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How to derive this formula, please? Why is it true?

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$\mu$ is an emission coefficient

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The full radiative transfer equation is,

$$ \frac{1}{c}\frac{\partial}{\partial t}I_\nu + \hat{\Omega} \cdot \nabla I_\nu + (k_{\nu, s}+k_{\nu, a}) I_\nu = j_\nu + \frac{1}{4\pi}k_{\nu, s} \int_\Omega I_\nu d\Omega $$

In this notation $j_\nu$ is the emission coefficient.

Setting the absorption and scattering coefficients to zero yields,

$$ \frac{1}{c}\frac{\partial}{\partial t}I_\nu + \hat{\Omega} \cdot \nabla I_\nu = j_\nu $$

Reducing to one dimension,

$$ -\frac{1}{c}\frac{d I_\nu}{d t} + \cos(\theta)\frac{ d I_\nu}{d x} = j_\nu $$

Tidying up,

$$ \frac{ d I_\nu}{d x} = \frac{1}{\cos(\theta)c}\frac{d I_\nu}{d t} + j_\nu $$

Which is the same form as your equation if $\mu^{-1}=-\cos(\theta) c$ and $j_\nu=0$.

Moreover, your equation is consistent with an atmosphere that is not emitting radiation, nor absorbing it or scattering it! This is basically an equation describing the propagation of a beam of photons with angle $\theta$ to the x-axis travelling at speed $c$.

It also means that what you have written as “the emission coefficient” ($\mu$) is actually associated with the direction of the beam.

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  • $\begingroup$ Thank you for your answer. $\mu$ is the substitution for cosine of an angle. $\endgroup$
    – Elena Greg
    Commented Apr 1, 2021 at 11:48

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