# mechanism responsible for red sunset

Let's say a plane EM wave passes through an air molecule. To explain scattering classically, you can consider an electron held to the rest of the molecule by a spring that makes a forced oscillation at the frequency of the wave. You can show that if the frequency of the wave is way smaller than the natural frequency of the electron (which is the case for visible light on air molecules), then the oscillating electron will emit an EM wave in phase with the incident wave, at the same frequency, in almost every direction, having an electric field whose amplitude depends on the direction and is proportional to the square of the frequency. This implies that blue light is scattered more than red light, which explains why the sky is blue. This is fine.

The explanation that sunsets are red because if blue scatters more, then a white beam of light from the sun has lost more blue and thus apppears red makes sense. However, it doesn't tell me what the mechanism responsible for reducing the E-field of the blue wave is! In my mind, the only way the E-field can decrease is by destructive interference, but the E-field produced by the molecule is in phase and it interferes constructively with the incident E-field! Even if you allow for damping (to take care of radiation resistance), you get a scattered E-field that is 90° out of phase with the incident E-field, which doesn't reduce the net E-field.

• The radiation from the sun comes from one direction and the shorter wavelengths get scattered in different directions and that's why the light path, when seen from the side, appears blueish. It's just geometric attenuation. Is that what you are looking for? In any case, interference doesn't produce attenuation. It will only diminish the intensity of the interfering waves in one location, but that has to be offset with a higher intensity in another. Mar 3, 2016 at 23:08
• It is scattered, not reduced. The sky above those to the west of you is blue because that is where the blue is going. Mar 3, 2016 at 23:14
• There is no loss of blue light as such. As the light is coming towards you some of the light is absorbed by the particles/molecules in the atmosphere and re-radiated in all directions, so there is less light coming towards you. The effect is most pronounced for blue light. Mar 3, 2016 at 23:16
• Let's forget about the sky for a moment. Let's say you have a monochromatic laser beam directed towards a light sensor. As you move the sensor away from the laser, the intensity read by the sensor will decrease because of scattering on air molecules (extinction). Since intensity is proportional to the square of the E-field amplitude of the EM wave, this means that the E-field amplitude has decreased in the beam. My question is : how? Mar 3, 2016 at 23:24

A simile to help your intuition: imagine an army of ants marching from A (the anthill where they live) to C (the cupboard where you keep the honey). At every point along the way there is a small probability $p$ that an ant will find an interesting scent, and starts wandering off. How may ants reach C? Clearly the answer will depend on how large $p$ is, and how far apart A and C. The ants don't disappear - they just stop walking in the right direction.
To address your comment: when a photon interacts with an atom, that atom becomes polarized. And just like a dipole antenna, the radiation from that oscillating dipole is in all directions (except directly along the direction of polarization). If we define a ray at an angle $\phi$ about the axis of polarization, and $\theta$ relative to that axis, then radiation is uniform in $\phi$ and follows $\sin\theta$ in the azimuthal direction. Meaning there are plenty of directions in which the light will be emitted with equal probability.