[Aavso-photometry] What are transformation coefficients??

[Jeff Hopkins]

Hi Trupti,

If you are the only person in the world doing photometry, you would 
not need to worry about the transformation coefficients for your 
system. Your system would be the standard.

Since that is not true and many observers would like to have there 
data be accurate with respect to others, a standard system was 
created. Your filters, telescope and detector will have different 
(slightly) characteristics, spectral response, than other systems. To 
level the playing field you must calibrate your equipment against a 
set of stars that have been measured and deemed standard 
stars/magnitudes.

While not simple, doing these measurements and calibrations to 
determine your transformation coefficients require no more than basic 
algebra.

There are two other factors that must be considered too. A standard 
star that doe not vary will produce different magnitudes at different 
air masses (distances from the zenith). To account for that nightly 
extinction in each band must be determined and used to adjust the 
magnitude to one that would be outside the Earth's atmosphere. There 
first order and second order extinction, but most people just set the 
second order to zero and ignore it. The last point and one that 
cancels if you do differential photometry is the system zero 
points(in each band). An 8" system with identical filter, detector 
and transformation coefficients will produce a very different 
magnitude than a 16" telescope for the same star and air mass and 
extinction. The zero points adjust for the sensitivity of the system.

Summary:

You get a number representing the brightness of an object.
(e.g., V filter= 300,000 ADU)

You take that number and take the log (base 10) of it and multiply if 
by -2.5 to create the raw magnitude.
-2.5 log (300,000)
V Raw magnitude= -13.69

You adjust for the system sensitivity by adding a zero point factor
(e.g., Zv = 19.09)

You adjust for first order extinction
(k'v * X; where k'v is the V band first order extinction coefficient 
and X is the air mass).
(e.g., k'v = 0.1821;   X= 1.002;  k'v * X = 0.18)

You then transform the data to the standard system with your V 
transformation coefficient (epsilon)
(e.g., epsilon = -.03)

The extraterrestrial instrumental magnitude (EIM) is then:

V EIM= -2.5 log(ADU Count) + epsilon(B -V) - k'v * X + Zv

The (B - V) is the B- V value for the star from a star catalogue.
(e.g., 0.15)

V EIM= -13.69 + (-0.03 * 0.15) - 0.18 + 19.09
V EIM=  -13.69 -0.0045 - 0.18 + 19.09
V EIM= 5.18

This gets more complicated for the UBRI bands, but is similar.

Jeff


At 10:52 -0700 03/12/2008, trupti ranka wrote:
>Hi all,
>   After all the discussion that is going on this group lately this 
>may seem a very naive question, but what do transformation coeffs 
>signify??. I read something about it in aavso's astronomical 
>photometry manual but didn't really get it.
>   Is it that: According to size of telescope and type of detector 
>the number of photons collected from a particular star will differ. 
>So for large telescope we will get larger counts and less for 
>smaller telescope. But since people with different instruments, 
>observing the same star should get same results , we transform this 
>absolute count to some standard count. But in differential 
>photometry since we take differece between the count of two stars, 
>why do we need transformation coefficients here?? shouldn't we get 
>same results irrespective of the instrument used (atleast under 
>ideal conditions).
>
>   Trupti

-- 
Jeff Hopkins
HPO SOFT
Counting Photons
http://www.hposoft.com/Astro/astro.html
Hopkins Phoenix Observatory
7812 West Clayton Drive
Phoenix, Arizona 85033-2439 U.S.A.
(623)849-5889
(623) 247-1190 (Fax)
www.hposoft.com