Atmospheric absorption or how to evaluate air pollution

Many here are fascinated by astrophotography and for the mighty it has become a hobby. However, your hobby can be a very interesting application and you can do a whole study on the basis of one photo. Today we will try to assess the degree of pollution in your city from a photo. Who are interested in the post, then welcome under cat.

Some 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, because of industrial enterprises or car exhausts. Due to all this, the radiation that passes through the atmosphere is greatly attenuated. And the longer the path that the light travels in the atmosphere, the stronger this attenuation is. 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 us try to derive the dependence of the long path of the light beam in the atmosphere on the height of the light source:

Apply the sine theorem to this triangle:

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

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

Here we used the fact that:

However, the length of the path in the atmosphere is not expressed in terms of length, but in dimensionless faces, that is, the path length of the light beam in the atmosphere is expressed in the heights of a homogeneous atmosphere and this unit is called the atmospheric mass (eng: airmass). Let k = R / H be the ratio of the Earth's radius and the height of a homogeneous atmosphere (k = 800)
Then in air masses our formula will look like:

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

Where:
m is the observed magnitude
m0 - star shine outside the atmosphere
Δm - atmospheric absorption in zenith in stellar magnitudes
L - air mass

A little theory about photometry

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

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

Where M0 is the atmospheric instrumental stellar magnitude.
Here is our takeover.
Thus, our main task has been reduced to finding absorption at the zenith Δm

Practice

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

Then we set camera parameters in Camera settings:


Then finally decode the 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 must select Analysis-> Aperture photometry. With a drop-down window, I advise you to simply agree and get to work. Instead of the cursor, you will have three circles and your task is to hover the center of such a cursor on the star and click. After clicking, the following data will appear in the Output window:

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 - brilliance in instrumental stellar magnitudes (for 0, such brilliance is taken whose intensity from one pixel is 1)
Background mean level = 2755.0 - background current in arbitrary units.
Next you need to open Stellarium and identify the star. This information should be recorded in any table, for example MS Excel.
I did the following:

In such a table should be recorded as much information about the star. It is obligatory - its catalog gloss (Cat mag), the gloss measured (Mag Image) and the height, which was determined by Stellarium (Alt). In order not to get confused, I recommend to record the number of the star by catalog (Star name), it is also desirable to record the intensity values ​​and the background values.
Then, the atmospheric mass (Airmass) is calculated by altitude for each star. 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 the stars at different altitudes, especially at low ones. After all, the more data, the more accurate the final result. Moreover, we did not make calibration frames and noise influenced the results of photometry to varying degrees. On the other hand, the stars are different in color, and as a result, the maximum of their radiation lies at different frequencies, and at different frequencies the absorption may differ significantly ...


Next, you need to build a graph of the dependence of the instrumental stellar magnitude (Dm) on the air mass (Airmass). Select the type of scatter diagram. Now, using the least squares method, we need to find the linear equation most suitable to 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 tick on the item “Show the equation on the diagram”
I got something like this:

As we see our equation: 0.3786x-16,651.
The coefficient is the atmospheric absorption at the zenith and it will be 0.38m, and the free term (-16.7) is an instrumental magnitude without absorption.
Graphics in gnuplot:
Dependence on air mass:

Height dependence:

Strictly speaking, we got a good grade, because the common average is 0.44m.

What does this give us?

Determine how many times the light is attenuated by the atmosphere according to the Pogson formula:

We get that the light is attenuated by 30%. That is, if from a pillar of atmosphere with a cross-section of 1 m2 to take motes and arrange them all close to each other, then their area will be 0.3 m2.
It is worth noting that the absorption of pure (without impurities) air is 0.2m. Thus, in our city the atmosphere weakens the light by 17% more than clean air ...

Conclusion

We did 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 only one image. If there is a series of images, then by adding them you can achieve even more accurate results ...