Download Chapter three: The properties of Stars

Chapter three: The properties of Stars
When we look up into the sky in a clear night, all of the stars locate at the
inner surface of a sphere called celestial sphere and they seem to be at same
distance from us. However this is just a projection effect. For the stars we can
see with our unaided eyes, the nearest star (Alpha Centauri) is about 4.365
light years whilst the furthest (stars in Andromeda galaxy) is 2.5 million light
years away. The closest star to the Solar System is Proxima Cen at 4.242 light
years. To tell how far a star is away, i.e. to measure the stellar distance
accurately is one of the most important tasks for astronomers.
The stellar distance can be measured by different methods, a very simple and
direct method is called parallax.
3.1 Measuring Stellar distances
3.1.1 What is parallax?
Now let’s try a simple experiment to decide the distance a pencil from you, do
the following steps:
 Hold out a pencil at your arm's length;
 Close each eye in turn, take the white board as a background, you find
the position of the pencil moved a distance on the white board.
d 
D
tan P
d
 You can calculate the distance d of the pencil from you by measuring half
of the angle P, subtended by half of the displacement of the position of the
pencil on the white board, caused by the two different views.
D is half the distance between your two eyes – which defines the baseline of
the distance measurement.
This is to say, to measure the distance of a remote object, we need to
measure the object from two positions some distance apart along a
baseline, then measure the change of the position of the object from two
1
different measurement referring to a background. This method of
measuring distances is called PARALLAX.
The further away the object, the greater the baseline is required. In the case of
stars, there is no perceptible change in angle from points across the Earth.
Therefore a considerable bigger baseline is required. Fortunately, there is one
for us—the distance across the Earth’s orbit around the Sun. Which requires
us to measure the distance of the Earth from the Sun – the Astronomical Unit.
3.1.2. Measuring the Astronomical Unit
Remember the 3rd Law of Kepler:
Where:
=
(
4
+
)
T – Period of Planet
a – Semi Major axis of Planet
G – gravitational constant
M – Mass of star (e.g. 1 Solar Mass; ~2*1030kg)
m – Mass of Planet (e.g. mjup=2*1027kg=1/1000MSUN)
Thus, for all planets and minor objects, their relative distances from the Sun
can be estimated from their easy to measure periods using:
=
Hence, all that is needed is ONE direct measurement of a distance in the Solar
System to determine all other distance and of course the Astronomical Unit.
- Early Attempts: Measure the angle between the Sun and the Half Moon to
triangulate the Sun’s distance. Requires to know the distance to the Moon
which can be measured via parallax. However: This is very tricky as the
angle between Sun and Half Moon is 89.853 degrees!
- Transit of Venus: Measure the parallax of Venus during one of its transits
of the Sun. Very difficult as it happens only every 110 yrs (two transits
2
within 8yrs) and measurement issues such as the black drop effect.
Transits of Mercury are more frequent, but it’s parallax angle (<30”) is
too small to obtain reliable results.
- Near Earth Asteroids: Direct measurements of the parallax of these
objects are possible when they are close to Earth. E.g. 433 Eros was used
in the early 1900th.
- Direct Radar travel time measurements are used since the 1960s. Radar
is bounced of the surface of Mars and Venus and the travel time and
known speed of light is used to estimate distances.
- Finally: As the above methods have reached very high precision: The
Astronomical Unit has now been defined as:
1AU = 149,597,870,700m
3.1.3 Astronomical use of Parallax
The two measurements are done with a 6 months gap. Though the majority of
stars are too far away to show any change in their observed positions from the
above two opposite positions. They can be taken as reference points, against
which, the movement in position of a relatively few nearby stars can be
measured, as shown in the above right figure.
The angular change in their position relative to their background stars,
coupled with our knowledge of the Earth’s orbit, can be used to find their
distance.
The definition of the Stellar Parallax: the angle subtended at a star by the
mean radius of the Earth's orbit around the Sun.
3
For the star in the above Figure, the parallax angle P is half the distance moved
by the star between the two photos.
P = 0.5’’ / 2 = 0.25’’ ( arc seconds)
The distance between the Sun and the star is : d = r / tan P ~ r / P, (|P| << 1)
where P is in radian, d is in the same unit as r. For the situation in the figure:
d = 1 AU / P
Because P is usually very small, and given by astronomers in arc second (‘’)or
less. We have to convert arc seconds P’’ to radian to use the above formula,
360o
360o
1 radian 
degree 
 3600' ' (arcseconds)
2
2
 2.06 105 ' ' (arcseconds)
1 ' ' arcseconds 
1
P' '
(
radian
)

P
'
'

(radian )
5
5
2.06  10
2.06  10
1 AU
d

P (in radian )
2.063  10 5  15  10 7 ( km )

p' '
( radian )
1 AU
P ''
2.06  10 5
1 pc  2.063  10 5  15  10 7 ( km )  3.086  1018 cm
if
let
d
1
( pc )
p' '
(3.1)
For the star in the previous figure, we can find : d = 1 / p’’(in arcseconds) = 1 /
0.25’’ = 4 pc , Therefore the star is four parsecs away .
3.1.3 About the parsec
From Eq.(3.1), we see that, when P=1’’, d = 1pc , 1pc = 3.086 1013km is the
distance of the object when it subtends a parallax angle of one arc second
to the observer on the surface of the Earth.
We can also find the relation between the parsec and another astronomical
length unit, the light year:
4
one
year  c 
light
one year
 3  10 5 ( km/s )  365  24  3600 ( s )  9 . 46  10 12 km
i.e., 1 pc = 3.086 1013km / 9.461012 km
=
3.26 light year.
Astronomers do NOT use light years!
Parallax will only work out to a few hundred (with GAIA a few 1000) parsec.
All other, larger distances have to be measured using objects of known
Luminosity and using:
=
4
This will be covered in next year’s course, PH507.
Knowing the distance of stars can only tell how far they are from us. How we
measure their other physical properties is the key task of modern astronomy.
The earliest astronomers observe stars by recording the their brightness,
obtained by using unaided eyes.
3.2 Stellar magnitudes
3.2.1 Catalogue of stars, apparent magnitude m
Efforts to quantify the brightness of stars dated back to the Greeks.
The brightest
m(mag.) =
B =
1
2.5125B6
2
3
4
The faintest star
5
6
2.5124B6 2.5123B6 2.5122B6 2.512B6 B6
160-130BC : Hipparchus grouped the stars visible to unaided eyes into six
magnitude groups, with the brightest the 1st magnitude (m) and the faintest,
the 6th magnitude, where m is the apparent magnitude and in units of mag..
1800-: Accurate measurement of stellar brightness found:
a) stars of a given magnitude m were approximately 2.512 times
brighter than those of the next fainter of magnitude m+1; i.e.,
B1 = 2.5125 B6 5 100
m
=1, brightness ratio =2.512;
b) stars of m =1 were about 100 times brighter than those of
m = 6, i.e.,
Which shows that the human eye has a logarithmic instead of linear response
to light.
1854: Norman Pogson: Extended the above result to calculate the
brightness ratio of any two stars of apparent magnitude difference of m= m1m2, by the following equation:
= 2.512
(3.2)
The brightness ratio between two stars whose apparent magnitude differs by
one magnitude is 2.512. Equation (3.2) can also be written in a better way as:
B2
 2 . 512
B1
 10
2
5 ( m1  m 2 )
5
( m1  m 2 )
5


( m1  m 2 )
5
( 2 . 512 5 )
( m1  m 2 )
10 2 . 5
 100
( m1  m 2 )
5
(3 .3)
Modern times: This apparent magnitude scale has been extended in both
directions, from m= -26.81 for the Sun to approximately m=29 for the faintest
object detectable. Vega has an apparent magnitude of 0.
*
One point to note is that the nominal apparent magnitude relate to the
brightness as observed with instrumentals having the same wavelength
response as our human eyes (visual band). We will see later that we can also
6
measure the apparent magnitudes in specific wavebands, such as blue, red, etc,
so that we can know the colour of a star.
( m1,v  m 2 ,v )
( m1,v  m 2 ,v )
B2 , v
5
2 .5
 100
 10
(3 .4 )
B1, v
Examples:
E.1: The apparent magnitudes of the Sun and the full Moon are -26.7 mag. and
-12.6 mag. Find how much brighter the Sun is than the Moon.
Sol: using Eq. (3.3):
Bsun
 10
Bmoon
( m Moom  m Sun )
2 .5
12 .6  26 .7
2 .5
 10
 10 5.64  436515 .83
The Sun is about 0.44 million times brighter than the full moon.
E2: A star has a brightness 10,000 times less than Vega. What is the apparent
magnitude of the star?
From the plot, we know mVega = 0, again using Eq. (3.3):
BVega
Bstar
( mstar mVega )
 10
2.5
 10000  mstar  mVega  2.5 log10000
mstar  10 mag.
In fact, even without using the formula, we can find the apparent magnitude of
the star, since we know 100 times of the ratio of the brightness corresponds to
5 apparent magnitude difference, 10000 = 100100 should corresponds to 10
apparent magnitude in difference, which is the apparent magnitude of the star,
because the apparent magnitude of the Vega star is 0.
About the apparent magnitude m:
The observed m of a star tells us nothing about its intrinsic brightness, for a
star appears bright in the sky can be either a faint star very close to our Sun,
or a far brighter star at a greater distance.
7
Next we will investigate how to measure the intrinsic brightness of a star
which is actually measured by the radiation flux F.
3.2.2 Brightness-the radiation flux F and Luminosity L
Let L be the total amount of light energy of all wavelengths released from the
surface of a star in unit of Js-1.
F is defined as the total amount of light energy of all wavelengths that crosses
a unit area perpendicular to the direction of the light’s travel in unit of J/m2s,
measured at the surface of the star.
a) Energy flux of a star
A star of luminosity L, its energy flux at a distance r is calculated as
L
F 
(3 .5 )
4 r 2
This is F  r -2  inverse square law for light flux, with unit J m-2 s-1
26
For the Sun, L  3.910 J / s its flux at the earth’s surface is measured as:
F
3.9 1026
4  (1AU ) 2

3.9 1026 ( J / s)
4  3.14  (1.5 1011(m))2
 1380J / s m2  1380watt / m2
which is called solar constant.
b) The relationship between F and m:
Using Eq.(3.3), replace B2/B1 by F2/F1 ,
F2
100(m1 m2 ) / 5 10(m1 m2 ) / 2.5
F1
F
F
m1  m2  2.5log10 2  2.5log10 1
F1
F2
(3.6)
This is called Pogson law.
Or for a star, the apparent magnitude can also be written as:
m   2 . 5 log 10 F  K
Where K is a constant or K = mref. + log10 Fref.
The definition of the zero point (the star Vega) and the resulting consequences
of this will be discussed next year in detail in PH507.
8
Note: in Eq.(3.6) ), (m1, F1) and (m2,F2) can be the results of:
-- two different measurements for same star at different distances;
or
-- Two measurements for two different stars at same distance.
Although we have related the apparent magnitude difference of two stars to
the ratio of their energy fluxes, Eq(3.3) or (3.6) cannot tell us which star is
intrinsically brighter. If we measure the apparent magnitude and the energy
flux of all stars at a fixed distance, their perceived relative brightness would be
a true indication of their relative energy fluxes.
Astronomers used this idea to define an absolute magnitude M of a star.
c) Absolute magnitude M
M is the apparent magnitude the star would have, if it is 10pc away from the
observer on the surface of the Earth, i.e., M = m(d=10pc). Next let’s see how M
is related to m at an any distance d from the observer.
Image two apparent magnitude measurements are done for a star, whose
luminosity is L
One is when the star is at a distance d1 =d from the Earth, the apparent
magnitude is m1=m, F1=F
the other is when the star is d2=10pc away from the Earth, m2 = M, F2 =
F(10pc).
Using Eq. (3.5), we have the ratio of the two brightness’s measured when the
star is the at distances of d1 and d2 from the Earth:
L
F10
F2
d
4  (10 pc ) 2


 (
)2
L
F1
F
10 pc
2
4 d
Using Eq. (3.6),
m1  m 2  2 .5 log 10
m  M  2 .5 log 10 (
  5  5 log 10 (
d
)
pc
(a )
F2
F1
d 2
d
d
)  5 log 10 (
)  5 log 10 ( )  5 log 10 (10 )
10 pc
10 pc
pc
(3 .7 )
9
When we derive these equations, we assumed that there is nothing between
the star and the observer to absorb the energy the star radiates. In reality,
there are molecular clouds and dust to absorb the star’s energy, to make the
star look dimmer to the observer on the surface of the Earth, we use a quantity
extinction A to describe this extinctive effect of cloud/dust on star light.
Eq.(3.7) can be rewritten as:
m  M  5  5 log(
d
)  A (3.8)
pc
Furthermore, if m and M is measured in specified spectrum band, e.g., visual
band, Eq. (6.7) can be rewritten as:
d
mv  Mv  5  5log( )  Av (3.9)
pc
In the following, we will use these derived Equations to solve some questions.
d) Worked Examples:
Example 1. Sirius has an absolute visual magnitude MV = 1.4 and a distance of
2.65 pc. What is its apparent mV ?
mv - MV = 5 log10 (d/pc) - 5  mv= 5 log10 (2.65) - 5 + 1.4 = -1.48(magnitudes.)
Example 2: A star has m=5.2 and M=4.3. How far away is it from the Earth
−
= 5log(
5.2 − 4.3 = 5
0.9 + 5 = 5
5.9
=
5
10 . =
= 15.1
10
) − 5;
−5
Example 3. Two stars, A and B, are observed with a photometer, producing
27.2 and 119 counts per second respectively.
(a) If the magnitude of star A is 5.3, what is the magnitude of star B?
(b) If the two stars are at same distance from the Earth, and are very close to
each other that they cannot been distinguished in observation, what is the
magnitude of the two stars combined?
(a)
Use: m1  m 2  2.5 log 10
F2
 2.5 log 10
F1
F1
F2
(3 .6 )
m(A) - m(B) = -2.5*log10(27.2/119) = 1.6 , m(B) = 5.3-1.6 = 3.7 (magnitudes)
(b) Q1: m(AB) m(a) +m(b) why?
Because: two stars together being taken as one star, the combined star should
look brighter than any of the two, when they are taken as two independent
stars, the m(AB) should be lower in value than m(A) and m(B), according to
the relation between m and F in Eq(3.5).
Q2: m(AB) m(a) - m(b) why?
m and F are not related linearly.
Sol: What we know it is definitely correct: F(AB)=F(A)+F(B), we can take the
combined star as one object, then star A or B as the second star, to using
Eq.(3.5).
The magnitude of the two stars combined will be
m(AB) – m(B) = -2.5*log10((27.2+119)/119) = -0.2
Since m(B) = 3.7,
m(total)=-0.2 + m(B)= 3.5 (magnitudes)
m(AB) < m(A) and m(B).
Example 4. A star behind a dense massive cloud, at a distance 10kpc, has an
apparent magnitude of 25. The extinction of the foreground dust to the star is
A = 6, Find the absolute M magnitude of the star.
Sol: The relationship between apparent and absolute magnitudes is
m – M = - 5 + 5 log10(r/pc) + A
M = m + 5 - 5 log10(r/pc) - A
= 19 + 5 – 5 log10(104) = 19 + 5 - 54 = 19 + 5 - 20 = 24 – 20 =4.0 (magnitudes)
11
Next, Let’s see what other knowledge about stars we can derive from their
radiation spectrum.
3.3
The properties of the spectrum of stars
3.3.1 The blackbody radiation spectrum of stars
At any temperature T > 0, an object emits light of all wavelengths with varying
degrees of efficiency.
An ideal emitter is an object that absorbs all of the light energy incident upon it,
and reradiates this energy with the characteristic spectrum called blackbody
radiation.
The radiation intensity of a blackbody radiation vs wavelength is shown in the
following figure:
We can see:
At a fixed temperature the radiation energy density changes with wavelength
non-monotonically and reaches maximum at a wavelength which we call it as
, for which we find:
when T = 5500K, max = 527nm, maxT  5500  527  2.9  10 3 mK
when T = 3500K, max = 829nm, maxT  3500  829  2.9  10 3 mK
The relation between the two quantities which is expressed as
a) Wien’s law
maxT  2.9  103 mK (3.10)
12
It is also found that the total energy rate is proportional to the 4 th power of
the black body temperature:
L  AT 4
( 3 . 11 )
  5.67 108 JK4m2s1
where A is the surface area of the star and
Based on Eq. (3.11) and the relation between luminosity L and energy flux F
(brightness), we can find how the brightness of a star is linked to its surface
temperature. This is called:
b) Stefan-Boltzmann law:
Further we can derive, the flux at the surface of the star, i.e., the surface flux:
=
=
=
(3.12)
By using the above two laws, we can derive a relation between the surface
temperature, the size (radius) and the luminosity of a star.
For a spherical star of radius R,
L  4  R 2 T 4
Since stars are not perfect blackbodies, we use Eq(3.11) to define the effective
temperature Te of a star’s surface,
L  4R 2Te 4
We can also find,
L
Lsun
(
(3.13)
R 2 Te 4
) (
)
Rsun Te, sun
What else we can know about stars from their spectrum? If we look at the sky
at night for a while, it is not difficult to know that stars can have different
colours. The colour of stars can range from red through to white and blue as
shown in the picture in the slide.
3.3.2 The colours of stars
The m and M measure the magnitude over all wavelengths of light emitted by a
star they are called bolometric magnitudes.
In practice, most detectors measure the radiant flux of a star only within a
certain wavelength region by the sensitivity of the detector.
13
The colour of a star may be precisely determined by using filters that transmit
the star’s light only within certain narrow wavelength bands.
In the standard UBV system, a star’s apparent magnitude is measured
through three filters and designed by three capital letters.
Filter name
Effective  (nm)
U (ultraviolet
apparent magnitude)
360  34
440  49
98
V (visual
apparent magnitude)
550  44.5
89
B (blue
apparent magnitude)
Effective bands
(nm)
68
* The effective wavelength is the wavelength to which the filter measurement
refers.
Using eq.(3.7) a star’s absolute colour magnitudes Mu, MB and Mv may be
determined if the distance d is known. In modern astronomy, the colour of a
star is described by a quantity called colour index.
The Colour index:
a) (U-B) colour index
It is defined as the difference between its ultraviolet and blue apparent
magnitudes, so is the same for (B-V) colour index.
mU  M U  5  5 log(
d
);
pc
d
);
pc
U  B  (M U  M B )  0
U  M U  5  5 log(
m B  M B  5  5 log(
d
);
pc
U  B  M U  M B (3.14 a )
B  M B  5  5 log(

d
)
pc
14
b) (B-V) colour index
In the same way, we can prove: B  V  M B  M V
( 3 . 14 b )
It shows that the colour index is independent of the star’s distance.
Example: There are two equal distant stars, one with colour index B-V =0, and
second B-V=5, which is bluer?
Stellar magnitude decreases with increasing brightness; consequently, a star
with a smaller (B-V) colour index is bluer than a star with a larger value of B-V.
c)The difference between a star’s bolometric magnitude and its visual
magnitudes is its bolometric correction BC:
d
d
mbol  Mbol  5log10(
)  5  5log( )
10pc
pc
We can find: BC=mbol – mV = Mbol - MV
d
d
mv  Mv  5log10(
)  5  5log( )
10pc
pc
(3.15)
Example:
Sirius, the brightest appearing star in the sky, has U= -1.50, B= - 1.46 and V= 1.46, and the bolometric correction BC=-0.09,
i) What is its colour indexes U-B and B-V?
ii) At what wavelengths is it the brightest (in terms of U, B, or V)?
iii) What is its apparent bolometric magnitude mbol ?
What else the spectrum of stars can tell us about star? – the chemicals.
15
3.3.3 Classification of the of Stellar Spectra
a) Harvard scheme
Astronomers arranged the spectra in a sequence which corresponds to their
temperatures Te, as shown in the following:
Type
Color
Approximate
Surface
Temperature
Main Characteristics
Examples
10 Lacertra
O
Blue
> 25,000 K
Singly ionized helium lines either
in emission or absorption. Strong
ultraviolet continuum.
B
Blue
11,000 - 25,000
Neutral helium lines in
absorption.
Rigel
Spica
Sirius
Vega
A
Blue
7,500 - 11,000
Hydrogen lines at maximum
strength for A0 stars, decreasing
thereafter.
F
Blue to
White
6,000 - 7,500
Metallic lines become noticeable.
Canopus
Procyon
G
White
to
Yellow
5,000 - 6,000
Solar-type spectra. Absorption
lines of neutral metallic atoms and
ions (e.g. once-ionized calcium)
grow in strength.
Sun
Capella
K
Orange
to Red
3,500 - 5,000
Metallic lines dominate. Weak
blue continuum.
Arcturus
Aldebaran
M
Red
< 3,500
Molecular bands of titanium oxide
noticeable.
Betelgeuse
Antares
The above classification system is called Harvard scheme
16
b) Subtypes:
Within each of these broad categories, Annie Jump Canon assigned subclasses
numbered 0 to 9. A star midway through the range between F0 and G0 would
be an F5 type star. The Sun is a G2 type star.
The Harvard scheme specifies only the surface temperature. A more precise
classification would also include the luminosity of the star
c) Luminosity class:
The standard scheme used for this is called the Yerkes classification (or MKK,
based on the initials of the authors William W. Morgan, Philip C. Keenan, and
Edith Kellman).
Ia
Most luminous supergiants
II
Luminous giants
Ib
III
IV
V
Less luminous supergiants
Normal giants
Subgiants
Main sequence stars (dwarfs)
Thus the Sun would be more fully specified as a G2V type star. The following
are some typical stellar spectrum.
17
Later on, astronomer use a diagram (L, Te) to specify the properties of stars.
3.3.4 Hertzsprung-Russell (HR) Diagram for stars near the Sun
For the main sequence stars, it may be partially understood in terms of the
luminosity for an object emitting thermal radiation: L ~ R 2T4
Schematically, stars fall into regions shown above with respect to the sun. We
can make an qualitative description on the relative size of a star on the
diagram to that of the solar radius.
18
--Stars cooler than, but brighter than the Sun
L
L sun
(
R
R sun
)2 (
Te
Te , sun
)4
 (
( 3 .13 )
R
R sun
)2 
L
T
( sun ) 4  1
L sun
T
These must be very large stars (Red Giants).
--Whereas stars hotter than the sun but less luminous must be quite small
( White Dwarfs), because
 (
R
R sun
)2 
L
T
( sun ) 4  1
L sun
T
Next we have a brief view on the life of a star.
3.4. The birth of stars
a) How are stars born?
Stars form in giant clouds of gas and dust as a consequence of gravitational
collapse.
19
The molecular cloud does not collapse into a single star. It fragments into
many clumps When a fragment of a molecular cloud reaches a critical mass,
it collapses to form a star.
But reaching this critical mass is not so easy and there are several ways for
molecular clouds to collapse:
We will discuss them in the 2nd/3nd year’s astronomy related courses. The
following is just a very simple picture on how a cloud collapses.
b). The Path to Collapse
i). Internal heating (T). When Cloud fragments collapse, potential energy is
converted to kinetic Energy. Gas particles speed up and collide, which makes
increases the temperature of the gas.
Because the collapsing cloud is rotating, due to conservation of angular
momentum, the collapse is inhibited in the plane of rotation.
ii) Disk formation -- The gravitational collapse is uninhibited along the axis of
rotation. The result of this asymmetric collapse is a flattened disk that is
rotating around the central condensed core.
20
iii). Planet Formation
The disk around the central core will fragment further, producing rings of
material. The particles in these rings can accrete together to form planets!
c). The signal for protostar formation
When the stellar surface T ~ 300 K, undergoing
continuous gravitational contraction makes the cloud
release part of its potential energy, which transfers
into thermal energy to heats the central core. (PE-->
KE), and raise its temperature. The core radiates like
a blackbody and the energy emitted is in the infrared.
Overview of the Building-up:
-- Collapse starts out in free fall controlled by gravity.
-- Central parts collapse more rapidly => central core becomes a protostar.
-- Core accretes material from the surrounding envelope
d. The New-Born Star
The protostar continues to collapse while the central core heats up to millions
of degrees. Fusion reactions start => A star is born
e. When does the Collapse stop ?
Collapse is halted by the pressures of the heated
gas and radiation pressure which counteract
gravity. Pressure is a force in outward direction,
gravity is force pulling star in. An equilibrium is
reached between opposing forces
Forces in balance => no collapse or expansion
21
f. How long does it take to form a star?
Mass
Time
15
0.16
(Msun)
5
2
1
0.5
(106 years)
0.7
8
30
100
The time for a star to reach the main-sequence varies with its mass
g. When is a new star visible?
After a star is born, nuclear reaction provides it the energy for radiation and
heats the gas and dust around it, eventually the gas and dust are pushed away,
the star is then becomes "visible." Prior to this it could be seen only in the
radio and the infrared
h. Entrance into the H-R Diagram
After a star becomes stable, we can find its position on H-R diagram. Which
point on H-R diagram it goes to depends on the mass of the star!
22
i) How long does a star live?
The mass of a star determines how long it will live.
Mass
Lifetime
1
~1010
10
~107
(Msun)
5
(years)
~108
Low mass stars evolve slowly so have longer lifetime while higher mass evolve
more dramatically, because of the strong self-gravity, has shorter lifetime.
3.5 The death of stars
While nuclear reactions inside the stars provide the energy resources for a star
to shine, and at the same time, the star evolves to the end of its life through
different routine, dependent on its mass.
H-R diagram can be used to trace out the evolution of a star from its birth to
death. However, the trajectory of a low mass star is very different from that of
a high mass star.
a) The fate of our Sun: The following is the evolutionary trajectory of our
Sun in H-R diagram, which shows that our Sun will end its life at point 9,
which is a white dwarf. We will investigate the details of the evolutionary
process in 2nd/3rd year astronomy/astrophysics modules.
23
b) The fate of massive stars: The following is the evolutionary of a mass (20
solar mass) star which will end it life as a black hole after a
supernova explosion.
The evolution of massive stars are more unpredictable and not yet understood
well. You will learn more about it in your higher year courses.
c) The fate of all of stars
The following lists the fate of stars of different mass.
Stellar mass
M < 1 Msun
Msun<M<8Msun
8Msun<M<20Msun
M > 20 Msun
Nature of
collapse
Slow
gravitational
contraction
Mild core
collapse
Fast core
collapse
Very fast core
collapse
Size of
radius (km)
---
Density
(g/cm3)
---
End product
7000
107
White Dwarf
4
1016
Black Hole
20
24
31014
---
Neutron Star
3.6 Black holes
In 1783, British amateur astronomer, Rev. John Mitchell, realized that
Newton's laws of gravity and motion implied that the more massive an object,
the greater the escape velocity. If you could somehow make something 500
times bigger than the Sun, but with the same density, he reasoned, even light
couldn't move fast enough to escape from it and it would never be seen.
a). Schwarzchild radius
Suppose we throw up a rock from ground, what is the minimum speed upward
the rock must have to escape the Earth’ gravity?
GMm
R

1
mv
2
2

2 GM
R
v 
Suppose the escape velocity of an object was equal to the speed of light.
v escape  c 
2 GM
2 GM
 Rs 
R
c2
( 3 . 16 )
Rs is called Schwarzchild radius.
Rs is defined as the radius an object must have to become a black hole.
If R < Rs: vescape > C  nothing can escape from the object, including light.
( since nothing can move with a speed higher than C).
Black Hole is an Ultimate Extreme Object:
-- Gravity is so strong that nothing, not even light, can escape.
-- Infalling matter is shredded by powerful tides and crushed to infinite
density.
-- Escape speed exceeds the speed of light
b). The size of black holes:
Rs(in km) = 3 x M (in Msun)
(3.17)
Object
Mass
(Msun)
Rs
Star
10
30 km
25
Star
3
9 km
Earth
3 x 10-6
9 mm
Sun
1
The density of black hole:
3 km
=
≈
(3.18)
More massive black holes are less dense. For a black hole with
M =Msun, = 2 × 10 /
, M=10Msun,
= 2 × 10 /
c) The Event Horizon
RS defines the "Event Horizon" surrounding the
black hole's singularity.
The event horizon is located at Rs.
Anything inside the event horizon is gone from
sight forever (nothing can escape).
d) Falling Into a Black Hole !
-- Recall that Running (or moving) clock is slower as it approaches the
event horizon. It is slower and slower at the event horizon, the clock
stops!
-- The gravitational redshift gets larger and larger as objects approach the
event horizon at the event horizon the redshift becomes infinite.
-- Tidal forces: Tidal forces are due to the difference in the gravitational
force across an object.
GMm
dF
2 GMm
F 


dr
r2
r3
 F  F ( rs  dr )  F ( rs )
or:
dF
1 d 2F
2
 F ( rs ) 
(

r
)
 ...  F ( rs )
r  rs  r 
r

r
s
dr
2 dr 2
dF
2 GMm


r

 26
r
( 3 . 19 )
r  rs
3
dr
rs
=
−→ 1
(The person is leaving , u is negative)
t moving 
t rest
1
u
2
  2
c2
dF
2 GMm

3
3
dr
r
Question:
If black holes are black, how can we hope to see them?
Answer:
Look for the effects of their gravity on their surroundings.
--Look for stars orbiting around an unseen massive object
--Look for X-rays emitted by gas that is superheated as it falls into a black
hole.
27