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stellar spectra evolution and energy sources

Learning objectives

  • Outcome

    Assign a stellar spectral class from surface temperature and from a set of absorption lines in its spectrum, using Wien's displacement law λmax⁡T=2.9×10−3\lambda_{\max} T = 2.9\times10^{-3} m·K.

  • Outcome

    Calculate a star's luminosity from L=4πR2σT4L = 4\pi R^2 \sigma T^4 and its radius from known luminosity and temperature, expressing the result in solar units.

  • Outcome

    Convert apparent magnitude to absolute magnitude using a parallax distance via M=m−5log⁡10(d/10 pc)M = m - 5\log_{10}(d/10\ \text{pc}) and place the star on the Hertzsprung–Russell diagram.

  • Outcome

    Estimate a main-sequence lifetime by computing the fusion energy budget from the mass defect E=Δmc2E = \Delta m c^2 and dividing it by the luminosity.

Understand the idea

Stellar Spectra, Evolution, and Energy Sources

Text

A star is a self-gravitating ball of gas whose emitting surface obeys the Stefan–Boltzmann law:

L=4πR2σT4.L = 4\pi R^2 \sigma T^4 .

A star's spectrum forms in its photosphere: absorption lines arise from atomic transitions between levels, and their strength depends on temperature and density through the Boltzmann distribution and the Saha equation.
Surface temperature therefore sets the spectral class O B A F G K MO\,B\,A\,F\,G\,K\,M, while Wien's displacement law λmax⁡T=2.9×10−3\lambda_{\max} T = 2.9\times10^{-3} m·K tells where the emission peak lies.
Observed flux and luminosity are linked by the inverse-square law F=L/(4πd2)F = L/(4\pi d^2), so the distance scale from the prerequisite topic is needed to convert apparent magnitude into absolute magnitude and recover the luminosity.
The energy source of stars is the nuclear fusion of helium from hydrogen (4 1H→4He+2e++2ν4\,{}^1\mathrm{H} \to {}^4\mathrm{He} + 2e^+ + 2\nu), in which the mass defect Δm\Delta m becomes energy through E=Δmc2E = \Delta m c^2.
Luminosity and temperature together place a star on the Hertzsprung–Russell diagram, where its position and mass determine its evolutionary path — from the main sequence to a giant, supergiant, or white dwarf.

Worked examples

Direct

Problem

A star has radius R=2 R⊙R = 2\,R_\odot and surface temperature T=10 000T = 10\,000 K. The Sun's temperature is T⊙=5800T_\odot = 5800 K. Find the star's luminosity in solar units and the wavelength of peak emission.

Solution

Luminosity. Write the Stefan–Boltzmann law for the star and for the Sun:

L=4πR2σT4,L⊙=4πR⊙2σT⊙4.L = 4\pi R^2 \sigma T^4, \qquad L_\odot = 4\pi R_\odot^2 \sigma T_\odot^4 .

The ratio gives

LL⊙=(RR⊙)2(TT⊙)4=22(100005800)4=4×8.84≈35.\frac{L}{L_\odot} = \left(\frac{R}{R_\odot}\right)^2 \left(\frac{T}{T_\odot}\right)^4 = 2^2 \left(\frac{10000}{5800}\right)^4 = 4 \times 8.84 \approx 35.

Peak wavelength. By Wien's law

λmax⁡=2.9×10−3 m⋅KT=2.9×10−310 000=2.9×10−7 m=290 nm.\lambda_{\max} = \frac{2.9\times10^{-3}\ \text{m·K}}{T} = \frac{2.9\times10^{-3}}{10\,000} = 2.9\times10^{-7}\ \text{m} = 290\ \text{nm}.

This lies in the ultraviolet: the star is hotter than the Sun and therefore bluish-white.

Application

Problem

From its spectrum the star is classified as B8 (T≈12 000T \approx 12\,000 K). Its apparent magnitude is m=5.0m = 5.0 and its annual parallax is 1010 milliarcseconds. Determine its absolute magnitude, luminosity, and radius, and place it on the Hertzsprung–Russell diagram.

Solution

Distance. The parallax is p=0.010′′p = 0.010'', so

d=1p=10.010=100 pc.d = \frac{1}{p} = \frac{1}{0.010} = 100\ \text{pc}.

Absolute magnitude.

M=m−5log⁡10 ⁣(d10)=5.0−5log⁡10(10)=5.0−5.0=0.0.M = m - 5\log_{10}\!\left(\frac{d}{10}\right) = 5.0 - 5\log_{10}(10) = 5.0 - 5.0 = 0.0 .

Luminosity. With M⊙=4.83M_\odot = 4.83:

LL⊙=10(M⊙−M)/2.5=104.83/2.5=101.93≈85.\frac{L}{L_\odot} = 10^{(M_\odot - M)/2.5} = 10^{4.83/2.5} = 10^{1.93} \approx 85 .

Radius. From L/L⊙=(R/R⊙)2(T/T⊙)4L/L_\odot = (R/R_\odot)^2 (T/T_\odot)^4:

RR⊙=85(580012 000)2=9.2×0.234≈2.2.\frac{R}{R_\odot} = \sqrt{85} \left(\frac{5800}{12\,000}\right)^2 = 9.2 \times 0.234 \approx 2.2 .

Position on the diagram. A blue star with L≈85 L⊙L \approx 85\,L_\odot and T=12 000T = 12\,000 K lies in the upper-left part of the main sequence (luminosity class V) — a moderately massive star that has not yet left the main sequence.

Transfer

Problem

Estimate the main-sequence lifetime of a 10 M⊙10\,M_\odot star, assuming that 10 % of its hydrogen is fused into helium and that the fusion of 4He{}^4\mathrm{He} from 1H{}^1\mathrm{H} converts 0.7 % of the mass into energy. Take the star's luminosity as 3000 L⊙3000\,L_\odot. Compare with the rough estimate t≈10 billion years×(M/M⊙)/(L/L⊙)t \approx 10\ \text{billion years} \times (M/M_\odot)/(L/L_\odot).

Solution

Fuel and energy. The star's mass is:

M=10×1.989×1030 kg=1.989×1031 kg.M = 10 \times 1.989\times10^{30}\ \text{kg} = 1.989\times10^{31}\ \text{kg}.

The reacting fraction is 0.100.10, and 0.0070.007 of that mass becomes energy, so

E=0.007×0.10×Mc2=7×10−4×1.989×1031×(3.0×108)2≈1.25×1045 J.E = 0.007 \times 0.10 \times M c^2 = 7\times10^{-4} \times 1.989\times10^{31} \times (3.0\times10^{8})^2 \approx 1.25\times10^{45}\ \text{J}.

Power.

L=3000×3.828×1026 W=1.15×1030 W.L = 3000 \times 3.828\times10^{26}\ \text{W} = 1.15\times10^{30}\ \text{W}.

Lifetime.

t=EL=1.25×10451.15×1030≈1.1×1015 s≈3.5×107 years≈35 million years.t = \frac{E}{L} = \frac{1.25\times10^{45}}{1.15\times10^{30}} \approx 1.1\times10^{15}\ \text{s} \approx 3.5\times10^{7}\ \text{years} \approx 35\ \text{million years}.

Check. The rough scaling gives t≈10 billion years×10/3000≈33t \approx 10\ \text{billion years} \times 10/3000 \approx 33 million years — agreement within the accuracy of the estimate. Conclusion: massive stars live hundreds of times shorter than the Sun despite having more fuel, because luminosity grows with mass much faster.

Practice

Text

Self-study plan (simulation only, no sky observation).

  1. Open a Hertzsprung–Russell diagram simulation (for example, the NAAP “Hertzsprung–Russell Diagram” lab) or use an archival dataset of 10 stars with known temperature, luminosity, and parallax.
  2. For each star, convert parallax to distance (d=1/pd = 1/p), then compute MM from M=m−5log⁡10(d/10)M = m - 5\log_{10}(d/10) and L/L⊙=10(M⊙−M)/2.5L/L_\odot = 10^{(M_\odot - M)/2.5}.
  3. Plot the points on a luminosity–temperature diagram (temperature increasing to the left) and mark which stars lie on the main sequence and which fall in the giant or white-dwarf regions.
  4. For one main-sequence star, use its simulated spectrum to list the dominant lines (hydrogen Balmer lines, Ca II lines, TiO bands) and assign the spectral class.
  5. Estimate that star's lifetime with E=0.007×0.10×Mc2E = 0.007 \times 0.10 \times M c^2 and t=E/Lt = E/L, and compare it with t≈10 billion years×(M/M⊙)/(L/L⊙)t \approx 10\ \text{billion years} \times (M/M_\odot)/(L/L_\odot).

Safety. Never point unprotected eyes, binoculars, or a telescope at the Sun without a proper solar filter. All spectra and photometry come from the simulation or archives, not from direct observation.

Visual reference

Purpose

Show the link between an observed spectrum (a set of absorption lines) and a star's position on the Hertzsprung–Russell diagram, so that temperature, luminosity, and evolutionary stage are read as one picture.

Representation

Diagram. The main panel is the Hertzsprung–Russell diagram: the vertical logarithmic axis carries luminosity L/L⊙L/L_\odot (from 10−410^{-4} to 10610^{6}), and the horizontal axis carries temperature decreasing to the right (from 40 00040\,000 K to 30003000 K), labelled with spectral classes O B A F G K MO\,B\,A\,F\,G\,K\,M. The main sequence, the giant branch, the supergiants, and the white dwarfs are drawn; the Sun is marked at L=1L = 1, T=5800T = 5800 K.

Inset. Three schematic spectra with absorption lines: class A shows strong hydrogen Balmer lines, class G shows Ca II lines, and class M shows broad TiO bands. Arrows connect each spectrum to the corresponding point on the diagram.

Additional. At the bottom, an arrow marks the evolutionary track of a massive star from the main sequence to supergiant, annotated “He synthesis from H: E=Δmc2E = \Delta m c^2”.

Real-world examples

A smartphone as a spectroscope: absorption lines in sunlight

Scenario

You hold a CD fragment over a smartphone camera and point it at the Sun or a fluorescent lamp. A rainbow crossed by dark lines appears on the screen. For a CD with track spacing d=1.6 μmd = 1.6\,\mu\text{m}, the first-order diffraction angle for green light λ=550 nm\lambda = 550\,\text{nm} is θ=arcsin⁡(λ/d)≈20∘\theta = \arcsin(\lambda/d) \approx 20^\circ. In the solar spectrum, the sodium doublet at 589.0 nm589.0\,\text{nm} and 589.6 nm589.6\,\text{nm} appears as a pair of dark lines.

Connection

This experiment reproduces the formation of Fraunhofer lines: the continuous spectrum of the hot, dense photosphere (T≈5800 KT \approx 5800\,\text{K}) is absorbed by cooler, thinner gas in the chromosphere. The wavelengths of the lines reveal the presence of sodium, while their depth and width give the temperature and density of the absorbing layer — the basis of stellar spectral classification.

The Sun as a fusion reactor: from energy to weather

Scenario

On a clear day, about 1000 W1000\,\text{W} arrives per square metre of Earth's surface. The Sun's luminosity L=3.8×1026 WL = 3.8\times10^{26}\,\text{W} at distance r=1.5×1011 mr = 1.5\times10^{11}\,\text{m} gives the solar constant S=L/(4πr2)≈1360 W/m2S = L/(4\pi r^2) \approx 1360\,\text{W/m}^2. This power is generated in the core at T≈1.5×107 KT \approx 1.5\times10^7\,\text{K} and density 1.5×105 kg/m31.5\times10^5\,\text{kg/m}^3: each second about 4.3×109 kg4.3\times10^9\,\text{kg} of mass is converted into energy via E=mc2E = mc^2, a mass defect of 0.7%0.7\% in the pp-chain.

Connection

This shows how the observed solar constant links a star's energy output to its source: the pp-chain converts four protons into a helium nucleus, releasing 4.3×10−12 J4.3\times10^{-12}\,\text{J} per reaction. The rate of these reactions maintains the luminosity and sets the star's main-sequence lifetime; the same energy drives Earth's weather and climate.

A gas stove: flame colour and emission lines

Scenario

The blue flame of a gas burner glows because excited CH and C$_2$ molecules emit lines at 430 nm430\,\text{nm} and 516 nm516\,\text{nm}. Sprinkling table salt (NaCl) into the flame turns it yellow: the sodium D lines at 589.0 nm589.0\,\text{nm} and 589.6 nm589.6\,\text{nm} dominate. A thermocouple reads a flame temperature of about 1300 K1300\,\text{K}.

Connection

This is an emission-line spectrum of a hot, low-density gas — the same type of spectrum used to classify Wolf–Rayet stars and to determine the chemical composition of stellar atmospheres. The wavelengths of the lines correspond to quantised atomic energy levels, and the relative line intensities give the temperature via the Boltzmann distribution.