Dr Matt Burleigh The Sun and the Stars

Dr Matt Burleigh The Sun and the Stars The Hertzsprung-Russell Diagram (E. Hertzsprung and H.N. Russell) Plot of surface temperature versus luminosity, or colour (e.g. B-V) versus absolute magnitude M, and various other combinations Hot Cool dim bright What are the physical mechanisms responsible for the L,T sequence?

Dr Matt Burleigh The Sun and the Stars The Harvard Classification scheme is an excitation-ionization scheme The physical properties of stellar atmospheres are determined by : i) Temperature, T ii) Pressure, P iii) chemical composition, 

Dr Matt Burleigh r = R r Centre (r = 0)  (r) P + dPPdr Hydrostatic Equilibrium A star is in hydrostatic equilibrium  no net force acting on gas (i.e. gravity balanced by pressure)

Dr Matt Burleigh The Sun and the Stars In hydrostatic equilibrium  no net force acting on gas (i.e. gravity balanced by pressure) Consider a slab of gas area A, thickness  r Force due to gravity F g, is given by This must be balanced by the upward pressure (remember P=F/A), that is rearrange, as  r  0, then where g is the surface gravity, (ie. gravitational acceleration at r=R) If we express the pressure, P in terms of the optical depth, , then  Gas pressure depends on opacity , and surface gravity g

Dr Matt Burleigh E.g. Sun’s central pressure What is the pressure at the centre of the Sun? First calculate average density: M  =1.989x10 30 kg, R  =6.96x10 8 m   (avg)  = 3M  /4  R  3 =1410 kgm -3 Then calculate pressure using Hydrostatic Equilibrium: Integrate from P c to P R  and from R=R 0 =0 to R= R , If surface pressure = 0, then P c ~ G M   (ave)  / R  = 2.7x10 14 Nm -2

Dr Matt Burleigh The Sun and the Stars We assume that particle densities are high enough that the gas is in thermodynamic equilibrium (i.e. all processes and their inverses occur at the same rate). The gas obeys the ideal gas law, Where the particle number density n is related to the mass density , and chemical composition  (the mean molecular weight), by For pure hydrogen,  =1 ionised hydrogen,  =1/2 (for ionized H there are roughly equal nos. of e- and p+, and m e << m p ) For stellar interiors, the gas is mostly ionised, and therefore in general Where, X is the mass fraction of H (mass fraction is the % by mass of Y “” He one species relative to the total. Ionized H gives two particles per m H ) Z “” metals (all other elements) k is Boltzmann’s constant (1.381  JK -1) m H is proton mass (mass of H atom) 1.67x kg

Dr Matt Burleigh The Sun and the Stars What is the temperature at the centre of the Sun? Combine ideal gas law, And mean molecular weight Re-arranged for n Gives Hence Use P c and  avg estimates, assume  ~ ½ (ie pure ionized hydrogen) Then T c ~ P c  m H /  (avg)  k ~ 1.2x 10 7 K Gas dissociated into ions & electrons but overall electrically neutral… a plasma

Dr Matt Burleigh The Sun and the Stars Light elements “burn” to form heavier elements Stellar cores have high enough T and  for nuclear fusion Work (after 1938) by Hans Bethe and Fred Hoyle  Energy release can be calculated from E=mc 2 –e.g. in H fusion 4 x 1 1 H atoms  1 x 4 2 He atom  Mass of 4 x 1 1 H atoms = 4 x x kg = x kg  Mass of 1 x 4 2 He atom = 1 x x kg  E = (4 x 1 1 H atoms He atom)c 2 = (6.6916x kg x kg)c 2 = 4x J Stellar Thermonuclear Reactions

Dr Matt Burleigh Stellar Thermonuclear Reactions In the Sun ~10% of its volume is at the T and  required for fusion Total energy available is… –Energy per reaction x (total mass available for fusion / mass in each reaction) = Energy per reaction x (0.1 x mass of Sun / mass of 4 x 1 1 H atoms) (where mass of Sun is 2x10 30 kg) Hence, E tot = 4.26x x (2x10 29 /6.6916x ) = 1.27x10 44 J Lifetime of Sun is then the time taken to fuse all the available mass, which is given by the total energy available E tot divided by the solar luminosity L  So, lifetime t sun = E tot /L  = 1.27x10 44 J / 3.9x10 26 Js -1  t sun ~ 3.3x10 17 s ~ yrs

Dr Matt Burleigh Stellar Thermonuclear Reactions Proton – proton chain (PPI, T < 2  10 7 K) 1.44MeV 5.49MeV 12.9MeV is deuterium, PPI occurs 91% of time in Sun, although there are other PP chains

Dr Matt Burleigh a b c PPI Chain

Dr Matt Burleigh CNO cycle Stellar Thermonuclear Reactions The Carbon cycle (CNO cycle) also converts H to He but requires a C nucleus as a catalyst Requires temperatures >1.6x10 7 K Occurs in Sun but minor compared to PPI More important fusion process for stellar masses >1.1M sun Since requires a C nucleus, only occurs in Pop I stars Second and fifth steps occur because 13 N and 15 O are unstable isotopes with half lives of only a few minutes

Dr Matt Burleigh CNO Cycle

Dr Matt Burleigh Stellar Thermonuclear Reactions At very high temperatures, ~10 8 K, other processes fuse Helium (alpha particles) into heavier elements Triple alpha process: The triple alpha process is first stage of helium burning in an evolved star that has left the main sequence In most massive stars, then burn C, Ne, O and Mg up to Fe