# Origin of Rayleigh scattering

Is Rayleigh scattering simply the elementary result of scattering theory, that, at low energies (long wavelengths) the scattering is dominated by $s$-wave scattering?

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S wave scattering means 0 angular momentum with respect to the scattering center, so I do not see how this is reconciled with the description in your link ( the small radiating dipole). –  anna v Jun 7 '13 at 15:32
I'm no particle physicist, so I can't comment on your analogy. What is true is that Rayleigh scattering is the first order (low index difference, small scatterer) term in the more general Mie scattering theory. See the section in Born and Wolf that I cite at physics.stackexchange.com/a/72955/26076. Maybe Mie theory can help you get insight and inspiration, althoug you might care to google Mie scattering because Born and Wolf's notation and style is getting rather "old" so I'm sure there are now better references. –  WetSavannaAnimal aka Rod Vance Aug 8 '13 at 7:16
@WetSavannaAnimalakaRodVance, (and annav) Your comments have been most useful. It is clear that in the low energy (frequency) limit the scattering is dominated by $p$-wave scattering rather than $s$ wave, since light is scattered by dipole moments. Therefore $s$ wave scattering is absent, and so the lowest partial wave scattering is indeed dipole ($p$-wave) scattering. Thanks! If one of you can post an answer related to this, I can accept it. –  QuantumDot Aug 10 '13 at 2:33
@annav The above comment addresses you, too. –  QuantumDot Aug 10 '13 at 2:34

Rayleigh scattering is the elastic scattering of electromagnetic waves (typically light) on neutral atoms or molecules (or other compound objects) without spin, in the regime where the electromagnetic wavelength is much larger than the atom or molecule's size. That is,

$$\gamma + \text{neutral}_{s=0}\rightarrow \gamma + \text{neutral}_{s=0}$$

One can speculate that this process takes also place with neutral, spinless, elementary particles if there is a Lorentz and gauge invariant term of dimension six in the Lagrangian of the form

$${1\over\Lambda^2}\Phi^{\dagger}\,\Phi\,F^2$$

where $F$ is the Faraday tensor. This goes beyond strict QED.

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Be careful, I believe you confound scattering of light-field (a wave) by particles (wave-atom scattering, the Rayleigh scattering is one of them when the wave-length is well larger than the atom) and the scattering of a particle onto a potential (representing some other particles for instance). The first one is classic, the second one is related to quantum mechanics. The theory is naturally developed onto $l$ angular momentum only for the second. So in short there is no per se Rayleigh scattering in Q. theory as far as I know.

The equivalent regime (not called Rayleigh scattering then) in quantum scattering when the $s$-wave component is dominant is for slow particles. See e.g. Landau and Lifshitz, Quantum theory, part I (Non relativistic theory) - 3-rd edition, available there. The slow particle problem is on section 130. It is not the same as the low energy scattering, that you could find on the following section 131.

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I thought that radiation scattering theory and non-relativisitc potential scattering theory were essentially mathematically equivalent. –  QuantumDot Jun 8 '13 at 3:46
@QuantumDot Well, they are to some extend, but the quantum theory has more parameters, and then more limiting cases than the classical theory. So you could distinguish between the scattering of low energy and slow particles. When it's about light, it is less clear for me, since photon has no (non-relativistic) mass. Say differently, would you like to say that Rayleigh scattering corresponds to slow photon ? Perhaps the Rayleigh regime is better understood in relativistic quantum mechanics. I don't know that, sorry. –  FraSchelle Jun 8 '13 at 6:53
Hmmm.. I was under the impression that if I reformulate potential scattering theory and radiation scattering theory entirely in terms of wavenumbers $k$. they would be 'close enough.' Alas, it seems that I must study radiation scattering theory more thoroughly to justify that.. –  QuantumDot Jun 8 '13 at 19:46

I'm no particle physicist, so I can't comment on your analogy. What is true is that Rayleigh scattering is the first order (low index difference, small scatterer) term in the more general Mie scattering theory. Again, not really being qualified to comment of particle scattering - I should think you could overcome at least partly the problem raised in other answers that light and atomic particles are orders of magnitude different in size with the full Mie theory: you can make your scatterer as big as you like relative to the wavelength.

A classic reference here is Born and Wolf "Principles of Optics" and I cite the section in the book with the Mie scattering treatment at http://physics.stackexchange.com/a/72955/26076. Maybe Mie theory can help you get insight and inspiration.

I have found another reference on the "math-physics-tutor.com" website:

http://www.math-physics-tutor.com/web_documents/multipole.pdf

whose treatment of Mie theory is rather tidier and somewhat kinder to read than Born and Wolf's. It doesn't go quite so far (Born and Wolf will give you full expressions for all scatterer sizes of all refractive indices, whether complex or real), but I should think if you want to get your head into the structure of the equations, this latter reference seems better. It shouldn't be too hard to generalize: they work the example of a perfectly conducting sphere, which may be enough for what you want.

They begin with a multipole expansion of a circularly polarized plane wave - see section 7.6 - the crucial equation here hwill be 7.128a (take heed that the left hand side of the equation should be $c \mathbf{B}$, not $\mathbf{B}$ (you may want to use units where $c = 1$ anyway). For the Rayleigh term, you'll just look at the $\ell = 1$ term anyway.

Section 7.7 "Mie Theory" gives you the appropriate forms for the scattered waves of all orders (relevant equation 7.130), then the waves inside the sphere are the same, but with first kind spherical Bessel functions (with no poles at the origin) instead of the outgoing Hankel functions in the scattered waves. Moreover here, of course, you'll need to replace $k$ by $k\,n$, where $n$ is the complex refractive index of the sphere. The great thing about this notation is that the incoming plane wave is split into different orders the colatitude modal index $\ell$, and different orders aren't coupled by a sphere. So you can just match up the boundary conditions for each term in the series separately. And, in any case, it's already done for you for the perfect conductor sphere in Eq 7.143.

I'm not too sure about the comments on s and p polarization - I'm guessing this has to do with Anna V's angular momentum comment. The incoming waves in the treatment just described are circularly polarized and you could rewrite Eq 7.130 so that instead of the separate $\alpha$s and $\beta$s, you have pure circular polarized terms running both right to left and left to right. This might let you see the angular momentum relationships more easily - although it is clear that, since the $\alpha$ and $\beta$ are not generally simply $\pm1$ times the other, that the circular polarization components are going to be mixed in complicated ways.

Hope this helps.

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There are at least two major differences betwen Rayleigh scattering and quantum s-wave scattering. One is that Rayleigh scattering can be fully described by classical electrodynamics, with no appeal to quantum mechanics. The second is that electrodynamic waves have a polarization, so they inherently can't be spherically symmetric.

To step back for a second, in classical electrodynamics, Rayleigh scattering comes from an oscillating electric dipole. Jackson works through this problem, arriving at: \begin{equation*} \frac{dP}{d\Omega} = \frac{c^2Z_0}{32\pi^2}k^4|(\mathbf{n}\times \mathbf{p}) \times \mathbf{n}|^2 \tag{9.22} \end{equation*} and \begin{equation*}\frac{dP}{d\Omega} = \frac{c^2Z_0}{32\pi^2}k^4|\mathbf{p}|^2\sin^2\theta \tag{9.23} \end{equation*} Equation 9.22 gives the general form for any set of phase relations among the components of $\mathbf{p}$; 9.23 is the specific case where they are all in phase. The electric dipole moment $\mathbf{p}$ oscillates with frequency $\omega = ck$. It then radiates power per unit area at the rate $\frac{dP}{d\Omega}$. In the coherent case of 9.23, $\theta$ is measured relative to the direction of $\mathbf{p}$. You can take this to be the z-axis. The presence of the $\theta$ factor makes this manifestly not spherically symmetric.

Jackson also works through the case of the oscillating electric monopole, in section 9.1. He shows that oscillations in the electric monopole do not contribute any radiation far away from the radiation source.

The quantum mechanical explanation of this is that there is no transition operator for the electric monopole, and that the [electric dipole operator] carries angular momentum. This is why you have selection rules for electric dipole transitions.

The other difference stems from that $|(\mathbf{n}\times \mathbf{p}) \times \mathbf{n}|^2$ term in equation 9.22. That describes the polarization state of the resulting wave. A given wave will have a specific polarization state, and that will depend on on the direction relative to $\mathbf{p}$. The quantum mechanical wave, being a scalar wave, will not have such a polarization.

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