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 2
Problem # 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 ≈ 1023 ρT −7/2 cm2 g−1 (ρ is in cgs).
We know that the radiation flux is given by
Frad ∼
caT 3
∇T
κρ
where we know that a is the radiation constant, c is the speed of light, T is the temperature, κ is the
opacity, which in our case is given by free-free absorption, ρ is the mass density and ∇T is the temperature
gradient. We have the following relationships
ρ∝
M
R3
∇T ∝
dT
TR − Tc
TC
∝
∝−
dR RR − RC
R
κ ∝ 1023 ρT −7/2
given these relationships we can find
Frad ∝
caT 15/2 R5
M2
we also know that the luminosity can be written as
L = 4πr2 Frad ⇒ Frad =
L
4πr2
which gives us
caT 15/2 R7
M2
we can find the temperature by using the virial theorem which can be written as
L∝
T≈
GMm p µ
3Rk
1
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 is
the 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
GM
= − 2 ρ
dr
r
M
ρ
Pc ∝
R
M
ρT ∝
ρ
R
M ∝ R
since T is constant, substituting this into the luminosity relationship yields
L ∝ M5
(c). Give a quantitative argument as to whether free-free opacity dominates electron scattering opacity
in 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
M
κ∝ρ ρ∝ 3 M∝R
R
thus we find
1
κ∝ 2
M
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 Tc ≃ 1.5 × 107 K. For the
conditions 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?
2
We know that the mean free path is given by
1
ne σ
l=
we know that for a completely ionized hydrogen gas that
ne ∼ n p ∼
ρ
mp
and the interaction cross section is given by
σ = πr2
where r is the Coulomb radius found comparing the thermal energy to the Coulomb energy
e2
∼ kT
r
r∼
e2
kT
σ∼
πe4
(kT )2
using these relationships we find the mean free path to be
mp
l∼
ρc π
kT
e2
2
and the collision time is given by
tcol =
l
ve
and the velocity can be found by using
3
1
kT = me v2 v =
2
2
s
3kT
me
thus the time is given as
te =
r
me m p
3kT ρc π
kT
e2
2
(b). What is the mean free path of an proton due to proton-proton Coulomb collisions? What is the
typical time between collisions? Which occurs more rapidly, electron-electron or proton-proton Coulomb
collisions?
The mean free path of proton-proton collisions would be the same as for the electron-electron collosion
because the gas is completely ionized. The mean free path is given by
mp
l∼
ρc π
kT
e2
2
The collision time would be the same except now that the mass is the mass of the proton not the
electron. i.e
tp =
r
mp mp
3kT ρc π
3
kT
e2
2
we can now see the collision times for the electron-electron collision occurs more rapidly due to the
mass being so much smaller.
te ≪ t p
(c). Which opacity is more important for photons, Thomson scattering or free-free absorption?
We know that
κT =
and
2σT
ne σT
=
∼ 0.80
ρc
mp
κF = 1023 ρT −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
2
1
e2
8π
l=
where σT =
= 6.65 × 10−25 cm2
ne σT
3 4πε0me c2
which yields
mp
∼ 8.3 × 10−3 cm
2ρσT
m p kT 2
∼ 8.9 × 10−7 cm
=
ρc π e2
l photon =
lelectron
The typical time for a photon collision is given by
t=
lp
∼ 2.8 × 10−13 s
c
(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
tdi f f =
R2 nk
thermal energy R2 nkT
R2 2ρk
∼
∼
∼
L
lc aT 4 lcaT 3
m p lcaT 3
we know that the average time for a photon to leave the star is given by
tdi f f
R2sun
∼
∼ 104 yr
l ph c
4
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 , and
the 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:
N5 (t) = N5 (0)e−λ5t and N8 (t) = N8 (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
T5 =
ln 2
ln 2
= 4.5 × 109 yrs
= 0.7 × 109 yrs T8 =
λ5
λ8
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 different
chemical 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 235 and 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 Pb206
nuclei. Show that this ratio is the same for all rock samples which were formed at the same time,
and that it is given by
N7 (t) − N7(0) N5 (t) eλ5t − 1
=
N6 (t) − N6(0) N8 (t) eλ8t − 1
We know that the ratio of the two isotopes can be written as
N7 (t) − N7(0) N5 (t) − N5(0)
=
N6 (t) − N6(0) N8 (t) − N8(0)
and given the first expression given in this problem, which can also be written as
N5 (0) = N5 (t)eλ5t and N8 (0) = N8 (t)eλ8t
substituting this into our previous expression yields
N7 (t) − N7(0) N5 (t) eλ5t − 1
=
N6 (t) − N6(0) N8 (t) eλ8t − 1
which is what we were asked to show.
• Consider a graph in which the measured abundances in the rock samples of Pb207 and Pb206 are
plotted, N7 (t) along the y-axis and N6 (t) on the x-axis. Show that a straight line will be obtained if
all the samples were formed at the same time.
5
We know that
N7 (t) =
where
N5 (t) eλ5t − 1
· N6 (t)
N8 (t) eλ8t − 1
N5 (t) eλ5t − 1
= constant
N8 (t) eλ8t − 1
• 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 N6 (t) so we just have to plug
in numbers
(a). t = 1 billion years.
We know that
λ5 ∼ 9.90 × 10−10 yr−1 λ8 ∼ 1.5 × 10−10 yr−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 get
eλ5t − 1
0.0071 · λ t
= 0.0715
e 8 −1
For 3 billion years we get
eλ5t − 1
= .231
0.0071 · λ t
e 8 −1
and finally for 5 billion years we get
0.0071 ·
eλ5t − 1
= .891
eλ8t − 1
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 Mr ≈ constant = M,
because rchanges, the gravitional acceleartion is not constant).
Write down the equations for hydrostatic equlibrium and energy transport by radiative diffusion. Use
these to calculate dPrad /dP, the change in radiatio pressure with pressure in the atmosphere. What does
this 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
1
Prad = aT 4
3
1 d
4
dPrad
= a (T 4 ) = T 4 ∇T
dr
3 dr
3
6
and we know that the radiation flux is given by
F=−
thus we can write
F=
4 caT 3
∇T
3 κρ
κρ
dPrad
=F
dr
c
dPrad c
dr κρ
the hydrostatic equilibrium equation is
dP
GM
=− 2 ρ
dr
r
deviding these two expressions yield
Fκr2
dPrad
=
dP
cGM
we know that the Flux and luminosity are related by
L = 4πr2 F
F=
L
4πr2
thus we find
dPrad
Lκ
=
dP
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
L
T4
L
∝
⇒
∝
Pg
M
ρT M
since we assumed that L and M are constant than this gives
T3 ∝ ρ
To show that P ∝ ρ4/3 we can also use scaling argument, we also know that the radiation pressure
scales as some constant times the gas pressure
Prad
L
∝
Pg
M
thus we find
but we know that
Pt = Pg + Prad ⇒ Pg = Pt − Prad
Pr
∝ 1 ⇒ Pr ∝ λPg
Pt − Pr
Pg ∝ ρT
T ∝ ρ1/3 Pg ∝ ρ4/3
thus we know that
λPg ∝ Pt − λPg
Pt ∝ 2λPg ∝ ρ4/3
7
Problem set 3
Problem # 1
(a). Show that heat transfer by radiative diffusion implies a non-zero gradient for the radiation pressure
which 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 assuming
that 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
Fr = −
4 aT 3
∇T
3 κρ
Fr ∝ Prad
knowing these relationships we can do
1
Prad = aT 4
3
dPr 4 3 dT
dT
3 1 dPr
= aT
⇒
=
dr
3
dr
dr
4 aT 3 dr
thus this implies that there is a non-zero gradient.
To show that
4πcGM
κ
we must begin with the equation derived from problem 4 in the last problem set, i.e
Lmax =
dPr
Lκ
=
dP
4πcGM
but since we know that
P = Pg + Pr
Pg ≪ Pr ⇒ P ≈ Pr
dPr
=1
dPr
and we find that
4πcGM
≈ 3.3 × 104Lsun
L=
κ
M
Msun
We cannot obtain a numerical estimate because we do not know tha mass. We could use Msun but this
would not be correct.
8
(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 Msun ) 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
L
=
Lsun
3
M
Msun
3
and substituting the Eddington luminosity for L we find that M is given by
M=
4πcGM
Lsun κT
1/2
3/2
Msun ≈ 180 Msun
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
0.1Rsun
m(r)
0.028Msun
Lr
24.2Lsun
Tr
2.2×107 K
ρ(r)
3.1 × 104kg m−3
κ
0.040 m2 kg−1
Using equation ?? from Phillips
L(r)
m(r)
=
crit
γ − 1 16πGc Pr
γ
κ P
and the following relationships
1
ρ(r)
Pr = aT 3 P ∼ Pg =
kb T
3
m̄
γ=
5
κT = 0.04 m2 /kg
3
we find
2 16πGc aT 3 m̄
5 κT 3ρ(r)kb
crit
W
W
> .07
0.175
kg
kg
L(r)
m(r)
=
which implies that the energy transport of this star is primarily due to convection.
Problem # 3
The 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 good
approximation 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 photosphere. Assume that Thompson scattering dominates the opacity.
9
Knowing that ℓ∼ h we can derive the following relationship
nσ = κρ ρ =
nσT
κT
ℓ=
1
kb T
=h=
nσ
m̄g
thus we find that the density is given by and assuming m̄ ≈ m p
ρ=
m̄g
≈ 8.26 × 10−4 g/cm3
kb κT T
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 × 10−31ρ1/2 T 9 cm2 g−1 . Using this (correct) opacity, repeat the estimate
from a) of the density at the photosphere of the sun.
Substituting the opacity given into the above expression yields
ρ
3/2
m̄g
=
2.5 × 10−13kT 10
b
⇒ ρ=
m̄g
2.5 × 10−31kb T 10
2/3
≈ 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
1
Fc = ρv3c
2
vc =
2F c
ρ
1/3
and knowing that
Fc =
L
4πr2
we find that the convective velocity is given by
vc =
2L
4πr2 ρ
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
tblob =
h
ℓ
≈
vc vc
we know that h which is the scale height of the sun is given by
h=
kT
= 1.7 × 107 cm
gm̄
10
thus we find that the blob timescale is
tblob ≈ 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 is
mostly 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 the
convection zone in comparison to the rest of the sun. For the reasons discussed in class, we can model
the convection zone as having P = Kργ 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 Mr = 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
GM
dP
= −ρ 2
dr
r
thus we find
P = Kρ
γ
1/γ
P
⇒ρ=
K
−3/5
GM
P
dP
= − 2 dr
K
r
and integrating over the following limits we find
Z P P
Prc
K
dP
−3/5
=−
Z R
GM
Rc
r2
dr
this integral yields
i
5 h 2/5
1
1
2/5
−
= GM
P − Pc
R Rc
2K 3/5
thus we find that the pressure is given by
P=
5/2
2 −3/5
1
1
2/5
+ Pc
K
GM
−
5
R Rc
To solve for the density we can just plug this solution into
3/2
3/5
1
2 −3/5
1
1
P
2/5
= 3/5
+ Pc
K
GM
−
ρ=
K
5
R Rc
K
11
and finally the temperature can be found by using
P = nkb T =
2ρ
kb T
mp
⇒ T (r) =
m pP
2ρkb
and substituting the P and ρ from the previous expressions we find
m p K 3/5 2 −3/5
1
1
2/5
T (r) =
+ Pc
K
GM
−
2kb
5
R Rc
b) In detailed solar models, the pressure at the base of the convection zone is ≈ 5.2 × 1013dyne/cm2
and the density is ρ ≈ 0.175 g cm−3 . Using your solution from a), estimate the radius of the base of the
convection zone Rc . Compare this to the correct answer of Rc ≈ 0.71Rsun
If we solve the density equation for Rc we find
i 5K 3/5
1 h
1
2/5
·
= − (ρK 3/2 )2/3 − Pc
Rc R
2GM
and plugging in values we find that
1
= 1.998 × 10−9m−1
Rc
⇒ Rc ≈ 5.11 × 108m = 0.72Rsun
c) In your model, what is the temperature of the sun at 0.99Rsun , 0.9Rsun , 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
m p K 3/5
2 −3/5
25
2/5
T (r = 0.99Rsun) =
≈ 4.1 × 104 K
− K
GMr
+ Pc
2kb
3
66 · Rsun
m p K 3/5
15
2 −3/5
2/5
≈ 5.1 × 105 K
+ Pc
− K
GMr
T (r = 0.90Rsun) =
2kb
3
18 · Rsun
T (r = 0.72Rsun) =
m p K 3/5 2/5
Pc ≈ 1.8 × 106 K
2kb
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 Fc ≈ 1/2ρv3c ≡ Fc,1 where vc is the characteristic velocity of the convective motions.
12
This is the KE flux carried by moving blobs. The other estimate is that Fc ≈ ρ∆Evc where ∆E is the
difference 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
∆ρ
ρb
ab = g
−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
1
W = mv2c =
2
Z l
0
thus we find that the convective velocity is given as
s
vc =
F · dl = aml
2g
∆ρ
l
ρ
which can also be expressed as
∆ρ
v2c
=
2gl
ρ
we can also write this as given that l ∼ H, thus
v2c = 2g
∆ρ kb T
∆ρ kb T
=2
ρ mg
ρ 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 = v2c
m ρ
2kb ∆ρ
and from the equation of state, which is given as
∆E =
1 kb ∆T
(∆ρ ·V )
φ m̄
where (∆ρ · V ) is the mass. Using these two expression and what we found from part (b) we can see
that
1 2 ∆E
∆E v2c
∆E
= vc =
∝
ρV
2φ
m̄
m
2
13
d) Combine your previous results to show that Fc,1 ≈ Fc,2 .
From (a) we know that
1
∆E
Fc,1 ≈ ρv3c Fc,2 ≈ ρ vc
2
m
using this and our solution we find
∆E
1
∆E
1 3
ρvc ≈ ρ vc ⇒ v2c ≈
2
m
2
m
Problem # 2
Estimate the convective velocity vc and the dimensionless entropy gradient (ds/dr)(H/c p) in the convection zones of 0.1 and 10 Msun 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
Fc =
but we know that
H ds 3/2
C p dr ρα3Cs3 thus
Fc ≈
H ds 1/2
vc = Cs C p dr L
4πR2
2/3
H ds L
1
=
C p dr 2
4πR ρ
Cs2
which reduces to
2/3
H ds mp
LR
C p dr = 3M∗
kb T
where
Cs2 =
kT
mp
ρ=
3M
4πR3
and for the convective velocity we find
H ds 1/2
LR 1/3
=
vc = Cs C p dr 3M
from Carrol and Ostley we find that for 10Msun and .1M sun we find that the radius, and luminosity are
approximately
M ≈ 10Msun R ≈ 6Rsun L ≈ 5700Lsun
M ≈ 0.1Msun R ≈ 0.2Rsun L ≈ .0034Lsun
given these values we find
M = 10Msun
M = 0.1Msun
H ds −6
vc ≈ 5.3 × 104 cm/s
C p dr ≈ 3.61 × 10
H ds −10
vc ≈ 698 cm/s
C p dr ≈ 5.8 × 10
14
Problem # 3 Polytropes
(a). The mass M of a star is given by
M=
Z R
0
4πr2 ρ(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
3M
an
ρc = ρ̄an =
4πR3
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 in
terms of the solution to the Lane-Emden equation).
Since we know that
Θ=
given these two relations we can find
ρ(r) = Θn ρc
ρ(r)
ρ
1/n
ξ=
r
a
r2 = a2 ξ2 a = R dr = Rdξ
and from the Lane-Amden equation we know
d
2 dΘ
ξ
= −ξ2 Θn
dξ
dξ
given these following relationships we find that
M=−
Z 1
0
Z 1
d
2 dΘ
4πR ξ Θ ρc dξ = −4πR ρc
ξ
dξ
3 2
n
3
0
dξ
dξ
thus we find that an is given by
1
an = −
3
Z 1
d
2 dΘ
ξ
dξ
0 dξ
dξ
and we can finally show that
ρc =
3M
an
4πR3
(b). Show that the central pressure of a polytrope can be written as
4πGρ2c a2
n+1
where a 6= an is the constant (with units of length) defined in lecture (note that the polytropic relation
−γ
P = Kργ can be used to write K = Pc ρc . Use this result and (a) to derive an expression for the central
pressure of a polytropic model of the form
GM 2
Pc =
cn
R4
Pc =
15
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
4/3
Pc = dn GM 2/3 ρc
where dn depends on an and cn . The values of an , dn , and dn can be determined by numerically solving the
Lane-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 dn .
We know that
K = Pc ρ−γ
c
and also
#
1/n−1 1/2
"
(n + 1)Kρc
a=
4πG
given these relationships we can now find
a2 4πG
1/n−1
1/n−1
= Pc ρ−γ
= Kρc
c ρc
n+1
but since we know that γ = 1/n + 1 we find
1/n−1
Pc ρ−γ
c ρc
−(1+1/n) 1/n−1
ρc
= Pc ρc
= Pc ρ−2
c
and finally we find
Pc =
a2 4πGρ2c
n+1
using this results we can now derive
Pc =
which becomes
a2 4πGρ2c
GM 2
= 4 cn
n+1
R
a2 4πGa2n
Pc =
n+1
3M
4πR3
looking at these two expressions we can see that
2
=
GM 2
cn
R4
9a2 a2n 1
cn =
4πR2 n + 1
now looking at
4/3
Pc = dn GM 2/3 ρc
and from (b) we find
2/3 4/3
a2 4πGρc ρc
Pc =
n+1
a2 4πG
=
n+1
16
an 3M
4πR3
2/3
4/3
ρc
which becomes
4/3
Pc = GM 2/3 ρc cn
4π
an
4/3
9−2/3
and after some fun algebra, which will be omitted here we find
dn = cn
4π
3an
4/3
(c). What are the values of dn for n = 3 and 1.5 polytropes, respectively?
using the above result we find
n = 3 an = 5.99 cn = 0.77 dn = 0.477
n = 3/2 an = 54.183 cn = 11.05 dn = 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 =
ρc
kb Tc
m̄
thus
1/3
dn GM 2/3 ρc m̄
Pc m̄
=
ρc kb
kb
and skipping some algebra we find that the central temperature is given by
Tc =
dn GM
Tc =
Rkb
3an
4π
1/3
mp
2
so all of the expressions can be written as
dn GM
Tc =
Rkb
3an
4π
1/3
mp
2
Pc =
GM 2
cn
R4
ρc =
3M
an
4πR3
(e). Calculate the central temperature, pressure, and density for γ = 4/3 (n = 3) and γ = 5/3 (n = 3/2)
polytropes for M = Msun 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=
we find
5
an = 5.99 cn = 0.77 dn = 0.477
3
ρc ≈ 8.44 g/cm3 Pc ≈ 8.67 × 1015 Tc ≈ 6.2 × 106 K
and for
γ=
4
an = 54.183 cn = 11.05 dn = 0.363
3
17
we find
ρc ≈ 77.26 g/cm3 Pc ≈ 1.24 × 1015 Tc ≈ 1.02 × 107 K
thus 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 Te f f ≈ constant. But
stars with M > 0.3Msun do not end up fully convective on the main sequence and so must go through
a phase of KH contraction in which energy transport is dominated by photons. Assume throughout this
problem that gas pressure dominates and that free-free ab- sorption 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/Msun )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
Msun
11/2 R
Rsun
−1/2
≈ Lrad ≈ −
1 GM 2 dR
2 R2 dt
since we are doing proportionalities we find
M2 R
∝ M 11/2 R−1/2
2
R tT
so we find that the radius scales as
R∝
plugging this into
1
M 7t 2
L ∝ M 11/2 R−1/2 ∝ M 9t
thus
L ∝ M 9t
and to find the temperature
L
∝ R2 Te4f f
Te f f ∝
18
L
R2
1/4
∝ M 23/4 t 5/4
thus
Te f f ∝ M 23/4 t 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 Msun ) at the time when energy transport
by 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 ≈ 0.2Lsun (M/Msun )4/7 (R/Rsun )2
if we set the free-free luminosity equal to the convective luminosity we find
Lsun
M
Msun
11/2 R
Rsun
−1/2
≈ 0.2Lsun (M/Msun )4/7 (R/Rsun )2
some algebra yields
R≈ 5
5/2
M
Msun
69/35
Rsun
(c). Sketch the paths of 1Msun 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 Te f f in K). Note that on
the main sequence a 3 Msun star has L ≈ 40Lsun and Te 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 ≈ 0.2Lsun
M
Msun
4/7 R
Rsun
2
≈5
2/5
· 0.2Lsun
which yields
L(Msun) ≈ 0.724Lsun
and a plot is given by
19
M
Msun
4/7 M
Msun
128/35
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 approach the main sequence - Kelvin Helmholz (KH) contraction.
(a). Argue that, for KH contraction to occur, the timescale
for KH contraction tKH must be longer than
p
the gravitational free-fall time of the cloud, t f f ≈ 1/ Ghρi, where hρi is the mean density of the cloud.
What happens if tKH < t f f ?
Since we know that
1
tf f ≈ p
Ghρi
tKH ≈
"
M
Msun
20
1/2
Rsun
R
#3
(2 × 107yrs)
froma a purely physical argument we know that things cannot fall faster than gravity can pull it. Thus
tKH ≫ t f f
and saying t f f ≫ tKH 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 relationships 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 tKH ≫ t f f .
(b). Estimate the critical radius Rc (inRsun) at which tKH ≈ t f f for a given cloud of mass M (in Msun ).
Assume, as we did in class, that the cloud is fully convective at early times. Show that for R < Rc , the
cloud undergoes KH contraction according to your criterion from a). Recall that the luminosity of a fully
convective star is L ≈ 0.2Lsun (M/Msun )4/7 (R/Rsun )2 .
We can find the critical radius by setting the above expression equal to each other, i.e
"
#3
1
M 1/2 Rsun
p
=
(2 × 107 yrs)
Msun
Rc
Ghρi
re-arranging this expression for Rc we find
M 1/2
Rsun (Ghρi)1/6 (2 × 107 yrs)1/3
Rc =
Msun
and letting
hρi =
3M
M
≈ 3
3
4πRc
Rc
we find
Rc ≈
2/3
7
1/3 G1/9
4/9 Rsun (2 × 10 yrs)
M
1/3
Msun
≈ M 4/9 7.6 × 10−3
if we put the mass off the sun we get
Rc ≈ 4.79 × 1010cm
Since we know that
t f f ∝ R3/2
tKH ∝
1
R3
we can see that
R ↑
R ↓
tKH ↓ t f f ↑
tKH ↑ t f f ↓
from these two relationships we can see that for KH contraction to occur that tKH ≫ t f f and also that
R < Rc .
(c). What is the central temperature of the (fully convective) cloud (in K) as a function of its mass M
(in Msun ) when R = Rc ?
We know that the temperature of a fully convective object is given by
dn GM 3an 1/3 m p
Tc =
Rkb
4π
2
21
and since we know that fully convective stars have a polytropic index of γ = 5/3. Knowing this we
find
n = 3 an = 5.99 dn = 0.477
thus
M
Rc
≈ 2.79 × 10−15 M 5/9
Tc ≈ 2.16 × 10−15
and if we want the temperature of a collapsing gas cloud with respect to Msun we get
Tc ≈ 1.9 × 104K
Problem # 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
Mj =
kb
Gm p
3/2
T 3/2
√
ρ
using the values given we find
M j ≈ 1.37 × 1042 g
and the mass of M13 is
M13 ≈ 0.5 × 106 · Msun ≈ 9.95 × 1038 g
and we can see that
M j ≫ M13
which 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 56 Fe 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
Etot =
Eb
× Nnucleon
nucleon
and the number of nucleons are given by
1 kg
M
≈
≈ 5.98 × 1026 nucleon
Nnucleon =
mp
mp
thus
Etot ≈ 5.27 × 1027 MeV
(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
22
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
EHe
EFe
LFe =
tHe
tFe
and since we know that these two luminosities are theoretically equal
LHe =
EHetFe = EFetHe
which gives
tHe =
6.4
EHe
tFe =
tFe ≈ 0.72tFe
EFe
8.5
thus we see that
tHe < tFe
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 4 He 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
rc =
e2 Z1 Z2
≈ 7.2 × 10−11 cm
E0
We know that the propbability for a particle-particle interaction is given by
1/2
E
− EG
P=e
where
EG = 2π2 α2 Z12 Z22 (mr c2 )
is the Gamow energy. For a proton-proton interaction we find
EG ≈ .493 MeV
thus the probability is given as
P ≈ 1.51 × 10−7
for a He-He interaction we find
e2 Z1 Z2
rc =
≈ 2.8 × 10−10 cm
E0
EG = 32π2 α2 2m p c2 ≈ 31.6 MeV
and the probabilty is given by
P ≈ 2.50 × 10−55
23
for the proton-He interaction we find
rc =
e2 Z1 Z2
≈ 1.44 × 10−10 cm
E0
EG ≈ 3.15 MeV
thus the probability is given by
P ≈ 5.8 × 10−18
(b) What energy E would be required for i) the two 4 He nuclei, ii) the proton and the 4 He nucleus, and
iii) two 12 C 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=
EG
(ln P)2
for the He-He interaction we find
EHe−He ≈ 0.125 MeV
for the proton-He interaction we find
E p−He ≈ .013 MeV
for the carbon-carbon interaction the Gamow energy is given by
EG ≈ 2592π2α2 6m p c2 ≈ 7.7 GeV
and thus
Ec−c ≈ 31 MeV
we know that
E
kb
T≈
so
THe−He ≈
1.45 × 109K
Tp−He ≈
1.5 × 108K
Tc−c ≈
3.6 × 1011K
Problem # 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 =
m1 r 1 + m2 r 2
m1 + m2
24
so the velocity would be
vcom =
m1 v1 + m2 v2
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 − vcom
vrel−2 = v2 − vcom
a bit of algebra yields
vrel−1 =
vrel−2 =
m2
(v1 − v2 )
m1 + m2
m1
(v2 − v1 )
m1 + m2
(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
1
1
m1 v2rel−1 + m2 v2rel−2
2
2
m1 m2
=
(m2 (v1 − v2 )2 + m1 (v2 − v1 )2 )
2(m1 + m2 )2
m1 m2
(v1 − v2 )2
=
2(m1 + m2 )
Ktot =
Ktot =
1
mr (v1 − v2 )2
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 Te f f ≈ 4000(L/Lsun )1/102 (M/Msun )7/51 K (I actually derived a
coefficient 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
25
L ≈ 0.2(M/Msun )4/7 (R/R)2sun
Lsun ≡ Lconv
I called this luminosity Lconv since it is derived from the properties of energy transport alone (convective interior + radiative atmosphere with H− opacity). The luminosity of a star is also given by
L f usion = 4πr2 ρε(T, ρ)dr
where ε is due to the proton-proton chain for low mass stars (this was given in lecture). As discussed in
class, 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), Tc (M), and L(Te f f ) (the HR
diagram) for fully convective stars, like we did for other examples in lecture. Approximate ε ∝ ρT β with
an appropriate choice of β (recall that low mass stars will have somewhat lower central temperatures than
the sun, closer to ≃ ×106 K, as you will see in part b).
We know that
L f us ∝ R3 ε(ρ, T ) ∝ R3 ρ2 T β
Lconv ∝ M 4/7 R2
we can find what β is by
EG 1/3
2
β=− +
3
4kT
given that we know what the temperature is and also what EG for p-p reaction
T ≈ 5 × 106 K
EG ≈ 500 keV
we find that
β = 5.92 ≈ 6.0
we also know
ρ∝
M
R3
we know that in steady state
L f usion = Lconv
thus we can find
M2
M R ∝R
Tc6 ⇒ T 6 ∝ M −5/21 R5/6
R6
we know from the Virial temperature, assuming gas pressure dominates
4/7 2
3
T∝
M
R
thus we find
Tc ∝ 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
26
with this we can now find what the effective temperature as a function of mass is, i.e
L ∝ R2 Te4f f
thus
Te4f f ∝
L
∝ M 4/7
R2
which yields
Te f f ∝ M 1/7
to find what the luminosity as a function of the effective temperature is (HR diagram)
M ∝ Te7f f
which yields
148/11
L ∝ Te f f
In a) you just determined a scaling relation between stars of different mass, but not the absolute values
of L, Te 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 a
polytrope, 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 not
asking you to prove equation (2). You will have to trust me. Note that for a typical value of β for the pp
chain, equation (2) says that L f usion ≃ 0.1εc M . This makes sense because fusion only takes place at the
center 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 Tc , 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). Note
that you should give expressions for Tc , ρc , and εc here, with constants and real units, not just scaling
relationships. So that the constants in front of your expressions are reasonable, please normalize M to
Msun and R to Rsun .
The general expressions given by HW 4 problem #3 are
dn GM
Tc =
Rkb
3an
4π
1/3
µm p
Pc =
GM 2
cn
R4
ρc =
we found that for a n = 3/2 polytrope
an = 5.99 cn = 0.77 dn = 0.477
27
3M
an
4πR3
thus we find that
m pGMsun
Tc = 0.322
kb Rsun
M/Msun
R/Rsun
Msun
ρc = 1.43 3
Rsun
M/Msun
(R/Rsun )3
plugging in all the constants yields
M/Msun
Tc ≈ 7.43 × 10 K
R/Rsun
6
we also know
M/Msun
ρc ≈ 8.41
(R/Rsun )3
−2/3 −15.7T7−1/3
εc = 5 × 105 ρc X 2 T7
e
we also know that we can approximate this as
β
εc = AρT7 X 2 = AρT76 X 2
setting this two expressions equal to each other we can find what A is, i.e lettingT ≈ 107 K we find
A = 5 × 105 e−15.7 ≈ 0.076
thus we find
εc ≈ 0.076ρcT76 X 2 ≈ 0.037ρcT76
substituting Tc and ρc gives
M
εc ≈ 0.053
Msun
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), Tc (M), and L(Te f f ) relations for fully convective stars. If you use the same β, your expressions here
should 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 star
from first principles. In doing this problem, remember that β is temperature dependent so make sure you
check 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 ≃
2.4εcM
≈ .09εcM
(3 + β)3/2
(β ≈ 6)
thus we know that
Lcon = L f usion
M
0.2
Msun
4/7 R
Rsun
2
Lsun
M
= 0.0047
Msun
7 Rsun
R
rearranging this we find
which yield
R
Rsun
11
Msun
= 0.023
Lsun
M
Msun
R
M 52/77
≈ 0.67
Rsun
Msun
28
52/7
9 M
Msun
Msun
to find for the temperature we can use
M/Msun
Tc ≈ 7.43 × 10 K
R/Rsun
6
using our previous resulst gives
M
Tc ≈ 1.1 × 10 K
Msun
7
25/77
to find for the luminosity we use
M
Lconv = 0.2
Msun
4/7 R
Rsun
2
Lsun
plugging in for the radius we find
148/77
Lsun
148/77
Lsun = 4πR2 σTe4f f
M
Lconv = 0.09
Msun
for the effective temperature we find
L = 4πR
2
σTe4f f
M
⇒ 0.09
Msun
which can be simplified to
M 148/77
M 104/77 4
18
0.09
Lsun = 1.56 × 10
Te f f
Msun
Msun
thus
Te f f
M
= 3868 K
Msun
which can also be written as
M
=
Msun
Te f f
3868 K
1/7
7
148/11
L(Te f f ) = 4.85 × 10−50 Lsun Te f f
d) What are your predicted luminosities, radii, and effective temperatures for main sequence stars with
M = 0.1 and 0.3Msun ? Compare your values to the values of L = 0.01Lsun , R = 0.3Rsun , and Te f f = 3450
K for M = 0.3Msun and L = 10−3 Lsun , R = 0.11Rsun, and Te f f = 3000 K for M = 0.1Msun that I found in a
graduate textbook (based on detailed models).
Given our relationships we find for
M = 0.1Msun L = 1.1 × 10−3Lsun R = 0.14Rsun Te f f = 2783 K
and for
M = 0.3Msun L = .009Lsun R = 0.298Rsun Te f f = 3256 K
29
Problem # 2 Very Massive Stars
Consider very massive stars with M ∼ 50 − 100Msun . Recall that I showed in lecture and you showed
on 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), Tc(M), and L(Te f f ) (the HR
diagram) for very massive stars, like we did for other examples in lecture.
Using radiative diffusion along with
1
Prad = aT 4
3
we also know
thus
and radiative diffusion says
dPr 4 3
= aT
dT
3
4
dT
GM
dP dP dT
=
= aT 3
= −ρ 2
dR dT dR 3
dR
R
ρM
dT
∝ 3 2
dR T R
dT
ρM
Lρ
∝
∝ 3 2
2
3
R T
dR T R
which gives us
L∝M
using the Virial theorem, where Prad dominates rather than Pgas we find
T4 M
∝
ρ
R
⇒ Tc ∝
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 ∝ MρT 18
where we chose β = 18 as a more appropriate value rather than the value given for the sun β = 20, this
is motivated by the fact that more massive stars have somewhat higher temperatures, thus reducing β. We
find
1 R3
T 18 ∝ ∝
ρ M
and using the result from Virial temperature we find
M 9 R3
∝
R18 M
thus we find
R ∝ M 10/21
and we also find for the central temperature
Tc ∝ M 1/42
30
and to find the effective temperature we know
Te4f f ∝
L
M
∝ 2 ∝ M 1/21
2
R
R
simplifying gives
Te f f ∝ M 1/84
and finally the luminosity as a function of Te f f is given by
L(Te f f ) ∝ Te84
ff
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 2Msun to 75% at 60Msun .
The condition for convection is given by
d ln T
1 Ptot L Lr /L
γ−1
≈
>
d ln P 4 Prad LEdd Mr /M
γ
since we know that
γ=
4
Ptot ≈ Prad
3
gives us
1 Lr M
1
>
4 LEdd Mr 4
which simplyfies to
Lr
Mr
>
Ledd
M
we know that in the limit that M → 150Msun Lr → LEdd ,
Mr
<1
M
which means that the fraction of the mass of the star that is undergoing convection approaches 1, which is
100% of the mass is undergoing convection. Its a little strange that stars that are much less massive than
the sun and the stars that are much more massive than the sun are both almost fully convective.
n=3/2 polytrope
The previous case yielded a result for a n=3 polytrope, we find that for
n = 3/2 polytrope
and in the limit where M → 150Msun
Mr 5 Lr
<
M
8 LEdd
Lr → Ledd
Mr 5
<
M
8
31
γ=
5
3
This seems rather strange in the sense that stars that are approximately 60Msun have a convective core
that encompasses 75% of the mass, which means that the convective core decreases after M > 60Msun ?
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
tMS ≈
Etot
LEdd
where
Etot = NQ
Q ≈ 7 MeV
where that is the total energy per reaction, we also know
Mr
M
=
(n = 3 polytrope)
mp mp
Mr 5 M
=
(n = 3/2 polytrope)
N =
mp 8 mp
N =
we also know that
4πcGM
κT
so we find the main-sequence lifetime to be given as
LEdd =
κT Q
(n = 3 polytrope)
m p 4πcG
5κT Q
≈
(n = 3/2 polytrope)
m p 32πcG
tMS ≈
tMS
we know that
κT ≈ 0.4 cm2 /g
Q ≈ 7 MeV ≈ 1.12 × 10−5 ergs
thus we find that the main-sequence lifetime for both types of polytropes are given by
2.21 × 106yr < tMS < 3.39 × 106yr
seems reasonable.
Problem set 8
Problem # 1 Fermi gas
32
a) Above what density is a gas of room temperature fermions degenerate? Below what temperature
would gas with the density of air be degenerate?
as
We know that if the density of the gas is ng ≥ nQ where nQ is the quantum concentration, nQ is defined
nQ ≡
2πm̄kT
h2
3/2
(1)
and for a gas at room temperature to be degenerate
ng ≥ nQ =
2πm̄kT
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 N2 . If we assume that the questio is only speaking about
free electrons we get
2πme kT 3/2
ng ≥ nQ =
≈ 1.25 × 1019 cm−3
2
h
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
= 2.52 × 1019 cm−3 = nQ
kT
and now using Equation 1 we find
2/3
T=
nQ h2
2πm̄kT
≈ 9.2 × 10−3 K
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
3
Eth ≈ kT ≈ .039 eV
2
We can write the Coulomb energy as
Ecoul ≈ Z 2 e2
we know
r ∼ n−1/3 ∼
33
1
r
ρ −1/3
m̄
thus
2 2
Ecoul ≈ Z e
ρ 1/3
≈ 12.43 keV
m̄
using ρ ≃ 10 g/cm3 and m̄ ≈ 100m p . The Fermi energy can be written as
Ef =
3
8π
2/3
h2
2me
47ρ
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
Ecoul ≫ E f > Eth
for 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 nD ≈ 2 × 10−5 nH .
Let’s focus on a low mass fully convective star undergoing KH contraction; such a star can be reasonably
well modeled as an n = 3/2 polytrope. Assume that the star has cosmic composition ( µ ≃ 0.6). Note that
in this problem, you should not use the approximation ε ∝ ρT β . Instead, you will need to keep the full
expression for ε.
a) What is the Gamow energy for D fusion? Write down the resulting thermally averaged cross-section
hσvi for D fusion.
The Gamow energy can be written as
EG = Z12 Z22
mr
MeV
mp
using Z1 = Z2 = 1 and mr = 32 m p we find the Gamow energy to be
EG ≈ .67 MeV
The thermally averaged cross-section is given as
1/6
2.6S(E)EG −3(EG /4kT )1/3
hσvi =
e
k2/3 T 2/3
using all the constants given and the Gamow energy we find
hσvi =
3.7 × 10−15 cm3 −3742(K/T )1/3
e
K−2/3 T 2/3 s
hσvi =
3.7 × 10−15 cm3 −3742(K/T )1/3
e
K−2/3 T 2/3 s
in terms of M and R we find
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 tc ≡ R/|dR/dt|
34
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
3 GM 2 dR = 0.2Lsun (M/Msun )4/7 (R/Rsun )2
L=
7 R2 dt thus we find
Msun 2
R 2
M 4/7
R 2
M
Rsun
Msun
Rsun
10/7 4
Msun
R
= 3.34 × 10−5cm/s
M
Rsun
dR
7 · 0.2 R2sun Lsun
=
dt
3 GM 2sun
and we can find the local contraction time to be
tc = Rsun
R
Rsun
1
M 10/7 Rsun 3
7
= 6.7 × 10 yr
(dR/dt)
Msun
R
As the star contracts the contraction time gets longer.
c) What is the lifetime tD 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 tD 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
tD =
l
1
=
v n p σv
which gives us
htD i =
but we also know
ρc =
thus we find
htDi =
we also know
µm p
1
=
n p hσvi ρc hσvi
3M
M
an = 1.43 3
3
4πR
R
µm p R3
µm p
1 R3
=
= 7.019 × 10−25 g
ρc hσvi 1.43hσviM
hσvi M
dn GM
Tc =
Rkb
3an
4π
1/3
µm p = 2.60 × 10−16
cm M
K
g R
thus we find
htD i = 7.019 × 10−25 g
3
1/3 R
1 R3
gs
= 1.89 × 10−10 3 T 2/3 e3742(K/T )
hσvi M
cm
M
35
written in terms of M and R we find
R
htD i = 1.23 × 10 s
Rsun
−6
7/3 Msun
M
1/3
1/3
e19.19((Msun /M)(R/Rsun ))
the deuteron lifetime gets shorter as the star contracts.
d)For any mass M show that there is a critical radius RD at which tD = tc . This represents the radius
(time) at which D starts to undergo significant fusion. Give the numerical value of RD for M = 0.03 and 0.1
Msun . For each of these two cases, also determine the central temperature of the star Tc and the D lifetime
tD when R = RD . Does D fusion occur before or after the star reaches the main sequence?
We know that
tD = tc
which yields
R
1.23 × 10 s
Rsun
−6
7/3 Msun
M
1/3
19.19((Msun /M)(R/Rsun ))1/3
e
M
= 6.7 × 10 yr
Msun
7
10/7 Rsun
R
3
and so we find
RD
Rsun
16/3
= 1.71 × 10
21
M
Msun
37/21
1/3
e−19.19((Msun /M)(R/Rsun ))
we can solve this numerically to find
RD = 0.44Rsun
RD = 1.11R
M = 0.03Msun
M = 0.1Msun
to find the central temperature we can use
M
Tc = 7.4 × 10 K
Msun
6
Rsun
RD
we find
Tc ≈ 5.0 × 105 K M = 0.03Msun
Tc ≈ 6.67 × 105K M = 0.1Msun
and to solve for the deuteron lifetime we find
RD 7/3 Msun 1/3 19.19((Msun /M)(R/Rsun ))1/3
−6
htD i = 1.23 × 10 s
e
Rsun
M
so we find
tD ≈ 4.67 × 106yr M = 0.03Msun
tD ≈ 4.1 × 105yr M = 0.1Msun
e) Can D fusion halt (at least temporarily) the KH contraction of the star? Explain your answer quantitatively.
36
Since we know that
ε=
Qrd
ρd
ρ d = md n d
we also know that
L = Mε =
rd =
nd
tD
MQ
MQrd
=
ρd
2m ptD
since we know that Q = 8.8 × 10−6ergs, we find
M(8.8 × 10−6ergs)
L=
2m ptD
so for M = .03Msun we find
L ≈ 5.37 × 1035 ergs/s
and for M = 0.1Msun we find
L ≈ 4.08 × 1037 ergs/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 = Kρ5/3 such an object can be modeled (rigorously) as an n = 3/2 polytrope.K is a constant that depends
on 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 depends
on the radius R and mass M of the object to derive the R(M) relation for degenerate objects (the radius
also depends on µe ). Note that you should give an expression with proper constants and not just a scaling
relationship. Normalize the mass M to Msun and the radius R to Rsun (this should sound pretty familiar by
now).
We know that
h2
Pdeg = Pc =
5me
rearranging this equation for ρc yields
−1/3
ρc
h2
=
5me
3
8π
2/3 3
8π
ρc
µe m p
2/3 5/3
1
µm p
4/3
= dn GM 2/3 ρc
5/3
1
dn GM 2/3
from the last problem set we showed
M
ρc = 8.41
Msun
Rsun
R
3
using this we find
R
M −1/3
−5/3
= 0.04µe
Rsun
Msun
b) 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 = MJ , we find
R ∼ 0.30Rsun
37
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 = Ecoul
then
1 e2
=
4πε0 r
but we know that
3
8π
2/3
h2 2/3
n
2me
1
∼ n1/3
r
thus
1 2 1/3
e n =
4πε0
2/3
3
8π
3
8π
2
h2 2/3
n
2me
so we find the density to be given by
n=
2me 2
e
4πε0 h2
3 ≈ 6.15 × 1028 m−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
Msun (M/Msun )
= nm p
(0.03Rsun)3 (M/Msun )−1
which can be simplified to
M
=
Msun
nm p (.03Rsun)3
1.43Msun
1/2
which can also be expressed as
M ≈ 190Mearth
using this we can now find the radius to be
R ≈ 0.361Rsun
Problem set 9
38
≈ 5.7 × 10−4
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 mechanical
distribution 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), but
the classical dist. fcn you end up with should, of course, be independent of h.
We know that the chemical potential is defined as
n
µ = mc + kT ln
gnQ
2
(2)
and that the quantum distribution function is defined as
n(p) =
g/h3
e(E p −µ)/kT ± 1
to show that in the classical regime
e(E p −µ)/kT ≫ 1
We can write Equation 1 as
nQ 1 (mc2 −µ)/kT
= e
n
g
2 −µ)/kT
e(mc
n ≪ nQ
≫1
thus the quantum distribution function can be written as
n(p) =
g/h3
e(E p −µ)/kT
We also know that
p2
E p = mc +
2m
2
nQ =
2πmkT
h2
3/2
and using equation 1 we find
n(p) =
g/h3
(mc2 +p2 /2m−mc2 −kT
e
ln(nQ /n))/kT
=
gn
h3 eE/kT n
Q
which after some simplification reduces to the Classical Boltzman distribution function
1
n(p) = n
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 this
formula is valid between m = 0.5Msun and m = 150Msun . Note that dN/dm has units of number of stars
per unit mass.
39
a) What is the ratio of the number of stars formed with masses within dm ≃ m1 of m1 and masses
within dm ≃ m2 of m2 ? What is the ratio of the number of 150Msun stars formed to the number of 0.5Msun
stars formed?
We know that the Initial Mass Function IMF is given as
dN
∝ m−α ∝ m−2.35
dm
α = 2.35
we also know that for dm ≃ m1 of m1 we find
1−α
dN(m1 ) = m−α
1 dm1 ≈ m1
we also know that for dm ≃ m2 of m2 we find
1−α
dN(m2 ) = m−α
2 dm2 ≈ m2
and the fraction is given by
dN(m1 )
=
dN(m2 )
m1
m2
1−α
≈ 2208 m1 = 0.5Msun m2 = 150Msun
b) Estimate the mass of a cloud M so that approximately one 150Msun 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
Ns ≈ 2208Nb
where Nb is for stars that are the number of ∼ 150Msun and Ns is for stars that have ∼ 0.5Msun and 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
hM∗ i ≈ 0.5Msun
Mcluster = Ns · hM∗ i ≈ 1104Msun
to find the density of the gas in which this cloud formed we can use the Jean’s density
3kT 3
3
≈ 3.14 × 10−22 kg m−3
ρJ =
4πM 2 2Gm̄
Problem # 3
A stellar atmosphere consists almost entirely of hydrogen. Assume that 50 % of the hydrogen molecules
are 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 H2 into hydrogen, i.e., the reaction
γ + H2 ←→ H + H .
We know that
γ + H2 ←→ H + H
and the Saha equation gives
µ(H2 ) ←→ 2µ(H)
40
but we know that the chemical potential is given by
nQ,H gH
nQ,K2 gH2
2
2
µ(H) = mH c − kT ln
µ(H2) = mH2 c − kT ln
nH
nH2
where
mH c2 = m p c2 + me c2 − χH
thus we find
mH2 c2 = 2m p c2 + 2me c2 − χH2 − 2χH
nQ,K2 gH2
mH2 c − kT ln
nH2
2
which becomes
χH
− 2 = ln
kT
"
nQ,H gH
2
= 2 mH c − kT ln
nH
nQ,H2 gH2
nH2
nH
nQ,H gH
2 #
we are given that
gH2 = gH = 1
nQ,H2 ≈ nQ,H ≈
thus we find
nH −χH /kT
nH2
=
e 2
= nH
nH
nQ
h2
2πmkT
2πmkT
h2
3/2
3/2
e−χH2 /kT
but we are given that
nH =
P
3kT
1
nH2
=
nH
2
h2
2πmkT
3/2
thus
P
1
=
2 3kT
e−χH2 /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 the
observed lines of H from stars. Recall that the energy levels of the H atom are given by E = −13.6/n2 eV
and the degeneracies are gn = 2n2 .
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 what
we are after is n p /n since an ionized hydrogen atom is just a proton. Your result for n p /n will depend
on 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 reason
this 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
41
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
ne n p ge g p
=
nH
gH
we know
2πmkT
h2
3/2
e−χ/kT
ne ∼ n p n = nH + n p χ ≈ −13.6 eV
thus
and so
n2p
=α
nH
α≡
2πmkT
h2
3/2
ge g p
=1
gH
e−χ/kT
n2p = αnH = α(n − n p)
which becomes a quadratic equation of the form
n2p + αn p − αn = 0
with the solution of n p being
√
−α ± α2 + 4αn
np =
2
and since we know that this must be a positive thus we will take the positive solution
√
−α + α2 + 4αn
np =
2
and finally we are looking for
√
n p −α + α2 + 4αn
=
n
2n
the plot is given by
42
np vs ntotal
0.8
np/n
0.6
0.4
0.2
0.0
5.0•103
1.0•104
1.5•104
Temperature
2.0•104
2.5•104
3.0•104
we can see that at a temperature of T ≃ 1.4 − 1.5 × 107K roughly ∼ 50% of the hydrogen atoms are
ionized.
b) Use your result from a) to calculate the fraction of all H atoms that have an electron in the n = 2
state of hydrogen. If n2 is the number density of atoms with electrons in the n = 2 state, then what we
are after here is n2 /n. You will need to use the Boltzmann factor in addition to your result from the Saha
equation in a). For densities appropriate to the photosphere of the sun, make a plot of n2 /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.
We know that
n − np
np
nH
=
= 1−
n
n
n
and from the Boltzman equation we know that
n2 g2 −(E2 −E1 )/kT
= e
n1 g1
we also know
nH = n1 + n2
and
what we are looking for is
→ n2 = nH − n1
→
nH
n1
= 1+
n2
n2
n1 −1
n2
= 1+
nH
n2
np n2
n2 nH
n1 −1 1−
=
= 1+
n
nH n
n2
n
43
thus we find the fraction of all H atoms that have an electron in the n = 2 state of hydrogen given by
!
√
np n2
−α + α2 + 4αn
n1 −1 g1 (E2 −E1 )/kT −1
1−
1−
= 1+ e
= 1+
n
n2
n
g2
2n
where α has been explicitily defined already. The plot is given by
n2 vs ntotal
4•10−5
np/n
3•10−5
2•10−5
1•10−5
0
1.0•104
1.5•104
2.0•104
2.5•104
Temperature
3.0•104
3.5•104
4.0•104
We can see that the fraction of hydrogen atomes in the energy state n = 2 peaks at ∼ 1.5 × 104 K.
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 Hα(n = 2 → 3)and Hβ(n = 2 → 4) lines of H?
Use your result from b) to explain why A stars show the most prominent Hα lines of hydrogen (relative to
more massive stars such as O stars and less massive stars such as M stars).
We know that
∆E = hν =
thus
λ=
and
hc
λ
hc
hc
=
∆E (E2 − E1 )
E=−
13.6 ev
n2
thus for the n = 2 → n = 3 transition we get
λ=
hc
≈ 656.3 nm
(3.4 − 1.51)eV
44
and for the n = 2 → n = 4 transition we get
λ=
hc
≈ 486.7 nm
(3.4 − 0.85)eV
From the plot given in part b) we can see that the fractional number of atoms in the n = 2 energy state
peaks 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 similar
for 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 and
the n = 2, 3, 4, .... states. What is the wavelength of the Lyα (n = 1 → 2) line of H? Roughly what fraction
of H atoms have electrons in the ground (n = 1) state of H in the atmosphere of an M-star? Would you
expect to see prominent Lyα lines from an M-star? Why or why not?
Using
λ=
we find
λ=
hc
hc
=
∆E (E2 − E1 )
hc
hc
=
≈ 121.6 nm
∆E (13.6 − 3.4)eV
We would not expect to see any Lyα 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 µ
is defined by P ≡ ρkT /µm p where P is the total ideal gas pressure (ions and electrons), while the electron
mean molecular weight µe is defined by ne ≡ ρ/µe m p .
Since we know that the total pressure is given by
ρkT
PT = PI + Pe =
mp
and thus
but we know that
1
1
+
µI µe
1
1
1
= +
µ µI µe
X
1
=
µI
A
XZ
1
=
µe
A
45
=
ρkT
µm p
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
A
1 A
µe =
(3)
µ=
X 1+Z
XZ
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
µe = 1
2
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 = 0.75 Y = 0.25 A = 4 Z = 2
thus
1
X Y
1
13
= + = (X +Y ) =
µI
A A A
16
Z
7
1
= (X +Y ) =
µe A
8
and so we find
µ=
8
16
µe =
27
7
iii) ionized He,
Since this is a pure fully ionized gas we can use Equation 1 with
X =1 A=4 Z=2
we find
µ=
4
µe = 2
3
iv) ionized O,
Using Equation 1 with
X = 1 A = 16 Z = 8
we find
µ=
16
9
µe = 2
v) ionized Fe
Using Equation 1 with
X = 1 A = 56 Z = 26
we find
µ=
56
27
µe =
46
28
13
b) Which gas has the largest ideal gas pressure? Which gas has the largest electron degeneracy pressure? Assume that ρ and T are the same in all cases.
Since we know that gas pressure goes as
1
µ
than the smallest µ will give us the highest pressure, thus the element that has the highest gas pressure
Pg ∝
is
µ=
1
2
Hydrogen gas
for degeneracy pressure we know
5/3
Pd ∝ ne
∝
1
µe
5/3
thus the gas that gives the highest degeneracy pressure is the one with the lowest value for µe and this
is
µe = 1
Hydrogen gas
Problem # 2 The Helium Main Sequence
In certain stages of stellar evolution, stars are largely composed of He and He fusion dominates the
stellar luminosity. One can approximate such stars as lying on a He main sequence. In this problem
we will calculate the properties of the He main sequence assuming that a star is composed of pure He,
that energy transport is via radiation, that electron scattering dominates the opacity, and that gas pressure
dominates. The energy generation rate for He fusing to Carbon is
ε = 5 × 1011 ρ2 T8−3 exp(−44/T8 ) ergs s−1 g−1
and 7.65 MeV is released converting 3 He nuclei into 1 C nucleus. Note that throughout this problem
you should not just give scaling (proportionality) laws for the desired relations; you should also determine
reasonable normalizations.
a) Calculate the relationship between mass M and luminosity L for the He main sequence.
For a star that has the given properties: energy transport is via radiation, electron scattering dominates
the opacity (σ = σT ), and that gas pressure dominates (P ∼ Pg ) we find the luminosity given as
L ∝ M 3 µ4 µe
This equation gives the evolution of the lumunosity on the MS as chemical composition changes. We
can scale this to the sun to find
µ 4
µe
M 3
L = Lsun
Msun
µsun
µe−sun
we also know that
4
µe = 2 µsun = 0.6 µe−sun ≈ 1.14
3
and so we find that the lumonosity for the Helium main sequence can be expressed as
µHe =
L ≈ 42.7Lsun
47
M
Msun
3
b) Estimate
the core temperature of a 1 solar mass He star. You do not need to do the full integral
R
L f usion = dMε , but can approximate this as L f usion ∼ 0.1Mε(r = 0).
In steady state we can express the lumunosity of energy transport be equal to the luminosity due to
fusion
L f usion ≈ Ltransport
where we can use part a) and the approximation given to find
42.7Lsun
M
Msun
3
= 0.1Mε = 0.1M5 × 1011 ρ2 T8−3 exp(−44/T8 ) ergs s−1 g−1
if we let
M = Msun ρ ≈ 150 g/cm3
we find
T83 e44/T8 = 1.36 × 1013
solving this numerically yields
Tc ≈ 1.5 × 108 K
T8 ≈ 1.52
c) Given your result for Tc for a 1 Msun star from b), calculate the power-law relation Tc (M) by imposing
the steady state requirement that L f usion = L photons and using ε ∝ ρα T β (where L photons is the energy carried
out of the star by photons from a).
Since we know that
L photons ∝ M 3
we find
L f usion ∝ Mε
M 2 ∝ ε ∝ ρα T β
where in part b) we are given
α=2
β = −3 +
44
≈ 26
T8
knowing that the density and temperature scale as
ρ∝
M
R3
T∝
M
R
gives
M ∝
2
M
R3
2 M
R
26
→ R ∝ M 13/16
using this along with our expression for the temperature gives
T∝
M
M
∝ 13/16 ∝ M 3/16
R M
to scale to a one solar mass star, from part b) we find
M
T ≈ 1.5 × 10 K
Msun
8
48
3/16
d) Use your results above to determine the R(M)and Te f f (L)relations for the He main sequence. Then
sketch the relative positions of the H & He main sequences in the HR diagram.
From the Virial temperature we know
Tc =
Gµm p M
k R
where k is Boltzmans constant, this expression can be scaled to the sun
Gµm p Msun
M
Rsun
Tc =
k Rsun Msun
R
using the result from c) and plugging in all the constants along with µ = 4/3 gives
M
1.5 × 10 K
Msun
8
3/16
M
= 3.08 × 10 K
Msun
7
Rsun
R
and we find
R
Rsun
M
= 0.21
Msun
13/16
To find the relationship between the luminosity and the effective temperature we can use
L = 4πR2 σTe4f f
but we know from our previous expression
2
R = (0.21Rsun)
2
M
Msun
13/8
plugging this in our expression of the luminosity gives
L = 4πσ(0.21Rsun)
2
M
Msun
13/8
Te4f f
now we can use the result from a)
M
Msun
=
L
42.6Lsun
1/3
using this we get
Lsun
L
Lsun
= 4πσ(0.21Rsun)
2
L
42.6Lsun
13/24
thus we find that
Te f f
L
≈ 2.1 × 10 K
Lsun
4
The HR diagram for this star is given by
49
11/96
Te4f f
which seems rather odd, We would expect this main sequence He buring star to be above the main
sequence line. This can be explained by our initial assumption that went into deriving this relationship.
We assumed that this was a pure ball of He gas.
e) At what mass does the luminosity of the star exceed the Eddington luminosity?
We know that the Eddington luminosity is given by
LEdd
4πcGM
=
κ
L f usion = 42.7Lsun
M
Msun
3
setting these two expressions equal to each other gives
4πcGMsun
κ
given the opacity defined as
κ=
thus
κ=
1
σT
µe m p
M
Msun
= 42.7Lsun
ne σT
ρ
µe = 2
M
Msun
3
ne =
ρ
µe m p
κ=
σT
≈ 0.2 g/cm
2m p
M ≥ 38.95Msun
is the mass that will exceed the Eddington luminosity
f ) What is the He main sequence lifetime as a function of stellar mass? Compare this to the corresponding H burning lifetime.
50
We know that the main sequence lifetime is given by
t=
E
L
E = NQ =
0.1M
0.1Msun M
(7.65MeV) =
(7.65MeV )
12m p
12m p Msun
thus
0.1Msun
t=
12m p (42.7Lsun )
Msun
M
2
Msun
(7.65MeV) ≈ 2.34 × 10 yrs
M
7
2
thus the main sequence lifetime is
Msun
t ≈ 2.34 × 10 yrs
M
7
2
this is much much shorter than the H burning lifetime which is t ∼ 1010 yrs for M ∼ 1Msun . Also can
be written as
tHe ∼ 0.2% the time of the main sequence Hydrogen burning
Problem # 3 The Thin Shell Instability
As we discussed in lecture, during several phases of stellar evolution, fusion takes place in a thin shell.
Consider such a shell located a distance Rs from the center of a star. The mass interior to Rs is M , the
mass of the shell itself is Mshell and the thickness of the shell is H ≪ dR ≪ R, where H is the scale-height
at radius Rs (recall that H is the distance over which the density, pressure, temperature, etc. change).
a) Use hydrostatic equilibrium to show that the pressure at the base of the shell is given by
P(Rs) ≃
GMMshell
4πR4s
HE gives
dP
GM
= −ρ 2
dr
R
which can be written in differential form
P(Rs + dR) − P(Rs)
GM
= −ρ 2
dR
Rs
but we know that
ρs =
Ms
Ms
≈
Vs
4πR2s dR
and since we know that H ≪ dR ≪ R (this comes from the definition of H) then P(Rs + dR) ≪ P(Rs)
than
GMMs
P(Rs)
=
dR
4πR4s dR
which simplifies to
P(Rs) =
GMMs
4πR4s
b) Use your result in a), together with the strong temperature dependence of fusion reactions, to explain
why fusion in a thin shell is unstable and will runaway, as in a bomb. Hint: How will P, ρ, T, and dR of the
shell change if there is a small perturbation that increases the amount of fusion in the shell?
51
Since we know that
M
dR3
If we apply a small pertubation that increases fusion then we know T ↑, ρ ↓, and dR ↑and since the
energy generation has such a high power temperature dependence we know that in order for this to be stable
than the density must decrease to compensate (assume constant pressure). But we can see that the density
dependence is a function of the radius and cannot decrease by 20 orders of magnitude to compensate, and
thus this becomes a runaway reaction, i.e like a bomb.
L ∝ Mρ2 T −44/T8
ρ∝
This unstable fusion occurs primarily when stars are on the asymptotic giant branch (fusion of He in
a thin shell outside a C/O core) and may be part of the reason that such stars lose so much mass on their
way to becoming white dwarfs.
Problem set 11
Problem # 1
Consider a 0.5 Msun WD. Approximate it as an n = 3/2 polytrope, reasonably appropriate since we
are below the Chandrasekhar mass. Estimate the ratio of the energy transported by photons (radiative
diffusion) to the energy transported by degenerate electrons (thermal conduction) at the center of the WD.
Scale the central temperature of the WD to 108 K, an appropriate number for a newly formed WD. Assume
that the opacity is due to electron scattering. Show that the energy transported by electron conduction
dominates that transported by photons.
We know that the radiative flux for photons is given by
4 acT 3
∇T = −κr ∇T
Fr = −
3 κ0 ρ
where κo is the opacity and κr is the conductivity (any process that transports energy). We also know that
flux due to thermal conduction of degenerate electrons is given by
Fdeg = −κdeg ∇T
where κdeg is the conductivity due to degenerate electrons, thus the ratio of the energy transported by
photons (radiative diffusion) to the energy transported by degenerate electrons (thermal conduction) is
given by
Fr
κr
=
Fdeg κdeg
we have defined the degenerate electron conductivity to be
κdeg ≃ κcls
EF
kT
3/2
=
k 2 h3 T ni
k 2 h3 T ρ c
=
32e4 m2e
32e2 m2e µi m p
52
ni =
ρc
µi m p
and the radiative diffusion conductivity to be
4 acT 3 4 acT 3 4 acT 3 µe m p
=
=
3 κo ρc
3 neσT
3 ρc σT
κr =
We also know that ρc (n = 3/2) polytrope equation is given by
ρc =
M
4M
an ≈ 1.43
3
3πR
R
an = 5.99
where the radius is given by
R ≈ 0.013Rsun
M
Msun
−1/3 µe −5/3
2
m
me
−1
≈ 0.016Rsun
if we use M = 0.5Msun and µe = 2. The ratio can be written as
128acT 2 e4 m2e m2p µe µi
κr
=
κdeg
3k2 h3 σT ρ2c
we also know
1
1
1
= +
µ µi µe
n
1
Xi
=∑
µi i=0 A
(4)
n
1
Xi Z
=∑
µe i=0 A
where we find
96
µe = 2
7
assuming 50% C and 50 % O. If we use a temperature of T ≈ 108 K on Equation 1 we find
µi =
κr
≈ 0.084 ⇒ κdeg ≈ 11.9κr
κdeg
Thus we can see that the energy transported by electron conduction dominates this process.
Problem # 2
Assume that stars are formed with the Salpeter initial mass function (dN/dM ∝ M −2.35 ) between 0.5
and 150 Msun , that stars with Mi < 8Msun become 0.5Msun WDs, that stars with 30Msun > Mi > 8Msun
become 1.4 Msun NSs, and that stars with Mi > 30Msun become 7Msun BHs (the typical WD, NS, and BH
masses chosen here are well-motivated observationally). Assume further that all NSs and BHs are formed
via SN explosions.
a) What fraction of stars undergo SN explosions at the end of their lives?
We know that the Salpeter initial mass function is given as
dN =
Z
M −2.35 dM
Thus the fraction of the stars that undergo SN explosions would be given by the sum of the fractions
of the stars that become NS and BH, this is given by
R 150
SN explosions = R8150
0.5
M −2.35 dM
M −2.35 dM
53
= 2.3%
b) What fraction of stars will become WDs? NSs? BHs?
The fraction of stars that become WD is given by
R8
M −2.35 dM
WD stars = R 0.5
= 97.7%
150 −2.35
dM
0.5 M
The fraction of NS is given by
R 30
−2.35 dM
8 M
NS stars = R 150
= 1.97%
−2.35 dM
M
0.5
and the fraction of BH is given by
R 150
BH = R30
150
0.5
M −2.35 dM
M −2.35 dM
= 0.35%
c) Estimate the fraction of the mass of a stellar population that is returned to the interstellar medium
(via stellar winds or explosions) after 10 Gyrs. You do not need to do a rigorous, accurate to many
significant digits, calculation.
We know that the total mass of a specific type of star can be calculated using
M = dNM dN = AM −2.35 dM
where A is a normalization constant
MT = A
Z b
a
M −1.35 dM
we also know that the total mass that is redistributed to the ISM can only come from the fraction of stars
that have M ≥ Msun and for WD this means that the total mass that is redistributed to the ISM (mass loss)
is given by
Mloss = A
this becomes
1
Mloss = A
0.35
Z b
1
a0.35
a
M −2.35 (M − Mend )dM
−
1
b0.35
Mend
−
1.35
1
a1.35
−
1
b1.35
(5)
we chose integration limits motivated by the knowledge that the sun M = Msun has an approximate lifetime
of 10 Gyr, thus stars wiuth this mass range are the only ones contributing to this enrichment of the ISM.
The mass fraction is given by
Mloss
M f rac =
Mtot
where
Z b
A
1
1
−1.35
Mtot = A M
dM =
−
(6)
.35 a.35 b.35
a
Using
a = 1Msun b = 8Msun Mend = 0.5Msun
on Equation 2 and
a = 0.5Msun b = 8Msun
54
on Equation 3 gives
M f rac (W D) ≈ 48%
This tells us that approximately 48 % of the mass from these stars is given back to the ISM, since we know
that this population comprises 97 % then the total contribution to the ISM will be
f rac(W D) = 0.48 ∗ 97.7 ≈ 47%
For NS we can say
a = 8 b = 30 Mend = 1.4Msun
Using this in equations 2 and 3 we find
M f rac (NS) = 65%
this tells us that approximately 65 % of this mass is given to the ISM and since we know that this population
comprises 1.97 % then the total mass fraction contribution to the ISM will be
f rac(NS) = .65 ∗ .0197 ≈ 1.2%
and finally For BH we can sa
a = 30 b = 150 Mend = 7Msun
Using this in equations 2 and 3 we find
M f rac (BH) = 87.5%
this tells us that approximately 87.5 % of this mass is given to the ISM and since we know that this
population comprises 0.3 % then the total mass fraction contribution to the ISM will be
f rac(NS) = .875 ∗ .003 ≈ .2%
Thus we can conclude that the total fraction of the mass that is redistributed to the ISM is given by
f racT ≈ 48.4%
Problem # 3
Consider a white dwarf with a mass of M = 0.5Msun and an effective temperature of 104 K.
a) Estimate the radius, luminosity, central temperature, and age of the WD. You are free to use any of
the results on WD cooling quoted in lecture.
We know that the radius of a WD can be estimated using
M −1/3 µe −5/3 m −1
≈ 0.016Rsun ≈ 1.11 × 109 cm
R ≈ 0.013Rsun
Msun
2
me
The luminosity can be calculated using
L = 4πσR2 Te4f f = 5.83 × 1030erg s−1 = 2.3 × 10−3Lsun
The central temperature is given by
L Msun
Tc = 10 K
5Lsun M
8
2/7
55
= 1.35 × 107 K
and finally the age can be calculated using
L Msun
t = 10 yrs
5Lsun M
6
−5/7
= 1.49 × 108yrs
b) Estimate the thickness of the photosphere of the WD. What is the number density in the photosphere? Assume for simplicity that the opacity in the photosphere is approximately equal to the electron
scattering opacity.
We know that the thickness of the photosphere is given by the scale height
R2 kT
R2 kT
=
≈ 1.5 × 104cm ≈ 1 × 10−5 RW D
h=
m̄GM µH m p GM
we used µH ≈ 1 since we assume that the atmosphere of the WD is comprised primarily of hydrogen. We
know that the mean free path is equal to the scale height in the photosphere, thus
l=
thus
n=
1
=h
nσ
1
≈ 9.75 × 1019 cm−3
hσT
c) Use the Saha equation for the ionization of hydrogen to estimate the temperature at which hydrogen
is 1/2 ionized at the surface of a WD. Is this larger or smaller than the temperature at which hydrogen is
1/2 ionized for photospheric densities appropriate to MS stars?
We know that the Saha equation can be written as
n p ne g p ge
=
nQ,ee−χ/kTe f f
nH
gH
we know that in a gas of 1/2 ionized hydrogen ne = n p = nH and also g p = 1 ge = 2 gH = 2 and χ =
13.6 eV. The quantum density is given as
nQ,e =
3/2
2πme kT
h2
thus we can write the Saha equation as
nH =
2πme k
h2
3/2
3/2
Te f f e−13.6/kTe f f
we know that for every hydrogen atom there are both a proton and electron, thus the hydrogen density
accounts for 1/3 the total density
n
nH = =
3
2πme k
h2
3/2
3/2
Te f f e−13.6/kTe f f
Plugging in all the constants yield
5 /T
ef f
3/2
7.43 × 10−5Te f f e−1.57×10
56
−1 = 0
solving this numerically yieds a temperature of
Te f f ≈ 2.6 × 104K
which is a higher than the 1/2 ionization temperature for the solar photosphere T ∼ 1.3 × 104K. This is
due to the differences in the densities.
Problem # 4
Consider the late stages of evolution of a 25Msun star. Focus on the core which has a mass ∼ 1Msun
and a radius ∼ 108 cm. The star’s photon luminosity is 3 × 105 Lsun . You might find some of the numbers
in Table 4.2 of Phillips useful.
a) Estimate the temperature at which cooling by neutrinos (which are optically thin and leave the core
directly) exceeds cooling by photons (which random walk out of the star). Use the expression for the
neutrino luminosity from class. At what stage of nuclear fusion (H, He, C, O, Ne, Si, ....) does neutrino
cooling become dominant?
We know that the luminosity given by the neutrinos is given by
Lν ≈ 1012 T93
Rc
−2
10 Rsun
3
e−11.9/T9 Lsun
and the luminosity of the photons is given by
L ph = 3 × 105 Lsun
we can find the temperatures at which this are approximately equal by setting these expressions equal to
each other and solving it numerically, i.e
Lν = L ph
1012 T93
Rc
−2
10 Rsun
3
e−11.9/T9 Lsun = 3 × 105 Lsun
this simplifies to
3.3 × 106 T93 e−11.9/T9 − 1 = 0
solving this numerically yields
T9 ≈ .794 ⇒ T ≈ 7.93 × 108K
Neutrino cooling becomes important after Helium fusion and before Carbon fusion due to the temperatures
given in table 4.2 of Phillips for these reactions.
b) If neutrino cooling were unimportant (and thus the photon luminosity determined the energy lost by
the star), estimate the time it would take the ≃ 1Msun core of the star to fuse from 20 Ne to 56 Fe. Compare
this to the true time of about 1.5 years (from Phillip’s Table) set by neutrino cooling. Assume that the
luminosity of the star is independent of time and that fusion of heavy elements releases ≃ 0.7 MeV per
nucleon (≃ 10 times less the fusion of H to He because the binding energies of heavy nuclei are closer to
each other.
We know that the time is given by
t=
NQ
E
=
L ph L ph
57
we know that the number of particles are given by
N=
Msun
mp
and the energy released is
Q ≈ 0.7 MeV/nucleon
thus we find that the time is
t=
Msun Q
≈ 3.6 × 104 yrs
m p L ph
We can see that neutrino cooling is very effective in “killing” a star. Due to the fact that the star would
exist for much longer if it were not for those pesky neutrinos.
Problem set 12
Problem # 1
The Energy needed to dissociate one 4 He nucleus into two neutrons and two protons is Q = 28.3 MeV.
Derive an expression relating the numbers of 4 He nuclei, neutrons and protons coexisting at a temperature
T in an equilibrium set up by the reactions
γ +4 He ⇋ 2n + 2p
Calculate the temperature for 50 % dissociation when the density is 1012 kg m−3 . [Note :This is a simple
example of nuclear statistical equilibrium (NSE) discussed in class, i.e., the balance of nuclei determined
when nuclear reactions go both ways at high temperatures (because photons have enough energy to photodisintegrate nuclei into their more basic constituents).] In addition to calculating the temperature for 50
% dissociation of He, also show explicitly that at high temperatures, NSE favors the nuclei being broken
apart (n and p in this case) while at low temperatures it favors nuclei being bound (He in this case).
We can use the Saha equation, which is given by
µ(γ) + µ(4He) = µ(2n) + µ(2p)
which can also be written as
g p nQ,p 2
gn nQ,n 2
gHe nQ,He
2
2
2
= 2mn c − kT ln
+ 2m p c − kT ln
mHe c − kT ln
nHe
nn
np
58
rearranging this equation yields
c2 (mHe − 2mn − 2m p ) = kT ln
gHe nQ,He
nHe
n2n n2p
(gn nQ,n )2 (g p nQ,p )2
!!
we are given the binding energy needed to dissociate a helium atom
c2 (mHe − 2mn − 2m p ) = −28.3 MeV
thus
n2n n2p
=
nHe
gHe
(gn g p )2
nQ,He
(n2Q,n n2Q,p
!
e−28.4MeV/kT
we are also given
gHe = 1 gn = g p = 2
and the quantum concentration is defined as
nQ,A =
2πmA kT
h2
3/2
and if we assume for simplicity
nQ,n = nQ,p =
it follows that
2πm p kT
h2
3/2
nQ,He =
8πm pkT
h2
3/2
n4Q,p −28.4MeV/kT
n2n n2p
=
e
nHe
nQ,He
inserting the expressions for the quantum density gives
n2n n2p
2πm p kT 9/2 −28.4MeV/kT
e
= 16(8)
nHe
h2
(7)
if we assume that this gas is 50% dissociated gives us
1
2
nn = n p = n nHe = n
5
5
for every helium nuclei there are two neutrons and two protons, giving a total of five particles. We also
know that
mHe + 2mn + 2m p 8
ρ
5 ρ
n=
m̄ =
≈ mp n =
m̄
5
5
8 mp
Using this along with plugging in all the constants into Equation 1 gives us
9/2
7.84 × 1021 T12 e−0.328/T12 = 1 T12 =
solving this equation numerically yields a temperature of
T12 = 0.0109 T ≈ 1.09 × 1010 K
59
T
1012 K
To show explicitly that at high temperatures, NSE favors the nuclei being broken apart (n and p in this
case) while at low temperatures it favors nuclei being bound (He in this case) we must consider Equation
1 along with assuming
n = nHe + nn + n p ≈ nHe + 2n p
we can see that
nHe ≈ n − 2n p
putting this into Equation 1 gives
n2n n2p
2πm pkT 9/2 −28.4MeV/kT
e
= 16(8)
n − 2n p
h2
(8)
If we consider the case where T → ∞ (very high temperatures) we can see that Equation 2 goes to infinity,
this only happens if the denominator is 0
1
n − 2n p = nHe = 0 n = n p
2
thus there are no bound nucleus only protons and neutrons in equal numbers. If we now consider the case
where T → 0 we can see that Equation 2 goes to 0 this can only happen in
n p = nn = 0 n = nHe
and thus it favors nuclei being bound.
Problem # 2
Compare the total energy released by a 25 Msun star during (a) its pre-main sequence evolution (KH
contraction), (b) its time on the MS, (c) its post-main-sequence-evolution, and (d) the supernova explosion
to form a neutron star.
To calculate the total energy released during the pre-main sequence we can assume can just calculate
the total gravitational energy relesed from contraction in the star. We know that the energy for a bound
system is given by the Virial theorem as
U
E ∼−
2
is given by
GM 2
U
∆E ≈ |Ei | − |EF | ≈ |EF | ≈ − ∼
2
2RMS
where RMS is given by the main sequence radius which is defined as
RMS ≈ Rsun
M
Msun
6/7
since we know that the mass of this star is
M = 25Msun
we find the radius to be
R = 15.78Rsun
60
and the gravitational energy, which is the total energy is
G(25Msun )2
≈ 7.45 × 1049 ergs
2 · 15.78Rsun
∆E pre−MS =
To calculate the total energy of a 25Msun star during the main sequence we need to multiply total
luminosity by the total time that the star spends on the main sequence
∆E = LMStMS
where the luminosity is given by
L = Lsun
M
Msun
3.5
= 7.8 × 104Lsun
and the time a star spends o the main sequence is given as
tMS = 10
10
M
Msun
−2.5
yrs = 3.2 × 106 yrs
this gives the total energy as
∆EMS ≈ 3.08 × 1052 ergs
we can compare this to the value derived by nuclear energetics, we can say that the total energy is given
by
M
∆E ≈ 0.1NQHe = 0.1 QHe
mp
where we know
M = 25Msun QHe ≈ 7 MeV/nucleon
Where QHe is the nuclear binding energy per nucleon of helium. We find the energy to be given as
∆E ≈ 3.33 × 1052 ergs
which is almost them.
To find the total energy during the post main sequence stage we need to consider the energy to be given
as
∆E = 0.5N∆Q
where ∆Q is the difference in the binding energy of iron to the binding energy of helium and we assumed
that 50% of the star will undergo fusion. The binding energy per nucleon of iron and helium are QFe ≈
8.78 MeV/nucleon and QHe ≈ 7 MeV/nucleon, thus
N=
25Msun
mp
∆Q ≈ 1.78 MeV/nucleon
and we find the total energy to be given as
∆E post−MS ≈ 4.24 × 1052ergs
61
To find the total energy released during the supernova explosion can be estimated using the same
equation as part a) except now the final radius is given by the radius of te neutron star.
∆E ≈
GM 2
2RNS
where the mass that we will consider will be the mass of the core
M ≈ 1.4Msun
R ≈ 10 km
thus we find the energy to be
∆ESN ≈ 2.59 × 1053ergs
this energy release is much greater than all other processes.
Problem # 3
Consider an ideal degenerate gas of electrons, protons and neutrons, and the equilibrium established
by the reactions
n → p + e− + ν¯e and e− + p → n + νe
Assume equal numbers of electrons and protons and assume that the density is so high that all the degenerate particles are ultra-relativistic. Show that the number densities of the particles are in the ratio
ne : n p : nn = 1 : 1 : 8
Using the Saha equation, along with the knowledge that all of the particles are now reletavistic we find
µ(n) + µ(νe) = µ(p) + µ(e−)
and we know that the Fermi momentum is given by
pF =
3n
8π
1/3
h
and the Fermi energy for relativistic particles is
ε F = pF c
thus we find
3ne
8π
1/3
3n p
hc +
8π
1/3
3nn
hc −
8π
1/3
hc = mn c2 − m p c2 ≈ 0
where we made the assumption that mn c2 − m p c2 ≈ 0. We are also told that
ne = n p
thus we find
1/3
2n p = nn
1
ne = n p = nn
8
Problem # 4
62
Assume that a hot, bloated neutron star emits thermal neutrino radiation from a surface of radius R at
an effective temperature equal to TE . Assume that three types of massless, or nearly massless, neutrinos,
νe , νν , ντ and their antiparticles, are emmited in equal numbers, in thermal equilibrium with zero chemical
potential. Show that the luminosity is given by
Lν =
21 4
σTE 4πR2
8
where σ is Stefan’s constant. Find an expression for the average energy for a neutrino in this radiation.
[Hint: Look back at Chapter 2 and reconsider Problem 2.5]
If we refer to Philipps problem 2.5 we find that the energy density of fermions is given by
7
uF = aT 4
8
(9)
and we know that the energy density of a photon is
u p = aT 4
(10)
the differences in these two expressions comes from solving the following two integrals
3 Z ∞
Z
kT
x2
1 ∞
N(p)d
p
=
8π
dx
n=
−x
V 0
hc
0 e ±1
this is the number density of particles with momentum p and p + d p, the ± is to differentiate between
bosons and fermions. Since neutrinos are fermions. And the energy density is given as
3 Z ∞ 3
Z
1 ∞
kT
x
u=
ε p N(p)d p = 8π
kT
dx
x
V 0
hc
0 e ±1
from Equationquations 3 and 4 we can see that the solution to the energy density for a fermion is given by
7
uF = aT 4
8
this was using the assumptions that the polarization of the fermion is 2, but we know that the polarization
of neutrinos is 1. We also need to take into account the 6 different species of neutrinos, thus for neutrinos
we find that the energy density is given as
uν =
6 7 4 21 4
aT = aT
28
8
we also know that Stefan’s constant is defined as
σ=
ac
4
thus the energy density is now given as
21 4 4
σT
8 c
but we know that the flux due to the neutrinos is given by
uν =
Fν =
21 4
σT
8
63
where the factor of 4/c was taking care of the fact the intensity radiated at a particular frequency is c/4
times the photon energy density at this frequency. We know that the luminosity of neutrinos is given by
Lν = Fν 4πR2
and we just derived the flux for neutrinos, thus the luminosity is
Lν =
21 4
σT 4πR2
8 E
Problem # 5
In this problem we will calculate the properties of the neutrinos emitted by a newly formed neutron
star (a “proto-NS”). The neutron star is formed during a supernova explosion and its gravitational binding
energy ENS is released in the form of neutrinos on a timescale tKH , so that the neutrino luminosity of
the NS is L = ENS /tKH . Assume that the NS has a mass of 1.4Msun and approximate it as an n = 3/2
polytrope supported exclusively by neutron degeneracy pressure. The initial central temperature of the NS
is ≃ 1011 K.
a) Calculate the radius and central density ρc of the NS.
We know that the relationship between the radius and the mass of a NS is given by
M
RNS = 15 km
Msun
−1/3
and since we know that M = 1.4Msun we find the radius to be approximately
RNS = 13.41 km
the mass density can be found by using the central density of a n = 3/2 polytrop which is given as
ρc =
3M
an
4πR3
an = 5.99 n = 3/2 polytrope
thus the central density is
ρc ≈ 1.43
M
R3NS
given the mass and radius we find
ρc ≈ 1.65 × 1015 g/cm−3
b) Show that the neutrinos are initially degenerate in the core of the NS. Use this fact to estimate the
typical energy Eν of a neutrino in the core of the NS. The neutrinos are relativistic.
we know that
Eν ≈ EF (e)
where
EF (e) =
we know that
3ne
8π
1/3
1
1 ρc
ne ≈ nn ≈
8
8 mp
64
hc
thus we find the Fermi energy to be
Eν ≈ EF (e) ≈
3nn
64π
1/3
hc ≈ 303 MeV
to show that they are degenerate we just need to show
Eν ≥ ET
where ET is the thermal energy, we find that for relativistic particles is
ET ≈ 3kT ≈ 25.8 MeV
thus we can see that
Eν ≥ ET
c) The cross section for neutrino’s interacting with matter is
σν ≃ 10
−44
Eν
me c2
2
cm2
Estimate the optical depth τ = nσν R ≃ R/ℓν of the NS to neutrinos, where ℓν is the neutrino mean free
path in the core of the NS.
Since we know that the optical depth is given by
τ = nσν RNS
where we have defined n to be the total number density
n≈
ρc
≈ 9.86 × 1038cm−3 RNS ≈ 13.41 km me c2 ≈ 0.508 MeV
mp
we find
τ ≈ 4.71 × 106
and we know that the neutrino mean free path is given by
ℓν =
RNS
≈ 0.284 cm
τ
d) The timescale tKH for the NS to radiate away its binding energy in neutrinos is the time for the
neutrinos to random walk out of the NS. Use your result from c) to estimate the time tKH and the neutrino
luminosity Lν of the NS.
we know that the time it takes for a neutrino to random walk out of the NS is given by
R2NS R2NS
=
t=
vℓν
cℓν
given the radius of the NS and the mean free path we find the time to be
t ≈ 210 s
65
we know that the luminosity is given as
Eν
t
using the neutrino energy in part b) and the time we find that the neutrino luminosity is
Lν =
Lν ≈ 9.19 × 1050 ergs/s
e) Look at Problem 4 Note Phillips’ hints at the back of the book for problems 6.3 and 2.5.
f ) Use your results from d) and problem 4 to calculate the effective temperature of the neutrino radiation (in K) and the average energy of a neutrino emitted by the newly formed NS (in MeV). For comparison
to the results you have calculated in this problem, the observed timescale of neutrino emission was ≃ 10 s
for SN 1987A and the typical neutrino energy was ≃20 MeV.
The effective temperature is given as
TE =
8Lν
21 · 4πR2σ
1/4
≈ 1.48 × 1010 K
the energy of a neutrino can be calculated by knowing the energy density and the number density. The
energy is given by
uν
Eν ≈
nν
where the number density is given by Philips equation 2.42 with a modification coming fro the fact that
we are dealing with fermions
6
8πk3
n = bT 3 where b = 1.803 × 3 3 = 45.48K−3 cm−3
2
h c
thus we know that the energy of a neutrino is given by
Eν =
21 a
T ≈ 6.24 MeV
8 b
66