[Aavso-photometry] Multi-color photometry

[Jeff Hopkins]

Color transformation coefficients are a function of your equipment 
and are used to transform your data to a standard. That way ten 
people with ten different systems observing the same object can have 
their data combined and meaningful.

You do not and cannot transform from one band to another.

To obtain V color transformations coefficients you need to observe in 
both B and V bands and determine the coefficients.

Few if any professionals will be interested in data that has not been 
taken with proper filters and reduced using color transformation 
coefficients.

You do not transform from one band to another. The (B-V) is used to 
determine the V value during the data reduction.

V= v + epsilon *(B-V) +Zv

(B-V)= mu * (b-v) + Zbv

Here extinction is ignored. Z are the zero points, epsilon is the V 
color transformation coefficient and mu the (B-V) color 
transformation coefficient. V and B are extraterrestrial magnitudes 
and b and v raw data.

Jeff

At 10:03 -0700 11/3/05, Justin Pryzby wrote:
>On Thu, Nov 03, 2005 at 03:37:41AM -0600, Brian C. Barnes wrote:
>>  I'm just starting to get into doing photometry with filters other than "V",
>I'm new too, lets see if I can say anything useful..
>
>>  1) Assume I have B, V, and R filtered images of M67 along with B, V, and R
>>  images of a target. I can create transformation coefficients for B-V, and
>>  using those, I can calculate B and V. I can also create a transformation
>>  coefficient for V-R, and I can then calculate R by calculating V-R and using
>>  the V value from the B-V calculation, or I can calculate V-R and then V and
>>  R individually just like I did with B and V. This would give me one B value,
>>  2 V values, and 1 R value. I can then average the 2 V values to get a final
>>  V value. Is there an advantage to calculating this extra V value and doing
>>  the averaging?
>It seems to me that if you *have* B, V, and R data, then there is no
>reason to compute transformation to those bands.  Transforming from
>one band to another is an approximation, of course, and real data is
>always better.  Also, transformations assume blackbody-type emission,
>an assumption which some objects violate.
>
>The only reason one transforms from one band to another is if you
>don't have data in that band, and you want to compare with values from
>an observation or star list which does have data in that band (but no
>data from the bands for which you do have data).
>
>It is my understanding that even when you do observe with a filter,
>you still have to do some correction to put yourself on a some
>standard system.  Of course, the 1st order correction is just an
>offset ("zeropoint").  You necessarily have to add *something* to the
>log(object_flux/exposure_time) to go from instrumental units to any
>semblance of standard units.  Then you might make some small
>correction to convert from "your filter" magnitude to "standard
>filter" mag (which is ideally zero, but never happens).
>
>As best as I can tell, all of photometric calibration is iteratively
>figuring out the source of the largest remaining trend between your
>predicted mag value and the accepted values, and then fitting a linear
>least-square slope to a line such as to optimally correct for that
>trend.  Stop iterating when the slope of the line is sufficiently
>small such that the largest correction you would make for your dataset
>is smaller than your tolerance for systematic error.
>
>The transformation formulas are either empirical observations specific
>to someone else's instrument, or derived from some theoretical
>calculation.  Instead of computing data points for which you already
>have *real* data, it would make much more sense to compute empirical
>transformation formulas for your own equipment (if for nothing else,
>than to figure out how to do it).
>
>I'll stop now, hopefully someone with experience will jump in.
>
>--
>Clear skies,
>Justin
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-- 
Jeff Hopkins
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