Difference: 20160214MSciPhotometryAnalysisObs3part2 (2 vs. 3)

Revision 319 Feb 2016 - ElenaCukanovaite

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Testing the fitting

  • First we wanted to test that the function we are fitting is actually normalised.
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$\nu_{\rm{tot}}$ 1.71e+06
$\chi^2$ 3222
Degrees of freedom 3591
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  • Found its magnitude using two methods, integration of the Gaussian and multiplication of two parameters:
  \begin{equation} \rm{METHOD \ 1} \ \rm{counts} = \int_{0}^{60} f_G(x,y,\vec{\theta}) dx dy = \int_{0}^{60} \frac{\alpha}{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\} \ dx \ dy  \end{equation}  \begin{equation} \rm{METHOD \ 2} \ \rm{counts} = \alpha \times \nu_{tot}  \end{equation}  \begin{equation} \rm{instrumental \ magnitude} = -2.5 \log{\left(\frac{\rm{counts}}{\rm{exposure \ time}}\right)}  \end{equation}  <br />
  • Method 1 gave magnitude = -8.4647927462328827
  • Method 2 gave magnitude = -8.4647927466101169
  • So they are equivalent to 9 decimal places.
  • Therefore we shall use the second method as integration might slow the analysing down, also it would be easier to find the errors using method 2.
 

Matching stars between images

  • Used Pythagoras to estimate the closest star image to image. Only used stars whose $\chi^2_{\rm reduced}$ were between 0.5 < $\chi^2_{\rm reduced}$ < 1.5
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