# Why does the scattering cross section equal to the sum over all differential cross section; including the incident angle?

According to Beer Lambert's law, the intensity of light passing through a homogeneous medium diminishes at a rate proportional to the incident intensity; i.e.

$$\frac{dI(s)}{ds} = -I(s)\sigma\, ,$$

where $$s$$ is the parameter for the length of path taken by the light, and $$\sigma$$ is the volume extinction coefficient. Now assuming that the extinction happens purely from scattering, furthermore, assume that only single scattering events take place. The usual method to determining $$\sigma$$ is to sum the differential scattering cross sections $$d\sigma$$,

$$\sigma = \int d\sigma = \oint\frac{d\sigma}{d\Omega}\, d\Omega = \int_0^{2\pi}\, d\phi\, \int_0^\pi \sin\theta\, d\theta \frac{d\sigma}{d\Omega}(\phi,\theta)\, ,$$

where the last equality was taken from wiki (https://en.wikipedia.org/wiki/Cross_section_(physics)).

My question is, assuming that the incident light is coming in from an angle of $$\theta=\pi$$, why do we include $$\theta=0$$ in the integral? Isn't it the case that any light that scatters into the $$\theta=0$$ angle contributes to the intensity $$I(s)$$? Instead, shouldn't scattering cross section the equation be the one shown below?

$$\sigma = \int^{2\pi}_0d\phi\, \lim_{\Theta\rightarrow 0} \int_\Theta^\pi\sin\theta\, \frac{d\sigma}{d\Omega}(\phi, \theta)$$

Consider a plane wave propagating inside a homogeneous medium. Its distribution of directions will be

$$\sigma_I(\vec\omega)=I \delta(\vec\omega-\vec\omega_0),$$

where $$\vec\omega_0$$ is the direction of its wave vector, and $$\delta$$ is the Dirac delta distribution. Any single scattering on a scattering center $$\vec r$$ will result in an outgoing wave, which cannot have a finite number of finite-amplitude$$^\dagger$$ delta-like singularities, because the scattering center itself is finite, and due to this finite size any sharp outgoing wave packet will diverge due to diffraction.

This means that integral of this outgoing wave over any solid angle $$\varepsilon$$ will have the limit of $$0$$ as $$\varepsilon\to0$$. On the other hand, integrating the incoming wave around $$\vec\omega_0$$ will yield a nonzero limit as the integration domain shrinks to zero (due to the properties of Dirac delta distribution). As your last equation only contains the former wave, but not the latter, this equation would be equivalent to the first one.

This will continue to be true when multiple scattering orders are taken into account. The only case where this analysis may break down is when the medium is organized: e.g. when it's a crystal or a quasi-crystal. In this case the interference between scattered fields (of all scattering orders) of individual scatterers will result in an outgoing wave with delta-like peaks in distribution of directions. But in this case Beer–Lambert law will be inapplicable anyway: $$\sigma$$ may be finite while extinction due to scattering is zero.

$$^\dagger$$By amplitude of a delta-like singularity $$A\delta(x)$$ I mean its factor $$A$$.

• Thank you for the answer, just out of curiosity, was there any literature you were referring to when you penned this answer? I would love to read it – Tian Oct 19 '19 at 7:14
• @Tian unfortunately I don't have any particular literature to recommend (I'd love to have something specific to read myself :) ). – Ruslan Oct 19 '19 at 7:22

But none of the radiation is scattered exactly at any particular angle. You always have to integrate over a range of solid angle. Therefore none of the scattered radiation precisely contributes to $$I(s)$$.

Of course you can modify the Beer-Lambert law to add a source term to the RHS of your equation to account for the contribution of scattered light to the specific intensity. Something like $$\frac{dI(s)}{ds} = -\sigma I + \sigma J,$$ where $$J$$ is the mean value of $$I$$ over all solid angles. You should do this when the optical depth to scattering becomes significant and it becomes a trickier equation to solve.