Stellar Structure: General Principles and Nuclear Physics

Now want to understand how stars "work".
e.g

• 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?

This is "stellar structure" problem

Some general comments first: There are 3 time scales involved with stars

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:

   T ∼ π (R³/GM)1/2    ∼ 4000s ∼ 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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2. Kelvin-Helmholtz time:

   How long would the sun run on gravity?

   Power density of sun is

   L/M = 4*1026/2*1030 = 2*10-4 W/kg 
   

   (ball-park figure is 1 watt/tonne)

   ---

   
   

   Can think of sun as being built up by successive layers of material. Mass of "core" m = 4/3 π r³ρ Mass of "shell" δm = 4πr²ρδr Hence P.E. for adding this shell is: V = G/r mδm = G/r 4/3πr³ρ 4πr²δrρ = 16/3π² G ρ² r⁴ δr Now think of the sun being built up as a series of layers: V = 16/3 π² G ρ² ∫₀r r⁴ dr = 16/15 π² G ρ² R5 However the total mass M = 4π R³ ρ so V = 3/5 G/R M² ∼ 3x1041 J

   
   

   ---

   This gives a gravitational lifetime of the sun of
   

   V/L = 3x1041J / 3.9x1026W
   
   ∼ 7.7x1014 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)

   ---

3. Einstein timescale

   In principle,

   E = Mc² = 2x1030 x (3x108)² = 1.8x1047 J, for a lifetime of 1.8x1047 J / 3.9x1026 W = 4.6x1020s  or 14.6x1012 yrs 
   

   (which is plenty!)

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

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

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 δm = 4πr² ρ(r) δr so dm = 4πr² ρ(r) dr
Hydrostatic equilibrium: Shell must be stable, so forces must balance, Grav. force on shell FG = -G m(r) (Mass of shell) / r² =-G m(r) 4πr²ρ(r)δr / r² Pressure must balance this. δP = P(r+δr) - P(r) (must be negative) Total force due to pressure FP = 4πr² δP δP=-G m(r) ρ(r)δr / r² or dP = -G m(r) ρ(r) dr r² (- 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:

   P V = N R T
   

   or (more useful for us)

   P =  k ρ T / μm
   

   • 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 μ^-1 = 2X + ³/₄Y + ¹/₂Z ∼ 1.6
     
   • X = mass fraction of H
     
   • Y = .......................He
     
   • Z = .......................metals

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2. 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)
   

   P = 1/3 σ c T⁴
   

   Which depends strongly on temp, but not pressure

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3. Degeneracy Pressure
   Electrons satisfy exclusion principle: cannot put more than one into same quantum state. Electron pressure
   

   P = K ρ5/3/m₀ ∼ 1012 ρ5/3m₀ Pa
   

   Note this does not depend on temperature at all, but depends very strongly on density

All three occur in all stars, but

• 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 F(r) = σT(r)⁴ Energy radiated inwards: F(r+δr) = σT⁴(r+δr) So the difference is δF = σ 4T³ δT

This energy must be absorbed by the layer: amount absorbed δF
∝ 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)

δF = -κ(r) ρ(r) F(r) dr

Combining

δF = σ 4T³ δT = -κ(r) ρ(r) F(r) δr

L(r) = 4πr² F(r) 

dT = - κ(r) ρ(r)  L(r) 
dr       16 π σ  r² T³

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

κ = C Z(1+X) ρ / T7/2

(i.e. it doesn't depend on wavelength)

Much harder is convection: if the opacity is very high, the gas will just heat up at the bottom, which will make it less dense than cooler gas above

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. If the actual temp gradient in a star is greater than this, then convection will set in. P1V1γ = P2V2γ and PV = NkT So T1P11/γ - 1 = T2P21/γ - 1 Hence: T = const P1 - 1/γ dT/dR = const (1-1/γ) P-1/γ dP/dR = (1-1/γ) T/P dP/dR

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

dL  = 4π r² ρ(r) ε(r)
dr

In summary

1. dL  = 4π r² ρ(r) ε(r)
   dr
   

   Luminosity/Energy generation

2. dT = -3 κ(r) ρ(r)L(r)  
   dr    64 π σ  r² T³
   

   Energy transport

3. dP = -G m(r) ρ(r)
   dr         r²
   

   Hydrostatic Equilibrium

4. dm = 4πr² ρ(r)
   dr
   

   Mass-density
5. Eqn. of State:

   P =  k ρ T / μm
   

6. Opacity:

   κ=C Z (1+X) ρ / T7/2
   

7. Energy generation:

   ε(r) = a (T/T₀)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!

A very simplified model which can be solved by hand (or preferably by computer algebra!) is to take

T

ρ = ρ₀(1-r²/R²) This vanishes at the surface of the sun (as it must). Then m(r) = ∫₀r 4πr²ρ(r) dr can be found exactly and m(R₀) = M₀ ⇒ ρ₀ = 15M₀ / 8πR³ ∼ 3500 (too low)
Then P = ∫₀r G m(r) ρ(r) / r² dr + P₀ gives P₀ = 15 GM₀² / 16π R⁴

/P>

T = μ m P / k ρ gives T₀ = μ m G M₀ / k 2 R ∼ 7.2x106 K (a bit too low)

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

and fit the result to the total luminosity of the sun

This is a star in the steady state. What happens as a function of time is that ε(r,T) changes: e.g.

low temp burning involves H

ε ∝ (density of H)xT⁴

once the H is consumed via H ⇒ He , star must go over to He burning

ε ∝ (density of He)xT15