CAPSTONE Lecture 7 Stellar types and the HR diagram July 13, 2011

1. CAPSTONE Lecture 7 Stellar types and the HR diagram July 13, 2011 CAPSTONE.lecture 7, 07.13.2011

2. Problems • The average period of a certain asteroid is about 76 years. If it is in a circular orbit, how far is it from the Sun and where does that place it with respect to the orbits of the major planets. (Find the values for the planets in the appendices to the text book.) • What is the angular diameter of a star identical to the Sun, at a distance of 1pc from Earth? 10pc? 100pc? To use the skinny triangle approximation, you must work in radians, but give your answer in arc seconds. (2x105 arc seconds per radian). What is the absolute magnitude of the stars at each location? What are the apparent magnitudes at each location? CAPSTONE.lecture 7, 07.13.2011 2

3. 3. A star just like the Sun is seen at a distance of 10pc. How far is it from Earth? 100pc? 1000pc? 10,000pc (the edge of the galaxy)? 100,000,000 pc (1 Mpc) (local group)? 5 Mpc? 4. In each of the last three cases, how many stars would it take of solar type to make galaxies in which the stars reside bright enough to see with your eye? CAPSTONE.lecture 7, 07.13.2011

4. Spectral types See text. Spectral types are categories which spectra of stars naturally fall into. Patterns in the atomic and molecular and ionic lines from the atmospheres of the stars were easily noticed and the stars were arranged in order of appearance. They system of spectral types was perfected by W. W. Morgan at Yerkes Observatory, University of Chicago, around 1942. CAPSTONE.lecture 7, 07.13.2011

5. Physical basis for spectral types • Key species: H, He, He+, Ca+, Ca0 • The temperature of the outer layers determines the ionization state • Saha equation, equilibrium in ionization states depends on temperature. • N r+1Ne / Nr = {[c2(2mkT)3/2]/h3} e-/kT • log (Nr+1Pe / Nr)= -( 5040)/T+(5 log T)/2 – 0.48 • log Nr+1/Nr= -(5040 )/T + (5 log T)/2 – log Pe.- 0.48. • [There is an extra power of T on the rhs because Ne in the earlier equation is converted to pressure—P=nkT, the equivalent, in the kinetic theory of gasses, of Boyle’s law that you used in high school chemistry, PV=NRT or • P=(N/V)RT] CAPSTONE.lecture 7, 07.13.2011

6. Equilibrium • If  =13.6 eV for neutral hydrogen (H I), the right hand side (rhs) is a positive number for 10,000 oK, and most of the hyrogen is ionized (that is, in stage r+1, not r). • At 3000 oK, the rhs is a negative number and most of the material is neutral hydrogen (only a fraction is ionized). • Recall that there is a distribution of electron energies in a gas of given T(and average energies). • Thus, even when the value of kT is low, there are electrons with enough energy to cause ionization. • Roughly speaking, kT for a gas can be 1/10 of the value of the ionization potential considered and still produce ionization. kT =1 eV for gas at 10,000 oK, and yet there are enough high energy electrons to ionize H (13.6 eV). CAPSTONE.lecture 7, 07.13.2011

7. Spectral type table of atoms and ions CAPSTONE.lecture 7, 07.13.2011

8. 10/18/09 PS119a.Lecture8.HR diagram 8 CAPSTONE.lecture 7, 07.13.2011

9. 10/18/09 PS119a.Lecture8.HR diagram 9 CAPSTONE.lecture 7, 07.13.2011

10. CAPSTONE.lecture 7, 07.13.2011

11. Analysis of binaries • The need now is to get as many stars as possible defined in terms of mass, temperature and luminosity. • Stars in the same part of the HR diagram that come from binaries are the same as other stars that fall near them in luminosity and temperature • In clusters, spectra can be taken of all stars. CAPSTONE.lecture 7, 07.13.2011

12. Knowing a few stars by absolute magnitude in clusters allows us to use the same distance for all cluster stars and to place millions of stars in the HR diagram. • This then allows one to calibrate spectral signatures of luminosity (the H lines are not so broad in giants as in dwarfs) in any stars. • Then, any star can have luminosity and temperature determined and fit into the HR diagram. CAPSTONE.lecture 7, 07.13.2011

13. Review of binaries • So, everything depends on a few binaries • Astronomical observations: Obtain orbits, center of mass, periods, radial velocity (Doppler), temperatures, distances, apparent magnitudes, spectral types. • With parallax, separation (AU)=(arcsec) x distance(pc). [ l(AU)=(arcsec)/ CAPSTONE.lecture 7, 07.13.2011

14. Follows from the skinny triangle and conversion of separation in pc to AU and angle from radians to arcsec. • r1/r2=v1/v2=m2/m1, r1+r2=R • P, total separation ==>M • m1=M/[1+(r1/r2)], use v1/v2 if cannot resolve. • From distance and luminosity, the stellar radius follows. • Composition follows from measuring absorption lines in the spectra. CAPSTONE.lecture 7, 07.13.2011

15. Spectroscopic parallax • Once we know the spectral type, even with no parallax, we can use the HR diagram to determine absolute mag. • Then, the distance follows from mV-MV=5log r-5 + AV • AV is the extinction due to dust that must be corrected for. • Once we know absolute mag., we can get luminosity • M1-Mo = -2.5 log (L1/Lo) (using the Sun as reference) • Divide by -2.5 on both sides and exponentiate • L1/Lo= 10-0.4(M1-M0) • Normally express luminosities in terms of “solar” value, • just as we express mass in terms of solar masses. CAPSTONE.lecture 7, 07.13.2011

16. Stellar radii, densities, “g” • L1=4R12T14, Lo=4Ro2To4 • R1/Ro=( L1/Lo)0.5(To/T1)2. • mean density is Mass/volume • g=GM/R, all known. Empirically, g=104 cm/sec2 on the main sequence. CAPSTONE.lecture 7, 07.13.2011