[Aavso-photometry] Reality Check - Basic Concepts

[arne]

David Trowbridge wrote:
> I have a number of questions I'll need to resolve before I'll have much
> confidence in my data processing techniques. 
> 
>  
> 
> I've been imaging the Nova V5558 Sgr with a V filter on an ST-7 through a
> Tak 106 refractor. Before chart 1804-18 was released, I had been using 5
> stars in area Landolt SA-110 for comparison. With the charts, I've been
> having some luck with differential photometry but without them I had major
> problems with all-sky.
> 
>  
> 
> Consider the AAVSO light curve for V5558 Sgr (http://tinyurl.com/3bh74o) 
> 
>  
> 
> Question #1: In general, do the data points in this light curve represent
> the results of observers who have been doing all-sky photometry and using
> standards such as Landolt, M67, etc? Are the values in this graph exactly as
> submitted by various observers, or have they been normalized in some way
> after they were submitted?
> 
>  
> 
> Say I examine the graph at the date of my last observation (May 30, JD
> 2454250.8965). I see that a "reasonable" observation of V magnitude would be
> approximately 8.72 by interpolation.
> 
>  
> 
> 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.

>  
> 
> After reducing and combining my images, I've been using the Single Image
> Photometry Tool in AIP4Win (v. 2.1.10) to obtain a Raw Instrumental
> Magnitude, v for V5558 and the Differential Magnitudes (V-C) for each
> comparison star. I then get a predicted standard magnitude V by adding each
> of the (V-C) values to the corresponding chart magnitudes Cn.
> 
>  
> 
> Question #2: How do I use the ensemble data generated in AIP4Win after
> clicking on several comparison stars? I know that this would give a more
> accurate result, but I don't know how to get it. There is a (V-Ens) value
> reported, but what number (an ensemble "standard magnitude" from the chart)
> do I add it to in order to obtain V? 
> 
You need to read how the (V-Ens) value is calculated.  My understanding,
and others correct me if I'm wrong, is that the "Ens" number is the sum
of all of the comparison star fluxes, converted to a magnitude.  To
figure out what "chart magnitude" to use for such a pseudo star, you
have to convert each comparison star's chart magnitude to a flux
(10^(-0.4*mag)), add these together, and convert back to a magnitude.
That is the "chart magnitude Cens" that you would add to convert
(V-Ens) back to a standard magnitude.  Personally, until you get
traditional (V-C) photometry working correctly, I would avoid ensemble
photometry like the plague.  Don't get confused.

>  
> 
> When I try to do all-sky photometry using the Landolt stars, my results are
> way off.
> 
>  
> 
> Question #3: Are the published values for Landolt standard magnitudes and
> color indexes normalized for air mass = 1 (at the Zenith), or are they air
> mass = 0 (above the atmosphere)?
> 
air mass 0 (above the atmosphere).  This is always done for standardized
photometry to remove all traces of atmospheric extinction.  Air mass = 1
values for you will be different than air mass = 1 values for someone
on Mauna Kea.
>  
> 
> I have data for the Landolt field SA-110 with two different filters, R and V
> at two different times, bracketing my exposures of V5558, so I can compute
> an atmospheric extinction coefficient, K for each of those filters. 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