astro160 problems solutions philipsquestion2.4pg40

8/20/2019 Astro160 Problems Solutions Philipsquestion2.4pg40

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Astro 160: The Physics of Stars

Served by Roger Griffith

Nutritional Facts:

Serving size: 1 Semester (16 weeks)

Servings per container: many problems and solutions

Problem set 2Problem # 1

In class, I derived the relationship between the luminosity and mass of stars under the assumption that

energy is transported by radiative diffusion and that the opacity is due to Thomson scattering. We will

carry out many related estimates so it is important to become familiar with this process. Consider a star

in hydrostatic equilibrium in which energy transport is by radiative diffusion. The star is composed of

ionozed hydrogen and is supported primarily by gas pressure.(a). Derive an order of magnitude estimate of the luminosity L of a star of mass M and radius R if the

opacity is due to free-free absorption, fo which κ ≈ 1023ρT −7/2 cm2g−1 (ρ is in cgs).

We know that the radiation flux is given by

F rad ∼ caT 3

κρ ∇T

where we know that a is the radiation constant, c is the speed of light, T is the temperature, κ is theopacity, which in our case is given by free-free absorption, ρ is the mass density and ∇T is the temperaturegradient. We have the following relationships

ρ ∝ M

R3 ∇T ∝

dT

dR ∝

T R−T c R R− RC ∝−

T C

R κ ∝ 1023ρT −7/2

given these relationships we can find

F rad ∝ caT 15/2 R5

M 2

we also know that the luminosity can be written as

L = 4πr 2F rad ⇒ F rad = L4πr 2

which gives us

L ∝ caT 15/2 R7

M 2

we can find the temperature by using the virial theorem which can be written as

T ≈ GMm p µ3 Rk

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where k is now the boltzman constant. Substituting this expression into the above equation yields

L ∝ caR−1/2 M 11/2

Gm p

k

15/2

this gives us an order of magnitude estimate of the luminosity of a star with mass M and radius R.

(b). If all stars have roughly the same central temperature, and are supported by gas pressure, what isthe mass-luminosity scaling (proportianality) relationship for stars?

we now know that the luminosity scales as

L ∝ M 11/2 R−1/2

we can find the relationship between the mass M and the radius R of a star by using hydrostatic

equilibrium.

dP

dr = −GM

r 2 ρ

Pc ∝ M R

ρ

ρT ∝ M

R ρ

M ∝ R

since T is constant, substituting this into the luminosity relationship yields

L ∝ M 5

(c). Give a quantitative argument as to whether free-free opacity dominates electron scattering opacityin stars more massive that the sun or in stars less massive that the sun.

We can solve this problem by looking at the defenition for the opacity in free-free absorption, which

can be written as with T constant

κ ∝ ρ ρ ∝ M

R3 M ∝ R

thus we find

κ ∝ 1

M 2

this expression tells us that the lower the mass of the star the higher the opacity, thus in lower mass

stars the free-free opacity dominates.

Problem # 2

The central density and temperature of the sun are ρc ≃ 150 g cm3 and T c ≃ 1.5× 107 K. For theconditions at the center of the sun, answer the following questions. Assume that the sun is composed

solely of ionized hydrogen.

(a). What is the mean free path of an electron due to electron-electron Coulomb collisions? What is

the typical time between collisions?

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we can now see the collision times for the electron-electron collision occurs more rapidly due to the

mass being so much smaller.

t e ≪ t p

(c). Which opacity is more important for photons, Thomson scattering or free-free absorption?

We know that

κ T = neσT

ρc=

2σT m p

∼ 0.80

and

κ F = 1023ρT −7/2 ∼ 1.15

free-free absorption dominates the opacity for photons in this case? not sure why this is. We know that

Thomson scattering is the primary way that photons move the energy out.

(d). What is the mean free path of a photon? How does this compare to the mean free path of an

electron (this should give you a feel for why photons are far more effective at moving energy around in

stars)? What is the typical time between photon absorptions/scattering?

we know that the mean free path of a photon is given by

l = 1

neσT where σT =

8π

3

e2

4πε0mec2

2= 6.65×10−25 cm2

which yields

l photon =

m p

2ρσT ∼ 8.3×10−3

cm

lelectron = m p

ρcπ

kT

e2

2∼ 8.9×10−7 cm

The typical time for a photon collision is given by

t = l p

c∼ 2.8×10−13 s

(e). For a photon undergoing a random walk because absorption/scattering, how long would it take to

move a distance Rsun given the results in (d)? For comparison, it would take 2.3 seconds moving at the

speed of light to travel a distance Rsun in the absence of scattering/absorption.

We know that the diffusion time can be acquired with

t d if f = thermal energy

L ∼ R

2

lc

nkT

aT 4 ∼ R

2nk

lcaT 3 ∼ R

22ρk

m plcaT 3

we know that the average time for a photon to leave the star is given by

t d if f ∼ R2sun

l phc ∼ 104 yr

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Problem # 3

How old is the sun? In this problem we illustrate how the naturally occuring radioactive isotopes of

uranium, U 235 and U 238 can be used to determine the age of the rocks. Both isotopes decay via a sequence

of α-decays and β-decays to form stabel isotopes of lead: the decay chain of U 235 ends up with Pb207, andthe decay chain of U 238 ends up with Pb206. As a result, the number of uranium nuclei in a rock decays

exponetially with time in accord with:

N 5(t ) = N 5(0)e−λ5t and N 8(t ) = N 8(0)e−λ8t

To avoid clutter, the last digit of the mass number of the isotope has been used as a subscript label. The

decay constants λ5 and λ8 for the two isotopes corresponds to half-lives of

T 5 = ln 2

λ5= 0.7×109 yrs T 8 = ln 2

λ8= 4.5×109 yrs

The magnitudes of these half-lives are ideally suitable to the determination of the ages of the rocks

which are over a billion years old. Now consider a set of rock samples which were formed at the same

time, but with different chemical compositions. They differ in chemical composition because differentchemical elements are affected differently by the processes of rock formation. However rock formation

processes do not favour one isotope over another. For example, on formation, the relative abundances of

U 235 and U 238 should be the same in every sample. But these abundances will change with time as the

deacy of U 235and U 238 produce nuclei of Pb207 and Pb206.

• Consider the ratio of the increase in the number of Pb207 nuclei relative to the increase of Pb206nuclei. Show that this ratio is the same for all rock samples which were formed at the same time,

and that it is given by

N 7(t )− N 7(0) N 6(t )

− N 6(0)

= N 5(t )

N 8(t )

eλ5t −1eλ8t

−1

We know that the ratio of the two isotopes can be written as

N 7(t )− N 7(0) N 6(t )− N 6(0) =

N 5(t )− N 5(0) N 8(t )− N 8(0)

and given the first expression given in this problem, which can also be written as

N 5(0) = N 5(t )eλ5t and N 8(0) = N 8(t )e

λ8t

substituting this into our previous expression yields

N 7(t )− N 7(0) N 6(t )− N 6(0) =

N 5(t )

N 8(t )

eλ5t −1eλ8t −1

which is what we were asked to show.

• Consider a graph in which the measured abundances in the rock samples of Pb 207 and Pb206 areplotted, N 7(t ) along the y-axis and N 6(t ) on the x-axis. Show that a straight line will be obtained if all the samples were formed at the same time.

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We know that

N 7(t ) = N 5(t )

N 8(t )

eλ5t −1eλ8t −1 · N 6(t )

where N 5(t )

N 8(t )

eλ5t −1eλ8t

−1

= constant

• Given that the current ratio of naturraly occurring U 235 to U 238 is 0.0071, evaluate the gradient of the straight line for rock samples of age (a) 1 billion years, (b) 3 billion years and (c) 5 billion years.

We know that the gradient of the straight line is just the constant in front of N 6(t ) so we just have to plugin numbers

(a). t = 1 billion years.We know that

λ5 ∼ 9.90×10−10yr−1 λ8 ∼ 1.5×10−10yr−1

given these and the fact that we know the ratio between U 235 and U 238 we can find the gradient, for 1

bilion years we get0.0071 · e

λ5t −1eλ8t −1 = 0.0715

For 3 billion years we get

0.0071 · eλ5t −1

eλ8t −1 = .231and finally for 5 billion years we get

0.0071 · eλ5t −1

eλ8t −1 = .891

Problem # 4 Radiative Atmospheres

In this problem we will solve for the structure of the outer part of a star assuming that energy is

transported solely by radiative diffusion (which is not the case in the sun, but is the case in stars more

massive than the sun). The star has a mass M and a luminosity L. Assume that the luminosity and mass

are approximately constant at the large radii of interest, that gas pressure dominates, and that the opacity is

due to electron scattering. Do not assume that the atmosphere is thin (i.e even though M r ≈ constant = M ,because r changes, the gravitional acceleartion is not constant).

Write down the equations for hydrostatic equlibrium and energy transport by radiative diffusion. Use

these to calculate d Prad /dP, the change in radiatio pressure with pressure in the atmosphere. What doesthis result imply for how the ratio of gas pressure to radiation pressure changes as a function of the

distance in the atmosphere? Show that your result for dPrad /dP implies that ρ ∝ T 3 and P ∝ ρ4/3 for

radiative atmospheres (in the language that we will use in the next week, this means that the radiative part

of the star is an n=3 polytrope).

since we know what the radiation pressure is we can find what the change is with respect to r

Prad = 1

3aT 4

dPrad

dr =

1

3a

d

dr (T 4) =

4

3T 4∇T

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and we know that the radiation flux is given by

F =−43

caT 3

κρ ∇T

thus we can write

F = dPrad

dr

c

κρ

dPrad

dr = F

κρ

c

the hydrostatic equilibrium equation is

dP

dr =−GM

r 2 ρ

deviding these two expressions yield

dPrad

dP =

Fκ r 2

cGM

we know that the Flux and luminosity are related by

L = 4πr 2F F = L4πr 2

thus we finddPrad

dP=

Lκ

4πcGM

this result implies that the ratio of the gas pressure to radiation pressure is independent of the distance

in the atmosphere. To show ρ ∝ T 3 we can just use scaling arguments

Prad

Pg∝

L

M ⇒ T

4

ρT ∝

L

M

since we assumed that L and M are constant than this gives

T 3 ∝ ρ

To show that P ∝ ρ4/3 we can also use scaling argument, we also know that the radiation pressurescales as some constant times the gas pressure

Prad

Pg∝

L

M Pt = Pg + Prad ⇒ Pg = Pt −Prad

thus we find

Pr

Pt −Pr ∝ 1 ⇒ Pr ∝ λPgbut we know that

Pg ∝ ρT T ∝ ρ1/3 Pg ∝ ρ

4/3

thus we know that

λPg ∝ Pt −λPgPt ∝ 2λPg ∝ ρ

4/3

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Problem set 3

Problem # 1

(a). Show that heat transfer by radiative diffusion implies a non-zero gradient for the radiation pressurewhich is proportional to the radient heat flux. Bearing in mind that the magnitude of the force per unit

volume in a fluid due to the pressure is equal to the pressure gradient, find the radient heat flux density

which can, by itself, support the atmosphere of a star with surface gravity g. Hence show that a star of

mass M has a maximum luminosity given by

Lmax = 4πcGM

κ

where κ is the opacity near the surface. Obtain a numerical estimate for this luminosity by assumingthat the surface is hot enough for the opacity to be dominated by electron scattering. (This maximum

luminosity is called the Eddington luminosity.

To show that the heat transfer by radiative defusion implies a non-zero gradient we must begin with

F r = −43

aT 3

κρ ∇T F r ∝ Prad

knowing these relationships we can do

Prad = 1

3aT 4

dPr

dr =

4

3aT 3

dT

dr ⇒ dT

dr =

3

4

1

aT 3dPr

dr

thus this implies that there is a non-zero gradient.

To show that

Lmax = 4πcGM

κ

we must begin with the equation derived from problem 4 in the last problem set, i.e

dPr

dP =

Lκ

4πcGM

but since we know that

P = Pg + Pr Pg≪ Pr ⇒ P ≈ Pr dPr

dPr = 1

and we find that

L = 4πcGM

κ ≈ 3.3×104 Lsun

M

M sun

We cannot obtain a numerical estimate because we do not know tha mass. We could use M sun but this

would not be correct.

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(b). Assume that radiative diffusion dominates energy transport in stars and that the opacity is due to

Thomson scattering. Use a scaling argument to estimate the mass M (in M sun) at which the luminosity of

a star is ≈ Ledd .We can do an order of magnitude estimate with respect to the sun by

L ∝

M 3 L

Lsun = M

M sun3

and substituting the Eddington luminosity for L we find that M is given by

M =

4πcGM

Lsunκ T

1/2 M

3/2sun ≈ 180 M sun

Problem # 2

The physical quantities near the center of a star are given in the following table. Neglecting radiation

pressure and assuming the average gas particle mass m̄ is 0.7 amu, determine whether energy transport is

convective or radiative.

r m(r ) Lr T r ρ(r ) κ

0.1 Rsun 0.028 M sun 24.2 Lsun 2.2×107K 3.1×104kg m−3 0.040 m2kg−1

Using equation ?? from Phillips L(r )

m(r )

crit

= γ −1

γ

16πGc

κ

Pr

P

and the following relationships

Pr = 1

3aT 3 P ∼ Pg = ρ(r )

m̄k bT γ =

5

3κ T = 0.04 m

2/kg

we find L(r )

m(r )

crit

= 2

5

16πGc

κ T

aT 3m̄

3ρ(r )k b

0.175 W

kg > .07

W

kg

which implies that the energy transport of this star is primarily due to convection.

Problem # 3The surface of a star (the “photosphere”) is the place where the mean free path of the photons ℓ is

comparable to the scale-height h of the atmosphere . At smaller radii (deeper in the star), the density is

higher and ℓ ≪ h , which implies that the photons bounce around many times; at larger radii ρ is smaller,ℓ ≫ h, and the photons are rarely absorbed and so travel on straight lines to us. Thus ℓ ≈ h is a goodapproximation to the place in the atmosphere of a star where most of the light we see originates.

a) The temperature at the photosphere of the sun is 5800 K. Estimate the mass density ρ in the photo-sphere. Assume that Thompson scattering dominates the opacity.

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Knowing that ℓ∼ h we can derive the following relationship

nσ = κρ ρ = nσT

κ T ℓ =

1

nσ = h =

k bT

m̄g

thus we find that the density is given by and assuming m̄ ≈ m p

ρ = m̄gk bκ T T

≈ 8.26×10−4 g/cm3

b) In reality, the surface of the sun is so low that hydrogen is primarily neutral. There are thus not that

many free electrons to Thompson scatter off of. The opacity at the surface of the sun is instead due to the

H − ion and is given by κ ≈ 2.5× 10−31ρ1/2T 9cm2g−1. Using this (correct) opacity, repeat the estimatefrom a) of the density at the photosphere of the sun.

Substituting the opacity given into the above expression yields

ρ3/2 = m̄g

2.5×10−13kT 10b ⇒ ρ = m̄g

2.5×10−31k bT 102/3

≈9.8

×10−

8 g/cm3

c) Just beneath the photosphere, energy is transported by convection, not radiation, for the reasons

discussed in class (in fact, the photosphere is the place where photons travel so freely out of the star that

energy transport by radiation finally dominates over convection). Estimate the convective velocity near

the photosphere given your density from b).

The convective heat flux is given by

F c = 1

2ρv3c vc =

2F c

ρ 1/3

and knowing that

F c = L

4πr 2

we find that the convective velocity is given by

vc =

2 L

4πr 2ρ

1/3≈ 1.09×106 cm/s

d) What is the characteristic timescale for convective ”blobs” to move around near the pho-tosphere?

How does this compare to the observed timescale for granulation on the surface of the sun, which was a

few min in the movie we watched in class?

since we know that the characteristic time scale is given by

t blob = ℓ

vc≈ h

vc

we know that h which is the scale height of the sun is given by

h = kT

g m̄ = 1.7×107 cm

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thus we find that the blob timescale is

t blob ≈ 15.6 s

which is a lot shorter than the timescale given by the movie which was approximately 2 minutes.

e) Is the assumption ds/dr ≈ 0 valid near the surface of the sun? Why or why not?

Since we know that the temperature gradient near the surface of the sun is very high and energy ismostly transported by photons impies that we cannot make the assumption ds/dr ≈ 0 .

Problem # 4 Convective atmospheres

In HW 2, you calculated the structure of a stellar atmosphere in which energy is transported by radiative

diffusion; you showed that such an atmosphere satisfies P ∝ ρ4/3 . Here we will consider the problem of a convective atmosphere, which is much more relevant to sun-like stars. For simplicity, assume that the

atmosphere is composed of fully ionized hydrogen. The solar convection zone contains very little mass

(only ≈ 2of the mass of the sun). Thus, let’s consider a model in which we neglect the mass of theconvection zone in comparison to the rest of the sun. For the reasons discussed in class, we can model

the convection zone as having P = K ργ with γ = 5/3 and K a constant. Rc is the radius of the base of the

convection zone.a) Solve for the density, temperature, and pressure as a function of radius in the convection zone. Do

not assume that the convection zone is thin (i.e., even though M r = constant = M , because r changes

significantly in the convection zone, do not assume that the gravitational acceleration is constant).

To solve for the Pressure we can begin with

dP

dr =−ρGM

r 2 P = K ργ ⇒ ρ =

P

K

1/γ

thus we find

dPP

K −3/5

= −GM

r 2 dr

and integrating over the following limits we find

Z PPrc

P

K

dP

−3/5=−

Z R Rc

GM

r 2 dr

this integral yields5

2K 3/5

P2/5−P2/5c

= GM

1

R− 1

Rc

thus we find that the pressure is given by

P =

2

5K −3/5GM

1

R− 1

Rc

+ P

2/5c

5/2

To solve for the density we can just plug this solution into

ρ =

P

K

3/5=

1

K 3/5

2

5K −3/5GM

1

R− 1

Rc

+ P

2/5c

3/2

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and finally the temperature can be found by using

P = nk bT = 2ρ

m pk bT ⇒ T (r ) = m pP

2ρk b

and substituting the P and ρ from the previous expressions we find

T (r ) = m pK 3/52k b

25

K −3/5GM

1 R− 1

Rc

+ P2/5c

b) In detailed solar models, the pressure at the base of the convection zone is ≈ 5.2× 1013dyne/cm2and the density is ρ ≈ 0.175 g cm−3. Using your solution from a), estimate the radius of the base of theconvection zone Rc . Compare this to the correct answer of Rc ≈ 0.71 Rsun

If we solve the density equation for Rc we find

1

Rc=

1

R−

(ρK 3/2)2/3−P2/5c · 5K

3/5

2GM

and plugging in values we find that

1

Rc= 1.998×10−9m−1 ⇒ Rc ≈ 5.11×108m = 0.72 Rsun

c) In your model, what is the temperature of the sun at 0.99 Rsun, 0.9 Rsun, and at the base of the solar

convection zone. This gives you a good sense of how quickly the temperature rises from its surface value

of ≈ 5800 K as one enters the interior of the sun.To find the temperature as a function of radius we would use the temperature equation derived from

part (a). i.e

T (r = 0.99 Rsun) = m pK

3/5

2k b

−2

3K −3/5GM r

25

66 · Rsun + P2/5c

≈ 4.1×104 K

T (r = 0.90 Rsun) = m pK

3/5

2k b

−2

3K −3/5GM r

15

18 · Rsun + P2/5c

≈ 5.1×105 K

T (r = 0.72 Rsun) = m pK

3/5

2k bP

2/5c ≈ 1.8×106 K

Problem set 4

Problem # 1

I mentioned in class that there are two ways to estimate the energy carried by convection. The first is

that the energy flux is F c ≈ 1/2ρv3c ≡ F c,1 where vc is the characteristic velocity of the convective motions.

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This is the KE flux carried by moving blobs. The other estimate is that F c ≈ ρ∆ Evc where ∆ E is thedifference in the thermal energy of a rising hot blob (or sinking cool blob) relative to the background star

(where E is per unit mass). I claimed in lecture that these two expressions are equivalent, to order of

magnitude (which is the accuracy of mixing length theory). In this problem, you will prove my claim.

(a). Calculate the acceleration a due to buoyancy of a rising hot blob (or sinking cool blob) in terms of

the fractional density difference ∆ρ/ρ relative to the background star. Don’t worry about the sign of the

acceleration or ∆ρ?, just their magnitudes.

We know that the accelaration of the blob due to bouyancy is given by

ab = g

ρbρ∗−1

= g

∆ρ

ρ

since ρb ≈ ρ∗.(b). Use (a) to calculate the convective velocity vc in terms of ∆ρ/ρ. Recall that in lecture we estimated

vc using the work done by the buoyancy force.

We know that the work done by the bouyancy force can be found by

W = 1

2mv2c =

Z l0

F ·d l = aml

thus we find that the convective velocity is given as

vc =

2g

∆ρ

ρ l

which can also be expressed as

v2c

2gl =

∆ρ

ρwe can also write this as given that l ∼ H , thus

v2c = 2g∆ρ

ρ

k bT

mg = 2

∆ρ

ρ

k bT

m

(c). Use (b) to calculate ∆ E , the difference in the thermal energy (per unit mass) of a rising hot blob(or sinking cool blob) relative to the background star, in terms of vc.

We can write the last expression as

T = v2

c

m

2k b

ρ

∆ρ

and from the equation of state, which is given as

∆ E = 1

φ

k b∆T

m̄ (∆ρ ·V )

where (∆ρ ·V ) is the mass. Using these two expression and what we found from part (b) we can seethat

∆ E

ρV =

1

2φv2c =

∆ E

m̄

∆ E

m ∝

v2c

2

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d) Combine your previous results to show that F c,1 ≈ F c,2 .From (a) we know that

F c,1 ≈ 12

ρv3c F c,2 ≈ ρ∆ E

m vc

using this and our solution we find

1

2 ρv3c ≈ ρ

∆ E

m vc ⇒ 1

2v2c ≈

∆ E

m

Problem # 2

Estimate the convective velocity vc and the dimensionless entropy gradient (ds/dr )( H /c p) in the con-vection zones of 0.1 and 10 M sun stars. Assume that the material undergoing convection is at about the

mean density of the star and that gas pressure dominates. You can either use a scaling argument to estimate

the density, temperature, luminosity, etc. of such stars or look up in a book (e.g., Carrol & Ostlie) any

properties of 0.1 and 10M? stars that you need to make your estimate (e.g., radius and luminosity). But

you can’t just look up vc and (ds/dr )( H /c p).

From class we know that

F c = ρα3C 3s

H C p dsdr 3/2 vc = C s

H C p dsdr 1/2

but we know that

F c ≈ L4π R2

thus H C pds

dr

=

L

4π R2ρ

2/31

C 2s

which reduces to

H C pds

dr = LR3 M ∗

2/3m p

k bT

where

C 2s = kT

m pρ =

3 M

4π R3

and for the convective velocity we find

vc = C s

H C pds

dr

1/2

=

LR

3 M

1/3from Carrol and Ostley we find that for 10 M sun and .1 M sun we find that the radius, and luminosity are

approximately M ≈ 10 M sun R ≈ 6 Rsun L ≈ 5700 Lsun M ≈ 0.1 M sun R ≈ 0.2 Rsun L ≈ .0034 Lsun

given these values we find

M = 10 M sun

H C pds

dr

≈ 3.61×10−6 vc ≈ 5.3×104 cm/s M = 0.1 M sun

H

C p

ds

dr

≈ 5.8×10−10 vc ≈ 698 cm/s

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Problem # 3 Polytropes

(a). The mass M of a star is given by

M =Z R

04πr 2ρ(r )dr

Use the Lane-Emden equation for polytropes, and the dimensionless density and radius defined in lecture,

to rewrite this in terms of the central density of the star as

ρc = ρ̄an =

3 M

4π R3

an

where an is a dimensionless number, the ratio of the central density to the mean density of the star. an is

a function that you should determine that depends only on the solution to the Lane-Emden equation (you

cannot actual evaluate an in general without numerically solving for θ[ζ], so your answer will just be interms of the solution to the Lane-Emden equation).

Since we know that

Θ = ρ(r )ρ

1/n

ξ = r

a

given these two relations we can find

ρ(r ) = Θnρc r 2 = a2ξ2 a = R dr = Rd ξ

and from the Lane-Amden equation we know

d

d ξ

ξ2

d Θ

d ξ

= −ξ2Θn

given these following relationships we find that

M = −Z

1

04π R3ξ2Θnρcd ξ = −4π R3ρc

Z 1

0

d d ξ

ξ2 d Θ

d ξ

d ξ

thus we find that an is given by

an = −13

Z 10

d

d ξ

ξ2

d Θ

d ξ

d ξ

and we can finally show that

ρc = 3 M

4π R3an

(b). Show that the central pressure of a polytrope can be written as

Pc = 4πGρ2ca

2

n + 1

where a = an is the constant (with units of length) defined in lecture (note that the polytropic relationP = K ργ can be used to write K = Pcρ

−γ c . Use this result and (a) to derive an expression for the central

pressure of a polytropic model of the form

Pc =

GM 2

R4

cn

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where cn is again a dimensionless function that you should write down. Also show that the central

pressure of a polytrope can be written as

Pc = d nGM 2/3ρ

4/3c

where d n depends on an and cn . The values of an, d n, and d n can be determined by numerically solving theLane-Emden equation. The most useful cases for our purposes are γ = 4/3 (n = 3) and γ = 5/3 (n = 3/2)polytropes. For n = 1.5, an = 5.99 and cn = 0.77 while for n = 3, an = 54.183 and cn = 11.05. We will use

these quite a bit during this course. Note how, as mentioned in class, the results for the central pressure and

density of polytropes above are very similar to what you would get from an order of magnitude estimate,

except that for polytropes we get an exact correct numerical factor given by an , cn and d n .

We know that

K = Pcρ−γ c

and also

a =

(n + 1)K ρ

1/n−1c

4πG

1/2

given these relationships we can now find

a24πG

n + 1 = K ρ

1/n−1c = Pcρ

−γ c ρ

1/n−1c

but since we know that γ = 1/n + 1 we find

Pcρ−γ c ρ

1/n−1c = Pcρ

−(1+1/n)c ρ

1/n−1c = Pcρ

−2c

and finally we find

Pc =

a24πGρ2c

n + 1

using this results we can now derive

Pc = a24πGρ2c

n + 1 =

GM 2

R4 cn

which becomes

Pc = a24πGa2n

n + 1

3 M

4π R3

2=

GM 2

R4 cn

looking at these two expressions we can see that

cn = 9a2a2n

4π R21

n + 1

now looking at

Pc = d nGM 2/3ρ

4/3c

and from (b) we find

Pc = a24πGρ

2/3c ρ

4/3c

n + 1=

a24πG

n + 1

an3 M

4π R3

2/3ρ

4/3c

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which becomes

Pc = GM 2/3ρ

4/3c cn

4π

an

4/39−2/3

and after some fun algebra, which will be omitted here we find

d n = cn 4π

3an4/3

(c). What are the values of d n for n = 3 and 1.5 polytropes, respectively?

using the above result we find

n = 3 an = 5.99 cn = 0.77 d n = 0.477

n = 3/2 an = 54.183 cn = 11.05 d n = 0.363

(d). Use your expressions for the central pressure and density to give an expression for the central

temperature of a polytrope. Assume gas pressure dominates.

to find these expression we will assume that gas pressure dominates, i.e

Pc = ρcm̄

k bT c

thus

T c = Pc m̄

ρck b=

d nGM 2/3ρ

1/3c m̄

k b

and skipping some algebra we find that the central temperature is given by

T c = d nGM

Rk b3an

4π

1/3m p

2

so all of the expressions can be written as

T c = d nGM

Rk b

3an

4π

1/3m p

2Pc =

GM 2

R4 cn ρc =

3 M

4π R3an

(e). Calculate the central temperature, pressure, and density for γ = 4/3 (n = 3) and γ = 5/3 (n = 3/2)polytropes for M = M sun and R = Rsun (i.e., for the sun). Assume fully ionized hydrogen for simplicity.Which polytrope better approximates the true interior temperature, pressure, and density of the sun? Can

you explain physically why this is the case?

using the above expressions for temperature, density and pressure we find for

γ = 5

3 an = 5.99 cn = 0.77 d n = 0.477

we find

ρc ≈ 8.44 g/cm3 Pc ≈ 8.67×1015 T c ≈ 6.2×106Kand for

γ = 4

3 an = 54.183 cn = 11.05 d n = 0.363

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we find

ρc ≈ 77.26 g/cm3 Pc ≈ 1.24×1015 T c ≈ 1.02×107Kthus we can see that the γ = 4/3 polytrope best represents the values observed in the sun, this is mainly

due to the fact that the center of the sun is radiative and not convective. Since we now know that

P ∝ ρ4/3 radiative

P ∝ ρ5/3 convective

Problem # 4

Consider a pre-main sequence “star” (gas cloud) of mass M undergoing Kelvin-Helmholz contraction.

In class, we showed that fully convective stars move down the Hayashi line with T e f f ≈ constant. Butstars with M > 0.3 M sun do not end up fully convective on the main sequence and so must go througha phase of KH contraction in which energy transport is dominated by photons. Assume throughout this

problem that gas pressure dominates and that free-free absorption dominates the opacity (because the

temperature is lower during KH contraction than on the main sequence, free-free absorption tends to

be even more important). Motivated by HW #2 Problem 1, assume that the luminosity of a star in

which photons carry the energy out and the opacity is dominated by free-free absorption is given by L ≈ Lsun( M / M sun)11/2( R/ Rsun)−1/2 .

(a). Determine how the radius, luminosity, and effective temperature vary as a function of time and

mass M for a radiative star undergoing KH contraction. Don’t worry about the constants in these relations;

all you need to calculate are proportionalities (i.e., how do the various quantities depend on time and mass

M ). Do the luminosity and effective temperature increase or decrease as the star contracts?

since we know that

Pg > Pr κ = κ f f

thus we know that

L f f ≈ Lsun

M

M sun

11/2

R

Rsun

−1/2 ≈ Lrad ≈−12

GM 2

R2dR

dt

since we are doing proportionalities we find

M 2

R2t

R

T ∝ M 11/2 R−1/2

so we find that the radius scales as

R ∝ 1

M 7t 2

plugging this into L ∝ M 11/2 R−1/2 ∝ M 9t

thus

L ∝ M 9t

and to find the temperature

L ∝ R2T 4e f f T e f f ∝

L

R2

1/4∝ M 23/4t 5/4

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thus

T e f f ∝ M 23/4t 5/4

We can see that as time and mass increase the luminosity increases as well as the effective temperature.

(b). Estimate the radius of a star (in Rsun) of a given mass M (in M sun) at the time when energy transportby photons takes over from convection during the KH phase. At what luminosity does this occur (again as

a function of mass M )? Assume based on lecture that the luminosity of a fully convective star is

L ≈ 0.2 Lsun( M / M sun)4/7( R/ Rsun)2

if we set the free-free luminosity equal to the convective luminosity we find

Lsun

M

M sun

11/2 R

Rsun

−1/2≈ 0.2 Lsun( M / M sun)4/7( R/ Rsun)2

some algebra yields

R ≈ 55/2 M

M sun69/35

Rsun

(c). Sketch the paths of 1 M sun pre-stellar gas clouds during their KH contraction phase in the HR

diagram. Include both the convective and radiative parts of their evolution and the correct transition point

between the two based on b). Be sure to properly label your axes ( L in Lsun and T e f f in K). Note that on

the main sequence a 3 M sun star has L ≈ 40 Lsun and T e f f ≈ 10000 K (you know the values for the sun).The KH contraction phase ends when the star contracts to the point where its luminosity and temperature

have these values.

since we now have a relationship for the radius we can find what the luminosity is by

L ≈ 0.2 Lsun

M

M sun

4/7

R

Rsun

2 ≈ 52/5 ·0.2 Lsun

M

M sun

4/7

M

M sun

128/35

which yields

L( M sun) ≈ 0.724 Lsunand a plot is given by

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Figure 1: We plot the path that a 1 solar mass star would trace when moving from the Hayashi track

to the main sequence.

Problem set 5

Problem # 1

In lecture we discussed the slow, nearly hydrostatic, contraction of pre-stellar gas clouds as they ap-

proach the main sequence - Kelvin Helmholz (KH) contraction.(a). Argue that, for KH contraction to occur, the timescale for KH contraction t KH must be longer than

the gravitational free-fall time of the cloud, t f f ≈ 1/

Gρ, where ρ is the mean density of the cloud.What happens if t KH < t f f ?

Since we know that

t f f ≈ 1 Gρ t KH ≈

M

M sun

1/2 Rsun

R

3(2×107yrs)

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froma a purely physical argument we know that things cannot fall faster than gravity can pull it. Thus

t KH ≫ t f f and saying t f f ≫ t KH would be an unphysical statement due to the knowledge we have about gravity.

We also know that when things collapse that the radius gets smaller, hence collapse and from the relation-

ships of time in both free-fall and Kelvin-Helmholtz contraction we can see that as R

↓that t f f

↓thus the

only possible solution is that t KH ≫ t f f .(b). Estimate the critical radius Rc(inRsun) at which t KH ≈ t f f for a given cloud of mass M (in M sun).

Assume, as we did in class, that the cloud is fully convective at early times. Show that for R


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and since we know that fully convective stars have a polytropic index of γ = 5/3. Knowing this wefind

n = 3 an = 5.99 d n = 0.477

thus

T c

≈ 2.16

×10−15

M

Rc≈ 2.79×10−15 M 5/9

and if we want the temperature of a collapsing gas cloud with respect to M sun we get

T c ≈ 1.9×104KProblem # 2

The globular cluster M13 in Hercules contains about 0.5 million stars with an average mass of about

half the solar mass. Use Jeans criteria to check whether this cluster could have formed in the early universe

just after the time when the universe was cool enough for the electrons and nuclie to form neutral atoms;

at this time the density of the universe was ρ ≈ 10−27

kg m−3

and the temperature was T ≈ 104

K .Using the Jeans mass equation

M j =

k b

Gm p

3/2T 3/2√

ρ

using the values given we find

M j ≈ 1.37×1042 gand the mass of M13 is

M 13 ≈ 0.5×106 · M sun ≈ 9.95×1038 gand we can see that

M j ≫ M 13which means that this cluster could not have formed in the early universe. Things only collapse if the

mass is greater than the Jeans mass

Problem # 3

The binding energy per nucleon for 56Fe is 8.8 MeV per nucleon. Estimate the energy released per

kilogram of matter by the sequnce of reactions which fuse hydrogen to iron.

We know that the enery released will be given as

E tot = E b

nucleon× N nucleon

and the number of nucleons are given by

N nucleon = M

m p≈ 1 kg

m p≈ 5.98×1026nucleon

thus

E tot ≈ 5.27×1027MeV(b). Consider two hypothetical stars of the same mass M and the same luminosity L (that is constant in

time). The stars are initially pure hydrogen. In star A, fusion proceeds until the entire star is converted into

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He. In star B, fusion proceeds until the entire star is converted into Fe. Which star has a longer lifetime,

and by how much?

We know that

L He = E He

t He LFe =

E Fe

t Fe

and since we know that these two luminosities are theoretically equal

E Het Fe = E Fet He

which gives

t He = E He

E Fet Fe =

6.4

8.5t Fe ≈ 0.72t Fe

thus we see that

t He < t Fe

we can see that the time for all of the hydrogen to fuse into helium is less then the time for all of the

hydrogen to fuse into iron so the star that is converted to iron has a longer lifetime.

Problem # 4(a) What is the classical distance of closest approach for two protons with an energy of 2 keV (the

mean thermal energy at the center of the sun)? Estimate the probability that the protons tunnel through the

Coulomb barrier trying to keep them apart. Answer the same two questions for two 4He nuclei and for a

proton and a 4 He nucleus with the same energy of 2 keV.

The classical distance of closest approach is given by

r c = e2 Z 1 Z 2

E 0≈ 7.2×10−11cm

We know that the propbability for a particle-particle interaction is given by

P = e−

E G E

1/2

where

E G = 2π2α2 Z 21 Z

22 (mr c

2)

is the Gamow energy. For a proton-proton interaction we find

E G ≈ .493 MeVthus the probability is given as

P ≈ 1.51×10−7

for a He-He interaction we find

r c = e2 Z 1 Z 2

E 0≈ 2.8×10−10cm

E G = 32π2α22m pc

2 ≈ 31.6 MeVand the probabilty is given by

P ≈ 2.50×10−55

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for the proton-He interaction we find

r c = e2 Z 1 Z 2

E 0≈ 1.44×10−10cm

E G ≈ 3.15 MeVthus the probability is given by

P ≈ 5.8×10−18

(b) What energy E would be required for i) the two 4He nuclei, ii) the proton and the 4 He nucleus, and

iii) two 12C nuclei to have the same probability of penetrating the Coulomb barrier as the two protons?

For particles with energies equal to the mean thermal energy of the plasma, what temperatures do these

correspond to?

Since we know that

E = E G

(lnP)2

for the He-He interaction we find

E He− He ≈ 0.125 MeVfor the proton-He interaction we find

E p− He ≈ .013 MeVfor the carbon-carbon interaction the Gamow energy is given by

E G ≈ 2592π2α26m pc2 ≈ 7.7 GeVand thus

E c−c ≈ 31 MeVwe know that

T ≈ E k b

so

T He− He ≈ 1.45×109KT p− He ≈ 1.5×108K

T c−c ≈ 3.6×1011KProblem # 5

Calculations of nuclear reaction rates are done in the center of mass (COM) frame, so it is useful to

remember a few results about the COM. Consider two particles of mass m1 and m2 with positions x1 and

x2 and velocities v1 and v2 .

(a) .What is the velocity of the COM?

We kbnow that the center of mass is given by

com = m1r 1 + m2r 2

m1 + m2

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so the velocity would be

vcom = m1v1 + m2v2

m1 + m2

(b). What are the velocities of each of the two particles in the COM reference frame (i.e., in the frame

for which the COM is at the origin)?

We know that the relative velocities are given by

vrel−1 = v1−vcomvrel−2 = v2−vcom

a bit of algebra yields

vrel−1 = m2

m1 + m2(v1− v2)

vrel−2 = m1

m1 + m

2

(v2− v1)

(c). What is the total KE of the two particles in the COM frame? Show that this is equal to the KE of

the reduced mass moving at the relative velocity, as claimed in class.

We know that the total kinetic energy is given by

K tot = 1

2m1v

2rel−1 +

1

2m2v

2rel−2

= m1m2

2(m1 + m2)2(m2(v1− v2)2 + m1(v2− v1)2)

= m1m2

2(m1 + m2)(v1

−v2)

2

K tot = 1

2mr (v1− v2)2

where mr is the reduced mass.

Problem set 7

Problem # 1 The Main Sequence for Fully Convective Stars

In this problem we will determine the main sequence for fully convective low mass stars. We showed

in lecture that fully convective stars have T e f f ≈ 4000( L/ Lsun)1/102( M / M sun)7/51K (I actually derived acoefficient of 2600 K in lecture but commented that more detailed calculations get something similar but

with the coefficient closer to the value of 4000 K used here). We can also write this result as

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L ≈ 0.2( M / M sun)4/7( R/ R)2sun Lsun ≡ LconvI called this luminosity Lconv since it is derived from the properties of energy transport alone (convec-

tive interior + radiative atmosphere with H− opacity). The luminosity of a star is also given by

L f usion = 4πr 2ρε(T , ρ)dr

where ε is due to the proton-proton chain for low mass stars (this was given in lecture). As discussed inclass, the main sequence is determined by the requirement that the energy escaping the star (in this case by

convection) is equal to the energy generated in the star (in this case by pp fusion), i.e., that Lconv = L f usion.

a) Use scaling arguments to derive the power-law relations R( M ), L( M ), T c( M ), and L(T e f f ) (the HR

diagram) for fully convective stars, like we did for other examples in lecture. Approximate ε ∝ ρT β withan appropriate choice of β (recall that low mass stars will have somewhat lower central temperatures thanthe sun, closer to ≃×106 K, as you will see in part b).

We know that

Lconv ∝ M 4/7

R2

L f us ∝ R3

ε(ρ, T ) ∝ R3

ρ2

T β

we can find what β is by

β = −23

+

E G

4kT

1/3given that we know what the temperature is and also what E G for p-p reaction

E G ≈ 500 keV T ≈ 5×106Kwe find that

β = 5.92 ≈ 6.0

we also knowρ ∝

M

R3

we know that in steady state

L f usion = Lconv

thus we can find

M 4/7 R2 ∝ R3

M 2

R6

T 6c ⇒ T 6 ∝ M −5/21 R5/6

we know from the Virial temperature, assuming gas pressure dominates

T ∝ M

R

thus we find

T c ∝ M 25/77

knowing this we can now find

R ∝ M 52/77

with this and the relationship for the convective luminosity we find

L ∝ M 4/7 R2 ∝ M 148/77

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with this we can now find what the effective temperature as a function of mass is, i.e

L ∝ R2T 4e f f

thus

T 4e f f ∝ L

R2 ∝ M 4/7

which yields

T e f f ∝ M 1/7

to find what the luminosity as a function of the effective temperature is (HR diagram)

M ∝ T 7e f f

which yields

L ∝ T 148/11

e f f

In a) you just determined a scaling relation between stars of different mass, but not the absolute values

of L, T e f f , etc. In class, we did the latter by scaling to the sun. Note, however, that it is not reasonable

to estimate the properties of low mass stars by scaling from the properties of the sun, since the sun is not

a fully convective star! Instead we need to actually determine the structure of some fully convective star.

This is what we will do in the rest of the problem. We can significantly improve on the above scaling

arguments by using the fact that fully convective stars are n = 3/2 polytropes. It turns out that for apolytrope, in equation (1) can be Taylor expanded near the center to yield

L f usion

≃

2.4εc M

(3 + β)3/2

where I have again approximated ε ∝ ρT β and where εc is evaluated at the center of the star. I am notasking you to prove equation (2). You will have to trust me. Note that for a typical value of β for the ppchain, equation (2) says that L f usion ≃ 0.1εc M . This makes sense because fusion only takes place at thecenter of the star (not all of the mass participates).

b) Use the results for n = 3/2 polytropes from HW 4, Problem # 3, to write the central temperature of the star T c , central density ρc, and pp energy generation at the center of the star εc in terms of the mass

M and radius R. Assume X = 0.7 and µ = 0.6 (typical for stars just reaching the main sequence). Notethat you should give expressions for T c , ρc , and εc here, with constants and real units, not just scalingrelationships. So that the constants in front of your expressions are reasonable, please normalize M to

M sun and R to Rsun.

The general expressions given by HW 4 problem #3 are

T c = d nGM

Rk b

3an

4π

1/3 µm p Pc =

GM 2

R4 cn ρc =

3 M

4π R3an

we found that for a n = 3/2 polytrope

an = 5.99 cn = 0.77 d n = 0.477

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thus we find that

T c = 0.322m pGM sun

k b Rsun

M / M sun R/ Rsun

ρc = 1.43

M sun

R3sun

M / M sun( R/ Rsun)3

plugging in all the constants yields

T c ≈ 7.43×106K M / M sun R/ Rsun ρc ≈ 8.41 M / M sun( R/ Rsun)3

we also know

εc = 5×105ρc X 2T −2/37 e−15.7T −1/3

7

we also know that we can approximate this as

εc = AρT β

7 X 2 = AρT 67 X

2

setting this two expressions equal to each other we can find what A is, i.e lettingT ≈ 107K we find A = 5

×105e−15.7

≈0.076

thus we find

εc ≈ 0.076ρcT 67 X 2 ≈ 0.037ρcT 67substituting T c and ρc gives

εc ≈ 0.053

M

M sun

7 Rsun

R

9

c) Use equation (2), the results of b), and Lconv = L f usion on the main sequence to determine the R( M ), L( M ), T c( M ), and L(T e f f ) relations for fully convective stars. If you use the same β, your expressions hereshould be the same as in a) except that you should now be able to determine the absolute normalization

for R( M ), L( M ), etc., i.e., you have determined the true luminosity and radius of a ful ly convective starfrom first principles. In doing this problem, remember that β is temperature dependent so make sure youcheck that your value of β is reasonable given the resulting central temperature that you calculate.

Using the results from b and also

L f usion ≃ 2.4εc M (3 + β)3/2

≈ .09εc M (β≈ 6)

thus we know that

Lcon = L f usion

0.2

M M sun

4/7 R

Rsun

2 Lsun = 0.0047

M

M sun

7 Rsun

R

9 M

M sun

M sun

rearranging this we find R

Rsun

11= 0.023

M sun

Lsun

M

M sun

52/7which yield

R

Rsun≈ 0.67

M

M sun

52/77

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to find for the temperature we can use

T c ≈ 7.43×106K

M / M sun R/ Rsun

using our previous resulst gives

T c ≈ 1.1×107K

M

M sun

25/77

to find for the luminosity we use

Lconv = 0.2

M

M sun

4/7 R

Rsun

2 Lsun

plugging in for the radius we find

Lconv = 0.09 M

M sun148/77

Lsun

for the effective temperature we find

L = 4π R2σT 4e f f ⇒ 0.09

M

M sun

148/77 Lsun = 4π R

2σT 4e f f

which can be simplified to

0.09

M

M sun

148/77 Lsun = 1.56×1018

M

M sun

104/77T 4e f f

thus

T e f f = 3868 K

M

M sun

1/7which can also be written as

M

M sun=

T e f f

3868 K

7

L(T e f f ) = 4.85×10−50 LsunT 148/11e f f

d) What are your predicted luminosities, radii, and effective temperatures for main sequence stars with M = 0.1 and 0.3 M sun? Compare your values to the values of L = 0.01 Lsun, R = 0.3 Rsun, and T e f f = 3450K for M = 0.3 M sun and L = 10

−3 Lsun , R = 0.11 Rsun, and T e f f = 3000 K for M = 0.1 M sun that I found in agraduate textbook (based on detailed models).

Given our relationships we find for

M = 0.1 M sun L = 1.1×10−3 Lsun R = 0.14 Rsun T e f f = 2783 Kand for

M = 0.3 M sun L = .009 Lsun R = 0.298 Rsun T e f f = 3256 K

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Problem # 2 Very Massive Stars

Consider very massive stars with M ∼ 50−100 M sun. Recall that I showed in lecture and you showedon HW 3, Problem # 1, that in such stars, radiation pressure due to photons (a relativistic particle) is more

important than gas pressure. Fusion is by the CNO cycle. Assume for now that energy is transported

primarily by photons and that the opacity is due to Thomson scattering (reasonable for hot massive stars).

a) Use scaling arguments to derive the power-law relations R( M ), L( M ), T c( M ), and L(T

e f f ) (the HR

diagram) for very massive stars, like we did for other examples in lecture.

Using radiative diffusion along with

Prad = 1

3aT 4

dPr

dT =

4

3aT 3

we also knowdP

dR =

dP

dT

dT

dR =

4

3aT 3

dT

dR =−ρGM

R2

thusdT

dR

∝ ρ M

T 3

R2

and radiative diffusion says Lρ

R2T 3 ∝

dT

dR ∝

ρ M

T 3 R2

which gives us

L ∝ M

using the Virial theorem, where Prad dominates rather than Pgas we find

T 4

ρ ∝

M

R ⇒ T c ∝ M

1/2

R

where the left hand term is from the radiation pressure, but since we know that is an energy density

we must devide by the density to find what the energy is per particle. Now using the steady state for

luminosity we find

L ∝ M ρT 18

where we chose β = 18 as a more appropriate value rather than the value given for the sun β = 20, thisis motivated by the fact that more massive stars have somewhat higher temperatures, thus reducing β. Wefind

T 18 ∝ 1

ρ ∝

R3

M

and using the result from Virial temperature we find

M 9

R18 ∝

R3

M

thus we find

R ∝ M 10/21

and we also find for the central temperature

T c ∝ M 1/42

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and to find the effective temperature we know

T 4e f f ∝ L

R2 ∝

M

R2 ∝ M 1/21

simplifying gives

T e f f ∝ M 1/84

and finally the luminosity as a function of T e f f is given by

L(T e f f ) ∝ T 84

e f f

b) Estimate the fraction of the mass in the star that is undergoing convection (recall that fusion by

the CNO cycle is very concentrated at small radii because of the strong temperature dependence). For

comparison, detailed calculations show that the fraction of the mass that undergoes “core” convection

increases from 10 % at 2 M sun to 75% at 60 M sun.

The condition for convection is given by

d ln T

d lnP≈ 1

4

Ptot

Prad

L

L Ed d

Lr / L

M r / M >

γ −1γ

since we know that

γ = 4

3 Ptot ≈ Prad

gives us1

4

Lr

L Ed d

M

M r >

1

4

which simplyfies to

Lr Ledd

> M r M

we know that in the limit that M → 150 M sun Lr → L Ed d , M r

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This seems rather strange in the sense that stars that are approximately 60 M sun have a convective core

that encompasses 75% of the mass, which means that the convective core decreases after M > 60 M sun?

c) Calculate the main sequence lifetime of a very massive star as a function of its mass M . Be sure to

take into account the results of b).

We know that the main sequence lifetime of a star is given by

t MS ≈ E tot L Ed d

where

E tot = NQ Q ≈ 7 MeVwhere that is the total energy per reaction, we also know

N = M r

m p

= M

m p

(n = 3 polytrope)

N = M r

m p=

5

8

M

m p(n = 3/2 polytrope)

we also know that

L Ed d = 4πcGM

κ T

so we find the main-sequence lifetime to be given as

t MS ≈ κ T Qm p4πcG

(n = 3 polytrope)

t MS ≈ 5κ T Qm p32πcG

(n = 3/2 polytrope)

we know that

κ T ≈ 0.4 cm2/g Q ≈ 7 MeV ≈ 1.12×10−5 ergsthus we find that the main-sequence lifetime for both types of polytropes are given by

2.21×106yr


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a) Above what density is a gas of room temperature fermions degenerate? Below what temperature

would gas with the density of air be degenerate?

We know that if the density of the gas is ng ≥ nQ where nQ is the quantum concentration, nQ is definedas

nQ ≡2π m̄kT

h2 3/2

(1)

and for a gas at room temperature to be degenerate

ng ≥ nQ =

2π m̄kT

h2

3/2≈ 1.46×1026cm−3

where we used

T = 300K m̄ = 28m p

due to the fact that air is mostly composed of N 2. If we assume that the questio is only speaking aboutfree electrons we get

ng ≥ nQ = 2πmekT

h23/2 ≈ 1.25×1019cm−3

using the same temperature as before.

To find the temperature at which gas with a density of air would be degenerate can by using the above

expression, except now we must find what the density of air is at STP and use this, i.e

nair = P

kT = 2.52×1019cm−3 = nQ

and now using Equation 1 we find

T =n

2/3Q h

2

2π m̄kT ≈ 9.2×10−3K

using m̄ = 28m p

b) Compare the relative importance of the thermal energy, the electrostatic (Coulomb) energy between

electrons and ions, and electron degeneracy (electron Fermi energy) in room temperature silver ( Z = 47;ρ≃10g cm3 ). Which dominates?

We can write the thermal energy as

E th ≈ 32

kT ≈ .039 eV

We can write the Coulomb energy as

E coul ≈ Z 2e2 1r

we know

r ∼ n−1/3 ∼ ρ

m̄

−1/3

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thus

E coul ≈ Z 2e2 ρ

m̄

1/3≈ 12.43 keV

using ρ ≃ 10 g/cm3 and m̄ ≈ 100m p. The Fermi energy can be written as

E f = 38π

2/3h2

2me47ρ

m̄

2/3

≈75 eV

where we used m̄ ≈ 100m p and the 47 comes from the fact that there are 47 electrons in a silver atom.We can see that

E coul ≫ E f > E thfor room temperature silver.

Problem # 2 Deuterium Fusion in Contracting Protostars

Small amounts of Deuterium are made in the Big Bang. D is destroyed in the interiors of stars via

the reaction p + D →3 He + γ . The S value for D-burning is2.5× 10−4 keV-barn = 4×10−37 erg cm2 ,each reaction releases

≈5.5 MeV, and the cosmic abundance of D from the Big Bang is n D

≈2

×10−5n H .

Let’s focus on a low mass fully convective star undergoing KH contraction; such a star can be reasonablywell modeled as an n = 3/2 polytrope. Assume that the star has cosmic composition ( µ ≃ 0.6). Note thatin this problem, you should not use the approximation ε ∝ ρT β . Instead, you will need to keep the fullexpression for ε.

a) What is the Gamow energy for D fusion? Write down the resulting thermally averaged cross-section

σv for D fusion.The Gamow energy can be written as

E G = Z 21 Z

22

mr

m pMeV

using Z 1 = Z 2 = 1 and mr = 23

m p we find the Gamow energy to be

E G ≈ .67 MeV

The thermally averaged cross-section is given as

σv= 2.6S ( E ) E 1/6G

k 2/3T 2/3 e−3( E G/4kT )

1/3

using all the constants given and the Gamow energy we find

σv = 3.7×10−15K−2/3T 2/3

cm3

s e−3742(K/T )

1/3

in terms of M and R we find

σv= 3.7×10−15

K−2/3T 2/3cm3

se−3742(K/T )

1/3

b) In class we derived a quantitative model for the Kelvin-Helmholtz contraction of a low mass star as

it approaches the main sequence. Use these results to calculate the local contraction time t c ≡ R/|dR/dt |

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as a function of the mass and radius of the star. This is the amount of time that a star of a given mass M

spends at a given radius R. Does the contraction time get shorter or longer as the star contracts?

From lecture we derived the following relationship

L = 3

7

GM 2

R2

dR

dt

= 0.2 Lsun( M / M sun)4/7( R/ Rsun)

2

thus we find

dR

dt =

7 ·0.23

R2sun Lsun

GM 2sun

M sun

M

2 R

Rsun

2 M

M sun

4/7 R

Rsun

2

= 3.34×10−5cm/s

M sun

M

10/7 R

Rsun

4

and we can find the local contraction time to be

t c = Rsun R

Rsun 1

(dR/dt ) = 6.7×107

yr M

M sun10/7 Rsun

R3

As the star contracts the contraction time gets longer.

c) What is the lifetime t D of a D nucleus at the center of the star in terms of the local density and

temperature (the lifetime is the average time before a D nucleus is destroyed by fusion into 3 He)? Use the

properties of n = 3/2 polytropes to write t D as a function of M and R. Does the D lifetime get shorter or

longer as the star contracts?

We know that average lifetime of a deuteron is given by

t D =

l

v =

1

n pσv

which gives us

t D = 1n pσv =

µm p

ρcσvbut we also know

ρc = 3 M

4π R3an = 1.43

M

R3

thus we find

t D = µm pρcσv

= µm p R

3

1.43σv M

= 7.019×10−25g 1

σv

R3

M

we also know

T c = d nGM

Rk b

3an

4π

1/3 µm p = 2.60×10−16 cm

gK

M

R

thus we find

t D = 7.019×10−25g 1σv R3

M = 1.89×10−10 g s

cm3T 2/3e3742(K/T )

1/3 R3

M

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written in terms of M and R we find

t D = 1.23×10−6s

R

Rsun

7/3 M sun

M

1/3e19.19(( M sun/ M )( R/ Rsun))

1/3

the deuteron lifetime gets shorter as the star contracts.

d)For any mass M show that there is a critical radius R D at which t D = t c . This represents the radius(time) at which D starts to undergo significant fusion. Give the numerical value of R D for M = 0.03 and 0.1

M sun. For each of these two cases, also determine the central temperature of the star T c and the D lifetime

t D when R = R D . Does D fusion occur before or after the star reaches the main sequence?

We know that

t D = t c

which yields

1.23×10−6s

R

Rsun7/3

M sun

M 1/3

e19.19(( M sun/ M )( R/ Rsun))1/3

= 6.7×107yr

M

M sun10/7

Rsun

R 3

and so we find R D

Rsun

16/3= 1.71×1021

M

M sun

37/21e−19.19(( M sun/ M )( R/ Rsun))

1/3

we can solve this numerically to find

R D = 0.44 Rsun M = 0.03 M sun

R D = 1.11 R M = 0.1 M sun

to find the central temperature we can use

T c = 7.4×106K

M

M sun

Rsun

R D

we find

T c ≈ 5.0×105K M = 0.03 M sunT c ≈ 6.67×105K M = 0.1 M sun

and to solve for the deuteron lifetime we find

t D = 1.23×10−6s

R D

Rsun

7/3

M sun

M

1/3

e19.19(( M sun/ M )( R/ Rsun))1/3

so we find

t D ≈ 4.67×106yr M = 0.03 M sunt D ≈ 4.1×105yr M = 0.1 M sun

e) Can D fusion halt (at least temporarily) the KH contraction of the star? Explain your answer quan-

titatively.

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Since we know that

ε = Qr d

ρd ρd = md nd r d =

nd

t D

we also know that

L = M ε = MQr d

ρd =

MQ

2m pt D

since we know that Q = 8.8×10−6ergs, we find

L = M (8.8×10−6ergs)

2m pt D

so for M = .03 M sun we find L ≈ 5.37×1035ergs/s

and for M = 0.1 M sun we find L ≈ 4.08×1037ergs/s

thus we can see that for both of these stars deuteron fusion can stop the KH contraction temporarily.

Problem # 3 The R(M) Relation for Degenerate Objects

Consider an object supported entirely by the pressure of non-relativistic degenerate electrons. Because

P = K ρ5/3 such an object can be modeled (rigorously) as an n = 3/2 polytrope.K is a constant that dependson the electron mean molecular weight µe .

a) Use your results for how the central pressure Pc and density ρc of an n = 3/2 polytrope dependson the radius R and mass M of the object to derive the R( M ) relation for degenerate objects (the radiusalso depends on µe ). Note that you should give an expression with proper constants and not just a scaling

relationship. Normalize the mass M to M sun and the radius R to Rsun (this should sound pretty familiar by

now).

We know thatPdeg = Pc =

h2

5me

3

8π

2/3 ρc µem p

5/3= d nGM

2/3ρ4/3c

rearranging this equation for ρc yields

ρ−1/3c =

h2

5me

3

8π

2/3 1

µm p

5/31

d nGM 2/3

from the last problem set we showed

ρc = 8.41 M M sun

Rsun R

3

using this we find

R

Rsun= 0.04 µ

−5/3e

M

M sun

−1/3b) Use a) to estimate the radius of Jupiter. How does your result compare to the correct value?

Using part a) with µe ≈ 1.17 which is the value given for the sun on Google and M = M J , we find R ∼ 0.30 Rsun

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c) The results you have derived in a) should show that as M → 0, R →∞. This is not correct, however,because Coulomb interactions become important in the equation of state of low-mass objects (brown

dwarfs and planets). Estimate the density at which the Coulomb energy per particle becomes comparable

to the Fermi energy. What mass and radius does this correspond to? Explain why this is a very rough

estimate of the maximum radius of a degenerate object.

If we know

E f = E coul

then

1

4πε0

e2

r =

3

8π

2/3h2

2men2/3

but we know that1

r ∼ n1/3

thus

1

4πε0e2n1/3 =

3

8π2/3

h2

2men2/3

so we find the density to be given by

n =

2me

4πε0h2e23

3

8π

2≈ 6.15×1028m−3

to find the mass we can use

ρc = nm p = 1.43 M

R3

and using R from part b) we find

1.43

M sun( M / M sun)

(0.03 Rsun)3( M / M sun)−1 = nm p

which can be simplified to

M

M sun=

nm p(.03 Rsun)3

1.43 M sun

1/2≈ 5.7×10−4

which can also be expressed as

M ≈ 190 M earthusing this we can now find the radius to be

R ≈ 0.361 Rsun

Problem set 9

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Problem # 1

Use the chemical potential µ for a non-degenerate, non-relativistic gas (derived in class; also 2.21

in Phillips) to show that in the limit n ≪ nQ (the non-degenerate limit), the full quantum mechanicaldistribution function reduces to the classical Maxwell-Boltzmann distribution function. A good check that

you have things correct is that the QM dist. fcn you start with has some h′s in it (Planck’s constant), butthe classical dist. fcn you end up with should, of course, be independent of h.

We know that the chemical potential is defined as

µ = mc2 + kT ln

n

gnQ

(2)

and that the quantum distribution function is defined as

n( p) = g/h3

e( E p− µ)/kT ±1to show that in the classical regime

e( E p− µ)/kT ≫ 1We can write Equation 1 as

nQ

n=

1

ge(mc

2− µ)/kT n ≪ nQ e(mc2− µ)/kT ≫ 1

thus the quantum distribution function can be written as

n( p) = g/h3

e( E p− µ)/kT

We also know that

E p = mc2 +

p2

2m nQ =

2πmkT

h2

3/2and using equation 1 we find

n( p) = g/h3

e(mc2+ p2/2m−mc2−kT ln(nQ/n))/kT =

gn

h3e E /kT nQ

which after some simplification reduces to the Classical Boltzman distribution function

n( p) = n 1

2πmkT 3/2

e− E /kT

Problem # 2

Consider a cloud of gas that has a total mass M . Assume that all of the gas in the cloud is converted

into stars with the initial mass function given in class dN /dm ∝ m−α where α = 2.35 and where thisformula is valid between m = 0.5 M sun and m = 150 M sun. Note that d N /dm has units of number of starsper unit mass.

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a) What is the ratio of the number of stars formed with masses within dm ≃ m1 of m1 and masseswithin dm≃m2 of m2 ? What is the ratio of the number of 150 M sun stars formed to the number of 0.5 M sunstars formed?

We know that the Initial Mass Function IMF is given as

dN

dm

∝ m−α ∝ m−2.35 α = 2.35

we also know that for dm ≃ m1 of m1we finddN (m1) = m

−α1 dm1 ≈m1−α1

we also know that for dm ≃ m2 of m2we finddN (m2) = m

−α2 dm2 ≈m1−α2

and the fraction is given by

dN (m1)dN (m2)

=

m1m2

1−α ≈ 2208 m1 = 0.5 M sun m2 = 150 M sunb) Estimate the mass of a cloud M so that approximately one 150 M sun star forms in the cloud. If the

temperature of the cloud at the time of formation was 10 K, what was the density of the gas out of which

the cloud formed?

From part a) we found that

N s ≈ 2208 N bwhere N b is for stars that are the number of ∼ 150 M sun and N s is for stars that have ∼ 0.5 M sunand to

get the total mass we must multiply the total number of small stars to the average mass of the stars, and

from lecture we are told that M ∗ ≈ 0.5 M sun

M cluster = N s · M ∗ ≈ 1104 M sunto find the density of the gas in which this cloud formed we can use the Jean’s density

ρ J = 3

4π M 2

3kT

2G m̄

3≈ 3.14×10−22kg m−3

Problem # 3

A stellar atmosphere consists almost entirely of hydrogen. Assume that 50 % of the hydrogen moleculesare dissociated into atoms and that the pressure is 100 Pa. Given that the binding energy of the hydrogen

molecule is 4.48 eV, estimate the temperature. Set all degeneracies to 1. As the hint at the back of the book

suggests, you should derive the Saha equation for the dissociation of H 2 into hydrogen, i.e., the reaction

γ + H 2 ←→ H + H .We know that

γ + H 2 ←→ H + H and the Saha equation gives

µ( H 2)←→ 2 µ( H )

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but we know that the chemical potential is given by

µ( H ) = m H c2−kT ln

nQ, H g H

n H

µ( H 2) = m H 2c

2−kT ln

nQ,K 2g H 2n H 2

where

m H c2 = m pc

2 + mec2

−χ H m H 2 c

2 = 2m pc2 + 2mec

2

−χ H 2

−2χ H

thus we find

m H 2c2−kT ln

nQ,K 2 g H 2

n H 2

= 2

m H c

2−kT ln

nQ, H g H

n H

which becomes

−χ H 2kT

= ln

nQ, H 2 g H 2

n H 2

n H

nQ, H g H

2

we are given that

g H 2 = g H = 1 nQ, H 2 ≈ nQ, H ≈2πmkT

h2 3/2

thus we find

n H 2n H

= n H

nQe−χ H 2 /kT = n H

h2

2πmkT

3/2e−χ H 2 /kT

but we are given that

n H = P

3kT

n H 2n H

= 1

2

thus

1

2 =

P

3kT

h2

2πmkT

3/2e−χ H 2 /kT

this can only be solved analyticaly, we find that the temperature is given by

T ≈ 2260 K

Problem # 4 Lines from Hydrogen

Consider a pure hydrogen gas. In this problem we will calculate the fraction of H atoms that have

an electron in the n = 2 state (a result I plotted in class), and use that to understand some aspects of theobserved lines of H from stars. Recall that the energy levels of the H atom are given by E = −13.6/n2 eVand the degeneracies are gn = 2n

2 .

a) Use the Saha equation to solve for the fraction of hydrogen atoms that are ionized as a function of

temperature T . If n is the total number density of hydrogen atoms (both neutral and ionized) then whatwe are after is n p/n since an ionized hydrogen atom is just a proton. Your result for n p/n will dependon n (because, as discussed in class, the ionization of a gas depends weakly on density in addition to the

primary dependence on temperature). For densities appropriate to the photosphere of the sun, make a plot

of n p/n as a function of temperature T . If you are familiar with graphing using IDL, Mathematica, etc.feel free to use that. Otherwise, you can just plug values into your calculator and make the plot by hand. In

your calculation, assume that all of the neutral hydrogen atoms are in the n = 1 (ground) state. The reasonthis is an ok approximation is as follows. According to the reasoning in class, which you will confirm

here, Hydrogen is 1/2 ionized at T ≃ 1.5×104 K. At that temperature, nearly all of the neutral H atoms

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are in the ground state (check it if you don’t believe me!), so for temperatures at which H is largely neutral

(T ≤ 1.5×104 K), it is reasonable to say that almost everything is in the ground state.We know from the Saha equation

nen p

n H =

geg p

g H 2πmkT

h2 3/2

e−χ/kT

we know

ne ∼ n p n = n H + n p χ ≈−13.6 eV geg pg H

= 1

thusn2 p

n H = α α≡

2πmkT

h2

3/2e−χ/kT

and so

n2 p = αn H = α(n−n p)which becomes a quadratic equation of the form

n2 p + αn p−αn = 0

with the solution of n p being

n p = −α±

√ α2 + 4αn

2

and since we know that this must be a positive thus we will take the positive solution

n p = −α +

√ α2 + 4αn

2

and finally we are looking forn p

n = −α +√ α2 + 4αn

2n

the plot is given by

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np vs ntotal

5.0•103 1.0•10

4 1.5•104 2.0•10

4 2.5•104 3.0•10

4

Temperature

0.0

0.2

0.4

0.6

0.8

n p

/ n

we can see that at a temperature of T ≃ 1.4−1.5×107K roughly ∼ 50% of the hydrogen atoms areionized.

b) Use your result from a) to calculate the fraction of all H atoms that have an electron in the n = 2state of hydrogen. If n2 is the number density of atoms with electrons in the n = 2 state, then what weare after here is n2/n. You will need to use the Boltzmann factor in addition to your result from the Sahaequation in a). For densities appropriate to the photosphere of the sun, make a plot of n2/n as a functionof temperature T . If you are familiar with graphing using IDL, Mathematica, etc. feel free to use that.

Otherwise, you can just plug values into your calculator and make the plot by hand.

We know thatn H

n =

n−n pn

= 1− n pn

and from the Boltzman equation we know that

n2

n1=

g2

g1e−( E 2− E 1)/kT

we also known H = n1 + n2 → n2 = n H −n1 → n H

n2= 1 +

n1

n2

andn2

n H =

1 +

n1

n2

−1what we are looking for is

n2

n=

n2

n H

n H

n=

1 +

n1

n2

−11− n p

n

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thus we find the fraction of all H atoms that have an electron in the n = 2 state of hydrogen given by

n2

n=

1 +

n1

n2

−11− n p

n

=

1 +

g1

g2e( E 2− E 1)/kT

−11− −α +

√ α2 + 4αn

2n

where α has been explicitily defined already. The plot is given by

n2 vs ntotal

1.0•104 1.5•10

4 2.0•104 2.5•10

4 3.0•104 3.5•10

4 4.0•104

Temperature

0

1•10−5

2•10−5

3•10−5

4•10−5

n p

/ n

We can see that the fraction of hydrogen atomes in the energy state n = 2 peaks at ∼ 1.5×104K.c) The Balmer lines of hydrogen are produced by transitions between the n = 2 states of Hydrogen and

the n = 3, 4, .... states. What are the wavelengths of the H α(n = 2 → 3)and H β(n = 2 → 4) lines of H?Use your result from b) to explain why A stars show the most prominent H α lines of hydrogen (relative to

more massive stars such as O stars and less massive stars such as M stars).

We know that

∆ E = h ν = hc

λ

thus

λ = hc

∆ E =

hc

( E 2− E 1)and

E = −13.6 evn2

thus for the n = 2 → n = 3 transition we get

λ = hc

(3.4−1.51)eV ≈ 656.3 nm

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and for the n = 2 → n = 4 transition we get

λ = hc

(3.4−0.85)eV ≈ 486.7 nm

From the plot given in part b) we can see that the fractional number of atoms in the n = 2 energy statepeaks at around 1.5

×104K, which is approximately the surface temperature of A stars, we can also see

that for O type stars that have surface temperatures much greater than 15,000 K that there are ≈ 0% of hydrogen atoms in the n = 2 energy state, most of the atoms are already ionized. The situation is similarfor M stars that have surface temperatures that are much lower than 15,000 K. We can see that at these

temperatures there are approximately 0 atoms with electrons in the n = 2 energy state.

d) The Lyman lines of hydrogen are produced by transitions between the n = 1 states of Hydrogen andthe n = 2, 3, 4, .... states. What is the wavelength of the Lyα(n = 1 → 2) line of H? Roughly what fractionof H atoms have electrons in the ground (n = 1) state of H in the atmosphere of an M-star? Would youexpect to see prominent Lyα lines from an M-star? Why or why not?

Using

λ =

hc

∆ E =

hc

( E 2− E 1)we find

λ = hc

∆ E =

hc

(13.6−3.4)eV ≈ 121.6 nm

We would not expect to see any L yα lines from M stars, even though all of the hydrogen atoms are in

the ground state, there is not enough thermal energy to excite the electrons from n = 1 to n = 2.

Problem set 10

Problem # 1

Consider a gas with total mass density ρ and temperature T . Recall that the mean molecular weight µis defined by P ≡ ρkT / µm p where P is the total ideal gas pressure (ions and electrons), while the electronmean molecular weight µe is defined by ne ≡ ρ/ µem p .

Since we know that the total pressure is given by

PT = P I + Pe = ρkT

m p

1

µ I +

1

µe

=

ρkT

µm p

and thus1

µ=

1

µ I +

1

µe

but we know that1

µ I =

X

A

1

µe=

X Z

A

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where X is the mass fraction of the species, Z is the number of electrons, and A is the atomic number

of the species, thus we find

µ = 1

X

A

1 + Z µe =

A

X Z (3)

a) What are the values of µ and µe if the gas consists of

i) ionized H,

Since we know that

X = 1 A = 1 Z = 1

then

µ = 1

2 µe = 1

ii) 75 % (by mass) ionized H and 25 % (by mass) ionized He,

We have to treat this case seperately since we have two species contributing to the mean molecular

weight

X =

astro160 problems solutions philipsquestion2.4pg40

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