[Aavso-photometry] Reality Check - Basic Concepts (Transformation Coefficients from Standard Fields)

[David Trowbridge]

Arne,

Your discussion of the physical meaning of these coefficients is extremely
illuminating. That's really what I needed: a back to basics discussion of
what the transformation equations mean. Also, I have obtained your book and
I've been reading it carefully. It's very helpful.

I did take a look at the transformation coefficients I was getting during
the NGC6811 campaign and they look pretty consistent from August through
October. These all used the same three stars in Landolt Field, SA 111. 

My most recent attempt to get the coefficients used 5 stars in Field SA 110.
The coefficient Eps in the equation V = v + Eps*(V-R) + zeta_v came out much
more negative (-0.129) than it had in any of the previous analyses on Field
SA 111 (mean 0.008, StdDev 0.049). Would one expect the coefficients to vary
this much from one standard field to another, or is there a likely mistake
hidden in my calculations?

Thank you,

David


> When doing differential photometry using five individual comparison stars
in
> chart 1804-18, I do get "reasonable" values around 8.713 (std dev 0.105). 
>
This is a very high standard deviation for such a bright star.  Can you give
us the actual instrumental magnitudes for each star?  You *are*
flatfielding,
right?  If you are using the comp stars given through VSP, and transforming
your results, you should have errors between (V-C1), (V-C2) etc. that are
in the 0.01mag range.

> Also, using five stars in the Landolt field, I can compute transformation
> coefficients and zero points. 
> 
>  
> 
> In equation (V-R) = Eta*(v-r) + zeta_vr, I get Eta = 1.056 and zeta_vr =
> 0.360.
> 
these are probably ok
>  
> 
> In equation (V-v) = Eps*(V-R) + zeta_v, I get Eps = -0.129 and zeta_v =
> 17.114
> 
This coefficient is a little large.  Now let's think about this and make
sure both equations make sense.  For V, the equation is saying that
Vtrue = Vins + zeropoint + coefficient*color
With your negative coefficient, this means that as the star gets red,
you are subtracting from the instrumental magnitude, making that magnitude
"brighter".  This means your V filter is slightly blueward of the standard,
so that all red stars are being measured fainter than their real values.

For the (V-R) equation, since your V filter is blueward of the standard V,
if your R filter were smack-on the standard, your color index would be
larger than standard.  From the equation:
(V-R)true = zeropoint + coefficient*(v-r)ins
and with your positive coefficient, this means that your color index
is actually a bit smaller than standard (you have to increase its size to
match the standard).  This means the R filter is also blueward of the
standard bandpass (to reduce the wavelength difference), and in fact,
has to be even further off of the standard than your V filter is.

Neither of these facts (V filter blueward of standard, R filter blueward
of standard) is exceptional; no filter/CCD combination exactly matches
the Johnson/Cousins standards, but you hope that you are pretty close
to the standard.  In some cases, you should consider quadratic terms in
addition to these linear equations.  For now, I'd stick with what you
are doing.
>  
> 
> I then go to my reduced and combined image of V5558 Sgr and use the first
> equation to come up with a standard color index (V-R).
> 
>  
> 
> I get (V-R) = 0.895
> 
>  
> 
> Then I use the second equation to compute a standard magnitude for V5558
Sgr
> and I get, V = 8.980. Also, R = 8.085
> 
>  
> 
> I know that I still need to take into account the different air masses of
my
> Landolt image and my V5558 image, but I'm confused as to how to do that.
> 
>  
> 
> Question #4: Does the number I have computed so far represent the
magnitude
> of V5558 AS IF IT WERE IN THE SAME FRAME as the Landolt image? 
> 
This is correct - you have neglected the extinction terms in the equations
above.
>  
> 
> If so, can I write Vactual = Vasif + Kv*(Xactual - Xasif) ?
> 
sorta.  The problems here are the following:
(1) you need to get signs right.  Again, think of what you are doing.
The extinction coefficient is always positive (more extinction at
higher airmass), and your measured magnitude will always be fainter
at higher airmass than if you measured it overhead.  So the extinction
term in the equation above should *subtract* from Vinstrumental to
make it brighter.  Your version would add.
(2) you don't indicate how you came up with Kv=0.4204 - your two
measures of the SA110 field?  If so, over what airmass range did you
determine this value?  Remember, it is really important for all-sky
calculations, so you need to get it right.
(3) how photometric was your night?  If the two Landolt measures were
under good conditions, but the high airmass Sgr observation was down
in the muck, you wouldn't get good results.
(4) I highly recommend that you do not attempt all-sky measures at
airmass 2.56 until you've been doing this for a long time.  Lots
of things happen when you get below X=2.
>  
> 
> Proceeding in blissful ignorance, I get:
> 
> Vasif = 8.980 [The standard V magnitude we found for Nova V5558]
> 
> Xactual = 2.56 [The actual air mass of V5558 at the time the V-filtered
> images were taken]
> 
> Xasif = 1.49 [The air mass of the Landolt frame and the air mass of V5558
as
> if it were in that frame]
> 
> Kv = 0.4204 [The slope of the graph of instrumental magnitude, v vs. air
> mass, X]
> 
>  
> 
> Using the equation above, I get,
> 
>  
> 
> Vactual = 9.4310  which is obviously wrong!
> 
>  
> 
> Question #5: What am I doing wrong here???
> 
>  
> 
> Now is the time I could really use some guidance from someone with more
> experience. Thank you in advance.
> 
The way I'd recommend beginners to start is with differential
photometry - measuring the target star and a comparison star in
the same field.  If transformation is important, determine the
transformation coefficients from a single Landolt field, like
SA110-503; I'd do this for several fields on several nights.  Don't
worry about extinction or zeropoints - what you want are those
color-index coefficients.  Then apply the transformation to the
differential measures you obtained, and again don't try extinction
or zeropoint - these subtract out when working in a single field.
This is discussed in our original book (Astronomical Photometry).

All-sky measures are inherently not difficult, but the devil is
in the details.  Start with the easy stuff.
Arne
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