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Most Basic Observations Of Sun
Chapter 10
The Interiors of Stars
Sun has maintained
(i) its size,
(ii) energy it emits every second,
(iii) its surface temperature
all this while.
• Sun is a self-gravitating objects − each part of the sun
exerts force on each other part of the sun.
• Gravity always attracts
There must be other kinds of forces acting against Gravity to
prevent the sun from collapsing onto itself.
These inward and outward forces must be in perfect balance!
Hydrostatic Equilibrium
• Sun is radiating the same amount of energy every
second − 3 x 1026 Joule
It must be generating the right amount of energy every
second.
Energy Generation Rate
It must be making sure that each of its layer transports just
the right amount of energy to maintain this energy output.
Energy Transport Rate
• Sun is maintaining the temperature of its surface, and is
also maintaining some fixed temperature at each of its
radial distance from the Sun.
Since the transfer of fixed (energy) radiation from the center to
the surface in each layer would depend crucially on how
temperature in each layer varies, it must maintain the required
temperature gradient in all layers.
Temperature Gradient
• Mass conservation
Stellar Structure Equations
Hydrostatic Equilibrium
Mass Conservation
Energy Transport
Hydrostatic Equilibrium
Ex. 10.1.1 Make a crude estimate of the pressure at the
center of the Sun using the Hydrostatic Equilibrium.
Rigorous calculation shows the pressure is
(
)
Pressure Equation of State
Completely neutral gas,
Completely ionized gas,
In terms of mass fractions, for a neutral gas
In terms of mass fractions, for a completely ionized gas
Go through P. 293
Ex. 10.1.2 Make an estimate of the temperature at the
center of the Sun using the Gas pressure term alone and
using the value for pressure we derived in the previous
example.
Stellar Energy Sources
Gravitational Energy
Kelvin-Helmholtz Timescale
Assuming a spherical star of Radius R and M, the
gravitational potential energy of the star,
Kelvin-Helmholtz time scale ~ 107 years
Too short! From dating of meteorites we know that our solar
system is about 4.5 billion year old.
Nuclear Time Scale
Total Mass of four (4) Hydrogen atoms
Mass of one Helium atom
Ex. 10.3.2 Assuming Sun was all Hydrogen, and that
inner 10% of mass is hot enough to convert Hydrogen
into Helium, how long Sun can produce its current
luminosity?
~ 1010 year
Reasonable!
A proton experiences repulsive Coulomb force and
attractive nuclear force due to another proton.
Temperatures
required
for
the
proton
to
have
sufficient energy to cross the Coulomb barrier
Considering
Quantum
tunneling,
temperatures
required for the proton to be in the range of
attractive nuclear force
Gamow Peak –
Reaction Rate of Nuclear Processes
Reaction rate in power-law (two particle interaction)
r0 is a constant, Xi and Xx are mass fractions of incident
and target nuclei, α´ and β are determined from reaction
rates.
Energy liberated per kilogram of material per second,
ε0 is the energy liberated per reaction.
In the form of power-law,
α = α’-1. By summing εix for all reactions occurring inside
stars, we find the total energy generation rate. It has
units of W kg-1.
Stellar Nucleosynthesis
Conservation Laws for Nuclear Reactions
In any nuclear reaction, total charge, total leptons, total
nucleons must be conserved.
by
A
ZX
Nuclei will be represented
, where Z is the number of protons, A is the
number of nucleons, and X is the chemical symbol of the
element.
The Proton-Proton Chains
0.42 MeV
PP I
5.49 MeV
12.89 MeV
Totally 26.22 MeV!
PP II
e- e+ annihilation :
1.02 MeV
PP III
where fpp is the pp chain screening factor, ψpp is a correction
factor that accounts for three chains of pp, and Cpp ≈ 1
involves higher order correction terms, T6 = T/106 K.
Written in power-law form near T = 1.5 X 107 K
The CNO cycle
Figure Copyright: http://csep10.phys.utk.edu/astr162/lect/energy/pp-cno.gif
Triple Alpha Process - Helium Burning
Written as a power-law centered around T = 108 K
Figure Source:
http://en.wikibooks.org/wiki/Introduction_to_Astrophysics/Main_Sequence
_Stars
Carbon and Oxygen Burning
Hydrogen fuses into Helium,
Helium into Carbon, Carbon into
Oxygen, Oxygen into Neon….
Nuclear reactions have strong
dependence on temperatures.
Heavier the element
(1) Higher the temperatures
required for its fusion
(2) Stronger becomes the
dependence (of energy
production) on temperature.
Can we go on making heavier
elements out of lighter ones
and generate energy in the
process?
OR
Is there a limit to this chain?
Binding Energy Per Nucleon
Radiative Temperature Gradient
Condition for Convection
Stellar Structure Equations
Hydrostatic Equilibrium
Mass Conservation
Energy Transport
Constitutive Relations
Boundary Conditions
Stellar Mass-Luminosity Relation
Comparison of Numerical Model and Dimensional Analysis (DA)
Figure: 3.2 A. Choudhuri (2010)
Stellar Mass-Luminosity Relation
Based on Observational Data of Binary stars
A linear fit gives a slope of ~ 3.7 (not too different from 3)!
Stellar Luminosity-Effective Temperature Relation
Comparison of Numerical Model and Dimensional Analysis (DA)
Solid:
Model
Dashed : DA
Stellar Luminosity-Effective Temperature Relation
Based on Observational Data of Nearby stars
Figure: 3.6 A. Choudhuri (2010)
A linear fit gives a slope of ~ 5.6 (not too different from 6)!
Limits on Stellar Mass
Lower Limit on a Stellar Mass
If it is not hot enough,
it is not good enough to
be termed a star!
~ 0.072 Mʘ below which it cannot fuse Hydrogen.
Upper Limit on a Stellar Mass
If it is too hot,
it must shed some mass!
~ 90 Mʘ above which it can hardly stay bound.
Eddington Luminosity
Chapter 11
The Sun
Changes on Main Sequence
Internal Structure of the Sun
Effect of Nucleosynthesis H, He, and 23He with radial distance
Temperature, Pressure with
radial distance
Energy Produced with
radial distance
Mass and Density with
radial distance
Critical Temp. Gradient (Convec.)
with radial distance
Is (Are) there any direct
evidence(s) of our stellar
structure theory?
• Solar Neutrinos
• Solar Oscillations
The Solar Neutrino Problem
In 1970, Raymond Davis first designed an experiment
to measure the solar neutrino flux − basically a HUGE
underground tank (~ one and a half km below the
ground) containing LARGE amount (~377,000 liters)
of cleaning fluid (C2Cl4) .
Homestake Gold Mine (Ray Davis)
Result of 24 years of Homestake Experiment (1970-1994)
It turned out that the average neutrino flux detected at the
Homestake (~ 2.5 SNU) was smaller by a factor of three than
expected from the Sun.
Super-Kamiokande Experiment
This underground Japanese experiment contained
50,000 tons of pure water . The blue Cerenkov light
emitted by electrons would be detected by the 11,200
inwardly-directed photomultiplier tubes.
The number of neutrinos detected were still less by a factor of
two.
The other experiments like Soviet American Gallium
Experiment (SAGE) and the Gran Sasso Underground
Laboraroty in Italy (GALLEX) which measured the lowenergy neutrinos produced in majority in p-p chain
confirmed the discrepancy in the neutron flux.
Sun in different wavelengths
Image Credit: NASA’s Solar Dynamics Observatory
Photosphere
• What are the typical temperatures here ?
4400 K (outermost layer) – 9400 K (innermost layer)
• What are the typical densities here ?
~ 10-3 kg/m3 (innermost layer)
• What is happening here ?
Zooming-in on the photosphere
Just under the surface, we see “granules”, ~1500 km in size,
lasting for 8-10 min before dissolving.
A few thousand km below the surface, we see “supergranules”,
~30,000 km in size, lasting for about a day before dissolving.
Henry Benard Convection Cells
Top-View (Real)
Image Source: Wiki
Side-View (Simulation)
Image Source: Wiki
Solar Convection
Wiggles on the „blue‟ side of the line is coming from the bright
regions on convection tiles, implying a radial velocity of 0.4 km s-1,
those on the „red‟ side are coming from the dark regions.
The “life-times” of this granules tell us about the time a convective
cell take to cover a distance one mixing length (back and forth).
The Chromosphere
• How much is the “thickness” of this region ?
~ 1600 km
• What are the typical temperatures here ?
4400 K (innermost layer) – 10000 K (outermost layer)
• What are the typical densities here ?
~ 10-7 kg/m3
• What is happening here ?
Chromosphere in Hα Filter
Image Credit: Dave Tyler
http://www.daystarfilters.com/filtergram.shtml
The Transition Region
• How much is the “thickness” of this region ?
~ 100 km
• What are the typical temperatures here ?
10000 K (innermost layer) – 106 K
• What are the typical densities here ?
~ 10-16 kg/m3
• What is happening here ?
The Corona
• How much is the “thickness” of this region ?
~ Does not have a well-defined outer boundary
• What are the typical temperatures here ?
~106 K
• What are the typical densities here ?
~ 10-17 kg/m3
• What is happening here ?
K Corona – White light produced due to scattering of
photospheric radiation by free electrons.
In the
region 1 Rʘ to 2.3 Rʘ
F Corona – Scattering of photospheric radiation due
to dust grains beyond 2.3 Rʘ.
E Corona – Between the K Corona, and the F Corona.
Source of emission lines produced due to highly
ionized atoms through out the Corona.
Forbidden
lines.
Radio emission, X-ray emission from Corona
Coronal Holes
Sun in X-ray
Image Credit: http://osoncje.tripod.com/planeti/Sun/x-ray_sun.jpg
Solar Magnetic Field
Solar Wind
Stars emit not only radiation, emit matter too!
Van Allen Radiation Belts
Image Source: Wiki
Alaska Aurora Borealis
Image Source: United States Air Force photo by Senior Airman Joshua Strang
The Parker Wind Model
Solar wind means Sun is not able to hold on to some of its mass
--- charged particles escape Sun’s gravity and fly away from Sun.
Considering again a balance between pressure and gravity, if the
pressure is too high (since temperatures are too high), and as
the gravity falls with inverse square of the distance,
particles would escape the Sun.
some
Eugene Parker developed a
simple model of Solar wind considering isothermal wind (1958).
For typical temperatures and number densities of inner corona,
T = 1.5 x 106 K, n0 = 3 x 1013 m-3 at about 1.4 Rʘ. Using our
expression from Parker Wind Model, we get,
This
doesn‟t
match
with
interstellar
pressures
and
number
densities, in general. The assumption of Hydrostatic Equilibrium is
incorrect!
What heats the upper atmosphere
to such high temperatures?
Still an unsolved problem! We only have guesses, and are no
where close to the answer!
Two candidates:
• Acoustic heating
• Magnetic heating
Acoustic Heating
Temperature in the Chromosphere changes only by a factor of two,
but the density changes by 4 orders of magnitude.
Hence,
eventually acoustic waves propagate faster than local sound speed,
and a shock is developed, which heats the gas.
Sunspots
Image Credit: Charles Chandler New Jersey Institute of Technology's New Solar Telescope
New Dalton Minimum. CLIMATE CHANGE AND ITS CAUSES – by Nicola Scafetta (2010)
Assignments:
10.4, 10.7, 10.9, 10.13, 10.14, 10.21, 10.22, 10.23