Spectroscopy and Spectral Lines

Spectroscopy is the single most information-rich technique in astrophysics: disperse the light of a star, nebula, or galaxy through a prism or grating and it fans into a spectrum, a smooth continuum crossed by sharp bright or dark features — spectral lines. Each line sits at a wavelength set by a single quantum transition, and because those rest wavelengths are fixed by atomic physics and identical in a campus laboratory and in a galaxy billions of light-years away, they are unforgeable chemical fingerprints. The physics splits cleanly into two questions. Which lines appear is answered by atomic structure — the Bohr model and its quantum refinement give the discrete energy levels and hence the Rydberg wavelengths of the Lyman, Balmer, and Paschen series of hydrogen. How strong they are is a problem in statistical mechanics: the Boltzmann equation fixes the populations of excited levels, the Saha equation fixes the balance between ionization stages, and their product explains the century-old puzzle of why hydrogen's Balmer lines are strongest not in the hottest O stars but near A-type stars at about 9500 K. Once a line is measured, three of its properties become three separate instruments: the shift of its center reads a line-of-sight velocity through the Doppler effect — the engine of exoplanet hunting, galaxy rotation curves, and cosmological redshift; its integrated depth, the equivalent width, reads an abundance through the curve of growth; and its profile shape reads temperature, turbulence, pressure, and rotation. A single well-measured line can thus constrain composition, temperature, density, gravity, and motion at once.

Key takeaways

Spectroscopy spreads light by wavelength, exposing emission and absorption lines whose positions, depths, and shapes encode a source's composition, temperature, pressure, and motion.
Line wavelengths come from quantized atomic transitions fixed by quantum mechanics — the Rydberg formula gives hydrogen's Balmer series (Hα at 656.3 nm) — so a line's position identifies the species unambiguously.
Line strengths, read as equivalent widths, are set by the Boltzmann and Saha equations, whose product peaks the hydrogen Balmer lines near T ≈ 9500 K (A-type stars), not in the hottest stars.
Doppler shifts of line centers measure radial velocities and cosmological redshift; thermal, pressure, and rotational broadening of the profile diagnose the physical conditions.
The MK classification OBAFGKM is built directly on line ratios, giving a two-dimensional temperature-plus-luminosity map of stars.
Key idea: a spectral line is a transition between two quantized energy levels at a fixed rest wavelength — its presence names a species, its strength (via Boltzmann + Saha) measures temperature and abundance, and its shift and shape (via Doppler and broadening) measure motion and physical conditions.

The core insight

A spectral line is a transition between two quantized energy levels at a fixed rest wavelength; its presence identifies a species, its strength (via Boltzmann + Saha) measures temperature and abundance, and its shift and shape (via Doppler and broadening) measure motion and physical conditions.

Where this lives today

Spectroscopy is still the workhorse of observational astrophysics, and the same equations now drive facilities the early spectroscopists could not have imagined. Precision radial velocities from stabilized spectrographs such as ESPRESSO on the VLT measure stellar wobbles below 1 m/s — a shift of one part in $10^{9}$ in a line's wavelength — to detect Earth-mass exoplanets, a direct application of equation (6) pushed to its limit. Massive redshift surveys — SDSS, DESI, and the coming 4MOST and MOONS instruments — read redshift from identified lines (the 4000 Å break, Ca II H and K, Lyman-α) in tens of millions of galaxies to map the cosmic web and constrain dark energy through baryon acoustic oscillations. In stellar physics, surveys like APOGEE and GALAH fit equivalent widths and Voigt profiles against model atmospheres — Boltzmann and Saha populations wrapped in radiative transfer — to measure abundances in hundreds of thousands of stars, reconstructing the chemical assembly of the Milky Way ("Galactic archaeology"). At the frontier, JWST detects redshifted rest-optical lines from galaxies at $z>10$ , and transmission spectroscopy catches atomic lines imprinted on starlight filtered through exoplanet atmospheres. Every one rests on the premise Fraunhofer stumbled onto: a line's rest wavelength is a universal constant, so any deviation is physics.

The physical picture

Think of every atom as an instrument that can sound only certain notes: the allowed energy levels are the notes, and a spectral line is the sound of an electron jumping between two of them. Because every hydrogen atom in the Universe carries the same ladder of notes, seeing the Balmer sequence imprinted on a distant galaxy's light says immediately "hydrogen is here." The first figure draws that ladder for hydrogen and labels the real jumps: the visible Balmer series lands on the second rung, the ultraviolet Lyman series lands on the ground rung, and the wavelengths that come out are exactly the ones a laboratory discharge tube emits.

Figure

Hydrogen levels Eₙ = −13.6 eV / n²: downward jumps landing on n = 2 make the visible Balmer series — Hα 656.3 nm, Hβ 486.1 nm — while jumps landing on n = 1 make the ultraviolet Lyman series such as Lyα 121.6 nm, and jumps to n = 3 make the infrared Paschen series

A spectrum is a bar code. The positions of the bars say what is present, the darkness of the bars says how much and under what conditions, and a uniform stretch of the whole code says how fast the source recedes. But line strength is a competition, not a headcount: Boltzmann statistics decide what fraction of atoms sit in the exact lower level that can absorb the photon, and the Saha equation decides what fraction of the element is even in the right ionization stage. Temperature turns both knobs at once, which is why every line has a temperature at which it is strongest. The second figure is the classic resolution of the puzzle that stumped the first generation of spectroscopists — why hydrogen, the most abundant element, shows its strongest visible lines in middle-temperature A stars rather than the hottest ones.

Figure

Balmer line strength versus temperature from the Boltzmann × Saha product: rising excitation into n = 2 fights falling neutral fraction, and the two combine to a sharp peak near T ≈ 9500 K — the temperature of A0 stars — so the strongest hydrogen lines appear at intermediate, not maximal, temperature

Finally, a line profile is a fingerprint of motion. Random thermal jostling and small-scale turbulence give a Gaussian core (Doppler broadening); the finite lifetime of a level and perturbing collisions give broad Lorentzian wings (natural and pressure broadening); coherent rotation smears the whole profile symmetrically. The convolution of the Gaussian and Lorentzian is the Voigt profile that real stellar lines follow.

The core equations

The wavelengths of hydrogen's lines follow a single empirical law that the Bohr model later explained exactly. It is the anchor of the whole subject: memorize it and you can predict every hydrogen line from two integers.

\frac{1}{\lambda} = R_{\mathrm{H}}\left(\frac{1}{n_1^{2}} - \frac{1}{n_2^{2}}\right) \qquad (1)

where: $\lambda$ — vacuum wavelength of the emitted or absorbed photon; $R_{\mathrm{H}}$ — Rydberg constant for hydrogen, $1.0968\times10^{7}\,\mathrm{m^{-1}}$ ; $n_1,\,n_2$ — principal quantum numbers of the lower and upper levels ( $n_2>n_1$ ). Setting $n_1=2$ gives the visible Balmer series; $n_1=1$ gives the ultraviolet Lyman series. The Rydberg law is a special case of the hydrogenic energy ladder, which also fixes the levels of one-electron ions like He II and sets the tiny isotope shift between hydrogen and deuterium through the reduced mass.

E = -13.6\,Z^{2}\,\frac{\mu}{m_e}\,\frac{1}{n^{2}}\;\mathrm{eV} \qquad (2)

where: $Z$ — nuclear charge (1 for H, 2 for He II); $\mu$ — reduced mass of the electron–nucleus system; $m_e$ — electron mass; $n$ — principal quantum number. The factor $Z^2$ scales every level, so He II lines fall at one quarter the wavelength of the matching hydrogen lines. Which lines can even form depends on how the atoms are distributed over their levels. In thermal equilibrium the ratio of populations in any two bound levels is a Boltzmann factor weighted by the level degeneracies.

\frac{N_b}{N_a} = \frac{g_b}{g_a}\,e^{-(E_b-E_a)/kT} \qquad (3)

where: $N_a,\,N_b$ — number densities of atoms in the lower level $a$ and upper level $b$ ; $g_a,\,g_b$ — statistical weights (degeneracies), $g_n=2n^2$ for hydrogen; $E_b-E_a$ — excitation energy; $k$ — Boltzmann constant, $1.381\times10^{-23}\,\mathrm{J\,K^{-1}}$ ; $T$ — excitation temperature (the kinetic temperature in LTE). The exponential is why cool gas keeps almost all its atoms in the ground level. Boltzmann only redistributes atoms of a given ion. The Saha equation goes one level higher and fixes how the element splits between successive ionization stages, and it is the reason temperature — not abundance — controls which lines a star shows.

\frac{N_{i+1}}{N_i} = \frac{2Z_{i+1}}{n_e Z_i}\left(\frac{2\pi m_e kT}{h^{2}}\right)^{3/2} e^{-\chi_i/kT} \qquad (4)

where: $N_{i+1},\,N_i$ — number densities in ionization stages $i+1$ and $i$ ; $Z_{i+1},\,Z_i$ — partition functions of the two stages; $n_e$ — free-electron number density; $m_e$ — electron mass, $9.109\times10^{-31}\,\mathrm{kg}$ ; $\chi_i$ — ionization energy from stage $i$ (13.6 eV for H I); $h$ — Planck constant, $6.626\times10^{-34}\,\mathrm{J\,s}$ . The $n_e$ in the denominator is why line ratios depend on pressure and hence on stellar luminosity. The measurable strength of a line is not its depth but its integrated area, expressed as an equivalent width — the width of a totally black rectangle that removes the same flux from the continuum.

W_\lambda = \int \frac{F_c - F_\lambda}{F_c}\,\mathrm{d}\lambda \qquad (5)

where: $W_\lambda$ — equivalent width, in units of wavelength; $F_c$ — continuum flux level just outside the line; $F_\lambda$ — observed flux inside the line at wavelength $\lambda$ . A stronger line removes more area and so has a larger $W_\lambda$ regardless of how the flux is distributed between core and wings. A shift of the line center from its rest wavelength measures the line-of-sight velocity. For everyday speeds the fractional shift is simply the velocity in units of the speed of light.

\frac{\Delta\lambda}{\lambda_0} = \frac{v_r}{c} \quad (|v_r|\ll c) \qquad (6)

where: $\Delta\lambda=\lambda_{\rm obs}-\lambda_0$ — the observed shift of the line center; $\lambda_0$ — rest (laboratory) wavelength; $v_r$ — radial velocity, positive for recession; $c$ — speed of light, $2.998\times10^{8}\,\mathrm{m\,s^{-1}}$ . A redshift ( $\Delta\lambda>0$ ) means the source is receding; a blueshift means it approaches. At high speed the classical formula fails and the special-relativistic Doppler formula takes over; in cosmology the same observed ratio is reinterpreted as a stretching of space itself.

1+z = \frac{\lambda_{\rm obs}}{\lambda_{\rm emit}} = \sqrt{\frac{1+v/c}{1-v/c}} \qquad (7)

where: $z$ — redshift; $\lambda_{\rm obs}$ — observed wavelength; $\lambda_{\rm emit}$ — emitted rest-frame wavelength; $v$ — recession speed in the special-relativistic radial case. Cosmologically $1+z=a_0/a$ , the ratio of the scale factor now to its value at emission — not a local velocity. The width of a line's Gaussian core is set by the spread of line-of-sight velocities from thermal motion and microturbulence.

\Delta\lambda_{\rm D} = \frac{\lambda_0}{c}\sqrt{\frac{2kT}{m} + \xi^{2}} \qquad (8)

where: $\Delta\lambda_{\rm D}$ — the $1/e$ Doppler half-width of the Gaussian core; $m$ — mass of the absorbing atom; $\xi$ — microturbulent velocity; $T$ — gas temperature. Light atoms (hydrogen) give wider thermal cores than heavy atoms (iron) at the same temperature. The core rides on Lorentzian wings whose narrowest, irreducible contribution comes from the finite lifetime of the levels themselves.

\Delta\lambda_{\rm nat} = \frac{\lambda_0^{2}}{2\pi c}\left(\frac{1}{\tau_a}+\frac{1}{\tau_b}\right) \qquad (9)

where: $\Delta\lambda_{\rm nat}$ — Lorentzian natural line width; $\tau_a,\,\tau_b$ — mean lifetimes of the lower and upper levels; $\lambda_0$ — rest wavelength. This is the quantum floor: even an isolated, motionless atom emits a line of nonzero width.

Derivation AFrom Bohr's quantized orbits to the Rydberg formula

The goal here is to show that the purely empirical Rydberg formula, written down to fit hydrogen's wavelengths before anyone understood atoms, drops straight out of a single physical assumption — that the electron's energy is quantized as $E_n=-13.6\,\mathrm{eV}/n^2$ . We take that ladder as given from the Bohr model, impose energy conservation on a jump between two rungs, and convert the photon energy to a wavelength, recovering equation (1) with the Rydberg constant expressed in fundamental constants. Along the way the Balmer series appears as nothing more special than "all jumps that land on the second rung."

Step 1. Write the hydrogen energy levels as $E_n=-13.6\,\mathrm{eV}/n^2$ , measured downward from the ionization limit at $E_\infty=0$ . Why: quantization is the whole content of the model — a continuum of orbits would give a continuum of emitted wavelengths and no lines at all, so the discrete ladder is what makes spectra discrete.

Step 2. Impose energy conservation on a downward transition: the emitted photon carries exactly the energy difference, $h\nu = E_{n_2}-E_{n_1}$ , and an upward transition absorbs a photon of the same energy. Why: this ties a wavelength to a pair of levels, so a measured line becomes a direct readout of two quantum numbers.

Step 3. Substitute $\nu=c/\lambda$ and the level formula to get $hc/\lambda = 13.6\,\mathrm{eV}\,(1/n_1^2 - 1/n_2^2)$ , then divide through by $hc$ to reach $1/\lambda = R_{\mathrm H}(1/n_1^2 - 1/n_2^2)$ with $R_{\mathrm H}=13.6\,\mathrm{eV}/hc = 1.0968\times10^7\,\mathrm{m^{-1}}$ . Why: this identifies the once-mysterious Rydberg constant with a combination of fundamental constants, turning an empirical fit into a prediction.

Step 4. Set $n_1=2$ : the shortest jump $n_2=3$ gives Hα at $1/\lambda=R_{\mathrm H}(1/4-1/9)$ , or $\lambda=656.3\,\mathrm{nm}$ , and $n_2=4$ gives Hβ at $486.1\,\mathrm{nm}$ ; the series converges to a head at $\lambda=364.6\,\mathrm{nm}$ as $n_2\to\infty$ . Why: these are exactly the wavelengths measured in a discharge tube and seen in stellar spectra, so the derivation is verified against the laboratory to four figures.

Go deeper · Derivation B · 2 minWhy the hydrogen Balmer lines peak near A-type stars

Derivation BWhy the hydrogen Balmer lines peak near A-type stars

The central puzzle of early stellar spectroscopy was that hydrogen — overwhelmingly the most abundant element — shows its strongest visible (Balmer) lines not in the hottest O stars but in intermediate A stars near 9500 K. The goal of this derivation is to resolve that by combining two of the core equations: Boltzmann's law, which sets what fraction of neutral hydrogen atoms have their electron in the n = 2 level that Balmer absorption starts from, and Saha's equation, which sets what fraction of hydrogen remains neutral at all. Their product is a rising factor times a falling factor, and any such product is peaked. The result is a specific temperature, which we then compare to the observed spectral sequence.

Step 1. Balmer absorption needs a neutral hydrogen atom with its electron already in n = 2, so first count excitation into that level with Boltzmann: $N_2/N_1 = (g_2/g_1)\,e^{-(E_2-E_1)/kT} = 4\,e^{-10.2\,\mathrm{eV}/kT}$ , using $g_1=2$ , $g_2=8$ , and a 10.2 eV gap. Why: at low temperature this exponential is minuscule, so almost no atoms are in n = 2 and there is nothing to make Balmer absorption — the lines are weak from below.

Step 2. Now count how much hydrogen is still neutral with Saha: $N_{\rm II}/N_{\rm I} = (2Z_{\rm II}/n_e Z_{\rm I})(2\pi m_e kT/h^2)^{3/2}e^{-13.6\,\mathrm{eV}/kT}$ , which climbs steeply once $kT$ approaches the 13.6 eV ionization energy. Why: above about $10^4$ K hydrogen ionizes wholesale, stripping away the neutral atoms Balmer absorption requires — the lines are killed from above.

Step 3. Multiply the two competing fractions to get the share of all hydrogen sitting in the exact absorbing state, $N_2/N_{\rm total} = \big(\tfrac{N_2/N_1}{1+N_2/N_1}\big)\big(\tfrac{1}{1+N_{\rm II}/N_{\rm I}}\big)$ , a rising factor times a falling one. Why: the product of a monotonically rising and a monotonically falling function is unavoidably single-peaked, so the physics guarantees a temperature of maximum Balmer strength.

Step 4. Evaluate the product at photospheric electron pressures and locate its maximum: it peaks at $T_{\rm peak}\approx 9500\,\mathrm{K}$ , precisely the temperature of A0 stars where Balmer lines are observed to be strongest. Why: matching an ab-initio peak to the observed spectral sequence confirms that line strength is governed by excitation and ionization, not by abundance, and settles the classic puzzle.

Go deeper · Derivation C · 1 minEquivalent width and the linear curve of growth

Derivation CEquivalent width and the linear curve of growth

Abundance analysis rests on turning a measured line area into a number of atoms, and the cleanest regime for that is the weak-line limit, where the equivalent width is directly proportional to the column of absorbers. The goal of this derivation is to establish that proportionality from the definition of equivalent width and the radiative-transfer relation between emergent flux and optical depth. We write the emergent flux for a simple absorbing slab, substitute it into the equivalent-width integral, linearize for small optical depth, and integrate the absorption cross-section, which the oscillator strength fixes. The payoff is the linear branch of the curve of growth — the working relation behind stellar abundances.

Step 1. For a slab of line optical depth $\tau_\lambda$ the emergent flux is $F_\lambda/F_c = e^{-\tau_\lambda}$ . Why: this is the Beer–Lambert attenuation applied wavelength by wavelength, and it is what connects an atomic cross-section to an observable depression in the spectrum.

Step 2. Substitute into the definition (5) to get $W_\lambda = \int (1-e^{-\tau_\lambda})\,\mathrm{d}\lambda$ . Why: the equivalent width is the area of the normalized dip, so it is the integral of one minus the transmitted fraction across the whole profile.

Step 3. In the optically thin limit $\tau_\lambda\ll 1$ , expand $1-e^{-\tau_\lambda}\approx\tau_\lambda$ , so $W_\lambda \approx \int \tau_\lambda\,\mathrm{d}\lambda$ . Why: when the line is weak, doubling the absorbers doubles the depth everywhere, so the dip grows in strict proportion — the origin of the linear regime.

Step 4. Evaluate the integrated cross-section, $\int \tau_\lambda\,\mathrm{d}\lambda = N_a\,(\pi e^2/m_e c^2)\,f\,\lambda_0^2$ , giving $W_\lambda \approx (\pi e^2/m_e c^2)\,N_a f \lambda_0^2$ . Why: the classical oscillator strength $f$ fixes the area under the absorption profile, so $W_\lambda\propto N_a f$ — measure the width, know the column, and the abundance follows on this linear branch.

Limiting cases

How lines and populations behave in the physically important regimes:

Case	Result	Meaning
Cool gas	$N_{n\ge2}\to0$ (Boltzmann)	almost all atoms sit in the ground level, so only ground-state (e.g. Lyman) absorption appears; Balmer lines are absent in the coolest stars
Hot gas, hydrogen ionized	$T\gg10^{4}\,\mathrm{K}\Rightarrow$ Balmer vanishes	few neutral atoms remain, so Balmer strength peaks near A stars (≈9500 K), not in the hottest O stars
Optically thin line	$W_\lambda\propto N$ (linear branch)	few absorbers: equivalent width tracks column density directly — the cleanest regime for abundances
Saturated line	$W_\lambda\propto\sqrt{\ln N}$ (flat branch)	the core is black; adding absorbers barely widens it, so strength becomes nearly insensitive to abundance
Damping-dominated	$W_\lambda\propto\sqrt{N}$ (square-root branch)	very strong lines grow through their Lorentzian wings and sense the damping constant (pressure, radiative)
Non-relativistic shift	$z\approx v_r/c$	the everyday radial-velocity regime, where equations (6) and (7) agree

Common trapThe strongest lines mark the most abundant element

They do not. The single most common mistake in reading a spectrum is to equate line strength with abundance. The strongest hydrogen (Balmer) lines occur in A stars near 9500 K — not in the hottest stars, and not in the most hydrogen-rich ones. Line strength is set by the fraction of atoms in the specific lower level of the transition, which the Boltzmann and Saha equations fix through temperature and pressure. A species can be enormously abundant yet show weak lines because its atoms are ionized away, or are all in the ground state and cannot make the transition in question. The Sun is roughly 90% hydrogen by number, yet its visible spectrum is dominated by Ca II H and K and by iron lines, because at 5800 K only a tiny fraction of hydrogen sits in n = 2. Always run the population through Boltzmann and Saha before converting a width into an abundance.

Worked examples

1 · Predicting hydrogen's visible lines (confidence builder)

Use the Rydberg formula to find the first two Balmer lines and the Balmer series limit, and confirm they land in the visible.

Step 1. Hα is the jump $n_2=3\to n_1=2$ : $1/\lambda = R_{\mathrm H}(1/2^2 - 1/3^2) = 1.0968\times10^7\,(0.25-0.1111)\,\mathrm{m^{-1}} = 1.523\times10^6\,\mathrm{m^{-1}}$ , so $\lambda = \mathbf{656.5\ nm}$ — deep red, matching the laboratory value 656.3 nm to four figures.

Step 2. Hβ is $n_2=4\to2$ : $1/\lambda = R_{\mathrm H}(0.25-0.0625) = 2.057\times10^6\,\mathrm{m^{-1}}$ , giving $\lambda = \mathbf{486.3\ nm}$ — blue-green, the classic value 486.1 nm.

Step 3. The series limit is $n_2\to\infty$ : $1/\lambda = R_{\mathrm H}\times 0.25$ , so $\lambda = \mathbf{364.6\ nm}$ in the near-ultraviolet. Beyond this "Balmer jump" the lines merge into a continuum — a break in the spectrum that itself serves as a temperature and redshift diagnostic. Hα = 656.5 nm, Hβ = 486.3 nm, series limit = 364.6 nm, all reproducing the laboratory hydrogen spectrum.

2 · Thermal broadening in the solar photosphere (fully worked)

Estimate the Doppler width of Hα in gas at T = 6000 K (ignoring turbulence), and compare it to the natural width. Which mechanism dominates?

Step 1. The most probable thermal speed is the Doppler b-parameter $b=\sqrt{2kT/m_{\rm H}}$ . With $k=1.381\times10^{-23}\,\mathrm{J\,K^{-1}}$ , $T=6000\,\mathrm{K}$ , and $m_{\rm H}=1.673\times10^{-27}\,\mathrm{kg}$ : $2kT/m_{\rm H}=9.91\times10^{7}\,\mathrm{m^2\,s^{-2}}$ , so $b=\mathbf{9.95\ km/s}$ .

Step 2. The Gaussian $1/e$ half-width follows from equation (8) with $\xi=0$ : $\Delta\lambda_{\rm D}=(\lambda_0/c)\,b = (656.3\times10^{-9}/2.998\times10^{8})(9.95\times10^{3})\,\mathrm{m} = 2.18\times10^{-11}\,\mathrm{m} = \mathbf{0.0218\ nm}$ . The full width at half maximum is $2\sqrt{\ln 2}\,\Delta\lambda_{\rm D}=\mathbf{0.036\ nm}$ .

Step 3. For the natural width take a level lifetime $\tau\approx1.6\times10^{-8}\,\mathrm{s}$ : $\Delta\lambda_{\rm nat}=\lambda_0^2/(2\pi c\,\tau) = (656.3\times10^{-9})^2/(2\pi\cdot2.998\times10^{8}\cdot1.6\times10^{-8}) = \mathbf{1.4\times10^{-5}\ nm}$ .

Step 4. The thermal Doppler width beats the natural width by a factor of about 1500, so the core of a photospheric hydrogen line is thermally broadened; the natural width only matters far out in the damping wings of very strong lines. Doppler width ≈ 0.022 nm dominates the natural width ≈ 0.000014 nm by ~1500×.

3 · Reading a galaxy's recession from a shifted line

A galaxy's Hβ line, rest wavelength 486.1 nm, is observed at 500.0 nm. Find its redshift and recession velocity, using both the classical and relativistic formulas.

Step 1. The redshift is defined by the wavelength ratio: $z=\Delta\lambda/\lambda_0 = (500.0-486.1)/486.1 = \mathbf{0.0286}$ . The wavelength increased, so the galaxy recedes.

Step 2. The classical estimate is $v\approx cz = 0.0286\times2.998\times10^{5}\,\mathrm{km/s} = \mathbf{8570\ km/s}$ .

Step 3. The relativistic formula (7) gives $\beta=\dfrac{(1+z)^2-1}{(1+z)^2+1}=\dfrac{1.0286^2-1}{1.0286^2+1}=0.0282$ , so $v=\mathbf{8450\ km/s}$ — about 1.4% below the classical value, the leading relativistic correction. At a Hubble constant $H_0\approx70\,\mathrm{km\,s^{-1}\,Mpc^{-1}}$ this implies a distance of order $v/H_0\approx120\,\mathrm{Mpc}$ . z = 0.0286; classical v = 8570 km/s, relativistic v = 8450 km/s.

Physics of the portrait

Spectral classification came before its physics: astronomers sorted stars by their line patterns for decades before anyone could say what set the patterns. The bridge was built by a young researcher who took a new equation from statistical physics and turned stellar spectra into a thermometer — and, almost incidentally, discovered what stars are made of.

Origins · 1 minCecilia Payne-Gaposchkin and how this was discovered

Cecilia Payne-Gaposchkin

1900–1979 · stellar spectroscopy and the composition of stars

Cecilia Payne-Gaposchkin wrote what has been called the most brilliant PhD thesis ever in astronomy. Arriving at Harvard in 1923, she inherited the Observatory's vast catalogue of stellar spectra — the OBAFGKM sequence that Annie Jump Cannon had classified by eye — and set out to explain it with physics rather than description. The tool she needed was brand new: Meghnad Saha's 1920 ionization equation, which she combined with Boltzmann's excitation law to show that the changing pattern of lines along the spectral sequence is a pure temperature effect, not a change in composition. Running the analysis through to its conclusion, her 1925 dissertation Stellar Atmospheres reached a startling result — that hydrogen and helium are overwhelmingly the most abundant elements in stars, hydrogen by a factor of about a million over the metals. The claim was so at odds with the prevailing belief that stars matched the Earth's composition that senior astronomer Henry Norris Russell pressed her to call it "almost certainly not real"; within a few years he derived the same result independently and it was vindicated. Her work established the quantitative link between a star's temperature and its spectrum that underlies all modern stellar astrophysics, and made the abundance of hydrogen — the fact that governs stellar structure, nucleosynthesis, and cosmology — a measured number rather than an assumption.

History

1814

— Fraunhofer maps and labels the dark absorption lines crossing the solar spectrum

1859

— Kirchhoff and Bunsen found spectrum analysis and state the three radiation laws

1885

— Balmer finds the empirical formula for hydrogen's visible lines

1920

— Saha derives the ionization equation and applies it to stellar spectra

1925

— Payne shows stars are made chiefly of hydrogen using Saha and Boltzmann

Going further

Texts. Carroll & Ostlie, An Introduction to Modern Astrophysics, 2nd ed., Ch. 5 (Interaction of Light and Matter) and Ch. 8 (Classification of Stellar Spectra) · Rybicki & Lightman, Radiative Processes in Astrophysics, Ch. 1, 9–10 · Gray, The Observation and Analysis of Stellar Photospheres, 3rd ed. (equivalent widths and the curve of growth) · Mo, van den Bosch & White, Galaxy Formation and Evolution (redshifts and line diagnostics) · Ryden, Introduction to Cosmology, 2nd ed. · Dodelson, Modern Cosmology (recombination and line formation) · Binney & Tremaine, Galactic Dynamics, 2nd ed. (radial velocities and kinematics).

Provenance. Every relation here is a standard textbook result and has been recomputed for this article: (1) the Rydberg/Balmer wavelengths from Bohr levels (Hα = 656.5 nm, Hβ = 486.3 nm, limit 364.6 nm); (2) the Boltzmann × Saha product for hydrogen peaking near T ≈ 9500 K (Carroll & Ostlie Ch. 8, Example 8.1.4); (3) the linear curve-of-growth relation $W_\lambda\propto N_a f\lambda_0^2$ (Gray Ch. 16; Rybicki & Lightman §10.6); (4) the thermal Doppler width of Hα at 6000 K (0.022 nm) versus the natural width (~ $10^{-5}$ nm); (5) the classical and relativistic Doppler/redshift formulas (Ryden Ch. 2; Dodelson). Constants ( $R_{\mathrm H}$ , $k$ , $h$ , $m_e$ , $c$ ) are CODATA/standard values.

Prerequisites. Luminosity and Radiant Flux · Color Indices and Bolometric Corrections · Photometric Systems and Filters. Leads to. The Cosmic Distance Ladder · Standard Candles · Stellar Parallax and the Parsec.

Ask the tutor: ask me to walk through the Boltzmann × Saha calculation for any element and temperature, or to turn a measured equivalent width or line shift into an abundance or a velocity.

What is the difference between emission and absorption lines?

Both come from the same atomic transitions; what differs is the geometry, captured by Kirchhoff's three laws. A hot, dense body (a stellar interior) radiates a smooth continuum. If you look at that continuum through a cooler, thinner gas, atoms in the gas absorb photons at their transition wavelengths and scatter them out of the beam, leaving dark absorption lines — the case for a normal stellar photosphere. If instead you look at a hot, rarefied gas with no bright background behind it — an emission nebula, say — its atoms emit at those same wavelengths, giving bright emission lines on a dark background. The identical line can appear in absorption or emission depending only on whether there is a hotter continuum source behind the gas.

Why does the equivalent width saturate instead of growing forever?

On the linear (optically thin) branch the line is weak, so doubling the number of absorbers doubles the depression everywhere and the equivalent width grows in proportion. But once the core of the line goes optically thick, the center is already black — it has removed all the flux it can, and adding more absorbers cannot make it darker. Further growth then comes only from the shallow flanks of the core, so the equivalent width creeps up only as $\sqrt{\ln N}$ : the flat part of the curve of growth. This is why saturated lines are poor abundance indicators. Only when the line becomes so strong that its Lorentzian damping wings take over does $W_\lambda$ resume rising, now as $\sqrt{N}$ .

How can one line diagnose so many different things at once?

Because the line carries several independent pieces of information in different attributes. Its central wavelength, compared to the laboratory value, gives the radial velocity through the Doppler shift. Its integrated depth (equivalent width), run through Boltzmann and Saha and a model atmosphere, gives an abundance. Its Gaussian core width gives the temperature plus turbulence. Its Lorentzian wing strength gives the pressure and hence the surface gravity, which is why luminosity class shows up in line widths. Even the splitting or polarization of the line encodes the magnetic field through the Zeeman effect. A single high-resolution profile is therefore a multi-parameter measurement, which is what makes spectroscopy so much more powerful than photometry.

Is cosmological redshift really a Doppler shift?

Not exactly, and the distinction matters at large distances. A local Doppler shift comes from relative motion through space and is described by the special-relativistic formula (7). Cosmological redshift instead comes from the expansion of space itself between emission and observation: $1+z = a_0/a$ , the ratio of the cosmic scale factor now to its value when the light left. The photon's wavelength stretches along with space during its journey. For small $z$ the two pictures give the same numerical answer, $z\approx v/c$ , which is why "recession velocity" is a useful shorthand nearby. But at high redshift the naive velocity can exceed $c$ without violating relativity, because it is not a motion through space at all — it is the geometry of an expanding universe.

Why do heavier elements make narrower thermal lines than hydrogen?

The thermal Doppler width scales as $\sqrt{2kT/m}$ , so at a given temperature it shrinks with the square root of the atomic mass. All species share the same kinetic temperature, but heavier atoms move more slowly for that shared energy — an iron atom (mass 56) moves about $\sqrt{56}\approx7.5$ times slower than a hydrogen atom. Its lines therefore have thermal cores about 7.5 times narrower. This mass dependence is genuinely useful: comparing the widths of lines from light and heavy species in the same gas lets you separate the thermal contribution (which depends on mass) from turbulence and rotation (which do not), and so measure the temperature and the turbulent velocity independently.

What sets the OBAFGKM order of the spectral sequence?

Temperature, decreasing from O to M. The letters are a historical accident — they were originally ordered alphabetically by hydrogen-line strength (A had the strongest Balmer lines) and only later reshuffled into a physical temperature sequence, which is why the order looks scrambled. O stars (over 30000 K) are hot enough to show ionized helium and only weak hydrogen; A stars near 9500 K show the strongest Balmer lines; G stars like the Sun (5800 K) show strong Ca II H and K plus many metal lines; and cool M stars (~3000 K) show molecular bands such as TiO because their atmospheres are cool enough for molecules to survive. Adding a luminosity class I–V from the pressure-sensitive line widths completes the two-dimensional Morgan–Keenan classification.