FoRK classic: Why is the sky blue?

I Find Karma (
Wed, 4 Feb 1998 23:39:49 -0800

Rohit used to joke that the final quiz you get asked before you're
allowed to leave Caltech is to discuss why the sky is blue. Here, my
friend Matt McIrvin gives an easy-to-understand yet detailed answer.
-- Adam

> Why is the sky blue?
> A semi-detailed explanation
> By Matt McIrvin
> No other question so strongly evokes images of parents shrugging their
> shoulders in bewilderment when kids ask it. (It isn't the champion in
> the blushing and stammering category, but I believe it leads the pack
> in bewilderment.) Popular books on science often simply explain that
> air molecules preferentially scatter blue light from the sun, but stop
> there. I thought that it might be interesting to provide a more
> detailed, but not tremendously mathematical, explanation of why this
> is so.
> 1. The basic idea: dipole scattering
> Light is an electromagnetic wave. If you stand in one spot as a light
> wave passes by, there will be an oscillating electric field and an
> oscillating magnetic field, which are perpendicular to each other. If
> the light is in the range of frequencies that we can see, then the
> frequency of the vibration affects the color of the light. The
> color-vision receptors in our eyes, the cones, are of three types:
> "blue" receptors that respond to light over a broad range of high
> frequencies, "green" receptors that respond to medium frequencies, and
> "red" receptors that respond to low frequencies. The ranges of
> sensitivity of the receptors overlap considerably, but they have their
> maximum sensitivities at different frequencies. The perceived color
> depends (among other things) on the relative strengths of the signals
> from these receptors.
> Molecules are usually electrically neutral, but they are made of
> charged objects: their atoms consist of negatively charged electrons
> and positively charged nuclei. If there is an electric field at the
> position of an atom, the nucleus will move a short distance in the
> direction of the field and the electrons will move the other way, and
> the atom will become a "dipole": the positive and negative charge will
> be centered around different places. A molecule made of such atoms
> will acquire its own electric field, something like the magnetic field
> of a bar magnet.
> A dipole's electric field falls off more rapidly with distance than it
> would if the molecule had a net electric charge. This is because at
> large distances, the fields from the positive and the negative charge
> tend to cancel each other out, as the difference between their average
> positions becomes less important.
> However, if the dipole is made to oscillate-- that is, if the positive
> and negative charge wiggle back and forth, out of phase with each
> other-- then the molecule can produce electromagnetic radiation of its
> own, for reasons I'll explain below. This is how air molecules scatter
> light: the oscillating electric field of the incoming wave makes the
> molecules develop oscillating dipoles, which in turn give off
> radiation.
> The radiation destructively interferes with the incoming wave in the
> forward direction. The original wave is lessened in intensity, and new
> waves move out in all other directions, so that overall energy is
> conserved (this requirement is sometimes called the "optical
> theorem"). The net effect is that light energy that was moving in a
> straight line from the sun ends up traveling in some other direction.
> Since sunlight appears white but the sky is a robin's-egg blue, it
> must be that the scattered light excites our blue-sensing cones more,
> and our red-sensing cones less, than the original sunlight. The
> distribution of frequencies in the scattered light must be biased
> toward high frequencies. Why is this?
> 2. Retarded potentials
> Scalar and vector potentials
> In the theory of electromagnetic radiation, it is not so convenient to
> work with the electric and magnetic fields directly, except for simple
> plane waves. It is more convenient to use the "scalar potential" and
> "vector potential."
> You are probably already familiar with the scalar potential: in many
> situations, it is just the same thing as voltage. A 5-volt battery has
> a scalar potential difference of 5 volts between its terminals. The
> electric field, in static situations (given the usual potential
> conventions of electrostatics), is just given by the spatial rate of
> change of the scalar potential, and it points "downhill" toward
> regions of lower electric potential.
> There is also a "vector potential" that has to do with magnetism. This
> is a quantity with a magnitude and a direction: a vector. In static
> situations, the magnetic field is related in a somewhat complicated
> way to the rates of change of the vector potential in various
> directions: essentially, it has to do with the extent to which the
> vector potential swirls around a given point.
> If the potentials are changing with time, as in radiation, then the
> relation between the potentials and the fields is more
> complicated. But in either case, in size, the electric and magnetic
> fields are proportional to the rates of change of the potentials in
> space and time.
> Potentials, charges, and currents
> Now, if the potentials are defined in a certain way (what the pros
> will recognize as a "covariant gauge"), the potential due to a certain
> charge and current distribution is related to the charges and currents
> in an extremely simple way.
> Suppose there is a point charge somewhere in space, which moves
> around. Then the scalar potential at some other place is directly
> proportional to the charge, and inversely proportional to the distance
> to the charge.
> But it is not the distance to the place where the charge is now; it is
> the distance to the place where the charge was, at such a time that a
> signal traveling at the speed of light from the position of the charge
> is just now getting to the place where we're calculating the
> potential. The news about where the particle is travels at a finite
> speed, the speed of light. This is called a "retarded potential,"
> meaning "delayed," because it responds to the charge's position with a
> speed-of-light delay.
> If there is more than just a point charge, then the scalar potential
> can be calculated by adding up the retarded potential of each little
> bit of charge.
> The vector potential is related in exactly the same way to the
> currents. Each little piece of current creates a retarded vector
> potential that is proportional to current and inversely proportional
> to distance, and the news about where the current is travels at the
> speed of light.
> 3. The potentials of an oscillating dipole
> Now consider a molecule that is a dipole. For simplicity, model the
> molecular dipole as a pair of opposite point charges, separated by a
> short distance. (Really, the positive charge consists of a couple of
> nuclei and the negative charge is a spread-out cloud of electrons, and
> the dipole comes from the separation between their average positions;
> but idealizing the molecule as a pair of point charges doesn't change
> the analysis in any substantive way, as long as the molecule is
> small.)
> If the dipole is not changing, then at large distances, the scalar
> potential due to one end of the dipole and the scalar potential due to
> the other end will tend to cancel each other out, since the distance
> to the two charges is almost the same. So the scalar potential will
> fall off faster with distance than it does for a single charge.
> But the news about the charge only travels at the speed of light! If
> we are slightly closer to one end of the dipole than to the other,
> then the potential here depends on the charge at the near end of the
> dipole at some previous time, and the charge at the far end of the
> dipole a short time before that. So if the charges are moving back and
> forth at a high speed, the cancellation between the ends of the dipole
> will be less complete. For instance, the scalar potential here could
> depend on the charge at the near end at a time when it was positive,
> but the charge at the far end at a time when the negative charge had
> not yet gotten all the way there.
> If the dipole is much smaller than the wavelength of the light (and
> air molecules are thousands of times smaller than the wavelengths of
> visible light), the cancellation becomes linearly less complete as the
> frequency of the oscillation increases. So at large distances, where
> the scalar potential of the static dipole would be negligible, the
> scalar potential due to an oscillating dipole goes up linearly with
> the frequency.
> How about the vector potential? That's easier to figure out. It also
> varies linearly with frequency, because it's proportional to the
> current-- and the faster the charges are moving, the more current
> there is.
> The potentials that are produced reverse direction as the dipole
> reverses direction. If the dipole wiggles back and forth, then
> oscillating waves of potentials move out from the dipole at the speed
> of light, with a strength proportional to the frequency of the
> wiggle. The higher the frequency, the shorter the waves, because they
> have less time to get out of the way before the dipole changes
> direction.
> 4. Radiation fields
> Now, the electric and magnetic fields are proportional to various
> rates of change of the potentials, in space and time. They get a
> factor of frequency from the sizes of the potentials; but they also
> get another factor of the frequency from the fact that the shorter a
> wave is, the faster it varies in space; and the higher its frequency
> is, the faster it varies in time. So the fields are proportional to
> the square of the frequency.
> But we are not done yet! The important thing is how much power is
> transmitted by the wave, and that is proportional to the product of
> the electric field and the magnetic field. So the power density in the
> wave goes up as the fourth power of the frequency.
> Therefore, the spectrum of the radiated light, and the scattered light
> from an induced dipole, will be very strongly peaked at high
> frequencies, or short wavelengths. There are things I have neglected
> here, such as the fact that sometimes, there are resonant frequencies
> at which the charge particularly likes to oscillate, which are
> determined by the quantum mechanics of the molecule. However, in this
> case, for the most part we can safely neglect resonance here when
> determining the overall shape of the spectrum.
> A full analysis would also take into account the fact that the
> electromagnetic field is quantized; the energy comes in photons. But
> that turns out not to affect the fourth-power dependence of the
> spectrum on frequency.
> This sort of scattering is called Rayleigh scattering, after Lord
> Rayleigh, who first worked it out for a very small classical dipole.
> 5. The sky, the sunset, and a Martian postscript
> If the dipole is a molecular dipole created by an electromagnetic wave
> from the sun impinging on an air molecule, then it is the higher
> frequencies that will be primarily scattered in different directions,
> and removed from the incoming wave. Lower frequencies will be
> scattered as well, but not as much; the scattered power goes like the
> fourth power of the frequency.
> The atmosphere does not absorb much light at visible wavelengths; the
> dominant effect is dipole scattering. The scattered light will be
> biased toward the high frequencies. If you look at a part of the sky
> where the sun is not, then your eyes are receiving the scattered
> light. That light excites the blue- and green- sensing cones in your
> retinas much more than the red-sensing cones (the largest amount of
> power is coming in at the frequencies covered by the blue cones, but
> they are less sensitive than the green ones). The result is the
> beautiful turquoise color of a clear sky-- and of Earth, when seen
> from space.
> The scattering is also responsible for the color of the sun at
> sunset. Since the emitted radiation removes energy from the incoming
> waves by destructive interference (thereby conserving the overall
> energy), the higher frequencies are preferentially depleted from the
> unscattered beam. At sunset, the sun is shining at a grazing angle
> through an unusually thick layer of air, so the depletion is
> particularly pronounced, and the sun appears yellow, orange or red
> rather than the usual blinding white.
> Notice that this argument depends very little on the composition of
> the atmosphere. Any clear atmosphere of more or less Earthlike size
> and density, lit by a sun whose light appears more or less white,
> would result in a blue sky.
> The recent color pictures from Mars Pathfinder are a spectacular
> reminder that the sky is not blue on Mars. Instead, it has colors that
> have been described as everything from "orange-pink" to "gray-tan", as
> was discovered in the 1970s by the Viking landers. This is because the
> atmosphere of Mars is very thin and dusty, and atmospheric light
> scattering is dominated not by the molecules of gas (in the case of
> Mars, mostly carbon dioxide) but by suspended dust particles. These
> are larger than the wavelengths of visible light, and they are
> reddened by iron oxide, like Martian soil. It's not just Rayleigh
> scattering, so the power spectrum is different.


Karma police, arrest this man.
-- Radiohead