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 hydroge n 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
caT 3
Frad
∼
∇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 ∝ dR RR
− Tc TC − RC ∝ − R
κ ∝ 10 23 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
2
L = 4πr Frad
⇒ Frad = 4πr
2
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
pµ ≈ GMm 3Rk
1
where k is now the boltzman constant. Substituting this expression into the above equation yields
L ∝ caR−1/2 M 11/2
15/2
Gm p k
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 equilibrium.
dP dr
M and the radius R of a star by using hydrostatic
− GM ρ r
=
2
M Pc ∝ ρ R M ρT ∝ ρ R M ∝ R
since T is constant, substituting this into the luminosity relationship yields 5
L∝M (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 M∝R κ∝ρ ρ∝ R3 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 cm 3 and Tc 1.5 107 K. For the conditions at the cente r 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 electro n-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 ioniz ed hydrogen gas that
ne
∼ n p ∼ mρp
and the interaction cross section is given by 2
σ = πr where r is the Coulomb radius found comparing the thermal energy to the Coulomb energy
e2 r
2
∼ kT
r
4
∼ kTe
σ
πe ∼ (kT )
2
using these relationships we find the mean free path to be
l
∼
mp
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 2 2
v=
3kT
me
thus the time is given as
te =
me m p 3kT ρc π
2
kT e2
(b). What is the mean free path of an proton due to proton-proton Co ulomb 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 ioniz ed. The mean free path is given by
l
∼
mp
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 =
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 tp
≪
(c). Which opacity is more important for photons, Thomson scatteri ng or free-free absorption? We know that
ne σT
κT =
=
ρc
2σT mp
∼ 0.80
and
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 fre e path of a photon? How does thi s compare to the mean fre e path of an electron (this should give you a feel for why photons are far more effective at moving energy around in stars)? What is the typical time between photon absorptions/scattering? we know that the mean free path of a photon is given by
l=
1
ne σT
where σT =
3
which yields
l photon = lelectron =
mp
2ρσT
mp
ρc π
2
e2 4πε0me c2
8π
= 6.65
∼ 8.3 × 10−
25
cm2
cm
2
∼ kT e2
3
× 10−
× 10−
7
8.9
cm
The typical time for a photon collision is given by
t=
lp c
∼ 2.8 × 10−
13
s
(e). For a photon undergoing a random walk because absorpt ion/scattering, how long would it take to move a distance R sun given the results in (d)? For comparison, it would take 2.3 seconds movi ng at the speed of light to travel a distance R sun in the absence of scattering/absorption. We know that the diffusion time can be acquired with
tdif f =
thermal energy
2
2
R nk ∼ Rlc nkT ∼ lcaT ∼ aT 4
L
3
R2 2ρk m p lcaT 3
we know that the average time for a photon to leave the star is given by 2
tdif f
∼ Rl phsunc ∼ 10 4
4
yr
Problem # 3 How old is the sun? In this problem we illustrate how the naturall y occuring radioact ive 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 isotop es of lead: the decay chain of U 235 ends up with Pb207 , and the decay chain of U 238 ends up with Pb 206 . 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 λ 5 and λ8 for the two isotopes corresponds to half-lives of
T5 =
ln 2
λ5
= 0.7
× 10
9
yrs T8 =
ln 2
λ8
= 4 .5
× 10
9
yrs
The magnitudes of these half-lives are ideally suitable to the determination of the ages of the rocks which are over a billion year s old. Now consider a set of rock samples whic h were formed at the same time, but with different chemical compositions. They differ in chemical composition because different chemical elements are affected differ ently by the processes of rock formation. However rock formation processes do not favour one isotope over another. For example, on formation, the relative abundances of U 235 and U 238 should be the same in every sample. But these abundances will change with time as the deacy of U 235and U 238 produce nuclei of Pb 207 and Pb 206 . 207 nuclei relative to the increase of Pb 206 nuclei. Show that this ratio is the same for all rock samples which were formed at the same time,
• Consider the ratio of the increase in the number of Pb and that it is given by
N7 (t ) N6 (t )
− N (0) = N (t ) eλ t − 1 − N (0) N (t ) eλ t − 1 7
5
6
8
5 8
We know that the ratio of the two isotopes can be written as
N7 (t ) N6 (t )
− N (0) = N (t ) − N (0) − N (0) N (t ) − N (0) 7
5
5
6
8
8
and given the first expression give n 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 ) N6 (t )
N5 (t ) eλ5t
N7(0)
1
− N (0) = N (t ) eλ t − 1 8
6
8
which is what we were asked to show. 207 and Pb 206 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.
• Consider a graph in which the measured abundances in the rock samples of Pb
5
We know that
N7 (t ) = where
N5 (t ) eλ5t N8 (t ) eλ8t
N5 (t ) eλ5t N8 (t ) eλ8t
− 1 · N (t ) −1 6
− 1 = constant −1
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.
• Given that the current ratio of naturraly occurring
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 = 0.0715 eλ8t 1
·
For 3 billion years we get
− −
0.0071
λ t · eeλ t −− 11 = .231
0.0071
λ t · eeλ t −− 11 = .891
5
and finally for 5 billion years we get
8
5 8
Problem # 4 Radiative Atmospheres In this problem we will solve for the structure of the outer part of a star assuming that energy is transported solely by radiative diffusion (which is not the case in the sun, but is the case in stars more massive than the sun). The star has a mass M and a luminosity L . Assume that the luminosity and mass are approximately constant at the large radii of interest, that gas pressure dominates, and that the opacity is due to electron scattering. Do not assume that the atmosphere is thin (i.e even though M r constant = M , because r changes, the gravitional acceleartion is not constant).
≈
Write down the equations for hydrostatic equlibrium and energy transport by radiative diffusion. Use these to calculate d Prad /dP , the change in radiatio pressure with pressure in the atmosphere. What 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 resul t 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
Prad =
1 3
aT 4
1 d 4 dPrad = a (T 4 ) = T 4 ∇T dr 3 dr 3
6
r
and we know that the radiation flux is given by F= thus we can write
− 43 caT κρ
∇T
dPrad κρ =F dr c
dPrad c dr κρ
F=
3
the hydrostatic equilibrium equation is
dP
GM
dr = deviding these two expressions yield
−r
2
ρ
dPrad F κr 2 = dP cGM we know that the Flux and luminosity are related by L = 4πr 2 F
L 4π r 2
F=
thus we find
dPrad Lκ = 4πcGM dP this result implies that the ratio of the gas pressure to radiation pressure is independen t of the distance in the atmosphere. To show ρ ∝ T 3 we can just use scaling argument s Prad L ∝ Pg M
⇒
T4 L ∝ 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 time s the gas pressure
Prad L ∝ Pg M
Pt = Pg + Prad
thus we find
Pr Pt
but we know that
∝1
P ∝ λP
− Pr
Pg ∝ ρ T
⇒ Pg = Pt − Prad
⇒
r
T ∝ ρ1/3
Pg ∝ ρ 4/3
thus we know that
λPg ∝ Pt
− λPg
Pt ∝ 2λPg ∝ ρ4/3 7
g
Problem set 3
Problem # 1 (a). Show that heat tran sfer by radiative diffusion implies a non-zero gradient for the radiation pressu re which is proporti onal 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 4πcGM
Lmax =
κ
where κ is the opacity near the surface. Obtain a numerical estimate for this luminosity by assuming that the surfa ce is hot enough for the opaci ty to be dominated by electron scattering. (This maximu m luminosity is called the Eddington luminosity. To show that the heat transfer by radiative defusion implies a non-zero gradient we must begin with 3
Fr =
− 43 aTκρ ∇T
Fr ∝ Prad
knowing these relationships we can do
dPr = 4 aT 3 dT dr 3 dr
Prad = 1 aT 4 3
⇒ dT =3 1 dr 4 aT
3
dPr dr
thus this implies that there is a non-zero gradient. To show that
Lmax =
4πcGM
κ
we must begin with the equation derived from problem 4 in the last problem set, i.e
dPr Lκ = dP 4πcGM but since we know that
P = Pg + Pr
Pg ≪ Pr
⇒ P ≈ Pr
dPr =1 dPr
and we find that
L=
4 πcGM
κ
4
≈ 3.3 × 10 Lsun
M Msun
We cannot obtain a numerical estimate because we do not know tha mass. We could use M sun but this would not be correct.
8
(b). Assume that radiat ive diffusion dominates energy transport in stars and that the opacity is due to Thomson scattering. Use a scaling argument to estimate the mass M (in M sun ) at which the luminosity of a star is Ledd .
≈
We can do an order of magnitude estimate with respect to the sun by
L = Lsun
L ∝ M3
3
M Msun
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 assu ming the average gas particle mass m ¯ 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
κ
× 10 kg m− 4
3
0.040 m 2 kg−1
Using equation ?? from Phillips
L( r ) m(r )
and the following relationships
Pr =
1 3
aT 3
=
γ
P
− 1 16πGc Pr γ
crit
∼ Pg = ρm(¯r) kbT
κ
γ=
5 3
P
κT = 0.04 m 2 /kg
we find
L(r) m(r)
0.175
2 16πGc aT 3 m ¯ 5 κT 3ρ(r)kb W > .07 kg
=
crit
W kg
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 photon s are rarely absor bed and so travel on strai ght 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 srcinates.
≫
≪
≈
a) The temperature at the photosphere of the sun is 5800 K. Estimate the mass density sphere. Assume that Thompson scattering dominates the opacity.
9
ρ in the photo-
Knowing that ℓ
∼ h we can derive the following relationship nσ = κρ ρ =
nσT
ℓ=
κT
1 kb T =h= nσ mg ¯
≈ mp
thus we find that the dens ity is given by and assuming m ¯
ρ=
¯ mg
≈ 8.26 × 10−
4
kb κT T
g/cm3
b) In reality, the surface of the sun is so low that hydrogen is primari ly 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 (corre ct) opacity, repeat the estimate from a) of the density at the photosphere of the sun.
≈ ×
Substituting the opacity given into the above expression yields
ρ3/2 =
mg ¯ 2. 5
× 10−
⇒ ρ=
13 kT 10 b
2. 5
×
mg ¯ 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 convec tion). 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
1/3
ρ
and knowing that
Fc =
L 4π r 2
we find that the convective velocity is given by
vc =
2L 4π r 2 ρ
1/3
≈ 1.09 × 10
6
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 =
ℓ vc
≈ vhc
we know that h which is the scale height of the sun is given by
h=
kT = 1 .7 ¯ gm 10
× 10
7
cm
thus we find that the blob timescale is
≈ 15.6 s
tblob
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 ∝ ρ 4/3 . 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 hydroge n. The solar convect ion zone contains very little mass (only 2of the mass of the sun). Thus, let’s consider a model in which we negle ct the mass of the convection zone in comparison to the rest of the sun. For the reasons discu ssed in class, we can model the convection zone as having P = K ργ with γ = 5/3 and K a constant. Rc is the radius of the base of the convection zone.
≈
a) Solve for the density, temperature, and pressure as a function of radius in the convection zone. Do not assume that the convection zone is thin (i.e., even though 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
dP
= ρ
dr
P = K ργ
2
dP
K
⇒
−
3/5
P K
1/γ
P
ρ=
r
−
thus we find
GM
=
− GM dr r
=
−
2
and integrating over the following limits we find
Z
P Prc
P K
dP
−3/5
Z
R Rc
GM dr r2
this integral yields 5 2K 3/5 thus we find that the pressure is given by
P=
2 5
P
2/5
−
2/5 Pc
K −3/5 GM
1
= GM
1 − R Rc
1
R
1
− Rc
5/2
2/5
+ Pc
To solve for the density we can just plug this solution into
ρ=
P K
3/5
=
1
K 3/5
2 −3/5 1 K GM 5 R
11
− R1c
2/5
+ Pc
3/2
and finally the temperature can be found by using 2ρ kb T mp
P = nkb T =
m pP
T (r) =
⇒
2 ρ kb
and substituting the P and ρ from the previous expressions we find
T (r ) =
m p K 3/5 2 −3/5 1 K GM 2kb 5 R
− R1c
2/5
+ Pc
b) In detailed solar models, the pressure at the base of the convection zone is 5 2 1013dyne/cm2 . the base of the and the density is ρ 0.175 g cm −3 . Using your solution from a), estimate the radius of convection zone R c . Compare this to the correct answer of R c 0.71Rsun
≈
≈ ×
≈
If we solve the density equati on for Rc we find 1 1 = Rc R
−
and plugging in values we find that 1 = 1.998 Rc
(ρK 3/2 )2/3
3/5
− Pc / · 52KGM 2 5
× 10− m− ⇒ Rc ≈ 5.11 × 10 m = 0.72Rsun 9
1
8
c) In your model, what is the temperature of the sun at 0.99 Rsun, 0.9 Rsun , and at the base of the solar convection zone. This gives you a good sense of how quickly the temperature rises from its surface value of 5800 K as one enters the interior of the sun.
≈
To find the temperature as a function of radius we would use the temperature equation derived from part (a). i.e
m p K 3/5 T (r = 0.99Rsun) = 2 kb
− −
2 3
T (r = 0.90Rsun) =
m p K 3/5 2 kb
T (r = 0.72Rsun) =
m p K 3/5 2/5 Pc 2 kb
K −3/5 GMr
25
·
66 Rsun
2/5 + Pc
2 −3/5 15 2/5 K GMr + Pc 3 18 Rsun
·
≈ 1.8 × 10
6
≈ ≈
4. 1
× 10
4
K
5. 1
× 10
5
K
K
Problem set 4
Problem # 1 I mentioned in class that there are two ways to estimate the energy carried by convection. The first is that the energy flux is Fc 1/2ρv3c Fc,1 where v c 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 equiv alent, to order of magnitude (which is the accuracy of mixing length theor y). 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 accelara tion of the blob due to bouyancy is given by
ab = g since ρ b
≈ ρ∗.
ρb ρ
∗− 1
∆ρ = g 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 = mvc2 = 2
Z
l
F d l = aml 0
·
thus we find that the convective velocity is given as
vc =
2g
∆ρ l ρ
which can also be expressed as
v2c ∆ρ = 2gl ρ we can also write this as given that l
∼ H , thus
vc2 = 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 = vc2
m ρ 2kb ∆ρ
and from the equation of state, which is given as
∆E =
1 kb ∆T (∆ρ V ) φ m¯
·
where ( ∆ρ V ) is the mass. Using these two expression and what we found from part (b) we can see that 1 ∆E ∆E ∆E vc2 = vc2 = ∝ 2φ ¯ 2 ρV m m
·
13
≈ Fc,
d) Combine your previous results to show that Fc,1 From (a) we know that
Fc,1
≈ 12 ρvc
3
2
.
≈ ρ ∆mE vc
Fc,2
using this and our solution we find 1 3 ρv 2 c
≈ ρ ∆mE vc ⇒
≈ ∆mE
1 2 v 2 c
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 materia l undergoing convection is at about the mean density of the star and that gas pressure domi nates. 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 = ρα3Cs3 but we know that
3 2 H ds / C p dr
Fc
vc = Cs
≈ 4πLR
which reduces to
C p dr
2
Cs2 =
LR 3M
kT mp
ρ=
1 2
4πR ρ
Cs
∗
H ds = C p dr
where
2/3
L
=
1 2 H ds / C p dr
2
thus
H ds
2/3
mp kb T
3M 4π R 3
and for the convective velocity we find
vc = Cs
H ds Cp dr
1/2
1/3
LR 3M
=
from Carrol and Ostley we find that for 10 Msun and . 1M sun we find that the radius, and luminosity are approximately
M M
≈ ≈
10Msun R 0.1Msun R
≈ 6Rsun L ≈ 5700Lsun ≈ 0.2Rsun L ≈ .0034Lsun
given these values we find
M = 10Msun M = 0.1Msun
H ds C p dr
≈ ≈
H ds C p dr
× 10−
6
vc
≈ 5.3 × 10
× 10−
10
vc
≈ 698 cm/s
3.61 5. 8 14
4
cm/s
Problem # 3 Polytropes (a). The mass M of a star is given by
M=
Z
R
4πr2 ρ(r )dr
0
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 4πR3
ρc = ρ¯ an =
an
where a n 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 a n in general without numerically solving for θ [ζ], so your answer will just be in terms of the solution to the Lane-Emden equati on).
Since we know that
Θ= given these two relations we can find
ρ(r ) = Θn ρc
ρ(r ) ρ
1/n
ξ=
r a
r 2 = a2 ξ2 a = R dr = Rd ξ
and from the Lane-Amden equation we know
d
ξ2
dΘ
= ξ2 Θn
dξ dξ given these following relationships we find that M=
−
Z
1 0
4πR3 ξ2 Θn ρc d ξ =
− 3
−4πR ρc
Z
1 0
d dξ
ξ2
dΘ dξ
dξ
thus we find that a n is given by
an =
− 13
1
d d 0 ξ
Z
ξ2
dΘ 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 = a n is the constant (with units of length) defined in lecture (note that the polytropic relation −γ P = K ργ can be used to write K = Pc ρc . Use this result and (a) to derive an expr ession for the cen tral pressure of a polytropic model of the form
Pc =
Pc =
GM 2 R4
15
cn
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, a n = 5.99 and c n = 0.77 while for n = 3, a n = 54.183 and c n = 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 a , c and d . n
We know that γ K = Pc ρ− c
and also
−
1/n 1
( n + 1) K ρ c a= 4πG
1/2
given these relationships we can now find
a2 4 π G 1/n−1 γ 1/n−1 = K ρc = Pc ρ− c ρc n+1 but since we know that γ = 1/n + 1 we find
−
1/n 1
γ Pc ρ− c ρc
−(1+1/n) ρ1/n−1 = P ρ−2 c c
= Pc ρc
c
and finally we find
Pc =
a2 4πGρc2 n+1
using this results we can now derive
Pc = which becomes
Pc =
a2 4πGρ2c GM 2 = 4 cn n+1 R
a2 4πGan2 n+1
2
3M 4π R 3
=
GM 2 cn R4
looking at these two expressio ns we can see that
cn
9a2 an2 1 = 4πR2 n + 1
now looking at 4/3
Pc = dn GM 2/3 ρc and from (b) we find 2/3 4/3
Pc =
a2 4πGρc ρc n+1
=
a2 4πG n+1
16
2/3
an 3 M 4π R 3
4/3
ρc
n
n
which becomes 4/3
Pc = GM 2/3 ρc cn
4π
4/3
an
9−2/3
and after some fun algebra, which will be omitted here we find
d n = cn
4π 3 an
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 domina tes. to find these expression we will assume that gas pressure dominates, i.e
Pc =
ρc m¯
kb Tc
thus
1/3
Tc =
¯ ¯ Pc m d n GM 2/3 ρc m = ρ c kb kb
and skipping some algebra we find that the central temperature is given by
Tc =
dn GM Rkb
3 an 4π
1/3
mp 2
so all of the expressions can be written as
Tc =
dn GM Rkb
3 an 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 = M sun and R = Rsun (i.e., for the sun). Assume fully ioniz ed hydrogen for simplici ty. Which polytrope better approximate s the true interior temper ature, pressure , and density of the sun? Can you explain physica lly why this is the case? using the above expressio ns for temperature, density and pressure we find for 5 γ = an = 5.99 cn = 0.77 dn = 0 .477 3 we find
ρc
3
≈ 8.44 g/cm
and for
γ=
Pc
≈ 8.67 × 10
15
Tc
4 an = 54.183 cn = 11.05 dn = 0.363 3 17
6
≈ 6.2 × 10 K
we find
ρc
3
≈ 77.26 g/cm
Pc
≈ 1.24 × 10
15
Tc
7
≈ 1.02 × 10 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
5/3
P ∝ ρ
convective
Problem # 4 Consider a pre-main “star” (gas massthe M Hayashi undergoing contraction. In class, we showed that sequence fully convective starscloud) moveof down lineKelvin-Helmholz 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 transpo rt 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 = κf f
thus we know that
Lf f
≈ Lsun
− 11/2
M Msun
R Rsun
1/2
≈ Lrad ≈ − 12 GM R 2
since we are doing proportionalities we find
M2 R ∝ M 11/2 R−1/2 R2 t T so we find that the radius scales as
R∝
1 M 7t 2
plugging this into
L ∝ M 11/2 R−1/2 ∝ M 9t thus
L ∝ M 9t and to find the temperature
L ∝ R2 Te4f f
Te f f ∝
18
L R2
1/4
∝ M 23/4 t 5/4
2
dR dt
thus
Te f f ∝ M 23/4 t 5/4 We can see that as time and mass increas e 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) / (R/Rsun) 4 7
2
if we set the free-free luminosity equal to the convective luminosity we find
Lsun
− M Msun
11/2
1/2
R Rsun
≈ 0.2Lsun(M/Msun) / (R/Rsun) 4 7
2
some algebra yields
R
≈5
5/2
69/35
M Msun
Rsun
(c). Sketch the paths of 1 Msun 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 L sun and Te f f in K). Note that on the main sequence a 3 M sun 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
4/7
2
≈ M Msun
R Rsun
5
2/5
· 0.2Lsun
which yields
L(Msun)
≈ 0.724Lsun
and a plot is given by
19
4/7
M Msun
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 the gravitational free-fall time of the cloud, t f f 1/ G ρ , where ρ is the mean density of the cloud. What happens if t KH < t f f ?
≈
Since we know that
tf f
≈
1
G ρ
tKH
≈
M Msun
20
1/2
Rsun R
3
(2
7
× 10 yrs)
froma a purely physical argument we know that things cannot fall faster than gravity can pull it. Thus
≫ tf f
tKH
≫
and saying t f f tKH would be an unphysical statement due to the knowledge we have about gravit y. 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 t KH tf f .
↓
≫
↓
(b). Estimate the critical radius R c (inRsun) at which t KH t f f for a given cloud of mass M (in M sun ). Assume, as we did in class, that the cloud is fully convective at early times. Show that for R < R c , 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
G ρ
1/2
M Msun
=
Rsun Rc
(2
107 yrs)
re-arranging this expression for R c we find
Rc =
1/2
M Msun
Rsun(G ρ )1/6 (2
107 yrs)1/3
and letting
ρ = 43πMR ≈ RM 3 c
we find
2/3
Rc
1/3
7
≈ M / Rsun (2 × 10 /yrs) 4 9
3 c
G1/9
1 3 Msun
≈ M / 7.6 × 10− 4 9
3
if we put the mass off the sun we get
Rc
≈ 4.79 × 10
10
cm
Since we know that
t f f ∝ R3/2
tKH ∝
1 R3
we can see that
R R
↑ ↓
tKH tKH
↓ ↑
tf f tf f
↑ ↓
from these two relationships we can see that for KH contraction to occur that R < Rc .
t KH
≫ t f f and also that
(c). What is the central temperature of the (fully convective) cloud (in K) as a function of its mass (in M sun ) when R = Rc ? We know that the temperature of a fully convective object is given by
Tc =
dn GM Rkb
3 an 4π
21
1/3
mp 2
M
γ = 5 /3. Knowing this we
and since we know that fully convective stars have a polytropic index of find n = 3 an = 5.99 dn = 0.477 thus 15 M
× 10− 2.79 × 10− 2.16
≈ ≈
Tc
Rc
15
M 5/9
and if we want the temperature of a collapsing gas cloud with respect to
M sun we get
4
≈ 1.9 × 10 K
Tc
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 criter ia 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
Mj
≈ 1.37 × 10
42
g
and the mass of M13 is
M13
6
≈ 0.5 × 10 · Msun ≈ 9.95 × 10
38
g
and we can see that
Mj
≫M
13
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 = and the number of nucleons are given by M Nnucleon = mp
Eb nucleon
× Nnucleon
≈ 1mkgp ≈ 5.98 × 10
26
nucleon
thus
Etot
≈ 5.27 × 10
27
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
LHe =
EHe tHe
LFe =
EFe tFe
and since we know that these two luminosities are theoretically equal
EHe tFe = EFetHe which gives
EHe 6. 4 tHe = E tFe = 8.5 tFe Fe
≈ 0.72tFe
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 distan ce of closest approach is given by
rc
e2 Z1 Z2 = E0
11
≈ 7.2 × 10−
cm
We know that the propbability for a particle-particle interaction is given by
P=e
−
EG E
1/2
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
rc =
e2 Z1 Z2 E0
≈ 2.8 × 10−
EG = 32π2 α2 2m p c2
≈ 2.50 × 10−
55
23
cm
≈ 31.6 MeV
and the probabilty is given by
P
10
for the proton-He interaction we find
rc =
e 2 Z1 Z2 E0 EG
≈ 1.44 × 10−
10
cm
≈ 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
EG
E=
(ln P)2
for the He-He interaction we find
EHe −He
≈ 0.125 MeV
E p−He
≈ .013 MeV
for the proton-He interaction we find
for the carbon-carbon interaction the Gamow energy is given by
EG
2 2
2
≈ 2592π α 6m pc ≈ 7.7 GeV
and thus
Ec−c
≈ 31 MeV
we know that
T
≈ kEb
so
THe −He Tp−He Tc−c
≈ ≈ ≈
9
1.45
× 10 K 1.5 × 10 K 3.6 × 10 K 8
11
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 m 1 and m 2 with positions x1 and x2 and velocities v1 and v 2 . (a) .What is the velocity of the COM? We kbnow that the center of mass is given by com =
m1 r1 + m2 r2 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 srcin)? We know that the relative velocities are given by
vrel −1 = v1 vrel
= v2
2
−
a bit of algebra yields
− vcom vcom
−
vrel −1 =
m2 ( v1 m1 + m2
−v )
vrel −2 =
m1 ( v2 m1 + m2
−v )
2
1
(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
Ktot
= = =
Ktot
=
1 1 m1 v2rel −1 + m2 v2rel −2 2 2
m1 m2 (m2 (v1 v2 )2 + m1 (v2 2(m1 + m2 )2 m1 m2 (v1 v2 )2 2(m1 + m2 )
−
−v ) ) 1
2
−
1 2
mr (v1
2
−v ) 2
where m r is the reduced mass.
Problem set 7
Problem # 1 The Ma in 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) / (R/R)sun 4 7
2
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 = 4πr 2 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 L conv = 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 convect ive 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
β=
2
−3 +
1/3
EG 4kT
given that we know what the temperature is and also what E G for p-p reaction
EG
≈ 500 keV
T
6
≈ 5 × 10 K
we find that
β = 5.92 we also know
ρ∝
≈ 6.0
M R3
we know that in steady state
L f usion = Lconv thus we can find
M
4/7 2
R ∝R
3
M2 R6
Tc6
⇒T
6
∝ M −5/21 R5/6
we know from the Virial temperature, assuming gas pressure dominates
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 latte r by scaling to the sun. Note, however, that it is not reasonab le 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 convec tive star. This is what we will do in the rest of the probl em. We can signi ficantly improve on the above scaling arguments by using the fact that fully convective stars are n = 3 /2 polytropes. It turns out tha t for a polytrope, in equation (1) can be Taylor expanded near the center to yield 2.4εc M L f usion (3 + β)3/2
≃
where I have again approximated ε ∝ ρ T 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 ρ c , and pp energy generation at the center of the star ε c in terms of the mass M and radius R . Assume X = 0 .7 and µ = 0.6 (typical for stars just reachi ng the main sequence). Note that you should give expressions for Tc , ρ c , and ε c here, with constants and real units, not just scaling relationships. So that the constants in front of your expressions are reasonable, please normal ize M to Msun and R to R sun . The general expressi ons given by HW 4 problem #3 are
Tc =
dn GM Rkb
1/3
3 an 4π
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
Tc = 0.322
m pGMsun kb Rsun
M /Msun R/Rsun
ρc = 1.43
Msun 3 Rsun
M /Msun (R/Rsun)3
plugging in all the constants yields
Tc
6
≈ 7.43 × 10 K
M /Msun R/Rsun
ρc
we also know 105 ρc X 2 T
εc = 5
≈ 8.41
M /Msun (R/Rsun )3
−2/3 e−15.7T7−1/3
7
×
we also know that we can approximate this as β
εc = AρT7 X 2 = AρT76X 2 setting this two expressions equal to each other we can find what A is, i.e lettingT
A=5
7
≈ 10 K we find
× 10 e− . ≈ 0.076 5
15 7
6 7
2
thus we find
εc substituting Tc and ρc gives
6 7
≈ 0.076ρcT X ≈ 0.037ρcT
εc
≈ 0.053
7
9
M Msun
Rsun R
equation results of b), and L on the main sequence to determine the R(M ), =L L(Mc)),Use Tc (M ), and L(2), (Te fthe 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. conv
f usion
Using the results from b and also
L f usion
≃ (32+.4εβc)M/ ≈ .09εcM 3 2
(β
≈ 6)
thus we know that
Lcon = L f usion 0. 2
M
4/7
Msun
2
R
Lsun = 0.0047
Rsun
Msun
rearranging this we find
R Rsun
which yield
11
R Rsun
M
= 0.023
≈ 0.67
7
Rsun R
9
M Msun
Msun Lsun
28
M Msun
M Msun
52/77
52/7
Msun
to find for the temperature we can use 6
Tc
≈ 7.43 × 10 K
Tc
≈ 1.1 × 10 K
M /Msun R/Rsun
using our previous resulst gives 7
25/77
M Msun
to find for the luminosity we use
Lconv = 0.2
4/7
2
M Msun
R Rsun
Lsun
plugging in for the radius we find
Lconv = 0.09
M Msun
148/77
148/77
Lsun
for the effective temperature we find
L = 4 πR2 σTe4f f
⇒ 0.09
M Msun
Lsun = 4πR2 σTe4f f
which can be simplified to 0.09
M Msun
148/77
Lsun = 1.56
× 10
18
104/77
M Msun
Te4f f
thus
Te f f = 3868 K which can also be written as
M = Msun L(Te f f ) = 4.85
M Msun
Te f f 3868 K
× 10−
50
1/7
7
148/11
Lsun Te f f
d) What are your predicted luminosities, radii, and effective temperatures for main sequence stars with
M and ? Compare −3your values to the values of L = 0.01Lsun, R = 0.3Rsun, and Te f f = 3450 = 0M .1 = .3Msun K for 0.0 3M sun and L = 10 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− Lsun 3
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, radiati on pressure due to photons (a relati vistic particle) is more important than gas pressu re. Fusion is by the CNO cycle. Assume for now that ener gy is transported primarily by photons and that the opacity is due to Thomson scatterin g (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
Prad = 1 aT 4 3 we also know
dPr = 4 aT 3 dT 3
dP dP dT 4 dT = = aT 3 = 3 dR dT dR dR
thus
−ρ GM R 2
dT ρM ∝ 3 2 dR T R
and radiative diffusion says
Lρ dT ρM ∝ ∝ 3 2 R2 T 3 dR T R which gives us
L∝M using the Virial theorem, where Prad dominates rather than Pgas we find
T4
ρ
∝
M R
1/2
⇒ Tc ∝ MR
where the left hand term is from the radiation pressure, but since we know that is an energy density we must devid e by the densi ty to find what the ener gy is per particle. Now using the steady stat e for luminosity we find L ∝ M ρT 18 where we chose β = 18 as a more appropriate value rather than the value given for the sun β = 20, 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
M9 R3 ∝ 18 R 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 ∝ ∝ M 1/21 R2 R2
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 concentrat ed at small radii because of the strong temperature dependence). For comparison, detailed calculations show that the fraction of the mass that undergoes “core” convection increases from 10 % at 2 Msun to 75% at 60 Msun . The condition for convection is given by
d ln T d ln P
L Lr /L γ −1 tot ≈ 14 PPrad > LEd d Mr /M γ
since we know that
γ=
4 3
Ptot
≈ Prad
gives us 1 Lr M 1 4 LEd d Mr > 4 which simplyfies to
Lr Mr > Ledd M we know that in the limit that M
→ 150Msun
Lr
→ LEd d ,
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 convec tive. n=3/2 polytrope The previous case yielded a result for a n=3 polytrope, we find that for
n = 3/2 polytrope 5 Lr Mr < 8 LEd d M and in the limit where M
→ 150Msun
Lr
→ Ledd Mr 5 < M 8 31
γ=
5 3
This seems rather strange in the sense that stars that are approximately 60 Msun 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
Etot
tMS
≈ LEd d Etot = NQ Q ≈ 7 MeV
where
where that is the total energy per reaction, we also know
Mr M = (n = 3 polytrope) mp mp Mr 5 M N = = (n = 3/2 polytrope) mp 8 mp N =
we also know that
LEd d =
4πcGM
κT
so we find the main-sequenc e lifetime to be given as
tMS
≈
tMS
≈
κT Q (n = 3 polytrope) m p 4πcG 5κ T Q (n = 3/2 polytrope) m p 32πcG
we know that
κT
2
≈ 0.4 cm /g
Q
≈ 7 MeV ≈ 1.12 × 10−
5
ergs
thus we find that the main-sequen ce lifetime for both types of polytropes are given by 2.21
6
6
× 10 yr < tMS < 3.39 × 10 yr
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? We know that if the density of the gas is ng as
nQ
≥ nQ where nQ is the quantum concentration, nQ is defined
≡
2πmkT ¯
h2
and for a gas at room temperature to be degenerate
where we used
≥
h
(1)
1026cm−3
1.46
2
3/2
3/2
2πmkT ¯
nQ =
ng
T =300K
≈
×
¯m = 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 3/2 2πme kT 1.25 1019 cm−3 ng nQ = h2
≥
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 kT
× 10
19
cm−3 = nQ
and now using Equation 1 we find 2/3
T=
nQ h2 2πmkT ¯
≈ 9.2 × 10− K 3
using m ¯ = 28 m 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 cm 3 ). Which dominates?
≃
We can write the thermal energy as 3
Eth
≈ 2 kT ≈ .039 eV
We can write the Coulomb energy as
Ecoul
2 21
≈Z e
we know
r
∼ n− / ∼ 1 3
33
r
− ρ
m ¯
1/3
thus 2 2
≈Z e
Ecoul using ρ
≃ 10 g/cm
3
and m ¯
1/3
m ¯
≈ 12.43 keV
≈ 100m p. The Fermi energy can be written as Ef =
where we used m ¯ We can see that
ρ
3 8π
2/3
h2 2m e
47ρ m ¯
2/3
≈ 75 eV
≈ 100m p and the 47 comes from the fact that there are 47 electrons in a silver atom. E >E
E coul
for room temperature silver.
f
≫
th
Problem # 2 Deuterium Fusion in Contracting Protostars Small amoun ts of Deuterium are made in the Big Bang. D is destroyed in the interi ors 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 cm 2 , 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 ∝ ρ T β . Instead, you will need to ke ep the full expression for ε.
→
×
≈
×
≈ ×
≃
a) What is the Gamow energy for D fusion? Write down the resulting thermally averaged cross-section
σv for D fusion. The Gamow energy can be written as
EG = Z12 Z22
mr MeV mp
using Z1 = Z2 = 1 and m r = 23 m p we find the Gamow energy to be
EG
≈ .67 MeV
The thermally averaged cross-section is given as 1/6
σv = 2.6kS/(ET)E/G 2 3
e−3(EG /4kT )
2 3
1/3
using all the constants given and the Gamow energy we find 3.7
σv =
K−
10−15 cm3 3742(K/T )1/3 T 2/3 s e−
2/3
×
in terms of M and R we find
−15 cm3 −3742(K/T )1/3 e
σv = 3K.7−×/ 10T / 2 3
2 3
s
b) In class we derived a quantitative model for the Kelvin-Helmholtz contraction of a low mass star as it approaches the main sequence. Use these results to calculate the local contraction time t c R/ dR /dt
≡ |
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 spends at a given radius R . Does the contracti on time get shorter or longer as the star contracts?
M
From lecture we derived the following relationship
L=
3 GM 2 dR = 0.2Lsun (M /Msun )4/7 (R/Rsun)2 7 R2 dt
thus we find
dR
7 0.2 R2sun Lsun
·
=
dt
3
2 GM sun
M Msun M
2
R Rsun 10/7
4/7
M
2
R
Msun
Rsun
3.34 × 10−5cm/s
=
2
Msun
R Rsun
4
and we can find the local contraction time to be
tc = Rsun
R Rsun
1
= 6. 7
(dR /dt )
7
× 10 yr
10/7
M Msun
Rsun R
3
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 lifetim e is the average time before a D nucleus is destroyed by fusion into 3 He)? Use the properties of n = 3/2 polytropes to write t D as a function of M and R . Does the D lifetime get shorter or longer as the star contracts? We know that average lifetime of a deuteron is given by
tD =
1 l = v n p σv
which gives us p tD = n p1σv = ρµm c σv
but we also know
ρc =
3M 4π R 3
thus we find
M R3
an = 1.43
3
3
µm p R 1 R p tD = ρµm = = 7.019 × 10− g 1.43σvM σv M c σv we also know
Tc =
dn GM Rkb
1/3
3a n 4π
25
µm p = 2.60
16 cm
× 10−
g
K
M R
thus we find 3
tD = 7.019 × 10− g σ1v RM = 1.89 × 10− 25
10
35
g s 2/3 3742(K /T )1/3 R3 T e cm3 M
written in terms of M and R we find 7/3
R Rsun
tD = 1.23 × 10− s 6
Msun M
1/3
1/3
e19.19((Msun /M)(R/Rsun ))
the deuteron lifeti me gets shorter as the star contracts. d)For any mass M show that there is a critical radius R D at which t D = 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 determin e the central temperatur e 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 1.23
× 10− s 6
7/3
R Rsun
Msun M
1/3
1 /3
e19.19((Msun /M )(R/Rsun ))
= 6 .7
7
× 10 yr
10/7
M Msun
Rsun R
3
and so we find
RD Rsun
16/3
= 1.71
× 10
21
37/21
M Msun
1 /3
e−19.19((Msun /M )(R/Rsun ))
we can solve this numeric ally to find
RD = 0.44Rsun
M = 0.03Msun
RD = 1.11R
M = 0.1Msun
to find the central temperature we can use
Tc = 7.4
6
× 10 K
M Msun
Rsun RD
we find
≈ ≈
Tc Tc
5
5. 0
× 10 K M = 0.03Msun 6.67 × 10 K M = 0.1Msun 5
and to solve for the deuteron lifetime we find 6
tD
= 1.23 × 10 so we find
tD tD
RD Rsun
7/3
Msun M
1/3
− s
≈ ≈
6
4.67
× 10 yr 4.1 × 10 yr 5
19.19((Msun /M )(R/Rsun ))1/3
e
M = 0.03Msun 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
Qrd
ε=
ρ d = m d nd
ρd
we also know that
L = Mε = since we know that Q = 8.8
MQr d
ρd
=
rd =
nd tD
MQ 2m ptD
× 10− ergs, we find 6
L=
M ( 8. 8
× 10− ergs) 6
2m ptD
so for M = .03Msun we find
L
≈ 5.37 × 10
35
ergs/s
L
≈ 4.08 × 10
37
ergs/s
and for M = 0.1Msun we find
thus we can see that for both of these stars deuteron fusion can stop the KH contraction tempor arily. Problem # 3 The R(M) Relation for Degenerate Objects Consider an object supported entirely by the pressure of non-relativistic degenerate electrons. Because P = K ρ5/3 such an object can be modeled (rigorously) as an n = 3/2 polytrope.K is a constant that depends on the electron mean molecular weight µ e . a) Use your results for how the central pressure Pc and density ρ c 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 express ion with proper consta nts and not just a scaling relationship. Normalize the mass M to M sun and the radius R to Rsun (this should sound pretty familiar by now). We know that
h2 Pdeg = Pc = 5m e rearranging this equation for ρ c yields
−1/3 =
ρc
2/3
h2 5m e
3 8π
ρc
5/3
µe m p
2/3
3 8π
1 µm p
4/3
= dn GM 2/3 ρc
5/3
1
dn GM 2/3
from the last problem set we showed
ρc = 8.41
M Msun
Rsun R
3
using this we find
R −5/3 = 0.04µe Rsun
− M Msun
1/3
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 R ∼ 0.30Rsun 37
M = MJ , we find
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 ene rgy. What mass and radius does thi s corres pond to? Explain why this is a very rough estimate of the maximum radiu s of a degenerate object.
→
→
If we know
E f = Ecoul then 1 e2 4πε0 r
but we know that
2/3
3
=
8π
∼n /
1 3
r thus 1 4πε0
n2/3
2m e
1
h2
e2 n1/3 =
2/3
3 8π
h2 2/3 n 2me
so we find the density to be given by
n=
3
2m e 4πε0 h
e2 2
2
≈ 3
8π
6.15
to find the mass we can use
ρc = nm p = 1.43
× 10
28
m−3
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 0.361R
R
sun
≈
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 gn
µ = mc2 + kT ln
(2)
Q
and that the quantum distribution function is defined as
g/h3 n( p) = (E −µ)/kT e p
±1
to show that in the classical regime
e(E p −µ)/kT
≫1
We can write Equation 1 as 2 nQ 1 = e(mc −µ)/kT n g
n
e(mc
≪ nQ
2
−µ)/kT ≫ 1
thus the quantum distribution function can be written as
/h3 n( p) = (Eg− e p µ)/kT We also know that
E p = mc2 +
p2 2m
nQ =
2πmkT
h2
3/2
and using equation 1 we find
n ( p) =
g/h3 e
(mc2 + p2 /2m
−mc2 −kT ln(nQ /n))/kT
=
gn h3 eE /kT nQ
which after some simplification reduces to the Classical Boltzman distribution function 1
n( p) = n
Problem # 2
2πmkT
3/2
e−E /kT
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 = 150 Msun . Note that d N /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 m 2 ? What is the ratio of the number of 150 Msun 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 we also know that for dm
≃m
1
α = 2.35
of m 1 we find α dN (m1 ) = m − 1 dm 1
we also know that for dm
≃m
2
m11−α
≈
of m 2 we find α dN (m2 ) = m − 2 dm 2
≈ m −α 1 2
and the fraction is given by
dN (m1 ) = dN (m2 )
1 α
− ≈ m1 m2
2208 m1 = 0 .5Msun m2 = 150Msun
b) Estimate the mass of a cloud M so that approximately one 150 Msun star forms in the cloud. If the temperature of the cloud at the time of formation was 10 K, what was the density of the gas out of which the cloud formed? From part a) we found that
N
2208N
s
b ∼≈150Msun and Ns is for stars that have ∼ 0.5Msunand to
where N b is for stars that are the number of get the total mass we must multiply the total number of small stars to the average mass of the stars, and from lecture we are told that M∗ 0.5Msun
≈ Mcluster = Ns ·M∗ ≈ 1104Msun
to find the density of the gas in which this cloud formed we can use the Jean’s density
ρJ =
3 4π M 2
3
≈ 3 kT
2G m ¯
3.14
× 10−
22
kg m −3
Problem # 3 A stellar atmosphere consists almost entirely of hydrogen. Assume that 50 % of the hydrogen molecules are dissociated andthe that the pressureSet is all 100degeneracies Pa. Given that theAs binding theofhydrogen molecule is 4.48into eV,atoms estimate temperature. to 1. the hintenergy at the of back the book suggests, you should derive the Saha equation for the dissociation of H 2 into hydrogen, i.e., the reaction H +H . γ + H2
←→
We know that
γ + H2
←→ H + H
µ(H2 )
←→ 2µ(H )
and the Saha equation gives
40
but we know that the chemical potential is given by
µ(H ) = m H c2
− kT ln
nQ , H g H nH
µ(H2 ) = m H2 c2
− kT ln
nQ,K2 gH2 nH2
where
mH c2 = m p c2 + me c2 thus we find
mH2 c2 which becomes
− kT ln
χH2
− kT
mH2 c2 = 2m p c2 + 2me c2
− χH
nQ,K2 gH2 nH2
= ln
= 2 mH c2
nQ,H2 gH2 nH2
2
nQ,H gH nH
− kT ln
nH nQ,H gH
− χH − 2χH
2
we are given that
gH2 = gH = 1
nQ,H2
−
≈ nQ , H ≈
thus we find
nH2 nH −χH /kT 2 = e = nH nH nQ
h2 2πmkT
2πmkT
3/2
h2
3/2
e χH2 /kT
but we are given that
nH =
P 3kT
nH2 1 = 2 nH
thus 1 P = 2 3kT
h2 2πmkT
3/2
e−χH2 /kT
this can only be solved analyticaly, we find that the temperature is given by
T
≈ 2260 K
Problem # 4 Lines f rom Hydrogen Consider a pure hydr ogen gas. In this proble m we will calc ulate 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 g n = 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 hydrog en 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 appro priate to the photosphere of the sun, make a plot of n p /n as a function of temperature T . If you are familiar with graphin g using IDL, Mathematic a, 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 follo ws. According to the reasoning in class, which you will confi rm 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 everyth ing is in the ground state.
≤ ×
We know from the Saha equation
ne n p ge g p = nH gH
2πmkT
h2
3/2
e−χ/kT
we know
ne
∼ np
thus
n2p nH
n = nH + n p χ
=α
α
≡
≈ −13.6 eV
2πmkT h2
3/2
ge g p =1 gH
e−χ/kT
and so
n2p = αnH = α(n
− n p)
which becomes a quadratic equatio n of the form
n2p + αn p with the solution of n p being
− αn = 0
np =
−α ± √α
np =
−α + √α
2 + 4αn
2 and since we know that this must be a positive thus we will take the positive solu tion
and finally we are looking for
np = n
2
2 + 4αn
−α + √α
2 + 4αn
2n
the plot is given by
42
np vs ntotal
0.8
0.6 n /
p
n
0.4
0.2
0.0 5.0 103 •
we can see that at a temperature of T ionized.
1.0 104
1.5 104 Temperature
•
•
2.0 104
2.5 104
•
•
3.0 104 •
7
≃ 1.4 − 1.5 × 10 K roughly ∼ 50% of the hydrogen atoms are
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 n 2 /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 n 2 /n as a function of temperature T . If you are fami liar with graphi ng using IDL, Math ematica, 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 nH = =1 n n and from the Boltzman equation we know that
− nnp
n2 g 2 −(E2 −E1 )/kT = e n1 g1 we also know
nH nH = n 1 + n2
and
→n
2
n2 = nH what we are looking for is
n2 n2 nH = = n nH n
= nH
−n → n 1
2
n1
= 1 + n2
− 1+
n1 n2
1
− − 1+ 43
n1 n2
1
1
np n
n = 2 state of hydrogen given by
thus we find the fraction of all H atoms that have an electron in the
n2 = n
− − 1+
n1 n2
1
1
np n
=
1+
g 1 (E2 −E1 )/kT e g2
√
− − − 1
1
α + α2 + 4αn 2n
where α has been explicitily defined already. The plot is given by n2 vs ntotal
4 10−5 •
3 10−5 •
n /
p
n
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 •
We can see that the fraction of hydrogen atomes in the energy state n = 2 peaks at
4.0 104 •
4
∼ 1.5 × 10 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 H α(n = 2 3)and H β (n = 2 4) lines of H? Use your result from b) to explain why A stars show the most prominent 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 =
hc
λ
thus
hc λ = hc = (E2 E1 ) ∆E
−
and
E= thus for the n = 2
− 13.n6 ev 2
→ n = 3 transition we get λ=
hc
( 3. 4
− 1.51)eV ≈ 656.3 nm 44
→
and for the n = 2
→ n = 4 transition we get λ=
( 3. 4
hc 0.85)eV
−
≈ 486.7 nm
From the plot given in part b) we can see that the fractional number of atoms in the n = 2 energy 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
∆E
=
hc hc = E ( E E1 ) ∆ 2
−
hc
(13.6
− 3.4)eV ≈ 121.6 nm
We would not expect to see any L yα lines from M stars, even though all of the hydrogen atoms are in the ground state, there is not enough thermal energ y 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 µ e is defined by n e ρ/µe m p .
≡
≡
Since we know that the total pressure is given by
kT PT = PI + Pe = ρ mp and thus
1
µ but we know that
=
1 X = µI A
1 1 + µI µe
1 1 + µI µ e 1 XZ = µe A 45
kT µm p
=ρ
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 1 A A (3) µ= µe = X 1+Z XZ a) What are the values of µ and µ e if the gas consists of i) ionized H, Since we know that
X=1 A=1 Z=1 then
µ= 1 2
µe = 1
ii) 75 % (by mass) ionized H and 25 % (by mass) ionized He, We have to treat this case seperately since we have two species contributing to the mean molecular weight X = 0.75 Y = 0.25 A = 4 Z = 2 thus 1 1 13 X Y = + = (X + Y ) = 16 µI A A A
1 7 Z = (X + Y ) = 8 µe A
and so we find
µ=
16 27
µe =
8 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 16
µ= 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
Pg ∝
1 µ
than the smallest µ will give us the highest pressure, thus the element that has the highest gas pressure is
µ=
1 2
Hydrogen gas
for degeneracy pressure we know 5/3
Pd ∝ ne
∝
5/3
1
µe
thus the gas that gives the highest degeneracy pressure is the one with the lowest value for
µ e 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 appro ximate such star s as lying on a He main sequ ence. In this probl em 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 dominat es the opacity, and that gas pressure dominates. The energy genera tion rate for He fusing to Carbon is
ε=5
× 10
11 ρ2 T 3 exp( 8
−
−44/T ) ergs s − g− 1
8
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 µ 4 µe This equation giv es the evolution of the lumunosity on the MS as chemical composition changes. We can scale this to the sun to find
L
L
= we also know that
M sun
3
Msun
µ
4
µsun
µe µe
− sun
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) Estima te the core temper ature of a 1 solar mass He star. You do not need to do the full integral L f usion = d M ε , but can approximate this as L f usion 0.1M ε(r = 0).
R
∼
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
3
M Msun
= 0 .1M ε = 0.1M 5
× 10
ρ T8−3 exp( 44/T8 ) ergs s −1 g−1
11 2
−
if we let 3
M = Msun ρ
≈ 150 g/cm
we find
T83e 44/T8 = 1.36
× 10
13
solving this numerically yields
≈ 1.52
T8
Tc
8
≈ 1.5 × 10 K
c) Given your result for Tc f or 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
∝ M3
L f usion
∝ Mε
we find
M 2 ∝ ε ∝ ρα T β where in part b) we are given
α=2
β=
−3 + 44 ≈ 26 T 8
knowing that the density and temperature scale as
ρ∝
M R3
T∝
M R
gives 2
M ∝
2
M R3
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
T
8
≈ 1.5 × 10 K 48
3/16
M Msun
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 k Rsun
Tc =
M Msun
Rsun R
×
µ = 4/3 gives
using the result from c) and plugging in all the constants along with 1. 5
8
× 10 K
3/16
M Msun
M Msun
107 K
= 3.08
Rsun R
and we find
R Rsun
M Msun
= 0.21
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
13/8
M Msun
plugging this in our expression of the luminosity gives 2
L = 4πσ(0.21Rsun)
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 13/24
Lsun thus we find that
L Lsun
= 4πσ(0.21Rsun)2
Te f f
4
≈ 2.1 × 10 K
The HR diagram for this star is given by
49
L 42.6Lsun
11/96
L Lsun
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 relationshi p. 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
L f usion = 42.7Lsun
κ
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
ne =
ρ µe = 2 M
M Msun
3
ρ µe m p
T κ= σ
2m p
≈ 0.2 g/cm
≥ 38.95Msun
is the mass that will exceed the Eddington luminosity f ) What is the He main seque nce life time 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
t=
0.1Msun 12m p (42.7Lsun)
thus the main sequence lifetime is
2
Msun M
t
≈
2.34
(7.65MeV)
×
107 yrs
7
≈ 2.34 × 10 yrs
Msun M
2
Msun M
2
this is much much shorter than the H burning lifetime which is t 1010 yrs for M be written as tHe 0.2% the time of the main sequence Hydrogen burnin g
∼
∼ 1Msun. Also can
∼
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 R s from the cente r of a star. The mass inte rior 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
GMM shell
P(Rs )
≃
4πRs4
dP = dr
−ρ GM R
HE gives
2
which can be written in differential form
P(Rs + dR ) dR but we know that
ρs = and since we know that H than
− P(Rs) = −ρ GM Rs2
Ms Vs
≈ 4πMR sdR 2 s
≪ dR ≪ R (this comes from the definition of H ) then P(Rs + dR ) ≪ P(Rs) P(Rs ) dR
GMM s
= 4πR4s dR
which simplifies to
P(Rs ) =
GMM s 4πRs4
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 dR 3 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 pressu re). 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 ∝ M ρ2 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 Chandrase khar mass. Estimate the ratio of the energy transpo rted by photons (radiati ve diffusion) to the energy transported by degenerate electrons (thermal conduction) at the center of the WD. Scale the central temperature of the WD to 10 8 K, an appropriate number for a newly formed WD. Assume that the opacity is due to electron scatte ring. Show that the energy transpo rted by electron conduct ion dominates that transported by photons. We know that the radiative flux for photons is given by
Fr =
4 acT 3
− 3 κ0 ρ
∇T =
−κr ∇T
where κo is the opacity and κ r 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 given by(radiative diffusion) to the energy transported by degenerate electrons (thermal conduction) is 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 = 4 2 32e me 32e2 m2e µi m p 52
ni =
ρc µi m p
and the radiative diffusion conductivity to be
κr =
4 acT 3 4 acT 3 4 acT 3 µe m p = = 3 κo ρc 3 ne σT 3 ρc σT
We also know that ρc (n = 3/2) polytrope equation is given by
ρc =
4M an 3π R 3
≈ 1.43 MR
an = 5.99
where the radius is given by
R
≈ 0.013Rsun
− − − ≈ M Msun
1/3
5/3
µe 2
1
m me
0.016Rsun
if we use M = 0.5Msun and µ e = 2. The ratio can be written as 128acT 2 e4 me2 m2p µe µi κr = 3k2 h3 σT ρc2 κdeg we also know
1
µ
=
1
µi
+
1
1
µe
µi
where we find
µi =
n
Xi A i=0
=∑ 96 7
(4)
n
1
µe
Xi Z A i=0
=∑
µe = 2 8
assuming 50% C and 50 % O. If we use a temperature of T
κr κdeg
≈ 10 K on Equation 1 we find ⇒ κdeg ≈ 11.9κr
≈ 0.084
Thus we can see that the energy transport ed by electron conduction dominat es 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 < 8 Msun become 0.5 Msun WDs, that stars with 30 Msun > M i > 8 Msun become 1.4 M sun NSs, and that stars with M i > 30 Msun become 7 Msun 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 SN explosions =
150 M 2.35 dM 8 150 2.35 dM 0.5 M
− −
R R
53
= 2.3%
b) What fraction of stars will become WDs? NSs? BHs? The fraction of stars that become WD is given by 8 2.35 dM 0.5 M 150 2.35 dM M 0.5
− −
R WD stars = R
= 97.7%
The fraction of NS is given by 30 8 150 0.5
R M− dM NS stars = = 1.97% R M− dM and the fraction of BH is given by R M− dM BH = R = 0.35% M − dM 2.35
2.35
150 30 150 0. 5
2.35 2.35
c) Estimate the fraction of the mass of a stellar population that is returned to the interstellar medium (via stel lar winds or explos ions) 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
b
MT = A
Z
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
1
a0.35
b a
M −2.35 (M 1
−b .
0 35
− Mend )dM
− Mend 1.35
1
1
a1.35
−b .
1 35
(5)
we chose inte gration 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
Mtot = A
Z
b a
M −1.35dM =
1 A .35 a.35
− b1.
35
Using
a = 1Msun b = 8Msun Mend = 0.5Msun on Equation 2 and
a = 0.5Msun b = 8Msun 54
(6)
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 populatio n comprises 97 % then the total contribut ion 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 contrib ution 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 contrib ution 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 10 4 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 0.013Rsun
R
≈
M
−1/3
Msun
µe
−5/3
m me
2
The luminosity can be calculated using
L = 4πσR2 Te4f f = 5.83
× 10
30
−1
0.016Rsun
≈
≈
erg s −1 = 2.3
× 10− Lsun 3
The central temperature is given by
Tc = 108 K
L Msun 5Lsun M
2/7
55
= 1.35
7
× 10 K
109 cm
1.11
×
and finally the age can be calculated using
t = 106 yrs
L Msun 5Lsun M
−
5/7
= 1.49
8
× 10 yrs
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
h = R2 kT = R2 kT mGM ¯ µH m p GM
≈ 1.5 × 10 cm ≈ 1 × 10− RW D 4
5
≈
we used µ H 1 since we assume that the atmospher e 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 hσT
1 =h nσ
≈ 9.75 × 10
19
cm−3
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,e e−χ/kTe f f nH gH n e = n p = n H and also g p = 1 ge = 2 gH = 2 and χ =
we know that in a gas of 1/2 ionized hydrogen 13.6 eV. The quantum densit y is given as
nQ , e =
2πme kT h2
3/2
thus we can write the Saha equation as
nH =
2πme k
3/2
h2
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
nH =
3/2
× 10− Te f/f e − .
×
n = 3
2πme k h2
3/2
Te f f e −13.6/kTe f f
Plugging in all the constants yield 7.43
5
3 2
1 57 105 /Te f f
56
−1 = 0
solving this numerically yieds a temperature of 4
≈ 2.6 × 10 K
Te f f
T
which is a higher than the 1/2 ionization temperature for the solar photosphere due to the differences in the densities. Problem # 4
4
∼ 1.3 × 10 K. This is ∼
Consider the late stages of evolution of a 25 Msun star. Focus on the core whic h 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 (whi ch 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
≈ 10
12
T93
Rc 10−2 Rsun
3
e−11.9/T9 Lsun
and the luminosity of the photons is given by
L ph = 3
5
× 10 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 10−2 Rsun
3
e−11.9/T9 Lsun = 3
5
× 10 Lsun
this simplifies to 3.3
× 10 T e− 6
3 9
11.9/T9
−1 = 0
solving this numerically yields
T9
8
≈ .794 ⇒ T ≈ 7.93 × 10 K
Neutrino cooling becomes important after Helium fusion and be fore 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=
E NQ = L ph L ph
57
we know that the number of particles are given by
N=
Msun mp
and the energy released is
Q thus we find that the time is
t=
≈ 0.7 MeV/nucleon Msun Q m p L ph
4
≈ 3.6 × 10 yrs
We can see that neutrino cooli ng 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 + 2 p Calculate the temperature for 50 % dissociation when the density is 10 12 kg m−3 . [Not e :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(2 p) which can also be written as
mHe c2
− kT ln
gHe nQ,He nHe
= 2 m n c2
− kT ln 58
gn nQ , n nn
2
+ 2 m p c2
− kT ln
g p nQ , p np
2
rearranging this equation yields 2
c (mHe
− 2mn − 2m p) = kT ln
gHe nQ,He nHe
nn2 n2p
(gn nQ,n )2 (g p nQ, p )2
we are given the binding energy needed to dissociate a helium atom
c2 (mHe thus
n2n n2p
− 2mn − 2m p) = −28.3 MeV nQ,He
gHe
nHe = we are also given
( gn g p ) 2
28.4MeV /kT
2 n2 (nQ ,n Q, p
e
−
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 = n Q , p =
2πm p kT
h2
3/2
nQ,He =
8πm pkT
h2
3/2
it follows that
n2n n2p = n4Q, p e−28.4MeV/kT nHe nQ,He inserting the expressions for the quantum density gives
nn2 n2p nHe
= 16(8)
2πm p kT
h2
9/2
e−28.4MeV/kT
(7)
if we assume that this gas is 50% dissociated gives us
nn = n p =
2 1 n nHe = n 5 5
for every helium nuclei there are two neutrons and two protons, givi ng a total of five particle s. We also know that mHe + 2mn + 2m p 8 5 ρ ρ n m¯ m n p 5 5 = m¯ = = 8 mp Using this along with plugging in all the constants into Equati on 1 gives us
≈
7.84
× 10
21
9/2
T12 e−0.328/T12 = 1 T12 =
solving this equation numerically yields a temperature of
T12 = 0.0109
T
≈ 1.09 × 10
59
10
K
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
≈ n − 2n p
nHe putting this into Equation 1 gives
n2n n2p n
2n p
9/2
2πm pkT
= 16(8)
h
2
e−28.4MeV/kT
(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
n
− 2n p = nHe = 0
1 n = np 2
thus there are no bound nucleus only proton s 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 M sun 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 gravit ational energy relesed from contrac tion 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 U GM 2 Ei EF EF ∆E 2 2RMS
∼−
≈ | |−| | ≈ | | ≈ − ∼
where RMS is given by the main sequence radiu s which is defined as
R MS
since we know that the mass of this star is
M
R
≈
sun
Msun
M = 25Msun we find the radius to be
R = 15.78Rsun
60
6/7
and the gravitational energy, which is the total energy is
G(25Msun )2 2 15.78Rsun
∆E pre−MS =
≈ 7.45 × 10
·
49
ergs
To calculate the total energy of a 25 Msun star during the main sequence we need to multiply total luminosity by the total time that the star spends on the main sequence
∆E = LMS tMS where the luminosity is given by
L = Lsun
M Msun
3. 5
= 7 .8
4
× 10 Lsun
and the time a star spends o the main sequence is given as
tMS = 1010
− M Msun
2.5
yrs = 3.2
6
× 10 yrs
this gives the total energy as
∆EMS
≈ 3.08 × 10
52
ergs
we can compare this to the value derived by nuclear energetics, we can say that the total energy is given by M ∆E 0.1NQ He = 0.1 QHe mp where we know M = 25Msun QHe 7 MeV/nucleon
≈
≈
Where Q He is the nuclear binding ene rgy per nucleon of helium. We find the energy to be given as
∆E
≈ 3.33 × 10
52
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 Q Fe 8.78 MeV/nucleon and
Q He
≈ 7 MeV/nucleon, thus 25Msun N= ∆Q ≈ 1.78 MeV/nucleon mp
and we find the total energy to be given as
∆E post −MS
≈ 4.24 × 10 61
52
ergs
≈
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. 2
≈ 2GM RNS
∆E
where the mass that we will consider will be the mass of the core
M
≈ 1.4Msun
≈ 10 km
R
thus we find the energy to be
∆ESN
≈ 2.59 × 10
53 ergs
this energy releas e 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 : n n = 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
1/3
h
8π
and the Fermi energy for relativistic particles is
εF = p F c thus we find
3 ne 8π
1/3
hc +
3n p 8π
2
where we made the assumption that m n c
1/3
hc 2
−
1/3
3 nn 8π
hc = mn c2
− m pc ≈ 0. We are also told that ne = n p
thus we find 1/3
2n p
= nn
ne = n p =
Problem # 4
62
1 nn 8
2
− m pc ≈ 0
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 21 σTE4 4πR2 8
Lν =
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 Probl em 2.5] If we refer to Philipps problem 2.5 we find that the energy density of fermions is given by
uF =
7 8
aT 4
(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
n=
1
V
Z
∞
N ( p) d p = 8 π 0
3
Z kT hc
x2
∞ 0
dx e−x ± 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
u= 1 V
Z
∞ε 0
p N ( p) d p
= 8π kT
hc
kT
Z
∞ 0
x3 dx ex 1
±
from Equationquations 3 and 4 we can see that the solution to the energy densit y for a fermion is given by
uF =
7 8
aT 4
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 =
67 28
aT 4 =
21 8
we also know that Stefan’s constant is defined as
σ = ac 4
thus the energy density is now given as 21 4 σT 4 8 c but we know that the flux due to the neutrinos is given by
uν =
Fν =
21 σT 4 8 63
aT 4
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 σTE4 4πR2 8
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 = E NS /tKH . Assume that the NS ha s 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
RNS = 15 km
− M 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
ρc =
3M an 4πR3
n = 3/2 polytrop which is given as
an = 5.99 n = 3/2 polytrope
thus the central density is
ρc
≈ 1.43 RM
3 NS
given the mass and radius we find
ρc
≈ 1.65 × 10
15
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 E ν of a neutrino in the core of the NS. The neutrinos are relati vistic. we know that
Eν
≈ EF (e)
where
E F ( e) = we know that
ne
3 ne 8π
1/3
≈ 18 nn ≈ 18 mρcp 64
hc
thus we find the Fermi energy to be
Eν
≈ E F ( e) ≈
3 nn
1/3
64π
hc
≈ 303 MeV
to show that they are degenerate we just need to show
≥ ET
Eν
where ET is the thermal energy, we find that for relativistic particles is
ET
≈ 3kT ≈ 25.8 MeV
thus we can see that
≥ ET
Eν
c) The cross section for neutrino’s interacting with matter is
σν Estimate the optical depth τ = nσν R path in the core of the NS.
≃ 10−
44
Eν m e c2
2
cm2
≃ R/ℓν of the NS to neutrinos, where ℓν is the neutrino mean free
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
≈ mρcp ≈ 9.86 × 10
38
cm−3 RNS
≈ 13.41 km
me c2
≈ 0.508 MeV
we find
τ
≈ 4.71 × 10
6
and we know that the neutrino mean free path is given by
ℓν =
RNS
τ
≈ 0.284 cm
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 t KH 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 2 RNS R2 = NS vℓ ν cℓ ν
t=
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 × 10
50
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 =
8 Lν 21 4πR2 σ
·
1/4
≈ 1.48 × 10
10
K
the energy of a neutrino can be calculated by knowing the ener gy 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
n = bT 3 where b = 6 1.803 2
× 8hπck
3 3 3
= 45.48K−3 cm−3
thus we know that the energy of a neutrino is given by
Eν =
21 a 8 b
T
≈ 6.24 MeV
66
