Most of the red-blue pairs, both the bright HIP pairs and the fainter SDSS pairs, are clustered around 00 Dec. Question: How much airmass change does one need to adaquately determine secondary extinction coefficiens when at 31 degrees north latitude? For example, can I pick a pair at 00 Dec?

Thanks, Ed

Hi Ed,

The normal coefficient is on the order of 0.03. It is calculated by

delta_V ~ k"*X*delta_C

so you plot the magnitude difference of your red-blue pair as a function of airmass*delta_color. Since the slope is only 0.03 instead of the normal 0.2 or so for first order extinction, this coefficient is much harder to determine correctly. The two ways to help in the fit are (a) select a pair with the widest color difference; and/or (b) observe over the widest airmass difference. Usually it is hard to get more than one magnitude of color difference, so the airmass is the biggest contributor to the quality of the fit. I usually observe from X=1.2 to X=2.5-3, and do it for multiple stars over multiple nights and average the slope results. Note that the second order coefficient doesn't change rapidly, so this only has to be done rarely.

In theory, you could take all the calibrated stars in your image and perform this calculation for every pair. However, this doesn't buy you a whole lot, because you still have a limited set of airmass values - the pairs are correlated since they are from the same set of images.

Good luck! Including second-order effects like this can get rid of many of the systematic differences between observers.

Arne

Ed:

For PEP calibration we try to make measurements over a range of 50 degrees of altitude. My most recent value was -0.033.

Tom

Arne et al:

When Gordon and I looked into this coefficient for our transformation/extinction presentation in England last summer, we concluded that using a default value of -0.03 was generally better than trying to make what is a difficult measure. Available papers indicated that this value was commonly calculated/confirmed by many!?

I like to measure things BUT it appears that the error far outweighs the value!?

Arne: Can you provide a conclusive statement about whether it has a negative or positive sign. I think it is a matter of exactly how the equation is set up/used?

BTW, we observed that even though most people ignore this second order extinction coefficient, it can be significant but only when measuring very red stars in a B filter? We also observed that 2nd order coefficients for other redder filters are effectively zero and can be assumed to be such?.

Ken

Second order extinction arises because the first order extinction decreases as a function of wavelength, and therefore is non-constant across a wide bandpass like a Johnson/Cousins or Sloan filter. For wide bandpasses, the blue end of the filter's passband is extincted more than the red end. The result is that a blue star shows more extinction, since it has more flux in the blue, than does a red star. So a red star will appear brighter than average, and therefore must be decreased in brightness in order to obtain its extra-atmospheric value (we always want magnitudes outside of the earth's atmosphere for comparison purposes, and always want magnitudes to be calculated for the central wavelength of the filter bandpass).

The basic equation to derive a magnitude outside of the earth's atmosphere, which is what we want, is

m_o = m -k'X -k"XC

where k' is the first order, and k" the second order extinction coefficients, and C is the appropriate color index for that star. k' is positive (typically 0.2-0.3 for V), which means you decrease the magnitude (increase the brightness) to determine the extra-atmospheric value - the atmosphere makes things fainter at ground level. k" is usually negative with color when the equation is written as above, so that red stars need to have their magnitude increased (decrease in brightness) because they overall suffer less extinction. The inverted magnitude system makes intuitive thinking difficult, so follow it through.

Note that first order extinction is a function of wavelength, and it changes most rapidly in the blue. So blue magnitudes are most strongly affected by 2nd order extinction, because extinction varies the most across those passbands.

Note also that the coefficient is about -0.03 or -0.04 for B using the equation above, about 10x smaller than the first order coefficient. So it is 10x harder to determine this coefficient. You have to use a pristene night, so that your red-blue pair can be observed for several hours with with quality photometry. You can certainly use a star pair at large Declination, but then it takes longer to cover a wide airmass range, and so your sky has to remain stable for that entire period.

Second order extinction is important in differential photometry because it does not subtract out of the difference equation. With poor notation, an example for calculating the b exo-atmospheric difference for two stars at the same airmass is

(b_o1-b_o2) = [b1 - k'X -k"XC1] - [b2 -k'X -k"XC2]

(b_o1-b_o2) = (b1-b2) - k"X(C1-C2)

If C1=C2, all of this reduces to

b_o1-b_o2 = b1-b2

yay! It is when the colors of the two stars don't match that you run into problems. Let's say that one star is has a color index one magnitude different than the other. Let's also assume that k'"= -0.03, and that we are observing at an airmass of 2. An example would be observing an A0 star with (B-V)=0 using a late G star with (B-V) = 1. Then

(B1-B2) = (b1-b2) -0.03*(1-0)*2.0

(B1-B2) = (b1-b2) - 0.06

So about 0.03mag from the zenithal (X=1) value. The large color dependence means that, even if you exactly match your filter system to the Johnson system, 2nd order extinction remains important if the two stars in differential photometry have different color. You will have an offset compared to other observers, much like the difference in transformation if your system does not match Johnson. Luckily, the 2nd order coefficients tend to be small; see Table 2 in Landolt's paper:

http://aspbooks.org/publications/364/027.pdf

A couple of other points. This is a systematic effect, based primarily on airmass and comparison-star color difference from the target. If everyone uses the same comparison star for a target, and observes at the same airmass, you wouldn't see the difference on the overlayed plot. If the star is at a far southern declination, so that it is commonly observed at high airmass for northern observers and low airmass for southern observers, then you should see a north/south discrepancy if 2nd order is neglected. If a star is observed in a time series from the zenith to as low as you can go, you should see a systematic trend of the target with time. In fact, time series in photometric weather can be a great way to get your 2nd order coefficient, using two constant stars in the field.

I hope that detail helps. As Ken mentions, the 2nd order coefficient is usually insignificant for VRcIc, and can be ignored for those passbands. It is important for B, and gets really weird for U, which is a different story.

Arne

Arne:

There is a puzzle about second-order extinction. It is caused by the atmosphere, but we measure it only occasionally, implying that atmospheric variation doesn't affect it much. I have held the theory that when we determine changes in k'', we are actually measuring changes in the instrumental response. Eg: if we gain extra sensitivity at the short-wavelength end of the B band, we will experience more second-order extinction. Is this correct?

Tom

Many of the brighter pairs in the lists of blue-red pairs are either too bright or too low. Or, are they? What is the minimum X-value for a good second-order estimate? x=1.4? X=1.2?

It would help encourage second-order estimates is there was a list of blue-red pairs in the +20-+40 Dec range with Vmags of 9-12.

Just wishing, Ed

Ed:

What you need is a large *range* of airmasses. If your horizon only permits observations down to X=2, you want to start near X=1.

Ed:

Try the AAVSO Standard Fields (M67, NGC7790). Arne's photometry is second to Landolt fields. There are lots of comps with a wide color range because that helps for Transform Coefficients. Pick out two (or more pairs) with the widest range?.

Run them from meridian which for us will give a small X since they are quite far north, to as low as you can go (~25 degrees altitude). If it is at your zenith, the X will not change very fast at the start. That will tell you something about your error?

See if you get the typical average of -0.03 for k"b and zero (or statistically zero) for the others. I have found this hard to do and gave up. ;-( Tom C has apparently been able to get good values for his PEP. Perhaps he can chime in about whether it is difficult of not?

You can update your transform coeffs and 1st order extinction coeffs as well!

The main issue is how good the night is over hours of time for us mere mortals not on a tall mountain? I find that the errors are often so bad that using the default values is better?? Sinful/argumentative thought?

Ken

Ken,

Yes M67 would be a definate possibility. I think I have an interesting pair in Melotte 111. One problem with the standard charts that I have encountered is that the AUID is (apparently) not in VSX. This means that I cannot correlate the AUID with other nomenclature. For example, I think one of the blue stars is a 102 (there are several 102s). This seems to be Tycho 01989-014-1. The other is a 10.15 NSV 5608 (notation: deleted from GVCS) named j12250.7+244023. These seem to be 2mass 12251215+2535569 and 12250880+2536121.

Things would be easier if I could access a UBVI catalog for Melotte 111 stars for overaly on an image from Aladin. I failed to find one in VizieR, butI admit I didn't look very hard; hints would be appreciated. I'll keep looking. The problem with Melotte 111 is that it is huge, the benefit of Melotte 111 is that it is huge. [added in edit: Too large for reasonable FOVs to cover, but with lots of secondary standards to consider.]

M67, as you suggest, seems to be the best bet. I can fit it into my FOV and have already done two recent nights for new transforms, but not at different air masses. Now all I need is clear skies.

Thanks, Ed

Arne, what do you think about using e.g. Sloan stars for 2nd order calibration? Because SDSS stars are fainter than many (or all?) classic blue-red pairs, they are abundant, one can search for them as well. SDSS stars should have fairly good magnitudes established.

Yes, their zeropoint accuracy is +-0.03

^{m}or so when converted to Johnson-Cousins, but for 2nd order extinction, variation of one pair in colour index (even in worst case scenario) is +-0.06^{m}, for two pairs (please correct me if I'm wrong) is in worst case scenario +-0.12^{m}. According your example that is ~10x smaller discrepancy compared to not using 2nd order extinction coefficients at all.However, because there are so many Sloan stars, IMHO statistical approach could do the trick - e.g. just observing fields with either constant airmass or altitude steps and selecting from there constant stars with large colour difference. Of course... there is a weak point - possible variability, which again could be overcome by statistics.

I haven't done such test but that would be interesting - here in Estonia, we don't have many truly photometric nights and celestial equator culminates at airmass ~2...

Best wishes,

Tõnis

Brian Warner presents a list of Blue-Red Pairs in his

Practical Guide. It does not help Tonis as it is +/- 05 DEC. However, Brian outlines how he obtained the data: onine query of the Sloan data via the Sloan data query utility. I don't speak SQL, but I suspect someone on this forum does. I don't know how easy it would be to generate such a list for, say +/- 2-5 DEC for, say +30 or +40 DEC, but I figure if Brian can do this, perhaps others can also, given that they have a bigger brain than I have. So, there seems to be a mechanism. The tutorial for findining close pairs:https://www.sdss.org/dr15/tutorials/pairs/

Ed

ps, I'll be looking into the SQL tutorial to expand my brain, but I am hoping that someone more knowledgable is interested in generating a list.

pps. Brian was kind enough to search for his original code, no luck, so its up to "us."

Answering Tom: 2nd order is 2nd order. It is 10x smaller than first order, and so changes in first order result in small variations in 2nd order. It is the slope of the first order extinction that is most important, and when extinction changes, the basic curve vs. wavelength is just shifted up or down for the most part.

Regarding new sets of 2nd order pairs, Tõnis has a good idea in using Sloan. However, there are several problems in practice. First, SDSS saturates at 14th magnitude, so these pairs would be very faint for use in determining the coefficients, with long exposures. Second, the red star may be ok at V, but would be even fainter at B and possibly saturated at converted Rc/Ic. APASS is probably your best bet for finding good red/blue pairs. What you want is to filter the catalog, choosing stars that are, say, within 5arcmin of each other, about the same Vmag, and without contaminating companions.

As Tom also mentioned, you want to observe over a wide airmass range. Whether you start at X=1.0 or X=1.5 makes little difference; it is the total range that helps you in fitting the line. Ken mentions both NGC7790 and M67, which are higher declination targets than the equatorial standards and which are suitable for 2nd order determination. Likewise in the south, you have the two open clusters NGC 1252 and NGC3532 with good photometry and are at decent southern declinations.

Don't be discouraged when trying to determine 2nd order extinction. It is similar in difficulty to observing an exoplanet transit IMHO. As Ken suggests, using typical coefficients takes care of most of the issue.

Arne