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

enter image description here

$\mu$ is an emission coefficient


1 Answer 1


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.

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.