Stellar Interiors ASTR 2110 Sarazin Interior of Sun
Test #1 Monday, October 9, 11-11:50 am Ruffner G006 (classroom) You may not consult the text, your notes, or any other materials or any person Bring pencils, paper, calculator ~2/3 Quantitative Problems (like homework problems) ~1/3 Qualitative Questions Multiple Choice, Short Answer, Fill In the Blank questions No essay questions
Test #1 (Cont.) Equation/Formula Card: You may bring one 3x5 inch index card with equations and formulae written on both sides. DO NOT LIST pc, AU, M, L, R DO NOT INCLUDE ANY QUALITATIVE MATERIAL (text, etc.)
Material: Test #1 (Cont.) Chapters: Preface, 1-3, 5-7, 13, 19.3 Introduction, Coordinates & Time, Motions of Planets, Early Astronomy (Greeks Renaissance), Kepler s Laws, Newton s Laws, Gravity, Light, Telescopes, Doppler Effect, Basic Stellar Properties, Binary Stars, the Sun, Atomic Physics (Qualitative Only) Homeworks 1-5 Know pc, AU, M, L, R
Test #1 (Cont.) No problem set week of October 2 9 to allow study for test Review Session: Discussion session Friday, October 6, 3-4 pm
Stellar Interiors ASTR 2110 Sarazin Interior of Sun
HR Diagram
Cluster HR Diagrams As clusters age Upper main sequence disappears O -> B -> A -> F -> G Giants, supergiants, and white dwarfs appear Conclude that Stars start on main sequence Main sequence = sequence due to intrinsic differences in star Giants, supergiants, white dwarfs are due to aging = stellar evolution
Theory of Stellar Interiors Assume 1. Steady-State, Equilibrium Sun has shone for 5 x 10 9 years L ~ constant for > 3 x 10 9 years 2. Stars are round (spheres) r dr V = 4π 3 r3 dv = 4πr 2 dr volume of shell within star
Differential Equations & Boundary Values Equations for interior of star involve derivatives of physical quantities Differential Equations In general, need both differential equation and the value of the quantity at one point Boundary value: Value of quantity in differential equation at one end or another (at a boundary) of the range of the independent variable.
Differential Equations & Boundary Values Example: dy dx = x2 Integrate 0 x 1 dy dx = x 2 dx dx y = 1 3 x3 + c c = constant Boundary value: y(0) = 1 c =1 y = 1 3 x3 +1 Could have given y(1) or any value
Hydrostatic Equilibrium dp dr = GM(r)ρ Hydrostatic equilibrium r 2 Boundary Condition: P(R) = 0 outer edge M(r) mass interior to r M(r) = r 0 ρ 4πr 2 dr dm dr = 4πr2 ρ Boundary Conditions: M(0) = 0, M(R) = M *
Hydrostatic Equilibrium What if not satisfied? Sun as example No pressure balancing gravity KE = ΔPE v ~ GM / R ~ 400 km/s t ~ R v ~ 103 sec ~ 30min Very fast punishment if HE law violated!
Stellar Pulsations What is natural frequency of star? Period of stellar pulsations? KE = ΔPE v ~ GM / R ~ 400 km/s t ~ R v ~ R GM / R = R 3 GM M ρ = 4π R 3 / 3 ~ M R 3 P pul ~ 1/ G ρ pulsation period
Due to motions of particles Pressure (again) Pressure = force / area
Pressure Forces (again) p = momentum Force = dp dt = Δp atom Δp atom = 2 p = 2mv x x atoms time = nv (area) x n atoms / volume P ~ np x v x ~ nm v x 2 atoms time P = nkt = ρkt µm p ideal gas law atom wall
Radiation Pressure p = hν / c photons time ~ nc(area) P = 2np x v x ~ nc(area)hν / c (area) P = 4 3 σt 4 c ~ energy time area 1 c ~ σt 4 c
Degeneracy Pressure When T 0, does P 0? Yes for gas, radiation pressure Degeneracy Pressure è Exclusion Principle + Uncertainty Principle Uncertainty Princ. è smaller region è higher momentum Exclusion Princ. è one particle / region è higher density è higher momentum
Degeneracy Pressure Consider a small volume Uncertainty Principle Δx Δp x ΔyΔp y Δy Δx Δz ΔzΔp z Multiply together (Δx ΔyΔz)(Δp x Δp y Δp z ) 3 Exclusion princ. è spin ½ particles è can t be in same place Uncertainty princ. è can t tell unless > Δx, etc.
Degeneracy Pressure V min = Δx ΔyΔz = minimum volume/particle V min (Δp x Δp y Δp z ) 3 Δp x < p V min 3 / p 3 n =1/V min p 3 / 3 one particle/(minimum volume) ρ µ d p 3 / 3, µ d = mass per degenerate particle
Degeneracy Pressure n p 3 / 3 ρ µ d p 3 / 3 P npv Here, do relativistic case, v c P (p 3 / 3 )pc = p 4 c / 3 p ( ρ / µ ) 1/3 d " P c$ # ρ µ d % ' & 4/3 Solve for p
Equation of State P(ρ,T ) "equation of state" P = P gas + P rad + P deg = ρkt + 4σT 4 µm p 3c + 2 m " $ # ρ % ' & µ d 5/3 Non-relativistic High T, low ρ High ρ, low T
Equation of State
Equation of State P gas most important except P rad important in very hot stars P deg important in white dwarfs, neutron stars, and giant star cores P = ρkt µm p 1 µ 2X + 3 4 Y + 1 2 Z ionized gas
Central Temperatures in Stars Estimate T c Mean Value Theorem of Calculus dp dr = P(R * ) P c R * 0 = P c R * P c dp dr = GM(r)ρ hydrostatic equil. P r 2 0 M(r) M *, M(r) M * / 2 0 r R *, r R * / 2 ρ c ρ 0, ρ ρ c / 2 r R *
Central Temperatures in Stars dp dr dp dr = P c R * = GM(r)ρ r 2 hydrostatic equil. P c G(M / 2)(ρ / 2) * c R * (R * / 2) 2 ρ c kt c GM * 2 µm p R * R * ρ c T c µm p k GM * 1.5 x 10 7 K (M/M ) (R/R ) R -1 *
Central Temperatures in Stars T c µm p k GM * 1.5 x 10 7 K (M/M ) (R/R ) R -1 * Center of Sun, other stars VERY hot Follows directly from requiring that pressure balance weight of star
Thermal Equilibrium Energies in every volume must balance E out = E in + ΔE source ε energy source/time/mass L out = L in +ε ΔM L( r #) = L(r)+ερ 4πr 2 Δr L( r #) L(r) = 4πr 2 ερ Δr dl dr = 4πr2 ερ boundary condition L(0) = 0 r L out r
Energy produced in center radiated at edge Energy Transport Energy Transport Conduction (only in WDs and NSs) Convection (if T changes rapidly) Radiation How?
Δr =1/α =1/ (κρ) L out = 4πr 2 σt 4 L in = 4πr 2 σ ( T ") 4 L = 4πr 2 σ $ % T 4 " ΔT 4 dt 4 T 4 & ' dr Δr 4T 3 dt dr L(r) = 4πr 2 16σ T 3 dt 3 κρ dr Energy Transport Consider shell, one optical depth = mean free path thick 1 κρ T Δr solve for dt/dr T r r
Energy Transport L(r) = 4πr 2 16σ T 3 dt 3 κρ dr dt dr = min " $ # $ % $ 3κρ 1 64πσ r 2 T L(r) 3 2 5 T P dp dr solve for dt/dr radiation convection
Luminosity Estimate L(r) = 4πr 2 16σ 3 Replace dt dr T 3 dt κρ dr dt dr = T eff T c R T T c r R L(r) L ρ M / R 3 L ~ R 2 σ T 4 c κρr Sun: T c ~ 10 7 K, κρ ~ 1 / cm L~ 10 34 erg/s ~ Great! L T c R since T c >> T eff
Mass-Luminosity Relation T c M R ρ M / R 3 L ~ R 2 σ T 4 c κρr R2 (M / R) 4 κ(m / R 3 )R M 3 κ L M 3 if κ ~ constant
Virial Theorem: Energies in Stars PE = 2 KE PE potential energy, KE kinetic energy E = KE Star stable, static, so <> = constant value Star static, so what is KE? KE = 3 2 NkT = TE thermal energy PE = 2 TE E = TE
Energies in Stars PE = 2 TE E = TE TE TE 0 E E PE PE Stars have negative heat capacity If you add energy (heat), they get cooler!!
Thermal and Gravitational Energy If PE changes ½ Δ PE è star ½ Δ PE è light Available energy = ½ PE = TE Sun: E rad = ½ G M 2 / R L = 4 x 10 33 erg/sec Lifetime (Kelvin time)? t K = E rad / L = 2 x 10 48 erg = 5 x 10 14 sec = 2 x 10 7 years
Thermal and Gravitational Energy t K << 10 billion years Energy of Sun? Not gravity! Not heat! (If Sun turned off, it would continue to shine for 10 million years!)
Energy Source for Stars For Sun, need total energy E = L t Sun = L x (10 10 years) ~ 10 51 erg N atoms = / m p ~ 10 57 atoms M E/atom ~ 10-6 erg ~ 10 6 ev = 1 MeV per atom E/mass ~ 1 Mton TNT equiv. / ton of Sun Chemistry, Electricity, Semiconductors, etc. All such processes involve energy levels in atoms è 1 ev/atom TNT è 1 ton TNT equiv. / ton
Energy Source for Stars Not Chemistry, Electricity, Semiconductors, etc. Nuclear Energy E/atom ~ 10 MeV ~ 10 7 ev (Note: this is still only ~1% of mc 2, rest mass) 1 Mton H-bomb weighs ~ 200 lbs. Stars are powered by nuclear energy!!