a brief Introduction of astrophysics to UG

introduction to the astrophysics in undergraduate level

1. 1. Blackbody radiation

2. Con$nuous spectra and blackbody radia$on
•A blackbody is an idealized case
of a hot, dense object.
•The figure shows the
con<nuous spectrum produced
by a blackbody at different
temperatures.
•Planck provided the first
explana<on of blackbody
radia<on by assuming that
atoms in the blackbody have
evenly spaced energy levels
and emit photons by jumping
from one energy level down to
the next one.
CHAPTER 39SUMMARY
De Broglie waves and electron diffraction:Electrons and
other particles have wave properties. A particle’s wave-
length depends on its momentum in the same way as for
photons. A nonrelativistic electron accelerated from rest
through a potential difference has a wavelength
given by Eq. (39.3). Electron microscopes use the very
small wavelengths of fast-moving electrons to make
images with resolution thousands of times finer than is
possible with visible light. (See Examples 39.1–39.3.)
V
ba
(39.1)
(39.2)
(39.3)l=
h
p
=
h
22meV
ba
E=hƒ
l=
h
p
=
h
mv
The nuclear atom:The Rutherford scattering experiments show that most of an atom’s mass and all
of its positive charge are concentrated in a tiny, dense nucleus at the center of the atom. (See
Example 39.4.)
Atomic line spectra and energy levels:The energies of
atoms are quantized: They can have only certain definite
values, called energy levels. When an atom makes a
transition from an energy level to a lower level it
emits a photon of energy The same photon can
be absorbed by an atom in the lower energy level, which
excites the atom to the upper level. (See Example 39.5.)
E
i-E
f.
E
f,E
i
(39.5)hƒ=
hc
l
=E
i-E
f
The Bohr model:In the Bohr model of the hydrogen
atom, the permitted values of angular momentum are
integral multiples of The integer multiplier nis
called the principal quantum number for the level. The
orbital radii are proportional to and the orbital speeds
are proportional to The energy levels of the hydro-
gen atom are given by Eq. (39.15), where Ris the
Rydberg constant. (See Example 39.6.)
1 > n.
n
2
h > 2p. (39.6)
(39.8)
(39.10)
(39.9)
(39.15)
1n=1, 2, 3,Á2
E
n=-
hcR
n
2
=-
13.60 eV
n
2
v
n=
1
P
0

e
2
2nh
=
2.19*10
6
m > s
n
=n
2
15.29*10
-11
m2
r
n=P
0

n
2
h
2
pme
2
=n
2
a
0
1n=1, 2, 3,Á2
L
n=mv
nr
n=n
h
2p
Incident electron
waves in phase
Scattered electron
waves in phase
50°
d
Atom on
crystal
surface
l
Nucleus
a
E
Ef
i
hf 5 E
i
2 E
f
i
f
v
n
M, 1e
r
n
F
Proton
Electron
m, 2e
The laser:The laser operates on the principle of stimulated emission, by which many photons with
identical wavelength and phase are emitted. Laser operation requires a nonequilibrium condition
called a population inversion, in which more atoms are in a higher-energy state than are in a lower-
energy state.
A*
A*
A*
n
1
n
0
Spontaneous
emission
Blackbody radiation:The total radiated intensity (aver-
age power radiated per area) from a blackbody surface
is proportional to the fourth power of the absolute tem-
perature T. The quantity
is called the Stefan–Boltzmann constant. The wave-
length at which a blackbody radiates most strongly
is inversely proportional to T. The Planck radiation law
gives the spectral emittance (intensity per wave-
length interval in blackbody radiation). (See Examples
39.7 and 39.8.)
I1l2
l
m
s=5.67*10
-8
W > m
2#
K
4
(Stefan–Boltzmann law)(39.19)
(Wien displacement law)(39.21)
(Planck radiation law)(39.24)
I1l2=
2phc
2
l
5
1e
hc > lkT
-12
l
mT=2.90*10
-3
m#
K
I=sT
4
0123456
1
2
3
4
2000 K
1750 K
1250 K
l (mm)
I(l) (10
11
W/m
3
)
1318

3. Physical meaning of !(#)
!(#)
!"/$
!
=&
"!
"
"
'()(≈'(×(
!−(
#
=power of light with wavelength ∈#
!,#
"radiated by the black body per unit area
(
CHAPTER 39SUMMARY
De Broglie waves and electron diffraction:Electrons and
other particles have wave properties. A particle’s wave-
length depends on its momentum in the same way as for
photons. A nonrelativistic electron accelerated from rest
through a potential difference has a wavelength
given by Eq. (39.3). Electron microscopes use the very
small wavelengths of fast-moving electrons to make
images with resolution thousands of times finer than is
possible with visible light. (See Examples 39.1–39.3.)
V
ba
(39.1)
(39.2)
(39.3)l=
h
p
=
h
22meV
ba
E=hƒ
l=
h
p
=
h
mv
The nuclear atom:The Rutherford scattering experiments show that most of an atom’s mass and all
of its positive charge are concentrated in a tiny, dense nucleus at the center of the atom. (See
Example 39.4.)
Atomic line spectra and energy levels:The energies of
atoms are quantized: They can have only certain definite
values, called energy levels. When an atom makes a
transition from an energy level to a lower level it
emits a photon of energy The same photon can
be absorbed by an atom in the lower energy level, which
excites the atom to the upper level. (See Example 39.5.)
E
i-E
f.
E
f,E
i
(39.5)hƒ=
hc
l
=E
i-E
f
The Bohr model:In the Bohr model of the hydrogen
atom, the permitted values of angular momentum are
integral multiples of The integer multiplier nis
called the principal quantum number for the level. The
orbital radii are proportional to and the orbital speeds
are proportional to The energy levels of the hydro-
gen atom are given by Eq. (39.15), where Ris the
Rydberg constant. (See Example 39.6.)
1 > n.
n
2
h > 2p. (39.6)
(39.8)
(39.10)
(39.9)
(39.15)
1n=1, 2, 3,Á2
E
n=-
hcR
n
2
=-
13.60 eV
n
2
v
n=
1
P
0

e
2
2nh
=
2.19*10
6
m > s
n
=n
2
15.29*10
-11
m2
r
n=P
0

n
2
h
2
pme
2
=n
2
a
0
1n=1, 2, 3,Á2
L
n=mv
nr
n=n
h
2p
Incident electron
waves in phase
Scattered electron
waves in phase
50°
d
Atom on
crystal
surface
l
Nucleus
a
E
Ef
i
hf 5 E
i
2 E
f
i
f
v
n
M, 1e
r
n
F
Proton
Electron
m, 2e
The laser:The laser operates on the principle of stimulated emission, by which many photons with
identical wavelength and phase are emitted. Laser operation requires a nonequilibrium condition
called a population inversion, in which more atoms are in a higher-energy state than are in a lower-
energy state.
A*
A*
A*
n
1
n
0
Spontaneous
emission
Blackbody radiation:The total radiated intensity (aver-
age power radiated per area) from a blackbody surface
is proportional to the fourth power of the absolute tem-
perature T. The quantity
is called the Stefan–Boltzmann constant. The wave-
length at which a blackbody radiates most strongly
is inversely proportional to T. The Planck radiation law
gives the spectral emittance (intensity per wave-
length interval in blackbody radiation). (See Examples
39.7 and 39.8.)
I1l2
l
m
s=5.67*10
-8
W > m
2#
K
4
(Stefan–Boltzmann law)(39.19)
(Wien displacement law)(39.21)
(Planck radiation law)(39.24)
I1l2=
2phc
2
l
5
1e
hc > lkT
-12
l
mT=2.90*10
-3
m#
K
I=sT
4
0123456
1
2
3
4
2000 K
1750 K
1250 K
l (mm)
I(l) (10
11
W/m
3
)
1318
#$
%
≪*
&+⇒-
!"
#$
%
&
≈1+
#$
%'
%(
⇒!#∝
1
#
)
ℎ3
#
≫*
&+
⇒!#∝-
*
#$
%'
%(

4. •The spectral emi.ance I(λ) for radia5on from a blackbody has a peak whose
wavelength depends on temperature:
Example:
Surface temperature of our Sun is -=58001⇒(
$≈500nm (green)
•We can obtain the Stefan–Boltzmann law for a blackbody by integra5ng I(λ) over
ALLwavelengths to find the total radiated intensity:
'=&
%
&
'()(=
24
'
5
(
157
!
ℎ
)
-
(
=9-
(
where 9=5.6704×10
*+
W/m
2
· K
4
is the Stefan–Boltzmann constant.
Con9nuous spectra and blackbody radia9on

5. 39.6The Uncertainty Principle Revisited
The discovery of the dual wave–particle nature of matter forces us to reevaluate
the kinematic language we use to describe the position and motion of a particle.
In classical Newtonian mechanics we think of a particle as a point. We can
describe its location and state of motion at any instant with three spatial coordi-
nates and three components of velocity. But because matter also has a wave
aspect, when we look at the behavior on a small enough scale—comparable to
the de Broglie wavelength of the particle—we can no longer use the Newtonian
description. Certainly no Newtonian particle would undergo diffraction like elec-
trons do (Section 39.1).
1314CHAPTER 39Particles Behaving as Waves
Example 39.7Light from the sun
To a good approximation, the sun’s surface is a blackbody with a
surface temperature of 5800 K. (We are ignoring the absorption
produced by the sun’s atmosphere, shown in Fig. 39.9.) (a) At what
wavelength does the sun emit most strongly? (b) What is the total
radiated power per unit surface area?
SOLUTION
IDENTIFY and SET UP: Our target variables are the peak-intensity
wavelength and the radiated power per area I.Hence we’ll use
the Wien displacement law, Eq. (39.21) (which relates to the
blackbody temperature T), and the Stefan–Boltzmann law, Eq.
(39.19) (which relates to T).
EXECUTE:(a) From Eq. (39.21),
=0.500*10
-6
m=500 nm
l
m=
2.90*10
-3
m#
K
T
=
2.90*10
-3
m#
K
5800 K
I
l
m
l
m
(b) From Eq. (39.19),
EVALUATE: The 500-nm wavelength found in part (a) is near the
middle of the visible spectrum. This should not be a surprise: The
human eye evolved to take maximum advantage of natural light.
The enormous value found in part (b) is the
intensity at the surfaceof the sun, a sphere of radius .
When this radiated energy reaches the earth, away,
the intensity has decreased by the factor
to the still-impressive .1.4 kW>m
2
10
11
m24
2
=2.15*10
-5
316.96*10
8
m2>11.50*
1.50*10
11
m
6.96*10
8
m
I=64.2 MW>m
2
=6.42*10
7
W>m
2
=64.2 MW>m
2
I=sT
4
=15.67*10
-8
W>m
2#
K
4
215800 K2
4
Example 39.8A slice of sunlight
Find the power per unit area radiated from the sun’s surface in the
wavelength range 600.0 to 605.0 nm.
SOLUTION
IDENTIFY and SET UP: This question concerns the power emitted
by a blackbody over a narrow range of wavelengths, and so
involves the spectral emittance given by the Planck radiation
law, Eq. (39.24). This requires that we find the area under the
curve between 600.0 and 605.0 nm. We’ll approximatethis area as
the product of the height of the curve at the median wavelength
and the width of the interval, From
Example 39.7, K.
EXECUTE:To obtain the height of the curve at
we first evaluate the quantity
in Eq. (39.24) and then substitute the result into Eq. (39.24):
hc
lkT
=
16.626*10
-34
J#
s212.998*10
8
m>s2
16.025*10
-7
m211.381*10
-23
J>K215800 K2
=4.116
hc>lkT6.025*10
-7
m,602.5 nm=
l=I1l2
T=5800
¢l=5.0 nm.l=602.5 nm
I1l2
I1l2
The intensity in the 5.0-nm range from 600.0 to 605.0 nm is then
approximately
EVALUATE: In part (b) of Example 39.7, we found the power
radiated per unit area by the sun at allwavelengths to be
; here we have found that the power radiated per
unitarea in the wavelength range from 600 to 605 nm is
about 0.6% of the total.0.39 MW>m
2
,
I1l2¢l=
64.2 MW>m
2
I=
=3.9*10
5
W>m
2
=0.39 MW>m
2
I1l2¢l=17.81*10
13
W>m
3
215.0*10
-9
m2
=7.81*10
13
W>m
3
I1l2=
2p16.626*10
-34
J#
s212.998*10
8
m>s2
2
16.025*10
-7
m2
5
1e
4.116
-12
Test Your Understanding of Section 39.5 (a) Does a blackbody at 2000 K
emit x rays? (b) Does it emit radio waves? !

6. 39.6The Uncertainty Principle Revisited
The discovery of the dual wave–particle nature of matter forces us to reevaluate
the kinematic language we use to describe the position and motion of a particle.
In classical Newtonian mechanics we think of a particle as a point. We can
describe its location and state of motion at any instant with three spatial coordi-
nates and three components of velocity. But because matter also has a wave
aspect, when we look at the behavior on a small enough scale—comparable to
the de Broglie wavelength of the particle—we can no longer use the Newtonian
description. Certainly no Newtonian particle would undergo diffraction like elec-
trons do (Section 39.1).
1314CHAPTER 39Particles Behaving as Waves
Example 39.7Light from the sun
To a good approximation, the sun’s surface is a blackbody with a
surface temperature of 5800 K. (We are ignoring the absorption
produced by the sun’s atmosphere, shown in Fig. 39.9.) (a) At what
wavelength does the sun emit most strongly? (b) What is the total
radiated power per unit surface area?
SOLUTION
IDENTIFY and SET UP: Our target variables are the peak-intensity
wavelength and the radiated power per area I.Hence we’ll use
the Wien displacement law, Eq. (39.21) (which relates to the
blackbody temperature T), and the Stefan–Boltzmann law, Eq.
(39.19) (which relates to T).
EXECUTE:(a) From Eq. (39.21),
=0.500*10
-6
m=500 nm
l
m=
2.90*10
-3
m#
K
T
=
2.90*10
-3
m#
K
5800 K
I
l
m
l
m
(b) From Eq. (39.19),
EVALUATE: The 500-nm wavelength found in part (a) is near the
middle of the visible spectrum. This should not be a surprise: The
human eye evolved to take maximum advantage of natural light.
The enormous value found in part (b) is the
intensity at the surfaceof the sun, a sphere of radius .
When this radiated energy reaches the earth, away,
the intensity has decreased by the factor
to the still-impressive .1.4 kW>m
2
10
11
m24
2
=2.15*10
-5
316.96*10
8
m2>11.50*
1.50*10
11
m
6.96*10
8
m
I=64.2 MW>m
2
=6.42*10
7
W>m
2
=64.2 MW>m
2
I=sT
4
=15.67*10
-8
W>m
2#
K
4
215800 K2
4
Example 39.8A slice of sunlight
Find the power per unit area radiated from the sun’s surface in the
wavelength range 600.0 to 605.0 nm.
SOLUTION
IDENTIFY and SET UP: This question concerns the power emitted
by a blackbody over a narrow range of wavelengths, and so
involves the spectral emittance given by the Planck radiation
law, Eq. (39.24). This requires that we find the area under the
curve between 600.0 and 605.0 nm. We’ll approximatethis area as
the product of the height of the curve at the median wavelength
and the width of the interval, From
Example 39.7, K.
EXECUTE:To obtain the height of the curve at
we first evaluate the quantity
in Eq. (39.24) and then substitute the result into Eq. (39.24):
hc
lkT
=
16.626*10
-34
J#
s212.998*10
8
m>s2
16.025*10
-7
m211.381*10
-23
J>K215800 K2
=4.116
hc>lkT6.025*10
-7
m,602.5 nm=
l=I1l2
T=5800
¢l=5.0 nm.l=602.5 nm
I1l2
I1l2
The intensity in the 5.0-nm range from 600.0 to 605.0 nm is then
approximately
EVALUATE: In part (b) of Example 39.7, we found the power
radiated per unit area by the sun at allwavelengths to be
; here we have found that the power radiated per
unitarea in the wavelength range from 600 to 605 nm is
about 0.6% of the total.0.39 MW>m
2
,
I1l2¢l=
64.2 MW>m
2
I=
=3.9*10
5
W>m
2
=0.39 MW>m
2
I1l2¢l=17.81*10
13
W>m
3
215.0*10
-9
m2
=7.81*10
13
W>m
3
I1l2=
2p16.626*10
-34
J#
s212.998*10
8
m>s2
2
16.025*10
-7
m2
5
1e
4.116
-12
Test Your Understanding of Section 39.5 (a) Does a blackbody at 2000 K
emit x rays? (b) Does it emit radio waves? !

7. 2. Distance
Measurement of stellar parameters

8. Parallax
•The apparent displacement of a nearby object against a distant fixed
background from two different viewpoints.

9. Stellar Parallax
•The apparent position shift of a star as the Earth moves from one
side of its orbit to the other (the largest separation of two
viewpoints possibly from the Earth)

10. 5=
1
2
7+8⇒9

11. Distances to the nearer stars can be determined by parallax, the apparent
shift of a star against the background stars observed as the Earth moves
along its orbit
Stellar Parallax and Distance
Relation between a star’s distance and its parallax
!=
1
$
!=distance to a star, in parsec
$=parallax angle of that star, in arcseconds
1$%=3.26*+
1$%=206,265/0=3.09×10
:;
34

12. •All known star have parallax angles less than one arcsec(1”),
meaning their distance more than 1 parsec (pc)
•Stellar parallaxes can only be measured for stars within a few
hundred parsecs (~100000stars)
•The closest star Proxima Centauri (比鄰星) has a parallax angle
of 0.772 arcsec
•!=
:
<
⇒!=
:
=.??@ABCDEC
⇒!=1.30$%
•!=1.30pc=4.24ly
•Therefore, the closest star is 4.24 light years away
Stellar Parallax

13. 3. Stellar Temperatures and Stellar Types
Measurement of stellar parameters

14. The emi8ed spectrum of a star is largely determined by the temperature in the
outermost “surface” (photosphere), where the photos can escape a star without
further absorp?on or sca8ering.
photosphere
Blackbody radiation

15. Blackbody radiation
A blackbody reflects no light. When it is kept at temperature <, it
emits EM-waves from each unit area of the surface with a spectrum
given by,
7 Stars

(A) Blackbody radiation

A blackbody reflects no light. When it is kept at temperature T, it emits EM-waves from each
unit area of the surface with a spectrum given below.
1
12
)(
/5
2


Tkhc
B
e
hc
I
O
O
S
O
(12)

Here O is the wavelength, h is the Planck constant, kB is the
Boltzmann constant. (Recall thermodynamics).
Total power/(unit area) =
4
0
)( TdI VOO ?
f
(13),
where V = 5.67 x 10
-8
W/(m
2
K
4
)
The wavelength at which I(O) reaches maximum is
Omax = (3.0 x 10
6
)/T (nm) (14)
For a grey body
)()()())(1()( OOHOOO
BBG IIRI
(15)
Where R(O) is its reflectivity, and H(O) is called emissivity.

(B) Luminosity of a star (L)
Stars can be regarded as blackbody as far as their light emission is concerned. The total
energy per second emitted by a star is called its luminosity L. According to Eq. (13),
24
4 RTLSV
(16),
where R is the star radius and T its surface temperature.
ThelighWinWenViW\S(Po\nWing?VYecWoU)aWadiVWancer from a
star of luminosity L satisfies
2
4
)(
r
L
rS
S

(17).
So light intensity decreases with 1/r
2
. If L can somehow be
determined, then the distance r of the star can be too, since S
can always be measured.

(C) Determine distance of stars

Some basic units (for convenience):
• 1 astronomical unit (AU) = Sun-Earth distance = 1.5 x 10
11
m.
• 1 light year = 365 x 24 x 60 x 60 x C = 9.46 x 10
15
m = 63,000 AU
• 1 arc second = S /(60*60*180) = 4.85 x 10
-6
radian



The small angle formula
d=LT (18)


T d
L
!=F
+
,
!#9#=
2G
-
*
)
153
"
ℎ
.
+
)
=I+
)
CHAPTER 39SUMMARY
De Broglie waves and electron diffraction:Electrons and
other particles have wave properties. A particle’s wave-
length depends on its momentum in the same way as for
photons. A nonrelativistic electron accelerated from rest
through a potential difference has a wavelength
given by Eq. (39.3). Electron microscopes use the very
small wavelengths of fast-moving electrons to make
images with resolution thousands of times finer than is
possible with visible light. (See Examples 39.1–39.3.)
V
ba
(39.1)
(39.2)
(39.3)l=
h
p
=
h
22meV
ba
E=hƒ
l=
h
p
=
h
mv
The nuclear atom:The Rutherford scattering experiments show that most of an atom’s mass and all
of its positive charge are concentrated in a tiny, dense nucleus at the center of the atom. (See
Example 39.4.)
Atomic line spectra and energy levels:The energies of
atoms are quantized: They can have only certain definite
values, called energy levels. When an atom makes a
transition from an energy level to a lower level it
emits a photon of energy The same photon can
be absorbed by an atom in the lower energy level, which
excites the atom to the upper level. (See Example 39.5.)
E
i-E
f.
E
f,E
i
(39.5)hƒ=
hc
l
=E
i-E
f
The Bohr model:In the Bohr model of the hydrogen
atom, the permitted values of angular momentum are
integral multiples of The integer multiplier nis
called the principal quantum number for the level. The
orbital radii are proportional to and the orbital speeds
are proportional to The energy levels of the hydro-
gen atom are given by Eq. (39.15), where Ris the
Rydberg constant. (See Example 39.6.)
1 > n.
n
2
h > 2p. (39.6)
(39.8)
(39.10)
(39.9)
(39.15)
1n=1, 2, 3,Á2
E
n=-
hcR
n
2
=-
13.60 eV
n
2
v
n=
1
P
0

e
2
2nh
=
2.19*10
6
m > s
n
=n
2
15.29*10
-11
m2
r
n=P
0

n
2
h
2
pme
2
=n
2
a
0
1n=1, 2, 3,Á2
L
n=mv
nr
n=n
h
2p
Incident electron
waves in phase
Scattered electron
waves in phase
50°
d
Atom on
crystal
surface
l
Nucleus
a
E
Ef
i
hf 5 E
i
2 E
f
i
f
v
n
M, 1e
r
n
F
Proton
Electron
m, 2e
The laser:The laser operates on the principle of stimulated emission, by which many photons with
identical wavelength and phase are emitted. Laser operation requires a nonequilibrium condition
called a population inversion, in which more atoms are in a higher-energy state than are in a lower-
energy state.
A*
A*
A*
n
1
n
0
Spontaneous
emission
Blackbody radiation:The total radiated intensity (aver-
age power radiated per area) from a blackbody surface
is proportional to the fourth power of the absolute tem-
perature T. The quantity
is called the Stefan–Boltzmann constant. The wave-
length at which a blackbody radiates most strongly
is inversely proportional to T. The Planck radiation law
gives the spectral emittance (intensity per wave-
length interval in blackbody radiation). (See Examples
39.7 and 39.8.)
I1l2
l
m
s=5.67*10
-8
W > m
2#
K
4
(Stefan–Boltzmann law)(39.19)
(Wien displacement law)(39.21)
(Planck radiation law)(39.24)
I1l2=
2phc
2
l
5
1e
hc > lkT
-12
l
mT=2.90*10
-3
m#
K
I=sT
4
0123456
1
2
3
4
2000 K
1750 K
1250 K
l (mm)
I(l) (10
11
W/m
3
)
1318

16. A starʼs color depends on its surface temperature

17. Photometry, Filters and Color Ratios
•Photometry measures the apparent brightness of a star
•Standard filters, such as U (Ultraviolet), B (Blue) and V (Visual, yellow-green)
filters,
•Color ratios of a star are the ratios of brightness values obtained through different
filters
•These ratios are a good measure of the starʼs surface temperature; this is an easy way
to get temperature
J
/J
&

18. Surface temperature and Color Ratio

20. The spectra reveal more about stars
•Stellar spectroscopy
•Stars are classified into
different spectral types
according to the strength or
weakness of the hydrogen
Balmer lines in the starʼs
spectrum

21. Stellar SpectrumK
0:M=2→M=3

22. Classic Spectral Types
•O B A F G K M
•(Oh, Be A Fine Girl, Kiss Me!) (mnemonic)
•Spectral type is directly related to surface temperature
Example: For the coolest star (type M), all hydrogen atoms are in
the ground state. There is no ==2excited hydrogen for the >
P.
•From O to M, the temperature decreases
•O type, the hoSest, blue color, Temp ~ 25000 K
•M type, the coolest, red color, Temp ~ 3000 K
•Sub-classes, e.g. B0, B1…B9, A0, A1…A9
•The Sun is a G2 type of star (temp. 5800 K)

23. Classic Spectral Types
The spectral class and type of a star is directly related to its
surface temperature: O stars are the hottest and M stars are
the coolest

24. More Types: Brown dwarf stars
•Most brown dwarfs are in even cooler spectral classes
called L and T
•Unlike true stars, brown dwarfs are too small to sustain
thermonuclear fusion

25. Luminosity, Radius, and Surface Temperature
%=4()
!
*+
"
•>= star’s luminosity, in watts (i.e. total power radiated by a star)
•?= star’s radius, in meters
•9= Stefan-Boltzmann constant = 5.67×10
*+
Wm
*!
K
*(
.
•-=star’s surface temperature, in Kelvins.

26. 4. Mass
Measurement of stellar parameters

27. Binary Systems and Measurement of Mass
Astrometric binary
Spectroscopic binary
basicastro4 October 26, 2006
24 CHAPTER 2
Figure 2.9Left:A binary system, viewed pole-on, with its members in circular orbits with
physical radiir1andr2around their common center of mass.Right:The ap-
pearance of the system when viewed as a visual binary, with the orbital plane
inclined by an angleito the line of sight, and orbital radii subtending angles on
the skyθ1andθ2. The circular orbits now appear as ellipses, with minor axes
foreshortened bycosi.
circular orbits. The center of mass of two spherical masses is at the point between
them where
r1M1=r2M2, (2.30)
withM1andM2being the masses andr1andr2their respective distances to the
center of mass (see Fig. 2.9, left). Thus, ifa=r1+r2is the separation between
the masses,
r1=
M2
M1
(a−r1), (2.31)
or
r1=
M2
M1+M2
a, (2.32)
and
r2=
M1
M1+M2
a. (2.33)
Each of the masses is subject to the mutual gravitational attraction, and as a result
orbits the center of mass with an angular frequencyω. The equation of motion for
the first mass is then
M1ω
2
r1=
GM1M2
a
2
, (2.34)
withGthe gravitational constant. After substitution ofr1from Eq. 2.32, this be-
comes Kepler’s Law:
ω
2
=
G(M1+M2)
a
3
. (2.35)
A simple example in which Kepler’s law can be used to determine a stellar mass is
in the case of the Sun. The mass of the Earth is negligible compared to the Sun, so
M⊙≈
ω
2
a
3
G
=
4π
2
a
3
τ
2
G
, (2.36)
a
C
!
C
#

28. Spectroscopic binary

29. Two-body problem
basicastro4 October 26, 2006
24 CHAPTER 2
Figure 2.9Left:A binary system, viewed pole-on, with its members in circular orbits with
physical radiir1andr2around their common center of mass.Right:The ap-
pearance of the system when viewed as a visual binary, with the orbital plane
inclined by an angleito the line of sight, and orbital radii subtending angles on
the skyθ1andθ2. The circular orbits now appear as ellipses, with minor axes
foreshortened bycosi.
circular orbits. The center of mass of two spherical masses is at the point between
them where
r1M1=r2M2, (2.30)
withM1andM2being the masses andr1andr2their respective distances to the
center of mass (see Fig. 2.9, left). Thus, ifa=r1+r2is the separation between
the masses,
r1=
M2
M1
(a−r1), (2.31)
or
r1=
M2
M1+M2
a, (2.32)
and
r2=
M1
M1+M2
a. (2.33)
Each of the masses is subject to the mutual gravitational attraction, and as a result
orbits the center of mass with an angular frequencyω. The equation of motion for
the first mass is then
M1ω
2
r1=
GM1M2
a
2
, (2.34)
withGthe gravitational constant. After substitution ofr1from Eq. 2.32, this be-
comes Kepler’s Law:
ω
2
=
G(M1+M2)
a
3
. (2.35)
A simple example in which Kepler’s law can be used to determine a stellar mass is
in the case of the Sun. The mass of the Earth is negligible compared to the Sun, so
M⊙≈
ω
2
a
3
G
=
4π
2
a
3
τ
2
G
, (2.36)
basicastro4 October 26, 2006
24 CHAPTER 2
Figure 2.9Left:A binary system, viewed pole-on, with its members in circular orbits with
physical radiir1andr2around their common center of mass.Right:The ap-
pearance of the system when viewed as a visual binary, with the orbital plane
inclined by an angleito the line of sight, and orbital radii subtending angles on
the skyθ1andθ2. The circular orbits now appear as ellipses, with minor axes
foreshortened bycosi.
circular orbits. The center of mass of two spherical masses is at the point between
them where
r1M1=r2M2, (2.30)
withM1andM2being the masses andr1andr2their respective distances to the
center of mass (see Fig. 2.9, left). Thus, ifa=r1+r2is the separation between
the masses,
r1=
M2
M1
(a−r1), (2.31)
or
r1=
M2
M1+M2
a, (2.32)
and
r2=
M1
M1+M2
a. (2.33)
Each of the masses is subject to the mutual gravitational attraction, and as a result
orbits the center of mass with an angular frequencyω. The equation of motion for
the first mass is then
M1ω
2
r1=
GM1M2
a
2
, (2.34)
withGthe gravitational constant. After substitution ofr1from Eq. 2.32, this be-
comes Kepler’s Law:
ω
2
=
G(M1+M2)
a
3
. (2.35)
A simple example in which Kepler’s law can be used to determine a stellar mass is
in the case of the Sun. The mass of the Earth is negligible compared to the Sun, so
M⊙≈
ω
2
a
3
G
=
4π
2
a
3
τ
2
G
, (2.36)
basicastro4 October 26, 2006
24 CHAPTER 2
Figure 2.9Left:A binary system, viewed pole-on, with its members in circular orbits with
physical radiir1andr2around their common center of mass.Right:The ap-
pearance of the system when viewed as a visual binary, with the orbital plane
inclined by an angleito the line of sight, and orbital radii subtending angles on
the skyθ1andθ2. The circular orbits now appear as ellipses, with minor axes
foreshortened bycosi.
circular orbits. The center of mass of two spherical masses is at the point between
them where
r1M1=r2M2, (2.30)
withM1andM2being the masses andr1andr2their respective distances to the
center of mass (see Fig. 2.9, left). Thus, ifa=r1+r2is the separation between
the masses,
r1=
M2
M1
(a−r1), (2.31)
or
r1=
M2
M1+M2
a, (2.32)
and
r2=
M1
M1+M2
a. (2.33)
Each of the masses is subject to the mutual gravitational attraction, and as a result
orbits the center of mass with an angular frequencyω. The equation of motion for
the first mass is then
M1ω
2
r1=
GM1M2
a
2
, (2.34)
withGthe gravitational constant. After substitution ofr1from Eq. 2.32, this be-
comes Kepler’s Law:
ω
2
=
G(M1+M2)
a
3
. (2.35)
A simple example in which Kepler’s law can be used to determine a stellar mass is
in the case of the Sun. The mass of the Earth is negligible compared to the Sun, so
M⊙≈
ω
2
a
3
G
=
4π
2
a
3
τ
2
G
, (2.36)
basicastro4 October 26, 2006
24 CHAPTER 2
Figure 2.9Left:A binary system, viewed pole-on, with its members in circular orbits with
physical radiir1andr2around their common center of mass.Right:The ap-
pearance of the system when viewed as a visual binary, with the orbital plane
inclined by an angleito the line of sight, and orbital radii subtending angles on
the skyθ1andθ2. The circular orbits now appear as ellipses, with minor axes
foreshortened bycosi.
circular orbits. The center of mass of two spherical masses is at the point between
them where
r1M1=r2M2, (2.30)
withM1andM2being the masses andr1andr2their respective distances to the
center of mass (see Fig. 2.9, left). Thus, ifa=r1+r2is the separation between
the masses,
r1=
M2
M1
(a−r1), (2.31)
or
r1=
M2
M1+M2
a, (2.32)
and
r2=
M1
M1+M2
a. (2.33)
Each of the masses is subject to the mutual gravitational attraction, and as a result
orbits the center of mass with an angular frequencyω. The equation of motion for
the first mass is then
M1ω
2
r1=
GM1M2
a
2
, (2.34)
withGthe gravitational constant. After substitution ofr1from Eq. 2.32, this be-
comes Kepler’s Law:
ω
2
=
G(M1+M2)
a
3
. (2.35)
A simple example in which Kepler’s law can be used to determine a stellar mass is
in the case of the Sun. The mass of the Earth is negligible compared to the Sun, so
M⊙≈
ω
2
a
3
G
=
4π
2
a
3
τ
2
G
, (2.36)
basicastro4 October 26, 2006
24 CHAPTER 2
Figure 2.9Left:A binary system, viewed pole-on, with its members in circular orbits with
physical radiir1andr2around their common center of mass.Right:The ap-
pearance of the system when viewed as a visual binary, with the orbital plane
inclined by an angleito the line of sight, and orbital radii subtending angles on
the skyθ1andθ2. The circular orbits now appear as ellipses, with minor axes
foreshortened bycosi.
circular orbits. The center of mass of two spherical masses is at the point between
them where
r1M1=r2M2, (2.30)
withM1andM2being the masses andr1andr2their respective distances to the
center of mass (see Fig. 2.9, left). Thus, ifa=r1+r2is the separation between
the masses,
r1=
M2
M1
(a−r1), (2.31)
or
r1=
M2
M1+M2
a, (2.32)
and
r2=
M1
M1+M2
a. (2.33)
Each of the masses is subject to the mutual gravitational attraction, and as a result
orbits the center of mass with an angular frequencyω. The equation of motion for
the first mass is then
M1ω
2
r1=
GM1M2
a
2
, (2.34)
withGthe gravitational constant. After substitution ofr1from Eq. 2.32, this be-
comes Kepler’s Law:
ω
2
=
G(M1+M2)
a
3
. (2.35)
A simple example in which Kepler’s law can be used to determine a stellar mass is
in the case of the Sun. The mass of the Earth is negligible compared to the Sun, so
M⊙≈
ω
2
a
3
G
=
4π
2
a
3
τ
2
G
, (2.36)
a
Two stars orbit around their center of mass
We have
Newton's 2
nd
law

30. Binary star systems: stellar masses
•The masses can be computed from measurements of the orbital period and
orbital size of the system
•The mass ratio of D
1and D
2is inversely proportional to the distance of
stars to the center of mass
?
:+?
@=
4A
@
B
C
;
<
@
This formula is a generalized format of Kepler’s 3
rd
law
a = semimajor axis of one star’s orbit around the other
T= orbital periods

31. We can determine the mass D
1,D
2individually if we can measure F
#,F
!and -
(visual binary star)
In other cases, there are binary star that is spa5ally unresolved but revealed as a
binary by its spectrum (spectroscopic binary)
•we can measure their orbital veloci5es via the Doppler effect and
C=
24F
-
•Hence by measuring C
1,C
2and -, we can get F
#,F
!and hence the masses.
basicastro4 October 26, 2006
STARS: BASIC OBSERVATIONS 23
Figure 2.7 Schematic view of an eclipsing binary system (left), and its total brightness as a
function of time (right). Numbers indicate the corresponding points on the orbit
and in the so-called “light curve”.
Figure 2.8 Schematic example of the spectrum of a “double-lined spectroscopic binary”.
Each of the absorption lines in the spectrum appears twice, Doppler-shifted to
longer and shorter wavelengths, respectively, as a result of the orbital motion of
the binary members about their center of mass. During the orbital period, each
absorption line oscillates back and forth about the restframe wavelengthλ0.
spatially unresolved) is inclined enough (i.e., close enough to edge-on) to our line
of sight that each of the members periodically eclipses the other. The presence of a
binary will then be revealed if the light from the system is monitored as a function
of time. During each orbital period, the brightness of the system will undergo two
“dips” (see Fig. 2.7), each corresponding to the eclipse of one star by the other. The
depth of the dips will depend on the relative sizes and luminosities of the two stars.
Aspectroscopic binaryis a spatially unresolved pair that is revealed as a binary
by its spectrum. For example, the observed photospheric absorption spectrum may
be the superposition of the spectra of two different types of stars. Alternatively,
even if the members are of the same type, their orbital velocities,v, may cause
large enough Doppler shifts,∆λ/λ=v/c, in the wavelengths of absorption, to
produce distinct lines, with shifts that oscillate periodically during each orbit (see
Fig. 2.8). Sometimes, one of the members may be too faint, or devoid of strong
absorption lines, to be detectable in the combined spectrum, but its presence will
still be revealed by the periodically changing Doppler shifts of the brighter star.
In fact, this is the very method by which planets orbiting other stars have been
discovered in recent years (see below).
To see how binaries sometimes allow stellar mass determination, let us review
some aspects of the Keplerian two-body problem, in which two masses orbit their
common center of mass in elliptical trajectories. For simplicity, let us consider only

32. 5. The Hertzsprung-Russell Diagram
HR Diagram

33. •The HR diagram is a graph plotting
the absolute magnitudes of stars
against their spectral types—or,
equivalently, their luminosities
against surface temperatures
•Luminosity is the total power
radiated by a star
D=4AE
@
F<
Q
•There are patterns
HR diagram

34. •The size can be denoted by
dotted lines:
0.001E
!"#GH1000E
!"#
D=4AE
@
F<
Q
HR diagram

35. •Main Sequence: the band stretching diagonally
from top-left (high luminosity and high surface
temperature) to bottom-right (low luminosity
and low surface temperature)
•90% stars in this band
•The Sun is one of main sequence stars
•Hydrogen burning as energy source
•Cool stars are faint and small; hot stars
are bright and large
HR diagram

36. •Giants
•upper-right side
•Luminous (100 –1000 L
sun)
•Cool (3000 to 6000 K)
•Large size (10 –100 R
sun)
•Supergiants
•Most upper-right side
•Luminous (10000 -100000 L
sun)
•Cool (3000 to 6000 K)
•Huge (1000 R
sun)
•White Dwarfs
•Lower-middle
•Dim (0.01 L
sun)
•Hot (10000 K)
•Small (0.01 R
sun)
HR diagram

37. Mass-Luminosity Relation for Main-Sequence Stars
The greater the mass of a
main-sequence star, the
greater its luminosity.
Roughly speaking,
D∝?
P
where J≈3−5.

38. •Masses from 0.2 M
Θto 60 M
Θ
•The greater the mass the greater
the luminosity
•The greater the surface
temperature the greater the
radius
Mass-Luminosity Relation for Main-Sequence Stars

39. 6. The Birth of Stars -Nebula

40. Guiding Ques=ons
1.Why do astronomers think that stars evolve?
2.What kind of maSer exists in the spaces between the
stars?
3.In what kind of nebulae do new stars form?
4.What steps are involved in forming a star like the Sun?
5.When a star forms, why does it end up with only a
frac<on of the available maSer?
6.What do star clusters tell us about the forma<on of stars?
7.Where in the Galaxy does star forma<on take place?
8.How can the death of one star trigger the birth of many
other stars?

41. Stars Evolve
•Stars shine by thermonuclear reactions
•They have a finite life, because the hydrogen fuel will be
exhausted
•A year to a star is like a second to a human
•To understand the evolution of stars, the approach is to piece
together the information gathered for many stars that are at
different evolution stages

42. Interstellar Medium and Nebulae
•The space between stars is filled with a thin gas and dust
particles
•Interstellar gas and dust pervade the Galaxy
•Nebula: a cloud of concentrated interstellar gas and dust; 10
4
to 10
9
particles per cubic centimeter

43. Emission Nebula or H II region
•Emission nebulae are glowing
clouds of gas
•They are found near hot,
luminous stars of spectral types
O and B
•They are powered by ultraviolet
light that they absorb from
nearby hot stars
•They are composed of ionized
hydrogen atoms; the so calledH
II region.
•They emit light through a process
called recombination: free
electrons get back to form
neutron hydrogen; similar to
fluorescence
•They glow red (Hαemission)

44. Dark Nebula
•Dark nebulae are so
dense that they are
opaque
•They appear as dark blobs
against a background of
distant stars

45. Reflection Nebulae: dust scattering
•Reflection nebulae are produced
when starlight is reflected from dust
grains in the interstellar medium,
producing a characteristic bluish
glow
•Short wavelength blue lights are
scattered more efficient that red
lights (like our blue sky)

46. Interstellar Extinction
•Remote stars seem to be dimmer than would be expected from their
distance alone
•extinctionis the absorption and scattering of EM radiation by dust and
gas between stars and the observer

47. Interstellar Reddening
•Reddening depends on distance; the more distant, the redder

48. Interstellar Reddening
•Remote stars are also reddened as they pass through the
interstellar medium, because the blue component of their
star light is scaSered and absorbed by interstellar dust

49. Rayleigh scatteringRayleigh scattering
•  Scattering of light off air molecules is called Rayleigh Scattering
•  Involves particles much smaller than the wavelength of incident light
•  Responsible for the blue color of clear sky
Atmospheric composition: N
2
(78%), O
2
(21%), Ar (1%)
Size of N
2
molecule: 0.31 nm
Size of O
2
molecule: 0.29 nm
Size of Ar molecule: 0.3 nm
Visible wavelengths ~400-700 nm

50. Distribution of Interstellar Gas and Dust
•The interstellar gas and dust are confined to the plane of the galaxy

51. Protostars form in cold, dark nebulae
•Protostar: the clump formed
from dense and cold nebula
under gravitational contraction
•The protostar contracts, because
the pressure inside is too low to
support all the mass.
•As a protostar grows by the
gravitational accretion of gases,
Kelvin-Helmholtz contraction
causes it to heat and begin
glowing

52. Protostarsform in cold, dark nebulae
Bokglobulesareisolatedandrela?velysmalldarknebulae,containingdensecosmic
dustandgasfromwhichstarforma?onmaytakeplace

53. The Orion Nebula
•Observing in the infrared, we see many new young stars in the
nebula still enshrouded in clouds of gas and dust. The visible light
still cannot pass through the clouds, however.
•Behind the Orion Nebula lies a great molecular cloud where more
stars are being produced.
53
Visible light
image
Infrared
image

54. Eagle Nebula
Radiation from bright
stars evaporating the
dust and expose the
evaporating gaseous
globules (EGGs)
containing protostars
A newborn star

55. 7. Protostar

56. The Road Towards Stardom
•When a gas cloud contracts, gravity
accelerates the molecules. This heats up the
gas cloud. A protostaris formed, yetremains
hidden in the dust cloud.
•When a protostar becomes hot enough in the
core, nuclear fusionstarts.
•Then it becomes a star.
•The energy produced causes the star to blow
away the surrounding cloud of gas and dust –
the cocoon.
•At the birth line, the star becomes visible.
•The process is faster for stars with more mass.
56

57. Dynamics of Gravitational Contraction
Assume spherical distribution, no rotation. Then we can derive
(next slide)
MN
!
@
N
!G
@
=−MNON+−
!P
!N
This shows that the gas will:
•contract when the gravitational field is stronger than the
pressure gradient;
•expand when the pressure gradient is stronger than the
gravitational field;
•reach equilibriumwhen the gravitational field is equal to the
pressure gradient.
57

58. Gravity vs Pressure gradient
58
rr+D
r
()Pr AD
()Pr r A+D D
mD
[ ]
dr
dP
r
rGm
V
m
dt
rd
V
m
Ar
dr
dP
r
mrGm
dt
rd
m
ArrPrP
r
mrGm
dt
rd
m
-
D
D
-=
D
D
Þ
DD-
D
-=DÞ
DD+-+
D
-=D
22
2
22
2
22
2
)(
)(
)()(
)(
GF
)
!
F
)H
!
=−GFIF+−
)K
)F

59. Stage 1: Free Fall
59
We first consider the case that there is nopressure.
Then the masses collapse towards the center like in a free fall.
This is a good approximation at the early stage of star formation.
Consider a spherical shell with radius renclosing mass m
0.
Assume m
0constant and r= r
0initially.
By energy conservation,
1
2
#$
#%
!
=
'(
"
$
−
'(
"
$
"
Time for free fall to the center
%##=*
$3
"
#%
#$
#$=*
"
$3
2'(
"
$
−
2'(
"
$
"
%&/!
#$

60. Free-Fall Time
60
Let +=$/$". Then
%##=
$
"
(
2'(
"
*
"
&
+
1−+
&/!
#+
Let +=sin
!
0. Then the integral becomes
)
!
and
%##=
1
!
$
"
(
8'(
"
The Gme depends on ("/$
"
(
. Let 3=("/(41$
"
(
/3)be the average
density of the maHer enclosed. Then
%##=
31
32'3

61. Application to Interstellar Cloud
61
When an interstellar cloud of molecular hydrogen collapses at an early stage,
energy is dissipated by:
•Radiation(it is transparent)
•Dissociationof hydrogen molecules into hydrogen atoms
•Ionizationof hydrogen atoms into protons and electrons
Since the energy dissipated in the form of random thermal motion is still
negligible, the effect of temperature and pressure can be neglected. The
free-fall approximation is relevant.
Remark: After hydrogen has been ionized, the cloud becomes opaque.
Radiation is trapped as internal thermal motion
Þtemperature and pressure rises
The cloud will approach hydrostatic equilibrium.

62. Stage 2: Hydrostatic Equilibrium
62
In hydrostatic equilibrium, we can derive
the virial theoremrelating pressure with
energy density:
8=−
1
3
9*+
:
where
8= volume-averaged pressure
9
*+= gravitational potential energy
9
*+=−*
"
,
'($
$
#(
:=volume

63. Deriva$on of the VirialTheorem
63
At equilibrium,
-
4
$
-.
4
=0. The pressure gradient is balanced by the
gravitational field:
−
#8
#$
=3$
'($
$
!
Multiplying both sides by 41$
(
and integrating,
−*
"
+
41$
(
#8
#$
#$=*
"
+
'($3$
$
!
41$
(
#$
Integrating by parts, and identifying #(=3$41$
!
#$,
−41$
(
<8
"
+
+3*
"
+
8$41$
!
#$=*
"
,
'($
$
#(
⇒38:=−9*+

64. Physical Origin of the Pressure
64
The pressure originates from the thermal motion of the particles.
Kinetic theory (see next slide for a review):
8=
?
3
〈⃗B⋅⃗D〉
where ?is the number density.
Valid for classical and quantum gases of non-relativistic and relativistic
particles.
Consider interstellar clouds. Particles are classical and non-relativistic.
⃗B⋅⃗D=(D
!
=29
/0
where 9
/0=translational kinetic energy
8=
!
(
of the translational kinetic energy density
Remark: This relation is important for the stabilityof the star.

65. Review of Kinetic Theory
65
Consider a cubical box of sides Lcontaining Npar5cles.
Let ⃗C=(C
,,C
-,C
.)be the velocity components of a par5cle.
Let ⃗O=(O
,,O
-,O
.)be the momentum components of the par5cle.
Time between two collisions on a wall perpendicular to the x-axis =
!/
0#
.
Momentum change in a collision = 2p
x.
Rate of change of momentum =
!1#
!//0
#
=
1#0#
/
According to Newton’s law, force = rate of change of momentum.
Pressure on the wall:
K=
PO
,C
,
>
)
=
P⃗O⋅⃗C
3S
=
T
3
⃗O⋅⃗C
vp×===
3
1
zzyyxx
vpvpvp

66. Equilibrium for a Gas of NR Par$cles
66
Virial theorem: K=−
#
)
3$%
4
Kinetic theory: K=
!
)
3&'
4
This implies U
56+VU
78=W
For cold hydrogen atoms, there are no internal degrees of freedom.
X
9:9=X
;<+X
=3
Implications:
X
9:9=−X
=3and X
9:9=
#
!
X
;<
•If the cloud evolves slowly and remains close to hydrostatic equilibrium,
then when the total energy is lost, the kinetic energy increases Þhotter Þ
increase in pressure opposing contraction Þstable equilibrium.
•If there is nuclear fusion at the center, the total energy increases. E
KE
decreases and E
GRincreases Þthe cloud expands and cools.

67. Additional Degrees of Freedom
67
If a gas has other degrees of freedom besides translation (e.g. rotation and vibration), then
the internal energy becomes
R
56=S
T
2
*
&+
For example,
helium: f= 3 (3 translational + 0 rotational)
oxygen: f= 5 (3 translational + 2 rotational)
methane: f= 6 (3 translational + 3 rotational)
This implies U
/=
7
"
T*
&and U
8=U
/+T*
&
Adiabatic index: V=
9
'
9
(
=
7:"
7
For example, V=
-
.
for helium, V=
;
-
for oxygen, V=
)
.
for methane.
In adiabatic processes,
WX
<
=constant

68. Only part of the internal energy is kinetic energy. Among the fdegrees of freedom, only 3 of
them are kinetic. Hence
R
=>=
3
S
R
56
Since V=
7:"
7
⇒S=
"
<*!
and R
=>=
.
"
V−1R
56.
Virial theorem: W=−
!
.
>
)*
/
and kinetic theory: W=
"
.
>
+,
/
=
<*!>
-.
/
Combining,
0=R?@+3V−1R56
R(A(=R56+R?@=−3V−4R56
•Check: for hydrogen atoms, V=5/3, we recover the total energy: R
(A(=−R
56
•If V>4/3, we have 3V−4>0.
•energy loss Þinternal energy increaseÞhot Þopposing pressure Þstable equilibrium
Adiaba$c index and Equilibrium
68

69. Conditions for Gravitational Collapse
69
•Magnitude of PE >Internal KE
X
;<>X
=3
•Assume
1.Uniform temperature -
2.Consists of hydrogen
3.Radius=?, Mass =D, contains Pparticles with average mass Z$
•Rough estimation:
X
;<=−&
$
$
[$F
F
)$=−\
[D
!
?
Forroughestimation,take\=1and
X
=3=
3
2
P5-.
Wehave
⇒
[D
!
?
>
3
2
P5-

70. 70
[D
!
?
>
3
2
P5-
SinceP=
>
?$
, we have
;>
<
>
)@9
!?$
and
D>
35-
2[Z$
?≡D
A(Jeanmass)
To derive the collapse condition for averaged density, we eliminate ?,
̅G≡
D
4
3
4?
)
⇒?=
3D
44̅G
#
)
⇒D>
35-
2[Z$
3D
44̅G
#/)
̅G>
3
44
35
2[Z$
)
-
)
D
!
≡G
A(Jeandensity)
Easiertohave gravitationalcollapsewhenDis large.
Example: A cloud of molecular hydrogen with -=201
D=2×10
))
5I=1000D
BCD
G
A=10
*!!
kg⋅m
*)
≈10
'
molecules/$
)

71. 71
•Since r
Jµ1/M
2
, the condensation of gas
clouds should take place in several stages.
•First, a massive extended gas cloud contracts
(thousands of solar masses).
•Consider a local region of the gas cloud.
Since its mass is less than M, its local Jeans
density is high.
•However, when the density increases due to
contraction, it exceeds the local Jeans
density, and the local region can contract
independently.
•Ultimately, the cloud fragment into masses
comparable to solar mass Þprotostar

72. Contrac$on of a Protostar
72
A protostar contracts freely, unopposed by internal pressure, since the energy loss is
NOTconverted to random thermal motion. (temperature doesn’t increase in this
stage)
This continues until a substantial fraction of hydrogen molecules are dissociatedand
the hydrogen atoms are ionized.
PE →Dissociation of o
!molecules and ionization of H atoms
•Dissociation energy p
E=4.5eV
•Ionization energy p
F=13.6eV
•Energy needed
•Energy provided by PE when the star contracts from ?
#to ?
!(it takes about
20000yr for D=D
GCD)
[D
!
?
!
−
[D
!
?
#
≈
D
2$
H
p
E+
D
$
H
p
F
?
!≪?
#⇒?
!≈10
##
$

73. 2
2
Most hydrogen ionized Plasma Opaque
Thermal energy of and increases
Pressure increases Hydrostatic
What temperature?
3
23
2
2
By virial thm 2 0
1
1
KE
HH
GR D I
HH
KE GR
ep
MM
EkTkT
mm
GM M M
E
Rmm
EE
kT
ee
®®
®
®®
»´´ ´=
éù
»- »-+
êú
ëû
+=
»[ ]2 2.6 eV 30000K
2
Note: This temperature is independent of the mass of the protostar
and is too low for nuclear fusion
Hydrostatic? Why we have star?
DI
ee+» «
73
+
fusion
>10
;
K

74. 78
This is not a real equilibrium.
Temperature higher than surrounding Slow radiation loss
Contracts quasi-hydrostatically 10 to 10 yr
Virial theorem Half of PE loss Internal thermal KE

®
Þ®
!
Half of PE loss Radiation loss from surface
Temperature of the protostar increases
®
\
Hot enough for nuclear fusion
Energy gain by fusion
Energy loss by radiation
Hydrostatic equilibrium
Star
=
®
Cold dense gas
Supported by degenerate electrons
Temperature low, no fusion
Brown dwarf
74
»
h
BC+ih
DE=j
⇒kl
FG=−
m
i
kl
HI

75. Condi$on for Stardom
75
Besides hydrostatic pressure, gravitational contraction can also be opposed by a cold,
dense gas of electrons, since they exhibit quantum mechanical properties. At low
temperature they obey Pauli exclusion principle and are known as a degenerategas.
V/N> V
Q
V/N»V
Q
Quantum mechanical de Broglie wavelength
#=
ℎ
5
Attemperature+,
5
"
2n
J
≈
1
2
*+⇒5=n*+⇒#=
ℎ
n*+
Classical mechanics is valid when the average
separation between particles ois much longer than the
de Broglie wavelength,
o
.
=
pn
q
≫#
.
No significant overlap between electron wavefunctions.

76. 76
an es'mate for the maximum
average internal temperature
reached by a contrac'ng
protostar
Classical mechanics valid when q≪
KL
%
/
≈pn
L
0'(
/
1
#
/
pn=average mass of a particle =0.5amu
R
?@=−
rs
"
t
,R
=>=
3
2
T*+
And thevirialtheoremgives
*+≈
rspn
3t
≈rpns
"
.q
!
.
Therefore, classical mechanics breaks down when
*+≳
r
"
pn
M
.n
J
ℎ
"
s
)/.
For nuclear fusion to start, consider whether the temperature can reach 10
;
v
Þminimum mass required to ignite fusion ≈0.08s
OPQ
If mass too small, contraction is opposed by electron degeneracy pressure before fusion Þ
brown dwarf