Comparing stars between different filters

  • In order to compare the quality of our data, we looked at the data with the largest signal to noise ratio, in this case data taken with a green filter.
  • For this, we found the largest and smallest stars, as determined by the amount of pixels which constituted them.
  • The coordinates of the stars were recorded, so they can later be compared to the same stars in other filters.
  • For the fitting the backgrounds were removed, therefore the error on the no background data is the following (assuming error is Poissonian):
 \begin{equation} \rm{Data \ with \ background \ removed} = \sqrt{(\Delta \rm{Original \ data})^2 + (\Delta \rm{Background})^2}\end{equation}  \begin{equation} \rm{Data \ with \ background \ removed} = \sqrt{ \rm{Original \ data} + \rm{Background}}\end{equation}  <br />
  • Below fits can be seen for the maximum and minimum size stars for all filters:

elena_no_br.png elena_r_b.png
elena_b_b.png elena_g_b.png
elena_no_d.png elena_r_d.png
elena_b_d.png elena_g_d.png
  • The plots are projections (they are summed in rows and columns), so to plot the errors the errors were summed in the following way:
 \begin{equation} \rm{Error \ projection} = \sqrt{\sum_i^{Number \ of \ pixels} \Delta p_i^2} \end{equation}
  • where the $\Delta p_i$ is the error on the data point (pixel value).

Introductory $\chi^2$ estimation

  • Then set about finding $\chi^2$ for the above stars for all filters.

  Visual Brightest Star Visual Dimmest Star Green Filter Brightest Star Green Filter Dimmest Star Red Filter Brightest Star Red Filter Dimmest Star} Blue Filter Brightest Star Blue Filter Dimmest Star
$\rm{Degrees \ of \ Freedom}$ $3593$ $3593$ $3593$ $3593$ $3593$ $3593$ $3593$ $3593$
$\chi^2$ $3250402$ $1746$ $1602258$ $2001$ $869411$ $2080$ $3172054$ $2242$
$\rm{reduced} \chi^2$ $904$ $0.486$ $445$ $0.557$ $241$ $0.579$ $882$ $0.624$
$\rho$ $0.107 \pm 0.00518$ $0.118 \pm 0.0316$ $0.0286 \pm 0.00530$ $0.106 \pm 0.0485$ $0.0631 \pm 0.00473$ $0.0715 \pm 0.0294$ $0.00456 \pm 0.00625$ $0.0423 \pm 0.0358$
$a$ $9399820\pm 43459$ $18450 \pm 441$ $9242014 \pm 42579$ $14155 \pm 522$ $12354155 \pm 51521$ $25730 \pm 573$ $12945440 \pm 70680$ $20255 \pm 547$
$x_0$ $30.1 \pm 0.0201$ $30.3 \pm 0.0867$ $30.7 \pm 0.0248$ $31.0 \pm 0.158$ $30.5 \pm 0.0195$ $30.0 \pm 0.0855$ $30.3 \pm 0.0263$ $30.4 \pm 0.108 $
$y_0$ $31.7 \pm 0.0218$ $29.9 \pm 0.0902$ $31.7 \pm 0.021$ $30.4 \pm 0.130$ $30.8 \pm 0.0204$ $30.4 \pm 0.0904$ $31.6 \pm 0.0246$ $30.1 \pm 0.104 $
$\sigma_x$ $4.18 \pm 0.0162$ $3.76 \pm 0.0870$ $5.16 \pm 0.020$ $4.51 \pm 0.160$ $4.51 \pm 0.0157$ $4.01 \pm 0.0861$ $4.66 \pm 0.0215$ $4.19 \pm 0.109 $
$\sigma_y$ $4.51 \pm 0.0178$ $3.91 \pm 0.0907$ $4.41 \pm 0.0178$ $3.71\pm 0.131$ $4.69 \pm 0.0166$ $4.25 \pm 0.0909$ $4.34 \pm 0.0205$ $4.06 \pm 0.105 $
$d$ $159 \pm 8.36$ $-4.96 \pm 0.373$ $140 \pm 8.68$ $-3.69 \pm 0.416$ $173 \pm 9.505$ $-2.30 \pm 0.448$ $255 \pm 12.2$ $1.52 \pm 0.430 $
  • In general, a good $\chi^2$ should be around equal to the number of degrees of freedom,

  • From the plots it is obvious that the Poisson errors are not correct as they are too small for the brightest star.

  • Also the brightest stars are poorly fitted and therefore have high $\chi^2$ in comparison to dimmest stars.

Testing curve_fit for better error estimation

  • In order to better obtain errors from curve_fit (as it tends to give outlandish errors), we set out to create a Gaussian of known parameters, and then fit a general Gaussian curve to this.
 

Fractional Error When Guessing Incorectly

Fractional Error When Guessing Exactly
Correlation Coefficient $0.015$ 0
Amplitude $1.87 \times 10^{-10}$ 0
X Centre $0.09$ 0
Y Centre $0.01$ 0
$\sigma_x$ $1.25 \times 10^{-10}$ 0
$\sigma_y$ $1.25 \times 10^{-10}$ 0
Offset $1.66 \times 10^{-10}$ 0

Setting the threshold

  • We decided to use our mode image to set the threshold.
  • The mode image is shown below:
mode_sec.png
  • We took the Poissonian error (the square root) of each section.
  • We used the error as our threshold by setting the threshold:
 \begin{equation} \rm{threshold} = 3 \times \rm{Poisson \ error}\end{equation}  <br />

Standard deviation of the stars

  • We previously tried to show that all stars in the image have the same standard deviations.
  • To find our standard deviations we did the following:
 \begin{equation} \sigma_x = \sqrt{\frac{\sum_{i = 1}^N (x_i - x_{\rm center})^2 p_i}{\sum_{i=1}^N p_i}}\end{equation}  \begin{equation} \sigma_y = \sqrt{\frac{\sum_{i = 1}^N (y_i - y_{\rm center})^2p_i}{\sum_{i=1}^N p_i}}\end{equation}
  • We cropped an area of 20x20 pixels around each star and found the standard deviations within the area.
  • Histograms below are of the ratio of $\frac{\sigma_x}{\sigma_y}$ for Location 1 images:
stan_dev_n_t.png
  • $sigma_x$ for Location 1 images:
sigma_x_loc_1.png
  • Ratio of $\frac{\sigma_x}{\sigma_y}$ for Location 2 images:
location_2.png
  • $sigma_x$ for Location 2 images:
sigma_x_loc_2.png
  • Therefore for both locations we decided to set the requirement that an objects is only a star if its ratio:
 \begin{equation} 0.9 &amp;lt; \frac{\sigma_x}{\sigma_y} &amp;lt; 1.1  \end{equation}  <br />

Monte Carlo simulation of a 2D star field

  • We have done some prelimenary work on the simulation.
  • To generate the positions of stars we uniformly generated x and y pixel positions.
  • To create a background we generated Poisson distributed random variables around the mean of 160, which corresponded to a typical value for images of Location 1.
  • For each position we wanted to create a Gaussian distribution that represents a star centered on the star's x,y position.
  \begin{equation} f(x,y,\mu_x,\mu_y,\sigma_x,\sigma_y,\rho,a,d) = \frac{a}{2\pi\sigma_x\sigma_y\sqrt{(1-\rho^2)}} exp\Bigg\{-\frac{1}{2(1-\rho^2)}\Bigg[\bigg(\frac{x-\mu_x}{\sigma_x}\bigg)^2 + \bigg(\frac{y-\mu_y}{\sigma_y}\bigg)^2 -2\rho\bigg(\frac{x-\mu_x}{\sigma_x}\bigg)\bigg(\frac{y-\mu_y}{\sigma_y}\bigg)\Bigg]\Bigg\}  \end{equation}  <br />
  • As the standard deviations should be the same for each star we decided to generate only one value for $\sigma$ for all stars. We also assumed that the standard deviations are the same in both directions so we generated only one $\sigma$ for $\sigma_x$ and $\sigma_y$.
  • To experiment we uniformly generated values for $a$ and $\rho$ for each individual star.
  • We then for each pixel in the image calculated the Gaussian value at that particular pixel for each star.
  • Resulting image for 50 stars is shown below:
monte_carlo.png

-- ElenaCukanovaite - 11 Jan 2016

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Topic revision: r11 - 14 Jan 2016 - ElenaCukanovaite

 
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