Stellar Structure. Observationally, we can determine: Can we explain all these observations?

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Terence Hood
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1 Stellar Structure Observationally, we can determine: Flux Mass Distance Luminosity Temperature Radius Spectral Type Composition Can we explain all these observations?

2 Stellar Structure Plan: Use our general knowledge of physics to derive differential equations governing stellar structure First some assumptions: 1. Stars are spherically symmetric => all parameters are functions of r only 2. Stars are in equilibrium, i.e., parameters are not functions of time OK as long as we are in Main Sequence phase 3. Chemical composition is uniform throughout the star, at least at the start

3 What creates the main sequence? L T 6

4 What causes the Mass-Luminosity relation? L M 4 L M 2.3

5 What accounts for stellar abundances?

6 What is the source of stellar luminosity?

7 What holds up the stars against gravity?

8 Hydrostatic Equilibrium

9 What makes the stars shine? The answer comes from Einstein s most famous equation, E = mc 2, as part of his work on relativity As shown at right, when 4 protons merge to form a 4 He nucleus (also known as an alpha (α) particle), mass is converted to energy This process is known as nuclear fusion This mechanism is capable of allowing the Sun to shine at its current rate for 10 billion years

10 Four forces of nature

11 Four forces of nature Force Relative Strength Range Gravity 1 EM Strong nuclear Weak nuclear m m

12 Maxwell-Boltzmann Velocity Distribution

13 Quantum tunneling

14 Maxwell-Boltzmann Velocity Distribution

15 Temperature dependence of nuclear reactions Maxwell-Boltzmann Tunneling probability

16 Nuclear Fusion (pp chain) Particles in the Sun s core collide two at a time, so the reaction 4p 4 He + energy requires multiple steps 1. First two protons combine to form deuterium ( 2 H), namely a proton and a neutron;; since electric charge (and something called lepton number) must be conserved, a positron (e + ) and a neutrino (ν) are emitted 2. Then each deuterium nucleus gets an additional proton to make 3 He 3. Finally, two 3 He nuclei fuse to form 4 He plus two protons

17 If the mass of a proton (in energy units) is 938 MeV/c 2, and the mass of a helium nucleus is 3726 MeV/c 2 then what mass fraction is converted to energy during fusion? A 25% B 2.7% C 0.7% D 0.007%

18 Nuclear Fusion The energy released can be calculated from E = mc 2 : m p = kg 4m p = kg m He = kg Δm = kg E = Δmc 2 = ( kg)( m/s) 2 = J = 26.4 MeV For comparison, the typical chemical reaction releases a few ev

$Nuclear Fusion Thus the fraction of mass converted to energy is: Δm m = 0.047 10 27 kg 6.690 10 24 kg = 0.007 (0.7%) 1 kg H 993 g He + 7 g converted to energy 4.$

19 Nuclear Fusion Thus the fraction of mass converted to energy is: Δm m = kg kg = (0.7%) 1 kg H 993 g He + 7 g converted to energy J sounds small, but this reaction happens ~10 38 times/second 4 10 ( 12+38) J = J is released each second or L = W = 400,000,000,000,000,000,000,000,000 W 600,000,000 tons of H is converted to 596,000,000 tons of He + energy each second!

Age of the Sun To determine the approximate age of the Sun, we compare the energy store (mass) to the rate of energy usage Since only 10% of the Hydrogen in the Sun is in the core, and therefore hot

20 Age of the Sun To determine the approximate age of the Sun, we compare the energy store (mass) to the rate of energy usage Since only 10% of the Hydrogen in the Sun is in the core, and therefore hot enough to undergo fusion, the age can be found as t (0.007)(0.1)M c 2 L (0.007)(0.1)( kg)( m/s) W s 10 billion years Since t solar sys ~ billion years, the Sun is about halfway through its lifetime

CNO Cycle At higher temperatures, a different mechanism exists to fuse H He, the CNO cycle This cycle requires higher temperatures so that that the H nuclei can get close enough to the C and N nuclei

21 CNO Cycle At higher temperatures, a different mechanism exists to fuse H He, the CNO cycle This cycle requires higher temperatures so that that the H nuclei can get close enough to the C and N nuclei to fuse;; Carbon, nitrogen, and oxygen act as catalysts, and are not consumed In p + p D + e + + ν a coincidence of a collision plus a weak interaction must occur Since that coincidence is not required in the CNO cycle, reaction 4p 4 He can run much faster than in the pp chain at high temperatures ε pp T 4 v. ε CNO T 20 They cross over around T~18x10 6 K

22 CNO v PP Temperature dependence

23 Energy Transport To find T(r), consider how energy is transported in a star 1. Conduction (collisions of particles) - works well for solids, poorly for gases (atoms too far apart) 2. Convection (boiling) - occurs when the thermal gradient gets too steep (usually near the surface) 3. Radiative transfer - most effective in most stars

Convection v. radiative energy transport Radiative transport occurs as a random walk. Photons travel about 2 cm in the solar interior before they are absorbed and reemitted in a random direction.

24 Convection v. radiative energy transport Radiative transport occurs as a random walk. Photons travel about 2 cm in the solar interior before they are absorbed and reemitted in a random direction. It takes, on average, about of these steps for a photon to reach the surface and escape. In the process, γ-rays and X-rays are degraded into multiple visible photons. Moving at the speed of light, this requires on the order of 1000s of years! κ If becomes too large, then radiative transport requires a very high temperature gradient, and convection (boiling) can become more efficient 1500 km

25 Interior structure of the Sun

26 Dependent Variables: Mass Luminosity Temperature Pressure Density Stellar Structure Independent Variable: Radius Additional Input: Composition

27 Equations of Stellar Structure dp dr = GM(r)ρ(r) r 2 dt dr = 3κ (r)ρ(r)l(r) 64πr 2 σt 3 (r) dm dr = 4πr2 ρ(r) dl dr = 4πr2 ρ(r)ε(r) P = ρ(r) kt (r) m H µ(r) ; µ = ρ m H n =mean mass

28 Solar structure L M T ρ

29 Russell-Vogt Theorem The mass and composition of a star uniquely determine its radius, luminosity, and internal structure, as well as its subsequent evolution Theoretical HR diagram from stellar models

30 Mass-Luminosity relation L M 4 L M 2.3

31 How does the main sequence lifetime (the time it takes a star to exhaust its nuclear fuel in the core) depend on stellar mass? (Hint: assume an average M-L relation of L M 3 ) A t MS M 2 B C D t MS M 1 t MS M - 1 t MS M - 2

Lifetimes of Main Sequence Stars Recall t 10 billion years lifetime M M 1 for high-mass stars lifetime M 4 3 M M 1 for high-mass stars 4 3 M so t(10 M ) (10 billion) (1/10) 3 (10 billion)/1000

32 Lifetimes of Main Sequence Stars Recall t 10 billion years lifetime M M 1 for high-mass stars lifetime M 4 3 M M 1 for high-mass stars 4 3 M so t(10 M ) (10 billion) (1/10) 3 (10 billion)/ million years so lifetime M M 1 for low-mass stars lifetime M 2 1 M M 1 for low-mass stars 2 1 M t(0.1 M ) (10 billion) (1/0.1) 10 (10 billion) trillion years 70 (age of the Universe)

33 Luminosity-Temperature relation L T 6

11/19/08. Gravitational equilibrium: The outward push of pressure balances the inward pull of gravity. Weight of upper layers compresses lower layers

11/19/08. Gravitational equilibrium: The outward push of pressure balances the inward pull of gravity. Weight of upper layers compresses lower layers Gravitational equilibrium: The outward push of pressure balances the inward pull of gravity Weight of upper layers compresses lower layers Gravitational equilibrium: Energy provided by fusion maintains