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Stellar Atmospheres

""	© Michael Richmond

Radiation-Matter Interactions

Three closely related ideas fo Black body radiation:

Energy Density
Radiation Pressure
Radiated flux

all derive from Planck function:

\color{red}{ B_\lambda \left( T \right)d\lambda = \frac{{2hc^2 }}{{\lambda ^5 }}\frac{{d\lambda }}{{e^{hc/\lambda kT} - 1}}}

\color{red}{ B_\nu \left( T \right)d\nu = \frac{{2h\nu ^3 }}{{c^2 }}\frac{{d\nu }}{{e^{h\nu /kT} - 1}}}

Radiation emitted in time \color{red}{{\Delta t}} is

\color{red}{ \frac{{\Delta E}}{{\Delta t\Delta \nu }} = \int {I_\nu \Delta A\cos \left( \theta \right)} d\Omega }

(Note \color{red}{I_\nu = B_\nu } for isotropic BB radiation, but this is more general)

so specific flux (i.e flux for given frequency: W m^-2Hz^-1)

\color{red}{ F_\nu = \frac{{\Delta E}}{{\Delta t\Delta \nu \Delta A}} = \int {I_\nu \cos \left( \theta \right)} d\Omega }
For radiation, can assume that all radiation is from inside star \color{red}{0 \le \theta \le \frac{\pi }{2}} so angular integral becomes
\color{red}{ F_\nu = I_\nu \int\limits_0^{2\pi } {d\varphi } \int\limits_0^{\pi /2} {\cos \left( \theta \right)} d\left( {\cos \theta } \right) = \pi I_\nu }

Can integrate this:

\color{red}{ u = \frac{{hc}}{{\lambda kT}} = \frac{{h\nu }}{{kT}} \Rightarrow \lambda = \frac{{hc}}{{ukT}},\nu = \frac{{kTu}}{h}}

and use

\color{red}{ \int\limits_0^\infty {\frac{{u^3 du}}{{e^u - 1}}} = \frac{{\pi ^4 }}{{15}}}

to obtain Stefan-Boltzmann:

\color{red}{ u = \sigma T^4 ,\sigma = \frac{{2\pi ^5 k^4 }}{{15c^2 h^3 }}}

(note S-B const. \color{red}{ \sigma = 5.67 \times 10^{ - 8} Wm^{ - 2} K^{ - 4} } is derived quantity)

Energy Density

Consider a small area dA in a cavity at uniform temp T:

radiation energy density u must be in equilibrium with walls.

Small area is in thermal equil.: energy emitted

\color{red}{ U_e = 2dA\sigma T^4 }

Must be equal to the incident energy: travels at c, so total energy absorbed is
\color{red}{ U_a = \frac{u}{c}\left\langle {\cos \theta } \right\rangle dA}
Hence total energy density is
\color{red}{ u = \frac{{4\sigma T^4 }}{c} = aT^4 }

BB radiation pressure

Consider a 1-photon gas colliding elastically with wall. the wall will produce a force on the wall, which is the pressure of the gas.

Change in momentum when molecule hits wall

\color{red}{ \Delta p = - 2p}

This provides an impulse to the wall. If we have a distance L between the walls, time between the collisions with one wall is

\color{red}{ t = \frac{{2L}}{v},v = c}

Hence average force = change in momentum in one second

\color{red}{ F = \frac{{\Delta p}}{t} = \frac{{pc}}{L}}

In 3-D?

Now suppose we have N photons present:Pressure

\color{red}{ P = \frac{F}{A} = \frac{{NE}}{{L^2 L}} = nE}

Can regard each dimension as separate

\color{red}{ \left\langle {p_x } \right\rangle = \left\langle {p_y } \right\rangle = \left\langle {p_z } \right\rangle = \frac{{\left\langle p \right\rangle }}{{\sqrt 3 }} = \frac{{\left\langle E \right\rangle }}{{\sqrt 3 c}}}

Only one component gets reflected:

\color{red}{ \left\langle {\Delta p} \right\rangle = - 2\left\langle {p_x } \right\rangle }

also

\color{red}{ \left\langle v \right\rangle = \frac{c}{{\sqrt 3 }}}

so pressure
\color{red}{ P = \frac{{2n\left\langle E \right\rangle }}{{\sqrt 3 c}}\frac{c}{{2L\sqrt 3 }}\frac{1}{{L^2 }} = \frac{1}{3}n\left\langle E \right\rangle = \frac{u}{3}}
(note this works for any relativistic particle, including ν's.)

Radiation Pressure

Previous argument is for Local Thermodynamic Equilibrium (LTE): however we often encounter asymmetric situation: e.g. photons radiated from photosphere of star will provided pressure on outer layes.

Momentum of photon

\color{red}{ p = \frac{E}{c} = \frac{{h\upsilon }}{c}}

Hence total pressure of star:

\color{red}{ P_{rad} = \frac{1}{c}\frac{{L_{tot} }}{{4\pi r^2 }}}

Where does the pressure come from?

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

Gas Pressure
Radiation pressure (usually small)
Degeneracy pressure (later)

Gas pressure:

\color{red}{ PV = NRT}

or (more useful for us)

\color{red}{ P = \frac{{k\rho T}}{{\mu m_H }}}

m_H = mass of H, k = Boltzmann constant,
\color{red}{m_Y = \mu m_H } is mean atomic weight: if star was pure H, m_Y = 1, but ionized H would have m_Y = 1/2, (since electrons form part of gas, but have m∼ 0)
. .................pure ionized He, m_Y = ⁴/₃ (why?)
In general \color{red}{\frac{1}{{m_Y }} = 2X + \frac{3}{4}Y + \frac{1}{2}Z \sim 1.6}
X = mass fraction of H
Y = .......................He
Z = .......................metals

A stellar atmosphere must be in equilm: Gravity must balance pressure

\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)

\color{red}{ \frac{{dP}}{{dr}} \approx - g\rho \left( r \right),g = \frac{{GM}}{{R^2 }}}

where g is the surface gravity" of the star.

\color{red}{ \rho = \frac{{P m_Y }}{{kT}}}

combining gives

\color{red}{ \frac{{dP}}{{dr}} = - g\frac{{Pm_Y }}{{kT}}}

with obvious solution if T is constant:

\color{red}{ P \propto e^{ - \frac{r}{H}} }

Hence equilibrium requires a "Scale height"

\color{red}{ H \approx 300km}

(equivalent on earth is about 8 km: what is pressure on top of Everest?) Note this 300 km is small compared to the size of the sun, so const. temp approx. is OK.

Mean free path

Gas molecules interact via (almost classical) collisions, GIves cross-section

\color{red}{ \sigma _{sc} = \pi \left( {2a_0 } \right)^2 }

gives a mean-free-path (mfp): average distance between collisions.

\color{red}{ l = \frac{1}{{n\sigma }}}

(n is number density: note dimensions!). For H gas at .01 kg m^-3 (roughly standard temp & pressure) n ~6x10²⁴

\color{red}{ n \sim 4 \times 10^{24} ,\sigma \sim 3.5 \times 10^{ - 20} \Rightarrow l \sim 1\mu }

note this is very small compared to scale height: LTE is good!

Opacity

Need to quantify absorption of light by gas

Pass light through thin layer: obviously intensity lost depends on

Original intensity I_λ
Thickness of layer ds
Specific properties of gas via opacity κ_λ, defined to be X-sect for absorbing photons/kg
Density ρ of gas;

Then

\color{red}{ dI_\lambda = - \kappa _\lambda I_\lambda \rho ds}

Hence passing through a thick layer has intensity
\color{red}{ \int\limits_{I_0 }^{I_\lambda } {\frac{{dI_\lambda }}{{I_\lambda }}} = - \int_0^s {\kappa _\lambda \rho ds} }
giving
\color{red}{ I_\lambda = I_0 e^{ - \kappa _\lambda \rho s} }

This is mean-free-path (mfp): distance between photon-atom collisions.

\color{red}{ l = \frac{1}{{\kappa _\lambda \rho }} = \frac{1}{{n\sigma }}}

\color{red}{ \kappa _\lambda = \frac{{n\sigma }}{\rho }}

gives units of opacity: m²kg^-1

Optical depth:
\color{red}{ \tau_\lambda = - \int_0^s {\kappa _\lambda \rho ds} }
so if \color{red}{\tau_\lambda \ll 1} atmosphere is optically thin (high chance of photon passing through.
Optical depth ≈ # of mfp's
Note that 50% of photons will escape if \color{red}{\tau _\lambda \approx \frac{2}{3}} i.e. when we look down on star, the "surface" corresponds to an optical depth of 2/3

Can combine pressure and optical depth (since both vary with density in same way) so

\color{red}{ \frac{{dP}}{{d\tau }} = \frac{g}{\kappa } \Rightarrow P \approx \frac{{g\tau }}{\kappa }}

This leads to very different properties for stars of different sizes>

\color{red}{ \frac{{g_{red\_giant} }}{{g_{white\_dwarf} }} = \frac{{M_{RG} }}{{R_{RG} ^2 }}\frac{{R_{wd} ^2 }}{{M_{wd} }} \sim 10^{ - 6} }

i.e. pressures on RG's are very small, so spectral lines are not broadened

Sources of opacity

Bound-bound transitions: i.e. atomic excitations. γ is often absorbed and re-emitted with no energy loss, so can be regraded as scatterin. More often, 2 or more γ's will be emitted
Bound-free transitions: i.e. photo-ionization. X-sect roughly geometric:
Free-free: e.g. Bremsstrahlung
Electron scattering aka Thomson scattering

Thomson scattering

In fully ionized plasma \color{red}{\gamma + e^ - \Rightarrow \gamma + e^ - } dominates so force on electron is

\color{red}{ F = P\sigma _{e} }

Mean free path

\color{red}{ \lambda = \frac{1}{{n_e \sigma _e }}}

n_e = n_p is electron density.

σ_e Thomson X-sect

\color{red}{ \sigma _e = \frac{{8\pi }}{3}\left( {\frac{{e^2 }}{{4\pi \varepsilon _0 mc^2 }}} \right)^2 = 6.65 \times 10^{ - 29} m^2 }

This gives an opacity

\color{red}{ \kappa _{es} = \frac{{n_e \sigma _e }}{\rho } \approx .02\left( {1 + X} \right)}

No density dependence (why) and no temp dep.

How do we calculate κ(r,λ,T)?

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

\color{red}{ \kappa = C\frac{{Z\left( {1 + X} \right)\rho }}{{T^{7/2} }}}

(X is fraction of H, Z is fraction of metals) (i.e. it doesn't depend on wavelength). C ∼ 10²¹ in SI units

Can get an average value for photosphere: \color{red}{\kappa \sim 0.03m^2 kg^{ - 1} } need to define Rosseland mean opacity

\color{red}{ \bar \kappa = \overline {\kappa _{es} + \kappa _{bb} + \kappa _{ff} + \kappa _{bf} } }

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!!)

Note how rapid variation is.

Limb Darkening

Sun appears darker near edge.

Because light always goes through same optical depth, we are seeing a "cooler" sun at edges

Now looks at the interior of stars: stellar modelling