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--- wiki:astronomy:observational_astronomy:data_reduction_toa [2025/01/07 14:21] – Roy Prouty
+++ wiki:astronomy:observational_astronomy:data_reduction_toa [2025/02/05 20:33] (current) – Roy Prouty
@@ Line 1: / Line 1: @@
 ===== Data Reduction III: Atmospheric Extinction =====
-WRITE THIS
+==== Recall the Pogson Equation====
+From the definition of a [[wiki:astronomy:magnitude|magnitude]]
-==== Instrument Magnitude ====
-Now invent an instrument-filter magnitude.
-Recall the definition of a [[wiki:astronomy:magnitude|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 ====
@@ Line 34: / Line 21: @@
 $$dI = -\kappa(x)I(x)\sec{(z)}dx$$
-From here, we can note that $\frac{dI}{I} = d\ln{I}$ OR we can invent an integrating factor $\exp{(-\int\kappa(x)dx)}$. In either case, 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 ($B$).
+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 ($B$).
-=== With Differential Natural Logarithm ===
 Rearrange:
-$$\frac{dI}{I} = d\ln{I} = -\kappa(x)\sec{(z)}dz$$
+$$d\ln{I} = -\kappa(x)\sec{(z)}dz$$
 Integrate:
-$$\int_T^B\frac{dI}{I} = \int_T^Bd\ln{I} = \int_T^B-\kappa(x)\sec{(z)}dx$$
+$$\int_T^Bd\ln{I} = \int_T^B-\kappa(x)\sec{(z)}dx$$
 We are left with:
 $$\ln{I(x=T)} - \ln{I(x=B)} = -\sec{(z)}\int_T^B\kappa(x)dx$$
@@ Line 52: / Line 38: @@
 $$\frac{I(x=T)}{I(x=B)} = e^{-\sec{(z)}K}$$
-=== With Integrating Factor ===
+==== Top-Of-Atmosphere Magnitude ====
-Rearrange:
+From here, we must trust that $I \propto F$ in some way. See [[wiki:astronomy:radiative_quantities|Radiative Quantities]] if that trust is hard to come by.
-$$\frac{dI}{dx} = -\kappa(x)\sec{(z)}I(x)$$
-Identifying the $-\kappa(x)\sec{(z)}$ as the "pre-factor" to the sought-after $I(x)$, we can invent the integrating factor:
-$$e^{-\int_T^B\kappa(x)\sec{(z)}dx}$$
-Choose to call the $\int_T^B\kappa(x)dx}$ as $K$:
-$$e^{-\K\sec{(z)}}$$
-Now multiply through by the integrating factor:
-$$e^{-\K\sec{(z)}}\frac{dI}{dx} = -e^{-\K\sec{(z)}}\kappa(x)\sec{(z)}I(x)$$
-Rearrange again:
-$$e^{-\K\sec{(z)}}\frac{dI}{dx} + e^{-\K\sec{(z)}}\kappa(x)\sec{(z)}I(x) = 0 $$
-Note that this matches the form of a product rule result:
-$$\frac{d}{dx}\Bigl(e^{-\K\sec{(z)}}I(x)\Bigr) = 0$$
+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:
-Sources
+$$m_T - m_B = -2.5(-\sec{(z)}K\cdot\log_{10}e)$$
-  - [[https://slittlefair.staff.shef.ac.uk/teaching/phy217/lectures/principles/l04/]]
+$$m_T - m_B = 2.5\sec{(z)}K\cdot\log_{10}e$$
+$$m_T = 2.5\sec{(z)}K\cdot\log_{10}e + m_B$$
+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 be 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.
+==== Measuring Atmospheric Extinction ====
+Take a look at the final result from above again:
+$$m_T = 2.5\sec{(z)}K\cdot\log_{10}e + m_B$$
+Choose to rearrange this in the form of a standard $y=mx+b$ linear form with $m_B$ as the dependent variable:
+$$m_B = -K\Biggl(2.5\sec{(z)}\cdot\log_{10}e\Biggr) + m_T$$
+With this, we can see that on a graph of $m_B$ vs $2.5\sec{(z)}\cdot\log_{10}e$, the magnitude of the slope will be $K$ and the y-intercept will be $m_T$.\\ \\
+Therefore, in order to measure $K$, we must observe a source for which $m_T$ is constant throughout the session and we must observe this source at a variety of zenith-angles, $z$. The top-of-telescope instrument magnitudes resulting from enough observations at differing zenith angles should trace a line. A fit to that line will result in a y-intercept of the top-of-atmosphere magnitude for the observed source and a slope that can serve as a reasonable estimate of the atmosphere-integrated extinction.\\ \\
+It's important to to note that the source we aim to understand as a part of the overall observing stands to be different from the source we observe to measure the extinction. Further, the source we observe to measure extinction should be non-variable in nature and appear in the same region of the sky as the source in question. This source we observe the measure extinction should be called the "extinction star".
+----
-{{tag>[not-done]}}
+{{tag>[done]}}