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.

  • $\begingroup$ 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. $\endgroup$
    – CuriousOne
    Mar 3, 2016 at 23:08
  • $\begingroup$ It is scattered, not reduced. The sky above those to the west of you is blue because that is where the blue is going. $\endgroup$
    – Jon Custer
    Mar 3, 2016 at 23:14
  • $\begingroup$ 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. $\endgroup$
    – Farcher
    Mar 3, 2016 at 23:16
  • $\begingroup$ 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? $\endgroup$
    – courno
    Mar 3, 2016 at 23:24

2 Answers 2


The air molecules scatter a fraction of the incident light - instead of it all traveling in the same direction, it ends up distributed in all directions. So at every point along the light beam, a few photons get "hijacked".

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.

If the red ants are more focused than the blue ants, you may start out with an equal mix; but more red ants arrive at C, because they didn't get distracted.

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.

  • $\begingroup$ I have no problem with the explanation in terms of photons. What I am looking for is a classical explanation, in terms of the electric field of the EM wave in the incident direction being reduced. $\endgroup$
    – courno
    Mar 4, 2016 at 1:24
  • 1
    $\begingroup$ @courno - Classical, you say? EM fields, you say? Try "Strutt, J. "On the transmission of light through an atmosphere containing small particles in suspension, and on the origin of the blue of the sky," Philosophical Magazine, series 5, vol. 47, pages 375-394 which is Rayleigh's complete explanation, or farside.ph.utexas.edu/teaching/em/lectures/node97.html for a somewhat shorter, modern version. $\endgroup$ Mar 4, 2016 at 1:56

The sky is blue. This is because white light, from the sun, does a small amount of scattering in air, and the scattered light comes from all portions of the sky, not just the direction of the sun. The scattering is called Compton scattering, and this effect is strongest for the blue end of the spectrum. So, when the sun is setting, it goes through, not just a mile or so of atmosphere (like at noon), but tens or hundreds of miles of atmosphere. The blue, green, part of the yellow light all scatter and the visible disk of the sun now appears red.


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