ASTR 1040 Recitation: Stellar Structure Ryan Orvedahl Department - PowerPoint PPT Presentation

ASTR 1040 Recitation: Stellar Structure Ryan Orvedahl Department of Astrophysical and Planetary Sciences University of Colorado at Boulder Boulder, CO 80309 ryan.orvedahl@colorado.edu February 12, 2014

This Week MIDTERM: Thurs Feb 13 (regular class time, 9:30 am) Review Session: Wed Feb 12 (5:00 - 7:00 pm) Observing Session: Web Feb 12 (7:30 pm) R. Orvedahl (CU Boulder) Stellar Structure Feb 12 2 / 16

Today’s Schedule Comments on Homework How to Build a Stellar Structure Model R. Orvedahl (CU Boulder) Stellar Structure Feb 12 3 / 16

How To Build A Star What physics do you need to build a star? R. Orvedahl (CU Boulder) Stellar Structure Feb 12 4 / 16

How To Build A Star What physics do you need to build a star? Gravity vs. Pressure R. Orvedahl (CU Boulder) Stellar Structure Feb 12 4 / 16

How To Build A Star What physics do you need to build a star? Gravity vs. Pressure Nuclear Reactions R. Orvedahl (CU Boulder) Stellar Structure Feb 12 4 / 16

How To Build A Star What physics do you need to build a star? Gravity vs. Pressure Nuclear Reactions Energy Transport R. Orvedahl (CU Boulder) Stellar Structure Feb 12 4 / 16

How To Build A Star What physics do you need to build a star? Gravity vs. Pressure Nuclear Reactions Energy Transport Equation of State R. Orvedahl (CU Boulder) Stellar Structure Feb 12 4 / 16

Gravity vs. Pressure Hydrostatic Balance: dr = − GM r ( r ) ρ ( r ) dP r 2 How much mass? � r 4 π r 2 ρ ( r ) dr M r ( r ) = 0 R. Orvedahl (CU Boulder) Stellar Structure Feb 12 5 / 16

Gravity vs. Pressure Q: What is the source of the pressure gradient outside the core in the equation for Hydrostatic Equilibrium? R. Orvedahl (CU Boulder) Stellar Structure Feb 12 6 / 16

Gravity vs. Pressure Q: What is the source of the pressure gradient outside the core in the equation for Hydrostatic Equilibrium? A: Energy transport mechanisms such as radiative diffusion or convection Radiation exerts a pressure P rad = aT 4 / 3 R. Orvedahl (CU Boulder) Stellar Structure Feb 12 6 / 16

Nuclear Reactions Reaction rates, r i , j r i , j ≈ r 0 X i X j ρ α +1 T β Energy released / kg / sec, ǫ i , j ǫ i , j = E 0 ( E 0 = Energy / Rx) ρ r i , j Combine the two equations 0 X i X j ρ α T β ǫ i , j = ǫ ′ R. Orvedahl (CU Boulder) Stellar Structure Feb 12 7 / 16

Nuclear Reactions Luminosity ∝ Energy released dL = ǫ dm where ǫ = ǫ nuc + ǫ grav is the total energy released / kg / sec by all reactions and gravity dm = dM r = ρ ( r ) dV = 4 π r 2 ρ ( r ) dr dL r = ǫ dM r = 4 π r 2 ρ ( r ) ǫ dr ⇒ dL r dr = 4 π r 2 ρ ( r ) ǫ R. Orvedahl (CU Boulder) Stellar Structure Feb 12 8 / 16

Nuclear Reactions dL r dr = 4 π r 2 ρ ( r ) ǫ = 4 π r 2 ǫ 0 ρ α +1 T β Reaction Name α β P-P Chain 1 4 CNO Cycle 1 15 Triple- α 2 40 R. Orvedahl (CU Boulder) Stellar Structure Feb 12 9 / 16

Energy Transport Remember radiation exerts a pressure P rad = aT 4 / 3 ⇒ dP rad = 4 3 aT 3 dT dr dr From Radiation Transport Theory = − ¯ κρ dP rad c F rad dr Combine equations κρ ¯ dT dr = − 3 L r T 3 F rad , where F rad = 4 ac 4 π r 2 Get T in terms of L r dr = − 3 ¯ κρ dT L r 4 ac T 3 4 π r 2 R. Orvedahl (CU Boulder) Stellar Structure Feb 12 10 / 16

What Equations Do We Have So Far? dr = − GM r ( r ) ρ ( r ) dP 1 r 2 # Eqns = 1, # Variables = 3: P ( r ), M r ( r ), ρ ( r ) R. Orvedahl (CU Boulder) Stellar Structure Feb 12 11 / 16

What Equations Do We Have So Far? dr = − GM r ( r ) ρ ( r ) dP 1 r 2 # Eqns = 1, # Variables = 3: P ( r ), M r ( r ), ρ ( r ) dM r ( r ) = 4 π r 2 ρ ( r ) 2 dr Eqns = 2, Vars = 3: P ( r ), M r ( r ), ρ ( r ) R. Orvedahl (CU Boulder) Stellar Structure Feb 12 11 / 16

What Equations Do We Have So Far? dr = − GM r ( r ) ρ ( r ) dP 1 r 2 # Eqns = 1, # Variables = 3: P ( r ), M r ( r ), ρ ( r ) dM r ( r ) = 4 π r 2 ρ ( r ) 2 dr Eqns = 2, Vars = 3: P ( r ), M r ( r ), ρ ( r ) dL r dr = 4 π r 2 ρ ( r ) ǫ = 4 π r 2 ǫ 0 ρ α +1 T β 3 Eqns = 3, Vars = 5: P ( r ), M r ( r ), ρ ( r ), L r ( r ), T ( r ) R. Orvedahl (CU Boulder) Stellar Structure Feb 12 11 / 16

What Equations Do We Have So Far? dr = − GM r ( r ) ρ ( r ) dP 1 r 2 # Eqns = 1, # Variables = 3: P ( r ), M r ( r ), ρ ( r ) dM r ( r ) = 4 π r 2 ρ ( r ) 2 dr Eqns = 2, Vars = 3: P ( r ), M r ( r ), ρ ( r ) dL r dr = 4 π r 2 ρ ( r ) ǫ = 4 π r 2 ǫ 0 ρ α +1 T β 3 Eqns = 3, Vars = 5: P ( r ), M r ( r ), ρ ( r ), L r ( r ), T ( r ) dr = − 3 ¯ κρ dT L r 4 T 3 4 π r 2 4 ac Eqns = 4, Vars = 5: P ( r ), M r ( r ), ρ ( r ), L r ( r ), T ( r ) R. Orvedahl (CU Boulder) Stellar Structure Feb 12 11 / 16

Equation of State Gives the pressure in terms of density and temperature P = P ( ρ, T ) Ideal Gas: P = ρ kT m , m = mean atomic mass ¯ ¯ m ¯ For the Sun: µ = m H ≈ 1 . 6 Or Electron Degenerate Matter: � 5 / 3 �� Z P = (3 π 2 ) 2 / 3 � 2 ρ � 5 m e A m H Or ... R. Orvedahl (CU Boulder) Stellar Structure Feb 12 12 / 16

Final Set of Equaions dr = − GM r ( r ) ρ ( r ) dP 1 r 2 dM r ( r ) = 4 π r 2 ρ ( r ) 2 dr dL r dr = 4 π r 2 ρ ( r ) ǫ = 4 π r 2 ǫ 0 ρ α +1 T β 3 ¯ κρ dT dr = − 3 L r 4 4 ac T 3 4 π r 2 P = ρ kT 5 ¯ m Still cannot solve without Boundary Conditions ... R. Orvedahl (CU Boulder) Stellar Structure Feb 12 13 / 16

Boundary Conditions M r → 0 as r → 0 L r → 0 as r → 0 ρ → 0 as r → R ∗ T → T eff as r → R ∗ P → 0 as r → R ∗ Now we can solve the system R. Orvedahl (CU Boulder) Stellar Structure Feb 12 14 / 16

Numerically Integrate the System R. Orvedahl (CU Boulder) Stellar Structure Feb 12 15 / 16

Numerically Integrate the System R. Orvedahl (CU Boulder) Stellar Structure Feb 12 16 / 16