Radiation Kirchoff’s Laws

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Presentation transcript:

Radiation Kirchoff’s Laws Light emitted by a blackbody, a hot opaque body, or hot dense gas produces a continuous spectrum A hot transparent gas produces an emission line spectrum A cool transparent gas in front of a blackbody produces an absorption spectrum

Visually Kirchoff’s Laws

Kirchoff’s Laws Top: solar spectrum Bottom: iron emission line spectrum What can you conclude about the Sun’s chemical composition from this comparison?

Spectra Spectroscopy is the study of spectral lines, and it’s very useful in astronomy. Over 80% of all astrophysical information is contained in spectra not in images. chemical composition (spectral lines) distance (redshift) velocity (Doppler effect) temperature (shape and type of spectral lines) density (ratios of spectral line intensities)

The Planck Function Blackbody Curves What are the units for Blambda? = h = m^2 Kg s-1 * m^2 s^-2 * m ^-5

What is a Blackbody? A perfect emitter and absorber of radiation; for example, black asphalt. Stars are to a large part a good approximation to a blackbody.

Just to Convince You From Silva and Cornell, 1992, ApJS 81, 865.

Interesting Results From the Planck Function! Wien’s Law: λmax = 0.2898 / T (λ in cm) Sun T = 5780 K ==> λmax = 5.01(10-5) cm = 5010 Å Stefan-Boltzman Law: L = σT4 Total Luminosity L = 4πR2σT4 σ = 5.67(10-5) ergs s-1 cm-2 K-4 Solar Luminosity = 3.862(1033) ergs s-1 R = 6.960(1010) cm L = 3.85(1033) ergs s-1! ==> It works!

Stellar Properties Mass Temperature Luminosity Composition

Spectral Types O Stars: Strong Helium Lines (35000K) B Stars: Moderate Hydrogen lines and no Helium lines (20000K) - Rigel A Stars: Strong Hydrogen lines (12000K) -Vega F Stars: Moderate Hydrogen lines with ionized metal lines (7500K) - Procyon G Stars: Weak Hydrogen lines with strong neutral metal lines (5500K) - The Sun K Stars: Very strong neutral metal with some molecules (4500K) - Arcturus M Stars: Strong molecular features (3500K) - Betelgeuse

Stellar Mass O Stars: 30 30 B Stars: 10 10 A Stars: 5 10 5-15 A Difficult Proposition at Best Dwarfs Giants Supergiants O Stars: 30 30 B Stars: 10 10 A Stars: 5 10 5-15 F Stars: 3 5 5-15 G Stars: 1 3 5-15 K Stars: 0.7 1.5 5-15 M Stars: 0.3 1 5-15

Determination of Stellar Mass A Very Difficult Problem Mass of the Sun: From Kepler’s 3rd Law Kepler’s 3rd Law: P2 = ka3 P = Period of Planet a = semimajor axis of planet k = constant (4π2/GM) G = Gravitional Constant

Determination of Stellar Mass Eclipsing Binaries It is NOT possible to directly determine the mass of a single star. Our only available tool is Kepler’s Third Law. P2 = ka3 where P= Period of the system k = 4π2/G(M1 + M2) a = Separation of the components Why Eclipsing Binaries? Inclination of the Orbit!

Composition Hydrogen 1012 70% Helium 1011 28% Lithium 101.00 The Sun Is the Common Reference Point Number Mass Hydrogen 1012 70% Helium 1011 28% Lithium 101.00 Carbon 108.55 Nitrogen 107.99 Oxygen 108.75 Silicon 107.55 Calcium 106.36 Iron 107.50 Barium 102.13

A Stellar High Resolution Spectrum