Student Solutions Manual for Stars and Stellar Processes 1st Edition by Mike Guidry

Student Solutions Manual for Stars and Stellar Processes 1st Edition by Mike Guidry

Student Solutions Manual for Stars and Stellar Processes 1st Edition by Mike Guidry

Student Solutions Manual: Stars and Stellar Processes Mike Guidry This document gives the solutions for select problems (those marked with ***) at the ends of chapters for the first edition of Stars and Stellar Processes

by Mike Guidry (Cambridge University Press, 2019). Unless otherwise indicated, literature references, equation numbers, figure references, table references, and section numbers refer to the print version of that book.

Student Solutions Manual for Stars and Stellar Processes 1st Edition by Mike Guidry

Student Solutions Manual for Stars and Stellar Processes 1st Edition by Mike Guidry

1

Some Properties of Stars 1.1 Sirius is 8.6 ly = 2.6 pc away, corresponding to a parallax angle p = 1/d( pc) = 1/2.6 = 0.38′′.

1.4 The total energy flux F emitted by a blackbody at temperature T can be expressed as ∫ ∞ 1 2 F = 2π c h dλ.

0 λ 5(ehc/λ kT − 1) Using the substitutions u = hc/λ kT

λ = hc/ukT

the integral can be evaluated giving kT

2

F = 2π c h =

dλ =

4∫ ∞

u2kT

du,

u3 eu − 1

0

hc 2π c2hk4T 4

−hc

du

ζ(4)Γ(4) h4c4 2π 5k4 4 = T ≡ σ T 4, 15h3c2 where the properties of the Riemann zeta function ζ(x) ζ(x) ≡

1 Γ(x)

∫ ∞ 0

ux−1 du eu − 1

and the gamma function Γ(x) have been used, and σ is the Stefan–Boltzmann constant.

1.7 From the following diagram, F

F1 r1

v1

v 2

2

r2

CM

circular motion about the center of mass requires that |F1| =

m1v2

1 =

r1

Gm1m2 (r1 + r2)

2

|F2| =

m2v2

Gm1m2

r2

(r1 + r2)

2 =

2

.

The angular velocity is ω = |ω| = v/r and the period is P = 2π/ω = 2π r/v, from which v2 = 4π 2r2/P2. Insert this into the above equation for |F 1| and solve for P2, 4π 2 2 2 r1 P = (r1 + r2) .

G m2 1

Some Properties of Stars

2

Impose the center of mass condition r1 = m2(r1 + r2)/(m1 + m2) and rearrange to give m1 + m2 =

4π 2 (r1 + r2)3 P2

G

for the sum of the masses.

1.8 In Kepler’s third law the factor 4π 2/G is 4π 2

8

= 5.9152 × 10 gs2cm−3 = 1.0M⊙ yr2 AU−3.

G Thus, if time is measured in years, masses in solar masses, and distances in AU, the factor 4π 2/G = 1 and from Kepler’s law m1 + m2 =

(r1 + r2)3

.

P2 This reduces to P2 = r3 if m1 >> m2 = 1M⊙ and r1 + r2 ≃ r2 ≡ r for circular orbits.

1.13 From the Planck law (1.10) the intensity is Bλ (T ) =

1 2hc2 5 λ ehc/λ kT − 1 .

Find the wavelength corresponding to the maximum by setting the !derivative equal to zero: dB 2hc2 −5(ehc/λkT − 1) + hc ehc/λkT λ kT = 6 = 0. dλ λ (ehc/λ kT − 1)2 Letting x ≡hc/λ kT , this is equivalent to xex/(ex −1) − 5 = 0. This equation doesn’t have a solution in terms of elementary functions; it may be solved numerically (for example, Maple, Mathematica, or Newton–Raphson iteration) to give x ∼ 4.965114. Thus λ = λmax =

hc = 0.201405 xkT

hc kT

=

0.2898 K cm T

= 2.898 × 10 K nm T 6

for the wavelength corresponding to the maximum of the Planck distribution (Wien law).

Student Solutions Manual for Stars and Stellar Processes 1st Edition by Mike Guidry

2

The Hertzsprung–Russell Diagram 2.3 For absolute bolometric magnitude M = −5, the ratio of luminosities is L = 10(M⊙−M)/2.5, L⊙ from which L = 7870L⊙, using the absolute bolometric magnitude M⊙ = 4.74 for the Sun. Solving L = 4πσ R2T 4 for the radius R gives R = R⊙

T⊙ T

2

s L = 329, L⊙

where T is the effective surface temperature and where the luminosity computed above and an effective solar surface temperature T⊙ = 5780 K have been used. From the known radius of the Sun, this gives a radius of about 1.5 AU (approximately the radius of the orbit of Mars) and an average density of about 3.9 × 10−7 g cm−3. The large size and low density and surface temperature indicate that this star is a red supergiant.

2.9 This problem is adapted from a discussion in Carrol and Ostlie [68]. The Balmer absorption line is produced by excitation of neutral hydrogen with the electron initially in the first excited state. The Ca H and K lines are produced by excitation of singly-ionized Ca initially in the ground state. Thus, the number of neutral hydrogens with electrons in the first excited state must be compared with the number of Ca+ ions in their ground state.

In Example 2.3 it was estimated that in the solar surface H+/H≃ 10−4, so the hydrogen may be assumed to be all neutral. The amount of this neutral hydrogen in the first excited electronic state is given by the Boltzmann formula, 2(2)2 −(−3.4−(−13.6))kT n2 g2 −(E −E )/kT −9 2 1 = e = e = 5.5 × 10 , 2 2(1) n1 g1 where an excitation energy of 3.4 eV for the first electronic state and an ionization energy of 13.6 eV have been used, and we have assumed T = 5800 K. Thus, the fraction of the hydrogen capable of producing the Balmer absorption series is n2 = nT

n2 n1 + n2

nI nT

≃

n2 = 5.5 × 10−9. n1

Now let’s do a similar analysis for the Ca ions. From Eq. (2.4) the ratio of singly-ionized to neutral calcium in the solar surface is given by , n+ u+ 5 5040 log = log + log T − E (eV) − log P − 0.179 = 2.66, n0 3

u0

2

i

T

e

The Hertzsprung–Russell Diagram

4

where u+ = 2.30 and u0 = 1.32 have been used, the pressure is assumed to be Pe = 31.6 dyne cm−2 (corresponding to log Pe = 1.5), and T = 5800 K. Therefore, the ratio of Ca II to Ca I is nII n+ = = 457, nI n0 and almost all the calcium is in the first-ionized state (a similar check shows that there is little population of Ca III). The ratio of Ca+ in the first excited state to that in the ground state is n2 g = 2 e−(E2−E1)kT ≃ 3.9 × 10−3, n1 g1 where we’ve used E2— E1 = 3.12 eV, g1 = 2, g2 = 4, and T = 5800 K. Thus, in the solar surface almost all Ca is in the ground state of the Ca + ion. Carrying out a similar calculation as for hydrogen, n1 nT =

≃

n1 n1 + n2 1 1 + n2/n1

nII nT

=

n1 n1 + n2

nII/nI 1 + nII/nI

nII nI + nII

≃ 0.994.

Hence, almost all the Ca in the surface of the Sun is in the right state to produce the Ca K (and by similar analysis the Ca H) line, but only a fraction 5.5× 10−9 of the hydrogen is in the right state to produce the Balmer absorption series. Since the calcium abundance relative to hydrogen abundance in the Sun’s surface is about 2.2 × 10−6, the Ca K line should be about 2.2× 10−6 ≃ 400 5.5 × 10−9 times stronger than the hydrogen Balmer absorption series in the solar spectrum.

2.12 Ionization from the ground state requires 24 eV of energy and for transitions from the ground state to the continuum the largest wavelength that can be absorbed is λ = 1240.8

1 eV 24 eV

nm = 52 nm.

But the Sun produces almost no light at this or shorter wavelengths, so this transition cannot contribute significantly to continuum absorption. For ionization from the first excited state at 20 eV the largest wavelength that can contribute is 1 eV λ = 1240.8 nm = 310 nm, (24 − 20) eV where the Sun does produce some light. However, at solar temperatures the population of the state at 20 eV is very weak: n2 g2 = e−∆E/kT ≃ exp(−20 eV/0.517 eV) = 1.6 × 10−17, n1 g1 where we’ve assumed a temperature of 6000 K and that for purposes of estimation g2/g1 ≃ 1. Thus this transition also makes negligible contribution to the continuum absorption.

Student Solutions Manual for Stars and Stellar Processes 1st Edition by Mike Guidry

5

The Hertzsprung–Russell Diagram

2.15 From the definition nT ≡ n0 + n+ + n++, n0 n0 1 = = n n + nT n0 + n+ + n++ 1+ + ++ n0 n0 1 1 = = , n+ n+ n++ + ++ 1+ n + n n 1+ n 1+ n + 0 0 n0 n+ and similarly for n+/nT and n++/nT.

3

Stellar Equations of State 3.1 The heat capacity is C= At constant volume dV = 0 and

δQ δT

=

CV =

dU + PdV dT ∂U ∂T

.

.

V

Likewise, at constant pressure CP =

∂U ∂V +P ∂T P ∂T

∂V ∂T

= CV + P P

, P

where we’ve used that for an ideal gas U = constant× NkT and ∂U = ∂T P

∂U ∂T

= CV. V

But also for an ideal gas PV = NkT so that P

Nk ∂V =P = Nk, ∂T P P

implying that CP = CV + Nk for an ideal gas.

3.6 This problem is adapted from a derivation in Prialnik [176]. Consider an elastic collision with a surface: z y x

The change in momentum is entirely normal to the surface. By Newton’s 2nd law F = d p/dt and by Newton’s 3rd law the surface recoils; the impulse delivered to the surface in the interval ∆t is F∆t = −∆ p = 2pzkˆ = 2p cosθ, where kˆ is a unit vector along the z axis, normal to the surface. Let n(θ, p)dθ dp be the number density in a beam of particles incident on the surface with momenta in the intervalp to p + dp and direction in the interval θ to θ + dθ. Assuming an isotropic particle distribution, the number of particles in the solid angle dΩ = 2π sinθ dθ is proportional to the solid 6

Student Solutions Manual for Stars and Stellar Processes 1st Edition by Mike Guidry

Stellar Equations of State

7

angle, so n(θ, p)dθ dp

=

n(p)dp

dΩ 4π

2π sinθ dθ = 1 sinθ dθ, 2 4π

=

which implies that

1

n(θ, p)dθ dp =

n(p) sinθ dθ dp.

2 In the time interval ∆t the number of particles striking the surface will be ∆N = (number density) × (volume) = (n(θ, p)dθ dp)(v∆tdS cosθ), where dS is the area of incidence for the beam on the surface. Hence the momentum transfer in the time ∆t is δ p = (n(θ, p)dθ dpv∆tdS cosθ)(∆p) = (n(θ, p)dθ dpv∆tdS cosθ)(2p cosθ) = ( 21 sinθ dθ n(p)dpv∆tdS cosθ)(2p cosθ) = p sinθ cos2 θ vn(p)dpdθ∆tdS.

The differential pressure is the momentum transfer per unit time per unit surface area, δp

p sinθ cos2 θ vn(p)dpdθ∆tdS = vpn(p)dp sinθ cos2 θ dθ, ∆tdS ∆tdS and the total pressure is obtained by integrating over all angles and all momenta, dP =

=

∫ ∞

P=

0

∫ π/2

vpn(p) dp

0

∫

2

sinθ cos θ dθ =

1 ∞ vpn(p) dp, 3 0

which is Eq. (3.2).

3.7 Assuming classical velocities with p = mv and inserting the velocity distribution (see Eq. (3.4)) 3/2 2 m n(v)dv = n e−mv /2kT 4π v2dv 2π kT in the pressure integral (3.2) gives m m P=n 3 2π kT

3/2

(4π)

∫ ∞ 0

2

v 4e −mv /2kT dv.

Make the change of variables x = v2 and use that ∫∞ 0

xn e−ax dx =

Γ(n + 1)

5

3

1/2

Γ( 2 ) = 4 π

an+1

to show that the portion within square brackets in the above expression for P is just kT , proving that P = nkT . (See Problem 3.18 for a more general derivation.)

3.8 (a) The energy density is given by ∫ ∞

u=

0

n(ν)hν dν =

8πh c3

∫ ∞ 0

ν3 ehν/kT − 1

dν

Stellar Equations of State

8

where Eq. (3.69) was used. Changing variables to x = hν/kT gives u=

8πh

4∫ ∞

kT

c3

h

0

where the definite integral

∫∞ 0

x3 e −1 x

dx =

8π 5k4

4

15h3c3

4

T ≡ aT ,

x3 π4 dx = ex − 1 15

was used and the radiation density constant a is defined as a=

8π 5k4 15h3c3

= 7.5657 × 10

−15

erg cm−3 K−4,

which is related to the Stefan–Boltzmann constant σ by a = 4σ/c. (b) Requiring that n(ν)dν = n(p)dp and using E = pc = hν to change variables in Eq. (3.69) gives 8π p2 n(p)dp = 3 pc/kT dp. h e −1 (c) Inserting the result from part (b) into the pressure integral (3.2) yields P=

1

∫ ∞

3

0

vpn(p) dp =

8π 3c2h3

∫ ∞

(pc)3

0

epc/kT − 1

dp,

where pv = pc was used. Change variables from p to x = pc/kT and use ∫∞ 0

x3 π4 dx = e −1 15 x

to evaluate the integral, giving P=

8π(kT )4 3h3c3

∫ ∞ 0

x3 ex − 1

dx =

1

8π 5k4

3

15h3c3

4

1

4

T = 3 aT ,

Comparing P = 13aT 4 with u = aT 4 from part (a), the equation of state for the photon gas is P = 13 u.

3.10 For adiabatic processes δQ = 0 and the first law (3.7) reduces to dU = −PdV . But for an ideal gas PV = NkT so dU PdV PdV + VdP = NkdT = Nk = −Nk = (1 − γ)PdV CV CV where dU = CVdT was used in the second step, dU = −PdV in the third step, and in the final step γ = CP/CV and CP = CV + Nk for an ideal gas have been used. Rearranging this expression gives dV dP γ =− .

V P By substitution and application of the ideal gas law it is clear that P1−γ T γ = constant

PV γ = constant

TV γ−1 = constant

Student Solutions Manual for Stars and Stellar Processes 1st Edition by Mike Guidry

Stellar Equations of State

9

are solutions. The adiabatic speed of sound is vs =

B ρ

1/2

∂P ∂V

B ≡ −V

,

where B is the bulk modulus. Taking the adiabatic equation of state in the form P = KV −γ where K is a constant, and evaluating the partial derivative gives for the bulk modulus B = γP. Thus the adiabatic speed of sound is vs = (γP/ρ)1/2.

3.11 For polytropes δQ/δ T = c, where c is a constant. From the first law of thermodynamics (3.7) and Eq. (3.5) in differential form, dU = CVdT (see Problem 3.9), cdT −CVdT = PdV. Inserting the ideal gas law PV = NkT and using that CP = CV + Nk for an ideal gas gives dT

= (1 − γ)

T

dV

CP − c .

CV − c

γ≡

V

Integrating both sides of the differential equation gives ln T = lnV 1−γ + ln c1, where ln c1 is an integration constant. Thus the differential equation has a solution TV γ−1 = c1.

Other forms of this solution may be obtained from the ideal gas law. Substituting for T from the ideal gas law gives a solution PV γ = c2, where c2 is a constant, and substituting for V from the ideal gas law gives a solution T γ P1−γ = c3, with c3 another constant.

Summarizing, the differential equation has solutions TV γ−1 = c1

PV γ = c2

T γ P1−γ = c3.

For constant entropy δQ = TdS = c = 0 and γ = CP/CV, which is the adiabatic γ. If instead the process is isothermal, δ T /δQ = 1/c = 0, which implies that c → ∞ and that γ = 1.

3.14 For pure radiation P = 13aT 4 and U /V = u = aT 4 = 3P. From the definitions (3.39), Γ2 Γ2 − 1

=

dP/P dT /T

Γ1 =

dP/P dρ/ρ

=−

dP/P dV /V

Γ2 − 1 Γ3 − 1 = , Γ2 Γ1

where derivatives are evaluated at constant entropy and ρ ∝ V −1 was used. The first of these and dP dT =4 P T

Stellar Equations of State

10

following from P = 1 T 4 give Γ2 = 4 . From the expression for U /V and the first law applied 3

3

to adiabatic processes

dU = 4aT 3VdT + aT 4dV = −PdV. But from the expression for P we have aT 4 = 3P and 4aT 3dT = 3dP. Substituting gives dP 4 dV =− , P 3 V from which we find dP/P 4 Γ1 = − = . dV /V 3 Finally, the relationship Γ2 − 1 Γ3 − 1 = Γ2 Γ1 and the substitution Γ1 = Γ2 = 4 gives Γ3 = 4 .

3

3

3.16 This example is worked out in Vol. 1 of Bowers and Deeming [55]. For a gas plus radiation mixture the internal energy is U = CVT + aT 4V . Differentiate and use the first law assuming an adiabatic system to write dU + PdV = CVdT + 4aT 3VdT + aT 4dV + PdV = 0.

Rewrite the term CVdT as CVdT = CV

β PV

dT =

CV

β PVd ln T =

CP −CV

NkT

β PV γ −1

d ln T,

where we’ve used Eq. (3.75) and Nk = CP — CV

γ=

CP

dT

CV

T

= d ln T,

and rewrite the term aT 4dV as aT 4dV = 3(1 − β )PdV using Eq. (3.72) and Eq. (3.74). Now substitute and rearrange the original expression to obtain [β + 12(γ − 1)(1 − β )] d lnT + (γ − 1)(4 − 3β ) d lnV = 0, which may be rewritten as d ln T d ln T (γ − 1)(4 − 3β ) — = = , d lnV d lnρ β + 12(γ − 1)(1 − β ) which is Eq. (3.76). This result can be used to evaluate the Γi as a function of β and γ. For example, from the preceding expression Γ3 = 1 +

d ln T (4 − 3β )(γ − 1) = 1+ , d lnρ β + 12(1 − β )(γ − 1)

which is Eq. (3.79).

Student Solutions Manual for Stars and Stellar Processes 1st Edition by Mike Guidry

Stellar Equations of State

11

3.17 From Eq. (3.40), for a gas under adiabatic conditions PV Γ1 = a, where a is a constant.

The density ρ is proportional to V −1, so this may also be written as P = bρ Γ1 , where b is another constant. The adiabatic speed of sound is s s s dP q b ρ Γ1 Γ1 P v= = bΓ1ρ Γ1−1 = Γ1 , = ρ dρ ρ 1/2

and thus is proportional to Γ1 .

3.18 Let’s follow the solution of Phillips [169]. From the first part of Eq. (3.63) and Eqs. (3.54) and (3.46), the internal energy of the gas is U=

∫ ∞ 0

3

εp f (εp)g(p) dp

2

2

2 2

2 4

εp = p c + m c .

g(p)dp = 4π gs(V /h )p dp

The gas pressure may be defined by ∂U P=− =− ∂V

∫∞ 0

dεp f (εp)g(p) dp = − dV

∫ ∞ 0

dεp dp f (εp)g(p) dp, dp dV

where we’ve assumed εp to be a function of p, and p to be a function of V . By the uncertainty principle, in a volume V = L3 the momentum is p ∝ L ∝ V −1/3, so dp

1

≃ − 3V

−4/3

dV and from the energy expression

=

−V −1/3 3V

≃−

p 3V

,

dεp 1 2pc2 pc2 = = ε p ≃ vp, dp 2 (p2c2 + m2c4)1/2 where vp is the speed of a particle with momentum p. Inserting these results in the above integral for P gives ∫ 1 ∞ vp p f (εp)g(p) dp.

P= 3V 0 Now insert a Maxwell distribution f (εp) = exp[ − (εp− µ)/kT ] in the pressure integral to give ∫ ∞ 4π µ/kT 3 −ε /kT gs p e p vp dp.

P= 3e 3h 0 To evaluate the integral note that 1 −εp/kT 1 −εp/kT e dε p = − e v p dp, kT kT where dεp = vpdp from earlier has been used. Thus d(e−εp/kT ) = −

∫ ∞ 0

3 −ε p/kT

p e

vp dp =

−1 kT

∫ ∞

3

p d(e

0

−ε p /kT

)

p=∞

3

= −kT p e −ε /kT p=0 + 3kT = 3kT

∫ ∞ 0

∫ ∞

p

e −ε p/kT p2 dp,

0

e−ε p/kT p2 dp

Stellar Equations of State

12

where an integration by parts and that εp = ∞ if p = ∞ were used in the second line.

Inserting this result in the earlier expression ∫ for P gives kT µ/kT ∞ −ε /kT P= e e p g(p) dp.

V 0 But the total particle number is N=

∫ ∞ 0

f (εp)g(p) dp = e

µ/kT

∫∞ 0

e

−ε p/kT

g(p) dp.

Therefore, a comparison of the previous two expressions gives the ideal gas law NkT P= = nkT, V where n = N/V is the particle number density. Thus the ideal gas law is valid for arbitrary particle velocity, provided that the gas has a Maxwell–Boltzmann distribution.

3.20 By requiring that ne = ∑i nei = ρ NA/µe and utilizing Eq. (C.3), !−1 ! −1 1 Xi y Z y Z Y ≃ µe = ∑ i i i = ∑ i i , Ai i X + Y /2 + δ i where in the last step complete ionization (yi = 1) has been assumed, X is the mass fraction of hydrogen,Y is the mass fraction of helium, and δ is the contribution of metals. Assuming the metal concentration to be small, δ may be neglected, X + Y = 1, and 2 2 µe ≃ ≃ .

2X + Y X + 1 For the ZAMS concentrations (3.28) this gives µe ≃ 1.15.

3.21 The result follows from P=

NkT T CP CV T = (CP −CV) = −1 = (γ − 1)u, V V CV V

where Eqs. (3.11), (3.8), and (3.12), and that the heat capacities are constant for ideal gases, have been used.

Student Solutions Manual for Stars and Stellar Processes 1st Edition by Mike Guidry

4

Hydrostatic and Thermal Equilibrium 4.1 With no pressure gradient to oppose gravity, each mass element is accelerated by g(r) = Gm(r)/r2. Assume a shell initially at rest at r = r0 and enclosing mass m0, which remains constant in collapse. Conservation of energy requires 1 ms v 2 = 2

Gm0ms Gm0ms − , r r0

where ms is the shell mass, implying that s dr = ± 2Gm0 dt

1 r

−

1

,

r0

where v = dr/dt was used. Then the free-fall timescale to collapse to the center of the sphere is ∫ 0 dt tff = dr. r0 dr Substituting the preceding expression for dr/dt and changing variables to x = r/r0 allows the integral to be put in the form s ∫ 1 1/2 r03 x dx, tff = 2Gm0 0 1 − x which can be integrated using the substitution x = sin2 θ to give s s 3π g cm−3 −5 6 7 10 ≃ . × tff = yr, 32Gρ¯ ρ¯ where ρ¯ = m0/( 4 π r3) is the average density at the beginning of collapse.

3

0

4.3 Define the nth segment of the random walk by a vector→λn so that the vector defining the final point relative to the initial point after N steps is →L =→λ1 +→λ2 + . . . +→λN, as illustrated in the following diagram.

...

L 13

Hydrostatic and Thermal Equilibrium

14

Then the square of the vector L is → → → → 2 L2 = λ 21+ λ 2 + 2 . . . + λ +N2( λ1 · λ2 + λ1 · λ3 + . . .).

But for many steps the terms in parentheses must average to zero since the direction is random. Thus, 2

2

2

2

2

⟨L ⟩ ≃ λ1 + λ2 + . . . + λN ≃ Nλ , if each step is replaced by one of average length λ. To escape from the Sun requires that 2

2

⟨L ⟩ ≃ R⊙

−→

N≃

R2

⊙

λ2

.

The time for each step is λ/c, implying that the total time for diffusion to the surface is λ

λ R2

R2

⊙

τdiff ≃ N = ≃ ⊙. c c λ2 cλ Inserting an average mean free path of~ λ 0.1 − 0.5 cm gives diffusion times in the range of 10,000 to 100,000 years.

4.5 For an ideal gas, Eq. (3.3) implies that P/ρ = kT /µ. Therefore, 3P ρ

=

3kT µ

=

2( 3 NkT ) 2

Nµ

=

2U Nµ

=

2U M

,

where we’ve used that for an ideal gas U = 32NkT and that the total mass M for the star is just Nµ. Therefore, ∫M

3P

0

ρ

dm =

2U

∫ M

M

0

dm = 2U

for the right side of the virial theorem in Eq. (4.19).

4.8 This solution follows the discussion in Hansen, Kawaler, and Trimble [107]. (a) For an ideal gas the energy density is u = 3 nkT = 3 ρ NAkT /µ, so the total internal 2

energy is U = uV =

3ρVNAkT 2µ

2

=

3MNAkT 2µ

,

where M = ρV was used. (b) For a sphere of uniform density the gravitational potential energy is Ω = − 35GM2/R, so by the virial theorem U=−

Ω

3GM2 = 10R .

2

Student Solutions Manual for Stars and Stellar Processes 1st Edition by Mike Guidry

Hydrostatic and Thermal Equilibrium

15

1011

109

Relativistic

Temperature (K)

Relativistic

107

Electron degeneracy 105 10-4

.

F i g . 4.1

10-2

100

102

104

106

108

Density (g cm-3 )

Temperature versus density computed in Problem 4.8 assuming µ = 1 and µe = 1. Solid curves correspond to stars of different masses. Regions of high temperature and high density where a relativistic description is necessary are indicated. Stars to the right of the dotted curve are expected to be electron-degenerate. The break in this curve corresponds to the transition from non-relativistic to relativistic degenerate behavior. In a realistic calculation this break would be smoothed out. In stars above the dashed curve the radiation pressure is expected to be larger than the gas pressure.

Equating the two previous expressions for U and solving for T gives µGM G T= = (4π/3)1/3 µ M2/3ρ 1/3 5NAkR 5NAk 6

= 4.09 × 10 µ

M M⊙

2/3

ρ g cm−3

1/3

K,

where R = (M/ 34 πρ)1/3 was used in the second step. Some temperature versus density curves computed using this formula are plotted in Fig. 4.1 [this document]. (c) –(d) Special relativity will be important if T is high enough that average gas velocities approach c, which will be true if kT~mc2. For the electrons (typically the lightest particles and therefore the fastest), mec2 9 Trel ∼ k ∼ 5.9 × 10 K, where mec2 = 511 keV has been used for the rest mass of the electron. Special relativity is also important for the electrons if the density is high enough that the Fermi energy is comparable to the Compton wavelength [see Eq. (3.62)], since then the average electron velocities will also become large by uncertainty principle arguments. In Ch. 3 it was found

Hydrostatic and Thermal Equilibrium

16

that this typically occurs for electrons when densities exceed ~ 106 g cm−3. Regions expected to require a relativistic description because of high temperature and/or high density are indicated in Fig. 4.1. Degeneracy effects are likely if the Fermi energy is comparable to kT for the lightest fermions in the gas. The Fermi energy can be expressed generally as εf = mc2[(1 + x2)1/2 − 1], where x ≡pf/mc. As shown in Hansen, Kawaler, and Trimble Ch. 3, this leads to typical critical densities for degeneracy ρdeg ∼ 6 × 10−9µeT 3/2 g cm−3

(non-relativistic)

ρdeg ∼ 4.6 × 10−24µeT 3 g cm−3

(relativistic)

These equations are plotted as dotted curves in Fig. 4.1. Thus, even in high-temperature environments densities ∼ 106 g cm−3 or larger tend to produce degenerate electron gases. (e) For an ideal gas P = ρ kT /µ and for radiation P = 13aT 4. Equating the two pressures and solving for T gives Tcrit =

3k µa

1/3

ρ 1/3 ≃ 4 × 107

ρ 1 g cm−3

1/3

K

for the temperature at which the gas and radiation pressures are equal. The dashed curve in Fig. 4.1 corresponds to equality of radiation and gas pressure. Equating the above expression with the earlier general expression for temperature and solving for the mass giving T = Tcrit yields M M⊙

∼

10 µ

3/2

~ 30,

where the last step has used the assumption µ ∼ 1 amu. Thus radiation pressure is expected to exceed gas pressure for stars having masses of~ 30M⊙ or more, as suggested by the dashed curve in Fig. 4.1.

4.9 In the Eulerian picture a fluid quantity is determined at a fixed point in space and the time derivative of a fluid quantity q is just the partial derivative with respect to t, with spatial coordinates fixed: ∂q

= Lim

q(x, y, z,t + ∆t) − q(x, y, z,t)

. ∂t ∆t In the Lagrangian picture the corresponding time derivative is the total derivative: ∆t→0

Dq dq Lim q(x + ∆x, y + ∆y, z + ∆z,t + ∆t) − q(x, y, z,t) ≡ ≡ ∆t→0 Dt dt ∆t ∂ q dy ∂ q ∂ q dx ∂ q dz = + + + ∂ t ∂ x dt ∂ y dt ∂ z dt ∂q ∂q ∂q ∂q = + vx + vy + vz, ∂t ∂x ∂y ∂z where the local fluid velocity is v = (vx, vy, vz). Thus the formal relationship between the

Student Solutions Manual for Stars and Stellar Processes 1st Edition by Mike Guidry

Hydrostatic and Thermal Equilibrium

17

Eulerian (partial) derivative ∂/∂ t and the Lagrangian (total) derivative D/Dt is D Dt where ∇ =

=

∂ ∂t

+ v · ∇,

∂ ∂ ∂ , , . ∂x ∂y ∂z

4.10 The pressure of an ideal gas is 1 P = n⟨pv⟩, 3 where the brackets denote an average over particles in the gas. The relativistic energy E is defined through E2 = p2c2 + m2c4.

In the ultrarelativistic limit → v c, so that pc >> mc2 and E≃ pc, while the nonrelativistic limit (expand assuming pc << mc2) is given by E = mc2 +

p2 , 2m

where v = p/m. Identifying n ⟨pc ⟩as the ultrarelativistic kinetic energy density and n ⟨p2/2m ⟩ as the nonrelativistic kinetic energy density, for ultrarelativistic particles 1 1 Ekin 1 p = n⟨pc⟩ = = εkin, 3 3 V 3 and for nonrelativistic particles 1 p= n 3

2 = n 3

p2 m

p2 2m

=

2 Ekin 2 = εkin, 3 V 3

where Ekin is the total kinetic energy, V is the volume, and εkin is the kinetic energy density.

4.16 From Eqs. (3.3), (3.36), and (3.70), all three equations of state can be expressed in the form P = AραT β , where α ≥ 0 and β ≥ 0. Generally, dP

=

P

1

∂P

P

∂ρ

dρ +

∂P ∂T

dT

.

But from P = Aρα T β we have that P−1

∂P α = ∂ρ ρ

P−1

∂P β = , ∂T T

which leads to dρ dT dP +β , =α P T ρ where α and β are non-negative.

4.20 This problem is based on material discussed in Prialnik [176]. Let δ Q be the change

Hydrostatic and Thermal Equilibrium

18

in heat, q the rate of energy release per unit mass from nuclear reactions, and L(m) the luminosity at the Lagrangian mass coordinate m. Then δ Q = qdm δ t + L(m)δ t − L(m + dm)δ t.

Expanding using L(m + dm) = L(m) +

∂L ∂m

dm then gives

δ Q = qdmδ t −

∂L

dm δ t, ∂m which is Eq. (4.23). Heat is absorbed if δ Q > 0 and emitted if δ Q < 0. Utilizing dV /dm = ρ−1, the work is dV dm dm

δW = −Pδ(dV ) = −Pδ

= −Pδ

1 dm, ρ

which is Eq. (4.24). If δ dV < 0 (compression) energy is added to the shell and if δ dV > 0 (expansion) energy is removed from the shell.

4.21 Insertion of Eqs. (4.23) and (4.24) into Eq. (4.22) gives 1 ∂L dm = q − dmδ t ρ ∂m

dm δ u + Pδ

Dividing by δ tdm and taking the limit δ t → 0 then gives 1 ∂L = q− , ρ ∂m

du d +P dt dt which is Eq. (4.25).

4.22 Let’s evaluate the terms in Eq. (4.27), ∫ M 0

∫

M du d P dm + dt dt 0

1 ρ

dm =

∫ M 0

∫M

qdm −

0

∂L dm, ∂m

following a discussion in Prialnik [176]. In the first term on the lefthand side m and t are independent variables so the order of differentiation and integration may be switched to give ∫ M

∫

du dm = d M d udm = U ≡ U˙ , 0 dt dt 0 dt where U is the total internal energy. For the last two terms on the right side of the first equation, L0 ≡

∫ M

∫M

L≡

∂L

dm, ∂m where L is the total luminosity of the star and L0 is the part of the luminosity produced by nuclear reactions [see Eq. (4.26)]. In the second term on the lefthand side of the first equation d dt

0

qdm

0

1 d ∂V ∂V˙ = = , ρ dt ∂ m ∂m

Student Solutions Manual for Stars and Stellar Processes 1st Edition by Mike Guidry

Hydrostatic and Thermal Equilibrium

19

since dV /dm = ρ−1. Inserting this in the second term gives ∫ M

∫ M ∂V˙ M ˙ dm = PV 0 − dm = V˙ dP, ∂m 0 0 0 where an integration by parts has been used in the last step. But V˙ = 4π r2r˙ since dV = 4π r2dr, and the first term above on the right side vanishes because V˙ = 0 at the center and P = 0 at the surface of the star. Thus,

P

∫ M

1 ρ

d dt

P

∫

M 0

P

d ρ

dt

1 dm = 4π ∫

M 0

∂P r2r˙ dm. ∂m

Combining the preceding results gives ∫ M

2 ∂P 4π r r˙ dm = L0 − L −U.˙ ∂m 0 Next, multiply Eq. (4.14) by r˙ and integrate over the Lagrangian coordinate m for the entire star,

—

∫ M 0

r˙r¨dm = −4π

∫M

∂P r2r˙ dm − ∂m 0

∫ M 0

Gmr˙ dm. r2

For the integral on the left side, ∫ M 0

r˙r¨dm =

∫M 0

∂ ∂t

1 2 2 r˙

dm =

d dt

∫ M 0

1 2 ˙ 2 r˙ dm = K,

since the total kinetic energy of the star is ∫ M

1 r˙2 dm, K= 2 0 from Eq. (4.31). Furthermore, for the gravitational term ∫ M

∫

∫

M Gmr˙ d 1 d M Gm ˙ Gm dm = dm = dm = −Ω, dt r dt 0 r 0 r2 0 where Ω is the total gravitational energy defined in Eq. (4.20). Combining the preceding three equations gives

—

∫M

r2 r˙∂ P dm = K˙ + Ω. ˙ ∂m Thus from this equation and the 7th equation above, −4π

0

U˙ + K˙ + Ω̇ = L0 − L, which is Eq. (4.29).

5

Thermonuclear Reactions in Stars 5.1 The mass excesses for 1H and 4He are 7.28899 MeV and 2.42475 MeV, respectively, so the mass difference between 41H and 4He is 26.73 MeV/c2 or 0.0287 amu. The mass of four hydrogens is 4 × 1.008 amu = 4.032 amu = 6.6905 × 10−24 g, so the fraction of mass converted to energy is 0.0287 amu/4.032 amu = 0.0071, and since 6.6905×10−24 g of hydrogen yields 26.73 MeV of energy, from one gram of hydrogen we may obtain 1 24 18 × 26.73 MeV ≃ 4 × 10 MeV = 6.4 × 10 erg −24 6.6905 × 10 if the hydrogen is converted completely to helium.

5.3 Since the metal concentration is small it may be neglected in first approximation and from Eq. (3.19) assuming complete ionization, 1 µ≃ .

2YH + 3YHe When the Sun first formed YH = XH = 0.71

YHe = 14 XHe = 0.0675,

which gives µinit ≃ 0.62. At the present time YH = XH = 0.34

YHe = 14 XHe = 0.16

which gives µnow≃ 0.86. Equivalently, we may use the equations of Section 3.4.3 to reach the same conclusions.

5.4 Find the maximum E0 by taking the derivative of the factor given in Eq. (5.27) and set≃0 31 keV. ting it equal to zero. Inserting the numbers for the 12C(p, γ)13N reaction gives E If you plot the height of Eq. (5.27) with respect to temperature, you will find that it depends strongly on the temperature. Assuming the height to vary as a power law in temperature, h ∼ T n , you can extract n by comparing the logarithmic derivatives (dh/dT )/h for the actual and power law curves. You should find that the exponent n lies in the range 15–20.

This is not surprising, since this example corresponds to a reaction associated with the CNO cycle and the rates are very sensitive to temperature in that case.

5.5 Locate the maximum of the Gamow peak by setting the first derivative equal to zero.

Then approximate the Gamow peak by a gaussian having the same maximum and the same curvature at the maximum (match the second derivatives of the actual Gamow peak 20

Student Solutions Manual for Stars and Stellar Processes 1st Edition by Mike Guidry

Thermonuclear Reactions in Stars

21

and the gaussian at the maximum). The width of the gaussian gives Eq. (5.29). Then use this gaussian approximation to perform the integral in Eq. (5.26), with the result given in Eq. (5.31).

5.8 From Eq. (5.40) the temperature exponent is ∂ lnε ∂ ln T

ν=

ρ

=

∂ε ∂ T9

T9 ε

ρ

,

where T9 is the temperature in units of 109 K. Evaluate using the energy production rate (5.38) and the approximate reaction rate (6.1), and convert the final result to units of T6 = 103 × T9 to give 2 ν = 11.26T6 −1/3 − .

3 Inserting a temperature of T6 = 15 for the solar core gives ν ∼ 3.9.

5.9 Start at any point in the cycle and go all the way around once, keeping track of species changes at each step.

5.12 (a) Assume that the overall rate of solar energy production is governed by the rate of the initial (slowest) reaction in the PP chains, p + p→2H + νe + β +. From the second term of Eq. (D.4) the rate in s−1 for all two-body reactions is Y˙i = ∑η i ρ NA⟨ jk⟩YjYk = ∑η i ρ RjkYjYk, jk

jk

jk

jk

where the quantity Rjk ≡ NA⟨ jk⟩ is the rate of Eq. (D.3) provided by the ReacLib library in units of cm3 mol−1 s−1. For the 2-body reaction p + p → 2H + νe + β + the above equation reduces to the single term Y˙p = −ρYp2Rpp. Let’s take for conditions in the core of the present Sun (where hydrogen is depleted from its ZAMS value) T9 = 0.015

ρ = 150 g cm−3

Yp = 0.5.

Then from tabulated values (for example, interpolation in Fig. 6.1) the rate for the ratecontrolling step is about Rpp ∼ 8.8 × 10−20 cm3 mole−1 s−1.

Inserting these values gives a rate Y˙p = 3.3 × 10−18 s−1.

The rate per unit volume is then this quantity multiplied by the number density np = NAρYp ∼ 4.5 × 1025 cm−3, which gives npY˙p = 4.5 × 1025 MeV s−1cm−3,

Thermonuclear Reactions in Stars

22

and the power per unit volume is then P/V = npY˙pQ ≃ 594 W m−3, where we have assumed ~ Q 25 MeV for the PP chains. Alternatively, Eq. (6.1) yields directly a rate rpp ≃ 1.19 × 108 s−1 cm−3 = 2.4 × 1014 s−1 m−3 and a power per unit volume P/V = rpp Q = 476 W m−3, with the difference in the two estimates arising from somewhat different rates deduced from two different sources. (b) The observed solar luminosity is L⊙ = 3.828 × 1033 erg s−1.

Assuming this to be produced in the inner 10% of the solar radius of R⊙ = 6.96× 1010 cm gives an active volume of 1.41×1030 cm3 for the energy producing region. Dividing the luminosity by this volume gives a power production rate of P/V = 271 W m−3, which is in rough agreement with the preceding estimates.

Thus we may conclude that the rate of solar energy production by the PP chains is equivalent to that of only several 100-watt light bulbs distributed over each cubic meter of the solar core, indicating graphically that the Sun is a very low-density power source.

5.13 From mass tables the Q-value for the reaction 12C(p, γ)13N is Q = ∆p + ∆12C − ∆13N = (7288.969 + 0 − 5345.456)keV = 1.9435 MeV, where ∆ denotes mass excesses. From Eqs. (5.32) and (5.33), the cross section will be strongly-peaked at a center of mass energy [see also Fig. 6.5(b)] E = ECM = E∗ − Q = (2.37 − 1.9435) MeV = 426.5 keV because of resonance with the excited state in 13N at E∗ = 2.37 MeV. The corresponding laboratory proton energy is 1 + 12 ELab = ECM = 462 keV, 12 which corresponds to the strong peak in Fig. 5.8.

Student Solutions Manual for Stars and Stellar Processes 1st Edition by Mike Guidry

6

Stellar Burning Processes 6.1 These derivations follow the presentation in Phillips [169]. From Eq. (3.55), the chemical potential for the non-relativistic case is gsnQ n

µ = mc2 — kT ln

2πmkT

nQ ≡

3/2

.

h2

The condition for concentration equilibrium in 8Be ↔ 24He is µ8 = µ4 + µ4, where the mass number of the isotope is used as a label. Substitute the explicit form from above for µi into this equation, rearrange, take the exponential of both sides, and substitute (i)

nQ =

2πmikT h2

3/2

to give n8 = n42

h2 π m4kT

3/2

eQ/kT .

Taking the inverse of both sides and changing notation, this is the nuclear Saha equation (6.4). For the equilibrium 4He + 8Be↔ 12C∗ with the 12C excited state, an analogous derivation yields n∗12 3 3/2 = n4n8 2

h2 2π m4kT

3/2

∗

2

e−(m12−m4−m8)c /kT ,

which may be combined with the preceding result to give Eq. (6.5). These considerations determine the dynamical equilibrium 4He + 4He + 4He ↔ 4He + 8Be ↔ 12C∗.

Because the probability for decay to the 12C ground state is very small compared with the probability for decay back to 4He + 8Be, the decay of 12C∗ to the ground state may be assumed to be a small perturbation on the previously-derived equilibrium and the rate of production for the 12C ground state may be written as n∗12 dn12 dt

∼

, τ(12C∗ → 12C)

where τ is the mean life for the decay and n∗12 is assumed to be well approximated by the previous equilibrium equations. Inserting the equilibrium results derived above for n∗12 then gives Eq. (6.9).

23

Stellar Burning Processes

24

6.2 The nuclear statistical equilibrium (NSE) equation (6.4) may be written as 2

3/2

T K

nα = 5.31 × 1020 n(8Be)

3/2 1.065/T9

= 1.68 × 1034T9

e

9

e1.065×10 (K/T ) cm−3 cm−3,

after evaluating the constants and using Q =—91.78 keV. From Eq. (3.17), pure helium with a mass density of ρ≃ 106 g cm−3 implies an α-particle number density nα = 1.5× 1029 cm−3, if the 8Be concentration is assumed small. Thus, from the NSE equation n(8Be) = 6.7 × 10−9, nα which implies that n(8Be) ≃ 1021 cm−3 (justifying our earlier assumption that the abundance of 8Be is small).

6.4 This is a variation on Problem 5.7 and illustrated by the solution of that problem and the numerical solution in Box 6.1.

Student Solutions Manual for Stars and Stellar Processes 1st Edition by Mike Guidry

7

Energy Transport in Stars 7.1 Equations (7.4), (7.9), the Thomson cross section (7.10)–(7.11), and the assumption of complete ionization permit the ratio (7.13), Kr KT √ Pr mec2 5/2 ≃ = 3Z , Ke Ke Pe kT to be derived. Assuming a hydrogen plasma with temperature 6 × 106 K and a density of ρ = 1.4 g cm−3 gives KT 5 ≃ 2 × 10 Ke for the Sun. 7.2 From Eq. (3.18), the number density ni in terms of the mass fraction Xi is ni =

ρXi .

AiMu

Neglect metals as a first approximation (assume only hydrogen and helium). Let X ≡ XH and Y ≡XHe. Assuming complete ionization, the electron density from a species i is Zini.

Thus the total electron density is the sum of contributions from hydrogen and helium, ρY ρ ρX + ≃ (X + 1), ne ≃ 2Mu Mu 2Mu where metals were ignored so X + Y = 1. If metals are neglected the number density of ions is the sum of the number densities for hydrogen and helium, ρX ρY ρ 1 nI ≃ + = 2X + Y , Mu 4Mu 2Mu 2 which could be reduced further since with our assumptions X + Y ∼ 1.

7.4 This problem is adapted from a discussion in Ref. [169]. Assume a blob and surrounding medium of uniform composition as in the following figure: T + T P + P  + 

Gravity

r + r

r

25

T, P, 

T, P, 

Energy Transport in Stars

26

with the blob in equilibrium with its surroundings at radius r. What happens if the blob is displaced to r + ∆r? In accordance with the Schwarzschild criterion, the blob is assumed to always be in approximate pressure (and concentration) equilibrium, so that δ P = ∆P, and it is assumed that the timescale for vertical motion of the blob is sufficiently fast that little transfer of heat occurs between the blob and the surround medium as it rises from r to r + ∆r (adiabatic conditions). For an adiabatic ideal gas, P ∝ ργ , from which δρ 1 δP = . ρ γ P The blob will continue to rise at r + ∆r if δρ < ∆ρ, since it will be buoyant in the surrounding medium. Thus, convective instability is implied if ρ δP γ P

< ∆ρ.

From the ideal gas law, ∆P ∆T − , P T

∆ρ = ρ

which may be combined with the preceding equation to give 1 δP γ P

<

∆P

−

P

∆T T

.

Utilizing the assumption δ P = ∆P and reverting to differential notation, this condition for convective instability may be written finally as dT 1 < 1− dr γ

T dP , P dr

which is Eq. (7.33).

7.5 Assuming an ideal gas the adiabatic temperature gradient is dT dr

= 1− ad

1 γ

T dP cV = 1− P dr cP

T dP = P dr

cp − cV cP

T dP , P dr

where the specific heats cp and cv, which are the heat capacities per unit mass, cp =

CP

cV =

ρV

CV ρV

have been introduced and γ = CP/CV = cP/cV has been used. From Eq. (3.11), cp − cv =

CP ρV

−

CV ρV

=

Nk ρV

.

Also, from Eq. (4.10), T dP P dr

= −g ρ

T P

.

Student Solutions Manual for Stars and Stellar Processes 1st Edition by Mike Guidry

Energy Transport in Stars

27

Therefore, inserting this and the above expression for cP − cV gives NkT g g dT =− =− , dr ad PV cP cP where the ideal gas law (3.3) was used in the last step.

7.7 Use the Standard Solar Model given in Table 10.1 for the required data at R = 0.7R⊙ and assume that all energy is produced within this radius so that L(r) is just the surface luminosity. The specific heat (heat capacity per gram) is determined as cP =

5 NAk ≃ 3.47 × 108 erg K−1 mole−1, 2 µ

assuming a mean molecular weight of µ = 0.6. The gravitational acceleration is computed from the enclosed mass to be g = 5.4 × 104 cm s−2, giving an adiabatic gradient dT dr

= ad

−g = −1.56 × 10−4 K cm−1. cP

From Eq. (E.15) the superadiabatic gradient may be computed assuming pure convective transport and α = 1, giving " #2/3 2 dT g µM L(r) δ = ≃ −9.5 × 10−11 K cm−1. u 2 2 dr T k 2π r ρ cpα Therefore, the ratio of the superadiabatic gradient to the adiabatic gradient is δ(dT /dr) −9.5 × 10−11 −7 = = 6.1 × 10 , (dT /dr)ad −1.56 × 10−4 strongly justifying the assumption that when the adiabatic gradient is exceeded the transport switches to pure convection with a temperature gradient equal to the adiabatic gradient. s From Eq. (E.8), the convective velocity is estimated as v¯ =

αk µ Mu

T g

δ

dT dr

≃ 1 × 104 cm s−1.

The adiabatic speed of sound in the Sun at this radius is approximately s γP vs = ≃ 2.3 × 107 cm s−1 ρ assuming that γ =35 . Thus, v¯/v≃ s 4 ×10−4 and the convective velocity required to carry the full energy flux is much less than the sound speed, justifying the neglect of shock or acoustic phenomena in the mixing-length approximation. The scale height is Hp =

P ≃ 5.8 × 104 km. gρ

Therefore one may estimate an average time Hp t≃ ≃ 6.7 days v¯

Energy Transport in Stars

28

for a rising blob to traverse one scale height. These results have been summarized in Example 7.4.

7.13 (a) From Eq. (3.71) urad = aT 4 = 7.566 × 10−15 erg cm−3K−4 × T 4.

This gives an energy density of 6.1 × 1023 erg cm−3 for T9 = 3 and 7.6 × 1021 erg cm−3 for T9 = 1. (b) From Fig. 7.13, for a density of ρ = 106 g cm−3 it may be concluded that pair production is about an order of magnitude stronger than other neutrino processes for T9 = 1 and several orders of magnitude stronger for T9 = 3. Thus the Clayton formulas should be approximately valid in both cases. From a rough interpolation in Fig. 7.14 at a density of 106 g cm−3 ε/ρ(T9 = 3) ≃ 1 × 1014 erg g−1s−1

ε/ρ(T9 = 1) ≃ 5 × 107 erg g−1s−1.

Multiplying these by the mass density to convert to units of erg cm−3s−1 gives ε(T9 = 3) ≃ 1 × 1020 erg cm−3s−1

ε(T9 = 1) ≃ 5 × 1013 erg cm−3s−1,

which is near the values of 8.8 × 1019 erg cm−3s−1 and 3.5× 1013 erg cm−3s−1, respectively, computed from the approximate Clayton formulas. (c) The neutrinos will be unimpeded in leaving the star once produced so the neutrino energy loss rate is equal to the neutrino energy production rate. Taking the ratio of the radiation energy densities to the neutrino energy loss rates gives urad(T9 = 3) = 1.7 hr ε(T9 = 3)

urad(T9 = 1) = 4.8 yr. ε(T9 = 1)

From Eq. (4.37) and data in Table 2.2 the thermal adjustment timescale for an 18 M⊙ star may be estimated to be about 6.4× 104 yr. Thus as the temperature grows above T∼ 9 1, neutrino production has the capacity to transport all the thermal energy of the photons out of the star on a timescale that is small compared with the thermal adjustment timescale, and it is safe to conclude that neutrino cooling will be important under these conditions (see also Example 7.7 in this regard).

7.14 Evaluating constants in Eq. (7.47) gives L(r) m(r)

=

16πaGc γ − 1 γ 3κ

= 2.538 × 10−10

T4 P γ −1 γ

T K

4

cm2g−1 κ

dyn cm−2 P

erg g−1s−1

For the Sun we may assume γ = 53 . From the values for the Standard Solar Model given in Table 10.1, at the 10% radius of the Sun (R = 0.1R⊙) M = 0.076M⊙ T = 1.31 × 107 K P = 1.35 × 1017 dyn cm−2

L = 0.45L⊙,

Student Solutions Manual for Stars and Stellar Processes 1st Edition by Mike Guidry

Energy Transport in Stars

29

while from the opacities plotted in Fig. 7.2 we may estimate κ~ 1.05 cm2g−1. Inserting these values in the above formula for the critical luminosity gives L ∼ 21 erg g−1s−1 M crit for the volume enclosed by the 10% radius. From the Standard Solar Model data given above the actual luminosity of the Sun at the 10% radius is L = 0.45L⊙ = 1.7 ×1033 erg s−1, and the enclosed mass is M = 0.076M⊙ = 1.5 × 1032 g. Thus L M

= actual

1.7 × 1033 erg s−1 1.5 × 1032 g

= 11.3 erg g −1s−1.

Thus the actual luminosity is less than the critical luminosity and the inner part of the Sun should not be convective. However, the actual luminosity is within a factor of two of the critical luminosity and it may be expected that stars only a little more massive than the Sun with hotter core temperatures (and powered by the much more temperature-sensitive CNO cycle) will be susceptible to central convection. Indeed, calculations more sophisticated than this simple estimate that are illustrated in Fig. 7.11 support this conclusion.

8

Summary of Stellar Equations 8.3 Consider Eqs. (8.6)–(8.10). The radius R is defined through ξ1 (the first zero of the dimensionless radial variable ξ): s r r (n + 1)Kρc(1− n)/n (n + 1)K (1−n)/2n , ρc ξ1 a≡ ξ ≡ R= 4π G 4π G a where the equation of state is P = Kρ 1+1/n. The mass is obtained by integrating Eq. (4.4), M=

∫ R 0

4π r2ρ dr

= 4π a3

∫ ξ1 0

ξ 2ρ dξ

∫ ξ1

= 4π a3ρc 0 ξ 2θ n dξ ∫ ξ1

d dξ dθ = −4π a3ρc ξ 2 dξ 3

= −4π a ρc

0

ξ

2 dθ

dξ

dξ

ξ1

0

= 4π a3ρcξ 2|θ ′(ξ1)|, 1

where in the second step variables were changed using r = aξ, in the third step ρ = ρcθ n was inserted, in the fourth step 1 d θn = − 2 ξ dξ

dθ ξ2 dξ

from the equation in footnote 3 following Eq. (8.8) was used, and the last step used that θ ′(0) = 0 from Eq. (8.9). Substituting the above definition for the constant a then yields 3/2 (n + 1)K (3−n)/2n 2 M = 4π ρc ξ1 |θ ′(ξ1)|, 4π G which is Eq. (8.12).

8.6 From Eqs. (8.10) and (8.12), R = aξ1

M = 4π a3ρc −ξ 2

dθ dξ

. ξ=ξ1

Eliminating a between these equations leads to M = 4π

dθ R3 ρc −ξ 2 dξ ξ13

ξ=ξ1

30

Student Solutions Manual for Stars and Stellar Processes 1st Edition by Mike Guidry

Summary of Stellar Equations

31

which may be solved for ρc to give ρc = −

ξ1 3θ′(ξ1)

M 4 π R3

,

3

where θ ′(ξ1) ≡ (dθ/dξ)ξ1 .

8.9 Solve Eq. (8.13) for K and insert into Eq. (8.6) to give Pc =

(4π G)1/n n+1

GM ξ 12|θ ′(ξ1)|

(n−1)/n

R ξ1

(3−n)/n

(n+1)/n

ρc

.

But from Eq. (8.11), R and ρc are not independent. Using this to eliminate R from the preceding expression gives Eq. (8.14).

9

The Formation of Stars 9.1 For a fluid, the magnitude of the force per unit volume is given by the gradient of the pressure, implying Eq. (9.13) for the force per unit volume arising from the radiation pressure. Inserting the temperature gradient for radiation diffusion (7.43) gives Fr κρ L = .

V 4π r2c The magnitude of the gravitational force per unit volume is Fg Gm1M ρM = 2 =G 2 .

V rV r Equating the two forces Frad = Fg and solving the resulting expression for L gives the Eddington luminosity (9.14). Inserting numerical values and estimating the opacity by the Thomson formula (7.25) allows Eq. (9.14) to be expressed in the form (9.15). Assuming the most luminous stars to have L ≃ several × 106L⊙, and that they are radiating at the Eddington limit, permits Eq. (9.15) to be solved for M/M⊙. This gives an estimate of ~100 solar masses for the most massive stars that are stable against their own radiation pressure.

9.4 Jupiter is about 1000 times less massive than the Sun and orbits at a radius of 5.2 AU (let’s assume the orbit to be circular for this problem).

MJ

~1000 MJ d

d'

CM

Jupiter

Sun

Thus, the distance d of the Sun from the Jupiter–Sun center of mass is determined by solving 1000MJd = MJd′

d + d′ = 5.2 AU,

giving d = 5.2 ×10−3 AU. Half the corresponding shift in angle as the Sun orbits the center of mass that would be observed from a distance of 30 ly = 1.9 × 106 AU is given by α ≃ sinα =

5.2 × 10−3 AU

9

−4 ≃ 2.7 × 10− rad = 5.6 × 10 arc-sec.

1.9 × 106 AU If a coin has a radius of ∼1 cm, the distance at which it would subtend this same angle is 1 × 10−5 km r= ≃ 3700 km.

2.7 × 10−9 The period is given by Kepler’s 3rd law and is 11.9 years for Jupiter or the Sun to circle 32

Student Solutions Manual for Stars and Stellar Processes 1st Edition by Mike Guidry

The Formation of Stars

33

the center of mass. The circumference of the Sun’s orbit around the center of mass for the Sun– Jupiter system is C = 2π d = 3.27 × 10−2 AU, so the average velocity of the Sun in its orbit around the center of mass is given by the circumference divided by the period: v≃ 13 m s−1. This is the maximum radial velocity that could be detected (if the orbit is oriented favorably so that the radial velocity is along the line of sight). The corresponding fractional Doppler shift of spectral lines caused by this radial component of velocity is Max

∆λ λ

=

v = 4.3 × 10−8. c

9.6 From Eq. (4.20), the gravitational potential is given by the integral Ω = −4π G

∫R

m(r)ρ(r) 2 r dr

0 r ∫ R 2 2 = − 16π 2ρ0 G R 0 2

2

= −16π ρ0 GR

5

r 3 − r5 2 R2 — r rdr 3 5R2 5 4 5 2 2 GR =− 64π ρ0 2 15 · 21 7 15

,

where in the second line the result for m(r) from the solution of Problem 4.15 has been substituted. But from that solution the total mass is 8 M = πρ0R3, 15 and the quantity in parentheses in the last line of the above equation is just GM2/R. Thus Ω = −f

GM2 R

=−

5 GM2 7 R

,

and f = 57 for this case.

9.8 From the formulas in Box 9.3 vdisk =

r0 v0 rdisk

GM v2 disk = 2 rdisk , rdisk

where the first equation follows from conservation of angular momentum and the second from requiring a stable circular orbit. Combining these and solving for the disk radius rdisk =

r02v2 0 = 1.13 × 10−3 GM

r0 2 AU

M⊙ M

v0 km s−1

2

AU.

If one then takes v0 = 0.1 km s−1

r0 = 2000 AU

M = 1 M⊙,

this formula yields rdisk ≃ 45 AU, which is comparable to the radius of the Solar System.

The Formation of Stars

34

9.14 From the virial theorem (see Section 4.5) and Eq. (9.1), if the luminosity is assumed to derive entirely from gravitational contraction the total radiated energy is E=

∆Ω

f GM2 = − 2R , 2

where f is of order one. Assuming the luminosity L to derive from this gravitational energy, L=− from which dR dt

=−

dE f GM2 =− dt 2R

2LR f GM

R=− 2

1 t KH

R

dR/dt R

tKH ≡

, f GM2 2LR

.

Assuming tKH to be constant, this differential equation has the solution R = R0e−t/tKH , so setting R/R0 = e−1 gives t = tKH and the Kelvin–Helmholtz timescale is (approximately, since tKH is not constant) the time for the radius to decrease by a factor of e−1. The estimates in Section 4.9 give tKH ~107 yr for the Sun, so gravitational collapse cannot explain the luminosity of the Sun over its billions of years on the main sequence. Since tKH ∝ (LR)−1, it will increase as the protostar descends to the main sequence. Thus the Kelvin–Helmholtz timescale estimated using the ZAMS main sequence mass, radius, and luminosity also may be taken as a rough estimate of the total time for the corresponding protostar to collapse to the main sequence.

Student Solutions Manual for Stars and Stellar Processes 1st Edition by Mike Guidry

10

Life and Times on the Main Sequence 10.3 From Fig. 5.2, each of PP-I, PP-II, or PP-III must release two neutrinos since in 4He two protons must be converted to neutrons by weak interactions (remember 4H → that in PP-I the first two steps must act twice to produce two 3He ions for the third step).

The average energy release contributing to photon luminosity for a proton–proton chain is Q¯≃ 25 MeV (after subtracting the average energy carried off by neutrinos). Thus, one may estimate L⊙ Lν ≃ (2 neutrinos) × ¯ ≃ 1038 s−1 Q for the solar neutrino luminosity.

10.8 (a) From Table 10.1, most energy is produced inside about 20% of the solar radius and at R∼0.2R⊙ the temperature is∼ T ∼107 K. Viewing this inner 20% as a blackbody, the luminosity is L = 4π R2σ T 4 = 4π(0.2R⊙)2σ T 4 = 1.38 × 1045 erg s−1, which is much larger than the actual solar luminosity of 3.83 × 1033 erg s−1. Furthermore, kT = (8.617 × 105 eV K−1) × (107 K) = 0.86 keV, which is in the soft X-ray portion of the electromagnetic spectrum. Thus the solar furnace is wrapped in an opaque blanket that both reduces the luminosity (and thus lengthens the age) of the Sun, and degrades the solar spectrum from the X-ray to visible portion of the spectrum. (b) From Problem 4.3 the time for a random walk from the center to the surface is 2 R2 λ tdiff = N∆t = ⊙ = R⊙ , λ 2c λc where N is the number of scatterings and λ is the mean free path. The Thomson scattering formula (7.25) with X = 0.7 and assuming ρ = 1.4 g cm−3 yields a mean free path 1 1 ≃ ∼ 2 cm. λ= κ T ρ 0.2(1 + X )ρ

Thus a diffusion time to the surface may be estimated as R⊙2 ∼ 2600 yr. λc This is much shorter than the Kelvin–Helmholtz timescale (GM2 /R⊙) ⊙ tKH = ≃ 3 × 10 7 yr L⊙ tdiff =

35

Life and Times on the Main Sequence

36

(and also much shorter than the diffusion time if more realistic opacities are used; see Problems 4.3 and 10.9). The discrepancy between the two scales is because most of the internal energy of the Sun is stored in the gas, not in the photons; see Problem 10.10. Thus absorptive processes lead to a much longer time for each diffusion step and a much longer timescale than that estimated for diffusion using Thomson scattering.

10.9 The mean free path is given by λ = (κρ)−1. From Fig. 7.2 and Table 10.1 we may estimate that at the solar center and at 90% radius, respectively, κ 0 ∼ 1 cm2g−1

ρ0 ∼ 152 g cm−3

κ0.90 ∼ 100 cm2g−1

λ0 ∼ 6.6 × 10−3 cm

ρ0.90 ∼ 2.6 × 10−2 g cm−3

λ0.90 ∼ 0.38 cm.

We may estimate very roughly from Table 7.1 an average internal mean free path of λ ~ 10−2 cm, which implies a diffusion time (see Problem 4.3) 2 (6.96 × 1010 cm)2 tdiff = R ⊙ = = 1.6 × 1013 s, λc (10−2 cm)· (3 × 1010 cm s−1) or ~ 500, 000 years for energy released at the center to diffuse to the surface. Such estimates are obviously subject to considerable quantitative uncertainty but they illustrate that the time for solar energy to reach the surface of the Sun is orders of magnitude longer than the direct flight time for photons.

10.11 From Eqs. (D.10) and (D.11) for p + d → 3He + γ, 1 Y˙d = −(ρ RpdYp)Yd τpd = ≃ 1.7 s, ρ RpdYp where we have used T6 = 15

ρ = 150 g cm−3

Yp = 0.4,

and that from Fig. 6.1 Rpd ≃ 10−2 cm3 mol−1 s−1.

Likewise, for 3He + 3He → 4He + 2p τ3He-3He =

1 ρ R3He-3HeY3He

≃ 540, 000 yr,

where Y3He ∼ 4 × 10−6 was used and R3He-3He ∼ 10−10 cm3 mol−1 s−1 was estimated from Fig. 6.1.

10.12 Parameterizing r in terms of the optical depth τ, we may write 1 dP dP dP dr = =− , dτ dr dτ κρ dr where dτ = −κρ dr from Box 10.1 was used. But Eq. (4.10) gives dP/dr = −ρ g, so dP 1 = − dτ κρ

(−ρ g) =

g . κ

Student Solutions Manual for Stars and Stellar Processes 1st Edition by Mike Guidry

Life and Times on the Main Sequence

37

Note that τ is dimensionless but it measures length because there is a monotonic relationship between r and τ.

10.13 The energy density for the magnetic field and gravitational field may be estimated as GM2 B2 , εmag ≃ εgrav ≃ R4 2µ0 respectively, where M is the mass of the Sun, R is its radius, B is the strength of the magnetic field, and µ0 is the permeability of the vacuum. Taking the ratio using B~ 0.1 T gives εmag −12 B2R4 ≃ ≃ 7 × 10 , 2 εgrav µ0GM suggesting that gravity completely dominates magnetic fields in contribution to energy density in the Sun. 10.14 The actual electron number densities ne = 6.4 × 1025 cm−3 (at R = 0)

ne = 5 × 1024 cm−3 (at R = 0.3R⊙)

were given in the problem and from Table 10.1 the corresponding temperatures are T = 15.7 × 106 K (at R = 0)

T = 6.8 × 106 K (at R = 0.3R⊙).

From Eq. (3.51) the critical number density for electrons is nc =

2πmekT h2

3/2

= 2.415 × 1015

T K

3/2

cm−3,

which gives nc~1.5 ×1026 cm−3 at the center and nc∼ 4.3 ×1025 cm−3 at 30% radius.

Thus the actual electron density is less than half the critical density at the center and about 12% of critical at 30% radius in the present Sun.

11

Neutrino Flavor Oscillations 11.2 In h¯ = c = 1 units the energy is given by E2 = p2 + m2. Thus, (E1 + E2)(E1 − E2) E2 − E2 p2 + m2 − p2 − m2 1 2 2 E1 − E2 = 1 2 = = 1 E1 + E2 E1 + E2 E1 + E2 ∆m2 m2 − m2 m2 − m2 1 2 1 2 ≃ ≃ 2E = − 2E , E1 + E2 where we have defined E1 ∼ E2 ≡ E

∆m2 ≡ m22 − m21,

and coherent states have been assumed with p1 ∼ p2. The distance traveled in a time t is r ≃ t, and is essentially the same for ν1 and ν2 if the state is coherent. Therefore, from these results and the trigonometric identity 1 − cos2α = 2 sin2 α, 1 − cos(E2 − E1)t = 1 — cos

∆m2 t 2E

= 2 sin2

∆m2 t 4E

= 2 sin2

πr , L

where the oscillation length is L ≡ 4π E/∆m2. This permits the probabilities (11.7) to be rewritten in terms of the oscillation length L and the distance r traveled in a time t, which gives Eq. (11.9).

11.3 In a 2-flavor model the electron and muon neutrino flavor eigenstates are |νe⟩ = |ν1⟩ cosθ + |ν2⟩ sin θ

|νµ ⟩ = −|ν1⟩ sin θ + |ν2⟩ cosθ

in terms of the mass eigenstates |ν1⟩ and |ν2⟩. The pure νe state evolves after a time t to |ν(t)⟩ = |ν1(0)⟩ cosθ e−iE1t + |ν2(0)⟩ sin θ e−iE2t .

Use the first two equations to solve for the mass eigenstates, |ν1⟩ = cosθ|νe⟩ − sinθ|νµ ⟩

|ν2⟩ = sinθ|νe⟩ + cosθ|νµ ⟩,

and insert this into the equation for |ν(t)⟩ to give |ν(t)⟩ = cos2 θ e−iE1t + sin2 θ e−iE2t |νe⟩ + sinθ cosθ −e−iE1t + e−iE2t |νµ ⟩.

The probability after a time t to observe a νe if the initial state was a νe is then P(νe → νe,t) = |⟨νe|ν(t)⟩|2 = cos4 θ + sin4 θ + sin2 θ cos2 θ ei(E2−E1)t + e−i(E2−E1)t , 38

Student Solutions Manual for Stars and Stellar Processes 1st Edition by Mike Guidry

Neutrino Flavor Oscillations

39

where ⟨νe|νe⟩ = 1 and ⟨νe|νµ ⟩ = 0 have been employed. Using the identities eix + e−ix = 2 cos x

cos4 θ + sin4 θ = 1 — 2 sin2 θ cos2 θ

sinθ cosθ =

1 2

sin 2θ,

this finally may be simplified to 1 sin2 2θ (1 − cos[(E2 − E1)t]).

2 By analogous steps one obtains for the probability of observing a νµ if the initial state was a νe, 1 2 2 P(νe → νµ ,t) = |⟨νµ |ν(t)⟩| = sin 2θ (1 − cos[(E2 − E1)t]) , 2 which also could have been obtained immediately by the condition P(νe → νe,t) = 1 −

P(νe → νµ ,t) = 1 − P(νe → νe,t) implied by unitarity (the sum of the probabilities must be 1) in the 2-flavor space.

11.7 Letting ϕ = π r/L and averaging over one oscillation 0 → 2π, 2

Average sin ϕ =

1 2π

∫ 2π 0

sin 2 ϕ dϕ =

2π

1 1 1 1 ϕ − sin 2ϕ = .

4 2π 2 2 0

Using this average in Eq. (11.9) gives Eq. (11.12). The average formula is valid only if probabilities are averaged over an oscillation length or more. Using the formula in Example 11.2, for E = 5 MeV

∆m2 = 10−5 eV2

the oscillation length is L = 2.48 × 10

−3

E MeV

eV2 km = 1240 km, ∆E2

and Eq. (11.12) gives an averaged probability to remain an electron neutrino 1 P¯(νe → νe) = 1 − sin2 2θ = 0.793, 2 for θ = 20◦.

11.8 One must evaluate the expression Pclass(νe → νe) = P(νe → ν1)P(ν1 → νe) + P(νe → ν2)P(ν2 → νe) = |⟨νe(t)|ν1⟩⟨ν1|νe(0)⟩|2 + |⟨νe(t)|ν2⟩⟨ν2|νe(0)⟩|2, where the quantum interference term in Eq. (11.14) has been dropped to give the classical limit. From Eq. (11.2), ⟨ν1|νe(0)⟩ = cosθ

⟨ν2|νe(0)⟩ = sinθ,

while from Eqs. (11.3) and (11.4) ⟨νe(t)|ν1⟩ = cosθ e−iE1t

⟨νe(t)|ν2⟩ = sinθ e−iE2t ,

Neutrino Flavor Oscillations

40

which gives for the classical probability Pclass(νe → νe) = |cos2 θ e−iE1t |2 + | sin2 θ e−iE2t |2 = cos4 θ + sin4 θ = 1 − 12 sin2 2θ, where the trigonometric identities cos4 θ + sin4 θ = 1 − 2 sin2 θ cos2 θ

sinθ cosθ = 12 sin 2θ,

were used in the last step.

11.9 (a) Notice that since h¯ = 6.5827 × 10−22 MeV s and c = 2.99792 × 1010 cm s−1, h¯ c = 197.3 × 10−13 MeV cm, so setting h¯ = c = 1 in this equation implies that in these units 1 MeV = 5.068 × 1010 cm−1.

Since c = 1, this yields the equivalence ∆E2 = ∆m2c4 = ∆m2. Thus from Eq. (11.10), E 4π E MeV2 MeV−1 MeV MeV = 4π L = 4π E = 4π E = ∆m2

∆E2

= 4π

E MeV

= 2.48 × 10

−3

∆E2 MeV ∆E2 MeV2 MeV2 MeV2 1 MeV −1 MeV × 2 ∆E 5.068 × 1010 cm−1 E MeV

eV2 km, ∆E2

where is the last line we have for convenience made the conversions MeV2 → eV2 and cm → km. (b) In h¯ = c = 1 units the weak coupling constant is GF = 1.16637 × 10−5 GeV−2.

From h¯ c = 1 = 197.3 × 10−16 GeV cm it may be deduced that 1 GeV−2 = 3.8927 × 10−28 cm2 (see Appendix B). Thus for the weak coupling constant GF = 1.16637 × 10−5 GeV−2 = 4.5389 × 10−33 cm−1 cm3 = 8.955 × 10−44 MeV cm3, where we have used 1 cm−1 = 197.3 × 10−13 MeV in the last step. Therefore GF = 1.16637 × 10−5 GeV−2

Student Solutions Manual for Stars and Stellar Processes 1st Edition by Mike Guidry

Neutrino Flavor Oscillations

41

in h¯ = c = 1 units translates to GF = 8.955 × 10−44 MeV cm3 in more conventional units.

11.10 (a) Applying the results of Problem 11.5 to the neutrino mass matrix given in the exercise yields λ± =

1 1 4m2 M ± M 1+ 2 M 2 2

1/2

≃

1 1 2m2 M ± M 1+ 2 M 2 2

,

where in the last step the square root has been expanded assuming 4m2/M2 << 1. Thus the neutrino mass eigenvalues are λ+ ≃ M ≃ 1015 GeV

λ− ≃

−m2 M

≃ − 10−2 eV,

where 2m2/M2 has been ignored relative to 1 and the eigenvalues have been evaluated assuming m = 100 GeV and M = 1015 GeV. (b) To obtain the eigenvectors requires solving the original linear system AX = λ X with the eigenvalues λ just found. Choose a basis νL ≡

1 0

νR ≡

0 1

,

where νL corresponds to a pure left-handed light neutrino and νR corresponds to a massive right-handed neutrino. The states X will in general be linear combinations of these basis states X+ 1 0 a X≡ = aν + bν = a +b = .

L R X− 0 1 b Thus the linear system is AX =

0 m m M

a b

=λ

a b

,

implying that mb = λ a. Inserting λ = λ+ = M gives a = (m/M)b, so choosing b = 1 gives a = m/M and ignoring normalization the eigenvector associated with eigenvalue λ+ may be written as m X+ = νL + νR ≃ νR, M since m << M. Likewise, inserting λ = λ− = −m2/M gives b = −(m/M)a, and choosing a = 1 gives m X− = νL − νR ≃ νL. M Thus the eigenvector associated with λ− is approximately the left-handed neutrino of the Standard Model but with a tiny mass, and the eigenvector associated with λ+ is approximately a very massive right-handed neutrino.

These results constitute the simplest form of what is called the seesaw mechanism, which

42

Neutrino Flavor Oscillations

gets its name because the two eigenvalues are driven in opposite directions by increasing the value of the mass scale M. They give a possible (phenomenological) explanation for why 1. Observed neutrinos are left-handed and have tiny but non-zero masses. (These are characteristics of the lowest-mass eigenvalue.)

2. No right-handed neutrinos are observed. (The highest-mass eigenvalue is too massive to produce in accelerators.)

3. The neutrino mass scale is much smaller than the quark mass scale in each generation of the Standard Model. (The mass scale for the lightest neutrinos has been driven down by the seesaw mechanism.)

4. The weak interactions violate parity maximally. (The observed parity violation of the weak interactions is a low-energy phenomenon that would be removed at energies sufficiently high that the massive right-handed neutrinos could participate freely in the weak interactions.)

Although a more sophisticated treatment than that of our discussion is required to account plausibly for the full set of observational constraints, this simplest form of the seesaw mechanism illustrates the basic idea of such considerations.

Student Solutions Manual for Stars and Stellar Processes 1st Edition by Mike Guidry

12

Solar Neutrinos and the MSW Effect 12.2 From Eq. (12.15) after restoring h¯ and c, and using GF = 8.958 × 10−44 MeV cm3, √ √ 2π 2πh¯ c cm−3 27 km. ℓm = = = 9.785 × 10 ne GFne GFne Alternatively, in h¯ = c = 1 units GF = 1.1664 × 10−5 GeV−2

1 GeV−2 = 3.893 × 10−28 cm2,

and therefore we may write √ √ 2π 2π cm−2 ℓm = = GFne (1.1664 × 10−5)(3.893 × 10−28) ne 27 cm−3 km. = 9.785 × 10 ne The electron density at the center of the Sun is ne ∼ 6.3 × 1025 cm−3 (see Fig. 12.9), from which ℓm = 155 km. From Eq. (11.11) with E = 10 MeV

∆E2 = ∆m2c4 = 7.5 × 10−5 eV2,

the vacuum oscillation length is L = 2.48 × 10

−3

E MeV

eV2 km = 330.7 km. ∆E2

Therefore χ = L/ℓm = 330.7/155 = 2.13.

12.3 Setting the critical density nR equal to the central density n0 in Eq. (12.27) and solving e

for E gives Emin =

2 2 ∆E √ cos 2θ = 3.95 × 1030 ∆E 2 2G F n0e eV2

e

cm−3 n0e

cos 2θ MeV,

where ∆E2 = ∆m2c4. For the Sun, setting θ = 35◦

∆m2c4 = 7.5 × 10−5 eV2

ne0 ∼ 6.3 × 1025 cm−3

gives Emin = 1.6 MeV; see Fig. 12.9(b). This implies that the MSW resonance flavor conversion will be suppressed for the lowest-energy neutrinos produced by the Sun. For example, from the solar neutrino spectrum in Fig. 10.3 we may expect that the pp neutrinos from PP-I and the 7Be neutrinos will be unaffected by the MSW resonance, while the majority of 8B neutrinos will encounter the resonance condition. As will be discussed later, observation of this energy-dependent suppression is one piece of evidence favoring the 43

Solar Neutrinos and the MSW Effect

44

MSW mechanism over vacuum oscillations for flavor conversion of higher-energy solar neutrinos.

12.4 Using that in h¯ = c = 1 units GF = 1.16637 × 10−11 MeV−2, √ √ A = 2 2GFEne = 2 2GF = 3.30 × 10−11 = 2.53 × 10−31

E MeV ne cm−3 ne cm−3

E MeV E MeV

ne cm−3

MeV cm−3

MeV−1 cm−3 eV2,

where in the last step 1 cm−1 = 197.3 × 10−13 MeV was used. The mass density ρ and electron number density ne are related by ne =

ρYe

,

Mu where Ye is the electron fraction (ratio of electrons to nucleons) and Mu is the atomic mass unit. Replacing ne with ρ gives E MeV

A = 1.52 × 10−7

ρ Ye eV2. g/cm−3

For the center of the Sun the Standard Solar Model suggests ρ ∼ 150 g cm−3 and Ye = 12, so for a 10 MeV electron neutrino one obtains A = 1.14 × 10−4 eV2, which is larger than the vacuum mass-squared splitting between the two states with the largest components of νe and νµ in Fig. 11.3. Hence the charged-current interaction of νe with the medium could have a large effect on solar neutrino oscillations.

12.8 The mass matrix can be written as [43] M=

1 m12 + m22 + A 1 A − ∆m2 cos 2θ 0 + 2 2 2 0 2 m1 + m 2 + A 2 ∆m sin 2θ

∆m2 sin 2θ , 2 −A + ∆m cos 2θ

where we have introduced the definitions 2

2

2

∆m = m2 − m1

A = 2EV =

∆m2L ℓm

.

The first term is already diagonal so the second term must be diagonalized. Let D equal the determinant and T the trace of the matrix in the second term. But the matrix is traceless so from Problem 11.5 the eigenvalues of the r second term are λ± =

1 2

T±

1 4

T 2 − D = ±√ −D.

Student Solutions Manual for Stars and Stellar Processes 1st Edition by Mike Guidry

Solar Neutrinos and the MSW Effect

45

Evaluating the determinant gives 1 (A − ∆m2 cos 2θ) D = det 2 1 ∆m2 sin 2θ 2 1 =−

4

1

∆m2 sin 2θ

2 1 (−A + ∆m2 cos 2θ) 2

(∆m2 cos 2θ − A)2 + ∆m2 sin2 2θ

Thus, the two eigenvalues (in units of mass-squared) of the mass matrix at a fixed density are given by √ 1 λ ± = (m21 + m22 + A) ± −D 2 q 1 1 (∆m 2 cos 2θ − A) 2 + ∆m2 sin 2 2θ = (m12 + m22 + A) ± q 22 2 2 2 m +m ∆m 1 2 2 2 2 = 1 + χ ± ∆m (cos 2θ − χ) + sin 2θ, 2 2 2 where in the last step ∆m2 = m2 − m2 χ≡L = A 2 1 ℓm ∆m2 were used. This result is Eq. (12.16). The first term is the average mass in vacuum, the second term is the average interaction with the matter, and twice the third term gives the splitting of the two eigenvalues (which reduces to ∆m2 if the matter interaction strength χ = 0). For a given vacuum mixing angle θ the splitting is minimum at the density where χ = cos 2θ, which is the condition for the MSW resonance; see Eq. (12.26).

12.9 In the interest of compact notation let us define cm ≡ cosθm(t)

sm ≡ sinθm(t)

c ≡ cos 2θ

s ≡ sin 2θ.

Then Eqs. (12.14) and (12.18) inserted in Eq. (12.19) give cm sm

−sm cm

χ −c s

s c−χ

cm −sm

L E1 sm = cm π 0

0 .

E2

Evaluating the triple matrix product on the left gives 2 (χ − c) − smcms cm −smcms + sm2 (c − χ)

smcm(χ − c) + c2ms 2 −sm s − smcm(c − χ)

2s s mc mχ ( — c) − sm 2 +cm s − smcm(c − χ)

2 χ — c) + s c s sm m m ( +smcms + cm2 (c − χ)

=

L

E1

0

π

0

E2

,

and, upon restoring the trigonometric functions, cos 2θm(χ − cos 2θ) −sin 2θm cos 2θ sin 2θm(χ − cos 2θ) +cos 2θm sin 2θ

sin 2θm(χ − cos 2θ) +cos 2θm sin 2θ −cos 2θm(χ − cos 2θ) −sin 2θm sin 2θ

=

L E1 0

π

0 , E2

Solar Neutrinos and the MSW Effect

46

where we have used the identities cos2 θm − sin2 θm = cos 2θm

sinθm cosθm = 12 sin 2θm.

The relationship between θm and θ now may be established by equating the entries of the matrix on the left with the entries of those of the matrix on the right. Comparing the upper right matrix entries requires that sin 2θm(χ − cos 2θ) + cos2θm sin 2θ = 0, implying that tan 2θm =

sin 2θm cos 2θm

=

sin 2θ cos 2θ − χ

=

tan 2θ 1 − χ/ cos2θ

,

which is Eq. (12.20). As shown in Problem 12.6, sin2 2θ

sin2 2θm =

, (cos 2θ − χ)2 + sin2 2θ is an equivalent statement of this relationship. Note that in the limit of vanishing matter coupling (χ → 0) the matter angle θm reduces to the vacuum angle θ.

12.10 Since both the wavefunctions and Um depend on time, from Eq. (12.30) and the formula for the derivative of a product, ν1 ν1 Um m21 0 Um ν1 = iUm d ν1 + i d Um = .

2 dt ν2 dt ν2 dt ν2 2E 0 m2 ν2 Multiply from the left by U † , use the unitarity relation U †Um = 1, and rearrange to give i

d

m

i

d dt

ν1 = ν2 =

m

m21/2E 0

0 m22/2E −i θ˙m

m1 2/2E 2 i θ˙ m /2E m

2

†

— iUm

d Um dt

ν1 ν2

ν1 , ν2

where θ˙m≡ dθm/dt and Eq. (12.18) with basic trigonometric identities has been used to evaluate † iUm

d cosθm U = dt m sinθm

−sinθm cosθm

−sinθmθ˙m −cosθmθ˙m

cosθmθ˙m −sinθmθ˙m

=

0 i θ˙m . −i θ˙m 0

Finally, the constant multiple of the unit matrix 1 2 (m + m2) 0 1 2 4E 1 2 0 (m + m2) 2 4E 1 can be subtracted from the mass matrix without affecting observables to give d ν1 −∆m2/4E −i θ˙m ν1 i = i θ˙m ∆m2/4E ν2 dt ν2 which is Eq. (12.31).

Student Solutions Manual for Stars and Stellar Processes 1st Edition by Mike Guidry

Solar Neutrinos and the MSW Effect

47

12.11 Letting θm(1) ≡ θm(t1) and θm(2) ≡ θm(t2), the classical probability (ignoring interference by averaging over oscillations) given in Eq. (12.34) is cos2 θm(2) sin2 θm(2) P¯(νe → νe) = (1 0) sin2 θm(2) cos2 θm(2 ×

1 0 0 1

cos2 θm(1) sin2 θm(1)

sin2 θm(1) cos2(θm(1

1 0

= cos2 θm(2) cos2 θm(1) + sin2 θm(2) sin2 θm(1) = 12 [1 + cos2θm(t1) cos 2θm(t2)] , where in the last step the identities cos2 θm(2) cos2 θm(1) = 14[1 + cos2θm(2)][1 + cos2θm(1)] sin2 θm(2) sin2 θm(1) = 14[1 − cos2θm(2)][1 − cos2θm(1)]

have been used.

12.13 A solution of Eq. (12.3) for definite momentum is given by

|νi⟩ = e−iEit · e−ip·x

Ei =

q

p2 + m2i .

For ultrarelativistic particles | p| >> mi so

Ei = p2 1 +

m2i p2

1/2

≃ p+

mi2 m2i ≃ p+ , 2p 2E

where E ∼ p ≡ | p|. Thus, ν (t)⟩ ≃ e−ipt · e−i p·x · e−i(m 2/2E)t ≃ e−i(m 2/2E)t , i i | i where the initial exponential factors have been omitted in the last step because they provide only a common overall phase that does not affect the observables of the problem.

12.14 Inserting the explicit forms for the transformation matrices U and U † into (12.11)

Solar Neutrinos and the MSW Effect

48

gives i

d dt

νe νµ

= +

cosθ −sinθ

sinθ cosθ

0 0

νe νµ

V 0

0 0 0 ∆m2/2E

cosθ sinθ

−sinθ cosθ

∆m2 2 ∆m2 sin θ sin θ cosθ V 2E 2E = + 2 ∆m 0 ∆m2 sin θ cosθ cos2 θ 2E 2E ∆m2 0 sinθ cosθ V 0 = + 0 0 2E sinθ cosθ cos2 θ − sin2 θ V =

∆m2 sin 2θ 4E

∆m2 sin 2θ 4E ∆m2 cos 2 θ 2E

νe νµ

≡M

νe νµ

0 0 νe νµ

νe , νµ

where in the second step (∆m2/4E) sin2 θ times the× 2 2 unit matrix has been subtracted, in the final step the trigonometric identities sinθ cosθ = 21 sin 2θ

cos2 θ − sin2 θ = cos 2θ

have been used, and M is the mass matrix (in the flavor basis).

Student Solutions Manual for Stars and Stellar Processes 1st Edition by Mike Guidry

13

Evolution of Lower-mass Stars 13.1 (a) Proceeding as in Eq. (4.19) but with integration limits from 0 to M(Rc), the only new feature is that the boundary term in the integration by parts no longer vanishes because P /= 0 at the upper limit and we obtain ∫

∫

M(Rc) P 1 M(Rc) Gm dm = −PcVc + dm, 3 0 r ρ 0 where Vc = 4 π R3 is the volume of the isothermal sphere and Pc is the pressure at its outer 3

c

boundary. From Eq. (4.20), ∫

1 M(Rc) Gm 1 dm = − Ωc, 3 0 r 3 where Ωc is the gravitational potential energy associated with the enclosed mass at radius Rc. Therefore we arrive at PcVc −

∫ M(Rc) 0

P 1 dm = Ωc, ρ 3

for the virial theorem applied to an isothermal sphere in the interior of a star. (b) Because dV = dm/ρ, the result of part (a) may also be expressed as PcVc −

∫ Vc 0

1 PdV = Ωc, 3

and since for an ideal gas P = (γ — 1)u, where γ is the adiabatic exponent and u is the internal energy density [see Eq. (3.14)], one may also write ∫ M(Rc) 0

P ρ

dm = (γ − 1) = (γ − 1)

∫ M(Rc) ∫ Vc 0

u ρ

0

dm

udV = (γ − 1)Uc,

where Uc = uVc is the internal energy of the isothermal core. Therefore, 3(γ − 1)Uc − 3PcVc + Ωc = 0 is an alternative form of the virial theorem for a core of isothermal ideal gas. (c) Parameterize the gravitational potential by Ωc = − f

GMc2 Rc

,

where f is a parameter of order one accounting for the distribution of matter in the star (for example, f = 35 for a sphere of uniform density). Assuming the gas to be ideal and 49

Evolution of Lower-mass Stars

50

isothermal, P = ρ kTc/µc, where Tc is the constant temperature and µc is the average mass of a gas particle in the isothermal core. Therefore, ∫ Vc 0

∫

PdV =

kTc Vc kTc ρ dV = Mc. µc 0 µc

Substitute the preceding two equations and Vc = 4 π R3 into the final equation derived in c

3

part (a) to give Pc =

3 kTcMc 4π µcR3c

—

f GM2

c,

4π Rc4

where the first term is associated with the thermal energy of the isothermal core and the second term is gravitational. For specific values of Tc and Rc, as the core mass is increased the first term increases the pressure at the surface of the isothermal sphere while the second term decreases it. (d) The results of part (c) imply there is some value of Mc that maximizes Pc. This limit can be found by taking the derivative of the expression for Pc derived in part (c) with respect to Mc, equating it to zero, and solving for Rc to give Rc =

2 f GMcµc 3kTc

.

Substituting this value of Rc into the original expression for Pc gives

Pcmax =

81k4 64π f 3G3

Tc4 µ4M2 c

c

for the maximum pressure at the surface of the isothermal sphere. These results show that Pcmax scales as the inverse square of the mass of the core. This pressure must balance the pressure of the layers overlying the isothermal core if the star is to be stable, suggesting that there may be a maximum core mass beyond which the isothermal core can no longer support the weight of the overlying layers. This maximum pressure can be related to the total mass of the star and to the mass of the isothermal core by using a stellar model to give the pressure at the surface of the core exerted by the overlying envelope and requiring that for stability the actual pressure at the core surface be less than the critical pressure.

Such considerations with realistic stellar models give the restrictions discussed in Section 13.4.1 for the maximum mass of an isothermal core of ideal helium gas (the Schönberg– Chandrasekhar limit).

13.4 This solution follows the discussion of thin-shell instabilities in Prialnik [176]. Consider the thin shell illustrated in the following figure,

Student Solutions Manual for Stars and Stellar Processes 1st Edition by Mike Guidry

Evolution of Lower-mass Stars

51

m, T,  R

r

L=r-r0

which has a density ρ, temperature T , and total mass ∆m = 4π r02Lρ, with the thickness of the shell L = r − r0 being much less than the radius of the star R. (a) The pressure P in the shell is generated by the weight of the layers above so hydrostatic equilibrium expressed in Lagrangian coordinates gives ∫

M Gm dP Gm =− −→ P = − 4 dm, dm 4π r4 m0 4π r where M is the total mass of the star and m0 is the mass interior to the shell. (b) Let’s perturb the shell by shifting its radial coordinate by a small amount δ r/r0, leading to a small change in pressure δ P,

P → P+ δP

r → 1+

δr r. r0

The new pressure in the shell after the perturbation is P +δP = —

∫ M

Gm

m0 4π[(1 + δ r/r0)r] −4 ∫ M

= − 1+

δr r0

4

dm

Gm 4 dm ≃ m0 4π r

1−4

δr P, r0

where the factor preceding the integral in the second line has been expanded assuming that δ r/r0 << 1. Thus, solving for δ P and taking the continuum limit gives the differential equation dP dr = −4 .

P r0 The expression above for ∆m gives ρ = ∆m/(4π r02L), so that dρ dL dr r0 dr =− =− =− . ρ L L L r0 Solving for dr/r0 and inserting into the equation derived above for dP/P gives dP 4L dρ = .

P r0 ρ (c) From Problem 4.16 the equation of state can be expressed in the generic form dT dP dρ , = α +β T P ρ

52

Evolution of Lower-mass Stars

where α ≥ 0 and β ≥ 0. The two preceding equations may be combined to eliminate the pressure in favor of the temperature, dT 4L dρ = β −1 −α . ρ T r0 Thermal stability requires that if the shell expands (dρ < 0) the temperature must drop (dT < 0). Since β ≥0, the shell will be stable only if L > 14 αr0. But α is fixed by the equation of state so for a sufficiently thin shell the stability condition cannot be satisfied.

This is called the thin-shell instability; it may be significant for AGB stars since they often develop thin, unstable shell sources. Since this instability was derived for a general equation of state, it can occur even if the gas is not degenerate.

13.5 From the expression for AGB luminosity given in the problem and Eq. (9.15), LAGB (Mc/M⊙) − 0.5 = 1.8 × , LEdd (M/M⊙) where M is the total mass and Mc is the mass of the carbon–oxygen core. For M = 1M⊙ and Mc = 0.6M⊙ this gives LAGB ∼0.18LEdd. Thus, as the core mass grows beyond∼ 0.5M⊙ the luminosity becomes an appreciable fraction of the Eddington luminosity and it isn’t surprising that AGB stars tend to exhibit large surface mass loss.

Student Solutions Manual for Stars and Stellar Processes 1st Edition by Mike Guidry

14

Evolution of Higher-mass Stars 14.4 Estimate the cross section for individual nucleons from Eq. (7.48) as σnucleon ≃ 10−44

Eν mec 2

2

cm2 = 8.6 × 10−42 cm2,

where a neutrino energy of Eν = 15 MeV and an electron mass of mec2 = 0.511 MeV have been used. Then the coherent scattering cross section from iron may be estimated as σ ≡ σcoherent ≃ A2σnucleon = 2.7 × 10−38 cm2, where A = 56 for iron has been used. The corresponding mean free path is λ = 1/nσ, where n is the number density of iron nuclei and σ is the cross section for coherent scattering.

Assuming a pure iron core of density 1010 g cm−3, ρ n= = 1.07 × 1032 cm−3, 56 Mu which gives a mean free path λ≃3.5 × 105 cm. (Constant factors omitted from the above very rough estimates for cross sections can change this by 1–2 orders of magnitude.) The total mass in the core is 4 M ≃ π R3ρ ≃ 5.2 × 1036 g.

3 From Fig. 7.14(b), a neutrino energy production rate in the core of ε ≃ 1013 erg g−1s−1 or greater may be estimated. Thus, if in a very crude estimate one neglects the holdup time of the neutrinos in the core caused by their finite mean free paths, the total neutrino luminosity of the core is Lν ≃ εM ≃ 5 × 1049 erg s−1.

From a typical HR diagram a red supergiant is perhaps as much as 10 6 times more luminous than the Sun, giving a photon luminosity of about 1039 erg s−1. Thus neutrinos are potentially much more efficient at carrying away energy than photon emission in late-burning phases.

14.5 Assume from Eq. (7.48) that σν ≃ 10−44 53

Eν mec 2

2

cm2 ≃ 3.8 × 10−42 cm2,

Evolution of Higher-mass Stars

54

where we’ve used that the mass of the electron corresponds to 0.511 MeV and taken an average value Eν ∼ 10 MeV. The mean free path λ is ρ NA 1 n= , λ= nσν µ where n is the number density. Assuming ρ ∼ 1014 g cm−3 and µ ∼ 1, one may estimate λ ∼ 4.4 × 103 cm.

n ∼ 6 × 1037 cm−3

Actually σν is larger because of various coherent effects (see Problem 14.4) and the neutrino mean free path may become of order centimeters under these conditions.

14.6 This problem follows a discussion in Ref. [176] The total initial mass in a population of stars between the mass limits M1 and M2 is M0 =

∫ M2

M dN =

M1

∫ M2

ξ(M) dM,

M1

where ξ(M) is the initial mass function (see §9.12). Let us make the following assumptions for the fraction f of a star’s mass ejected in the course of stellar evolution: 1. Stars with initial M > MSN ≃ 8 M⊙ will explode as core collapse supernovae. Since for a typical supernova the ejected matter is much more massive than the neutron star left behind, as a rough estimate assume that f (M > MSN)~ 1. (If the core collapse leads to a black hole, little or no matter may be ejected, but such events should be much less common and will be neglected for this estimate.)

2. If MMS < M < MSN, assume a planetary nebula is ejected, leaving behind a white dwarf.

Thus, the fraction of ejected mass may be approximated as f (MMS < M < MSN) ~

M − MWD

, M where MWD is the average mass of a white dwarf in the evolved population.

3. If M < MMS, the star is of low mass and still on the main sequence, so approximate f (M < MMS) ∼ 0.

Thus the total ejected mass ∆M is given by ∆M =

∫ M2

ξ(M) f (M) dM

M1

=

∫ M2

ξ(M) dM +

MSN

=

∫ M2

MMS

∫ MSN MMS

ξ(M) dM − MWD

ξ(M)

∫M

M − MWD M

SN ξ(M)

MMS

M

dM + 0

dM.

Approximating ξ(M) by Eq. (9.16), evaluating the integrals, and assuming M1 = 0.1 M⊙

M2 = 100 M⊙

MWD = 0.6 M⊙

MMS = 0.7 M⊙

MSN = 8 M⊙,

Student Solutions Manual for Stars and Stellar Processes 1st Edition by Mike Guidry

55

Evolution of Higher-mass Stars

the above expressions give ∆M/M~ 0 0.32 for the fraction of initial mass returned to the interstellar medium.

14.7 We abridge a more extensive discussion that may be found in Prialnik [176]. The Sun is about 4.5 billion years old and ZSun~0.02. The galaxy is about 13 billion years old, so when the Sun was born the galaxy was about 8–9 billion years old. Given that the oldest observed stars in the galaxy have Zold ∼ 0.0003 and the youngest have Zyoung ∼ 0.04, ZSun Zyoung 0.02 0.04 ∼ ∼ 70 ∼ ∼ 2.

Zold 0.0003 ZSun 0.02 Assuming the metallicity of these stars to be representative of the metallicity of the galaxy at their time of formation, and for the oldest stars to be about the age of the galaxy and the youngest stars to have formed essentially today, it may be concluded that the metal content Z of the galaxy increased by a factor of ~70 in its first 8–9 billion years, but only by an additional factor of~2 in the next 4–5 billion years. Thus, the increase in metallicity was much larger early in the history of the galaxy than more recently.

15

Stellar Pulsations and Variability 15.3 In Lagrangian coordinates the equation of motion including possible deviations from hydrostatic equilibrium is obtained by rewriting Eq. (4.9) with mass as the independent variable (see Section 4.3.1): ∂P ∂m

=−

Gm 4π r4

−

1 ∂ 2r 4π r2 ∂ t2

.

Assume a small time-dependent variation in r and P around the equilibrium values, r(m,t) = r0(m)(1 + δ r(m,t))

P(m,t) = P0(m)(1 + δ P(m,t)),

insert into the previous differential equation, expand all factors containing 1 + δ P and 1 + δ r to first order in a binomial expansion, and neglect resulting terms of higher than linear order in δ r and δ P to give the desired result.

56

Student Solutions Manual for Stars and Stellar Processes 1st Edition by Mike Guidry

16

White Dwarfs and Neutron Stars 16.1 Assume a completely ionized sphere of radius R containing A nucleons and A electrons, with pressure dominated by the electrons but mass coming primarily from the nucleons. The average spacing between electrons is λ~ RA−1/3 and the characteristic momentum for particles confined to the corresponding volume is by uncertainty principle arguments h¯ h¯A1/3 h¯M1/3 pe ≃ ≃ ≃ , λ 1/3 R Rm p where mp is the average nucleon mass and the total mass of the star is M ≃ mpA. The corresponding energy for the assumed ultrarelativistic electrons is E≃e pec and the total Fermi energy of the electrons may be estimated as M Ef ≃ AEe ≃

mp

pec =

h¯McM1/3 1/3

=

h¯cM4/3 4/3

.

Rmp

Rm p m p

The gravitational energy produced by the nucleons is Eg ≃ −G

(mpA)2 R

= −G

M2 R

,

so the total energy can be estimated as E = Ef + Eg =

h¯cM4/3 4/3

Rmp

M2 −G

R

.

Both terms scale as R−1 but they have different mass dependence. The first term varies as M4/3 but the second term is quadratic in M. Thus at large enough M the second term dominates and the total energy will become negative. It is then energetically favorable to reduce R to make the energy even more negative: the system becomes unstable against gravitational collapse. The critical mass for this instability may be estimated by equating the two terms in the preceding equation, leading to !3/2 h̄c Mcrit ≃ ≃ 1M⊙, 4/3 Gm p which depends only on fundamental constants and mp. The corresponding critical number of electrons is M/mp ≃ 2 × 1057.

16.2 Assume for simplicity complete ionization with A nucleons and A electrons in a spherical radius R, with all pressure coming from electrons and all mass from the nucleons.

57

White Dwarfs and Neutron Stars

58

The mass is M~ mpA, where mp is the average mass of a nucleon. By the uncertainty principle, h¯ h¯A1/3 h¯M1/3 pf ≃ ≃ ≃ . λ 1/3 R Rm p The energy for a single electron is p2 Ee ≃ 2mf e ≃

h¯2M2/3

,

p me 2R2m2/3

and the total energy for all electrons is Ef = AEe =

M h¯2M2/3 = h̄ 2 M 5/3 . · 2m5/3m R2 mp 2R2m2/3 p me p e

Assuming all gravity to come from the nucleons, the total gravitational energy is EG ≃ 2 −GM /R. Thus the total energy is E = Ef + EG =

h¯2M5/3 5/3

2m p m eR

GM2 − R .

2

The equilibrium radius at fixed M may be found by setting dE/dR = 0 and solving for R, which gives 5/3

R=

(h̄2 /Gmp me) M1/3

19

= 7.7 × 10

g M

1/3

cm,

implying that MR3 = constant for a non-relativistic white dwarf. Inserting M = 0.6 M⊙ into the formula for the radius gives R∼7260 km. This may be compared with R∼9000 km inferred from Fig. 16.1, which was computed more accurately using a realistic equation of state and relativistic effects.

16.4 Since L = 4πσ R2T 4, L=

R R⊙

2

T T⊙

4

L⊙.

Using a radius R ∼ 10 km and a surface temperature T ∼ 106 K, the luminosity of a typical neutron star is L≃0.18 L⊙. From the Wien law (1.12), if T ∼106 K the spectrum peaks at about 2.9 nm, which is in the soft X-ray, extreme UV part of the spectrum. Thus, neutron stars are relatively bright to X-ray observatories like Chandra, as illustrated in Fig. 16.6.

16.11 The number density of 3He atoms is given by ρ n= = 1.6265 × 1028 m−3, AMu where ρ = 81 kg m−3 and A = 3 were used. The Fermi wavevector is then pf kf = = (3π 2n)1/3 = 7.66× 109 m−1 = 0.77Å−1. h¯

Student Solutions Manual for Stars and Stellar Processes 1st Edition by Mike Guidry

59

White Dwarfs and Neutron Stars

The Fermi energy measured relative to the ground state is h¯2k2 εf = 2mf ≃ 4.1 × 10−4 eV, where the bare helium-3 mass rather than its renormalized effective mass in the medium has been used (at zero pressure, the effective mass is about three times the mass of a free 3 He atom, but this difference doesn’t matter much for our estimates). The Fermi velocity is then h̄ k f vf = ∼ 1.62 × 104 cm s−1, m where again the bare 3He mass has been used. Thus, the Fermi velocity is highly nonrelativistic, justifying our use of non-relativistic formulas (which should come as no surprise). Finally, the superconducting transition temperature of around 2–3 mK implies an average energy E = kT ∼ 10−7 eV. This is far less than εf, so the dense Fermi liquid is highly degenerate.

17

Black Holes 17.1 From the Minkowski metric given in (17.6) −ds2 1 d τ2 = = (c2dt2 − dx2 − dy2 − dz2) c2( c2 " 2 2 1 dx 2 dz dy = dt 1 − 2 + + dt c dt dt

#)

2

v2 = 1 − 2 dt 2 . c

where v = [(dx/dt)2 + (dy/dt)2 + (dz/dt)2]1/2 is the magnitude of the velocity.

17.7 From Eq. (17.11) the Schwarzschild radius is 2GM M rs = = 2.95 km c2 M⊙ = 1.27 × 107 km = 0.085 AU = 18.2 R⊙, where M = 4.3×106 M⊙ was used. Then from small-angle trigonometry the angle subtended by the transverse size size of the black hole in the absence of gravitational lensing is θ ∼ 1 × 10−10 rad = 21.2 µarcsec at a distance of 8 kpc = 2.46 × 1017 km.

60

Student Solutions Manual for Stars and Stellar Processes 1st Edition by Mike Guidry

18

Accreting Binary Systems 18.2 At the accretion radius racc the gravitational potential from the neutron star of mass M is ∼ GM/racc and the kinetic energy per unit mass for the wind is 12v2. Equating these gives for the accretion radius 2GM racc = 2 . v If all mass from the companion wind within π r2acc is assumed captured, the accretion rate on the neutron star is Ṁacc =

2 G2M2 π racc Ṁwind = 2 4 Ṁ wind , 2 4π a av

where a is the separation of the two stars, the first factor in parentheses is the solid angle fraction subtended by the neutron star capture radius as seen from the companion, and Ṁ w in d is the mass loss rate from the companion. Taking as representative numbers a ∼ 50R⊙

v ∼ 103 km s−1

M ∼ 2M⊙

gives an accretion rate Ṁ ac c ∼ 5.7 × 10−5Ṁwind .

Thus an accretion rate in excess of 10−12 M⊙ yr−1 on the neutron star requires winds from the companion with Ṁ w i nd in excess of~ 10−8 M⊙ yr−1. Such large mass loss rates occur only for a few stars, such as the most massive main sequence stars and some supergiants.

18.3 The solution involves somewhat messy empirical estimates and may be found in Section 4.5 of Ref. [89]. Basically one finds that the circularization radius for the accretion disk is always larger than the radius of compact objects but not necessarily larger than the radius of more normal stars if the binary period is short enough, as for Algol.

18.4 (a) From the masses before and after in nuclear reactions burning hydrogen to helium one finds a typical efficiency for mass to energy conversion of η ∼ 0.007. Requiring that η ṁc 2 = 1047 erg s−1 = 3.2 × 1054 erg yr−1 yields that m˙≃255M⊙ yr−1 if this luminosity is supplied by hydrogen fusion. It is difficult to conjecture a mechanism consistent with observations that could account for this. (b) On the other hand, for black hole accretion the mass to energy conversion efficiency could lie in the range η ≃ 0.1 − 0.3. Then the luminosity could be sustained by accretion of as little as ~5 M⊙ yr−1, for which there are plausible mechanisms. The Eddington luminosity (maximum luminosity for which radiation pressure would not reverse accretion 61

Accreting Binary Systems

62

infall) is given by LEdd = 1.3 × 1038

M erg s−1.

M⊙

Equating this to the observed ∼ 1047 erg s−1 and solving for the mass of the central object gives M ≃ 7.7 × 108 M⊙. (c) Observed light variation on timescales of days argues by causality that the source has a maximum diameter of 1 light-day or about 170 AU (twice the diameter of the Solar System). Thus, it may be inferred on rather general grounds that the AGN central engine has of order 109M⊙ concentrated in a region not much larger than the Solar System. The most plausible explanation is a supermassive black hole. The Schwarzschild radius of a 109M⊙ black hole would be R ≃ 2.95

M km ≃ 3 × 109 km ≃ 20 AU, M⊙

which is approximately the radius of the orbit of Uranus.

18.6 From Eq. (18.11) the energy released through accretion is ∆Eacc =

GMm M = 1.327 × 1021 R M⊙

m g

km erg, R

where R is the radius and M the mass of the object onto which accretion is taking place, and m is the mass of the accreting material. The entries in Table 18.1 result from using in this formula m = 1 g, 1. M ∼ 1M⊙ and R ∼ 10 km for a neutron star, 2. M ∼ 1M⊙ and R ∼ 104 km for a white dwarf, 3. solar parameters for a normal star, 4. the radius of the innermost stable circular orbit for a Schwarzschild black hole, R = 3rs = 3 × 2.95

M M⊙

km,

where rs is the Schwarzschild radius and M = 10 M⊙ has been assumed, and the result from Problem 5.1 that fusion of hydrogen to helium releases×6 1018 erg g−1.

These considerations are simple and do not account for real-world effects such as accretion rates and efficiencies for extracting the released energy, but they indicate clearly that enormous power is potentially available from accretion onto highly compact objects.

18.10 Assuming the gravitational accretion energy to be absorbed by a surface layer that equilibrates at a temperature Tacc and radiates as a blackbody, it may be estimated that (see Box 1.3) 1/4 1/4 GMM˙ L = , Tacc = 4πσ R2 4πσ R3 where in the second step all luminosity is assumed to come from the accretion and GMM˙ L = Lacc = R

Student Solutions Manual for Stars and Stellar Processes 1st Edition by Mike Guidry

63

Accreting Binary Systems

has been taken from Eq. (18.12). The objects are compact so the accretion layer likely has a radius R comparable to that of the non-accreting object. Thus, we assume 1. For a representative white dwarf, R = 8000 km and M = 0.8 M⊙.

2. For a representative neutron star, R = 10 km and M = 1.5 M⊙.

3. For the 10M⊙ black hole the radius of the innermost stable circular orbit at R = 3rs = 3 × 2.95

M M⊙

km

is taken as the accretion radius (see Chs. 11 and 14 of Ref. [101]). This is of course not an actual surface, so we are arguing even more loosely here than in the rest of this problem.

Then the entries in Table 18.2 follow from using these numbers in Eq. (18.14) and in the preceding equation for Tacc. The present crude arguments give approximately correct numbers but Problem 18.8 addresses these issue in a more rigorous way, by assuming the accretion luminosity to arise from a disk in orbit around the compact object.

19

Nova Explosions and X-ray Bursts 19.2 Use tabulated rates (for example the Caughlan–Fowler or ReacLib rates described in Appendix D) to calculate the resonant and non-resonant contributions to the rate for Q = 4.628 MeV,

13N(p, γ)14O

for temperatures relevant to main-sequence CNO reactions. Compare this rate, which will be strongly temperature dependent, with the constant rate expected for the competing β decay assuming an appropriate density. The (p, γ) velocity-averaged rate using the Caughlan– Fowler compilation is [107] ⟨σ v⟩ ≃

6.71× 10−17 2/3 T9

1/3

e−15.2/T9 +

4.04× 10−19 3/2 T9

e−6.35/T9 cm3 s−1,

where the first term is non-resonant and the second term is resonant. The β -decay constant for 13N → 13 C+ e+ + νe is λ = 1.159 × 10−3 s−1.

To compare the two rates requires converting the (p, γ) rate⟨ σ v⟩ to units of s−1 by multiplying it by the proton number density ρ np = X , Mu where Mu is the atomic mass unit, X is the hydrogen mass fraction, and ρ is the mass density. For example, Fig. 19.1 [this document] compares these rates as a function of the temperature in units of 109 K for the case X = 0.7

ρ = 100 g cm−3

np = 4.2 × 1025 cm−3.

In this figure the various curves show the total (p, γ) rate, the separate contributions from the resonant and non-resonant terms (dashed lines), and the β -decay rate (dotted line), all in units of s−1. From this figure we see that the proton capture reaction begins to compete strongly with β -decay under these conditions when the temperature reaches T9 ∼ 0.1.

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Student Solutions Manual for Stars and Stellar Processes 1st Edition by Mike Guidry

Nova Explosions and X-ray Bursts

65

105

Rate (s-1)

100

Nonresonant

-decay

10-5

10-10

(p, ) Resonant

10-15

10-20 0.01

0.05

0.10

0.50

1.0

T9

.

Fig. 19.1

Rates as a function of temperature for radiative proton capture and β -decay of 13N assuming ρ = 100 g cm−3 and X = 0.7. See the solution for Problem 19.2.

20

Supernovae 20.2 From Eq. (4.16) the dynamical timescale is s s 3π gcm−3 τ= 2100 = ρ¯ 32Gρ¯

s,

where ρ¯ is the average density. Inserting ρ¯ =×5 109gcm−3 for the iron core at the beginning of collapse gives τ = 3×10−2 s. Thus the core responds to gravitational perturbations on a timescale of tens of milliseconds.

20.4 The 56Fe core will become gravitationally unstable when it reaches the Chandrasekhar mass. Let’s take this to be Mcore ~1.4M⊙ for our estimate. (Because of the composition difference of the iron core relative to a white dwarf, the actual Chandrasekhar mass for the core is typically in the range 1.2—1.3M⊙, but we are just making an estimate so we will neglect such details.) The number of 56Fe nuclei in this core is Mcore = 3 × 1055, 56Mu where Mu is the atomic mass unit. If each of these is photodisintegrated the total energy absorbed is NFe=

∆Ephoto ≃ 3 × 1055 Fe

124.4 MeV Fe

= 3.7 × 1057 MeV = 6 × 1051 erg.

This may be compared with approximately 3×1051 erg obtained in the more realistic simulation of Fig. 20.8(b). The total number of protons in this core is 26 times the number of 56Fe nuclei computed above, so Np ∼ (26)(3 × 1055) ∼ 7.8 × 1056.

The core is assumed electrically neutral so there will be an equal number of electrons, and if all electrons and protons undergo p + → e n + νe an equal number of neutrinos will be produced. Thus, if each neutrino carries off an average of 10 MeV the energy removed from the core by complete neutronization is ∆Eneutronize ≃ 7.8 × 1056 neutrinos

10 MeV neutrino

= 7.8 × 1057 MeV = 1.2 × 1052 erg.

Hence photodisintegration and neutronization each are capable of removing 10 51-1052 erg from the core quickly. If the Sun radiated at its present luminosity of 3.8 × 1033 erg s−1 66

Student Solutions Manual for Stars and Stellar Processes 1st Edition by Mike Guidry

67

Supernovae

for its main-sequence lifetime of about 10 billion years, the amount of energy released would be around 1051 erg. Thus, photodisintegration and neutronization of an iron core can remove (on a timescale of seconds) an amount of energy comparable to or greater than the energy that will be released by the Sun over its entire main-sequence life.

21

Gamma-ray Bursts 21.4 Evaluation of constants [see Eq. (7.10) for the Thomson cross section] allows Eq. (21.3) to be written as τ≃

f σTFD2 13 , = 7.7 × 10 f 2 4 δ t mec

D 3000 Mpc

2

F −7 10 erg cm−2

10 ms δt

2

.

Since f and the product of quantities in parentheses are of order one, the optical depth evaluated from the above expression is huge (~ 1014). This is inconsistent with the observed nonthermal spectrum for gamma-ray bursts, since a nonthermal spectrum typically requires a medium that is optically thin. The fallacy is that Eq. (21.3) is invalid for a gamma-ray burst because it must be modified to account for the ultrarelativistic kinematics of the burst.

When that is done, as in Eq. (21.4), the above expression is multiplied by a factor approximately equal to 1/γ4+2α, as discussed in Section 21.4.2. For α ~ 2 this will yield optical depths smaller than one for γ of order 100 or larger.

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Student Solutions Manual for Stars and Stellar Processes 1st Edition by Mike Guidry

22

Gravitational Waves and Stellar Evolution 22.1 From Fig. 22.5 the peak strain is h ∼ 10−21, so the displacement may be estimated as ∆L ∼ hL ∼ (10−21)(2 × 105 cm) = 2 × 10−16 cm for an arm length L = 2 km. This is some 1000 times smaller than the size of a proton in the atoms of the interferometer!

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