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wiki:astronomy:observational_astronomy:observational_astronomy:extinction [2023/11/14 12:47] – Roy Prouty | wiki:astronomy:observational_astronomy:observational_astronomy:extinction [2023/11/14 13:11] (current) – Roy Prouty | ||
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If the atmosphere was absent any radiant sources or sinks, there would be no change to the radiance along its optical path. See [[wiki: | If the atmosphere was absent any radiant sources or sinks, there would be no change to the radiance along its optical path. See [[wiki: | ||
- | $$dI = 0$$ | + | $$dI_\lambda |
+ | |||
+ | However, if the atmosphere has some extinction cross-section (that is, an ability to scatter or absorb radiant energy at this wavelength) due to the particles within it, there is now a sink along the optical path for radiance. The rate at which the radiance is removed from incidence is proportional to both the amount of radiance available and the length through the extincting medium (the atmosphere). | ||
+ | |||
+ | $$dI_\lambda \propto - I_\lambda \sec{z} dx$$ | ||
+ | |||
+ | We invent some constant of proportionality, | ||
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+ | $$dI_\lambda = - \alpha_\lambda(x)I_\lambda \sec{z} dx$$ | ||
+ | |||
+ | We can solve this equation by rearranging terms and integrating //from the top to the bottom// of the atmosphere. | ||
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+ | $$\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$$ | ||
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+ | Employing some algebra, we are left with the ratio of two radiances. | ||
+ | |||
+ | $$\frac{I_\lambda^b}{I_\lambda^t} = \exp{\Biggl(- \sec{z} \int_b^t \alpha_\lambda(x) dx\Biggr)}$$ | ||
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+ | From the page on the [[wiki: | ||
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+ | So blazing forward with this equivalence, | ||
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+ | $$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)$$ | ||
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+ | Finally, recalling that we want to calibrate to the top-of-the-atmosphere, | ||
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+ | $$m_t = m_b + 2.5 \log_{10}(e)\biggl(- \sec{z} \int_b^t \alpha_\lambda(x) dx\biggr)$$ | ||
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+ | 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 " | ||
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+ | 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:// | ||
+ |