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:

    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

    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
    Date06/19/14 23:53 ID image252  
    Star nameCat magAltMag imageIntensityBIDmAirmass
    20 boo4.841.07-11,128282532468-15,9281,520865865
    f boo5,444.38-10.35137982457-15.751,428837794
    14 boo5.537.5-10,545165182482-16,0451,64094179
    15 boo5.2535.1-10.525162252483-15,7751,736922288
    HIP 704005.132,4-10.645181182463-15,7451.863391058
    u boo4.0536,073-12.06666552516-16.111,696331832
    70 vir4.9532,2-11,021256162565-15,9711,873663569
    71 vir5.6529.85-10,324134762556-15,9742,005323345
    e vir5.1527-10,352138332589-15,5022.197418439
    eps vir2.8526.22-12,8641398372607-15,7142.25757501
    HIP 634206.724.72-8,94937992614-15,6492,384289105
    sigma vir4.7523.08-11,065266712585-15.8152,542206706
    84 vir5,425.5-10,561167612533-15,9612,316481972
    d2 vir5.221.18-10.612175672631-15,8122,756376852
    d1 vir5.5519.95-9,79280672639-15,3422.917077172
    c vir4.9514-10.826214032656-15,7764.092871581
    pi vir4.6513.82-10,764202092676-15,4144,144019169
    o vir4.116.17-11.44376602658-15.543,564544399
    6 com5.0522.67-10.65181962625-15.72,58533522
    12 vir5.8518.06-9,25547292647-15.1053,20695548
    11 com4.725.65-11,084271442610-15,7842,303928428
    24 Com A4.9528.23-11,226309292614-16,1762,109551775
    23 com4.9522.55-11,069267662596-16.0192.598260423
    31 com4.937.07-10.961242332546-15,8611,657141284
    beta com4.240.93-11,658460372523-15,8581,525134451
    37 com5.0541.35-11,107277282521-16,1571,51242676
    HIP 629726.2542.04-9.7780912500-16.021.492174262
    14 CVn5.245.68-10,579170502481-15,7791.396892826
    HIP 626415.8544.53-10.011101022476-15,8611.425039994
    HIP 645436.6544.6-9,46360992496-16,1131.423277903
    HIP 632677.1524.18-8,60527672613-15,7552,43386583
    HIP 63221 A7.523,23-8,81933712604-16,3192,526815663
    delta vir3.3518.95-12,289823592620-15,6393.063224414
    37 vir618.12-9,78381852620-15,7833.19682149
    33 vir6.418,017-9,46861292625-15,8683,214259678
    HIP 616585.6515.27-9,78982372643-15,4393,765690356
    HIP 616376.316,4-8,51825542647-14.8183,516648775
    HIP 608506.715.92-9.2952012658-15.993.618160686
    eta vir3.8510.57-11,084271512690-14,9345.357099392
    10 vir5.9511.18-7,4939942702-13,4435,077628047
    b vir5.3511.18-8,78832752706-14,1385,077628047
    HIP 588096.3513.28-8,54526182684-14.8954.305596468
    11 vir5.714.42-8,83734252677-14,5373.97839932
    17 vir6.4515.81-9.35255072657-15,8023,642283826

    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 ...

    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 ...

    Also popular now: