[Aavso-photometry] Ensemble photometry
Jim Roe
jroe at jamesroe.com
Fri Nov 23 10:37:18 EST 2007
I'm trying to understand and learn how to do "ensemble photometry"
because I have come to believe that it will produce more accurate and
precise results. Right? I have re-visited threads on the list,
Googled, read several papers but I have yet to find a clear, concise
definition of the technique. E.g.,
Honeycutt (PASP 104:435-440, June 1992) states: "A simple (and often
quite adequate) technique for CCD ensemble photometry is to calculate
the difference between the instrumental magnitude for the program star
and a comparison magnitude obtained from the sum of the intensities of
perhaps a dozen of the brighter stars which appear in each exposure of
the series." This doesn't seem right to me, perhaps it is a typo that
was mentioned in an earlier post.
In another paper on the web by Hayes-Gehrke
<http://www.astro.umd.edu/~avondale/extra/Variability.html>
it is stated "Ensemble averages make this technique even more reliable.
Given at least a few dozen stars in the image, the assumption is made
that the average magnitude of these stars would remain constant from
frame to frame, regardless of the type or behavior of the individual
stars. " This makes more sense to me.
Finally, The MaximDl manual says: "If you tag more than one reference
star, you must set the magnitude for each; the results will be
averaged." Does this, then, constitute "ensemble photometry?"
If so, and I would just as soon continue using Maxim given my $400
investment therein, several questions arise.
Which extra reference stars should I include? Presumably only from
those for which photometry exists, but sometimes (often) there are
photometry stars that are much fainter than the program star and, unless
I expose for good S/N on them, their measurement errors will be greater
and it would seem adding them to the "ensemble" will degrade the whole
system.
Would I consider all of the individual reference stars to be check stars
to be compared to the ensemble average?
Jim Roe [ROE]
Wentzville, Missouri
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