Light as Information

After completing this reading, you will be able to:

• Describe

  the electromagnetic spectrum and relate wavelength, frequency, and photon energy

• Identify

  the key ways light interacts with matter (absorption, emission, transmission, reflection, refraction, scattering) with astronomical examples

• Explain

  Rayleigh scattering qualitatively — blue skies and red sunsets

• Interpret

  what a blackbody is and read the Planck spectrum qualitatively

• Identify

  the limiting cases of the Planck function (Rayleigh-Jeans, Wien tail)

• Use

  Wien’s law to estimate temperature from peak wavelength

• Distinguish

  continuous, absorption, and emission spectra using Kirchhoff’s laws

• Explain

  the Doppler shift qualitatively and connect line shifts to radial motion

• Apply

  telescope scaling relations for collecting area and angular resolution

By the end of this reading you’ll be able to: use the continuum shape to infer temperature; use absorption and emission lines to infer composition; use line shifts to infer line-of-sight motion; and use telescope scalings to infer faintness reach and angular detail.

Part 1: Why Light Matters

Look up on a clear night. Every point of light you see is a message from across space and time. But what does that message contain?

In Lecture 1 we learned that astronomers can directly measure only four things: brightness, position, wavelength, and timing. Everything else — temperature, mass, radius, age, composition — must be inferred. Today we begin building the inference toolkit, starting with the question: what can we learn from a star’s color?

The answer involves one of the most important developments in physics: the birth of quantum mechanics. Max Planck’s attempt to explain how hot objects glow led to a revolution that reshaped our understanding of the universe — and the equation he derived, the Planck function, remains the foundation of stellar astrophysics. By the end of this reading you’ll be able to look at a star and estimate its surface temperature from its color, understand why hot stars are blue and cool stars are red, and have your first encounter with a professional-grade physics equation, along with strategies for making such equations less intimidating.

Part 2: The Electromagnetic Spectrum

Unit convention: we use CGS units throughout (cm, s, erg, K). For convenience, optical/UV wavelengths are often written in nm, where $1\ \text{nm} = 10^{-7}\ \text{cm}$.

Light as a Wave

Light is an electromagnetic wave — oscillating electric and magnetic fields that propagate through space at a constant speed.

Unlike mechanical waves (water ripples, sound, a vibrating string), electromagnetic waves don’t need a medium. A water wave is the water moving up and down; a sound wave is air molecules compressing and expanding. But light travels perfectly well through the vacuum of space — what’s oscillating are the electric and magnetic fields themselves.

Both wave types share the same vocabulary:

• Wavelength ($\lambda$): the distance between successive crests. For light, wavelengths range from kilometers (radio) down to sub-atomic scales (gamma rays).
• Frequency ($\nu$): the number of oscillations per second, in s$^{-1}$ (Hertz, Hz).
• Speed ($c$): in vacuum, all electromagnetic waves travel at $c = 2.998 \times 10^{10}\ \text{cm/s}$.

Nanometer

A unit of length, $1\ \mathrm{nm} = 10^{-9}\ \mathrm{m} = 10^{-7}\ \mathrm{cm}$ — roughly ten hydrogen-atom diameters, and the natural unit for optical and UV light. Visible light spans about 400-700 nm.

Optical and UV wavelengths are often written in nanometers. These three quantities are related by the fundamental wave equation:

All EM waves travel at the same speed $c$ in vacuum; they differ only in how they trade off wavelength against frequency. Since $c$ is constant, the two are inversely related, $\nu \propto \lambda^{-1}$: long wavelength means low frequency, short wavelength means high frequency.

Problem

Using $\nu = c\lambda^{-1}$: (1) if wavelength doubles, frequency does what? (2) If frequency triples, wavelength does what? (3) Radio waves have $\lambda \sim 1$ cm; visible light $\lambda \sim 5\times10^{-5}$ cm. How do their frequencies compare?

The Spectrum: From Radio to Gamma

The electromagnetic spectrum spans an enormous range of wavelengths, divided into named bands:

Key insight: different wavelengths reveal different physics. A galaxy looks completely different in radio versus X-rays because you’re seeing different physical components — cold gas versus hot plasma. This is why astronomers build telescopes for every part of the spectrum.

Light as Particles: Photon Energy

Light also behaves as particles called photons. Each photon carries a discrete amount of energy:

Here $h = 6.626 \times 10^{-27}$ erg·s is Planck’s constant. The energy depends on either frequency or wavelength (they’re linked by $c = \lambda\nu$). The story: $E \propto \lambda^{-1}$ — shorter wavelength means higher energy, which is why gamma rays can damage DNA while radio waves pass harmlessly through your body.

Problem

Using $E = hc\lambda^{-1}$: (1) if wavelength doubles, photon energy does what? (2) If wavelength is halved? (3) A 200 nm photon versus an 800 nm photon — how do their energies compare?

Pause & Predict

Same atmosphere, same sunlight - what's different? Commit to a guess before Part 3.

Why do you think the sky looks blue during the day but red/orange at sunset?

Part 3 answers this with Rayleigh scattering.

Part 3: How Light Interacts with Matter

Light carries information because it doesn’t just travel — it interacts. The fundamental interactions are absorption, emission, and scattering; at a boundary we also describe transmission (it gets through), reflection (it bounces), and refraction (it bends). Every spectrum you will ever interpret is some combination of these.

• Absorption: a photon is destroyed and its energy goes into the material (heat, or exciting an electron). Absorption is selective — that selectivity turns a smooth spectrum into one with informative “missing” wavelengths.
• Emission: matter loses energy by creating a photon. What sets the photon’s wavelength is the energy the material loses — which is why atoms produce lines at specific wavelengths.
• Transmission: light passes through without being absorbed. Glass transmits visible but absorbs UV; Earth’s atmosphere transmits visible and radio but absorbs most X-rays.

• Reflection: light bouncing off a surface. The fraction reflected (the albedo) encodes surface composition and texture.
• Refraction: bending as light crosses between media. Lenses focus by refraction; atmospheric refraction makes stars twinkle. It is wavelength-dependent — blue bends more than red, which is why prisms spread white light.

Albedo

The fraction of incident light a surface reflects. A bright icy moon has high albedo; dark rock has low albedo. It encodes information about surface composition and texture.

Scattering: Blue Skies and Red Sunsets

Scattering redirects light into new directions. When the scattering particles are much smaller than the wavelength (like air molecules), we get Rayleigh scattering.

Rayleigh scattering

Scattering of light by particles much smaller than its wavelength, with a steep $\sigma \propto \lambda^{-4}$ dependence. It makes the daytime sky blue, sunsets red, and reddens starlight passing through interstellar dust.

The $\lambda^{-4}$ dependence is steep: blue light ($\lambda \approx 450$ nm) scatters about $(650/450)^4 \approx 4$ times more than red ($\lambda \approx 650$ nm), so $\sigma_{450}/\sigma_{650} = (650/450)^4$.

Why the sky is blue: sunlight enters the atmosphere and meets countless N$_2$ and O$_2$ molecules. Short wavelengths scatter strongly in all directions, so when you look away from the Sun, much of what you see is scattered blue light arriving from all over the sky. Why sunsets are red: at sunset, light travels through much more atmosphere; along that long path the blue is preferentially scattered out of the direct beam, so the light that continues forward is redder. Same physics, different geometry.

Opacity and How Far Light Gets

Often we care less about “what is this made of?” than “how far can photons travel before being absorbed or scattered?” That idea is captured by opacity — how strongly matter blocks light at a given wavelength, usually highly wavelength-dependent. In Module 2 this gives a key idea: we only see to the depth where the material becomes optically thick. For a star, that boundary is the photosphere — where photons last scattered before escaping.

Opacity

A measure of how strongly matter absorbs or scatters light at a given wavelength. High opacity means light is blocked over a short distance. It is strongly wavelength-dependent — a cloud can be transparent in the optical but opaque in the UV.

Retrieval — topic: rayleigh-scattering▸

Part 4: Blackbody Radiation

What Is a Blackbody?

A blackbody is an idealized object that absorbs all radiation falling on it — no reflection, no transmission. When heated, it re-emits radiation with a spectrum that depends only on its temperature. The name is misleading: a cold blackbody looks black, but a hot one glows — the hotter, the brighter and bluer.

Blackbody

An idealized perfect absorber and emitter: it absorbs all incident radiation and, when heated, emits a spectrum (the Planck curve) that depends only on its temperature $T$. Stars are approximate blackbodies.

Stars are approximate blackbodies — their spectra deviate due to absorption lines, but the overall shape follows the blackbody curve. Other astronomical blackbodies: planets (Earth emits as a $\sim 255$ K blackbody in the infrared), the Cosmic Microwave Background (the most perfect blackbody ever measured, at 2.725 K), and accretion disks around black holes.

The Planck Spectrum: Qualitative Picture

When you heat an object it glows, and the color depends on temperature: a stovetop burner dull red ($\sim$700 K), an incandescent filament yellow-white ($\sim$2700 K), the Sun yellow ($\sim$5800 K), a welding arc blue-white ($\sim$6000+ K).

The blackbody spectra (Planck curves) show two patterns: (1) the peak shifts to shorter wavelengths as temperature increases — hot objects peak in the blue/UV, cool objects in the red/IR; (2) the total energy emitted increases dramatically with temperature — the area under the curve grows as $T^4$ (the Stefan-Boltzmann law, quantified in Module 2).

Pause & Predict #1

Think about the shape of the Planck curve, not just the peak location.

The blue star (8000 K) peaks in the ultraviolet. Why does it still appear blue to our eyes, rather than invisible?

The red star (3000 K) peaks in the infrared. Why does it still glow visibly red?

Check your reasoning against the answers below.

Multiple choice

An 8000 K star peaks in the ultraviolet, yet it still looks blue to your eyes (not invisible). Why?

Its Planck curve is broad — it emits plenty of visible light, and more blue than redIts peak is actually in the visible, not the UVUV light reddens to blue in our atmosphereAll hot stars look white

The Ultraviolet Catastrophe

Before 1900, classical physics gave the Rayleigh-Jeans law, $B_\lambda \propto T\lambda^{-4}$. It works at long wavelengths, but as $\lambda \to 0$ the intensity diverges — classical physics predicted any warm object should emit infinite energy in the ultraviolet. This ultraviolet catastrophe was a fundamental failure: ovens don’t emit infinite UV.

Planck’s Quantum Solution

In 1900 Max Planck found the fix — “an act of desperation.” He proposed that energy is emitted in discrete packets (quanta) with energy $E = h\nu$. To emit a short-wavelength (high-frequency) photon, an oscillator needs energy $h\nu$; if the available thermal energy ($\sim k_B T$) is less than $h\nu$, emission is suppressed — you can’t emit a fraction of a photon. This quantum suppression cuts off the catastrophe. It took Einstein (1905) to take quanta seriously, showing light itself comes in photons.

The Planck Function: Meeting a Real Equation

The complete blackbody spectrum is the Planck function. It looks intimidating — don’t panic; we read it piece by piece.

$B_\lambda(T)$ is the spectral radiance — power emitted per unit area, per unit solid angle, per unit wavelength. The $\lambda^{-5}$ factor would drive short wavelengths up, but the exponential term in the denominator suppresses them; the competition between the two creates the peak. You don’t need to compute it from scratch — you need to recognize its shape and know which term controls which regime.

Spectral radiance

The power a surface emits per unit area, per unit solid angle (“per direction”), per unit wavelength interval — what the Planck function $B_\lambda(T)$ gives. CGS units: $\mathrm{erg\,s^{-1}\,cm^{-2}\,sr^{-1}\,cm^{-1}}$.

The spectral radiance tells you how bright a surface looks in a given direction at each wavelength. A unit check confirms $[B_\lambda] = \mathrm{erg\,s^{-1}\,cm^{-2}\,sr^{-1}\,cm^{-1}}$ — power per area per solid angle per wavelength interval, exactly what “spectral radiance” should be.

Limiting Cases: Taming Complex Equations

The way to not be intimidated by a scary equation is to take its limiting cases — what happens when one term dominates.

Case 1 — long wavelengths (Rayleigh-Jeans). When $hc/\lambda k_B T \ll 1$, expand $e^x - 1 \approx x$, so

$B_\lambda \approx \frac{2hc^2}{\lambda^5}\cdot\frac{\lambda k_B T}{hc} = \frac{2ck_B T}{\lambda^4}.$

Classical physics works here because quantum effects are negligible when $h\nu \ll k_B T$.

Case 2 — short wavelengths (Wien tail). When $hc/\lambda k_B T \gg 1$, the $-1$ is negligible and

$B_\lambda \approx \frac{2hc^2}{\lambda^5}\,e^{-hc/\lambda k_B T}.$

The exponential cuts off short-wavelength emission — the quantum suppression that prevents the catastrophe. You can’t emit photons whose energy exceeds the available thermal energy.

Problem

Using $B_\lambda \propto T\lambda^{-4}$: (1) if $T$ doubles (fixed $\lambda$), intensity does what? (2) If $\lambda$ doubles (fixed $T$)? (3) Two blackbodies at the same $T$, one at $\lambda = 1$ cm and one at $\lambda = 0.5$ cm — how do their Rayleigh-Jeans intensities compare?

Part 5: Wien’s Displacement Law

The Peak Wavelength

The Planck function peaks at a wavelength that depends only on temperature:

Here $b = 0.2898\ \text{cm·K} = 2.898\times10^{6}\ \text{nm·K}$ is Wien’s displacement constant, and “peak wavelength” means the maximum of $B_\lambda$ (per unit wavelength). The story: $\lambda_{\text{peak}} \propto T^{-1}$ — hotter objects peak bluer, cooler objects redder. This is why color encodes temperature.

Problem

Using $\lambda_{\text{peak}} = bT^{-1}$: (1) if $T$ doubles, $\lambda_{\text{peak}}$ does what? (2) If $T$ halves? (3) Star A at 6000 K versus Star B at 3000 K — how do their peaks compare?

Worked Example 1The Sun's Surface Temperature

Problem

The Sun’s spectrum peaks at approximately 500 nm (green-yellow). What is its surface temperature?

Step·Invert Wien's law

$T = \frac{b}{\lambda_{\text{peak}}} = \frac{2.898 \times 10^{6}\ \text{nm·K}}{500\ \text{nm}} = 5796\ \text{K} \approx 5800\ \text{K}$

Dimensional check

$[T] = \dfrac{\text{cm·K}}{\text{cm}} = \text{K}$ ✓ (using nm·K / nm consistently).

Result

About 5800 K — determined from the color of sunlight alone, no thermometer and no spacecraft. This is astronomical inference in action.

Worked Example 2Comparing Betelgeuse and Rigel

Problem

Betelgeuse (a red supergiant) peaks near 830 nm; Rigel (a blue supergiant) peaks near 240 nm. Find their surface temperatures.

Step·Apply Wien's law to each

$T_{\text{Bet}} = \frac{2.898 \times 10^{6}\ \text{nm·K}}{830\ \text{nm}} \approx 3500\ \text{K}, \qquad T_{\text{Rigel}} = \frac{2.898 \times 10^{6}\ \text{nm·K}}{240\ \text{nm}} \approx 12{,}000\ \text{K}$

Dimensional check

nm·K divided by nm leaves K ✓ — a temperature, as required.

Result

Rigel is about $3.5\times$ hotter than Betelgeuse. In Orion you’re literally seeing temperature as color: the red shoulder (Betelgeuse) is cool, the blue foot (Rigel) is hot.

Worked Example 3The Cosmic Microwave Background

Problem

The CMB — the afterglow of the Big Bang — peaks at about $0.11$ cm. What is its temperature?

Step·Invert Wien's law (CGS)

$T_{\text{CMB}} = \frac{0.2898\ \text{cm·K}}{0.11\ \text{cm}} = 2.63\ \text{K} \approx 2.7\ \text{K}$

Dimensional check

cm·K divided by cm leaves K ✓.

Result

About 2.7 K — the temperature of the universe itself, cooling for 13.8 billion years as space expands. The CMB is the most perfect blackbody ever measured (COBE confirmed deviations $\lesssim 50$ ppm); Planck mapped its $\pm 0.0002$ K fluctuations, the seeds of all cosmic structure.

Pause & Predict

A nebula glows bright red at one specific wavelength (656.3 nm); a nearby star shows a smooth rainbow with dark lines. Think about what's needed to produce each spectrum type.

Which object produces an emission spectrum, and which an absorption spectrum?

What does that tell you about the physical conditions in each object?

Part 6 develops Kirchhoff's laws to answer this.

Part 6: Atoms and Spectral Lines

The Sun’s spectrum isn’t a smooth rainbow — it’s crossed by hundreds of dark absorption lines. They aren’t imperfections; they’re information. The key idea: atoms have quantized energy levels.

The Bohr Model

Modern quantum mechanics is deeper, but the Bohr model gets the essential story right and gives correct energies for hydrogen. The electron can only occupy discrete levels labeled by an integer $n = 1, 2, 3, \dots$

Bohr model

An early model of the atom in which an electron occupies discrete, quantized energy levels labeled by an integer $n$. It correctly predicts hydrogen’s energy levels and spectral lines, and supplies the intuition behind quantum atomic structure.

For hydrogen, the energy of level $n$ is:

The ground state ($n=1$) is the most tightly bound; higher levels have energies closer to zero (less bound, easier to ionize).

Photon Absorption and Emission

When an electron changes levels, the atom absorbs or emits a photon whose energy matches the difference:

$\Delta E = E_{\text{upper}} - E_{\text{lower}} = h\nu = \frac{hc}{\lambda}.$

An absorption line appears when photons of a particular energy are removed because electrons jump up; an emission line appears when a thin gas produces photons as electrons fall down.

Kirchhoff’s Laws: Three Types of Spectra

Kirchhoff’s laws identify three basic spectrum shapes, each from a different physical situation — this is how we know what stars are made of without sampling them.

Kirchhoff's laws

Three rules linking a spectrum’s appearance to its source: (1) a hot dense object gives a continuous spectrum; (2) a hot thin gas gives an emission-line spectrum; (3) cool gas in front of a hot continuous source gives an absorption-line spectrum.

A continuous spectrum comes from hot, dense matter (a blackbody-like source). An absorption spectrum arises when cool gas sits in front of a hot continuous source. An emission spectrum comes from hot, thin gas with no bright background. The remarkable thing: the same atoms produce both absorption and emission — what changes is the density/temperature structure and whether there’s a bright continuum behind the gas.

Continuous spectrum

A smooth spectrum spanning all wavelengths, with no lines — emitted by a hot, dense (blackbody-like) source such as a stellar interior.

Absorption spectrum

A continuous spectrum crossed by dark lines, produced when cooler gas in front of a hot continuous source removes photons at its characteristic wavelengths. Stellar spectra are absorption spectra.

Emission spectrum

Bright lines on a dark background, produced by a hot, thin gas with no bright continuum behind it — for example, an emission nebula.

Spectral lines reveal more than composition: which transitions are populated gives temperature; Doppler shifts give velocity; line widths give density/pressure — all developed in Module 2.

Retrieval — topic: kirchhoff-laws▸

Part 7: Real Stellar Spectra

Beyond the Ideal Blackbody

Stars are approximately blackbodies. Real spectra show two features: the overall shape follows a Planck curve (temperature, via Wien’s law), and absorption lines appear as dark notches (composition, via spectroscopy).

The spectrum of Altair (an A-type star) shows the smooth blackbody shape plus sharp absorption dips from hydrogen and other elements — two inference tools in one observation: the curve shape gives temperature, the line positions give composition.

Compare the Sun’s spectrum (approximately continuous, like a blackbody) with a fluorescent bulb (discrete emission lines): the shape tells you the emission mechanism — thermal versus non-thermal.

Doppler Shift: Motion in Spectral Lines

Spectral lines are also speedometers. If a source moves toward or away from us, its lines shift slightly, encoding the line-of-sight velocity — we measure motion using nothing but light.

For speeds much smaller than $c$, the non-relativistic Doppler shift is:

Here $\Delta\lambda = \lambda_{\text{obs}} - \lambda_0$, $\lambda_0$ is the rest wavelength, and $v_r$ is the radial (line-of-sight) velocity. If $v_r>0$ the lines redshift (longer $\lambda$); if $v_r$ is negative they blueshift. Accurate when $|v_r| \ll c$; very large cosmological redshifts need a relativistic treatment.

This tiny shift unlocks some of the biggest stories in astronomy: a star’s wobble reveals exoplanets; binary Doppler shifts give orbital speeds and masses; line shifts across a galaxy map its rotation (and point to dark matter); and overall redshift shows the universe is expanding.

Retrieval — equation: doppler-shift▸

Pause & Predict

JWST has a 6.5 m mirror; Hubble has a 2.4 m mirror. Think about how area and resolution scale with diameter.

How much more light-gathering power does JWST have than Hubble?

Which telescope can see finer detail at the same wavelength?

Part 8 makes both quantitative.

Part 8: Telescopes as Light Buckets

Telescopes extend our reach two ways: they collect more photons and resolve finer detail.

Collecting Area: Why Bigger Is Better

For a circular aperture, the collecting area is:

Collecting area

The light-gathering area of a telescope’s aperture, $A = \pi(D/2)^2$. The photon rate from a source scales with $A \propto D^2$, so aperture — not magnification — sets how faint an object a telescope can detect.

The photon rate scales with area, so the faintest detectable source scales as $D^2$: a mirror twice as wide gathers four times the light.

Problem

Using $A \propto D^2$: (1) if diameter doubles, area does what? (2) Triples? (3) JWST ($D = 6.5$ m) versus Hubble ($D = 2.4$ m)?

Angular Resolution: Why Bigger Is Sharper

Even with lots of light, you can’t see detail finer than your angular resolution. Diffraction sets the limit:

Angular resolution

The smallest angular separation a telescope can distinguish, set by diffraction: $\theta_{\min} \approx \lambda/D$ (the circular-aperture form is $1.22\,\lambda/D$). Larger apertures and shorter wavelengths give finer resolution.

For a circular aperture the precise form is $\theta \approx 1.22\,\lambda/D$; for scaling arguments we use $\lambda/D$. Bigger $D$ and shorter $\lambda$ both sharpen the image.

Problem

Using $\theta \approx \lambda D^{-1}$: (1) if diameter doubles (fixed $\lambda$), resolution does what? (2) If wavelength doubles (fixed $D$)? (3) Observing at $\lambda = 2\ \mu$m instead of $0.5\ \mu$m with the same telescope?

Ground-based optical telescopes often miss the diffraction limit because atmospheric turbulence blurs images — seeing. Adaptive optics corrects some blur in real time; space telescopes avoid the atmosphere entirely. Large modern telescopes are reflectors: mirrors can be supported from behind (even segmented), making huge apertures practical, while large lenses sag under their own weight and suffer chromatic aberration.

Seeing

The blurring of astronomical images by turbulence in Earth’s atmosphere, which makes ground-based resolution (typically $\sim 1$ arcsec) far worse than the diffraction limit. Adaptive optics or space telescopes overcome it.

Part 9: Synthesis — The Observable → Model → Inference Chain

We’ve followed the photon’s story end to end: how it’s produced, how it interacts, what it tells us, and how we catch it. The chapter’s whole method is the inference chain — measure something, apply a physical model, infer what you couldn’t measure directly. Two of these chains are clean enough to render explicitly:

The same pattern runs through the rest of the chapter:

What’s coming: next we’ll measure distance (parallax, standard candles, the inverse-square relation); combined with apparent brightness, that gives luminosity. Then we return to blackbody physics to connect temperature, luminosity, and radius (Stefan-Boltzmann), and deepen spectroscopy. Each week the inference chain grows longer — by the end of Module 2 you’ll characterize a star’s temperature, luminosity, radius, and mass entirely from its light.

Proportional Reasoning Practice

The equations here are all proportional relationships; internalizing them — without a calculator — is a crucial skill.

Problem

Two telescopes observe the same star — Telescope A: $D = 4$ m at $\lambda = 500$ nm; Telescope B: $D = 8$ m at $\lambda = 2000$ nm. (1) Which collects more light, and by what factor? (2) Which has better angular resolution? (3) For a $T = 5000$ K blackbody, which observes closer to the peak?

Quick Practice

Four no-calculator warm-ups — each answer is a single clean factor.

Numeric answer

If a wave’s wavelength decreases by a factor of 3, by what factor does its frequency change?

Your answer

Numeric answer

Two photons have wavelengths 500 nm and 1000 nm. By what factor is the 500 nm photon’s energy greater than the 1000 nm photon’s?

Your answer

Numeric answer

Rayleigh scattering scales as $\lambda^{-4}$. By what factor is scattering stronger at 400 nm than at 800 nm?

Your answer

Numeric answer

If a blackbody’s peak wavelength is cut in half, by what factor does its temperature change?

Your answer

The 14 graded practice problems for this lecture live in the companion practice set.

Reference Tables

Key Equations

Physical Constants

Temperature and Color

Glossary

Absorption spectrum

A continuous spectrum crossed by dark lines, produced when cooler gas in front of a hot continuous source removes photons at its characteristic wavelengths. Stellar spectra are absorption spectra.

Albedo

The fraction of incident light a surface reflects. A bright icy moon has high albedo; dark rock has low albedo. It encodes information about surface composition and texture.

Angular resolution

The smallest angular separation a telescope can distinguish, set by diffraction: $\theta_{\min} \approx \lambda/D$ (the circular-aperture form is $1.22\,\lambda/D$). Larger apertures and shorter wavelengths give finer resolution.

Blackbody

An idealized perfect absorber and emitter: it absorbs all incident radiation and, when heated, emits a spectrum (the Planck curve) that depends only on its temperature $T$. Stars are approximate blackbodies.

Bohr model

An early model of the atom in which an electron occupies discrete, quantized energy levels labeled by an integer $n$. It correctly predicts hydrogen’s energy levels and spectral lines, and supplies the intuition behind quantum atomic structure.

Collecting area

The light-gathering area of a telescope’s aperture, $A = \pi(D/2)^2$. The photon rate from a source scales with $A \propto D^2$, so aperture — not magnification — sets how faint an object a telescope can detect.

Continuous spectrum

A smooth spectrum spanning all wavelengths, with no lines — emitted by a hot, dense (blackbody-like) source such as a stellar interior.

Emission spectrum

Bright lines on a dark background, produced by a hot, thin gas with no bright continuum behind it — for example, an emission nebula.

Kirchhoff's laws

Three rules linking a spectrum’s appearance to its source: (1) a hot dense object gives a continuous spectrum; (2) a hot thin gas gives an emission-line spectrum; (3) cool gas in front of a hot continuous source gives an absorption-line spectrum.

Nanometer

A unit of length, $1\ \mathrm{nm} = 10^{-9}\ \mathrm{m} = 10^{-7}\ \mathrm{cm}$ — roughly ten hydrogen-atom diameters, and the natural unit for optical and UV light. Visible light spans about 400-700 nm.

Opacity

A measure of how strongly matter absorbs or scatters light at a given wavelength. High opacity means light is blocked over a short distance. It is strongly wavelength-dependent — a cloud can be transparent in the optical but opaque in the UV.

Rayleigh scattering

Scattering of light by particles much smaller than its wavelength, with a steep $\sigma \propto \lambda^{-4}$ dependence. It makes the daytime sky blue, sunsets red, and reddens starlight passing through interstellar dust.

Seeing

The blurring of astronomical images by turbulence in Earth’s atmosphere, which makes ground-based resolution (typically $\sim 1$ arcsec) far worse than the diffraction limit. Adaptive optics or space telescopes overcome it.

Spectral radiance

The power a surface emits per unit area, per unit solid angle (“per direction”), per unit wavelength interval — what the Planck function $B_\lambda(T)$ gives. CGS units: $\mathrm{erg\,s^{-1}\,cm^{-2}\,sr^{-1}\,cm^{-1}}$.