Thomson scattering and unpolarized light

In summary, the author suggests that unpolarized light can be obtained by superposing two waves with opposite polarization directions.
  • #1
Wox
70
0
Elastic scattering from a bound electron is classically described by considering the driven, damped harmonic oscillator model for the motion of a bound electron in a classical em-wave. The (non-relativistic) equation of motion is written as
[tex]
m\frac{d^{2}\bar{x}}{dt^{2}}=q\bar{E}(t,\bar{x})-m\omega_{0}^{2}\bar{x}-m\gamma\frac{d\bar{x}}{dt}
[/tex]
where the effect of the magnetic field is neglected. For a monochromatic plane wave [itex]\bar{E}(t,\bar{x})=\bar{E}_{0}e^{i(\bar{k\cdot\bar{x}}-\omega t)}[/itex] the solution is given by
[tex]
\bar{x}(t)=e^{-\frac{\gamma}{2}t}(C_{1}e^{Ct}+C_{2}e^{-Ct})-\frac{q\bar{E}_{0}}{m(\omega^{2}-\omega_{0}^{2}+ \omega\gamma i)}e^{i(\bar{k\cdot\bar{x}}-\omega t)}
[/tex]
where [itex]C=\sqrt{\frac{\gamma^{2}}{4}-\omega_{0}^{2}}[/itex], [itex]C_{1}[/itex] and [itex]C_{2}[/itex] constants depending on the boundary conditions, [itex]\omega_{0}[/itex] is the resonance frequency of the harmonic oscillator and [itex]\gamma[/itex] the damping coefficient. (The first part is known as the transient solution and the second as the steady state solution.)

Now what would be the solution for unpolarized light?
 
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  • #2
Wox said:
Elastic scattering from a bound electron is classically described by considering the driven, damped harmonic oscillator model for the motion of a bound electron in a classical em-wave. The (non-relativistic) equation of motion is written as
[tex]
m\frac{d^{2}\bar{x}}{dt^{2}}=q\bar{E}(t,\bar{x})-m\omega_{0}^{2}\bar{x}-m\gamma\frac{d\bar{x}}{dt}
[/tex]
where the effect of the magnetic field is neglected. For a monochromatic plane wave [itex]\bar{E}(t,\bar{x})=\bar{E}_{0}e^{i(\bar{k\cdot\bar{x}}-\omega t)}[/itex] the solution is given by
[tex]
\bar{x}(t)=e^{-\frac{\gamma}{2}t}(C_{1}e^{Ct}+C_{2}e^{-Ct})-\frac{q\bar{E}_{0}}{m(\omega^{2}-\omega_{0}^{2}+ \omega\gamma i)}e^{i(\bar{k\cdot\bar{x}}-\omega t)}
[/tex]
where [itex]C=\sqrt{\frac{\gamma^{2}}{4}-\omega_{0}^{2}}[/itex], [itex]C_{1}[/itex] and [itex]C_{2}[/itex] constants depending on the boundary conditions, [itex]\omega_{0}[/itex] is the resonance frequency of the harmonic oscillator and [itex]\gamma[/itex] the damping coefficient. (The first part is known as the transient solution and the second as the steady state solution.)

Now what would be the solution for unpolarized light?


I think that these sites seem to cover your questions well :

http://en.wikipedia.org/wiki/Thomson_scattering
http://fermi.la.asu.edu/PHY531/thomson/index.html
 
  • #3
The links you gave me follow the usual reasoning. First they derive the result (i.e. the scattered intensity or the cross section) by assuming a monochromatic plane wave. This starts by solving the equation of motion of the electron as given in my original post (for a free electron [itex]\omega_{0}=0[/itex] and [itex]\gamma=0[/itex]). But when extending this to unpolarized radiation, I'm lost. The idea I get when reading it is that I should consider a monochromatic plane wave, but now with a random orientation of the electric field vector in the plane perpendicular to the propagation. That would mean
[tex]
\bar{E}(t,\bar{x})=\bar{E}_{0}(t)e^{i(\bar{k}\cdot \bar{x}-\omega t)}
[/tex]
with [itex]\bar{E}_{0}(t)[/itex] a vector in the plane perpendicular, with a particular norm but with a random orientation. This then means we can use the solution for scattering of a monochromatic plane wave
[tex]
\bar{E}(t,\bar{x})=\bar{E}_{0}e^{i(\bar{k} \cdot\bar{x}-\omega t)}
[/tex]
up till the point of time averaging (since [itex]\bar{E}_{0}[/itex] is now time dependent). Two reasons why this is not valid:
1. I'm not sure the solution of the equation of motion of the electron is analog to the monochromatic plane wave, just replacing [itex]\bar{E}_{0}[/itex] by [itex]\bar{E}_{0}(t)[/itex]
2. I've been told by several physicists that this is not the correct way of representing unpolarized radiation
Therefore this reasoning is not valid and my question still stands.
 
  • #4
Two identical waves with opposite polarization directions result in unpolarized light. So i suggest to try the superposition of a monochromatic polarized plane wave and a second wave with a 180°-phase shift. so that the polarization vector is always in the oppotise direction, resulting in zero net poolarization. but its only a guess, maybe it helps. i will sit down to the problme for myself later. ^^ have fun!
 
  • #5
Unpolarized light is the superposition of many wavelets of different polarization.

The electric field vectors add linearly, so if in any small volume of space there were many many waves coexisting they would average out to a net result of zero electric field, i.e. no wave.

Therefore if there is a wave with non-wero amplitude within our small volume, then it must have a polarization.

Then how can you get unpolarized beams?

It is more useful to think of "unpolarized" as average over time or over a finite volume much larger than the wave length and thus much larger than the atom. On the atomic scale light is polarized.

One might make the argument that the electric field vector performs a random walk with non-zero result... in that case there will be a (smaller) resulting electric field vector and thus a resulting polarization.
 
  • #6
I'm reading "Statistical Optics" by Goodman to understand how unpolarized light can be rigorously treated and there is more to it than just the sum of some monochromatic plane waves.
 
  • #7
Born and Wolf treat it using the "coherency matrix" that has nothing to do with coherence; instead it is a density matrix that effectively provides a means of summing up individual plane, perfectly polarized waves.

One way to rigorously treat scattering of polarization is to use Jones matrices that allow you to calculate how the density matrix (or better the polarization matrix) transform in a scattering process. Jones matrices are derived from the scattering of perfectly polarized waves.

Transmission (including phase shifts and absoption) is just forward scattering and can be treated using the same formalism.

http://arxiv.org/abs/1106.4446
 

1. What is Thomson scattering?

Thomson scattering is a physical phenomenon where electromagnetic radiation, such as light, is scattered by charged particles. It is named after physicist J.J. Thomson, who discovered it in the late 19th century.

2. How is unpolarized light related to Thomson scattering?

Unpolarized light is light where the electric field oscillates in all directions perpendicular to the direction of propagation. In Thomson scattering, unpolarized light is scattered by charged particles in all directions, resulting in a depolarization of the scattered light.

3. What can Thomson scattering be used for?

Thomson scattering is used in a variety of applications, including plasma diagnostics, remote sensing of atmospheric properties, and studying the properties of materials. It is also used in astronomy to measure the temperature and density of hot gas in distant objects, such as stars and galaxies.

4. How does the energy of the scattered light relate to the energy of the incident light in Thomson scattering?

The energy of the scattered light is typically lower than the energy of the incident light in Thomson scattering. This is because some of the energy is transferred to the charged particles during the scattering process.

5. What is the difference between Thomson scattering and Rayleigh scattering?

The main difference between Thomson scattering and Rayleigh scattering is the size of the particles involved. Thomson scattering occurs when the size of the particle is much larger than the wavelength of the incident light, while Rayleigh scattering occurs when the size of the particle is much smaller than the wavelength of the incident light.

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