# Atmospheric absorption or how to assess air pollution

Many here are fond of astrophotography and for the mighty it has become a favorite pastime. However, you can find a very interesting application for your hobby and do a whole study on the basis of one photograph. Today we will try to estimate the degree of air pollution in your city from a photograph. Anyone interested in the post, then welcome to cat.

#### Bit of theory

Our atmosphere is not perfectly transparent: in addition to air there are gas and dust that get there in different ways, in cities, as a rule, due to industrial enterprises or car exhausts. Due to all this, the radiation that passes through the atmosphere is significantly attenuated. And the longer the path that light travels in the atmosphere, the stronger this attenuation. If you look at the figure below, it turns out that the path in the atmosphere is the longer, the lower the height of the light source:

Let's try to derive the dependence of the long path of a light beam in the atmosphere on the height of the light source:

We apply the sine theorem to this triangle:

Using the fact that A = 90 + h, where h is the height of the star, we get:

Further, using the first and last relation from the sine theorem, as well as the theorem on the sum of the angles in a triangle, we get:

Here we used the following:

However, the path length in the atmosphere is usually expressed not in dimension of length, but in dimensionless expressions, i.e. length The paths of a ray of light in the atmosphere are expressed in the heights of a homogeneous atmosphere, and such a unit is called the atmospheric mass (eng: airmass). Let k = R / H be the ratio of the radius of the Earth and the height of a homogeneous atmosphere (k = 800)
Then in the air masses our formula will take the form:

Now we need to understand how light is weakened in the atmosphere, depending on the air mass passed. The law that describes this is called Booger's law
in stellar magnitudes.the law will take a rather simple form:

Where:
m is the observed magnitude
m0 is the brightness of the star outside the atmosphere
Δm is the atmospheric absorption at the zenith in magnitudes
L is the air mass
###### A little theory about photometry

Now I’ll talk a little about photometry. When a star’s brightness is measured in a photograph, its magnitude m is given relative to the instrumental magnitude M:

Where m0 is the actual magnitude.
In turn, the instrumental magnitude M will be equal to:

Where M0 is the extra-atmospheric instrumental magnitude.
This is where our absorption hides.
Thus, our main task was to find the absorption at the zenith Δm
##### Practice

Now to practice. First, we need software for photometry. And it will be the workhorse of all astrophotographers - IRIS
The first thing we will do is decode raw.
First we set the working directory in File-> Settings.

Then we set the camera parameters in Camera settings:

Then finally we decode RAW: Digital photo -> Decode RAW files.
After decoding, press the Done button and the image will appear on the screen. Now we are ready for photometry.
You need to select Analysis-> Aperture photometry. I advise you to simply agree with the drop-down window and get to work. You will have three circles instead of the cursor and your task is to move the center of such a cursor to the star and click. After a click in the Output window, approximately the following data will appear:
Phot mode 3 - (979, 2553)
Pixel number in the inner circle = 197
Pixel number for background evaluation = 816
Intensity = 52348.0 - Magnitude = -11.797
Background mean level = 2755.0

We are interested in the last two lines:
Intensity = 52348.0 - intensity in arbitrary units
Magnitude = -11.797 - gloss in instrumental magnitudes (for 0 this brightness is taken whose intensity from one pixel is 1)
Background mean level = 2755.0 - background current in arbitrary units.
Next, you need to open the Stellarium and identify the star. This information should be entered in some table, for example MS Excel.
I did as follows:

In such a table should be entered as much information as possible about the star. Necessarily her catalog gloss (Cat mag), measured gloss (Mag Image) and height, which was determined by the Stellarium (Alt). In order not to get confused, I recommend recording the star number in the catalog (Star name), it is also advisable to record the intensity values ​​and background values.
Then, the atmospheric mass (Airmass) is calculated for each star by height. Then we find the instrumental magnitude Dm as the difference: Dm = Mag image-Cat mag
Please pay attention!It is necessary to enter as much data as possible about stars at different heights, especially at low. After all, the more data, the more accurate the final result. Moreover, we did not take calibration frames and the effects of photometry are affected to varying degrees by noise. On the other hand, Stars are different in color, and as a result, the maximum of their radiation lies at different frequencies, and at different frequencies the absorption can differ significantly ...
My details
 Date 06/19/14 23:53 ID image 252 Star name Cat mag Alt Mag image Intensity BI Dm Airmass 20 boo 4.8 41.07 -11,128 28253 2468 -15,928 1,520865865 f boo 5,4 44.38 -10.35 13798 2457 -15.75 1,428837794 14 boo 5.5 37.5 -10,545 16518 2482 -16,045 1,64094179 15 boo 5.25 35.1 -10.525 16225 2483 -15,775 1,736922288 HIP 70400 5.1 32,4 -10.645 18118 2463 -15,745 1.863391058 u boo 4.05 36,073 -12.06 66655 2516 -16.11 1,696331832 70 vir 4.95 32,2 -11,021 25616 2565 -15,971 1,873663569 71 vir 5.65 29.85 -10,324 13476 2556 -15,974 2,005323345 e vir 5.15 27 -10,352 13833 2589 -15,502 2.197418439 eps vir 2.85 26.22 -12,864 139837 2607 -15,714 2.25757501 HIP 63420 6.7 24.72 -8,949 3799 2614 -15,649 2,384289105 sigma vir 4.75 23.08 -11,065 26671 2585 -15.815 2,542206706 84 vir 5,4 25.5 -10,561 16761 2533 -15,961 2,316481972 d2 vir 5.2 21.18 -10.612 17567 2631 -15,812 2,756376852 d1 vir 5.55 19.95 -9,792 8067 2639 -15,342 2.917077172 c vir 4.95 14 -10.826 21403 2656 -15,776 4.092871581 pi vir 4.65 13.82 -10,764 20209 2676 -15,414 4,144019169 o vir 4.1 16.17 -11.44 37660 2658 -15.54 3,564544399 6 com 5.05 22.67 -10.65 18196 2625 -15.7 2,58533522 12 vir 5.85 18.06 -9,255 4729 2647 -15.105 3,20695548 11 com 4.7 25.65 -11,084 27144 2610 -15,784 2,303928428 24 Com A 4.95 28.23 -11,226 30929 2614 -16,176 2,109551775 23 com 4.95 22.55 -11,069 26766 2596 -16.019 2.598260423 31 com 4.9 37.07 -10.961 24233 2546 -15,861 1,657141284 beta com 4.2 40.93 -11,658 46037 2523 -15,858 1,525134451 37 com 5.05 41.35 -11,107 27728 2521 -16,157 1,51242676 HIP 62972 6.25 42.04 -9.77 8091 2500 -16.02 1.492174262 14 CVn 5.2 45.68 -10,579 17050 2481 -15,779 1.396892826 HIP 62641 5.85 44.53 -10.011 10102 2476 -15,861 1.425039994 HIP 64543 6.65 44.6 -9,463 6099 2496 -16,113 1.423277903 HIP 63267 7.15 24.18 -8,605 2767 2613 -15,755 2,43386583 HIP 63221 A 7.5 23,23 -8,819 3371 2604 -16,319 2,526815663 delta vir 3.35 18.95 -12,289 82359 2620 -15,639 3.063224414 37 vir 6 18.12 -9,783 8185 2620 -15,783 3.19682149 33 vir 6.4 18,017 -9,468 6129 2625 -15,868 3,214259678 HIP 61658 5.65 15.27 -9,789 8237 2643 -15,439 3,765690356 HIP 61637 6.3 16,4 -8,518 2554 2647 -14.818 3,516648775 HIP 60850 6.7 15.92 -9.29 5201 2658 -15.99 3.618160686 eta vir 3.85 10.57 -11,084 27151 2690 -14,934 5.357099392 10 vir 5.95 11.18 -7,493 994 2702 -13,443 5,077628047 b vir 5.35 11.18 -8,788 3275 2706 -14,138 5,077628047 HIP 58809 6.35 13.28 -8,545 2618 2684 -14.895 4.305596468 11 vir 5.7 14.42 -8,837 3425 2677 -14,537 3.97839932 17 vir 6.45 15.81 -9.352 5507 2657 -15,802 3,642283826

Next, you need to build a graph of the dependence of the instrumental magnitude (Dm) on the air mass (Airmass). Select the type of scatter plot. Now, using the least squares method, you need to find the linear equation that is most suitable for this graph. To do this, go to the top menu: Work with diagrams-> Layout-> Trend line-> Additional parameters of the trend line. Here we select the linear type and put a checkmark on the item “Show the equation in the diagram”
. I got a graph like this:

As we see our equation: 0.3786x-16.651.
The coefficient is atmospheric absorption at the zenith and it will be 0.38m, and the free term (-16.7) is an instrumental magnitude without absorption.
Charts in gnuplot:
Dependence on air mass:

Dependence on height:

Strictly speaking, we got a good mark, because The generally accepted average value is 0.44m.

#### What does this give us?

Let us determine how many times the light attenuates the atmosphere according to the Pogson formula:

We obtain that the light is attenuated by 30%. That is, if dust particles are taken from the atmospheric column with a cross section of 1 m2 and placed all of them close to each other, then their area will be 0.3 m2.
It should be noted that the absorption of clean (without impurities) air is 0.2m. Thus, in our city the atmosphere weakens the light by 17% more than clean air ...
##### Conclusion

We made a fairly simple analysis and did not go into complex processes, such as scattering or the dependence of absorption on the wavelength. However, we got a fairly accurate estimate using just one image. If there is a series of images, then adding them together you can achieve even more accurate results ...