Stellar Physics Flashcards
Method to find distance to nearer stars
Parallax
Luminosity of a star
It’s energy output per unit time
Measured in watts
If star radiates isotropically flux equal across sphere
L=4pir^2F
Visual magnitude
Magnitude in the visual part of the spectrum
If a star is hotter than its environment
It will cool down by re-radiating it’s energy
Rayleigh jean law for blackbodies
Good approximation at Lon wavelengths but radiance keeps increasing indefinitely at short wavelengths (UV catastrophe)
Failure of classical physics to explain thermal radiation
Wien’s law
Good approximation to the observed spectrum at short wavelength
Planck’s law
Assuming energy comes in discrete quanta
Fit data
What determines apparent colour of an object
Shape of spectrum and position of its peak
Atmospheric extinction
Light from sun obscured by atmosphere of Earth, which absorbs more than others.
Ignore in this course
Stefan Boltzmann law
Total luminosity per area is the spectral radiance integrated over solid angle and wavelength
j=L/A=sigma T^4
Total luminosity of a spherical star
Multiply Stefan Boltzmann law by surface area
L=4piR^2sigmaT^4
Stars have wide range of luminosity so helpful to use logs
Take logs of both side of luminosity equation
Plot graph of y=log10L against x=-log10T for some value r and get straight line
Lower radii lie on lines nearer bottom
HR diagram
80-90% of stars cluster in main sequence
Other branches of stars: white dwarfs, giants and supergiants
Temperature x axis decreases from left to right
Y axis is luminosity
Interpretation of HR diagram
Only certain combinations of L and T allowed
Most stars on main sequence
What tells us that stars move around HR diagram as they evolve
Clusters are stars at similar stages of their lives
Number of stars in each part proportional to duration of that stage of evolution
Main sequence
Most variation from top left to bottom right along a line of roughly constant radius
Top left blue stars hotter and more luminous
Two cut offs to the main sequence
At top: luminous stars blow material from their surface through radiation pressure naturally limiting their mass
At bottom: cool red stars not hot enough to begin nuclear reactions. Temperature in core too low
Estimate of time a star spends on main sequence
Lifetime=energy available/ luminosity
Most massive stars spend…
Least amount of time on main sequence
Giants and supergiants
Sit in top right of HR diagram
Large L but low T
Less populated so stars spend less time in this phase
Reach after main sequence
White dwarfs
Bottom left
Below main sequence so radius s smaller
None visible to naked eye
Are not powered by nuclear fusion
Photometric system
Divides spectrum into commonly used bands
Ultraviolet band centred 350nm, blue band 440nm, visible 550nm, red 600nm, near infrared 800nm
Filters placed over telescope to select a band
Colour index
Numerical difference in magnitudes between measurements made in two wavelength bands
Measurements made through two different filters eg B-V difference between magnitude in blue and visible band
The smaller the colour index (ie lower position on number scale that ranges from positive through zero into negative)
The more blue and hotter the star
Kirchoff’s first law
A hot and opaque solid, liquid or highly compressed gas emits a continuous black body spectrum with no spectral lines
Kirchoff’s second law
A hot, transparent gas illuminated by a continuum source produces a spectrum of bright emission lines
Kirchoff’s third law
If a continuous spectrum passes through a transparent gas at a lower temperature the cooler gas will absorb at characteristic wavelengths resulting in dark absorption lines
Harvard classification
Spectra of light from stars fell into natural categories based on strength of certain key line features
Every star has a letter that describes its colour, known as spectral class
Harvard classification scheme
Considers changes in other lines as well as hydrogen, gives sequence indicating source temperature
OBAFGKS
O hit and blue, m cool and red
O spectral class
Hottest blue stars
Few lines
Strong He II absorption lines
M spectral class
Coolest red stars
Spectra dominated by molecular absorption bands especially TiO
Strong metal absorption lines
Harvard subclasses
Each type is divided into 10 subclasses
These reflect gradual temperature change
eg for A0,…,A9
0 is hotter end
9is cooler end so O9 is next to B0
What does allocation of subtype depend on
Line strengths and ratios
How does the Harvard classification scheme not completely describe a star
Cannot distinguish between stars with the same temperature but different luminosities
Morgan keenan luminosity class
Established to add discrimination on the basis of luminosity
Ranges from I to VII
Class I
Hypergiant
Divided I-O
Through 1a, bright supergiant and 1b, dim supergiant
Classes II - V
Goes from bright giants down to main sequence dwarfs
Classes VI and VII
VI sub-dwarf
VII white dwarf
What are luminosity classes determined from
Mainly from observed width of spectral lines
Broadening of spectral lines
Several effects can cause broadening
High pressure and temperature causes atoms to collide more frequently which broadens spectral emission particularly in hot dense stars like white dwarfs
Mass luminosity relationship for main sequence stars
L/lsun ~ (M/m sun)^a
Value of a depends on fit used in data but approx 3<a<3.5
More massive, more luminous
Most massive in top left of HR
Explaining mass luminosity relationship
Massive stars have large gravitational compression of cores
For equilibrium, need high radiation pressure outwards
High thermal pressure provided by high temp in core
Nuclear reaction rate very sensitive to core temp so even slight change produces large change in luminosity
Implication of mass luminosity relationship
3-3.5 is big power
Great implications on how long star lives on main sequence
Massive stars have short lifetimes because they burn up fuel quicker
What does variable mean
Star’s flux changes over time
Observe by measuring changes in apparent magnitude
Real and apparent variation
Real: star itself changes
Apparent: eg something moves in front of star and blocks light partially or fully
Irregular and regular variation
Irregular: no particular pattern, can be sudden or random
Regular obviously opposite
Novae
Sometimes called cataclysmic variables
Flare in brightness irregularly
Luminosity can increase factor 1000 over period ~ a week
All novae exist in binary systems in which material transferred from one to another causing bright outburst
Grange point
Point in middle of binary system
Disk
Best way to distribute angular momentum
T Tauri stars
Class of irregular variables
Luminosity increases by factor 3 in few days
Very young powered by gravitational energy as they contract
Type II supernovae
End state of a very massive star
After nuclear fuel exhausted core collapses and outer layers blown off
Small dense neutron star remains surrounded by expanding spheres of circumstellar matter
Type 1a supernovae
Expect one type 1a SN every few decades
Types of supernovae
Classified according to observational features
1a all have nearly equal brightness - standard candles
Examples of regular variable stars
Cepheids
Cepheid variable stars
Very luminous giant or supergiant
Luminosity varies by factors up to 10
Variation repeats over periods between 1 and 100 days
Eg Polaris period ~4 days
Radial pulsation results in a regular pulsation of
Velocity of star’s surface
Effective temperature
Luminosity
Instability strip
Where cepheid variables sit in HR diagram
Lies at roughly right angles to main sequence towards giant branch
Stage on the way to being a giant
Cepheid period luminosity relation
Period of pulsation depends only on average luminosity of the star
Longer pulsation period, the more luminous the star
Two types of period magnitude relations
Type I - massive young cepheids: M=-1.8+2.4log10P
Type II - older smaller cepheids: same but 0.4 instead of 1.8
period in days
Cepheid variables as distance indicators
Observing cepheid, measure period of oscillation, find intrinsic luminosity from period luminosity relation
Measure flux, can use F=L/4pid^2 to calculate distance
4 main classes of binaries
Visual
Astrometric
Spectroscopic
Eclipsing
Visual binaries
With telescopes, possible to resolve two components
Astrometric binaries
Cannot resolve individual stars but where we see a periodic wobble of observed overall position
Spectroscopic binaries
Components are not resolvable but Doppler shits in spectral lines reveal there are two stars orbiting same centre of mass
Eclipsing binaries
Not resolvable but see periodic brightening and dimming
Binary system orbital analysis
For cases where one mass»_space; other, deal with orbit as if larger mass is stationary
Masses of components can be more comparable. Newton and Kepler still apply
Binaries: if the masses are equal
The centre of mass is halfway between them
Binaries: if two masses are different
Centre of mass closer to heavier object
m1/m2=r2/r1=a2/a1
Where a is semi major axis
How to find semi major axis
If measure period, angular separation of stars and distance to binary is known
Allows to find total mass of system
Visual binaries finding mass
m1a1=m2a2
For visual, can measure a1 and a2 so can find m1/m2
Since we know total mass, can deduce individual masses
Redshift and blueshift - spectroscopic binaries
Star moves away, redshifted
Towards, blueshifted
Eg if A blue shifted and B red shifted then A towards and B away
No Doppler shift during tangential motion
Speed toward/away
Reduced by inclination of orbital plane
l is inclination angle v=vtrue sinl
Finding combined mass for spectroscopic binaries
v1/v2=a1/a2=m2/m1
Using v=row and w=2pi/T
v1+v2=2pia/T
Gives m1+m2=T/2piG(v1+v2)^3
Stars will eclipse each other only if
We are viewing the system near edge on (l~90 degrees)
Eclipsing binary: most light when
No overlap
Plot of magnitude against time for eclipsing binaries
Get dip at secondary minimum and bigger dip at primary minimum
Primary minimum
Hotter star moves behind the cooler
Secondary minimum
Cooler star behind the hotter
If T2>T1
Then F’>F1 so secondary minimum
T2<T1
F’<F1 so primary minimum
In eclipsing, spectroscopic binaries we can get
Orbital period from either light curve or spectroscopy
Speed of stars in orbit from spectroscopy which can be used to find size of orbit
Enough info to calculate mass of two stars
Light curve allows us to compute size of each star, by measuring time of transit and combining with speed measured in Doppler shift
Vogt-Russell theorem
Structure of a star, in hydrostatic and thermal equilibrium, with all energy derived from nuclear reactions, is uniquely determined by its mass and the distribution of chemical elements throughout its interior
What do Kirchoff’s laws tell us?
the absorption lines indicate the presence of a cooler layer of diffuse gas, on top of a hot layer of denser gas
Structure of sun
(From inner to outer)
Interior, photosphere, chromosphere, transition region, corona
When we observe the sun, we see light from
Photosphere, chromosphere and corona
In descending order of contribution
Photosphere
Sun has no solid surface but becomes opaque to visible light at the photosphere
T varies throughout
Most light comes from 5800K region where density is 1000th of air
Sunspots
Photosphere marked by darker, cooler sunspots
Around 1500K cooler than rest of photosphere
Sunspots appear to move across suns disk, shows sun is rotating
Rotation of sun
Ball of gas so different regions rotate at different rates
Equatorial region rotate in about 24 days
Polar regions take more than 30 days
Solar flares
Extends hundreds of thousands of miles above surface
Originate in photosphere
Produce bursts of radiation across EM spectrum
Chromosphere
Get little light from chromosphere
Seen as dim reddish pink glow, only really seen during eclipse usually light too weak to be seen against photosphere
Corona
Very hot
Outer atmosphere extends as far as 3 solar radii
Only really seen from Earth during eclipse
Eventually leads to solar wind
Spectral line strength depends on
Number of atoms present and temperature of gas
Effects on lines due to bulk motion
Allow measurements of overall rotation and large scale expansion or contraction
Effects on lines due to random thermal motions of atoms in atmosphere
Allow measurements of temperature and pressure
Energy of nth level of hydrogen
E proportional to -1/n^2
- because binding energy
Transition between level m (higher) and level n
Photon would be emitted of energy corresponding to energy difference between states
deltaEmn prop. ((1/n^2-1/m^2)
Calculating wavelength of emitted photon
Using rydberg equation
wavelength must be positive so use n<m
Lyman series
Transitions from excited states to ground state
Balmer series
Transitions from m>2 to n=2
Emission
Electron drops to lower level
Photon emitted, of energy corresponding to e difference between transition levels
Absorption
e jumps to higher level
Photon of correct energy needed for this to happen
Strongest visible line from sun
Ha line
e going from n=2 to m=3
Boltzmann factor
Probability of an electron occupying energy state E
- tells us states with E»kbT are very improbable
If there are 8 possible states in n=2 shell
8 times the probability of finding an electron there
Can calculate ratio of population of atoms in two states
Rydberg equation for other elements
Fails to take account of electron orbital screening in beaver elements
Single electrons around heavier nuclei, can multiply by Z^2
Ionisation
Under extreme conditions
Lose electrons if absorb enough energy from photons
Can find ionisation energy by using Rydberg equation with m=infinity
Gives x=13.6/n^2 eV
How do ions form
Absorption: of a photon with at least x joules of energy
Collision: with another particle like an electron (scattering)
Ion formation by photon scattering
Average photon has E=kbT but some have higher energy
Only photons with energy > x can cause ionisation (same as higher chem)
Ion formation by scattering
Collisions occur with other particles. Similar distribution of energies
Average E=3/2kbT
To find proportionof atoms ionised, find equilibrium in: photon +atom (reversible arrows)electron+ion
Equilibrium is Saha equation
Saha equation
No of atoms/ no ions ~ 10^21T^3/2exp(-x\kbT) / no of electrons
Rule of thumb for Saha equation
50% ionisation occurs when kbT~x/18
If a gas gets too hot
All atoms may already by ionised
May not be any low level electrons available to absorb
May not be any high level electrons able to emit
If a gas gets too cool
Electrons may be in too low an energy state for a particular line
(Balmer needs n=2 for example)
Effects of temperature on helium lines
Little seen from sun in absorption although they are observed faintly in upper chromosphere during eclipses
Helium ions can be excited to a state which gives visual absorption lines if temp right
Metal lines (metal in Astronomy is anything after helium)
Very rare
Only dominate at low temperature where H and He frozen out
Relative abundances of metals in all stars is fairly similar
Abundance of metals relative to hydrogen is very different in some stars
Population I stars
Ratio of metals to H and He is similar to Sun
Young stars, generally made from material ejected from older stars
Formed late in evolution of galaxy and are found predominantly in galactic disk
Population II stars
Ratio of metals to H and He is 100 times less than that found in sun
‘Metal poor’
Old stars formed before galaxy was a disk and are found predominantly in galactic halo
Population III stars
Possible massive stars formed just after Big Bang
Obviously no metals around then
Molecular bands
Molecules can form in outer atmosphere of cool stars
Typical binding energy of molecule is 4-6eV
This means lines will fade if T>~5000K
We observe TiO, ZrO, CN and sometimes H2O
Solar energy requirements
Know mass,luminosity of sun
Age of solar system is ~4.6 billion years
Energy required to keep sun shining is E~tau L where tau is solar lifetime
Efficiency of sun’s energy
e=E/M> or = tauL/M
Units if j/kg
Candidates for source of suns energy
Chemical energy
Gravitational energy
Relativistic energy
Nuclear fusion
Chemical energy
Chemical reactions involve rearrangement of electrons
Units is electron volts
Comparing to mass per atom
Efficiency is around 100million j/kg so if chemical reactions were source of power, would only last Me/l =50000 year
Gravitational energy
Using gravitational energy for a proton coming from infinity to r E=GMm/r so e=Gm/R ~10^11 j/kg
Still not enough to power sun but considerable amount of energy
Relativistic energy
Using eff=mc^2/m gives 9x10^16 j/kg
tau=mc^2/L = 3x10^13 years
Easily long enough to keep sun bright
However, this conversion requires equal amount of matter and antimatter
Nuclear fusion
Very high temp and density, several light nuclei can fuse to form a single nucleus
Newly formed nucleus will have slightly lower mass
Delta m is released as fusion energy
Nuclear fusion in the sun
Four H nuclei (protons) fuse together to form a Helium-4 nucleus (alpha particle)
Energy released in the form of radiation (gamma rays) and neutrinos
Energy budget
DeltaE=deltam c^2
Calculating delta e is energy available in one H-He fusion reaction
Eff=deltaE/4mproton, tau=Msuneff/L gives 400 billion years
This is big but plausible if we assume not all of sun’s mass undergoes fusion
Interstellar gas cloud
Galaxy full of H and He from Big Bang and dust and gas from previous generations of stars
If not completely uniform, gravity will cause clumps to collapse
As it collapses, gas heats up(potential energy converted to heat energy)
If cloud is initially rotating even slightly
Rotation speed Increases as material collapses inwards
Spinning material
Flattens into protoplanetary disc with dense central part forming a protostar
Surrounding material accretes onto the protostar and as temp rises,
Gas becomes ionised
Eventually, if enough material accreted, temp becomes high enough to
Initiate nuclear fusion
Radiation from the hot starpishes material out until equilibrium is reached
How much energy would it take to unbind cloud
Approximate two half spheres each M/2, distance R apart
Use equation for potential
Assume uniform density
Ereleased=Einit-Efin
If very large at start assume Einit=~0 so energy available is Gm^2/R
Why do we not see many protostars
Lifetime if powered by left over heat from contraction is Kelvin helmholtz timescale
About 10^7 years for sun which is quite short
Brown dwarf
Pressure and temperature never high enough for nuclear fusion
Will shine for its kelvin helmholtz timescale
Hayashi tracks
Evolutionary path for protostar moves it around on HR diagram
For low mass stars, vertical motion downward until proto star hits main sequence
unstable during this phase
Henyey tracks
For higher mass stars
Continues to heat up, gas in cloud fully ionised
No atoms to provide atomic absorption lines so gas transparent in outer regions
Protostar slowly contracts until onset of nuclear reaction in core
Nearly horizontal track on HR (slight luminosity increase with large temperature increase)
Distinctive characteristics of pre main sequence stars
Unstable luminosity
Eject lots of gas
Surrounded by warm clouds
Bipolar outflows
Young stars
Jets of material in opposite directions extending over a distance ~light year
Indicate new star is gaining material from a surrounding accretion disk
Emission nebulae
Young stars often surrounded by emission nebulae
UV light from hot star sweeps out a gravity and ionises surrounding hydrogen
Recombination produces Ha
Virial theorem
Total kinetic energy of a stable, self gravitating mass distribution is negative one half of the total gravitational potential energy
What is virial theorem used for
Helps calculate the conditions that must exist for cloud collapse
Derivation of jeans mass
Number of particles with mass mp in cloud is N=Mc/mp
Potential energy of spherical cloud with uniform density
Gas so ask of each particle is 3/2kbT
Total Ekby multiplying by N
Use viral therorem
Cloud collapse if unbalanced so use <
Use initial density and sub in to find Mc
Jeans mass
Critical mass
A cloud with Mc>Mj will collapse under gravity
Cloud collapse temperature
Use virial theorem
3/2kbTM/mp=1/2 x 3/5 GM^2/R
What provides the energy needed to stabilise the cloud against further contraction
Temperature released by fusion
p-p chain reaction
- h1+H1 —> positron+neutrino+ H2 (positron denoted by beta)
- H1 + H2 —> He3 + gamma
- He3+He3 —> He4+2H1
Net reaction for p-p chain reaction
4H1 —> He4 + 2 positrons + 2 neutrinos + 2 gamma
What is needed for p-p chain reaction
Carbon
Acts as a catalyst but not used up
For CNO cycle to work at all
Carbon nitrogen and oxygen must be present which depends on the star type
When does pp chain dominate
Lower masses and temperature
When does CNO cycle dominate
Higher temperatures
Why is CNO more important for heavier stars
Energy production rate varies strongly with CNO cycle so is more important for heavier stars which have higher interior temperatures
Are we justified in modelling stars as gases
Yes
At high temperature hydrogen is ionised into protons and electrons (TINY compared to H)
In photosphere density of H tiny and mostly unionised
Density of neutral H much lower than maximally packed H, behaves like an ideal gas
Hydrostatic equilibrium
Inside a stable star, inward pull of gravity is balanced by the thermal pressure due to heat
Hydrostatic equilibrium equation derivation
In equilibrium, forces on each side balanced
Fpressure=Fgrav
Solving hydrostatic equilibrium equation
Differential equation
Solve to find pressure as a function of radius
Assume uniform density
Need boundary conditions: P(R)=0, at centre, r=0 let pressure be P0
Solve by integrating both sides
What would make a more accurate estimate of hydrostatic equilibrium
Density increase towards centre
Central temp in sun
Use ideal gas law
Rewrite in terms of P and R=Nakb
Simplify using Nana=M and divide by u (mean molecular weight)
More realistic models of stellar structure
Allow for temperature, pressure and density gradients inside star
Convection can occur instead of simple radiation
What transpires heat energy from core to surface in sun like stars
Both convection and radiation
Heat transfer in stars <0.5 solar masses
Convection
0.5-1.5 solar masses heat transfer
Radiation then convection
Heat transfer >1.5 solar masses
Convection then radiation
End states of stellar evolution
White dwarfs, neutron stars and black holes
When fuel runs out
Star will resume contracting since there is no longer a thermal pressure to balance gravity
Low mass stars
Stars dues elements up to carbon
After, because stellar mas low, core pressure due to self gravity is not enough to cause temp rise high enough for heavier elements
Carbon core collapses to form white dwarf, outer layers expelled to form planetary nebula
High mass stars
Continue fusion up to nickel and iron
Iron is most tightly bound atomic nucleus, no more energy available from fusion
Star collapses but much more massive, more gravitational energy can be released, more violent event - supernova
ZAMS
zero age main sequence mass
Supernovae are classified into different types depending on
Whether they show hydrogen in their spectrum
Type I have almost no H, type II do show presence of H
Type Ia supernovae are believed to be the result of
Accreting white dwarfs being pushed over the mass limit and undergoing further nuclear fusion in a runaway reaction that destroys the star
Types Ib, Ic and II result from
Collapse of high mass stars straight to a neutron star or black hole
SN with no H but has Si
Type Ia
SN with no H, no Si but does have He
Type Ib
SN with no H, Si or He
Ic
SN with H
Type II
(First thing to lo9 for are Balmer lines
What are type Ia SN characterised by
Silicon
Where are types Ib and Ic found
Only in spiral galaxies where there was recent star formation (implies they are due to massive short lived stars)
Onion like shell structure with fusion by products
Final stage is silicon burning, generates host of nuclei centred around 56Fe minimum of the 26 binding energy curve
Each stage less energy efficient (take shorter and shorter) eg H burning 107 years, Si ~days for star of 20 solar masses
Nuclear fuel exhausted in core, radiated energy can no longer be replaced so
Pressure falls, hydrostatic equilibrium lost, core rapidly contracts
Releases g potential energy, T rises allowing endothermic reactions
Photodisintegration
Earlier exoteric fusion reactions are reversed
Neutron capture
Endothermic reactions create elements heavier than Fe by neutron capture
Can only get elements heavier than Fe because of supernovae
Electron capture
Endothermic reactions absorb gravitational energy released by initial collapse, further contraction, further temp rise
Core can get hot enough for inverse beta decay
Causes extreme pressure drop in core
Core separates from outer envelope and goes into free fall
Collapse
Closely packed neutrons
Core stiffens due to degeneracy pressure and abruptly halts collapse
Sends shock wave back up through in falling material
Shock wave initiates further endothermic reactions and loses energy
Remant
Bouncing material and enormous neutrino flux blows outer layers off star into space, forming supernovae remnant
Why do SN remnants expand so quickly
Momentum transferred from lower layers to upper
Shock wave carries energy and momentum, encountering material of ever decreasing density
P=mv roughly constant p and smaller m so v big
Luminosity- energy source is
the release of gravitational potential energy by the contraction of the core
What do supernovae produce
Cosmic rays (extremely high energy particles, mostly protons)
Supernovae distribute
Heavier elements into space which are incorporated into new stars and planets
Pauli exclusion principle
Allows at most one fermion to occupy a given quantum state since no two fermions can have the same set of quantum numbers
Degenerate
At zero temperature all of lower states and none of higher states occupied
Heisenberg uncertainty principle
Particle confined to smaller and smaller volume in space will have a larger uncertainty in momentum
Degeneracy pressure
In stellar remnant, central pressure must be balanced by the cold pressure
Simplified model
Cubic meter of particles
Number density n so contains n particles
Pauli - different particles
Uncertainty in position of particles in simplified model cannot be
Larger than their physical separation
Spacing between each particle is delta x so volume
Delta x cubed
Delta x =
1/n1/3 so n=1/V
By HUP each particle has momentum in x direction of
Px ~ h bar/ delta x
Average kinetic energy of non relativistic particles
E=1/2mv^2 (sub in momentum)
*p^2 = 3px^2
Average quantum energy of electron
Sub in equation for momentum into Ek
Since me«mp
Ed»
Electron pressure dominates over proton pressure for white dwarfs
To find degeneracy pressure
Consider work done by changing volume of box by dx
