1 Why “Is” The Sky Blue?
Before we ask why the sky is blue, we should ask whether the sky
is blue. There are many times and places where the sky is, in fact,
not blue. There is some nifty physics that explains the blue part of
the story, which is our topic for today, but we should keep in mind
that it is not the whole story.
We shall consider the restricted case of standing
on the earth’s surface in the daytime
under a clear sky, without
clouds, dust, or pollution. This is sometimes a decent approximation
in real life. As we shall see, under such conditions we expect the
sky to be blue.
A crucial sub-goal will be to understand why the scattering depends on
the fourth power of the wavelength.
There are lots of pseudo-explanations out there that focus attention
on a piece of sky “one wavelength on a side”. However, I believe
any alleged explanation of that sort is wrong physics, and misses the
right physics, as will become clear below.
2 Executive Summary
Here is an “executive summary” that outlines where the discussion is
going. For details see section 3.
-
Divide the interaction region in to P zones and N zones. The
contributions from the P zones are roughly 180 degrees out of phase
with the contributions from the N zones, so if the zones were
balanced the amplitude of the scattered wave would be zero.
- Distribute the air molecules into these zones at random, like sand on
a checkerboard.
- Due to statistical fluctuations,
there will be some imbalance between P and N, and therefore some
scattering. The chance of an imbalance is independent of wavelength.
- The significance of an imbalance does, however, depend on
wavelength. It is a stronger source when the wavelength is small.
The scattering amplitude is proportional to k2, and the intensity
is proportional to k4.
3 Detailed Explanation
Consider the following diagram of the interaction:
.
. .
. . .
. . . .
. . . . .
. . . . .
. . . . . transmitted -->
| | | / |/ | / | /| / | | | | | | |
| | | | / |/ | / | /| / | | | | | |
| | | | | / |/ | / | /| / | | | | |
| | | | | | / |/ | / | /| / | | | |
| | | | | | | / |/ | / | /| / | | |
incident --> / / / / /
/ / / / /
/ / / / /
/ / / / /
/ / / / /
scattered / / / /
--> / / /
/ /
/
The lines represent the crests of the waves.
The dots don’t have much physical significance. They are just the
backwards extrapolation of the scattered beam. Loosely speaking, they
represent the direction the scattered beam “appears” to be coming
from. Ignore the dots if you like.
Here is the diagram again, with labels on some points in the
interaction region:
.
. .
. . .
. . . .
. . . . .
. . . . .
. . . . . transmitted -->
| | | P |/ | N | /| P | | | | | | |
| | | | P |/ | N | /| P | | | | | |
| | | | | P |/ | N | /| P | | | | |
| | | | | | P |/ | N | /| P | | | |
| | | | | | | P |/ | N | /| P | | |
incident --> / / / / /
/ / / / /
/ / / / /
/ / / / /
/ / / / /
scattered / / / /
--> / / /
/ /
/
- Let’s use this to identify a pattern in the index-deviations that will
result in strong scattering.
-
At each point P, a crest lines up with a crest. A positive
deviation in the index at this point will make a positive contribution to
the overall interaction.
- At each point N, a crest lines up with a trough. A negative
deviation in the index at this point will make a positive contribution to
the overall interaction.
- In fact, we can classify every point in the interaction region
(not just the special points mentioned in the previous item) according to
whether a positive or negative index-deviation (at the given point)
results in a positive
contribution to the desired interaction. This defines a notion of “P zones” and “N
zones”.
We have made some mildly arbitrary
choices about the relative phases. This results in no loss of generality.
We assume that the interaction region is transparent to zeroth
order. This is consistent with (and stronger than) previous assumptions.
- Note that the P zones collectively cover half the interaction
region, while the N zones cover the other half.
This assumes the interaction region is reasonably large relative to
lambda. This is consistent with previous assumptions.
This assumes the scattered beam does not coincide with the
transmitted beam. In the other case (i.e. forward scattering) this
half-and-half property does not hold, which is a good thing; the distinction allows us to
uphold the optical theorem, conservation of energy, and other good things.
- We could, if we wanted, use what we have just learned to
construct a fancy diffraction grating (actually a
hologram) that would create the strongest possible scattered beam.
We just put many scatterers in the P regions and
few (if any) scatterers in the N regions.
Note that this part of the argument is independent of wavelength, for
over the relevant range of wavelengths. In the atmosphere, the
interaction region is vastly larger than wavelength cubed. That means
there are a vast number of P regions, covering half of the
interaction region. If the wavelength is smaller, each P zone will
be smaller, but there will be more of them, and collectively they will
still cover 50% of the interaction region.
- Now the statistical question reduces to this: what is the chance that the
air will fluctuate into a configuration that has an extra-large number of
molecules in the P zones, and an extra-small number of molecules in the N
zones? Essentially we are talking about thermally-excited sound modes.
The amplitude of these excitations should be independent of frequency,
since the compressibility of air doesn’t depend on wavelength.
We are approximating the air as an ideal gas. This should be a very
good approximation.
Compressibility being independent of wavelength is another line of
argument supporting the previously-mentioned point that we should not
focus attention on a single region that has volume on the order of
wavelength cubed.
- If you don’t want to think of it in terms of compressibility, you can
instead use elementary notions of statistical mechanics. That is, we model
the distribution of air molecules as a random statistical
process. Distributing the air molecules into P and N regions is like
scattering sand particles at random onto a checkerboard. It doesn’t
matter how large are the black and white cells on the board. It doesn’t
matter whether they are even square or not. As long as half of the board
is black and half of the board is white, and as long as the sand-grains are
randomly and independently distributed, about half of them will land on
black squares and about half will land on white squares.
Applying this idea to the air: We have every reason to believe that
half of the interaction region is P-type, and half is N-type. We
have every reason to believe that air is an ideal gas, to a good
enough approximation, indeed more than good enough. Therefore it
does not matter how big the P-regions and N-regions are.
This point is worth emphasizing because there are widespread
misconceptions that somehow the light is being scattered by
“small” regions where the density differs from the average.
- The fluctuation in index of refraction is proportional to the
fluctuation in the number of atoms, according to the Clausius-Mossotti
relation.
This assumes the index is not too different from
unity. This is a good assumption for gases under ordinary conditions. If
you care about liquids, you could easily relax this assumption and redo the
following derivation, thereby obtaining a more general result.
- We can use a variant of the Born approximation to understand how a
fluctuation in refractive index produces scattering. In a uniform
medium we can write the wave equation as:
|
v2 (∂ / ∂ x)2 φ − (∂ / ∂ t)2 φ = 0
(1) |
where v is the speed of light in the medium,1 x is
position, and t is time. We call equation 1 the
“unperturbed” wave equation, for reasons that will be obvious in a
moment.
There is a one-to-one relationship between the index of refraction and
the speed of light in the medium. To first order, the relationship
is:
|
v = v0 / (1 + d)
(2) |
where (1 + d) is the index, 1 is the average index (not the
index of the vacuum) and d is the fractional deviation from the
average index, and v0 is the propagation speed associated with the
average index. We assume d is small compared to unity.
So in a slightly non-uniform medium we have, to first order in d:
|
v02 (1−2d) (∂ / ∂ x)2 φ
− (∂ / ∂ t)2 φ = 0
(3) |
and by re-arrangement:
|
v02 (∂ / ∂ x)2 φ
− (∂ / ∂ t)2 φ =
2d v02 (∂ / ∂ x)2 φ
(4) |
So, the term involving the index-deviation d can be moved to the RHS.
This term can be considered a driving force, i.e. a source
term added to the RHS of the unperturbed wave equation (equation 1).
We are assuming that the scattering is not too strong, and the
interaction region is not too overly huge, so that we imagine that
the light gets scattered at most once.
We account for light scattering from the incident beam into the scattered
beam, while ignoring any possible secondary scattering (out of the
scattered beam). This is called the first Born approximation.
For simplicity, we are ignoring polarization. It would introduce
some factors that depend on theta (the scattering angle). You can add them
in if you like.
For clarity, I left out the y and z variables in
equation 4. But you get the idea.
-
But wait, 2d is not the only factor on the RHS of
equation 4. The
source term is not 2d times the wavefunction, it is more like 2d times
the second derivative of the wavefunction. For periodic waves, the
scattered amplitude picks up a factor of k2 (or ω2), and the
scattered power picks up a factor of k4 (or ω4).
- This factor of k4 is not “on top of” the factor of k4 you would get
in the formula for scattering from a single molecule. It is the same
factor, derived in a slightly more general context. Specifically, you can
view the isolated atom as an isolated localized deviation in the index of
refraction of the vacuum.
4 Remarks
- There is only a rather narrow set of conditions that can give a planet
a blue sky. An atmosphere with too little depth and/or too little
density would have very little scattering of any kind, so the sky
would be black with only a slight tinge of deep blue. At the other
extreme, an atmosphere with too much depth and/or too much density
would have a lot of second-order and higher-order scattering,
violating some of the assumptions made in section 3 and
leading to a murky white sky.
- As Einstein pointed out in reference 1, the color of the
sky can be used to pin down the size of air molecules, within a rather
narrow range. The argument goes like this: Macroscopic benchtop
measurements tell us the overall refractive index of a parcel of air,
but they don’t tell us whether that is due to a smallish number of
highly refractive molecules, or a larger number of less refractive
molecules. However, the fluctuations in the refractive index do
depend on how many molecules there are. If there are N molecules in
the parcel, the fluctuations in the number will scale like
√(N). Meanwhile, the refraction per molecule must scale like
1/N, to be consistent with the macroscopic observations, so the
fluctuations in the index scale like √(1/N). That means that
if atoms were 100 times smaller than they really are, there would be
10 times less scattering in the atmosphere, and the sky would be
almost black in the daytime.
- Here is an analogy that might help: Consider an altocumulus
standing lenticular (ACSL) cloud, such as the one shown in figure 3. The cloud remains stationary over the
mountain as the air flows past the mountain and through the cloud. The
air is nearly transparent before entering the cloud, cloudy while
within the cloud, and nearly transparent again afterwards.
Furthermore, suppose a similar cloud produced a quarter inch of rain.
If you look at a sheet of water, a quarter inch thick, it is almost
perfectly transparent.
Each water molecule is very small, so it is not, by itself, a very
strong scatterer. On the other hand, they are very numerous, and if
they scatter coherently, the effect is quite large.
We get strong scattering from the cloud because there are tiny
droplets, a fraction of a micron across. All N molecules in a given
droplet scatter more-or-less coherently, in phase with each other, so
the scattered intensity is proportional to N squared rather than
simply N. Each droplet, however, is out of phase with other
droplets.
In the sheet of liquid water, the density is the same everywhere, so
in accordance with the Huygens construction, you get very strong
forward scattering. In other words, you get an index of refraction
but no sideways scattering.
The moral of the story is you can’t think in terms of scattering by
individual molecules. The intensity of the scattering you see from the
blue sky, or from the white cloud, is determined by the variations in the density.
5 References
-
-
Albert Einstein, “Theorie der
Opaleszenz von homogenen Flüssigkeiten und Flüssigkeitsgemischen
in der Nähe des kritischen Zustandes”, Annalen der Physik,
33,
1275–1298 (1910).
http://www.physik.uni-augsburg.de/annalen/history/einstein-papers/1910_33_1275-1298.pdf
English translation:
https://einsteinpapers.press.princeton.edu/vol3-trans/245
