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Physics 161
Homework 7 - Solutions
Wednesday November 16, 2011
Make sure your name is on every page, and please box your final answer. Because we will
be giving partial credit, be sure to attempt all the problems, even if you don’t finish them.
The homework is due at the beginning of class on Wednesday, November 23rd. Because
the solutions will be posted immediately after class, no late homeworks can be accepted! You
are welcome to ask questions during the discussion session or during office hours.
1. (a) Assuming that the surface temperature of the Sun is T = 5800 K, use Stefan’s
law to determine the rest mass lost per second to radiation by the Sun. Take the
Sun’s radius to be R = 7.0 × 108 meters.
(b) What fraction of the Sun’s rest mass is lost each year from electromagnetic radiation? Take the Sun’s rest mass to be M = 2.0 × 1030 kg.
————————————————————————————————————
Solution
(a) The intensity of radiation (in watts per square meter) from the Sun is I = P/A =
σT 4 , where σ is the Stefan-Boltzman constant and T is the temperature. Thus
2
σT 4 , where R is the radius of the Sun. Now, the
the power is P = AI = 4πR
power is P = Ė, and since the energy is coming from the rest mass, E = mc2 ,
then Ė = ṁc2 . Thus, the rate of mass loss is
2
σ 4
4πR
dm
=
T .
2
dt
c
So, plugging in the constants gives
2
4πR
σ 4 4π (7 × 108 )2 (5.67 × 10−8 )
dm
T =
=
(5800)4 = 4.39 × 109 kg/sec.
2
8
dt
c2
(3 × 10 )
(b) So, in one year the Sun burns through
∆m =
dm
· ∆t = 4.39 × 109 × 3.15 × 107 = 1.38 × 1017 kg.
dt
This is a fraction
∆m
1.38 × 1017
=
= 6.9 × 10−14 ,
30
M
2 × 10
meaning that the Sun loses roughly 10−13 % of its mass per year.
2. Emission lines of hydrogen, Hβ (n = 4 → 2 and λrest = 4861Å) are observed in the
spectrum of a spiral galaxy at redshift z = 0.9. The galaxy disk is inclined by 45◦ to
the line of sight (if the inclination was 0◦ then we’d see the galaxy from a top view,
while if it was 90◦ , we’d see it edge-on). The Hβ wavelength of lines from one side of
the galaxy are shifted to the blue by 5Å relative to the emission line from the center
of the galaxy, and to the red by 5Å on the other side.
(a) What is the galaxy’s rotation speed?
(b) Find the age of the Universe when the light was emitted from this galaxy.
————————————————————————————————————
Solution
(a) There are two Doppler shifts in this problem, one from the redshift z, and the
other from the rotational motion of the galaxy. The redshift from the galaxy
shifts the wavelength
λobs
= 1 + z = 1.9.
λemit
Thus, the observed wavelength to the center of the galaxy is λobs = (1 + z) λemit =
1.9λemit = 9236 Å.
Next, the rotation of the galaxy adds an additional redshift, depending on the
rotational velocity. This velocity also depends on the angle of orientation. If the
angle was zero, giving a top view of the galaxy, then we wouldn’t measure any
rotational velocity, since it would be rotating perpendicular to the line of sight.
If the angle was 90◦ , then we’d see the galaxy edge-on, and would measure the
full rotational velocity. Thus, we see that we care about the sine component of
the velocity, v sin θ.
Now, the Doppler effect in the light emitted from the rotational velocity is (taking
the specific case of the edge that is moving away from us),
r
c + v sin θ
λ0obs
=
,
λobs
c − v sin θ
where λ0obs is the wavelength seen at the edge of the galaxy, and λobs is the
wavelength at the center of the galaxy ((1 + z) λemit ). Since the rotation velocity
is much smaller that light we can expand
r
2
λ0obs
v
v
1 + sin θ = 1 + sin θ.
≈
λobs
c
c
So,
λ0obs − λobs
∆λ
v
=
= sin θ,
λobs
λobs
c
where ∆λ is the difference in wavelength between the edge and center of the
galaxy (5 Å). Now, λobs = (1 + z) λemit , and so, all together the velocity is
v=
∆λ
c
.
sin θ (1 + z) λemit
Plugging in the numbers gives v = 230 km/s (which shows that we were justified
in assuming v small in our derivation).
(b) The redshift is defined in terms of the scale factor as
1
a (t)
=
.
a0
1+z
The galaxy emitted light in the matter-dominated era, and so a ∼ t2/3 . Thus,
a (t)
=
a0
2/3
t
1
,
=
t0
1+z
where t0 is the current age of the Universe. Solving for the emission time, t we
find
t = t0 (1 + z)−3/2 .
Taking t0 = 13.7 billion years, then
t=
t0
(1 + z)3/2
=
13.7 × 109
(1 + 0.9)3/2
= 5.23 billion years.
3. We have seen that, for a star composed of a classical, nonrelativistic, ideal gas, Etotal =
tot
tot
Eth
+ Egrav = −Eth
, and therefore the star is bound. Repeat the derivation, but for a
classical relativistic gas of particles. Recall that the equation of state of a relativistic
total
, and therefore Etotal =
gas is P = 31 EVth . Show that, in this case, Egrav = −Eth
tot
Eth + Egrav = 0; i.e., the star is marginally bound. As a result, stars dominated by
radiation pressure are unstable.
————————————————————————————————————
Solution
The pressure of a relativistic gas is P = Eth /3V , which means that
hP i =
1 tot
E .
3V th
Furthermore, we’ve seen that
1 Egrav
,
3 V
tot
tot
= −Eth
. Since Etot = Eth
+ Egrav = −Egrav + Egrav = 0.
hP i = −
which tells us that Egrav
4. Because of the destabilizing influence of radiation pressure, the most massive stars
that can form are those in which the radiation pressure and the nonrelativistic kinetic
pressure are approximately equal. Estimate the mass of the most massive stars, as
follows:
(a) Assume that the gravitational binding energy of a star of mass M and radius R
is Egrav ∼ GN M 2 /R. Use the virial theorem
hP i = −
to show that
P ∼
4π
34
1/3
1 Egrav
,
3 V
GN M 2/3 ρ4/3 ,
where ρ is the typical density.
(b) Show that if the radiation pressure, Prad = 13 aT 4 , equals the kinetic pressure, then
the total pressure is
1/3 4/3
3
kB ρ
,
P =2
a
m̄
where m̄ = mH /2 is the mean mass.
(c) Equate the expressions for the pressure in parts (a) and (b), to obtain an expression for the maximal mass of a star. Find its value, in solar masses, assuming a
fully ionized hydrogen composition.
————————————————————————————————————
Solution
(a) The pressure
P ∼
GN M 2
.
3RV
Now, 1/V = ρ/M , while R = (3M/4πρ)1/3 , and so
1/3 1/3
GN M ρ 4πρ
4π
P ∼
=
GN M 2/3 ρ4/3 .
3
3M
34
(b) The total pressure is P = Pgrav + Prad =
pressures are equal then
ρkB T
m̄
+ 31 aT 4 . If the kinetic and radiation
ρkB T
1
= aT 4 ⇒ T =
m̄
3
3ρkB
m̄a
1/3
Then, the total pressure is
1
2
2
P = Pg + aT 4 = aT 4 = a
3
3
3
3ρkB
m̄a
4/3
1/3 4/3
3
kB ρ
=2
.
a
m̄
(c) Equating the two expressions gives
4π
34
1/3
GN M
2/3 4/3
ρ
1/3 4/3
3
kB ρ
∼2
.
a
m̄
4
This gives for the mass, after plugging in for a = 8π 5 kB
/15h3 c3 ,
s
3
14580 hc
= 2.27 × 1032 kg,
M∼
π 6 m4H GN
which is M ∼ 113M . So, the largest stars have a mass of roughly 100 solar
masses.
5. Suppose a star of total mass M and radius R has a density profile
r
ρ = ρC 1 −
,
R
where ρC is the central density (assume constant).
(a) Find M (r).
(b) Express the total mass M in terms of R and ρC .
(c) Solve for the pressure profile, P (r), with the boundary condition P (R) = 0.
————————————————————————————————————
Solution
(a) We can determine M (r) via the mass continuity equation,
r3
dM (r)
2
2
= 4πr ρ (r) = 4πρC r −
.
dr
R
Separating and integrating noting that M (r = 0) = 0, we find
Z
M (r)
Z
dM (r) = 4πρC
0
0
r
r3
2
dr r −
.
R
Integrating both sides gives
4π
3r
3
ρC r 1 −
M (r) =
.
3
4R
(b) Finding the total mass is very simple. We can either repeat the above derivation
in part (a), integrating to r = R, or just plug in r = R in to our result for M (r).
Then,
3
π
4π
3
ρC R 1 −
= ρC R3 .
M = M (R) =
3
4
3
(c) Now, to determine the pressure, P (r), we use the hydrostatic equilibrium equation,
dP (r)
GN M (r) ρ (r)
.
=−
dr
r2
The density ρ (r) was given, and we have just found M (r), so we find
dP (r)
dr
N 2 3
= − 4πG
ρC r 1 −
3r2
3r
×
4R
r
3r
− 4R
R
N 2
= − 4πG
ρC r 1 −
3
2
4πGN 2
= − 3 ρC r − 7r
4R
1 − Rr
2
+ 3r
R2
3r3
+ 4R2 .
Now, since we know that P (R) = 0, we integrate from r = r to r = R,
Z 0
Z
4πGN 2 R
7r2
3r3
dP (r) = −
ρC
dr r −
+
3
4R 4R2
P (r)
r
Integrating gives
4πGN 2
ρC
−P (r) = −
3
5 2 r2
7 r3
3 r4
R −
+
−
48
2
12 R 16 R2
,
or,
P (r) =
πGN ρ2C R2
5
2 r 2 7 r 3 1 r 4
−
+
−
.
36 3 R
9 R
4 R
Notice that the pressure at the center of the star (in this model) is
P (0) =
5πGN 2 2
ρ R .
36 C