In short

Scattering is what happens when a light wave meets a small particle and sets the particle's charges oscillating — the oscillating charges then re-radiate light in all directions. How strongly they do this depends on how small the particle is compared to the wavelength.

Rayleigh regime (a \ll \lambda, tiny particles like individual air molecules). The scattered intensity goes as

\boxed{\;I_\text{scattered} \;\propto\; \frac{1}{\lambda^4}\,.\;}

Blue light (\lambda \approx 450 nm) is scattered roughly (700/450)^4 \approx 5.8 times more strongly than red light (\lambda \approx 700 nm). That single fact explains why the sky is blue, why sunsets are red, and why the Sun looks yellow-white from Earth's surface but pure white from outside the atmosphere.

Tyndall (Mie) regime (a \sim \lambda, colloidal particles like dust, smoke, fog droplets). The scattered intensity is much weaker in its wavelength dependence — smog and mist scatter all colours nearly equally, which is why they appear white or grey, not blue.

Raman scattering (C.V. Raman, 1928). When a tiny fraction of the scattered photons exchange a quantum of vibrational energy with the molecule, they emerge with shifted wavelengths. Those shifts are a fingerprint of the molecule — the basis of an entire branch of chemistry.

Stand on a rooftop in Chennai at 11 am. Look straight up. The sky is blue — not the grey you would expect if sunlight passed through the atmosphere without doing anything, and not the black you would see on an airless planet. Something in the air is taking the Sun's white light and selectively redirecting its blue component sideways into your eye. That is scattering.

Now look straight at the Sun. It is yellow-white, not blue. The blue has been removed from the direct beam and sent sideways, leaving a beam that is slightly deficient in blue — yellowish. At noon the effect is subtle. Wait until 6 pm. The Sun is near the horizon; its light now traverses roughly forty times as much atmosphere on its way to you. More atmosphere means more scattering; most of the blue and a lot of the green have been removed from the direct beam by the time it reaches you. What is left is mainly red and orange. That is why the Sun at sunset is red, and why the clouds it lights up are golden. And the sky — at the place where the Sun itself is low — is no longer blue but a gradient of pink, orange, deeper blue, fading to violet at the zenith.

This entire palette comes from one equation.

The essential picture — scattering as re-radiation

A light wave is a travelling electromagnetic field. When it hits a molecule — say a nitrogen molecule in the atmosphere — its oscillating electric field pushes the electrons in the molecule back and forth at the same frequency. Those oscillating electrons are themselves little antennas; an oscillating charge radiates. So the molecule absorbs a bit of the passing wave and re-emits it, mostly in directions other than the original.

This is elastic scattering: the frequency of the re-emitted light is the same as the incoming frequency (no energy is lost to the molecule in the simplest case). But the direction is randomised. Multiply over the countless molecules along every line of sight through the atmosphere and you get a diffuse glow of scattered light reaching your eye from every direction — the sky.

The key question: how strongly does a molecule re-emit as a function of wavelength?

Deriving Rayleigh's 1/\lambda^4 law

Consider a small particle of radius a much smaller than the wavelength \lambda of the incoming light. The incoming light's electric field E oscillates at frequency \omega = 2\pi c/\lambda. Because a \ll \lambda, the field is essentially uniform across the whole particle at any instant — the particle sees a spatially uniform, temporally oscillating E.

Assumptions: the particle is small compared to the wavelength (a \ll \lambda). The particle is electrically neutral overall, with polarisable bound charges (so there is no conduction, just oscillation). The scattered intensity is measured far from the particle.

Step 1. The incoming field polarises the particle. For a small polarisable object in a uniform field, the induced electric dipole moment is

p(t) = \alpha\,E(t)

where \alpha is the particle's polarisability — a property of the particle that is essentially independent of frequency in the regime where the incoming light is far from any resonance of the particle.

Why: a polarisable object responds to an applied field by separating its positive and negative charges slightly. The amount of separation is proportional to the applied field (for small fields) — that proportionality constant is what \alpha measures.

Step 2. The dipole oscillates at the same frequency as the driving field:

p(t) = \alpha E_0 \cos(\omega t).

Step 3. An oscillating electric dipole radiates power. The standard result from electromagnetism says the power radiated by an oscillating dipole of amplitude p_0 at frequency \omega is

P_\text{rad} = \frac{\mu_0\,p_0^2\,\omega^4}{12\pi c}

Why: a charge undergoing acceleration \ddot x radiates power proportional to \ddot x^2 (the Larmor formula). For a dipole p = p_0\cos\omega t, the charge's acceleration amplitude is \omega^2 p_0/q, and the radiated power therefore contains \omega^4. This is a consequence of how time derivatives of an oscillation bring down powers of \omega. The numerical prefactor \mu_0/(12\pi c) comes from doing the full integration of the radiation pattern; the scaling is what matters here.

Step 4. Since p_0 = \alpha E_0, the radiated power from a single small particle is

P_\text{rad} = \frac{\mu_0\,\alpha^2\,E_0^2\,\omega^4}{12\pi c}.

Now convert \omega to \lambda using \omega = 2\pi c/\lambda:

\omega^4 = \frac{(2\pi c)^4}{\lambda^4}.

Substituting:

P_\text{rad} = \frac{\mu_0\,\alpha^2\,E_0^2\,(2\pi c)^4}{12\pi c\,\lambda^4}.

All the prefactor is a bunch of constants. The wavelength dependence is what matters:

\boxed{\;P_\text{rad} \;\propto\; \frac{1}{\lambda^4}\,.\;}

Step 5. The scattered intensity you measure is proportional to the power radiated per unit solid angle, which inherits the same 1/\lambda^4 scaling.

I_\text{scattered} \;\propto\; \frac{1}{\lambda^4}.

This is Rayleigh's law — the single most important equation in the physics of scattering. It says: the shorter the wavelength, the more strongly it scatters. Short wavelengths (blue, violet) are scattered much more than long ones (red).

How big is the effect?

Plug in the numbers. Take red light at \lambda_\text{red} = 700 nm and blue light at \lambda_\text{blue} = 450 nm:

\frac{I_\text{blue}}{I_\text{red}} = \left(\frac{\lambda_\text{red}}{\lambda_\text{blue}}\right)^4 = \left(\frac{700}{450}\right)^4 = (1.556)^4 \approx 5.85.

Why: (1.556)^2 = 2.42, and (2.42)^2 = 5.86. The fourth power is very sensitive — even modest wavelength differences produce large scattering differences.

Blue light is scattered about six times more strongly than red. Violet light (\lambda \approx 400 nm) is scattered (700/400)^4 = 9.4 times more strongly than red.

Rayleigh scattering intensity across the visible spectrumA smooth curve showing scattering intensity proportional to one over lambda to the fourth power, plotted against wavelength from 400 to 700 nanometres. Violet and blue at the left have much higher intensity than yellow, orange, and red on the right.wavelength $\lambda$ (nm)scattering intensity $\propto 1/\lambda^4$400450500550600650700×5.8 more than redbaseline
Rayleigh scattering intensity $\propto 1/\lambda^4$ across the visible spectrum, with the visible colour band as a guide. Violet at 400 nm is scattered $(700/400)^4 \approx 9.4$ times more strongly than red at 700 nm; blue at 450 nm is scattered about 5.8 times more strongly.

Explore the wavelength dependence

Drag the slider below to shift the wavelength \lambda between 400 nm (violet) and 700 nm (red). The relative scattering intensity (700/\lambda)^4 — normalised so that red scatters unity — redraws live.

Interactive: Rayleigh intensity as a function of wavelengthRayleigh scattering intensity curve one over lambda to the fourth power, plotted against wavelength from 400 to 700 nanometres. A draggable slider shows how the intensity changes with wavelength.wavelength $\lambda$ (nm)relative intensity4005006007000510drag the red point on the x-axis
Drag the red control to change the wavelength. The readout reports the scattering intensity relative to red (700 nm). Violet at 400 nm scatters nearly 10 times more strongly than red; yellow at 580 nm, only 2.1 times more.

Why the sky is blue — and why the Sun is not violet

The Rayleigh law tells you that violet is scattered most, blue second, green less, and red barely. If the scattering were the only factor, you would expect the sky to be violet. But the sky is unambiguously blue.

Three reasons dilute the violet:

  1. The Sun emits less violet than blue. The solar spectrum peaks in the green-yellow around 500 nm and falls off faster toward violet (400 nm) than toward blue (450 nm). There is simply less violet light to scatter.
  2. The atmosphere absorbs some of the shortest wavelengths. The ozone layer absorbs most of the ultraviolet and some of the deepest violet.
  3. Your eye is less sensitive to violet than to blue. The cones in your retina respond most strongly to wavelengths around 500 nm; violet at 400 nm triggers the "blue" cones only weakly.

The combined effect of all three factors shifts the peak of the perceived scattered light from the theoretical violet at 400 nm to the blue-cyan around 470 nm that you actually see.

The same physics explains why the Sun appears yellow-white from the surface. The direct sunlight that reaches your eye has lost its blue and violet components to scattering in every direction. What is left is the original solar spectrum minus its blue — which shifts the balance from a roughly uniform white toward yellow.

Why sunsets are red

At sunset the Sun sits low on the horizon. Its light now reaches you through a much longer path through the atmosphere — roughly 40 times longer at the horizon than straight up. On this long path, essentially all of the blue and most of the green light has been scattered out of the beam. What arrives is dominated by red and orange.

This is not just the Sun itself that reddens — the scattered light you see coming off clouds and the sky near the horizon is the same light that was scattered out of someone else's direct beam. The sunset sky near the zenith is dark blue (still fed by the normal blue-scattered beam from above), but the horizon sky is glowing orange from the long-path blue-depleted sunlight.

Geometric picture of why sunsets are redA schematic cross-section of Earth's atmosphere. At noon, sunlight travels a short path through the atmosphere to reach the observer. At sunset, the same sunlight travels a much longer path along the horizon. Blue wavelengths are scattered out over long paths, leaving the red wavelengths to reach the observer.atmosphere (thin layer around Earth)Earthobserver (noon view)short path (~1 unit)observer (sunset view)long path (~40 units)sunlightsunlight
At noon, sunlight takes the shortest path through the atmosphere to reach the observer — blue light is only mildly depleted and the Sun looks yellow-white. At sunset, sunlight grazes through perhaps $40$ times as much atmosphere before reaching the observer, stripping out almost all the blue and green and leaving red and orange.

The Tyndall effect — scattering by bigger particles

The Rayleigh formula assumes particles much smaller than the wavelength (a \ll \lambda — atomic or molecular scale, \sim 0.1–1 nm). When the scattering particles are larger — colloidal particles in a suspension, droplets in fog, smoke particles, dust — the physics changes.

In this regime (formally called Mie scattering, named after the physicist who solved the sphere-scattering problem exactly), the wavelength dependence weakens. For particles comparable to \lambda, the scattering intensity goes roughly as 1/\lambda^2 rather than 1/\lambda^4. For particles much larger than \lambda, it becomes nearly wavelength-independent — all colours scatter nearly equally.

This is why:

The name Tyndall effect refers specifically to this scattering by colloidal particles — first studied in the 19th century to distinguish true solutions (no scattering, no visible beam through the liquid) from colloids (strong scattering, a visible Tyndall cone). Every time you shine a torch through a glass of dilute milk and see a bluish beam, you are watching the Tyndall effect.

Delhi in winter. The winter smog over Delhi — a mixture of dust, soot, particulate matter, and condensed hydrocarbons with particle sizes in the 0.1–10 μm range — sits in the Mie regime. It scatters all wavelengths, which is why smoggy air looks grey-white rather than blue. It also makes headlights of cars appear as glowing cones (the Tyndall cone), because the side-scattered light reaches your eye from every point along the beam. In clear air you would not see the beam from the side at all; in Mie-regime aerosol you see the entire path of the light.

Raman scattering — when the photon trades energy

In the Rayleigh and Mie regimes, the scattered photon has the same wavelength as the incoming photon — elastic scattering. But in 1928, C.V. Raman and his student K.S. Krishnan in Kolkata discovered that a tiny fraction (about one part in 10^7) of the scattered light emerges with shifted wavelengths.

The physics: an incoming photon of energy hf_0 interacts with a molecule. Normally, the molecule gives it back unchanged. But sometimes the molecule is left in a slightly excited vibrational state with energy \Delta E — so the emerging photon carries energy hf_0 - \Delta E and has a slightly longer wavelength. Conversely, if the molecule was already vibrating and gives up that quantum to the photon, the emerging photon has energy hf_0 + \Delta E and a shorter wavelength. The pattern of shifts seen around the incoming frequency — two small satellite lines on either side of each spectral line, spaced by the molecule's vibrational quanta — is the Raman spectrum.

Each molecule has its own set of vibrational frequencies (it is a function of which atoms are in the molecule and how they are bonded). So the Raman spectrum is a fingerprint: point a laser at an unknown substance, record the Raman-scattered light, and the pattern of shifts identifies the molecule. This is the foundation of Raman spectroscopy, now a standard tool in chemistry, materials science, and forensic analysis.

The discovery won Raman the 1930 Nobel Prize in Physics — the first Asian and the first non-white Nobel laureate in the sciences. The experiment used ordinary sunlight filtered through a blue filter, a liquid sample, and a green filter on the detector side: if any light came through both, it was proof that something had shifted the wavelength from blue to green. That it worked with household-grade optics speaks to the robustness of the physics; that it was noticed at all speaks to the depth of Raman's patience with tiny effects.

Worked examples

Example 1: Blue versus red through the atmosphere

Sunlight contains roughly equal intensities of red (700 nm) and blue (450 nm) light at the top of the atmosphere. If the red light is reduced by a fraction \epsilon_\text{red} in traversing the atmosphere (direct path at noon), by what factor is the blue light reduced, assuming Rayleigh scattering is the dominant extinction mechanism and the effect on the red light is small?

Comparative extinction of red and blue through a column of atmosphereTwo parallel arrows of light, one red and one blue, entering a thick atmospheric slab on the left. Both emerge on the right, but the blue arrow is much thinner (more light removed) while the red arrow is nearly full width.atmosphere (column depth $L$)red inred out (95%)blue inblue out (~73%)
Red light passes through the atmosphere almost unscathed; blue is more than five times more strongly scattered and emerges substantially dimmed.

Step 1. Set up the scattering ratio.

Rayleigh's law gives the ratio of scattering intensities as

\frac{I_\text{blue, scattered}}{I_\text{red, scattered}} = \left(\frac{\lambda_\text{red}}{\lambda_\text{blue}}\right)^4 = \left(\frac{700}{450}\right)^4 \approx 5.85.

Why: scattering is what removes light from the forward beam. The fraction of light removed in passing through the atmosphere is proportional to the scattering cross-section, which is proportional to 1/\lambda^4.

Step 2. Translate "scattered amount" to "transmitted fraction". For a thin column, the transmitted fraction is approximately 1 - \epsilon, where \epsilon is the scattered-out fraction. If \epsilon_\text{red} is small (say 5%), then

\epsilon_\text{blue} \approx 5.85 \times \epsilon_\text{red} = 5.85 \times 0.05 = 0.29.

Why: for small extinctions, multiple-scattering corrections are negligible and extinction is linear in path length and scattering cross-section. This is the optically thin limit.

Step 3. Compute transmitted fractions.

  • Red transmitted: 1 - 0.05 = 0.95 \to 95\%.
  • Blue transmitted: 1 - 0.29 = 0.71 \to 71\%.

Why: a direct noon-time check. This is close to the measured ratio — by noon roughly 30% of the blue in direct sunlight has been scattered out, and the missing photons populate the blue sky you see in all other directions.

Step 4. Sunset amplification.

At sunset the atmospheric path is \sim 40 times longer, so \epsilon_\text{red} jumps to roughly 40 \times 0.05 = 2 — but we should not use the thin-atmosphere approximation any more. Using the correct exponential extinction I_\text{out} = I_\text{in}\,e^{-\tau} with optical depth \tau = 40 \times 0.05 = 2 for red:

  • Red transmitted: e^{-2} \approx 0.14 (\tau = 2)
  • Blue transmitted: e^{-5.85 \times 2} = e^{-11.7} \approx 8 \times 10^{-6}

Why: the blue is essentially gone at sunset — nine orders of magnitude reduction. Red dims by a factor of about 7. The remaining direct sunlight is red-orange by overwhelming majority, which is exactly what you see.

Result: At noon, ~95% of red and ~71% of blue reaches you directly. At sunset, ~14% of red and essentially zero blue.

What this shows: The 1/\lambda^4 law does not just "prefer" blue — the preference becomes extreme at large optical depths. The sunset's red is not a small shift; it is the total elimination of the other colours from the direct beam.

Example 2: Why the Tyndall cone in headlights on a smoggy night

A car headlight has a luminous flux of about 1000 lumens and radiates roughly uniformly through the forward hemisphere. On a clear night in Ooty, the beam is invisible from the side — you only see the light when the beam points at your eye. On a Delhi winter night with smog (aerosol particle size \sim 1 μm), the beam is visible as a glowing cone for tens of metres. Why the difference?

Tyndall cone in smoggy airLeft panel: a headlight on a clear night, beam invisible from the side, shown as a faint dashed cone. Right panel: same headlight on a smoggy night, beam visible as a bright scattering cone because aerosol particles side-scatter light into the observer's eye.clear night (Ooty)headlightnot visible from sidesmoggy night (Delhi)observerside-scattered light
On a clear night the headlight's beam is invisible from the side. On a smoggy night, every aerosol particle in the beam scatters a tiny fraction of the light sideways into your eye, painting the path of the beam visible.

Step 1. On a clear night, what's in the air?

Clean mountain air in Ooty contains mostly N₂ and O₂ molecules (\sim 0.3 nm in size) at number density n \sim 2.5 \times 10^{25} m⁻³. These are in the deep Rayleigh regime (a \ll \lambda), so they scatter — but the individual-molecule scattering cross-section is tiny, \sigma_\text{Rayleigh} \sim 10^{-31} m² for visible light. Over a 10 m beam path the fraction of light scattered sideways is n \sigma L \sim 2.5 \times 10^{25} \times 10^{-31} \times 10 \sim 2.5 \times 10^{-5}, or one photon in 40{,}000.

Why: you do notice molecular scattering eventually — that's why distant mountains turn blue — but over metres of clear air the effect is below the threshold of visibility against a dark background.

Step 2. On a Delhi winter smog night, what's in the air?

Delhi's winter PM_\text{2.5} concentrations routinely hit 300–500 μg/m³. For aerosol particles averaging \sim 1 μm diameter and density \sim 10^3 kg/m³, a single particle has mass \sim 5 \times 10^{-16} kg, so the number density is n_\text{aero} \sim 500 \times 10^{-9} / (5 \times 10^{-16}) \sim 10^9 particles/m³.

Why: this is 16 orders of magnitude fewer particles than the molecular number density, but each particle is millions of times larger — so what matters is the cross-section times the number density, not either alone.

Step 3. The cross-section for a 1 μm particle in the Mie regime is approximately its geometric cross-section, \pi r^2 \sim \pi (0.5 \times 10^{-6})^2 \sim 10^{-12} m² — nineteen orders of magnitude bigger than one molecule's Rayleigh cross-section.

The fraction of light side-scattered over a 10 m beam is now n_\text{aero} \sigma L \sim 10^9 \times 10^{-12} \times 10 \sim 10^{-2}, or one photon in 100.

Why: for the same 1000-lumen headlight, that is tens of lumens being scattered sideways per metre of beam — clearly visible to a dark-adapted eye against the dark sky.

Step 4. Colour.

Because the Mie regime is weakly wavelength-dependent, the Tyndall cone is roughly white — the same colour as the headlight itself. The sky itself on a smoggy night is not "blue" because Mie scattering has washed out the Rayleigh signature. That is why smoggy air looks dingy grey-white rather than deep blue.

Why: the Mie cross-section for 1 μm particles has a mild residual wavelength dependence that favours shorter wavelengths, but it is nothing like the 1/\lambda^4 of Rayleigh — the Tyndall cone is nearly achromatic.

Result: Rayleigh scattering by air molecules is \sim 2.5 \times 10^{-5} over 10 m — invisible. Tyndall/Mie scattering by smog aerosol is \sim 10^{-2} over 10 m — the visible glowing cone.

What this shows: The Tyndall cone is a visibility measurement of air quality. The brighter the cone, the more aerosol is in the air. In Switzerland or Ladakh on a clear night, headlights vanish immediately from the side; in Delhi or Beijing winter, they stretch for hundreds of metres. The same physics that turns the sky blue over a pristine ocean makes pollution visible in the beam of a car.

Common confusions

If you came here to understand why the sky is blue, the sunset is red, and what the Tyndall and Raman effects are, you have what you need. What follows is the exact scattering cross-section for a small dielectric sphere, the bridge to Mie theory, and the sense in which Raman scattering is a "vibrational spectroscopy at rest temperature."

The Rayleigh scattering cross-section

The scattered power from a single small dielectric sphere is

P_\text{scattered} = \sigma_\text{Rayleigh}\,I_\text{incident},

where the scattering cross-section for a sphere of radius a and relative permittivity \epsilon_r is

\sigma_\text{Rayleigh} = \frac{8\pi}{3}\left(\frac{2\pi}{\lambda}\right)^4 a^6 \left(\frac{\epsilon_r - 1}{\epsilon_r + 2}\right)^2.

Why: the polarisability of a small sphere is \alpha = 4\pi\epsilon_0 a^3 (\epsilon_r - 1)/(\epsilon_r + 2) (Clausius–Mossotti relation). Substituting into the dipole-radiation power formula and integrating over all directions gives the cross-section above. The \lambda^{-4} is visible in the (2\pi/\lambda)^4 factor; the a^6 dependence means tiny particles scatter much less than even modestly larger ones.

Two things worth noticing:

\sigma \propto a^6 in the Rayleigh regime. Double the particle radius and you multiply the cross-section by 64. This is why rare but relatively large aerosol particles dominate atmospheric scattering over the much more numerous small molecules — the large particles have cross-sections thousands of times bigger per particle.

\sigma \propto 1/\lambda^4. This is the kernel of everything. It comes purely from the \omega^4 in the Larmor formula for a small dipole.

For a nitrogen or oxygen molecule in air, \epsilon_r \approx 1.0006, a \approx 0.1 nm. Plugging in: \sigma \approx 5 \times 10^{-31} m² at \lambda = 550 nm. The mean free path of a blue photon in air is 1/(n\sigma) \sim 1/(2.5 \times 10^{25} \times 5 \times 10^{-31}) \sim 80 km — comparable to the thickness of the atmosphere. Which explains why the atmosphere is partly transparent and partly scattering: blue photons traverse the atmosphere with a roughly one-in-three probability of being scattered, while red photons (with cross-section 5.85\times smaller) traverse with a one-in-fifteen probability.

The size parameter and the transition to Mie

The key dimensionless number is the size parameter

x = \frac{2\pi a}{\lambda}.
  • x \ll 1: Rayleigh regime. \sigma \propto \lambda^{-4}, intensity highly colour-dependent.
  • x \sim 1: Mie regime. Complicated oscillations of \sigma with x; some weak residual colour-dependence.
  • x \gg 1: Geometric-optics regime. \sigma \approx 2\pi a^2 (not \pi a^2: a factor-of-2 enhancement from diffraction around the edge of the object, called the extinction paradox). No wavelength dependence at all.

A cloud droplet at 20 μm and \lambda = 550 nm has x \approx 230 — deep in the geometric regime. The droplet scatters all colours equally, plus an equal diffractive halo, giving the white appearance of clouds.

A 1-μm soot particle has x \approx 11 — still in the geometric regime but with some residual wavelength dependence. Fine smoke is faintly blue from behind (consistent with Rayleigh-like residual) but white from the side.

The full Mie solution gives \sigma(x) as an infinite series of spherical harmonics and Bessel functions — exactly solvable for a sphere but algebraically ugly. For non-spherical particles (ice crystals, pollen grains, volcanic ash), the scattering is even more complex, with distinctive angular patterns that atmospheric scientists use to infer particle shape and composition from light alone.

Raman scattering — a quantum calculation

In a quantum picture, Rayleigh scattering is a virtual process where a photon is absorbed and re-emitted at the same energy, leaving the molecule in its initial state. The molecule's electronic wave function is momentarily perturbed but returns to the ground state.

Raman scattering involves the same virtual absorption, but the molecule lands in a final vibrational state that is different from the initial one. Let |g, v=0\rangle be the ground electronic/vibrational state and |g, v=1\rangle be the first excited vibrational state (still in the electronic ground manifold). Photon absorption takes |g, v=0\rangle \to |\text{virtual}\rangle; re-emission takes |\text{virtual}\rangle \to |g, v=1\rangle. The emitted photon has energy

h f_\text{out} = h f_\text{in} - \Delta E_\text{vib}

where \Delta E_\text{vib} is the vibrational quantum. Since the frequency shift equals the molecule's vibrational frequency, measuring the Raman shift directly measures \Delta E_\text{vib}.

The intensity of Raman scattering is typically \sim 10^{-7} of the Rayleigh intensity — the process requires the molecule to change state, which is a much less likely outcome than the molecule returning unchanged. Modern Raman spectrometers overcome this by using narrow-band lasers, high-efficiency dispersive gratings, and sensitive CCD detectors that can pull out 10^{-7} signals from 10^{0} backgrounds. The result: a fingerprint spectrum of vibrational frequencies unique to every molecule.

Stokes and anti-Stokes lines. If the molecule is initially in |v=0\rangle and ends in |v=1\rangle, the emitted photon is at f_\text{in} - f_\text{vib} — called a Stokes line, lower energy than incident. If the molecule is initially in |v=1\rangle and ends in |v=0\rangle (possible in thermal equilibrium where some molecules are in excited vibrational states), the emitted photon is at f_\text{in} + f_\text{vib} — an anti-Stokes line, higher energy than incident.

The ratio of anti-Stokes to Stokes intensities gives the thermal population of the excited vibrational state, which is e^{-\Delta E_\text{vib}/k_B T}. Measuring the ratio gives the temperature — Raman thermometry. You can measure the temperature of a localised hotspot without touching it, just by illuminating it and looking at the ratio of two spectral lines. This is how surface-enhanced Raman spectroscopy probes temperature inside living cells.

The blue of oceans and the reddening of stars

Two related applications:

Deep oceans appear blue not primarily because of scattering — the dominant effect is absorption of red and infrared by water's O–H stretching modes (vibrations of the water molecule), leaving mostly blue in the transmitted light. This is complementary physics: ocean blue is an absorption effect, sky blue is a scattering effect. The sky also appears blue during the day because scattered blue dominates; after sunset the sky's blueness fades because there is no sunlight to scatter.

Stars look redder than they are. Light from a distant star traverses interstellar dust, which scatters shorter wavelengths preferentially (not quite Rayleigh, since dust grains are not tiny, but with a \lambda^{-1} to \lambda^{-2} dependence from actual dust sizes). Over the thousands of light-years between us and a distant star, this interstellar extinction removes blue light and leaves the star looking redder than its intrinsic colour. Astronomers call this reddening, and they correct for it routinely when inferring the temperatures of distant stars. The technique is a direct astronomical application of atmospheric scattering at a galactic scale.

Where this leads next