Download Ay123 Homework 1 Solutions

Ay123 Homework 1 Solutions
October 13, 2010
Problem 1
(10 pts) Assume a star obeys a linear density model so that
ρ(r) = ρc (1 − r/R)
where ρc is the central density and R is the radius of the star.
1. Find an expression for the central density in terms of R and the mass M
of the star.
The total mass M must be
Z
M
R
4πr2 ρ(r)dr
=
0
Z
R
4πr2 ρc (1 − r/R)dr
=
0
=
4πR3 ρc
=
π 3
R ρc
3
Z
1
x2 (1 − x)dx
0
Therefore,
ρc =
3M
= 4ρ̄
πR3
2. Use the equation of hydrostatic equilibrium and zero boundary conditions
to find the pressure as a function of radius. Your answer will be in the
form p(r) = pc × f (r/R), where f (x) is a function you will determine.
What is the dependence of the central pressure pc in terms of M and R?.
Express pc numerically with M and R in solar units.
First we need the mass enclosed within radius r:
3
Z r
r4
r
−
m(r) =
4πρ(r)r2 dr = 4πρc
3
4R
0
1
From the hydrostatic equilibrium equation,
dP
dr
=
p(r)
=
=
=
=
=
−Gm(r)ρ(r)
r2
Z R
−Gm(r)ρ(r)
dr
r2
r
Z R
4πGρ2c r3
r4 r
dr
−
1
−
r2
3
4R
R
r
Z
x 7x2
36GM 2 1
x3
−
+
dx
πR4
3
12
4
r/R
x=1
7x3
x4
36GM 2 x2
−
+
πR4
6
36
16 x=r/R
2
5GM
4
9x4
2
3
1−
6x − 7x +
4πR4
5
4
x=r/R
2
15
2
−4
dyne cm−2
Therefore, p(r) = pc f (x) where pc =5GM
rπR4 = 4.4×10 (M/M ) (R/R )
4
and f (x) = 1 − 54 6x2 − 7x3 + 9x4 . Note that this value of pc is much
smaller than the Sun’s central pressure, 2.4 × 1017 dyne cm−2 , since the
Sun is much more centrally concentrated.
3. Assuming the equation of state for an ideal gas, what is the central temperature Tc ?
ρ
kT , where n is the particle number density, mp is the
p = nkT = µm
p
proton mass, and µ ≈ 0.6 is the mean molecular weight (HKT 1.52).
Thus, the central temperature Tc is
Tc
µmp pc
kρc
µmp 5GM 2 πR3
=
k rπR4 3M
5µmp GM
=
12kR
= 5 × 106 (M/M )(R/R )−1 K
=
which is about a third of the actual value.
4. Find the ratio of the radiation pressure to the gas pressure at the center
of this star as a function of the total stellar mass (expressed in units of
M ). At what mass does the radiation pressure become comparable to
the ideal gas pressure?
2
The central radiation pressure is pγ,c = 13 aTc4 . This is 2.8×1012 (M/M )4 (R/R )−4 dyne cm−2 .
The ratio of radiation to total (part 1b) pressure is
pγ,c
pc
=
2.8 × 1012 (M/M )4 (R/R )−4 dyne cm−2
4.4 × 1015 (M/M )2 (R/R )−4 dyne cm−2
=
6.4 × 10−4 (M/M )2
This implies that gas pressure dominates up to M ∼ 30M . This is not
an exact result, since our formula for Tc assumes that radiation pressure
is negligible.
5. Write down an explicit expression for the total gravitational potential energy of this toy star, and verify that the virial theorem is exactly satisfied.
Be sure to discuss matter with a general equation of state, not just an
ideal monatomic nonrelativistic gas.
Virial Theorem, Form 1
The gravitational binding energy is (HKT 1.6)
Z M
Gm(r)
Ω = −
dm
r
0
Z R
Gm(r)
= −
4πρc (1 − r/R)r2 dr
r
0
Z R 3
r4
r
2
−
(1 − r/R) rdr
= −(4πρc ) G
3
4R
0
Z 144GM 2 1 x4
7x5
x6
= −
−
+
dx
R
3
12
4
0
26GM 2
= −
35R
Virial Theorem, Form 2
The stellar kinetic energy is (HKT 1.22)
Z
3 M p(r)
K =
dm
2 0 ρ(r)
Z
3 R
=
p(r)4πr2 dr
2 0
Z
3 R 5GM 2
4
9x4
2
3
=
1−
6x − 7x +
4πr2 dr
2 0 4πR4
5
4
x=r/R
Z 6GM 2 1 5x2
9x6
4
5
=
− 6x + 7x −
dx
R
4
4
0
3
=
13GM 2
35R
K = − 12 Ω, as expected from the virial theorem. Note that this result is fully
general and independent of the EOS.
Problem 2
From the result of problem1, we have
26 GM 2
35 R
(1)
4
L = 4πR2 σTeff
(2)
Ω=−
The definition of Teff is:
Since the star has not started nuclear reaction, the radiation energy is only
from the release of gravitational energy:
1
dΩ
2
13 GM 2
dR
35 R2
Z R
dR
−
R4
∞
τKH =
=
−Ldt
(3)
=
4
−4πR2 σTeff
dt
Z
4
4πσTeff
35 τKH
dt
2
GM 13 0
(4)
=
1 13 GM 2
8
2 35 4πσT 4 ' 10 yr
3R
eff
(5)
(6)
Problem 3
Use the virial theorem to estimate a typical value for the speed of sound deep
inside a star of mass M and radius R.
ln p
. From the
The sound speed cs is given by c2s = Γ1 ρp where Γ1 = dd ln
ρ
ad
virial theorem we know
Z
Z 2
p
cs
c2
Ω=3
dm = 3
dm ≈ 3 s M
ρ
Γ1
Γ1
Since Ω ≈
GM 2
R ,
c2s
cs
Γ1 GM
3R
≈ 3 × 107 (M/M )(R/R )−1 cm s−1
≈
4
Figure 1: Evolution track of proto-stars
Using this estimate, show that the time required for sound to cross the star
is P ∼ (Gρ̄)−1/2 . Hence, show that, for the fundamental p-mode of a star, the
pulsation period takes a value of order 1 hr.
P
2R
cs
=
1/2
3R
Γ1 GM
−1/2
GM
∼
R3
=
2R
∼ (Gρ̄)
=
−1/2
−1/2
0.91 (ρ̄/ρ̄ )
hours
Given that the luminosity L of a massive hydrogen-burning star varies with
mass roughly as L ∝ M 3 , show that, at fixed effective temperature, the fundamental pulsation period of such stars scales as P ∝ L7/12 .
At fixed Tef f , L ∝ R2 . Then
P
∝
M
R3
−1/2
5
= R3/2 M −1/2
∝
L3/4 L−1/6
=
L7/12
Problem 4
Sketch a plot of the stellar-remnant (i.e., white dwarf, neutron star, black hole)
mass as a function of the initial stellar mass.
There are very detailed models for finding the remnant mass given an initial
mass, but I will use rough numbers to keep it simple. The stellar remnant (Mr )
vs. initial mass (Mi ) relation is such that Mr ≈ ( M5i ) for 0.1M ≤ Mi ≤ 8M ,
Mr ≈ 1.4M for 8M ≤ Mi ≤ 60M . For Mi > 60M the relationship isn’t
very well known, but can be ignored because there are very few stars with such
high masses.
Assuming stars form with a Salpeter initial mass function - i.e., that the
number of stars in the mass interval M → M + dM is ξ(M )dM , with ξ(M ) ∝
M −2.35 , estimate the fraction of the mass of a stellar population that has been
returned to the interstellar medium 10 Gyr after this population was formed.
Comment briefly on the result.
Let the IMF have some normalization a, i.e. ξ(M ) = aM −2.35 . Then the
population has a mass
Z
Mtot,i
60M
=
Mi ξ(Mi )dMi
0.1M
Z
60M
= a
0.1M
=
5.71a
6
Mi−1.35 dMi
Note that you need a minimum and maximum mass, or the integral will be
ill-behaved.
In 10 Gyr, all stars with M & M will have evolved off the main sequence.
The mass locked up in low-mass stars or stellar remnants is
Z
Mtot,remain
1M
=
Z
0.1M
Z
=
1M
a
1M
Mi−1.35 dMi +
0.1M
=
8M
Mi ξ(Mi )dMi +
a
5
Z
Mi
ξ(Mi )dMi +
5
8M
Z
60M
1.4ξ(Mi )dMi
8M
Mi−1.35 ξ(Mi )dMi + 1.4a
1M
Z
60M
8M
3.9a
The amount returned to the ISM is 1.81a, or 32% of the original mass. Most of
the mass is still in main-sequence stars.
Problem 5
Prove that if the orbital plane of binaries are oriented randomly with respect to
the plane of the sky, that the average value of sin3 i is 0.59.
This problem is easiest to calculate from the point-of-view of the binary
system. Center a celestial sphere on the binary system, oriented such that the
systems angular momentum vector points North. The Solar systems location
on the sphere is described by two angles θ and φ as in spherical coordinates.
Note that the inclination angle i is equal to θ.
Assuming a random orientation, the probability of falling at a given θ and
φ is 1/4π. The probability of falling at a given i is
p(i) =
1
1
2π sin i = sin i
4π
2
The average value of sin3 i is then
Z
3 1 π 3
sin i =
sin i sin idi = 0.589
2 0
How has this result been useful in calibrating the relationship between stellar
mass and luminosity?
If the inclination is not known, then Keplers 3rd Law for binary stars takes
the form
(2π)2 a3
(M1 + M2 ) sin3 i =
G P2
Given luminosities measured through other means and a large enough sample,
this allows one to find the average mass for a star of a given luminosity. The
mass-luminosity relation follows.
7
1.4Mi−2.35 dMi
If the inclination is not known, then Keplers 3rd Law for binary stars takes the form
(M1 + M2 ) sin3 i =
(2π)2 a3
G P2
Given luminosities measured through other means and a large enough sample, this allows
one to find the average mass for a star of a given luminosity. The mass-luminosity relation
follows.
Problem 6
Because the problem was ambiguous about whether you should assume a perfect parallax
for parts a, d, and e, we accepted either interpretation. Since part c explicitly said “if the
uncertainty in the parallax is...”, ignoring parallax errors for part c counted against you.
a What is the angular size of each of the stars and of the semi-major axis? If you can
measure angles on the sky with a 1σ rms accuracy of 0.01 arcsec, what is the percentage
accuracy of the measurement of the semi-major axis and of the radius of each star?
From the parallax, we know d = 10 pc = 3 × 1019 cm.
The angular sizes of the star and semimajor axis are θ2R = 2R /d = 9.3 × 10−4 arcsec
and θa = 500R /d = 0.23 arcsec, respectively.
The measured sizes are d = θd /Π, where Π is the parallax, so the uncertainty in r is
s 2
σ 2
σθr
σr
Π
=
+
r
θr
Π
If you assume σθ2R = σθa = σΠ = 0.01 arcsec, then
σR
σ2R
=
= 1100%
R
2R
σa
= 11% (4.3% if assume precise parallax)
a
b Assume that the flux as a function of wavelength is given by a Planck function (i.e.,
it is a blackbody) with effective temperature Tef f = 5800 K. Assume that we have measurements of the flux ratio between log(ν) = 14.0 and 15.0 (where ν is given in Hz) that
9
have an accuracy of 10%. With what precision can we determine the Tef f of the stars
from these measurements?
Fν (1014 ) Hz
Fν (1015 ) Hz
3
2hν14
/c2 ehν15 /kT − 1
=
3
/c2 ehν14 /kT − 1
2hν15
3 hν15 /kT
ν14
e
−1
=
ν15
ehν14 /kT − 1
x ≡
To find the uncertainty in T from x, we need dx/dT :
hν15 hν14 ehν14 /kT
dx/dT = −x
−
kT 2
kT 2 ehν14 /kT − 1
hν15
≈ −x 2
kT
since ν14 = 0.1ν15 . Therefore, since σx =
dx
σ ,
dT T
kT σx
σT
≈
= 1.2%
T
hν15 x
c What is the uncertainty in the mass of the system if the uncertainty in the parallax is
taken as given in part a, 0.01 arcsec?
From Kepler’s Third Law,
GM
P
2π
2
= a3
4π 2 3
a
GP 2
12π 2 2
M
=
a =3
2
GP
a
σa
= 3
a
= 33%
M =
dM
da
σM
M
10
d What is the luminosity of each star calculated from the Tef f and the R, and what is the
uncertainty in the calculated luminosity of each star in this binary?
33
4
erg/s = L , as it should be.
L = 4πR2 σTef
f = 3.9 × 10
Since R and Tef f were determined independently,
dL
L
= 2
dR
R
dL
L
= 4
dTef f
Tef f
s
2 4σTef f 2
2σR
σL
=
+
L
R
Tef f
= 2200%
e We can measure the apparent flux at the Earth from this star to an accuracy of 5%.
Can we derive a more accurate luminosity and/or a more accurate radius from this
observation than from relying on the measured parallax and the Tef f determined in part
d?
Both the luminosity and the radius can be found much more accurately from the flux
L
F = 4πd
2:
dL
L
=
dF
F
dL
L
= −2
dΠ
sΠ
σ 2 2σ 2
σL
Π
F
=
+
L
F
Π
= 21%
4
2
From F = R2 σTef
f /d , and approximating Tef f and F as uncorrelated measurements,
dR
0.5R
=
dF
F
dR
2R
= −
dTef f
Tef f
11
dR
R
= −
dΠ
sΠ
2 σ 2 2σ
σR
σΠ 2
Tef f
F
=
+
+
R
2F
Tef f
Π
= 11% (3.5% if assume precise parallax)
This is why (almost) nobody tries to measure radii directly.
Problem 7
Show that for a polytropic equation of state:
p = Kργ
where K and γ are independent of radius, the gravitational potential energy of a star is given
by the expression
3GM 2
Ω=−
(5 − n)R
What is the implication of this result for the range of n applicable to stable stars?
By integrating by parts, we can show the potential is
Z
M
Gm(r)
dm
r
0
m=M Z M
Gm2 (r)
Gm2 (r) dr
= −
−
dm
2r
2r2 dm
0
m=0
Z R
Z
GM 2
Gm(r) m(r)
= −
−
dm0 dr
2
2R
2r
0
0
Z MZ R
2
GM
Gm(r0 ) 0
= −
−
dr dm
2R
2r2
0
r(m)
Ω = −
where the last transformation was made by interpreting the integral as that of a function
f (m0 , r) over a two-dimensional region bounded by the curve m(r) ≡ r(m) (note that, while
r and m0 are both measures of position within the star, they are independent variables as
far as the integral is concerned). The integrand is simply 21 g(r).
12