Lecture #11b-- Mie scattering theory
1) Rayleigh limit
Mie scattering begins in the Rayleigh limit, where the wavelength is
much larger than the radius of the (spherical) dielectric dust grain.
This is quite analogous with the Rayleigh limit of a bound electron,
which is also a low frequency (long wavelength) limit, wherein the
electric field may be treated as static from the point of view of
calculating the induced dipole moment.
Then this dipole moment simply oscillates in lock step with the
oscillating electric field.
The amplitude of the dipole moment oscillation doesn't depend on the
frequency of the field, in this Rayleigh quasi-static limit, so the
Larmor formula for the power radiation gives that the power is
proportional to the square of the dipole moment amplitude, times the
fourth power of the driving frequency (since it's the square of the
acceleration of the dipole moment).
This scaling continues roughly until the cross section equals the
geometric cross section of the grain, and then self-shadowing in the
grain limits the cross section to be larger.
It can be a little larger due to diffraction, which is a wave
coherence effect, but the peak is only about 4 times the geometric
cross section, and this comes when the wavelength equals 2 pi times
the radius of the dust grain.
2) Contrasting the large wavelength (Rayleigh) limit to when the
wavelength is of order the radius (Mie regime)
In addition to this transition, for a given wavelength,
from smaller to larger particles (i.e., from the Rayleigh to the
Mie regime),
which distinguishes the blue sky
and red sunset from the white glow around a foggy streetlight,
we also find a shift from symmetric fore/aft scattering to highly
forward scattering.
The sky is blue in all directions because Rayleigh scattering is not
highly forward scattered, but the glow of the streetlight does tightly
hug the lamp, because the Mie regime is highly forward scattered.
The latter is more relevant to dust extinction in the ISM, because the
dust grains tend to have sizes that cross through the visible spectrum.
If all dust particles were much smaller than that, interstellar reddening
would be much more pronounced.
3) The cause of interstellar reddening
So if the Rayleigh regime is not strictly applicable to ISM dust, why
is there reddening at all?
The main important feature of dust scattering is in the Mie regime is not
that this regime is roughly frequency independent, it is that this regime
is characterized by a cutting off of the dust absorption for wavelengths
that are long enough to pass into the Rayleigh regime.
So it is the connection between Mie and Rayleight regimes that is decisive,
not either regime by itself.
This means that for any two contrasting wavelengths,
say a blue vs. a red wavelength, what matters to the
reddening is the different numbers of
dust grains that are larger than the wavelength (or more correctly, for
whom 2 pi times the radius is larger than the wavelength).
Since there are generally a larger number of
smaller dust grains, it is significant that red wavelengths are larger
than a lot of the smaller dust grains, and this is why red light passes
better through dust (and infrared better still).
Hence the degree of reddening does not depend so much on the details of
the Mie cross sectional dependence on wavelength, it depends more on the
details of the particle size distribution.
That's why the study of
interstellar reddening is basically the study of dust particle
size distributions.