[Aavso-photometry] Differential vs. absolute magnitudes

[Jeff Hopkins]

Hi Michael,

I think we are in different worlds. While I do CCD photometry most of 
my work is with a single channel PMT based UBV photon counting 
photometer.

BTW, I did not say a B filter would reduce precision, I said  "You 
cannot increase your precision by adding a B filter. You might be 
able to increase your accuracy, however."

In my world error is signified by a standard deviation of three (or 
more) determined magnitudes (for a short series on long period 
stars). For me SNR is just that. There are other factors that can 
produce error and in some cases significantly larger error. Taking 
three magnitudes, averaging and taking the SD  of the three is a 
better indication, at least to me, about how good the data is. I do 
this even with CCD data. If the resulting SD is large, than it may 
point to something you are doing wrong, strange sky conditions or 
something else that can be corrected. An error from SNR will not show 
any of that.

I have no argument that a higher SNR should produce better data, but 
by itself SNR may not be a good indication as to the quality of the 
data.

Jeff

At 17:02 -0700 03/08/2008, Michael Newberry wrote:
>Jeff,
>
>I am not disagreeing with what you say, but this brings up another 
>point worth mentioning:
>
>It is not strictly true that a B filter would reduce the precision 
>of the photometry. Of course, one benefit of using a B filter would 
>be that the magnitude could be transformed to a standard system (at 
>least theoretically and if he also had a V filter to get the 
>transformation coefficients). But let's look at the issue of 
>magnitude precision with and without the B filter.
>
>The magnitude precision is the random error the magnitude 
>measurement which is simply related to the signal to noise ratio 
>("SNR") of the measurement. In equation form, this is Sigma(mag) = 
>1.0857 / SNR. Remember that we are measuring the object, not the 
>(object+sky), so SNR is calculated using the net signal of the 
>object without the sky whereas the noise includes noise from both 
>the object and the sky underneath the star. The signal and noise are 
>both summed inside the measuring aperture.
>
>The B filter does not pass the Sodium D lines near 5890A that comes 
>from both light pollution and the natural sky background. So if 
>someone has a heavily light polluted sky caused by low pressure 
>sodium vapor lamps, then a B filter would dark the sky quite a bit. 
>But the B filter reduces the energy from the star by an amount that 
>is fixed by the bandpass, and that might be around a factor of 4 on 
>a typical CCD.
>
>Two side notes: 1) The SNR is also affected by the aperture size 
>(good seeing vs bad seeing). The larger the aperture, the more 
>background is added to the starlight. 2) There is noise from the 
>CCD's dark current. Even though the dark current is subtracted, its 
>noise remains as part of the total noise underneath the star 
>profile. In the same way that sky light is subtracted but leaves sky 
>noise, the dark "light" is subtracted but leaves its dark noise. 
>Therefore, observing in poor seeing or with a warm CCD will increase 
>the noise and decrease the SNR. But I'm just talking about filters 
>here.
>
>As an example, the B filter might reduce the sky brightness by a 
>factor of 10, 20, or even more but the star would be reduced by 
>about 4x. It is true that the sky has to be pretty bright or the 
>seeing pretty bad for the SNR to actually be improved by using a B 
>filter, but it can happen. Let's look at some rough numbers. I am 
>not going to include all noise sources because the sky is bright and 
>its noise swamps these other sources.
>
>Consider a fairly light polluted urban sky so that, in the measuring 
>aperture, the sky is 4 magnitudes brighter than natural sky. Suppose 
>that, with the B filter, we have an exposure time and aperture size 
>that gives 1000e- from the star. Assume no dark current. We get the 
>following:
>
>B filter:
>Star = 1,000e- above sky
>Sky = 1,000e- (no light pollution)
>Star/Sky = 1.
>Noise = sqrt(1,000+1,000) = 45e-
>SNR = 1000 / 45 = 22.4, Sigma(mag) = 0.048.
>
>No filter:
>Star = 4x brighter without B filter = 4,000e- above sky
>Sky = 4 mag (40x) brighter = 40,000e-.
>Star/Sky = 0.1
>Noise = sqrt(40,000+4,000) = 210e-
>SNR = 4000 / 210 = 19.1, Sigma(mag) = 0.057.
>
>Just for grins, suppose we expose 10x longer or we combine 10 images 
>and measure that. Then we get a better result for both cases:
>
>B filter:
>SNR = 70.7, Sigma(mag) = 0.015
>
>No filter:
>SNR = 60.3, Sigma(mag) = 0.018
>
>BUT! The *ratio* of SNR obtained with and without the filter is 
>identical to the case for shorter exposure. Hmmm.... what does this 
>tell us? Answer: Exposing longer does not make the 
>filtered/unfiltered comparison any different but it does give us 
>higher precision in either case.
>
>As you can see, a measurement through the B filter can win when the 
>sky is rather bright. On the flip side, a brighter star will provide 
>more photons so the effect is not as dramatic. The critical factor 
>is the ratio of star/sky brightness inside the measuring aperture.
>
>What I have described is the principle behind the so-called 
>"nebular" filter. However it is not true that *any* filter will 
>improve the SNR as I have shown. The signal in a given bandpass 
>depends on the net transmission of the atmosphere, optical system, 
>and filter combined with the response of the CCD and the spectrum of 
>the star. Since the U filter is very dense and the CCD, atmosphere, 
>and optics also ge worse below 4000A, the sky would have to be 
>awfully bright at 5890A to win when using a U filter. But that is 
>not the case for the B filter.
>
>Michael Newberry


-- 
Jeff Hopkins
HPO SOFT
Counting Photons
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