wiki:astronomy:observational_astronomy:observational_astronomy:extinction

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wiki:astronomy:observational_astronomy:observational_astronomy:extinction [2023/11/14 12:40] – created Roy Proutywiki:astronomy:observational_astronomy:observational_astronomy:extinction [2023/11/14 13:11] (current) – Roy Prouty
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 ==== Atmospheric Extinction ==== ==== Atmospheric Extinction ====
-Atmospheric extinction is the term used to describe any removal of radiance from incidence. This removal can be due to scattering events that scatter radiance out of incidence or due to absorption events that directly remove the radiance via transformations of the radiant energy to another form.+Atmospheric extinction is the term used to describe any removal of radiance from incidence. This removal can be due to scattering events that scatter radiance out of incidence or due to absorption events that directly remove the radiance via transformations of the radiant energy to another form.\\ \\
  
  
 {{https://slittlefair.staff.shef.ac.uk/teaching/phy217/lectures/principles/l04/files/small_1257.png}} {{https://slittlefair.staff.shef.ac.uk/teaching/phy217/lectures/principles/l04/files/small_1257.png}}
  
-Account for extinction.+Consider the radiance incident on the observer from the top-right of the frame. The optical path is measured from the top of the image downward to the bottom of the image. We hope to recover the radiance (or flux or energy or counts) that would be detected at the so-called top-of-the-atmosphere.\\ \\ 
 + 
 +If the atmosphere was absent any radiant sources or sinks, there would be no change to the radiance along its optical path. See [[wiki:astronomy:radiative_quantities#spectral_radiance|spectral radiance]] for a refresher. 
 +$$dI_\lambda = 0$$ 
 + 
 +However, if the atmosphere has some extinction cross-section (that is, an ability to scatter or absorb radiant energy at this wavelength) due to the particles within it, there is now a sink along the optical path for radiance. The rate at which the radiance is removed from incidence is proportional to both the amount of radiance available and the length through the extincting medium (the atmosphere). 
 + 
 +$$dI_\lambda \propto - I_\lambda \sec{z} dx$$ 
 + 
 +We invent some constant of proportionality, $\alpha_\lambda(x)$ that depends on the wavelength at play as well as at the level in the atmosphere the radiance is traversing.\\ \\   
 + 
 +$$dI_\lambda = - \alpha_\lambda(x)I_\lambda \sec{z} dx$$ 
 + 
 +We can solve this equation by rearranging terms and integrating //from the top to the bottom// of the atmosphere. 
 + 
 +$$\int_b^t \frac{dI_\lambda}{I_\lambda} = - \sec{z} \int_b^t \alpha_\lambda(x) dx$$ 
 + 
 +From the definition of the natural logarithm: 
 +$$\ln{(I_\lambda^b)} - \ln{(I_\lambda^t)} = \ln{\Biggl[\frac{I_\lambda^b}{I_\lambda^t}\Biggr]} = - \sec{z} \int_b^t \alpha_\lambda(x) dx$$ 
 + 
 +Employing some algebra, we are left with the ratio of two radiances. 
 + 
 +$$\frac{I_\lambda^b}{I_\lambda^t} = \exp{\Biggl(- \sec{z} \int_b^t \alpha_\lambda(x) dx\Biggr)}$$ 
 + 
 + 
 +From the page on the [[wiki:astronomy:magnitude|magnitude scale]], we can see that this is almost immediately relatable to the Pogson Equation. As it turns out, the only difference between the spectral, radiant fluxes reported on that page and the spectral radiances reported above is that the units are different. They're different by an integration over solid angle. Since these measurements are all derived from pixels of the same size -- and therefore observing the same solid angle -- both the flux ratio and the radiance ratio should be identical!\\ \\ 
 + 
 +So blazing forward with this equivalence, we can re-write the Pogson Equation in terms of this atmospheric extinction relationship we've just derived. 
 + 
 +$$m_b-m_t = -2.5 \log_{10}\frac{I_\lambda^b}{I_\lambda^t} = -2.5 \log_{10}\Biggl[\exp{\Biggl(- \sec{z} \int_b^t \alpha_\lambda(x) dx\Biggr)}\Biggr]$$ 
 + 
 +Let's carefully take the log-base-10 of the exponential... 
 +$$m_b-m_t = -2.5 \log_{10}(e)\biggl(- \sec{z} \int_b^t \alpha_\lambda(x) dx\biggr)$$ 
 + 
 +Finally, recalling that we want to calibrate to the top-of-the-atmosphere, we find 
 + 
 +$$m_t = m_b + 2.5 \log_{10}(e)\biggl(- \sec{z} \int_b^t \alpha_\lambda(x) dx\biggr)$$ 
 + 
 +So we find that the top-of-the-atmosphere magnitude is our bottom-of-the-atmosphere = top-of-telescope magnitude with some additional magnitude that was "extincted" away from incidence as the radiance traversed the atmosphere.\\ \\ 
 + 
 +Note that the $F$ or $I$ used in the above equations are //not// counts. They must be some unit of power. So radiance, flux, energy per second (power), or even counts-per-second. 
 + 
 +---- 
 +Sources 
 +  - Lifted with minor modification from: https://slittlefair.staff.shef.ac.uk/teaching/phy241/lectures/l07/