Difference: 20160214MSciPhotometryAnalysisObs3part2 (1 vs. 4)

Revision 419 Feb 2016 - DavidHadden

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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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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

%BEGINLATEX%

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Revision 218 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.

Line: 72 to 72

$\nu_{\rm{tot}}$	1.71e+06
$\chi^2$	3222
Degrees of freedom	3591

Added:

>
>

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

$\begin{equation} \rm{distance}_{ij} = (x_i - x_j)^2 + (y_i - y_j)^2 \end{equation}$

Went through images and found the image with highest number of stars.
This was the base image that all other images were compared to.
It worked fine for images in focus and blue filter since they were taken close enough in time for individual stars to not have moved much.

In red filter however, there was double matching, when more than one star was matched to the same base star as shown below:

Improved the algortihm by:
- first calculate the minimum distances, some of them will be wronged as stars will be wrongly matched.
- Use those matches to work out the median x and y shifts.
- Then check between matches if the x and y shifts match the median x and y shifts to 1 pixel, if they do its a match.
Below is the same red image and matches:

HR diagram for Cr39

Using the matching and ML fits we got the following HR diagrams:

The stars in V-R HR that seem to have the same V-R are indexed 26 and 27. The star with V close to -8 is indexed 2, it appeared 3 times. Star with V close to -7.5 appeared 3 times.
The HR diagrams are terrible.
From SIMBAD they should look like this:

-- ElenaCukanovaite - 14 Feb 2016

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Revision 114 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.
The skewed Gaussian function is:

$\begin{equation} f_G(x,y,\mu_x,\mu_y,\sigma_x,\sigma_y,\rho, \beta_1, \beta_2) = \frac{1}{2\pi\sigma_x\sigma_y\sqrt{(1-\rho^2)}} \exp \Bigg\{-\frac{1}{2(1-\rho^2)}\Bigg[\bigg(\frac{x-\mu_x}{\sigma_x}\bigg)^2 + \bigg(\frac{y-\mu_y}{\sigma_y}\bigg)^2 -2\rho\bigg(\frac{x-\mu_x}{\sigma_x}\bigg)\bigg(\frac{y-\mu_y}{\sigma_y}\bigg)\Bigg]\Bigg\} \times \Bigg[1 + \rm{erf} \frac{\beta_x x + \beta_y y}{\sqrt{2}}\Bigg] \end{equation}$

Below are some skewed Gaussians for different skews:

A normalized function would have values between 0 and 1, but here the values are between 0 and 14. This is because the function is summed over all columns and rows.
If one looks at an individual column, it can be seen that the function is indeed normalised.
Middle column:

At the far columns one can see:

Also, the formula was extracted from this reference ( doi: 10.1093/biomet/83.4.715 ) which derives a normalized skewed bivariate Gaussian.
We wanted to look at a good star and see how the fit looks like in as much detail as possible, as the summation along rows and columns seems to distort the details of fitting.
The fit is done by minimizing the $\chi^2$ in the following way:

$\begin{equation} f(x,y,\vec{\theta}) = \frac{\alpha}{2\pi\sigma_x\sigma_y\sqrt{(1-\rho^2)}} \exp \Bigg\{-\frac{1}{2(1-\rho^2)}\Bigg[\bigg(\frac{x-\mu_x}{\sigma_x}\bigg)^2 + \bigg(\frac{y-\mu_y}{\sigma_y}\bigg)^2 -2\rho\bigg(\frac{x-\mu_x}{\sigma_x}\bigg)\bigg(\frac{y-\mu_y}{\sigma_y}\bigg)\Bigg]\Bigg\} \times \Bigg[1 + \rm{erf} \frac{\beta_x x + \beta_y y}{\sqrt{2}}\Bigg] + \frac{1-\alpha}{N*M} \end{equation} \begin{equation} \nu_{ij} = \nu_{\rm{tot}} \int \int_{ij} f(x,y,\vec{\theta}) dx dy \end{equation} \begin{equation} \chi^2_P = 2 \sum^{N,M}_{i, j=1} \left( n_{ij} \ln \frac{n_{ij}}{\nu_{ij}} + \nu_{ij} - n_{ij} \right) \end{equation} <br />$

So we fit for 9 parameters.
We do not integrate, instead we evaluate the function at the pixel and multiply by pixel area.

We fitted this star:

Column by column the fit looks like this:

The plot parameters are shown below:

$\rho$	0.0509
	33.5
	33.8
$\sigma_x$	5.021
$\sigma_y$	5.373
$\beta_x$	-1.184
$\beta_y$	-1.384
$\alpha$	0.0014
$\nu_{\rm{tot}}$	1.71e+06
$\chi^2$	3222
Degrees of freedom	3591

-- ElenaCukanovaite - 14 Feb 2016

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