> 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
> 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
> 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
> 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.