Download Physics 161 Homework 8 - Solutions Wednesday

Physics 161
Homework 8 - Solutions
Wednesday November 30, 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, December 7th. 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. How fast would a collapsed stellar core rotate if its initial spin rate was 1 revolution
per day, and its radius decreased from 10,000 km to 10 km? Estimate the maximum
rotational frequency of a white dwarf star, assuming that it has a mass equal to a solar
mass and radius of 1% of the solar radius.
————————————————————————————————————
Solution
As the star collapses, it conserves angular momentum. So, the angular momentum of
a sphere is
2
L = Iω = M R2 ω.
5
2
So, the initial angular momentum is Li = 5 M Ri2 ωi , and the final is Lf = 25 M Rf2 ωf .
Conservation requires
2
Ri
ωi .
ωf =
Rf
Taking ω = 2πν we have
2
Ri
νf =
νi .
Rf
So, we find
2
2
Ri
10000
νf =
νi =
× 1 = 106 rev/day.
Rf
10
The star will reach its maximum rotational frequency when the centripetal force on
a piece of the star of mass m at the edge of the star, FC = mω 2 R just balances the
gravitational force on it, FG ∼ GN mM/R2 . So,
r
GN mM
GN M
2
mω R ∼
⇒ω∼
.
2
R
R3
With M = M and R = 0.1R , then
s
r
GN M
6.672 × 10−11 × 2 × 1030
ω≈
=
≈ 0.63 rad/sec,
R3
(0.01 × 7 × 107 )3
which is about 8600 revolutions per day.
2. In a fully degenerate gas, all of the particles have energies lower than the Fermi energy.
For such a gas we found the relation between the density, ne = Ne /V , and the Fermi
momentum, pF ,
8π p3F
.
ne =
3 (2π~)3
√
(a) For a nonrelativistic electron gas, use the relation pF = 2me EF between the
Fermi momentum, the electron mass me , and the Fermi energy EF to express EF
in terms of ne and me .
(b) Estimate a characteristic ne under typical conditions inside a white dwarf. Using
the result of (a), and assuming a temperature of T = 107 K, evaluate numerically
the ratio ETH /EF , where ETH is the characteristic thermal energy of an electron
in a gas of temperature T , to see that the electrons inside a white dwarf are indeed
degenerate.
————————————————————————————————————
Solution
(a) We can solve for the Fermi momentum,
pF =
3
(2π~)3 ne
8π
1/3
.
The Fermi energy is EF = p2F /2me , so
1
EF =
2me
3
(2π~)3 ne
8π
2/3
.
(b) The number density of electrons can be expressed in terms of the total number
of electrons, Ne and the volume of the white dwarf, V , as ne = Ne /V . Now, as
we’ve seen, the total number of electrons is Ne = (Z/A) M /mp , where mp is the
proton mass. So, taking Z/A ≈ 1/2 for a white dwarf, we find
Ne
1 Z M
1 M
3 M
=
≈
=
.
ne =
V
V A mp
2V mp
8π mp R3
Taking R = 7 × 107 m for a typical white dwarf, we find
ne =
3 M
≈ 4.2 × 1032 m−3 .
3
8π mp R
This means that the Fermi momentum is
1/3
3
3
pF =
(2π~) ne
≈ 2.4 × 10−23 kg m/s.
8π
Thus, the Fermi energy is
Ef =
p2f
≈ 2, 040 eV.
2me
The thermal energy is Eth ∼ kB T ≈ 900 eV. Thus,
ETh
900
∼
= 0.44,
Ef
2040
meaning that the thermal energy is too low to raise the electrons up out of their
degeneracy.
3. Type Ia supernovae are probably thermonuclear explosion of accreting white dwarfs
that have approached or reached the Chandrasekhar limit.
(a) Use the virial theorem to obtain an expression for the mean pressure inside a
white dwarf of mass M and radius R.
(b) Use thepresult of (a)pto estimate, to an order of magnitude, the speed of sound
dP/dρ ∼
P/ρ, inside a white dwarf. In an accreting white dwarf
vS =
with a carbon core that has reached nuclear ignition temperatyure, a nuclear
binding “flame” emcompases the star at the sound velocity or faster. Within how
much time, in seconds, does the flame traverse the radius of the white dwarf,
assuming R = 104 km, and M = 1.4M ? Note that this sound-crossing timescale
is ∼ (GN ρ)−1/2 , which is also the free-fall time scale.
————————————————————————————————————
Solution
(a) The virial theorem says that the mean pressure inside a white dwarf is
hP i = −
1 Eg
,
3V
2
where the gravitational energy Eg = − 53 GNRM . Thus, the mean pressure is (dropping the brackets)
3 GN M 2
1 GN M 2
=
.
P =
5 RV
20π R4
(b) We can rewrite the pressure in terms of the density, ρ, noting that M = ρV =
4π
ρR3 , so
3
1 GN ρM
P =
.
3 R
p
p
Then, P/ρ = GN M/3R = vS . Now, the time scale to cross the star is
s
r
R
R3
1
R
∼R
=
∼√
.
t∼
vS
GN M
GN M
GN ρ
So, the time t ∼ (GN ρ)−1/2 .
Now, if we take R = 104 km, and M = 1.4M = 2.8 × 1030 kg, then
s
s
3
R
(107 )3
t∼
=
≈ 2.5 seconds.
GN M
6.672 × 10−11 × 2.8 × 1030
4. We have determined the radius of a white dwarf of mass M ,
RWD =
9π
4
2/3
~2
5/3
1/3
GN me mp M
5/3 1/3
Z
M
,
A
M
where me is the electron mass, mp is the proton mass, and Z/A is the ratio of the
atomic number to atomic mass number. In the white dwarf case, Z/A ≈ 1/2, while in
the neutron star case Z/A → 1. We have also determined the maximum mass for a
Z 2
M , with a numerical value of
white dwarf, MWD ∼ A
MWD = 1.4M .
Using these results, determine the (unstable) radius, RNS , and maximum mass, MNS ,
of a neutron star.
————————————————————————————————————
Solution
The difference between a white dwarf and neutron star is the source of the internal
pressure. In the case of the white dwarf, the pressure is supported by electron degeneracy pressure, while in the case of the neutron star it’s neutron degeneracy pressure.
Thus, we expect that, based on the expression for the radius of the white dwarf, we
should replace me → mn . Furthermore, we should set Z/A → 1. Thus, we have
RNS =
9π
4
2/3
~2
5/3
1/3
GN mn mp M
5/3 1/3
Z
M
.
A
M
Plugging in the constants gives
RNS
5/3 1/3
Z
M
= 12, 360
meters.
A
M
Taking Z/A → 1 gives
RNS = 12, 360
M
M
1/3
meters.
So, a 2 solar mass neutron star would have a radius of about 9.8 km.
Next, since MWD ∼ (Z/A)2 M with Z/A ≈ 1/2 for a white dwarf, giving a maximum
mass of 1.4M , we should expect that since Z/A → 1 for a neutron star, then the
maximum mass of a neutron star should be four times greater than for a white dwarf.
Thus, we should expect that
MNS = 5.6M .
General relativity adjusts things to about 5 solar masses, but we’ve still made a good
estimate.
5. In the comic book story All-Star Superman, Superman protects his Fortress of Solitude
using a key made of “neutron star material,” which he leaves lying on the doorstep,
reasoning that he alone would be able to lift it. Estimate the weight (in tons) of a key
made of neutron star material. Is such a key physically possible (i.e., could you bring
a small lump of this material to Earth)? Why or why not?
————————————————————————————————————
Solution
A neutron star is held together by neutron degeneracy pressure counteracting the
gravitational pressure, and is essentially a huge nucleus. The density of a neutron star
is, thus, the density of a nucleus! We can determine the density in two ways, either by
determining the density of a nucleus, or by simply dividing the mass of a neutron star
by its density (the two results should give the same order of magnitude, of course).
The mass of a neutron star is roughly 2 solar masses, and it’s radius is about 12 km.
Thus, it’s density is
ρ=
2 × 2 × 1030 kg
M
17
= 4
kg/m3 .
3 = 5.5 × 10
V
π (12, 000)
3
A nucleus has a density of ρ ∼ 3 × 1017 kg/m3 , and so the densities are comparable,
as expected!
We can estimate the volume of a key as about 1 cubic centimeter, or about 10−6 m3 .
Thus, the mass of the neutron star material key would be m 1012 kg! Since one ton is
907 kg (or about 1000 kg, or so), then the key would have a weight of about a billion
tons! This seems excessive, even for Superman to lift, but let’s suppose that he could...
Now, once again, the neutron star is composed of neutrons held together by gravitational pressure, supported by neutron degeneracy pressure. Since the star is in
equilibrium, the pressures have to be equal. Now, if the key was pulled out of the star,
then the gravitational pressure would be removed. So, the key now has a tremendous
internal pressure, with no counterbalanced external pressure (once Superman lets go
of it, let’s say), and so it would explode! Thus, whether or not Superman is the only
being that can lift the key is irrelevant, since it wouldn’t last long enough to put into
the lock!