# Change of intensity along path element - plane-parallel geometry

How to derive this formula, please? Why is it true? $$\mu$$ is an emission coefficient

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

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.

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