# I Mechanism for extinction by scattering?

1. Mar 3, 2016

### damosuz

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?

In my mind, the only way to decrease the E-field is if the air molecules produce an E-field that interferes destructively with the incident E-field. However, by treating air molecules as electrons oscillating on a spring, you get constructive interference for visible incident light.

2. Mar 3, 2016

### Staff: Mentor

How?
Note that the far field looks different from the near field.

Scattering is a quantum mechanical process, but you should get a qualitative picture with small antennas.

3. Mar 3, 2016

### damosuz

If light frequency is way smaller than resonance frequency of electron in molecule, electron will oscillate in phase with the force, thus accelerate 180° with the force. The EM wave produced by the molecule will then be 180° with the force, thus in phase with the incident E-field.

I agree that far field is different than near field, but should we observe the same extinction whether we put the light sensor close to the air molecules or not (by putting it at the end of a long vacuum tube)?

4. Mar 4, 2016

### Staff: Mentor

That is not in phase.
If the frequency of light is small, the electron will follow the force vector in phase, and the induced electric field is opposite to the external field.
Scattering will look completely different if your sensor is within 1-2 wavelengths of things that scatter.

5. Mar 4, 2016

### damosuz

Electron accelerates at 180° with the force, but force is 180° with incident E-field, so acceleration (and emitted E-field) is in phase with incident E-field.

6. Mar 4, 2016

### Staff: Mentor

No, it follows the force direction.

7. Mar 4, 2016

### damosuz

You mean it follows the force in the sense that its position will be positive if force is positive, right? Which means acceleration will be 180° with the force...

8. Mar 5, 2016

### Staff: Mentor

I get a different 180° count.

Anyway, what you suggest would violate conservation of energy. Even if scattering would be purely classical the phase has to be right.

9. Mar 6, 2016

### damosuz

I think I found the answer. It seems like you cannot get destructive interference with a single molecule, but if you consider a large plane of such molecules oscillating in phase, then you can show that the E-field produced by the plane will be at 180° with the velocity of the electrons. If you consider a damping force (which would be related to radiation reaction for a gas), then for f < f0 you get a scattered wave at 90° with the incident wave (giving the impression that light travels slower) and another at 180° (responsible for extinction).

10. Mar 9, 2016

### Jano L.

This is indeed the case if the frequency of the external field is much lower than the natural frequency of the oscillating electron.

It can be shown that the radiation component of the electric field due to such oscillating electron, at a place farther down the line of propagation of the external wave, has the same sign as the external electric field. Thus the electron increases the total electric intensity in this direction on this line, "downstream". On the other hand, the electric field due to the electron on a half-line upstream now has two travelling waves of opposite velocity and opposite phase at each point of half-line upstream. Thus the electron decreases the total electric intensity on this half-line.

In the following graph, field of external linearly polarized wave (electric field has only y component) coming from the left (violet), the radiation field of the charged particle at $x=0$ (green) and their sum (gray) are depicted on the line of propagation at time when the force on the particle is zero. Notice how the field gets decrease in front of the oscillating particle but increased behind it. This is only one-dimensional view of things though; there are other points of the surrounding space where the field strength may be higher or lower. This is significant only in the vicinity of the electron; far from it, the external wave dominates.