See discussions, stats, and author profiles for this publication at: https://www.researchgate.net/publication/353980622
Stellar Evolution & Death of Stars
Preprint · August 2021
DOI: 10.13140/RG.2.2.10790.86087/1
CITATIONS
READS
0
2,174
1 author:
Sagar J C
National Institute of Technology Karnataka
8 PUBLICATIONS 0 CITATIONS
SEE PROFILE
Some of the authors of this publication are also working on these related projects:
Comprehensive review of Physics of Gravitational waves and Functional elements of LIGO View project
Comprehensive review of Physics of Gravitational waves and Functional elements of LIGO View project
All content following this page was uploaded by Sagar J C on 20 January 2022.
The user has requested enhancement of the downloaded file.
Stellar Evolution & Death of Stars
Sagar J C
St. Joseph’s College, Bengaluru
19PCM21015
2nd year Bachelor of Science
Physics, Chemistry and Mathematics
2nd August 2021
ABSTRACT
Origin of our universe is one of the biggest mysteries to date. There are numerous theories that
point towards the ‘Big bang’ to be the origin. But still, no experiments proved the Big bang theory.
But what we can do to study the origin of the universe is by going backward in time and question
‘How did the universe begin to evolve? ’, ‘What processes gave rise to the present universe? ’, etc.
And one of such questions is ‘How are the stars formed? ’. By knowing how stars are born, and
their processes we can say what was the initial conditions of the universe and draw conclusions
on its origin. So in my review paper, I have studied the chronology of our universe according to
inflationary cosmology, then how the interstellar gas clouds aggregated to form protostars, stellar
birth, required nuclear physics to understand fusion, conditions for stellar birth, properties of a
star like pressure, mass profile, temperature, luminosity, magnitude system, etc. Analysis of the
HR diagram, Main sequence stars, the life cycle of different mass categories of stars, and finally
Supernova remnants like Neutron stars and Black holes.
i
Contents
1 The Beginning
1
2 Interstellar clouds and Hydrostatic Equilibrium
2.1 Equation of Hydrostatic equilibrium . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.2 The Protostar . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2
3
4
3 Nuclear Fusion in the core of stars
3.1 Nuclear Potential Energy and Nuclear Force . . . . . . . . . . . . . . . . . . . . . . .
5
6
4 Birth of the Star
4.1 The Proton-Proton fusion Chain . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
7
7
5 Properties of Stars
5.1 Virial theorem . . . . . . . . . . . . . . . . . . . . . . . . . . .
5.2 Total gravitational potential energy of a star . . . . . . . . . . .
5.3 Pressure inside a Star . . . . . . . . . . . . . . . . . . . . . . .
5.3.1 Pressure at the center . . . . . . . . . . . . . . . . . . .
5.3.2 Pressure as the function of distance from the centre . . .
5.4 Mass Profile of Star . . . . . . . . . . . . . . . . . . . . . . . . .
5.5 Temperature profile of star . . . . . . . . . . . . . . . . . . . . .
5.5.1 Temperature at the core . . . . . . . . . . . . . . . . . .
5.5.2 Temperature at the surface . . . . . . . . . . . . . . . .
5.5.3 Temperature as the function of distance from the center,
5.6 Luminosity of star . . . . . . . . . . . . . . . . . . . . . . . . .
5.6.1 Brightness vs Luminosity . . . . . . . . . . . . . . . . .
5.6.2 Magnitude system - Apparent and Absolute Magnitude
. .
. .
. .
. .
. .
. .
. .
. .
. .
‘r’
. .
. .
. .
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
6 Classification of Stars - HR Diagram
9
9
9
10
10
10
11
12
12
12
12
13
13
14
15
7 Stages of star
7.1 Main Sequence Star . . . . . . . . . . . .
7.2 Life cycle of Low or Average mass stars →
7.2.1 White Dwarfs . . . . . . . . . . . .
7.3 Life cycle of High Mass stars → ≥ 8 M .
. . . .
80 MJ
. . . .
. . . .
. . . . . . . .
to 8M . . .
. . . . . . . .
. . . . . . . .
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
16
16
17
17
18
8 Supernova remnant
19
8.1 Neutron star → mass of the remnant core is between 1.4M and 3M . . . . . . . . 19
8.2 Black hole → mass of the remnant core is greater than 3M . . . . . . . . . . . . . . 20
ii
1
The Beginning
Our universe is thought to have been originated from a process called ‘Big Bang’ and time should
have been started only after the Big bang. So, at Big bang, time, t = 0s and as time increased,
many events happened as the temperature decreased. In the below table, we can see the summary
of the different events. [1], [2], [3], [4], [5], [6].
Time interval
T
0 s − 10−43 s
1032 K
10−43 s - 10−36 s
1029 K
10−36 s - 10−32 s
1028 K
10−32 s - 10−12 s
1015 K
10−12 s - 10−6 s
1012 K
10−6 s - 1s
1010 K
1s - 10s
109 K
10s - 17 min
107 K
17 min - 377 kyr
> 4000K
18 kyr - 377 kyr
4000K
Chronology of our Universe
Main event
The four fundamental forces were unified to one
single force due to the very high temperature.
Theory of Everything tries to model this force.
Due to the decrease in temperature,
Gravitational force separated leaving behind the
Electro - Strong force, the Grand-Unified theory
tries to model this force.
The Strong force separated leaving behind the
Electro - Weak force. And the universe expanded
to 1026 times its initial volume, dispersing the
fundamental particles uniformly.
At this temperature the Electro - Weak forces
decouple finally and allows the formation of
Higgs Field. The Large Hadron Collider can
achieve till this temperature
Higg’s Field bestows mass to the particles. But
still the temperature is too high for the Quarks
to combine and form Hadrons. So the space is
filled with Quark - Gluon Plasma.
Quarks and Gluons combine to form the
Hadrons like Proton and Neutron.
All hadrons and anti-hadrons annihilate. But
a slight asymmetry made hadrons to dominate
over anti-hadrons. And all the matter that
we see today in our universe are those remaining hadrons. But the Leptons & Anti-Leptons
are in equilibrium as the photons can create
the electron-positron pairs at this temperature.
Thus the Leptons dominated the universe.
Baryons like Proton and Neutron fused to form
smaller nuclei like 11 H, 21 H, 32 He, 42 He and 73 Li.
The nuclei and electrons were scattered in the
universe in the form of ions in Plasmic state.
At this temperature, the nuclear fusions and
the nucleosynthesis stops and the ratio of H : He
ceased to be 3 : 1. And the large amount of
nuclear energy which was released was in the
form of photons, thus the energy of universe was
dominated by photons.
The electrons were captured by nuclei & formed
neutral ground state atoms. This released a
large amount photons which are now red shifted
to microwaves and exist as Cosmic Microwave
Background (CMB) in the present universe.
1
Epoch
Planck Epoch
Grand Unification
Epoch
Electro -Weak or
Inflationary
Epoch
The End of
Electro -Weak
Epoch
Quark Epoch
Hadron Epoch
Lepton Epoch
Big Bang
Nucleosynthesis
Photon Epoch
Recombination
2
Interstellar clouds and Hydrostatic Equilibrium
As we saw the chronology of the universe, the atoms which were formed during the recombination
era were scattered throughout the universe. But as there was no process that produces energy, the
temperature became 60K. By Einstein’s Theory of General Relativity, we know that any matter
with mass curves space-time, so even the Hydrogen and Helium gas curved the space-time, so as
time elapsed, the atoms came together and started forming large interstellar clouds. These clouds
formed patch-like structures in our universe and started to make filaments like structure of gases
where denser regions became more dense and rarer regions became more empty. As shown in the
below picture, we can how the large interstellar clouds formed dense patches of gas in the form of
filaments. The denser regions (nodes) are shown brighter and rarer regions are shown darker.
(a)
(b)
Figure 1: (a) Filament structure of the universe. Credits:- [7] (b) Image of NGC 7027 planetary
nebula with an illustration of the Helium Hydride molecule, where it was the first-ever molecule to
be formed in the early universe. Credits:- NASA/ESA/Hubble Processing: Judy Schmidt
These clouds get accumulated which have a mass of several solar masses (M ) and are very huge
which are several trillion kilometers long. Due to the action of gravity, the gas will be dragged
towards the center. As the cloud gets converted to a spherical ball and starts contracting, the
atoms present inside exert an outward force due to the repelling force of the outer electron of each
atom, thus creating internal pressure. And at a particular density and radius of the cloud, the
inward Gravitational force and outward force due to the internal pressure will be balanced and the
cloud is in a state of equilibrium called as ‘Hydrostatic equilibrium’
Figure 2: An Interstellar Cloud, The Rosette Nebula, constellation Monoceros. credits:- [8]
2
2.1
Equation of Hydrostatic equilibrium
Figure 3: Spherically symmetric interstellar cloud in Hydrostatic equilibrium.
Consider a spherical cloud of radius ‘R’. Since it has spherical symmetry, we can divide the cloud
into very thin shells where the inner radius of the shell is ‘r’ and the outer radius is ‘r + dr’. So,
the thickness of the shell is ‘dr’ and the surface area of the shell will be, S = 4πr2 . Then Volume
of the shell (dV ) is surface area times the thickness of the shell, hence dV = S × dr = 4πr2 dr. And
then the mass of the shell is given by its volume times density. So, dm = ρ(r)dV = 4πr2 ρ(r)dr
dm
= 4πr2 ρ(r)
(1)
dr
This is nothing but the Mass continuity equation. Here the density is taken to be a function of ‘r’
because it is not a constant because near the center, density will be higher and near the surface,
density will be lower. So, ρ(r) gives the value of density at a distance ‘r’ from the centre. Similarly,
the mass of the cloud inside the shell will be a function of ‘r’ because it will be density × volume
of shell, where density itself is a function of ‘r’. So, ‘m(r)’ will give the mass of the cloud present
inside the shell of radius ‘r’. Then the gravitational force exerted on the shell of mass ‘dm’ by the
cloud inside the shell of mass ‘m(r)’ is ,
∴ dm = 4πr2 ρ(r)dr
⇒
Gm(r)dm
Gm(r)
=
× 4πr2 ρ(r)dr = 4πGm(r)ρ(r)dr
(2)
r2
r2
There will be outward internal pressure, because of the photons which are traveling outward which
exert a pressure on the shell which is dP , where, dP = dFp /S, then we get the outward force to be,
dFg =
dFp = SdP = 4πr2 dP
(3)
When the spherical cloud is in equilibrium, then the net force acting on the shell has to be zero.
⇒
X
F = dFg + dFp = 0
Then substituting the values of dFg and dFp from Eq.(2) and Eq.(3) to the above equation we get,
4πGm(r)ρ(r)dr + 4πr2 dP = 0
Then by further rearrangement,
4πr2 dP = −4πGm(r)ρ(r)dr
dP
Gm(r)ρ(r)
=−
dr
r2
Thus, eq. 4 gives the equation for Hydrostatic equilibrium of the cloud.
3
(4)
2.2
The Protostar
If the mass of the cloud is more than a particular value called as the ‘Jean’s mass’, then the internal
pressure of cloud will be insufficient to prevent the collapse cloud under gravity [9]. Jean’s mass
‘Mj ’ and Jeans Density, ‘ρj ’ is given to be,
Mj =
3kB T R
2Gm
(5)
3 3kB T 3
(6)
4πM 2
2Gm
Jean’s mass of a cloud depends on the factors like its average temperature (T ), radius (R), and
average mass of a gas-particle(m) which will be in the order of ≈ 104 M . And Jean’s density of the
cloud depends on the average temperature, total mass (M ), and average mass of the gas-particle [10].
ρj =
If any interstellar cloud holds these conditions true, i.e., M > Mj and ρ > ρj , then the condition of
Hydrostatic equilibrium breaks down and the cloud will collapse, it gets compressed for millions of
years. The size of the cloud will reduce a lot and its size reduces to ≈ 10−4 times its initial value.
This makes the innermost part, i.e., the center of the cloud, very hot such that the speed of the
particles increases to a large value and creates enough amount of internal pressure which balances
the further gravitational collapse and reaches a condition of temporary Hydrostatic equilibrium.
This object which is in a temporary Hydrostatic equilibrium is called a ‘Protostar ’. The high temperature in the core of the protostar, makes the atoms lose its electrons and re-ionize them to form
a plasmic state of matter. This Epoch in the Big Bang is called as Re-ionization Epoch. The study
in some very distant galaxies which emit spectral lines of Lyman’s α series showed that the spectral
lines correspond to the ionized Hydrogen and Helium, indirectly proving the re-ionization of atoms
in protostars. [11].
The core of protostars will have a very high temperature, but it is not enough for Nuclear fusion
reactions to take place. Rather it still gathers mass from nearby interstellar gas medium, and slowly
increases its mass [12]. The below image shows how a protostar looks. The red glow is due to the
radiation that it emits due to the heat generated by the internal ions, due to the pressure. And it
shows the nearby interstellar gas from which the protostar gathers more mass.
Figure 4: Illustration of a Protostar. credits:- Chandra X-ray Observatory
4
3
Nuclear Fusion in the core of stars
The rest mass of free Neutron (mn ) & Proton (mp ) are 1.0086649 u & 1.0072764 u, respectively [13].
Consider the rest mass of a Helium−4 Nuclei, M42 He which is given to be, 4.0026032 u [14]. But the
Helium−4 nuclus has 2 protons and 2 neutrons. Then the additive mass of 4 He is,
m = 2 × mp + 2 × mn = 2 (1.0072764 u + 1.0086649 u) = 4.0318827 u
So we see that the additive mass of constituent particles corresponding to 42 He nuclei is higher than
the actual mass of 42 He nuclei. Thus there is some mass defect in He−4 Nuclei given by,
∆M = m − M42 He = 4.0318827 u − 4.0026032 u = 0.0292795 u
The energy of corresponding mass defect, ∆M can be calculated by Einstein’s Mass - Energy relation
which is E = ∆m c2 [15]. So that part of the mass of protons and neutrons which is converted
to energy is used in the binding of protons and neutrons, is called as Binding Energy of Nucleons,
denoted by Eb . So the mass defect in Helium nuclei is converted to the binding energy for Helium−4,
1
× 10−6 M eV
e
1
Eb = 0.0292795 × 1.660539 × 10−27 × (3 × 108 )2 ×
× 10−6 = 28.295 M eV
1.6021765 × 10−19
Eb = ∆M × c2 ×
So to disintegrate a 42 He nucleus, energy of 28.295 M eV has to be supplied to it. But to disintegrate
just 1 Nucleon (either proton or neutron), an energy corresponding to Eb /N has to be supplied,
which is called Binding Energy per Nucleon, ‘Ebn ’. So, for 42 He,
Ebn =
Eb
28.295 M eV
=
= 7.073 M eV
N
4
Thus to find binding energy per nucleon, Ebn of any nuclei, A
Z X, the formula is,
Ebn = Zmp + (A − Z) mn − MA X u × c2 ×
Z
10−6
in M eV
e
(7)
Where ‘Z’ is the number of protons, ‘mp ’ is the rest mass of a free proton, ‘A’ is the mass number,
‘mn ’ is the rest mass of a free neutron, ‘MA X ’ is the mass of the nuclei (detected experimentally),
Z
‘u’ is the atomic mass unit, ‘c’ is the speed of light, ‘e’ is the elementary charge.
Figure 5: Graph of Ebn vs A for various nuclei. credits:- [16]
5
Binding Energy per Nucleon, determines how stable a nucleus is, the higher the value of Ebn , higher
will be its stability. When a graph is plotted between Ebn vs Atomic mass number (A), then we get
the above curve where we see the Ebn values for various nuclei. The highest value of Ebn is for 56
26 F e
56
nucleus, which is 8.79 M eV . Moreover, for nuclei less heavy than 26 F e, their Ebn values roughly
increase, but for nuclei heavier than 56
26 F e, their values decrease gradually. So, when two nuclei
combine (fuse), such that the product is a lighter nucleus (A < 56), then that process liberates
energy, so those nuclear fusion reactions occur spontaneously. But if two nuclei combine, such that
the product is a heavier nucleus (A > 56), then that process consumes energy, so those nuclear
fusion reactions don’t occur spontaneously.
3.1
Nuclear Potential Energy and Nuclear Force
But, for nuclear fusion to occur, the nuclei have to overcome the ‘coulombic repulsion force’, as the
nuclei will be positively charged. But once that energy barrier is crossed, then the Strong Nuclear
Force acts on the nuclei.
Figure 6: Nuclear Potential energy Vs Distance between two nucleons
Figure 7: Variation of Nuclear force between two nucleons with the distance between them
The above graphs show the action of strong nuclear potential energy and nuclear force between two
nucleons as they come closer. As two nucleons come closer, the potential energy which is almost
zero becomes negative and reaches a minimum value at 0.8 f m and when the nucleons come closer
than 0.8 f m the potential energy starts increasing and it becomes positive after a certain value. So
nucleons reside in that potential well where the potential energy is minimum and the nuclear force
is attractive. But if potential energy is positive, then the nuclear force is repulsive. In the nuclear
force graph, which is got by the formula F = − dP
dr the negative value of the force is describing it
as an attractive force. So the minimum is at 1.2 f m which is the efficient distance between two
nucleons. So the typical distance between two nucleons is 1.2 f m [17], [18]. Thus for Nuclear fusion
to happen, the protons have to come very close to ≈ 1.2f m.
6
4
Birth of the Star
When the protostar is in temporary hydro-static equilibrium, it gathers mass continuously from the
nearby interstellar clouds. Due to an increase in mass the gravitational collapse starts when the
protostar gathers enough mass such that its mass exceeds 0.08 M [19] [20], then the gravitational
pull overcomes the internal pressure of protostar and it starts to collapse. Due to an increase in the
external pressure on the center of the protostar, the core temperature becomes very hot, such that
it provides sufficient kinetic energy for protons to fuse. For that to occur, the temperature has to
be very large so that the transnational speed of protons is very high and with enough kinetic energy
the protons can overcome the Coulombic repulsion between the protons. In the previous section we
saw that for the strong force to act, the distance between two particles has to be ≈ 1.2f m So by
coulomb’s law we get the force of repulsion between two protons which is at a distance of 1.2f m is:
Fc =
1 e2
1
(1.6021 × 10−19 )2
=
N = 160.2137 N
4πε0 r2
4π × 8.8541 × 10−12 (1.2 × 10−15 )2
The potential energy barrier created by this repulsive force is given by,
U = F r = 160.2137 × 1.2 × 10−15 J = 1.9225 × 10−13 J = 1.199 M eV
Then the Kinetic energy of protons to overcome this barrier has to be greater than ‘U ’. If we assume
that the temperature of the core provides sufficient kinetic energy for protons to cross the barrier,
then considering that the proton is having 3 degrees of freedom (3 directions for movement), the
temperature relationship is given by,
3
U = kB T
2
This way of calculation, corresponds to a temperature of 9.283 × 109 Kelvins, where the speed
of protons would be 1.5162 × 107 ms−1 . But in actual practice, nuclear fusion starts in the core
when the temperature reaches ≈ 107 Kelvins. This difference is because we neglected the overall
energy distribution of energy of protons and deviated from Maxwell’s distribution laws. But a
thorough calculation by considering all aspects, the core temperature of ≈ 107 Kelvins is sufficient
for a nuclear reaction to occur. [21]. So when the protostar collapses due to gravity, and if the
temperature of the core reaches ≈ 107 K, then the protons fuse through a set of nuclear fusions
called ‘The proton-proton chain’ and form a more stable ‘42 He’ nucleus by releasing energy. [22]
4.1
The Proton-Proton fusion Chain
Proton-Proton Chain or the pp Chain is the set of nuclear fusion reactions that produce the enormous
energy which is a primary source for stars. The step by step reaction is given below:
Step 1 Step 2 -
1H +
1
+ 11 H + →
e− + e+ →
Step 3 -
2H +
1
Step 4 -
3
2+
2 He
2H +
1
+ e+ + νe
2 × γ0
+ 11 H + →
⇒ E2 = 1.02 M eV
λ=2.42 pm
3 He2+
2
+ 32 He2+ →
⇒ E1 = 0.42 M eV
⇒ E3 = 5.49 M eV
4
2+
2 He
+ 2 11 H +







×2






⇒ E1 = 12.86 M eV
The overall proton-proton chain reaction can be written in this form where 4 protons and two will
fuse together to form a Helium-4 nuclei by releasing the energy of 26.7 M eV . [23]
4 11 He+ + 2e− →
4
2+
2 He
+ 2 νe
7
⇒ E = 26.7 M eV
Figure 8: Proton Proton Chain forming He-4 Nuclei. credits:- scienceatyourdoorstep.com
Thus the nuclear fusion in the core gives rise to a tremendous amount of energy which results in
the Birth of a self-gravitating luminous object powered by nuclear fusion called a ‘Star’. The birth
of a star produces so much energy that it creates a burst of radiation which initiates the ionization
of atoms and molecules in nearby clouds. Thus a large number of stars were formed in our universe
when it was ≈ 150 million years after the big bang [24]. Then due to gravity large number of stars
got collected in certain regions of space forming galaxies. Thus started a new era in the universe
called ‘The Stellar Era’, where the fusion in star cores became the main source of energy. [25]
Figure 9: Formation of Galaxies in Stellar Era
The nuclear fusion in the core of a star produces an enormous amount of energy such that it creates
enough outward pressure to balance the inward gravitational collapse. So the star will be in a state
of a Hydrostatic Equilibrium. governed by the equation,
dP
Gm(r)ρ
=−
dr
r2
8
(8)
5
5.1
Properties of Stars
Virial theorem
Virial theorem is a general equation that relates the total kinetic energy (T ) of a star that is in
a stable condition, i.e., Hydrostatic Equilibrium, bound by the potential forces (Fi ) to the total
potential energy (V ) of Star. This theorem explains most of the aspects of a star given by, [26]
1
1
T = − ΣFi · ri = − V
(9)
2
2
5.2
Total gravitational potential energy of a star
dr
R
C
r
Consider a spherical star of Mass ‘M ’ and radius ‘R’. To calculate the total gravitational potential
energy, i.e., the potential energy liberated while forming the star, let us consider the energy when a
new shell of particles is being deposited on the existing mass system and integrate the process from
the center of the star to the surface.
Initially, let the star already be having a radius of ‘r’ and a mass ‘m’, whose density is ‘ρ’. Then
let the mass ‘dm’ is being deposited on the original mass in the form of a shell, whose thickness is
given to be ‘dr’. By mass continuity equation we have,
dm
= 4πr2 ρ So, the mass deposited is, dm = 4πr2 ρ dr
dr
Then the gravitational potential energy between the shell and the original mass is given by,
(10)
G
G
m × dm = − m × 4πr2 ρdr = −4πGρmr dr
r
r
By using the value of ‘m’ to be equal to ρ × Volume, we get
(11)
dU = −
4πr3
16
dr = − π 2 Gρ2 r4 dr
3
3
Then by integrating the equation within the limits of r, 0 → R and U, 0 → U ,
dU = −4πGρr × ρ
Z R
16 2 2 4
16
16
R5
π Gρ r dr = − π 2 Gρ2
r4 dr = − π 2 Gρ2
3
3
3
5
0
0
0
M 2 R 5
16
Substituting the value of ρ back we get,
U = − π2G 4 3
3
5
3 πR
Z
U
(12)
Z
dU = U =
R
−
Thus for a Star, the total gravitational potential energy, ‘U ’ is given by,
U =−
3 GM 2
5 R
(13)
The −ve sign indicates that energy is released while the star-matter is being aggregated in the form
of Gravitational Energy since the System is attaining a state of −ve potential energy.
9
5.3
5.3.1
Pressure inside a Star
Pressure at the center
We can derive the relation between Pressure at any point in the star and distance of the point from
the center, ‘r’ by integrating the equation of Hydrostatic Equilibrium.
Gm(r)ρ
dP = −
dr
r2
on integrating we get
Psurf ace
Z
R
4πr3
3
R
−
dP =
0
Pcenter
substituting m(r) = ρ ×
Gm(r)ρ
dr
r2
and approximating ‘ρ0 to be a constant, we get,
Psurf ace − Pcenter =
G 4πr3
− 2 ρ2
dr =
r
3
R
4πGρ2
4πGρ2 h r2 iR
r dr = −
3
3
2 0
0
0
2πGρ2 R2
2πG M 2 2
3GM 2
Psurf ace − Pcenter = −
=−
R
=
−
4
3
3
3
8πR4
3 πR
Z
⇒
Z
Z
−
3GM 2
(14)
8πR4
In the above equation, ‘M ’ is the total mass of star, ‘R’ is the radius of the star. For sun, substituting
M = 1030 Kg, R = 108 m we would get the pressure at the center to be in the order of 1015 pascals.
Thus, we can calculate the approximate pressure at the center of a star.
Neglecting the surface pressure as it is very small, we get,
5.3.2
Pcenter =
Pressure as the function of distance from the centre
If we had chosen the limits of integration to be 0 → ‘r’ then the equation would be,
4πGρ2 h r2 ir
2πGρ2 r2
3GM 2 r2
3GM 2 r2
P (r) − Pcenter = −
=−
⇒
P
(r)
=
P
−
=−
center
3
2 0
3
8πR6
8πR6
Thus the Pressure at a distance ‘r’ from the center is,
P (r) =
3GM 2 2
(R − r2 )
8πR6
(15)
This result is however an approximation as ‘ρ’ is considered to be constant, but practically it isn’t.
Considering the variation of density and other internal effects the actual graph is given below.
10
5.4
Mass Profile of Star
According to the Mass-continuity equation (eq. 1), we have,
dm = 4πr2 ρ(r)dr
(16)
Integrating it using the condition, when r = 0 → m = 0 and approximating that ‘ρ(r)’ to be a
constant which is ρ
Z
Z
r3 dm = m = 4πr2 ρdr = 4πρ
+C
(17)
3
Using the Condition when r = 0 → m = 0, we get the value of ‘C’ to be ‘0’. Thus the equation
which relates the mass (m) of the star present within a shell of radius ‘r’ is,
4
(18)
m = πρr3
3
When we plot the above equation (Mass Vs Radius), we get the curve to be as shown below.
But in the real Scenario, that is not the case. As we have neglected ‘Variation of Density’, and
other internal effects (rotation, radiation, etc). So the actual graph looks like the one below [27].
11
5.5
5.5.1
Temperature profile of star
Temperature at the core
Considering the star material to be made of an ideal gas, following the equation, P = N kB T , where
N is the number density of the gas, we can find the Temperature of the core of a star.
Tcenter =
3GM 2
8πR4 N kB
(19)
Substituting the values of M = 1030 kg, R = 108 m, and N = 1029 m−3 we get the temperature of
the core of our sun to be in the order of 107 Kelvins.
5.5.2
Temperature at the surface
At the surface of the star since there is no fusion of any nuclei, the temperature at the surface has
to be mostly due to the conduction of heat through the star material from the core. An effective
way of measuring the surface temperature is by analyzing the wavelength of light produced by the
star having maximum spectral energy and finding the surface temperature by the formula [28],
Tsurf ace =
5.5.3
2.897771955 × 10−3
Kelvin
λmax
(20)
Temperature as the function of distance from the center, ‘r’
According to the Dalsgaard Model of Stars, Core has the highest temperature and as ‘r’ increases,
the temperature of star decreases. In the core, the decrease in temperature as ‘r’ increases is very
less. But in the next layer, i.e., the Radiative Zone, temperature decreases slightly as ‘r’ increases,
but in the outer convective zone, the temperature decreases significantly as ‘r’ increases. And finally
when r = R, i.e., in the surface of the star, temperature will be equal to surface temperature which
can be calculated by ‘Wein’s Displacement Law’ [29]
Figure 10: Temperature of star Vs Distance from the center. credits:- solarscience.msfc.nasa.gov
12
5.6
Luminosity of star
By plank’s distribution of black bodies, the total radiation energy density (dEλ ) within the range
(dλ) at a particular wavelength (λ), emitted by a black body of a particular temperature (T ) per
unit wavelength, per surface area, per unit time is given by [30],
Eλ dλ = dEλ =
2πhc2
dλ
hc
5
λ e λkB T − 1
(21)
Then, the Energy emitted for the whole range of wavelength (0 ≤ λ < ∞) we have to integrate the
above equation.
Z ∞
2πhc2
dλ
E=
(22)
hc
5
λ
λkB T
0
e
−1
If λ =
hc
−hc
, we get, dλ = 2
dx, when λ = 0 → x = ∞ and when λ = ∞ → x = 0,
xkB T
x kB T
Using the above substitutions, Eq.(22) can be written as,
Z
Z ∞
2πk 4 T 4 ∞ x3
2πk 4 T 4 π 4
2π 5 k 4 T 4
2πk 4 T 4 x3 dx
=
dx
=
×
=
E=
h3 c2
ex − 1
h3 c2
ex − 1
h3 c2
15
15h3 c2
0
0
(23)
Thus we get, the total radiation energy emitted per unit second per unit surface area (E) to be,
2π 4 k 4 T 4 = σT 4
→ Stephan’s Law
(24)
15h3 c2
Luminosity in general is the total amount of radiant energy emitted by an object per unit time.
Since stars behave similar to a black body, we can use Stephan’s Law to find its luminosity (L) by:
E=
L = σ × Surface Area × T 4 = 4πσR2 T 4
(25)
where, ‘σ’ is the Stephan’s constant (5.67037 × 10−8 W m−2 K −4 ), A = 4πR2 , ‘R’ is the radius of
the star and ‘T ’ is the surface temperature. Thus we can note that, more the size of the star or
more the temperature of the star, the more is its luminosity. i.e., if we consider two stars of the
same size, but of different temperatures, the star with more temperature will be more luminous and
if we consider two stars of the same temperature but different sizes, then the star with the bigger
size will be more luminous.
Sun has a surface temperature ≈ 5700 K and a radius ≈ 6.9634 × 108 m. Then using Eq.(25) sun’s
luminosity can be calculated to be 3.828 × 1026 Watts which is denoted by L .
5.6.1
Brightness vs Luminosity
During broad daylight, no other star other than the sun is visible to us. Because Sun is the brightest
object than other stars. Does that mean Sun is more Luminous than all other stars? No, because
the sun is very close to us than other stars which makes us assume that the sun is more bright.
So the distance of stars from the earth significantly changes its brightness, as if it is less luminous
than the sun. It will appear fainter as the distance is very large. Luminosity is an intrinsic property
of any object which doesn’t depend on the distance of the observer. Unlike luminosity, brightness
depends on the distance (d) between observer and source, which is measured in the following way,
Brightness =
⇒
Brightness ∝
2
Luminosity
4 R
=
σT
4πd2
d
1
(Brightness of an object follows inverse square law)
d2
13
(26)
5.6.2
Magnitude system - Apparent and Absolute Magnitude
In astronomy, magnitudes are used to quantitatively describe the brightness of a star. The higher
the magnitude of a star, the lesser is its brightness. There are two types of magnitudes.
• Apparent magnitude
• Absolute magnitude
Apparent Magnitude
Brightness of a star as we know depends directly on the Luminosity of the star and inversely on the
square of the distance between the observer (Earth) and the Star. This is also called the ‘Apparent
brightness’. The apparent magnitude scale on the other hand is a convenient way of comparing the
brightness of stars. The apparent magnitude of any star is based on the brightness of a standard
star α− Lyare, also known as ‘Vega’. Thus we assign star ‘Vega’ to have a magnitude of ‘0’ and
brightness (Ivega ). Thus apparent magnitude (mx ) of any star is defined as, [31] [32],
mx = 2.5 log10
I
vega
(27)
Ix
Thus, we see that any star brighter than Vega will have mx < 0 and any star fainter than Vega will
have mx > 0. Hence, the brighter the star, the lesser is its apparent magnitude. If the difference in
magnitude of two stars is ∆m = mb − ma , where mb > ma then,
I
I
I vega
vega
a
∆m = mb − ma = 2.5 log10
− 2.5 log10
= 2.5 log10
Ib
Ia
Ib
Ia
⇒
(28)
= 100.4∆m or Ia = Ib × 2.512∆m
Ib
Apparent magnitude of sun is −26.74, which means it is 4.97 × 1010 times brighter than Vega star.
i.e., there have to be 4.97 × 1010 Vega stars in our sky for it to have the same brightness as our sun.
Absolute Magnitude
The absolute magnitude of a star is the quantitative measure of actual brightness or Luminosity
(L) of a star. Absolute magnitude of a star is defined as that magnitude of the star if it is placed
at a standard distance of 10 pc (parsecs) away. Where 1 pc = 3.26156 light years. So, if there is a
star at a distance ‘D’ (in pc) with an apparent magnitude of (mx ), but now if it is placed at a distance of 10 pc, then that magnitude brightness which will appear to us is‘Absolute magnitude’ (Mx ).
Then the difference between the magnitudes, ∆m = mx − Mx gives us the ratio of the brightness of
the star when it is 10pc away (Iabs ) to the brightness of the star when it is in its original position
(Iapp ). Thus by Eq.(29), we get, Iabs = Iapp × 2.512∆m .
L
L
L
L
Using the relation Iabs =
and Iapp =
we get,
= 2.512∆m ×
4π(10pc)2
4πD2
4π(10pc)2
4πD2
D 2
D Simplifying further,
2.512∆m =
or
∆m log10 (2.512) = 2 log10
10 pc
10pc
Finally we get,
∆m = 5 (log10 (D) − 1)
or
Mx = mx − 5 (log10 (D) − 1)
(29)
We can also find the distance of a star from earth (in ‘pc’) by solving the above equation for ‘D’,
√
D = 10 2.512∆m
(30)
If‘Msun ’ is the absolute magnitude of sun and ‘L ’ is solar luminosity. Then by knowing the absolute
magnitude (Mx ) of any star, we can find its luminosity (Lx ) using the following equation.
Lx = 100.4(Msun −Mx ) L
14
(31)
6
Classification of Stars - HR Diagram
In our universe, a large number of stars of various ‘Colour’, ‘Mass’, ‘Size’ and ‘Surface temperatures’
exist and there exists an interesting Diagram that includes all the details of every star that exists
in our universe. That is called as ‘Hertzsprung Russell ’ (HR) Diagram which is shown below.
Figure 11: HR Diagram.
credits:- [33]
• In the HR diagram, luminosity (increases as Y increases) and absolute magnitude (decreases as
Y increases) is plotted in ‘Y’ axis, whereas the surface temperature (decreases as X increases)
and the peak wavelength of a star (increases as X increases) is plotted in the ‘X’ axis.
• So a very hot star (T > 10000 K) lies towards the left of the diagram and the peak wavelength
emitted is in the Blue region of the EM Spectrum. Stars with intermediate temperature
(5000K < T < 10000K) lie in the middle whose peak wavelength emitted is in the Yellow
region. Then, the stars with very low surface temperature (T < 5000 K) lie towards the right
of the diagram, whose peak wavelength emitted is in the Orange/ Red region.
• Stars which are in the ‘Top’ are more luminous and their absolute magnitude is very small,
whereas the stars in the ‘Bottom’ are less luminous whose absolute magnitude is very large.
Sun, which is in the yellow region, T ≈ 5700 K, L = 1 L and, absolute magnitude = 4.83,
lies on the right side of the middle region.
• Usually, more luminous stars have a higher temperature (Top-left) and less luminous stars
have a lower temperature (Bottom-right). But, for a star that is in the top-right, it has a
lower temperature, but high luminosity, then it should have a large radius (Giant stars) as
L ∝ R2 . And for a star that is in the bottom-left, it has a higher temperature, but low
luminosity, then it should have a small radius (white dwarf).
15
7
Stages of star
As seen earlier, if the interstellar cloud accumulates mass more than ‘Jean’s mass’ and if its density
is more than ‘Jean’s density’, then it will become a ‘Protostar ’. Then the protostar accumulates
mass over time from the surrounding nebulae and when its mass exceeds 0.08 M then, it collapses
due to gravity and thus its core temperature rises enough for Nuclear fusion to occur and the inward
gravitational force will be balanced by the outward radiation force (due to nuclear fusion) and this
forms a ‘Star ’. But if the protostar’s mass is less than 0.08 M or 80 MJ and if there were no
surrounding nebulae for it to gather further mass, then it remains as a protostar which is called a
Brown Dwarf and if the mass is less than 13 MJ then it is called as a Sub-Brown dwarf. [34]
7.1
Main Sequence Star
Protostars whose mass is more than 0.08 M will become a star which is the main sequence. Most of
the stars which we see today in our universe are this type of stars including our ‘Sun’. They are the
stars whose core is still fusing Hydrogen (1 H) into Helium (4 He). Stars spend their 90% of life in this
stage for Billions of years and are governed mainly by the equation of Hydrostatic equilibrium where
the inward gravitational force is balanced by the outward radiation force [35]. They are represented
by the green line in the above HR diagram. In the main sequence, the more luminous stars are
more massive, and emitted peak wavelength is in the blue region. The stars whose luminosity is
average have average mass and the peak wavelength is in the Yellow region. And the fainter stars
have very little mass and emitted peak wavelength is in the Red region. Thus these stars follow
a pattern where the Temperature, Luminosity, Radius, Mass, peak frequency emitted are directly
related by the empirical relations [36],
L ≈ 0.23 M 2.3 {M < 0.43 M }
L = M4
{0.43 M < M < 2 M }
L ≈ 1.4 M 3.5
{2 M < M < 55 M }
L ≈ 32000 M
{M > 55 M }
The life cycle of any star depends only on its mass deciding its life span and the last stage (Death)
which is known as the ‘Vogt–Russell theorem’ [31], [37]. Then for a star whose mass is M (in solar
masses) the total main sequence lifetime (τM S ) is defined by,
τM S = 1010 × M −2.5 years
(32)
Figure 12: Mass Vs Total main sequence life time relation for a star.
According to the above equation, the higher the mass of a star, the lower is its lifetime. That
is because, more the mass, the more should be its inward gravitational pull. But according to
Hydrostatic equilibrium, the outward radiation force generated through nuclear fusion has to be
more, so the rate of nuclear fusion should be higher. That is why the stars use up the fuel faster if
their mass is very high and will have a relatively very short total main sequence lifetime.
16
7.2
Life cycle of Low or Average mass stars → 80 MJ to 8M
Low and average mass stars have their main source of fuel to be the Hydrogen Fusion reaction
via the proton-proton chain in its core. Such stars continue to live in the main sequence stage for
billions of years maintaining a steady temperature, size, and luminosity. But as the abundance of
Hydrogen in the core decreases then there will be significant changes that will occur.
• Since the rate of Hydrogen fusion decreases, the outward radiation pressure decreases. Thus
the inward gravitational pull overcomes the outward force and the star starts to shrink.
• But as the core shrinks, core temperature builds up so high that the remaining Hydrogen burn
faster and that energy released makes the outer region of the star expand and increases the
surface area of the star thus increasing its luminosity but decreasing its surface temperature.
Thus the star is now a Red giant which can live for ≈ 1 Billion years.[37], [38]
• And finally after all the Hydrogen in the core is used, leaving behind only a very thin shell,
the core compresses even further. But it can’t go to a single point, because of the action of
electron degeneracy pressure.1 The outward force of electron degeneracy pressure prevents
further collapse of the core. [39]
• But due to extreme temperatures of 108 Kelvins, the core is now able to fuse Helium into
Carbon and Oxygen through the Triple α process or Helium Flash. [40]
Step 1 -
4
2 He
Step 2 -
4
2 He
+ 42 He → 84 Be
+
8
4 Be
→
12
6 C
(33)
16
• As the star is using up its final energy reserve converting 42 He to 12
6 C and 8 O, it shrinks
becoming blue and very hot. Then the core will collapse to a very small size, but that
pressure is not enough for the core to fuse Carbon and Oxygen as its mass is very small. The
star pulsates whose size decreases and increases periodically.
• And when the core has mostly Carbon and Oxygen with only very thin outer shells of Helium
and Hydrogen, the core collapses into a very dense object whereas the outer material of the
star expands so much that it goes into a second red giant phase [41] and finally the Outer
material forms a Planetary Nebula leaving behind a dense, hot but fainter core called as
White Dwarf whose mass will be less than 1.4 M (Chandrasekhar limit). [42]
7.2.1
White Dwarfs
White dwarfs are remnant cores of stars that are very hot whose temperature ranges from 104 to 107
K, but has a very less luminosity. That means they have to be very small according to Eq.(25).
Their mass will be roughly equal to the mass of the sun whereas the size will be similar to that of
Earth. Hence, they are highly dense ≈ 109 kg m−3 [43]. Because of their small size, they radiate
energy very slowly and be luminous (though very faint) for more than hundreds of billions of years
and finally become a black dwarf. But since the universe is only 13.8 Billion years old, there are no
known black dwarfs yet [44]. For White dwarfs, as the mass increases their radius decreases, unlike
a normal star. Because there is no nuclear fusion occurring in it to produce any outward force. A
white dwarf is stable only because of Electron Degeneracy pressure which is exerted by electrons
against gravity preventing further collapse. But electron degeneracy pressure will overcome gravity
only if its mass is less than 1.4 M [45]. But if the white dwarf accretes any mass from nearby
material or star, and if its mass exceeds ‘Chandrasekhar Limit’, then electrons won’t be able to
push against gravity and then collapse forming a Type I Supernova (see section 8 to know about
supernova) [46].
1
As electrons follow Pauli’s exclusion principle, they can’t be in the same energy state, thus they are forced to
occupy higher and higher energy states making them travel at higher speeds and create a net outward force.
17
7.3
Life cycle of High Mass stars → ≥ 8 M
The birth of such stars is similar to that of average mass stars. Except that their τM S will be much smaller
than average mass stars as said earlier because of the ‘Vogt–Russell theorem’. So hydrogen in the core fuse
very quickly (≈ 10 million years) and after the fuel runs out, the core shrinks due to gravity. But when
16
28
it gains enough pressure and temperature, 42 He will fuse to form 12
6 C, then it further fuses to 8 O, 14 Si,
32
16 S, and other heavier nuclei. This process creates so much energy that, the outer layers expand a lot and
the star is now in a Red Super Giant Stage. But if the mass of the star is greater than 40M then
the temperature in the core of such star will be so hot that, the surface temperature will be more than 10
thousand Kelvins and it appears blue, this stage is called a Blue super giant. Then upon further nuclear
fusion, once the core forms 56
26 S, as we know that this nucleus is the most stable than any other nuclei, further
fusion consumes energy rather than liberating it. So the fusion comes to a halt, when the whole core is made
of mostly (≥ 90%) iron-56 and there is no source of energy or outward pressure, so the star collapses due
to gravity overcoming even the electron degeneracy pressure where gravity fuses the protons and electrons
‘p+ + e− → n0 + νe ’ releasing a tremendous amount of heat and pressure in the form of a shock wave and
the whole collapsed star bounces of the core and becomes a Core Collapsed Supernovae. [47] [38]
Figure 13: Simulated image of a supernova. credits:- NASA, ESA, N. Smith, and J. Morse
Figure 14: Schematic representation of Life cycle of stars. credits:- Scioly.org
18
8
Supernova remnant
A Supernova releases so much energy and light that it can outshine its own galaxy. 90% of the emitted
particles are Neutrinos and the rest being cosmic rays and light. It is mainly in supernovae where most of
the heavy nuclei (A > 56) are synthesized. The two main types of supernova remnant core are:
8.1
Neutron star → mass of the remnant core is between 1.4M and 3M
• If the supernova remnant core left behind is more than 1.4 M , then the degenerate electron pressure
cannot overcome gravity. So, even electrons fuse with protons by a process called ‘Electron capture’
forming an object which is like one big giant nuclei full of neutrons held together not by the strong
force, rather mainly by gravity called as Neutron Star.
p+ + e− → n0 + νe
(34)
• Neutron stars will be formed by stars whose original mass would be between 8 M to 25 M [48].
Neutron stars are highly dense objects whose mass ranges between 1.4 M to 3 M , but they will
have a size in the order of just 10 Km to 20 Km. Thus, its density will be between 2 × 1017 Kg m−3 to
3 × 1017 Kg m−3 [49] which is close to the density of atomic nuclei.
• Neutrons are balanced by ‘Neutron Degeneracy Pressure’. This is similar to the degenerate pressure of
Electrons. Since neutrons are fermions, according to Pauli’s principle no two fermions can occupy the
same state, thus the neutrons in degenerate neutron gas are forced to occupy different energy states
and they exert a net outward pressure against gravity and balance the neutron stars [50].
• Neutron stars have a highly dense plasma of neutrons, where the core will be like a super-fluid. Some
scientists also theorize that, in the core of massive neutron stars, neutrons are broken into their
constituent quarks and the state of matter is like a ‘Quark-Gluon’ soup. But the crust of neutron
stars will have a thickness of ≈ 2 km which is hard and consist of solid nucleons in form of a lattice
and a sea of leptons (electrons and muons) flowing through the lattice. [51] [52]
• If the neutron star rotates about a fixed axis then it sends pulses of electromagnetic waves and they
are called ‘Pulsars’. And some neutron stars will have a very high magnetic field around 10 billion
Tesla and they are termed ‘Magnetars’ [53].
Figure 15: Neutron star
19
8.2
Black hole → mass of the remnant core is greater than 3M
If the mass of the remnant core of a supernova is more than 3 M then even the neutron degeneracy pressure
cannot resist the inward gravitational pull. So neutron degeneracy pressure breaks after 3 M also known
as the ‘Tolman–Oppenheimer–Volkoff’ (TOV) limit [54]. After this limit, the neutrons are also compressed
by gravity and since there isn’t any opposing force, the whole matter content will be compressed to a single
point with Zero volume and Infinite Density and forms a Black hole.
Figure 16: Computerized image of Black hole. Credits:- NASA, M.Kornmesser/N.Bartmann
• Singularity is the innermost or center of the Black hole where the whole mass is concentrated to a
single point where the space-time curvature is infinite that it doesn’t allow even light to escape [55].
• The black color part that we can see in the above image is ‘Event horizon’, which is like an envelop
that covers the singularity. Anything that enters the event horizon by crossing a boundary called
‘Innermost stable orbit’, can never escape from the black hole and goes directly into the singularity.
That is the reason why we can’t see the inside of event horizon and acts as a perfect black body. [56]
• When stars or other objects like stars come under the vicinity of a black hole, due to the high
gravitational effect the black hole just shreds the object and pulls it and this creates a hot, circularly
moving gas around the black hole called the ‘Accretion’ Disk. Most of the accreted matter goes into
the event horizon which increases the mass of the black hole and the size of its event horizon. [57]
• Then perpendicularly to the plane created by the accretion disk, there is a burst of very fast-moving
particulate matter or cosmic rays which are created when the gravitational pull and circular motion
of the particles come in the correct ratio at a specific distance from the black hole. [58] [59]
According to the ‘No hair theorem’, any black hole has only three fundamental properties which differentiate
one another, they are Mass, Angular momentum, and Charge. And all other properties (that metaphors
to no hair) are inaccessible to observers because of the perfect black body nature of the black hole [60].
Black holes are so fascinating objects that they form an entirely new branch of study in physics where the
scientists have dedicated themselves to understanding the nature of black holes. Still, we don’t know how
does singularity looks like, what is present inside of it, and what actually happens if something goes into it.
Conclusion
In this review paper, I came to know how the universe expanded, its chronology, When matter was created.
How it became huge gas clouds because of the action of gravity. Condition of Hydrostatic equillibrium in
stable systems like Protostar and Star. The sustained nuclear fusion in star core and the required conditions.
Also came to know about various properties of a star like pressure profile, temperature profile, mass profile,
luminosity, absolute and apparent magnitudes, etc and how they vary in a star. Main sequence stars and the
various mass categories for stellar life cycle. Lastly, how a star becomes supernova and turns into a white
dwarf (Mcore < 1.4 M ) or Neutron star (Mcore > 1.4 M ) or Black hole (Mcore > 3 M ) & their properties.
20
References
[1]
Johanna Karouby. “Topics in Early Universe Cosmology”. Juin 2012. Theses. Universite McGill,
July 2012. url: https://tel.archives-ouvertes.fr/tel-00727133.
[2]
David L Alles. The Evolution of the Universe. 2013. url: https://fire.biol.wwu.edu/trent/
alles/Cosmic_Evolution.pdf.
[3]
Prof. Jim Brau. Galaxies and the Expanding Universe, Chapter 27 - The Early Universe. url:
https://pages.uoregon.edu/jimbrau/astr123/Notes/Chapter27.html.
[4]
S Singh. The Origin of the Universe. Harper Perennial, 2005.
[5]
Eric J. Chaisson. Cosmic Evolution - Particulate. url: https : / / lweb . cfa . harvard . edu /
~ejchaisson/cosmic_evolution/docs/fr_1/fr_1_part3.html.
[6]
Wikipedia. Chronology of the universe. url: https://en.wikipedia.org/wiki/Chronology_of_
the_universe.
[7]
Andrew Pontzen and Hiranya Peiris. Illuminating illumination: what lights up the universe?
Aug. 2014. url: https://commons.wikimedia.org/wiki/Commons:Reusing_content_outside_
Wikimedia.
[8]
Nancy Atkinson. Astrophoto: Spectacular View of the Rosette Nebula. Sep 12, 2014. url: https:
//www.universetoday.com/114497/astrophoto-spectacular-view-of-the-rosette-nebula/.
[9]
Duncan Forgan and Ken Rice. “The Jeans mass as a fundamental measure of self-gravitating
disc fragmentation and initial fragment mass”. In: Monthly Notices of the Royal Astronomical
Society 417.3 (2011), pp. 1928–1937.
[10]
Website – Journey through astronomy course. The Jeans Collapse Criterion. url: http://
csep10.phys.utk.edu/OJTA2dev/ojta/c2c/starbirth/recipe/criterion_tl.html.
[11]
Mark Dijkstra. “Lyα Emitting Galaxies as a Probe of Reionisation”. In: Publications of the
Astronomical Society of Australia 31 (2014). issn: 1448-6083. doi: 10.1017/pasa.2014.33.
url: http://dx.doi.org/10.1017/pasa.2014.33.
[12]
Steven W. Stahler. “Protostars”. In: Encyclopedia of Astrobiology. Ed. by Muriel Gargaud
et al. Berlin, Heidelberg: Springer Berlin Heidelberg, 2011, pp. 1378–1385. isbn: 978-3-64211274-4. doi: 10.1007/978-3-642-11274-4_1304. url: https://doi.org/10.1007/978-3-64211274-4_1304.
[13]
Website - Sciencing.com. Kevin Beck & Lana Bandoim. How to Calculate the Mass of a Proton,
Neutron and Electron. url: https://sciencing.com/calculate-mass-proton-6223840.html.
[14]
Wikipedia. Helium-4. url: https://en.wikipedia.org/wiki/Helium-4.
[15]
John Stachel and Roberto Torretti. “Einstein’s first derivation of mass–energy equivalence”.
In: American Journal of Physics 50.8 (1982), pp. 760–763. doi: 10 . 1119 / 1 . 12764. eprint:
https://doi.org/10.1119/1.12764. url: https://doi.org/10.1119/1.12764.
[16]
OpenStax CNX. “Nuclear Binding Energy.” In: (Nov. 2020), p. 4549. url: https : / / phys .
libretexts.org/@go/page/4549.
[17]
‘Kikuchi’, ‘Mitsuru’, and et.al. “Fusion Physics”. In: (Sept. 2012).
[18]
Zhengdong Huang. “A Strong Force Potential Formula and the Classification of the Strong
Interaction”. In: OALib 05 (Jan. 2018), pp. 1–30. doi: 10.4236/oalib.1104187.
[19]
S. Schneider and T. Arny. “Readings: Schneider & Arny: Unit 60 Star Formation”. In: (), p. 6.
url: http://abyss.uoregon.edu/~js/ast122/lectures/lec13.html.
[20]
Philippe Andre, Derek Ward-Thompson, and Mary Barsony. “From Pre-stellar cores to protostars: The initial conditions of star formation”. In: arXiv preprint astro-ph/9903284 (1999).
[21]
Robert W. Conn. “Nuclear fusion”. In: Encyclopedia Britannica (July 2019), p. 16. url: https:
//www.britannica.com/science/nuclear-fusion.
21
[22]
VA Dergachev and GE Kocharov. “PROTON–PROTON CYCLE IN STARS IN THE TEMPERATURE RANGE 107 K”. In: Kosm. Luchi, No. 10, 177-96 (1969). (1969).
[23]
E. E. Salpeter. “Nuclear Reactions in the Stars. I. Proton-Proton Chain”. In: Phys. Rev. 88
(3 Nov. 1952), pp. 547–553. doi: 10.1103/PhysRev.88.547. url: https://link.aps.org/doi/10.
1103/PhysRev.88.547.
[24]
Richard B. Larson and Volker Bromm. “The First Stars in the Universe and Exceptionally
massive and bright, the earliest stars changed the course of cosmic history”. In: Scientific
American (Jan. 2009). url: https://www.scientificamerican.com/article/the-first-stars-inthe-un/.
[25]
Naoki Yoshida, Takashi Hosokawa, and Kazuyuki Omukai. “Formation of the first stars in the
universe”. In: Progress of Theoretical and Experimental Physics 2012.1 (Oct. 2012). 01A305.
issn: 2050-3911. doi: 10.1093/ptep/pts022. url: https://doi.org/10.1093/ptep/pts022.
[26]
Herbert Goldstein. Classical mechanics. Pearson Education India, 2011.
[27]
Tran Dai Nghiep and Pham Ngoc Dinh. “The mass profile in the solar-mass stars”. In: Nuclear
Physics A 722 (2003). doi: https://doi.org/10.1016/S0375- 9474(03)01409- X. url: https:
//www.sciencedirect.com/science/article/pii/S037594740301409X.
[28]
Biman Das. “Obtaining Wien’s displacement law from Planck’s law of radiation”. In: The
Physics Teacher 40.3 (2002), pp. 148–149.
[29]
Christensen-Dalsgaard and et. al. “The Current State of Solar Modeling”. In: Science (New
York, N.Y.) 272 (June 1996), pp. 1286–92. doi: 10.1126/science.272.5266.1286.
[30]
Luis J Boya. “The Thermal Radiation Formula of Planck (1900)”. In: (Mar. 2004).
[31]
A.B. Bhattacharya and S. Joardar. Astronomy and astrophysics. Hingham, Mass. :Infinity
Science Press, 2008, xv, 364 pages. isbn: 9781934015056.
[32]
E Böhm-Vitense. Introduction to Stellar Astrophysics. Vol. 1. Cambridge University Press,
1989, 255 pages. isbn: 0521348692.
[33]
Encyclopedia Britannica. “"Hertzsprung-Russell diagram."”. In: Britannica (June 2021). url:
https://www.britannica.com/science/Hertzsprung-Russell-diagram.
[34]
K. L. Luhman. “DISCOVERY OF A ∼ 250 K BROWN DWARF AT 2 pc FROM THE SUN”.
In: The Astrophysical Journal 786.2 (Apr. 2014), p. L18. issn: 2041-8213. doi: 10.1088/20418205/786/2/l18. url: http://dx.doi.org/10.1088/2041-8205/786/2/L18.
[35]
S. Woosley, A. Heger, and T. Weaver. “The evolution and explosion of massive stars”. In:
Reviews of Modern Physics 74(4) (1978), pp. 1015–1071.
[36]
G. Kuiper. “The Empirical Mass-Luminosity Relation.” In: The Astrophysical Journal 88
(1938), p. 472.
[37]
Bradley W Carroll and Dale A Ostlie. “An introduction to modern astrophysics and cosmology”. In: An introduction to modern astrophysics and cosmology/BW Carroll and DA Ostlie.
2nd edition. San Francisco: Pearson (2014), p. 1479.
[38]
Michael Zeilik and Stephen A. Gregory. “Introductory Astronomy and Astrophysics”. In:
BROOKS COLE publication: 4th Edition (1998).
[39]
Sami AL-Jaber. “Degenerate electron gas and star stabilization in D dimensions”. In: Nuovo
Cimento B Serie 123 (Jan. 2008). doi: 10.1393/ncb/i2008-10426-9.
[40]
Heinz Oberhummer, Rudolf Pichler, and Attila Csoto. “The triple-alpha process and its anthropic significance”. In: (Nov. 1998), p. 4.
[41]
I. -Juliana Sackmann, Arnold I. Boothroyd, and Kathleen E. Kraemer. “Our Sun. III. Present
and Future”. In: Astrophysics Journal (Nov. 1993). doi: 10.1086/173407. url: https://ui.
adsabs.harvard.edu/abs/1993ApJ...418..457S.
22
[42]
Gregory Laughlin, Peter Bodenheimer, and Fred C. Adams. “The End of the Main Sequence”.
In: The Astrophysical Journal 482.1 (June 1997), pp. 420–432. doi: 10.1086/304125. url:
https://doi.org/10.1086/304125.
[43]
NASA Website. White Dwarf Stars. url: https://imagine.gsfc.nasa.gov/science/objects/
dwarfs2.html.
[44]
Fred C. Adams and Gregory Laughlin. “A dying universe: the long-term fate and evolution
of astrophysical objects”. In: Reviews of Modern Physics 69.2 (Apr. 1997). doi: 10 . 1103 /
revmodphys.69.337. url: http://dx.doi.org/10.1103/RevModPhys.69.337.
[45]
Subrahmanyan Chandrasekhar. “The maximum mass of ideal white dwarfs”. In: Astrophys. J.
74 (1931), pp. 81–82. doi: 10.1086/143324.
[46]
George C. Jordan et al. “Failed Detonation Supernovae: Subluminous Low-Velocity Ia Supernovae”. In: The Astrophysical Journal 761.2 (Dec. 2012). doi: 10.1088/2041-8205/761/2/l23.
url: http://dx.doi.org/10.1088/2041-8205/761/2/L23.
[47]
J. Eldridge. “Massive stars in their death throes”. In: Philosophical transactions. Series A,
Mathematical, physical, and engineering sciences (Oct. 2008). doi: 10.1098/rsta.2008.0160.
[48]
A. Heger et al. “How Massive Single Stars End Their Life”. In: The Astrophysical Journal
591.1 (July 2003), pp. 288–300. issn: 1538-4357. doi: 10.1086/375341. url: http://dx.doi.
org/10.1086/375341.
[49]
NASA Website. Calculating a Neutron Star’s Density. url: https://heasarc.gsfc.nasa.gov/
docs/xte/learning_center/ASM/ns.html.
[50]
J. R. Oppenheimer and G. M. Volkoff. “On Massive Neutron Cores”. In: Phys. Rev. 55 (4 Feb.
1939), pp. 374–381. doi: 10.1103/PhysRev.55.374. url: https://link.aps.org/doi/10.1103/
PhysRev.55.374.
[51]
J. Lattimer and M Prakash. “The Physics of Neutron Stars”. In: Science (New York, N.Y.)
304 (May 2004), pp. 536–42. doi: 10.1126/science.1090720.
[52]
V S Beskin. “Radio pulsars: already fifty years!” In: Physics-Uspekhi 61.4 (Apr. 2018), pp. 353–
380. issn: 1468-4780. doi: 10.3367/ufne.2017.10.038216. url: http://dx.doi.org/10.3367/
UFNe.2017.10.038216.
[53]
Victoria M. Kaspi and Andrei M. Beloborodov. “Magnetars”. In: Annual Review of Astronomy
and Astrophysics 55.1 (Aug. 2017), pp. 261–301. issn: 1545-4282. doi: 10.1146/annurev-astro081915-023329. url: http://dx.doi.org/10.1146/annurev-astro-081915-023329.
[54]
Vassiliki Kalogera and Gordon Baym. “The Maximum Mass of a Neutron Star”. In: The
Astrophysical Journal 470.1 (Oct. 1996), pp. L61–L64. issn: 0004-637X. doi: 10.1086/310296.
url: http://dx.doi.org/10.1086/310296.
[55]
Robert M. Wald. General Relativity. Chicago, USA: Chicago Univ. Pr., 1984. doi: 10.7208/
chicago/9780226870373.001.0001.
[56]
Paul Davies. The new physics. Cambridge University Press, 1992. isbn: 978-0-521-43831-5.
url: https://www.google.co.in/books/edition/The_New_Physics/akb2FpZSGnMC?hl=en&
gbpv=1&pg=PA26&printsec=frontcover.
[57]
Ramesh Narayan, Shoji Kato, and Fumio Honma. “Global Structure and Dynamics of Advectiondominated Accretion Flows around Black Holes”. In: The Astrophysical Journal 476.1 (Feb.
1997), pp. 49–60. doi: 10.1086/303591. url: https://doi.org/10.1086/303591.
[58]
C Sivaram and Kenath Arun. “Relativistic Jets from Black Holes: A Unified Picture”. In:
arXiv preprint arXiv:0903.4005 (2009).
[59]
Shinji Koide, Kazunari Shibata, and Takahiro Kudoh. “Relativistic Jet Formation from Black
Hole Magnetized Accretion Disks: Method, Tests.” In: The Astrophysical Journal (Sept. 1999).
doi: 10.1086/307667. url: https://doi.org/10.1086/307667.
[60]
Kip S Thorne, Charles W Misner, and John Archibald Wheeler. Gravitation. Freeman, 2000.
23
View publication stats