Stellar Structure

Stellar StructureStellar StructureStellar StructureStellar Structure

Chapter 10Chapter 10

Stellar StructureStellar StructureStellar StructureStellar Structure

We know external properties of a star We know external properties of a star L, M, R, TL, M, R, Teffeff, (X,Y,Z), (X,Y,Z)

Apply basic physical principlesApply basic physical principles

From this, can we infer the From this, can we infer the internal structure?internal structure?

Stellar ForcesStellar ForcesStellar ForcesStellar Forces

P + dP pressure on top of cylinder, P on P + dP pressure on top of cylinder, P on bottombottomdr is height cylinder, dA its area and dm dr is height cylinder, dA its area and dm its massits massVolume of the cylinder is dV = dAdr Volume of the cylinder is dV = dAdr Mass of the cylinder is dm = Mass of the cylinder is dm = dAdr where dAdr where = = (r) is the gas density at the radius (r) is the gas density at the radius rrThe total mass inside radius r is MThe total mass inside radius r is Mr r

Gravitational forceGravitational force on volume element is on volume element is dF dFgg = -GM = -GMrrdm/rdm/r22 = -GM = -GMrrdAdr/rdAdr/r2 2 (- as (- as force directed to centre of star)force directed to centre of star)

Gravity and Gas PressureGravity and Gas PressureHydrostatic equilibrium (§10.1) - balance between Hydrostatic equilibrium (§10.1) - balance between gravity and gas pressuregravity and gas pressureConsider small cylinder located Consider small cylinder located

at radius, r, from centre of the at radius, r, from centre of the star:star:


Gravitational forceGravitational force on volume element on volume element is dFis dFgg = -GM = -GMrrdm/rdm/r22 = -GM = -GMrrdAdr/rdAdr/r22 (- as force directed to centre of star) (- as force directed to centre of star)

Gravity and Gas PressureGravity and Gas PressureHydrostatic equilibrium (§10.1) - balance between Hydrostatic equilibrium (§10.1) - balance between gravity and gas pressuregravity and gas pressureConsider small cylinder located Consider small cylinder located

at radius, r, from centre of the at radius, r, from centre of the star:star:

Equilibrium condition: the Equilibrium condition: the total force acting on volume is total force acting on volume is zero i.e.zero i.e. 0 = dF0 = dFgg + dF + dFpp = -GM = -GMrrdAdr/rdAdr/r22 - -dPdA or dPdA or 1.1. dP/dr = - GM dP/dr = - GMrr/r/r22 (Equation (Equation of Hydrostatic Equilibrium)of Hydrostatic Equilibrium)

Net Net pressure forcepressure force acting on acting on element is dFelement is dFpp = PdA - (P + = PdA - (P + dP)dA = -dPdA (dP is negative dP)dA = -dPdA (dP is negative as pressure decreases as pressure decreases outward)outward)

Let ma = FLet ma = Fgg - F - Fp p (m mass, a acceleration)(m mass, a acceleration)Let FLet Fpp = 1.00000001 F = 1.00000001 Fgg = (1 + 1 x 10 = (1 + 1 x 10-8-8)F)Fgg

Then acceleration at solar surface given by:Then acceleration at solar surface given by:ma = Fma = Fg g - F- Fpp = F = Fgg(1 - 1 - 1 x 10(1 - 1 - 1 x 10-8-8) = -10) = -10-8-8FFgg

FFgg = ma = GM = ma = GMsunsunm(10m(10-8-8)/R)/Rsunsun2 2

m cancels and putting in numbers m cancels and putting in numbers a = (270 m sa = (270 m s-2-2)(10)(10-8-8) = 2.7 x 10) = 2.7 x 10-6-6 m s m s-2-2

Displacement of solar surface in t = 100 days (8.6 x Displacement of solar surface in t = 100 days (8.6 x 10106 6 s) would be:s) would be:

d = 1/2 atd = 1/2 at2 2 = 2.0 x 10 = 2.0 x 1088 m = 0.29 solar radii! m = 0.29 solar radii!

So if equilibrium is unbalanced by only 1 part in So if equilibrium is unbalanced by only 1 part in 101088, Sun would grow (or shrink) by ~30% in a few months , Sun would grow (or shrink) by ~30% in a few months - clearly not observed- clearly not observed

30% change in radius would result in a 70% change in L 30% change in radius would result in a 70% change in L resulting in 18% change in global temperature of resulting in 18% change in global temperature of Earth!!Earth!!

Stellar ForcesStellar ForcesStellar ForcesStellar ForcesIs Sun in hydrostatic Is Sun in hydrostatic equilibrium?equilibrium?


RotationRotation (gives centripetal and coriolis forces) (gives centripetal and coriolis forces)

FFcentcent = m Ω = m Ω22r where Ω = angular velocity (radian sr where Ω = angular velocity (radian s-1-1))

At equator on stellar surface, FAt equator on stellar surface, Fcentcent = m Ω = m Ω22RR** and F and Fgg = = GMGM**m/Rm/R**

22

Thus, FThus, Fcentcent/F/Fgg = Ω = Ω22RR**33/GM/GM**

e.g. Sun: R = 7 x 10e.g. Sun: R = 7 x 1088 m, P m, Protrot = 25 days = 25 days ΩΩsunsun = 3 x 10 = 3 x 10-6-6 rad s rad s-1-1, ,

MMsunsun = 2 x 10 = 2 x 103030 kg kg FFcentcent/F/Fgg = 2 x 10 = 2 x 10-5-5

For FFor Fcentcent to be important, a star must be large and to be important, a star must be large and rotating rapidly. There are some examples of such rotating rapidly. There are some examples of such stars. e.g. Be stars which show emission lines.stars. e.g. Be stars which show emission lines.

Are other forces Are other forces important?important?


Radiation PressureRadiation Pressure Before we do this let’s estimate pressure at centre Before we do this let’s estimate pressure at centre of Sunof Sun

Crudely, dP/dr ~ Crudely, dP/dr ~ P/P/r ~ (Pr ~ (Pss - P - Pcc)/(R)/(Rss - R - Rcc) where P) where Pcc is the central pressure, Pis the central pressure, Pss is the surface pressure is the surface pressure (= 0), R(= 0), Rss is the surface radius and R is the surface radius and Rcc the radius at the radius at center = 0) center = 0)

So dP/dr ~ (0 - PSo dP/dr ~ (0 - Pcc)/(R)/(Rs s - 0) or dP/dr ~ - P- 0) or dP/dr ~ - Pcc /R /Rss

Now apply to the Sun. Substituting into Now apply to the Sun. Substituting into Hydrostatic equilibrium eq., Hydrostatic equilibrium eq., dP/dr = - GMdP/dr = - GMrr/r/r22,, givesgives

-P-Pcc/R/Rsunsun = -GM = -GMsunsunsunsun/R/Rsunsun22; or P; or Pcc = = GMGMsunsunsunsun/R/Rsunsun

((sunsun = 1.4 g cm = 1.4 g cm-3-3 = 1400 kg m = 1400 kg m-3; -3; RRsunsun = 7 x 10 = 7 x 1088 m) m)

PPcc = 2.7 x 10 = 2.7 x 101414 N m N m-2 -2 (or Pa) (Surface Earth (or Pa) (Surface Earth 101055 Pa) Pa)



Radiation pressure, PRadiation pressure, Pradrad = (1/3)aT = (1/3)aT4 4 with a = 7.57 x 10with a = 7.57 x 10-16-16

J mJ m-3-3 K K-4 -4

At the Sun’s centre, T ~ 10At the Sun’s centre, T ~ 1077 K K P Pradrad = 7.6 x 10 = 7.6 x 1012 12 N mN m-2 -2

(Pa)(Pa)

This is a crude estimate but indicates that in stars This is a crude estimate but indicates that in stars like the sun, radiation pressure not important like the sun, radiation pressure not important (compare with gas pressure 2.5 x 10(compare with gas pressure 2.5 x 1016 16 Pa) - but it is Pa) - but it is important for hotter stars.important for hotter stars.

Are other forces important?Are other forces important?

Radiation PressureRadiation Pressure The value of PThe value of Pcc = 2.7 x 10 = 2.7 x 101414 N m N m-2 -2 (Pa)(Pa) for for

the pressure at the Sun’s centre is crude the pressure at the Sun’s centre is crude and too low by about 2 orders of magnitude. and too low by about 2 orders of magnitude. It has not taken into account the It has not taken into account the increased density near the Sun’s centre. increased density near the Sun’s centre. Theoretical models give a value of 2.5 x Theoretical models give a value of 2.5 x 10101616 Pa Pa


Magnetic PressureMagnetic Pressure

PPmagmag = H = H22/8/8 where H = field strength in Gauss (cgs) or where H = field strength in Gauss (cgs) or Tesla (SI) Tesla (SI) (1 T = 10(1 T = 1044 Gauss) Gauss)At the centre of the Sun:At the centre of the Sun:

PPgasgas ~ 2.5 x 10 ~ 2.5 x 101616 Pa = 2.5 x 10 Pa = 2.5 x 101717 dyne cm dyne cm-2-2 we need ~10we need ~1099 Gauss at centre for P Gauss at centre for Pmagmag to be to be

importantimportantAt base of photosphere (“surface of Sun”):At base of photosphere (“surface of Sun”):

PPgasgas ~ 10 ~ 1055 dynes cm dynes cm-2 -2 need fields of ~ 10need fields of ~ 103 3 G for PG for Pmagmag to be important to be important Measured value of PMeasured value of Pmagmag at the surface is ~ 1 G at the surface is ~ 1 G

However, there are stars with surface fields of many kG However, there are stars with surface fields of many kG and even giga G (magnetic white dwarfs) and even giga G (magnetic white dwarfs)

Bottom line: For normal stars like the Sun, the only Bottom line: For normal stars like the Sun, the only force we need to consider, in the first approximation, force we need to consider, in the first approximation, is the force due to gas pressure.is the force due to gas pressure.


Mass Conservation, Energy ProductionMass Conservation, Energy ProductionMass Conservation, Energy ProductionMass Conservation, Energy Production

dL(r) = dL(r) = 44rr22(r)dr (r)dr 3.3. dL(r)/dr = 4dL(r)/dr = 4rr22(r)(r)(r) - Energy (r) - Energy Production Equation Production Equation

Here Here (r) > 0 only where T(r) is high enough to (r) > 0 only where T(r) is high enough to produce nuclear reactions produce nuclear reactions In Sun, In Sun, (r) > 0 when r < 0.2 R(r) > 0 when r < 0.2 Rsunsun

Mass of shell of thickness dr Mass of shell of thickness dr at radius r is dM(r) = at radius r is dM(r) = x V = x V = x 4 x 4rr22dr dr

Luminosity produced by shell Luminosity produced by shell of mass dM is dL = of mass dM is dL = dM, dM,

= total energy produced /mass/sec by all = total energy produced /mass/sec by all fusion nuclear reactions and gravityfusion nuclear reactions and gravity

2.2. dM(r)/dr = 4dM(r)/dr = 4rr22(r) - (r) - Equation Mass ConservationEquation Mass Conservation

Summary Summary (So Far)(So Far) Stellar Structure Stellar Structure EquationsEquations

Summary Summary (So Far)(So Far) Stellar Structure Stellar Structure EquationsEquations

1.1. dP/dr =dP/dr = - GM- GMrr(r)/r(r)/r22Equation of Equation of Hydrostatic Hydrostatic EquilibriumEquilibrium

2.2. dM(r)/dr = 4dM(r)/dr = 4rr22(r)(r) Equation of Mass Equation of Mass ConservationConservation

3.3. dL(r)/dr = 4dL(r)/dr = 4rr22(r)(r)(r)(r) Equation of Energy Equation of Energy ProductionProduction

4.4.

Temperature GradientTemperature GradientTemperature GradientTemperature Gradient

The fourth equation of stellar structure gives The fourth equation of stellar structure gives temperature change as function of radius r, i.e temperature change as function of radius r, i.e dT/dr.dT/dr.

In the interior of stars like the Sun, conduction In the interior of stars like the Sun, conduction of heat (by electrons) is very inefficient as of heat (by electrons) is very inefficient as electrons collide often with other particles.electrons collide often with other particles.

However, in white dwarfs and neutron stars, heat However, in white dwarfs and neutron stars, heat conduction is a very important means of energy conduction is a very important means of energy transport. In these stars, the mean free path of transport. In these stars, the mean free path of some electrons can be very long whereas the mean some electrons can be very long whereas the mean free path of their photons is extremely short. free path of their photons is extremely short.

Energy TransportEnergy Transport

Temperature GradientTemperature GradientTemperature GradientTemperature Gradient

A star that carries its energy outwards entirely by radiation is said to be in A star that carries its energy outwards entirely by radiation is said to be in radiative equilibrium - photons slowly DIFFUSE outwardradiative equilibrium - photons slowly DIFFUSE outward

Flux(at radius r) = LFlux(at radius r) = Lrr/4/4rr22 = -D dU = -D dUrr/dr where U/dr where Urr is energy density in radiation = aT is energy density in radiation = aT44

(a is radiation constant 7.6 x 10(a is radiation constant 7.6 x 10-16-16 J m J m-3-3 K K-4-4) and D = 1/3 ) and D = 1/3 c where c where is mean free is mean free path of photons). Need to know what fraction photons absorbed - defined through path of photons). Need to know what fraction photons absorbed - defined through so that so that dl gives fraction energy lost by absorption over distance dl (dl gives fraction energy lost by absorption over distance dl ( = 1, = 1, makesmakes

sense as sense as is the mean free path) units is the mean free path) units are m are m22/kg /kg So LSo Lrr/4/4rr22 = -(4/3) (acT = -(4/3) (acT33//) dT/dr or ) dT/dr or

4.4. dT/dr = (-3/4ac) ( dT/dr = (-3/4ac) (/T/T33) (L) (Lrr/4/4rr22) )

Energy TransportEnergy Transport

Thus, the majority of energy is Thus, the majority of energy is transported by radiation in interior most transported by radiation in interior most stars. Photons emitted in hot regions of a stars. Photons emitted in hot regions of a star are absorbed in cooler regions. star are absorbed in cooler regions.

Equation of StateEquation of State Equation of StateEquation of State

Expresses the dependence of P(pressure) on other Expresses the dependence of P(pressure) on other parameters.parameters.

Most common eqn. of state is the ideal gas law, Most common eqn. of state is the ideal gas law, P = P = NkTNkT

Where k = Boltzmann constant, N = # particles per unit volume, and Where k = Boltzmann constant, N = # particles per unit volume, and T = temperatureT = temperature

Holds at high accuracy for gases at low density.Holds at high accuracy for gases at low density. It is also accurate at high densities if the gas It is also accurate at high densities if the gas temperature is also high as in stellar interiors. temperature is also high as in stellar interiors.

Now we introduce the gas composition explicitly.Now we introduce the gas composition explicitly. Let X = mass fraction hydrogen in a star, Y same for Let X = mass fraction hydrogen in a star, Y same for He, and Z for everything else.He, and Z for everything else.

(We have seen that X = 0.73, Y=0.25, and Z = 0.02.) (We have seen that X = 0.73, Y=0.25, and Z = 0.02.) Of course X + Y + Z = 1.Of course X + Y + Z = 1.

Equation of StateEquation of State

Equation of StateEquation of State Equation of StateEquation of State

Assume gas is fully ionized so sum all items to get: Assume gas is fully ionized so sum all items to get: N = (2X + (3/4)Y + (1/2)Z)N = (2X + (3/4)Y + (1/2)Z)/m/mHH

Equation state then is:Equation state then is:P = (1/P = (1/)k)kT/mT/mHH with 1/ with 1/ = 2X + (3/4)Y + (1/2)Z = 2X + (3/4)Y + (1/2)Z

In the Sun, 1/In the Sun, 1/ = 2(0.73) + 3/4(0.25) + 1/2(0.02) = 1.658 = 2(0.73) + 3/4(0.25) + 1/2(0.02) = 1.658 so so = 0.60 ( = 0.60 ( is called the mean molecular weight) (eg is called the mean molecular weight) (eg for pure for pure H gas?)H gas?)

Equation of StateEquation of StateNow we tabulate the number of atoms and Now we tabulate the number of atoms and number corresponding electrons per unit number corresponding electrons per unit volume (here mvolume (here mHH is mass of the proton) is mass of the proton)

ElementElement HydrogeHydrogenn

HeliumHelium HeavierHeavier

No. atomsNo. atoms XX/m/mHH YY/4m/4mHH [Z[Z/Am/AmHH] ]

(usually (usually small)small)

No. No. electronselectrons

XX/m/mHH 2Y2Y/4m/4mHH 1/2 AZ1/2 AZ/Am/AmHH

Summary Stellar Structure EquationsSummary Stellar Structure Equations Summary Stellar Structure EquationsSummary Stellar Structure Equations

Plus boundary conditions: Plus boundary conditions: At centre - M(r) At centre - M(r) 0 as r 0 as r 0; L(r) 0; L(r) 0 as r 0 as r

0 0At surface - T(r), P(r), At surface - T(r), P(r), (r) (r) 0 as r 0 as r R R**

These equations produce the Standard Solar Model These equations produce the Standard Solar Model and the Mass - Luminosity Relationand the Mass - Luminosity Relation

1.1. dP/dr =dP/dr = - GM- GMrr(r)/r(r)/r22Equation of Equation of Hydrostatic Hydrostatic EquilibriumEquilibrium



4.4. dT/dr = (-3/4ac) dT/dr = (-3/4ac) (((r)/T(r)/T33) (L) (Lrr/4/4rr22)) Temperature GradientTemperature Gradient

P = (1/P = (1/)k)kT/mT/mHH Equation of StateEquation of State

Standard Solar ModelStandard Solar ModelStandard Solar ModelStandard Solar Model

Mass - Luminosity RelationMass - Luminosity Relation Mass - Luminosity RelationMass - Luminosity Relation

Density, Density, M/R M/R33 ….eqn. (1) ….eqn. (1)Substitute (1) into Hydrostatic Equil. Eqn., Substitute (1) into Hydrostatic Equil. Eqn., dP/dr =dP/dr = - - GMGMrr(r)/r(r)/r2 2 to getto get P P M M22/R/R4 4 …eqn. (2)…eqn. (2)

Use Eqn. State, Use Eqn. State, P = (1/P = (1/)k)kT/mT/mHH with eqn. (2) is T with eqn. (2) is T M/R …eqn. M/R …eqn.

(3)(3)Put (2) and (3) into Radiative Equilibrium Eq., Put (2) and (3) into Radiative Equilibrium Eq., dT/dr = (-3/4ac) (dT/dr = (-3/4ac) ((r)/T(r)/T33) (L) (Lrr/4/4rr22)) To get L To get L M M33 … eqn. (4) which is close to observed … eqn. (4) which is close to observed relationship, L relationship, L M M3.5 3.5

1.1. dP/dr =dP/dr = - GM- GMrr(r)/r(r)/r22 Equation of Hydrostatic Equation of Hydrostatic EquilibriumEquilibrium



4.4. dT/dr = (-3/4ac) dT/dr = (-3/4ac) (((r)/T(r)/T33) (L) (Lrr/4/4rr22)) Temperature GradientTemperature Gradient

P = (1/P = (1/)k)kT/mT/mHH Equation of StateEquation of State

Use the Equations of Stellar Structure to calculate Use the Equations of Stellar Structure to calculate how the luminosity of a star depends on its mass how the luminosity of a star depends on its mass (the Mass Luminosity Relation).(the Mass Luminosity Relation).

Central Temperature SunCentral Temperature Sun Central Temperature SunCentral Temperature Sun

Use ideal gas law, P = kT/mH Pc = ckTc/mH Tc = mHPc/ck

Approximate central density, c as <> ~ 1.4 g cm-3 = 1400 kg m-3

Take Pc = 2.7 x 1014 Pa from earlier estimate, and = 0.60

Tc ~ 1.4 x 107 K (models predict about 1.6 x 107 K)

Agreement is fortuitous since the pressure estimate used as Pc and the density estimate, <>, used as c are both too low by a factor of 100.

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Stellar Structure

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