Stellar Physics Flashcards

1
Q

Method to find distance to nearer stars

A

Parallax

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2
Q

Luminosity of a star

A

It’s energy output per unit time
Measured in watts
If star radiates isotropically flux equal across sphere
L=4pir^2F

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3
Q

Visual magnitude

A

Magnitude in the visual part of the spectrum

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4
Q

If a star is hotter than its environment

A

It will cool down by re-radiating it’s energy

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5
Q

Rayleigh jean law for blackbodies

A

Good approximation at Lon wavelengths but radiance keeps increasing indefinitely at short wavelengths (UV catastrophe)

Failure of classical physics to explain thermal radiation

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6
Q

Wien’s law

A

Good approximation to the observed spectrum at short wavelength

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7
Q

Planck’s law

A

Assuming energy comes in discrete quanta
Fit data

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8
Q

What determines apparent colour of an object

A

Shape of spectrum and position of its peak

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9
Q

Atmospheric extinction

A

Light from sun obscured by atmosphere of Earth, which absorbs more than others.
Ignore in this course

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10
Q

Stefan Boltzmann law

A

Total luminosity per area is the spectral radiance integrated over solid angle and wavelength

j=L/A=sigma T^4

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11
Q

Total luminosity of a spherical star

A

Multiply Stefan Boltzmann law by surface area

L=4piR^2sigmaT^4

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12
Q

Stars have wide range of luminosity so helpful to use logs

A

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

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13
Q

HR diagram

A

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

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14
Q

Interpretation of HR diagram

A

Only certain combinations of L and T allowed
Most stars on main sequence

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15
Q

What tells us that stars move around HR diagram as they evolve

A

Clusters are stars at similar stages of their lives
Number of stars in each part proportional to duration of that stage of evolution

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16
Q

Main sequence

A

Most variation from top left to bottom right along a line of roughly constant radius
Top left blue stars hotter and more luminous

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17
Q

Two cut offs to the main sequence

A

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

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18
Q

Estimate of time a star spends on main sequence

A

Lifetime=energy available/ luminosity

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19
Q

Most massive stars spend…

A

Least amount of time on main sequence

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20
Q

Giants and supergiants

A

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

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21
Q

White dwarfs

A

Bottom left
Below main sequence so radius s smaller
None visible to naked eye
Are not powered by nuclear fusion

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22
Q

Photometric system

A

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

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23
Q

Colour index

A

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

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24
Q

The smaller the colour index (ie lower position on number scale that ranges from positive through zero into negative)

A

The more blue and hotter the star

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25
Kirchoff’s first law
A hot and opaque solid, liquid or highly compressed gas emits a continuous black body spectrum with no spectral lines
26
Kirchoff’s second law
A hot, transparent gas illuminated by a continuum source produces a spectrum of bright emission lines
27
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
28
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
29
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
30
O spectral class
Hottest blue stars Few lines Strong He II absorption lines
31
M spectral class
Coolest red stars Spectra dominated by molecular absorption bands especially TiO Strong metal absorption lines
32
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
33
What does allocation of subtype depend on
Line strengths and ratios
34
How does the Harvard classification scheme not completely describe a star
Cannot distinguish between stars with the same temperature but different luminosities
35
Morgan keenan luminosity class
Established to add discrimination on the basis of luminosity Ranges from I to VII
36
Class I
Hypergiant Divided I-O Through 1a, bright supergiant and 1b, dim supergiant
37
Classes II - V
Goes from bright giants down to main sequence dwarfs
38
Classes VI and VII
VI sub-dwarf VII white dwarf
39
What are luminosity classes determined from
Mainly from observed width of spectral lines
40
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
41
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
42
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
43
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
44
What does variable mean
Star’s flux changes over time Observe by measuring changes in apparent magnitude
45
Real and apparent variation
Real: star itself changes Apparent: eg something moves in front of star and blocks light partially or fully
46
Irregular and regular variation
Irregular: no particular pattern, can be sudden or random Regular obviously opposite
47
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
48
Grange point
Point in middle of binary system
49
Disk
Best way to distribute angular momentum
50
T Tauri stars
Class of irregular variables Luminosity increases by factor 3 in few days Very young powered by gravitational energy as they contract
51
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
52
Type 1a supernovae
Expect one type 1a SN every few decades
53
Types of supernovae
Classified according to observational features 1a all have nearly equal brightness - standard candles
54
Examples of regular variable stars
Cepheids
55
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
56
Radial pulsation results in a regular pulsation of
Velocity of star’s surface Effective temperature Luminosity
57
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
58
Cepheid period luminosity relation
Period of pulsation depends only on average luminosity of the star Longer pulsation period, the more luminous the star
59
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**
60
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
61
4 main classes of binaries
Visual Astrometric Spectroscopic Eclipsing
62
Visual binaries
With telescopes, possible to resolve two components
63
Astrometric binaries
Cannot resolve individual stars but where we see a periodic wobble of observed overall position
64
Spectroscopic binaries
Components are not resolvable but Doppler shits in spectral lines reveal there are two stars orbiting same centre of mass
65
Eclipsing binaries
Not resolvable but see periodic brightening and dimming
66
Binary system orbital analysis
For cases where one mass >> other, deal with orbit as if larger mass is stationary Masses of components can be more comparable. Newton and Kepler still apply
67
Binaries: if the masses are equal
The centre of mass is halfway between them
68
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
69
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
70
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
71
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
72
Speed toward/away
Reduced by inclination of orbital plane l is inclination angle v=vtrue sinl
73
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
74
Stars will eclipse each other only if
We are viewing the system near edge on (l~90 degrees)
75
Eclipsing binary: most light when
No overlap
76
Plot of magnitude against time for eclipsing binaries
Get dip at secondary minimum and bigger dip at primary minimum
77
Primary minimum
Hotter star moves behind the cooler
78
Secondary minimum
Cooler star behind the hotter
79
If T2>T1
Then F’>F1 so secondary minimum
80
T2
F’
81
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
82
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
83
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
84
Structure of sun
(From inner to outer) Interior, photosphere, chromosphere, transition region, corona
85
When we observe the sun, we see light from
Photosphere, chromosphere and corona In descending order of contribution
86
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
87
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
88
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
89
Solar flares
Extends hundreds of thousands of miles above surface Originate in photosphere Produce bursts of radiation across EM spectrum
90
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
91
Corona
Very hot Outer atmosphere extends as far as 3 solar radii Only really seen from Earth during eclipse Eventually leads to solar wind
92
Spectral line strength depends on
Number of atoms present and temperature of gas
93
Effects on lines due to bulk motion
Allow measurements of overall rotation and large scale expansion or contraction
94
Effects on lines due to random thermal motions of atoms in atmosphere
Allow measurements of temperature and pressure
95
Energy of nth level of hydrogen
E proportional to -1/n^2 - because binding energy
96
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)
97
Calculating wavelength of emitted photon
Using rydberg equation *wavelength must be positive so use n
98
Lyman series
Transitions from excited states to ground state
99
Balmer series
Transitions from m>2 to n=2
100
Emission
Electron drops to lower level Photon emitted, of energy corresponding to e difference between transition levels
101
Absorption
e jumps to higher level Photon of correct energy needed for this to happen
102
Strongest visible line from sun
Ha line e going from n=2 to m=3
103
Boltzmann factor
Probability of an electron occupying energy state E - tells us states with E>>kbT are very improbable
104
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
105
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
106
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
107
How do ions form
Absorption: of a photon with at least x joules of energy Collision: with another particle like an electron (scattering)
108
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)
109
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
110
Saha equation
No of atoms/ no ions ~ 10^21T^3/2exp(-x\kbT) / no of electrons
111
Rule of thumb for Saha equation
50% ionisation occurs when kbT~x/18
112
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
113
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)
114
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
115
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
116
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
117
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
118
Population III stars
Possible massive stars formed just after Big Bang Obviously no metals around then
119
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
120
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
121
Efficiency of sun’s energy
e=E/M> or = tauL/M Units if j/kg
122
Candidates for source of suns energy
Chemical energy Gravitational energy Relativistic energy Nuclear fusion
123
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
124
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
125
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
126
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
127
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
128
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
129
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)
130
If cloud is initially rotating even slightly
Rotation speed Increases as material collapses inwards
131
Spinning material
Flattens into protoplanetary disc with dense central part forming a protostar
132
Surrounding material accretes onto the protostar and as temp rises,
Gas becomes ionised
133
Eventually, if enough material accreted, temp becomes high enough to
Initiate nuclear fusion Radiation from the hot starpishes material out until equilibrium is reached
134
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
135
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
136
Brown dwarf
Pressure and temperature never high enough for nuclear fusion Will shine for its kelvin helmholtz timescale
137
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**
138
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)
139
Distinctive characteristics of pre main sequence stars
Unstable luminosity Eject lots of gas Surrounded by warm clouds
140
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
141
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
142
Virial theorem
Total kinetic energy of a stable, self gravitating mass distribution is negative one half of the total gravitational potential energy
143
What is virial theorem used for
Helps calculate the conditions that must exist for cloud collapse
144
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
145
Jeans mass
Critical mass A cloud with Mc>Mj will collapse under gravity
146
Cloud collapse temperature
Use virial theorem 3/2kbTM/mp=1/2 x 3/5 GM^2/R
147
What provides the energy needed to stabilise the cloud against further contraction
Temperature released by fusion
148
p-p chain reaction
1. h1+H1 —> positron+neutrino+ H2 (positron denoted by beta) 2. H1 + H2 —> He3 + gamma 3. He3+He3 —> He4+2H1
149
Net reaction for p-p chain reaction
4H1 —> He4 + 2 positrons + 2 neutrinos + 2 gamma
150
What is needed for p-p chain reaction
Carbon Acts as a catalyst but not used up
151
For CNO cycle to work at all
Carbon nitrogen and oxygen must be present which depends on the star type
152
When does pp chain dominate
Lower masses and temperature
153
When does CNO cycle dominate
Higher temperatures
154
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
155
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
156
Hydrostatic equilibrium
Inside a stable star, inward pull of gravity is balanced by the thermal pressure due to heat
157
Hydrostatic equilibrium equation derivation
In equilibrium, forces on each side balanced Fpressure=Fgrav
158
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
159
What would make a more accurate estimate of hydrostatic equilibrium
Density increase towards centre
160
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)
161
More realistic models of stellar structure
Allow for temperature, pressure and density gradients inside star Convection can occur instead of simple radiation
162
What transpires heat energy from core to surface in sun like stars
Both convection and radiation
163
Heat transfer in stars <0.5 solar masses
Convection
164
0.5-1.5 solar masses heat transfer
Radiation then convection
165
Heat transfer >1.5 solar masses
Convection then radiation
166
End states of stellar evolution
White dwarfs, neutron stars and black holes
167
When fuel runs out
Star will resume contracting since there is no longer a thermal pressure to balance gravity
168
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
169
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
170
ZAMS
zero age main sequence mass
171
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
172
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
173
Types Ib, Ic and II result from
Collapse of high mass stars straight to a neutron star or black hole
174
SN with no H but has Si
Type Ia
175
SN with no H, no Si but does have He
Type Ib
176
SN with no H, Si or He
Ic
177
SN with H
Type II (First thing to lo9 for are Balmer lines
178
What are type Ia SN characterised by
Silicon
179
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)
180
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
181
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
182
Photodisintegration
Earlier exoteric fusion reactions are reversed
183
Neutron capture
Endothermic reactions create elements heavier than Fe by neutron capture Can only get elements heavier than Fe because of supernovae
184
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
185
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
186
Remant
Bouncing material and enormous neutrino flux blows outer layers off star into space, forming supernovae remnant
187
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
188
Luminosity- energy source is
the release of gravitational potential energy by the contraction of the core
189
What do supernovae produce
Cosmic rays (extremely high energy particles, mostly protons)
190
Supernovae distribute
Heavier elements into space which are incorporated into new stars and planets
191
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
192
Degenerate
At zero temperature all of lower states and none of higher states occupied
193
Heisenberg uncertainty principle
Particle confined to smaller and smaller volume in space will have a larger uncertainty in momentum
194
Degeneracy pressure
In stellar remnant, central pressure must be balanced by the cold pressure
195
Simplified model
Cubic meter of particles Number density n so contains n particles Pauli - different particles
196
Uncertainty in position of particles in simplified model cannot be
Larger than their physical separation
197
Spacing between each particle is delta x so volume
Delta x cubed
198
Delta x =
1/n1/3 so n=1/V
199
By HUP each particle has momentum in x direction of
Px ~ h bar/ delta x
200
Average kinetic energy of non relativistic particles
E=1/2mv^2 (sub in momentum) *p^2 = 3px^2
201
Average quantum energy of electron
Sub in equation for momentum into Ek
202
Since me<
Ed>> Electron pressure dominates over proton pressure for white dwarfs
203
To find degeneracy pressure
Consider work done by changing volume of box by dx