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Radiative transfer - how does light propagate through material? Iν1 Iν2 Iν is the specific intensity de ν da cos θdtdνdω Units ergs cm -2 s -1 Hz -1 steradian -1 energy passing through and area per time at a given frequency ν per solid angle We will talk about the full derivation when you re older

Three other important quantities derived from Iν J ν = 1 4π I ν dω 4π mean intensity over solid angle 0 same units as Iν. If Iν is isotropic, then Iν = Jν 4π F ν = I ν cos θdω Net flux ergs s -1 cm -2 Hz -1. This is the quantity you observe, 0 and the one that goes as 1/r 2 Radiation pressure. Momentum of a photon is E/c Pressure = momentum per unit time per unit area momentum flux in direction θ is dfν/c component of momentum flux normal to da is dfν/c cosθ So radiation pressure is P dynes cm -2 Hz -1 ν = 1 I ν cos 2 θdω c one cosθ to get component of momentum vector perpendicular to da, one to get effective area right da θ ˆl ˆn

Radiative transfer - how does light propagate through material? ds Iν1 Iν2 = Iν1 + diν When photons pass through material, Iν changes due to absorption emission scattering so diν = diν + - diν - - diν sc (Iν added by emission - Iν subtracted by absorption - Iν subtracted by scattering)

Emission diν + diν + = jνds jν is the volume emission coefficient: energy emitted in direction ˆl per volume dv per time dt per frequency interval dν per solid angle dω Absorption diν - Experimentally determined that diν - = ανiνds Microscopic picture: N absorbers/cm 3. Each absorber has a cross section for absorption σν cm 2. In general σν will be a function of frequency Assume randomly distributed, independent absorbers with σν 1/2 << mean interparticle distance N 1/3 Total absorbers in volume = NdAds Total area presented by absorbers = NdAds σν

Absorption diν - Experimentally determined that diν - = ανiνds Microscopic picture: N absorbers/cm 3. Each absorber has a cross section ˆl for absorption σν cm 2. In general σν will be a function of frequency Assume randomly distributed, independent absorbers with σν 1/2 << mean interparticle distance N 1/3 Total absorbers in volume = NdAds Total area presented by absorbers = NdAds σν so energy absorbed when light passes through the volume is de = I ν Nσ ν dadsdωdtdν or diν = I ν Nσ ν ds so α ν = Nσ ν σν is often derivable from theory

Equation of radiative transfer or di ν = di ν + diν = j ν ds α ν I ν ds di ν ds = j ν α ν I ν amount of Iν removed by absorption proportional to Iν amount of Iν added by emission independent of Iν after some integration you end up with I ν2 = I ν1 e ( R s2 s1 α νds) + s2 often see this in terms of optical depth instead of path length, since is easier to measure τ ν = dτ ν α ν ds s2 s1 α ν ds s1 j ν e ( R s2 s1 α νds) ds

Formulation with I ν2 = I ν1 e τ ν + much less cumbersome. If we use the source function Sν = jν/αν we can write even more simply di ν dτ ν = I ν + S ν s2 s1 j ν e τ ν ds > 1: optically thick, opaque. Typical photon will be absorbed. Any features in the spectrum entering the material will be lost as the photons are thermalized < 1: optically thin, transparent. Typical photon will traverse the material without being absorbed Iν > Sν : then diν/d ν < 0 Iν decreases along path Iν < Sν : then diν/d ν > 0 Iν increases along path If >> 1, Iν Sν

Mean free path- average distance a photon travels before being absorbed - distance through absorbing material corresponding to optical depth = 1 dτ ν = α ν dl so τ ν = α ν l ν = 1 or l ν = 1 α ν αν = nσν where σν = cross section for absorption and n is number density hence l ν = 1 nσ ν so if n increases lν decreases. If σν increases lν also decreases

Thermal Radiation - radiation emitted by matter in thermal equilibrium Iν = Bν: The Planck function, a universal function of ν and T. Sν becomes Bν Thermodynamic equilibrium has a very specific meaning. dt/dt = 0 is necessary but not sufficient. Detailed balance holds. Rate of every reaction = rate of inverse reaction If true we can describe The radiation field by the Planck function The ionization of atoms by the Saha equation The excitation of electrons in atoms by the Boltzmann distribution The velocity distribution of particles by the Maxwell-Boltzmann distribution All of these components will have the same temperature T

When detailed balance holds, i.e. when all four conditions are described by a single temperature, then the system is in thermodynamic equilibrium Since radiation and matter have the same temperature, they must be very strongly coupled. I.e. >> 1 A system can also be in a steady state - measurable quantities are constant in time, but Tionization Texcitation Tradiation Tkinetic (some may be equal, but not all) There is also local thermodynamic equilibrium (LTE) where the thermodynamic equilibrium conditions are met, but only locally (i.e. in a star). Good when meanfree path << than the distance over which T changes Steady state example: HII region - ionized gas around a hot star Radiation field is a blackbody at the temperature of the star (say 50,000-100,000K) gas cools to 10,000K (kinetic temperature of the electrons and the ionization temperature of H)

Question: Is this room in thermodynamic equilibrium?

Useful formulae: 1. The Boltzmann Equation - Boltzmann showed thatr the probability of finding an atom with an electron in an excited state n with energy χn above ground level decreases exponentially with χn and increases exponentially with T. N n N 1 = g n g 1 e χ n/kt where Nn = number of atoms in excited state/volume N1 = number of atoms in ground state per volume gn = 2n 2 is the statistical weight of level n = # of different angular momentum states in energy level n Example: In hydrogen gas of temperature 10,000K, what fraction of atoms are in energy level n=2? χ2 = 10.2 ev (to convert ev to kt multiply by 11,605). g2 = 8, g1=2 N2/N1 = 2.9x10-5

2. The Planck Function - emission from a blackbody B ν = 2hν3 c 2 1 e hν/kt 1 where h = planck s constant In terms of wavelength λ instead of frequency B λ = 2hc2 λ 5 1 e hc/λkt 1

2. The Planck Function - emission from a blackbody Two limits simplify the expression. For hν << kt we have the Rayleigh-Jeans Law: e hν kt 1 hν kt I ν(t ) = 2ν2 c 2 kt

2. The Planck Function - emission from a blackbody Two limits simplify the expression. For hν >> kt we have Wein s Law: 1 kt 1 1 e hν e hν kt I ν (T ) = 2hν3 c 2 e hν/kt

2. The Planck Function - emission from a blackbody - optically thick gas Monotonicity with temperature: If T1 > T2 then Bν(T1) > Bν(T2) for all frequencies Of 2 blackbody curves, the one with the higher temperature lies entirely above the other B ν (T ) T = T ( 2hν 3 /c 2 e hν/kt 1 ) = 2h2 ν 4 c 2 kt 2 e hν/kt e hν/kt 1 > 0 Wein displacement Law: simple expressions for the frequency of wavelength at which B(T) peaks. ν max T = 5.88 10 10 Hz deg 1 λ max T = 0.29 cm deg Note that the peaks of Bν and Bλ do not occur in the same place. This is because the relationship between an given wavelength interval with a frequency interval is nonlinear. λν = c so dλ/dν = -c/ν 2. This is in the conversion of to B ν (T ) T B λ (T ) T

3. Maxwell-Boltzmann distribution - describes the distribution of velocities in an ensemble of particles in thermal equilibrium with a temperature T f(v) = 4π ( m 2πkT ) 3/2 v 2 e mv2 2kT Where f(v) is the fraction of particles with a velocity between v,v+dv and m is the mass of an individual particle most probable speed: average speed: root-mean-square speed: v p = (2kT/m) 1/2 < v >= (8kT/πm) 1/2 v rms =< v 2 > 1/2 = (3kT/m) 1/2

4. The Saha Equation: gives the ratio of numbers of atoms in different ionization states N m+1 n e = 2 Z ( ) 3/2 m+1 2πme kt N m Z m h 3 e χ m/kt where ne is the number density of free electrons, Nm is the number of atoms in the m th ionization state (m = number of electrons lost -1. So neutral oxygen is OI, oxygen that has lost two electrons is OIII, etc.), Zm is the partition function of the m th ionization state (an atomic constant like the statistical weights in the Boltzmann equation), χm is the ionization energy of the m th ionization state