When I first came across this, I speculated this law is experimental. However, later I came to know that this is not the case. A concise answer, with some reasoning as to why the negative fourth power of wavelength is involved when the scattering particles are of the dimensions of the wavelengths, is desired. On a side note, I'd also like to know why the wavelength doesn't matter effectively when the scattering particles are larger.


I'm not sure where to start.

Whereas there is a Scalar potential for the Electric field, one can use a Vector Potential for the magnetic field.

The field of static fields typically drop off as $1/r^2$. The charge density is proportional to the square of the fields. So the energy density for static fields often drops off as $1/r^4$. Integrate the field over a spherical surface distant from the charges as in Gauss law, you will need to multiply that by a factor of $r^2$ before performing the surface integral. This gives you the flux, dropping off as $1/r^2$.

Now if your field drop off as $1/r$, then the flux arbitrarily for away would be independent of the radius.

This can happen when charge moves in certain ways. As an example, consider an ideal dipole oscillating at the origin.

$$\vec{A}=\frac{\mu_0}{4\pi r}\frac{d\vec{p}}{dt}$$

Where r is distance from origin, $\vec{A}$ is the vector potential and $\vec{p}$ is the possible time varying dipole moment.

You get the Magnetic Field by taking the curl of the Vector Potential, $\vec{B}=\nabla \times \vec{A}$.

Here $\vec{A}$ has the form $f\vec{u}$. By the identity, $\nabla \times \vec{A}=\nabla f\times \vec{u}+f\nabla \times \vec{u}$. The gradient of $f$ here is proportional to $1/r^2$. Since we are only interested in the radiation field far from the source, we ignore it since it goes to zero.

Since we are worried about the field far away from the source, the radiation field, and taking into account the Retarded time, we have $\nabla \times \frac{d\vec{p}}{dt}=\frac{-\hat{r}}{c}\times \frac{d^2\vec{p}}{dt^2}$

So the Magnetic Field far from the dipole is $$\vec{B}=\frac{-\mu_0\hat{r}}{4\pi c r}\times \frac{d^2\vec{p}}{dt^2}$$

Consider a simple dipole moment, $\vec{p}=p_0\sin{\omega t}\hat{z}$

Then the second derivative is proportional to $\omega^2$, and so then is the Magnetic Field.

Far from the source, the Electric field is : $\vec{E}=c\vec{B}\times \hat{r}$

So $\vec{E}$ is also proportional to $\omega^2$

The Poynting Vector measuring the "Energy Current of the EM field" is:

$$\vec{S}=\frac{1}{\mu_0}(\vec{E}\times \vec{B})$$.

If we set up a coordinate system with our $z$ axis parallel to the dipole moment we can set up the integral for the flux through an imaginary surface.

Specifically the flux a radius R away from the dipole is $\vec{S} \cdot \hat{r} dA$

where $dA$ is the area element of our spherical surface, $r^2 \sin{\phi}d\phi d\theta$.

$\vec{S}$ is proportional to $1/r^2$. So the factors cancel leading to a final expression. Since both the electric field and the magnetic field are proportional to $\omega ^2$, the flux is proportional to $\omega^4$.

Regarding different frequencies:

If we assume the electric field effect an electron in an atom is fairly constant along the diameter of the atom, then we can approximate the force on an electron with:

$m\ddot{x}=-kx-b\ddot{x}-eE_{0x}e^{i\omega t}$

This gives the position of the electron as $\vec{x}=\frac{e\vec{E_0}e^{i\omega t}}{m\omega^2 + b \omega +k}$

The denominator can be expressed as $m\omega^2 + b\omega +k=m(\omega-\omega_0)^2+c_0$ where $\omega_0=b/2m$.

This $\omega_0$ is the Natural Frequency of the electron attacked to the atom. m is the mass of the electron, b is the dampening factor, and k is the restoration constant representing the strength of the bond of the electron to the atom.

Given our $\vec{x}$, our dipole moment is $\vec{p}=-e\vec{x}$.

And remember our scattering flux is proportional to the second time derivative of the dipole moment. The denominator is independent of time, so it is constant.

Our denominator reaches a peak when $\omega$, the frequency of the applied electric field is $\omega_0$, the Natural Frequency due to the physical parameters b and k.

So we expect more intense scattering at preferred frequency/wavelength.

  • $\begingroup$ To understand, in number of photons there is not difference in scattering say, a blue and a red one. I never thought before of this, neglrcting that all treatment I have seen focused on Energy. Am I right? $\endgroup$ – Alchimista Apr 3 at 8:26
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    $\begingroup$ I added some details on why some frequencies would be preferred. $\endgroup$ – R. Romero Apr 3 at 15:28

When sunlight enters the upper atmosphere, a particular air molecule finds itself being washed over by an electromagnetic wave of frequency $f$. The molecule's charged particles (nuclei and electrons) act like oscillators being driven by an oscillating force, and respond by vibrating at the same frequency $f$. Energy is sucked out of the incoming beam of sunlight and converted into the kinetic energy of the oscillating particles. However, these particles are accelerating, so they act like little radio antennas that put the energy back out as spherical waves of light that spread out in all directions. An object oscillating at a frequency $f$ has an acceleration proportional to $f^2$, and an accelerating charged particle creates an electromagnetic wave whose fields are proportional to its acceleration, so the field of the reradiated spherical wave is proportional to $f^2$. The energy of a field is proportional to the square of the field, so the energy of the reradiated wave is proportional to $f^4$.

  • $\begingroup$ Nice answer. Actually I'm still a High School student, so I'm not completely exposed to the Mathematics that's used in rigorous proofs for this. I get this one though, precise. Upvoted. $\endgroup$ – Aabesh Ghosh Apr 3 at 18:48

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