Difference: 20160214MSciPhotometryAnalysisObs3part2 (2 vs. 3)

Revision 319 Feb 2016 - ElenaCukanovaite

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META TOPICPARENT	name="StewartBoogertPhotometry2015"

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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