[Aavso-photometry] Transform coefficients and two-filter measurements

[Michael Newberry]

Hi Michael,

As I mentioned, there are two issues:

1) differences in effective wavelength, which gives rise to the 
transformation coefficient being different from 1.0,  and

2) differences in bandpass shape.

Just by knowing whether the filters are bluer or redder (the effective 
wavelength) does not give you a definitive answer. Mismatches in the shapes 
of the bandpasses between your system and the standard filter/detector 
system can give nonlinearities in the transformation.

If the overall bandpass shapes are exactly matched, then a small 
transformation coefficient results, and the nonlinearities are "negligible". 
The main problem is the shapes of the bandpasses, which cannot be exactly 
matched to what Johnson and Cousins used. But the closer, the better.

One thing our study showed was that having a small transformation 
coefficient (because the effective wavelengths are closely matched) DOES NOT 
guarantee linear transformations across the range of spectral types.

To definitively answer the question, short of doing a full synthetic 
photometry analysis for your filters and CCD like we did, you would need 
observe stars across the range of spectral classes and determine 
transformations from them, then compare the coefficients you get. Or, in a 
nutshell, for the purposes of 0.02 magnitude photometry, just get the 
bandpass shapes and effective wavelengths as close to the system as you can.

Michael Newberry

----- Original Message ----- 
From: "Michael Koppelman" <michael at slackerastronomy.org>
To: "Michael Newberry" <mnewberry at mirametrics.com>
Cc: <tom_krajci at tularosa.net>; <tom.krajci at gmail.com>; 
<aavso-photometry at aavso.org>
Sent: Thursday, July 23, 2009 11:24 AM
Subject: Re: [Aavso-photometry] Transform coefficients and two-filter 
measurements


> Thank you Dr. Newberry.
>
> But to Tom's original question, I don't think there are 4 degrees of 
> freedom, there are only 2.
>
> So, not this:
>
> - B filter bandpass is bluer than ideal
> - B filter bandpass is redder than ideal
> - V filter bandpass is bluer than ideal
> - V filter bandpass is redder than ideal
>
> This:
>
> - B-V is bluer than ideal
> - B-V is redder than ideal
>
> Yes?
>
> M.
>
>
> On Jul 23, 2009, at 12:07 AM, Michael Newberry wrote:
>
>> Are you asking if it is possible for the graph of b-v vs B-V to have 
>> curvature? It so, the answer is yes---it depends on the filter/ detector 
>> bandpasses. The transformations are linear over the range  of spectral 
>> types if the bandpasses are an exact match to the  standard system. But 
>> as the filters move off the standard  bandpasses, in wavelength and 
>> shape, nonlinearities creep in.
>
>