1. Stellar Structure Chapter 10

2. Stellar Structure We know external properties of a star L, M, R, T eff, (X,Y,Z) Apply basic physical principles From this, can we infer the internal structure?

3. Stellar Forces ä P + dP pressure on top of cylinder, P on bottom ä dr is height cylinder, dA its area and dm its mass ä Volume of the cylinder is dV = dAdr ä Mass of the cylinder is dm =  dAdr where ä  =  (r) is the gas density at the radius r ä The total mass inside radius r is M r ä Gravitational force on volume element is dF g = -GM r dm/r 2 = -GM r  dAdr/r 2 (- as force directed to centre of star) Gravity and Gas Pressure Hydrostatic equilibrium (§10.1) - balance between gravity and gas pressure Consider small cylinder located at radius, r, from centre of the star:

4. Stellar Forces ä Gravitational force on volume element is dF g = -GM r dm/r 2 = -GM r  dAdr/r 2 (- as force directed to centre of star) Gravity and Gas Pressure Hydrostatic equilibrium (§10.1) - balance between gravity and gas pressure Consider small cylinder located at radius, r, from centre of the star: ä Equilibrium condition: the total force acting on volume is zero i.e. 0 = dF g + dF p = -GM r  dAdr/r 2 -dPdA or 0 = dF g + dF p = -GM r  dAdr/r 2 -dPdA or 1. dP/dr = - GM r  /r 2 (Equation of Hydrostatic Equilibrium) 1. dP/dr = - GM r  /r 2 (Equation of Hydrostatic Equilibrium) ä Net pressure force acting on element is dF p = PdA - (P + dP)dA = -dPdA (dP is negative as pressure decreases outward)

5. ä Let ma = F g - F p (m mass, a acceleration) ä Let F p = 1.00000001 F g = (1 + 1 x 10 -8 )F g ä Then acceleration at solar surface given by: ma = F g - F p = F g (1 - 1 - 1 x 10 -8 ) = -10 -8 F g F g = ma = GM sun m(10 -8 )/R sun 2 m cancels and putting in numbers a = (270 m s -2 )(10 -8 ) = 2.7 x 10 -6 m s -2 ä Displacement of solar surface in t = 100 days (8.6 x 10 6 s) would be: d = 1/2 at 2 = 2.0 x 10 8 m = 0.29 solar radii! ä So if equilibrium is unbalanced by only 1 part in 10 8, Sun would grow (or shrink) by ~30% in a few months - clearly not observed ä 30% change in radius would result in a 70% change in L resulting in 18% change in global temperature of Earth!! Stellar Forces Is Sun in hydrostatic equilibrium?

6. Stellar Forces Rotation (gives centripetal and coriolis forces) ä F cent = m Ω 2 r where Ω = angular velocity (radian s -1 ) ä At equator on stellar surface, F cent = m Ω 2 R * and F g = GM * m/R * 2 ä Thus, F cent /F g = Ω 2 R * 3 /GM * ä e.g. Sun: R = 7 x 10 8 m, P rot = 25 days  Ω sun = 3 x 10 -6 rad s -1,  Ω sun = 3 x 10 -6 rad s -1, M sun = 2 x 10 30 kg M sun = 2 x 10 30 kg  F cent /F g = 2 x 10 -5  F cent /F g = 2 x 10 -5 ä For F cent to be important, a star must be large and rotating rapidly. There are some examples of such stars. e.g. Be stars which show emission lines. Are other forces important?

7. Stellar Forces Radiation Pressure ä Before we do this let’s estimate pressure at centre of Sun ä Crudely, dP/dr ~  P/  r ~ (P s - P c )/(R s - R c ) where P c is the central pressure, P s is the surface pressure (= 0), R s is the surface radius and R c the radius at center = 0) ä So dP/dr ~ (0 - P c )/(R s - 0) or dP/dr ~ - P c /R s ä Now apply to the Sun. Substituting into Hydrostatic equilibrium eq., dP/dr = - GM r  /r 2, gives -P c /R sun = -GM sun  sun /R sun 2 ; or P c = GM sun  sun /R sun -P c /R sun = -GM sun  sun /R sun 2 ; or P c = GM sun  sun /R sun (  sun = 1.4 g cm -3 = 1400 kg m -3; R sun = 7 x 10 8 m)  P c = 2.7 x 10 14 N m -2 (or Pa) (Surface Earth 10 5 Pa) Are other forces important?

8. Stellar Forces ä Radiation pressure, P rad = (1/3)aT 4 with a = 7.57 x 10 -16 J m -3 K -4 ä At the Sun’s centre, T ~ 10 7 K  P rad = 7.6 x 10 12 N m -2 (Pa) ä This is a crude estimate but indicates that in stars like the sun, radiation pressure not important (compare with gas pressure 2.5 x 10 16 Pa) - but it is important for hotter stars. Are other forces important? Radiation Pressure ä The value of P c = 2.7 x 10 14 N m -2 (Pa) for the pressure at the Sun’s centre is crude and too low by about 2 orders of magnitude. It has not taken into account the increased density near the Sun’s centre. Theoretical models give a value of 2.5 x 10 16 Pa

9. Stellar Forces Magnetic Pressure ä P mag = H 2 /8  where H = field strength in Gauss (cgs) or Tesla (SI) (1 T = 10 4 Gauss) ä At the centre of the Sun: P gas ~ 2.5 x 10 16 Pa = 2.5 x 10 17 dyne cm -2 P gas ~ 2.5 x 10 16 Pa = 2.5 x 10 17 dyne cm -2  we need ~10 9 Gauss at centre for P mag to be important  we need ~10 9 Gauss at centre for P mag to be important ä At base of photosphere (“surface of Sun”): P gas ~ 10 5 dynes cm -2 P gas ~ 10 5 dynes cm -2  need fields of ~ 10 3 G for P mag to be important  need fields of ~ 10 3 G for P mag to be important Measured value of P mag at the surface is ~ 1 G ä However, there are stars with surface fields of many kG and even giga G (magnetic white dwarfs) Bottom line: For normal stars like the Sun, the only force we need to consider, in the first approximation, is the force due to gas pressure. Are other forces important?

10. Mass Conservation, Energy Production  dL(r) =  4  r 2  (r)dr  3. dL(r)/dr = 4  r 2  (r)  (r) - Energy Production Equation Here  (r) > 0 only where T(r) is high enough to produce nuclear reactions In Sun,  (r) > 0 when r 0 when r < 0.2 R sun Mass of shell of thickness dr at radius r is dM(r) =  x V =  x 4  r 2 dr  Luminosity produced by shell of mass dM is dL =  dM,  = total energy produced /mass/sec by all fusion nuclear reactions and gravity 2. dM(r)/dr = 4  r 2  (r) - Equation Mass Conservation

11. Summary (So Far) Stellar Structure Equations Summary (So Far) Stellar Structure Equations 1. dP/dr = - GM r  (r)/r 2 Equation of Hydrostatic Equilibrium 2. dM(r)/dr = 4  r 2  (r) Equation of Mass Conservation 3. dL(r)/dr = 4  r 2  (r)  (r) Equation of Energy Production 4.

12. Temperature Gradient ä The fourth equation of stellar structure gives temperature change as function of radius r, i.e dT/dr. ä In the interior of stars like the Sun, conduction of heat (by electrons) is very inefficient as electrons collide often with other particles. ä However, in white dwarfs and neutron stars, heat conduction is a very important means of energy transport. In these stars, the mean free path of some electrons can be very long whereas the mean free path of their photons is extremely short. Energy Transport

13. Temperature Gradient ä A star that carries its energy outwards entirely by radiation is said to be in radiative equilibrium - photons slowly DIFFUSE outward ä Flux(at radius r) = L r /4  r 2 = -D dU r /dr where U r is energy density in radiation = aT 4 (a is radiation constant 7.6 x 10 -16 J m -3 K -4 ) and D = 1/3 c where is mean free (a is radiation constant 7.6 x 10 -16 J m -3 K -4 ) and D = 1/3 c where 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  dl gives fraction energy lost by absorption over distance dl (  = 1, makes so that  dl gives fraction energy lost by absorption over distance dl (  = 1, makes sense as is the mean free path) units  are m 2 /kg sense as is the mean free path) units  are m 2 /kg So L r /4  r 2 = -(4/3) (acT 3 /  ) dT/dr or So L r /4  r 2 = -(4/3) (acT 3 /  ) dT/dr or 4. dT/dr = (-3/4ac) (  /T 3 ) (L r /4  r 2 ) Energy Transport ä Thus, the majority of energy is transported by radiation in interior most stars. Photons emitted in hot regions of a star are absorbed in cooler regions.

14. Equation of State Equation of State ä Expresses the dependence of P(pressure) on other parameters. ä Most common eqn. of state is the ideal gas law, P = NkT Where k = Boltzmann constant, N = # particles per unit volume, and T = temperature ä Holds at high accuracy for gases at low density. ä It is also accurate at high densities if the gas temperature is also high as in stellar interiors. ä Now we introduce the gas composition explicitly. ä Let X = mass fraction hydrogen in a star, Y same for He, and Z for everything else. ä (We have seen that X = 0.73, Y=0.25, and Z = 0.02.) Of course X + Y + Z = 1. Equation of State

15. Equation of State Equation of State ä Assume gas is fully ionized so sum all items to get: N = (2X + (3/4)Y + (1/2)Z)  /m H ä Equation state then is: P = (1/  )k  T/m H with 1/  = 2X + (3/4)Y + (1/2)Z ä In the Sun, 1/  = 2(0.73) + 3/4(0.25) + 1/2(0.02) = 1.658 so  = 0.60 (  is called the mean molecular weight) (eg  for pure H gas?) Equation of State Now we tabulate the number of atoms and number corresponding electrons per unit volume (here m H is mass of the proton) ElementHydrogenHeliumHeavier No. atoms X  /m H Y  /4m H [Z  /Am H ] (usually small) No. electrons X  /m H 2Y  /4m H 1/2 AZ  /Am H

16. Summary Stellar Structure Equations Summary Stellar Structure Equations ä Plus boundary conditions: At centre - M(r)  0 as r  0; L(r)  0 as r  0 At surface - T(r), P(r),  (r)  0 as r  R * ä These equations produce the Standard Solar Model and the Mass - Luminosity Relation 1. dP/dr = - GM r  (r)/r 2 Equation of Hydrostatic Equilibrium 2. dM(r)/dr = 4  r 2  (r) Equation of Mass Conservation 3. dL(r)/dr = 4  r 2  (r)  (r) Equation of Energy Production 4. dT/dr = (-3/4ac) (  (r)/T 3 ) (L r /4  r 2 ) Temperature Gradient P = (1/  )k  T/m H Equation of State

17. Standard Solar Model

18. Mass - Luminosity Relation Mass - Luminosity Relation ä Density,   M/R 3 ….eqn. (1) ä Substitute (1) into Hydrostatic Equil. Eqn., dP/dr = - GM r  (r)/r 2 to get P  M 2 /R 4 …eqn. (2) ä Use Eqn. State, P = (1/  )k  T/m H with eqn. (2) is T  M/R …eqn. (3) ä Put (2) and (3) into Radiative Equilibrium Eq., dT/dr = (-3/4ac) (  (r)/T 3 ) (L r /4  r 2 ) ä To get L  M 3 … eqn. (4) which is close to observed relationship, L  M 3.5 1. dP/dr = - GM r  (r)/r 2 Equation of Hydrostatic Equilibrium 2. dM(r)/dr = 4  r 2  (r) Equation of Mass Conservation 3. dL(r)/dr = 4  r 2  (r)  (r) Equation of Energy Production 4. dT/dr = (-3/4ac) (  (r)/T 3 ) (L r /4  r 2 ) Temperature Gradient P = (1/  )k  T/m H Equation of State ä Use the Equations of Stellar Structure to calculate how the luminosity of a star depends on its mass (the Mass Luminosity Relation).

19. Central Temperature Sun Central Temperature Sun Use ideal gas law, P =  kT/  m H  P c =  c kT c /  m H  T c =  m H P c /  c k Approximate central density,  c as ~ 1.4 g cm -3 = 1400 kg m -3 Take P c = 2.7 x 10 14 Pa from earlier estimate, and  = 0.60  T c ~ 1.4 x 10 7 K (models predict about 1.6 x 10 7 K) Agreement is fortuitous since the pressure estimate used as P c and the density estimate,, used as  c are both too low by a factor of 100.