 3.977635
##  [8] 4.550936 4.517116 4.126808 4.303796 4.558424 4.178025
  1. Calculate mean \(Z\) from sample in region B,C.
## [1] "Mean Z of sample in region B,C / Metallicity of Sun"
## [1] 0.06678291
  1. Compare with mean Z from all stars in the sky.
## [1] "Metallicity of all stars in the sky / Metallicity of Sun"
## [1] 0.06667711

Notice that they are about equal.