Stellar Structure

"*Hadyja HBV (BR) Don El Chall Harley FHP 16 March 2002 In foal to *Magic Dream CAHR - foal date 3/2/2011 Hadyja embodies the definitive traits of a Brazil mare - big, bold, grey mare and still enchantingly feminine with a proportional harmonious design and enrapturing charisma. Her sire, Don El Chall, world renowned for their superior type, noble elegance, perfectly balanced three-dimensional proportion and stellar structure.Ê

Stellar Structure

• What stops a star collapsing?
• How long will stars last?
• Why do stars switch from being main-sequence to red giant?
• How is surface related to core of star?
• What powers a star?
• This is "stellar structure" problem

Time Scales

1. Free-fall time:
   i.e. how long would it take for a particle to fall from the outside of a star to the centre if there was no resistance:
   \color{red}{ T = \pi \left( {\frac{{R^3 }}{{GN}}} \right)^{1/2} \sim 4000s \sim 1hr}
   (can get this either by doing the calculation correctly, or by dimensional analysis).

   Computer shows evolution to close to main-sequence in free-fall time!!!

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   • Kelvin-Helmholtz time: How long would the sun run on gravity?

     Power density of sun is \color{red}{ \frac{L}{M} = \frac{{4 \times 10^{26} }}{{2 \times 10^{30} }} \simeq 2 \times 10^{ - 4} W/kg} (ball-park figure is 1 watt/tonne) Now think of the sun being built up as a series of layers: \color{red}{ V\left( R \right) = \int_0^R {\frac{{Gm\left( r \right)\rho \left( r \right)4\pi r^2 }}{r}dr} = \frac{{16\pi ^2 G\rho ^2 }}{3}\int_0^R {r^4 dr} = \frac{{16\pi ^2 G\rho ^2 R^5 }}{{15}}} However the total mass \color{red}{ M = 4\pi R^3 \rho \Rightarrow V = \frac{3}{5}\frac{{GM^2 }}{R} \approx 3 \times 10^{41} J}

   • This gives a gravitational lifetime of the sun of
     \color{red}{\frac{V}{L} = \frac{{3 \times 10^{41} }}{{4 \times 10^{26} }}\~8 \times 10^{14} s} or 24 million years (!!)
   • This is long, but not long enough/ (Note: when calculated in ∼1860, this lifetime was reasonable, but geologists already suspected that some rocks were older)

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2. Einstein timescale

   In principle,

   \color{red}{ E = Mc^2 \approx 2 \times 10^{47} J}
   for a lifetime of 14.6x10¹² yrs (which is plenty!)

   In fact, process is about ∼0.1% efficient, so we get a reasonable life of ∼ 14x10⁹ yrs.

3. A note in passing: Chemical timescale:
4. Typical reaction gives ∼ 10⁸ J/kg Hence lifetime ∼ 10¹² s ∼ 10⁵ years
5. Now the details............

Stars: General Equations

At every point in star, there are six quantities (which depend on time as well) . Stellar structure problem is to relate them

Pressure P(r) Density ρ(r) Temperature T(r) Luminosity L(r) Mass "inside" m(r) Energy Production ε(r)
Mass-density is easy \color{red}{ \delta m = 4\pi r^2 \rho \left( r \right)\delta r} so \color{red}{ \frac{{dm}}{{dr}} = 4\pi r^2 \rho \left( r \right)} Note earlier calc was a special case ρ = constant

Hydrostatic equilibrium: Shell must be stable, so forces must balance, Grav. force on shell \color{red}{\begin{array}{l} F_G = - \frac{{Gm(r)M_{shell} }}{{r^2 }} \\ = - \frac{{Gm(r)4\pi r^2 \rho \left( r \right)\delta r}}{{r^2 }} \\ \end{array}} Pressure must balance this. \color{red}{ \delta P = P\left( {r + \delta r} \right) - P\left( r \right)} (must be negative) Total force due to pressure \color{red}{ F_P = 4\pi r^2 \delta P = - F_G } Hence \color{red}{ \delta P = - \frac{{Gm\left( r \right)\rho \left( r \right)\delta r}}{{r^2 }} \Rightarrow \frac{{dP}}{{dr}} = - \frac{{Gm\left( r \right)\rho \left( r \right)}}{{r^2 }}} (- sign means pressure decreases as we move outwards)

e.g. assuming density is constant, what is central temp and pressure of sun?

Where does the pressure come from?

Equation of state relates pressure to temp and density. Three possible ways:

1. Gas Pressure
2. Radiation pressure
3. Degeneracy pressure

1. Gas pressure:
   \color{red}{ PV = NRT}
   or (more useful for us)
   \color{red}{ P = \frac{{k\rho T}}{{\mu m_H }}}
   • m = mass of H, k = Boltzmann constant,
   • μ is mean atomic weight: if star was pure H, μ = 1
     
   • . .................pure ionised H, μ = ¹/₂ (since electrons form part of gas, but have m∼ 0)
     
   • In general
     \color{red}{ \frac{1}{\mu }^{} = 2X + \frac{3}{4}Y + \frac{1}{2}Z \sim 1.6}
   • X = mass fraction of H
     
   • Y = .......................He
     
   • Z = .......................metals

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   • Radiation pressure:
     photons form part of gas, but can be created and destroyed (i.e. γ density depends on temp: this is what the black body curve tells us)
     
     \color{red}{ P = \frac{1}{3}\sigma cT^4 }
     Which depends strongly on temp, but not density

   • Degeneracy Pressure
     Electrons satisfy exclusion principle: cannot put more than one into same quantum state. Electron pressure
     
     \color{red}{ P = \frac{{\kappa \rho ^{5/3} }}{{m_e }} \sim \frac{{10^{12} \rho ^{5/3} }}{{m_e }}}
     Note this does not depend on temperature at all, but depends very strongly on density

• Gas Pressure important in "normal" stars (M < 2M₀)
• Radiation pressure important in hot (i.e. massive) stars
• Degeneracy pressure totally unimportant at normal densities: need ρ ∼ 10⁹ kg.m^-3 i.e. white dwarfs (will return to this later)

Energy transport: i.e. how does energy move out from solar core?

1. Conduction: totally unimportant for any star, since pressures are low
   
2. Radiation
   
3. Convection

• Radiation: Temp difference across shell is δT Hence the energy/unit area radiated out \color{red}{ F\left( r \right) = \sigma T\left( r \right)^4 } Energy radiated inwards: \color{red}{ F\left( {r + \delta r} \right) = \sigma T\left( {r + \delta r} \right)^4 } So the difference is \color{red}{ \delta F = 4\sigma T\left( r \right)^3 \delta T}

• ∝ amount of radiation
  
• ∝ amount of material
  
• ∝ thickness of layer
  
• ∝ "lack of transparentness" = opacity κ(r,λ)
  
• (e.g. glass would have opacity 0 for optical γ's)
  

• Combine and use Luminosity = total outwards flux
  \color{red}{ L\left( r \right) = 4\pi r^2 F\left( r \right)}
  gives
  
  \color{red}{ \frac{{dT}}{{dr}} = - \frac{{\kappa \left( r \right)\rho \left( r \right)L\left( r \right)}}{{16\pi r^2 \sigma T\left( r \right)^3 }}}
• (actually we cheated slightly: the correct equation multiplies this by 4/3, due to an angular averaging).
• Note dT/dr is negative: the luminosity increases

How do we calculate κ(r,λ,T)?
Extremely difficult from first principles Only people who really care are bomb makers: hence only good codes are at Los Alamos and Livermore (and no, you can't borrow them!!)

A reasonable approx. that works for ionised gases over a large range is

Convection

• Imagine a bubble of gas which is slightly less dense than its surroundings: if it rises and the density decreases faster than the surrounding gas, it will continue to rise. Adiabatic temp gradient. \color{red}{ P_1 V^\gamma _1 = P_2 V^\gamma _2 } and \color{red}{ P_1 ^{\frac{1}{\gamma } - 1} T^{} _1 = P_2 ^{\frac{1}{\gamma } - 1} T^{} _2 } Hence: \color{red}{ T^{} = CP_{} ^{1 - \frac{1}{\gamma }} } so \color{red}{ \frac{{dT}}{{dr}} = C\left( {1 - \frac{1}{\gamma }} \right)\frac{1}{{P_{} ^{\frac{1}{\gamma }} }}\frac{{dP}}{{dR}} = C\left( {1 - \frac{1}{\gamma }} \right)\frac{T}{P}\frac{{dP}}{{dR}}} If the actual temp gradient in a star is greater than this, then convection will set in. Unfortunately, it doesn't tell you how much energy is moved around!

• Note same happens on earth on hot summer days: air heated near ground is adiabatically unstable, and rises => thunderstorms

Finally energy generation: obviously(?)

Equations In summary

• Mass-density
  \color{red}{ \frac{{dm}}{{dr}} = 4\pi r^2 \rho \left( r \right)}
• Eqn. of State:
  \color{red}{ P = \frac{{k\rho T}}{{\mu m_H }}}
• Luminosity/Energy generation
  \color{red}{ \frac{{dL}}{{dr}} = 4\pi r^2 \rho \left( r \right)\varepsilon \left( r \right)}
• Energy transport
  \color{red}{ \frac{{dT}}{{dr}} = - \frac{{3\kappa \left( r \right)\rho \left( r \right)L\left( r \right)}}{{64\pi \sigma r^2 T^3 }}}

• Hydrostatic Equilibrium
  \color{red}{ \frac{{dP}}{{dr}} = - \frac{{Gm(r)\rho \left( r \right)}}{{r^2 }}}
• Opacity:
  \color{red}{ \kappa \left( r \right) = \frac{{CZ\left( {1 + X} \right)\rho \left( r \right)}}{{T^{7/2} }}}
• Energy generation: (roughly for hydrogen burning)
  \color{red}{ \varepsilon \left( r \right) = a\left( {\frac{T}{{T_0 }}} \right)^4 }
• Note that stars are much simpler than (e.g..) human beings: on the other hand we still need a computer to solve for one!

Guess \color{red}{ \rho \left( r \right) = \rho _0 \left( {1 - \frac{r}{R}} \right)^2 } This vanishes at the surface of the sun (as it must). Then \color{red}{ m\left( r \right) = \int_0^r {4\pi r'^2 } \rho \left( {r'} \right)dr'} can be found exactly and m(R₀) = M₀ ⇒ \color{red}{ \rho _0 = \frac{{15M_0 }}{{8\pi R^3 }} \sim 3500} (too low)

Then< \color{red}{ P\left( r \right) = \int_0^r G \frac{{\rho \left( {r'} \right)m\left( {r'} \right)}}{{r'^2 }}dr' + P_0 } (what is P(Ro?) gives \color{red}{ P_0 = \frac{{15GM_0^2 }}{{16\pi R_0^4 }}}

Then \color{red}{ P = \frac{{k\rho T}}{{\mu m_H }}} gives \color{red}{ T_0 = \frac{{\mu m_H GM_0 }}{{2kR_0}} \sim 7.2 \times 10^6 K} (best estimate is 14x106K, so it's a bit low

One can get the luminosity from this (assuming H cycle)

• and fit the result to the total luminosity of the sun

T

• his is a star in the steady state.
• What happens as a function of time is that ε(r,T) changes: e.g.
  \color{red}{ \varepsilon \left( {r,T} \right) = k_H \rho _H T\left( r \right)^4 + k_{He} \left( {\rho _{He} } \right)^2 T\left( r \right)^{15} + ............}
• Star will always burn easiest nucleus
• low temp burning involves H
• once the H is consumed via H ⇒ He , star must go over to He burning

Nuclear reactions: long preamble

Forces

Strong force is the force that binds together nucleons to make nuclei, weak force is force that causes β-decay. Believe there are only 4 forces in nature

Note: "feels" means that this is what the force couples to: e.g. gravity does not care whether a particle is charged, only whether it has mass.

Range: if it is ∞ then F ∝ 1/r², else it cuts off at distance shown

Strength: roughly the relative strength of the forces at a distance of 1 fm.

Although (e.g.) strong force>>>E.M at 1 fm (=10^-15 m), it vanishes totally beyond 10^-14 m. E.M >>> Gravity, but it tends to cancel out since most matter is electrically neutral, whereas mass accumulates.

The weaker the force, the more particles feel it!

Particles

Strength also gives us (very roughly) the depth that a particle will penetrate matter without interacting: e.g. a proton will penetrate a few mm, a X-ray photon a few cm, a neutrino several parsecs!

Antiparticles:

For every particle, with given quantum numbers, there is a corresponding anti-particle with the properties flipped:

e.g. electron has charge -1.6x10^-19 Coulomb.
Positron has same mass, charge = 1.6x10^-19 C

Conservation laws

• Mass-energy conservation:
• Usually easier to quote elementary particle masses in terms of energy, measured in eV.
• e.g. "mass" of electron \color{red}{m_e c^2 = 511keV}
• Conservation of momentum:
• Usually (from point of view of astrophysics) means that reactions are: 2 bodies ⇒ 2 bodies
• e.g. \color{red}{p + n \to d + \gamma } is allowed
• \color{red}{p + n \to d} (where the d moves off with the extra energy) is forbidden.

Conserved Quantum Numbers

• Conservation of charge
  \color{red}{ n \to e^ - + \gamma }
  is forbidden: Algebraic sum of charges at end of reaction must = sum at start

  note

  \color{red}{ \gamma \to e^ + + e^ - }
  is allowed
• Conservation of baryon number:

  The process

  \color{red}{ p \to e^ + + \gamma }
  is not forbidden by anything: however experimentally it does not occur:

  No of (protons + neutrons) is always the same in a reaction:

  Easiest to say that p & n carry baryon number B = 1, rest carry 0

• Conservation of lepton number: Number of (electrons + neutrinos) is conserved

  L is lepton number = 1 for electrons,
  -1 for positrons

  (\color{red}{{\bar \nu }} means anti-neutrino: what is an anti-neutrino? one with L = -1).

Conservation of angular momentum ("spin"):

Mainly important because some particles carry spin ½:
e.g. n ⇒ p + e^- is not allowed since
½ ⇒ ½ + ½ requires creation of angular momentum.
Instead n ⇒ p + e^- + ν
½ ⇒ ½ + ½ + (-½)

A bit more subtle than this, since we need to have orbital angular momentum added in

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These conservation laws let us make up an extended particle table. The numbers are all conserved: e.g. why doesn't n ⇒ p e- γ happen?"

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Families

For later, we need to know something about "families". Easiest with leptons:

Lepton #	Charged lepton	Lepton mass	Neutrino	Sample Reaction
Le	e-	.511 MeV	νe	n ⇒ p + e- + ν̄e
Lμ	μ-	105MeV	νμ	μ- ⇒ e- + ν̄e+ νμ
Lτ	τ-	1784 MeV	ντ	τ- ⇒ μ- + ν̄μ+ ντ

• How do we know there aren't more? LEP shows we have exactly 3.00 kinds of neutrinos
• Why 3 families? We don't know!
• Why the masses? We don't know!
• How heavy are the neutrinos? Very small but not zero!

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Nuclear Physics

• The only extra ingredient we need for astro-nuclear physics is the stable nuclei and their binding energies: e.g. deuteron is stable: the state (pp) is not: Rule of thumb: stable nuclei need Number of protons ∼ number of neutrons

• If a nucleus has too many neutrons, one will decay via \color{red}{n \Rightarrow p + e^ - + \bar \nu }

  If it has too many protons, it will decay via \color{red}{p \Rightarrow n + e^ + + \nu }

• Note that there are no stable elements of A = 5 or A = 8

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As a rule of thumb, most reactions up to Fe are exothermic, any past that are endothermic

• Since stars get their energy via fusion reactions the most important curve is binding energy vs A
• (Put differently): fusion reactions can occur for light elements, since medium-heavy elements are more stable, fission reactions can occur for heavy elements.

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How do stars work?

• Why do stars need to be hot (or why is there no cold fusion?)

• There is a large repulsive E.M. force at large distances between two protons: only if r < 1 fm is the force attractive (fm=femtometre =fermi =10-15m)

• Height of this barrier is \color{red}{E = \frac{{kZ_1 eZ_2 e}}{{R^2 }}}
• Hence to "ignite" thermonuclear fusion, it would appear to need temp ~ 1MeV, for Z =1
• using E = ³/₂ kT, gives T ~ 8x10⁹ K.
• This is temp reached in H-bomb: however the sun can work more slowly! (Central temp is 13x10⁶ K)

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Quantum Mechanics tells us that particles can tunnel through barrier:

• For a particle of given energy. Prob of tunnelling P(E)~ exp(-1/√(E)) which gets larger for large E However there are very few nuclei with this energy Ignition Temp T₀~ 107K Rates depend very strongly on temp and charge of nuclei: e.g. for pp, R ~ T4

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Star starts off as H + ⁴He: what reactions can occur?

• \color{red}{ p + p \to ^2 He + \gamma }
  can't occur, ²He doesn't exist
• \color{red}{ p + p \to d + e^ + + \nu }
  is the initial process
• and is very slow (the ν has only weak interactions, and so the whole process must go by that)

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Once past this, the reactions are simple:

• \color{red}{ p + d \to ^3 He + \gamma }
  \color{red}{ d + ^3 He \to ^4 He + p}
• or various variations:
• Most important is the "executive summary"
  \color{red}{ 4p \Rightarrow ^4 He + 2e^ + + 2\nu + 27MeV }
• And this is what keeps us warm!

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Past H burning, there are various processes that build up heavier nuclei: the crunch comes at A = 8: ⁸Be is unstable (τ ∼ 10^-16 s) but Triple-α process occurs:

\color{red}{ ^4 He + ^4 He + ^4 He \Rightarrow \left[ {^8 Be} \right] + ^4 He \Rightarrow ^{12} C}

Since three particles are involved, Rate ∝ ρ² (Two body reactions ∝ ρ) Means only happens at high (>10⁸K) temps and very high pressures

Beyond 12C processes add whole nuclei until we get to Fe

Now we'll move to a larger scale and look at galaxies