What happens to the incident electromagnetic wave in Rayleigh scattering? In Rayleigh scattering an incident EM wave causes an induced dipole oscillation of an atom/molecule, which in turn causes radiation at the same frequency of the incident wave. But what happens to the original wave?
I have read quite a few pages on stackexchange and wiki articles on Rayleigh scattering but nowhere does it seem to be clearly stated. Is the wave expended in inducing the dipole? Is it cancelled by something? It makes 'classical' sense to me that as you can't get something for nothing it ought to be 'spent', but I don't know how.
Possibly related: What does it mean to be 'less strongly' scattered'?
The power radiated from the dipole is proportional to wavelength as  $P\propto\omega^4$. Red light from the sun is said to be less strongly or less efficiently scattered in the atmosphere. Now since I know that more red light passes by undeviated in direct sunlight, I'm led to think that 'less strongly scattered' means that more of the incident red light somehow survives the scattering event. I can interpret this two ways:


*

*Less of the incident red light wave is expended in inducing the oscillating dipole, compared with blue light. (Which perhaps explains why the radiated power is lower).

*Less red light is 'incident' in the first place - a higher wavelength makes the scattering event less likely.


Neither option feels right and I think my misunderstanding here is related to the above question.
(long-time reader, first-time poster)
 A: For a simple, single scattering, scenario: If the scattering cross-section is $\sigma$, then for Rayleigh scattering on very small particles, it is something like $\sigma \simeq \sigma_T (\omega/\omega_0)^4$, where $\omega_0$ is some characteristic resonant frequency and $\sigma_T = 6.6\times 10^{-29}$m$^2$ is the Thomson scattering cross-section.
If the number density of particles is $n$ per m$^3$, then when traversing a length $x$ through the scattering medium, then a fraction $\sim n \sigma x$ of the area of an incoming beam will be effectively blocked by scatterers (as long as $n \sigma x \ll 1$). More formally we can write a version of the radiative transfer equation.
$$ \frac{dI}{dx} = - n\sigma I$$
with a solution
$$ I = I_0 \exp (-n\sigma x)$$
Here $I/I_0 = \exp(-n\sigma x)$ will be the fraction of the beam intensity that emerges on the other side of the scattering medium.
Red light is less scattered than blue light because the resonant frequencies of the oxygen and nitrogen atoms in the atmosphere is somewhere in the ultraviolet, so $\omega \ll \omega_0$ and the cross-section of red light is much less than that for blue light. i.e. $\sigma$ is smaller for blue light, so $I/I_0$ will be larger.
A more correct interpretation than either your (1) or (2) would be that red light excites lower amplitude oscillations in the dipole which results in less scattered power.
