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Data Reduction III: Atmospheric Extinction
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Instrument Magnitude
Now invent an instrument-filter magnitude.
Recall the definition of a magnitude
$$m_1 - m_2 = -2.5\log_{10}\frac{F_1}{F_2}$$
Observe that the collection of source #2 terms amounts to some constant
By properties of logarithms: $$m_1 - m_2 = -2.5\log_{10}F_1 - 2.5\log_{10}F_2$$
Grouping all source #2 terms: $$m_1 = -2.5\log_{10}F_1 + (m_2 - 2.5\log_{10}F_2)$$
Identify $m_2 - 2.5\log_{10}F_2 = C$ as an arbitrary constant. $$m = -2.5\log_{10}F + C$$
Atmospheric Extinction
Imagine breaking the vertical extent of the atmosphere up into infinitely many thin, parallel slabs of varying composition and character. As a ray of light enters the atmosphere and traverses some path through these layers, the ray interacts with each layer.
The Beer-Lambert Law gives us the relationship between the intensity of light rays entering a medium and the intensity of the resultant outgoing light rays. If the medium can be characterized with some extinction coefficient $\kappa$ in units of $\frac{\%}{m}$, then the change in intensity for every unit length of the medium traversed is $\Delta{I} = - \kappa I(x=0)\cdot \Delta{x}$, where $I(x=0)$ can be taken as the intensity of the light ray as it enters the medium.
In differential form, and allowing for $\kappa$ to depend on vertical location: $$\frac{dI}{dx} = -\kappa(x)I(x)$$
In the diagram above, you can see that any angle relative to zenith ($z$) actually forces the light ray to traverse a longer path through the atmosphere by a factor of $\sec{(z)}$. So the Beer-Lambert Law tells us that the differential amount of intensity lost to the extinctive character of the layers of the atmosphere is:
$$dI = -\kappa(x)I(x)\sec{(z)}dx$$
From here, we can note that $\frac{dI}{I} = d\ln{I}$. We will integrate over the path of the light ray: From the top of the atmosphere ($T$) to the bottom of the atmosphere, i.e., the ground ($B$).
Rearrange: $$\frac{dI}{I} = d\ln{I} = -\kappa(x)\sec{(z)}dz$$ Integrate: $$\int_T^B\frac{dI}{I} = \int_T^Bd\ln{I} = \int_T^B-\kappa(x)\sec{(z)}dx$$ We are left with: $$\ln{I(x=T)} - \ln{I(x=B)} = -\sec{(z)}\int_T^B\kappa(x)dx$$ Note that the $\int_T^B\kappa(x)dx$ cannot be known without careful (likely *in situ*) measurements. So let's just call this $K$. $$\ln{I(x=T)} - \ln{I(x=B)} = -\sec{(z)}K$$ By some property of logarithms: $$\ln{\frac{I(x=T)}{I(x=B)}} = -\sec{(z)}K$$
And so:
$$\frac{I(x=T)}{I(x=B)} = e^{-\sec{(z)}K}$$
Top-Of-Atmosphere Magnitude
From here, we must trust that $I \propto F$ in some way. See Radiative Quantities if that trust is hard to come by.
Given this proportionality, we can say that $I_1$ is the top-of-atmosphere intensity and that $I_2$ is the intensity that we register on our detector (reduced to the top-of-telescope) intensity.
$$m_T - m_B = -2.5\log_{10}\frac{I_T}{I_B}$$
Given the results from above, this means that: $$m_T - m_B = -2.5\log_{10}e^{-\sec{(z)}K}$$
We can clean this up a little algebraically:
$$m_T - m_B = -2.5(-\sec{(z)}K\cdot\log_{10}e)$$
$$m_T - m_B = 2.5\sec{(z)}K\cdot\log_{10}e$$
$$m_T = 2.5\sec{(z)}K\cdot\log_{10}e + m_B$$
So here is the work-up for the top-of-atmosphere magnitude based on $m_B$, $\sec{(z)}$, & $K$.
- $m_B$ is the top-of-telescope instrument magnitude.
- $\sec{(z)}$ is the secant of the angle between zenith and the observed source
- $K$ is the atmosphere-integrated extinction
The first two are easy (ish) ! That last one, the atmosphere-integrated extinction is less easy. It must me derived either from measurements of the layer-by-layer extinction coefficient and numerically integrated OR we must find some way to measure the atmosphere-integrated extinction itself.