# Polarization by Scattering

Above this sentence is an image of the diagram and paragraph I have been reading about polarization via scattering.

Question: Light EM waves are transverse waves aka move perpendicular to the direction they are traveling. There can be no polarization in the scattered light parallel to the original ray because that would require the original ray to be a longitudinal wave. Can someone clarify why this is the case?

My thoughts

I was trying to relate the scattering of unpolarized light to how unscattered light behaves through two polarizers. From observation when the second polarizer is orthogonal to the first, then no polarized light can get through. Should these two phenomenons be observed as separate cases?

• So the short version of the question is why there are no longitudinal EM waves (in vacuum and air)? – user137289 Oct 24 '20 at 10:20
• No. Why is the reason there is no polarization in the scattered light parallel to the original ray when observing unpolarized light due to the fact that EM waves are transverse waves and the original ray would have to be longitudinal in order to see polarized light parallel ot the original ray? – WigbertPowrr Oct 24 '20 at 11:02
• The incident ray is a transverse wave (with a rapidly changing direction of the electric field). The forward scattered ray (I suppose that is what you mean with parallel ray) is then also unpolarized. As shown in the figure. I clearly do not understand what your question is. I would advise you to do the experiment. – user137289 Oct 24 '20 at 15:48

Keep in mind that the red arrows in the sketch represent electric field vectors which are perpendicular to the direction of propagation. These oscillating fields cause the molecules (or electrons) to oscillate in that same direction. Oscillating charges radiate strongly in a direction perpendicular to their direction of motion and not at all in a direction parallel to their motion. Therefore, if you observe a beam of light from the side, the scattered light will tend to be polarized in a direction which is perpendicular to the beam.

For unpolarized incident light, there is only one initial vector: $$\bf k$$, the wave vector of the light. The final state vector is $$\bf k'$$

The scattering plane is defined by $$\bf \hat k \times \hat k'$$, which defines an alignment for linear polarization:

$$\bf{\hat{\varepsilon}} \propto \pm ({\bf \hat k \times \hat k'})$$

Note that linear polarization is a tensor alignment, so it points in either direction, like an un-oriented plane. Of course, if $${\bf \hat k'} = {\bf \hat k}$$, then the cross product is zero, and there is no scattering plane.

You may ask about circular polarization in the forward direction, as it's defined relative to $$\bf k'$$. That would look like

$${\bf S \cdot \hat k'} = \alpha {\bf \hat k \cdot \hat k'}$$

where $${\bf S}$$ is the spin on the photon and $$\alpha$$ is some factor. Under a parity transformation, that would go to:

$${(+{\bf S}) \cdot (-{\bf \hat k'})} = \alpha (- {\bf \hat k}) \cdot (-{\bf \hat k'})$$

$${\bf S \cdot \hat k'} = -\alpha {\bf \hat k \cdot \hat k'}$$

so that a non-zero $$\alpha$$ violates parity symmetry.