astro 300b: jan. 24, 2011 optical depth eddington luminosity thermal radiation and thermal...
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Astro 300B: Jan. 24, 2011
Optical DepthEddington Luminosity
Thermal radiation and Thermal Equilibrium
Radiation pressure: why cos2?
Each photon of energy E=h has momentum h/c
Want the component of the momentum normalTo direction defined by dA, this will be
€
hν
ccosθ
Pressure = net momentum normal to dA/time/area
or the net energy x cos /c /time/area
The net energy = flux in direction n, i.e.
€
F = Iν∫ cosθdΩ
OPTICAL DEPTH
It’s useful to rewrite the transfer equation in terms of the optical depth:
dld l
lo
ldll )()( or
ldeljeLILIL
L
)( )()(
2
1
12
emergent incident
>1: optically thick opaque, typical photon will be absorbed
<1: optically thin transparent, typical photon will traverse the medium without being absorbed
Source FunctionS
jS
so that the Equation of Radiative Transfer is
SI
d
dI
Define
Which has solution
dSeeII )()0()(0
)(
If SI then 0
d
dISpecific intensity decreases along path
If SI then 0
d
dI Specific intensity increasesalong path
so SI If >>1, SI
Mean Free Path
The mean free path, is the average distance a photon travels beforebeing absorbed
Or in other words, the distance through the absorbing material corresponding to optical depth = 1
1Recall dld where = absorption coefficient
so 1 or
1
also N
#absorbers/Vol Cross-section for absorption
hence
N
1
N
1
Makes sense:
If N increases, decreases
If σν increases, decreases
Radiation Force: The Eddington Limitsee R&L Problem 1.4 and p. 15
• Photons carry momentum• When radiation is absorbed by a
medium, it therefore exerts a force upon it
c
F
Consider a source of radiation, with luminosity L (ergs/sec)And a piece of material a distance r from the source
Each photon absorbed imparts momentum = E / c
Specific flux = Fν ergs s-1 cm-2 Hz-1
Momentum flux = Fν / c momentum s-1 cm-2 Hz-1
Momentum imparted by absorbed photons =
Where = absorption coefficient, cm-1
c
F Momentum /area /time /Hz /pathlength through absorber
Now, area x pathlength = volume
so
c
F Is momentum/time /Hz /volume
But momentum /time = Force
So, integrating over frequency, the Force/volume imparted by the absorbed photons is
d
c
F
Likewise, in terms of the mass absorption coefficient, κ
dFc
f 1
An important application of this concept is
the Eddington Luminosity, or Eddington Limit
This is the maximum luminosity an object can have before it ejectshydrogen by radiation pressure
Eddington Luminosity c.f. Accretion onto a black hole
When does fgrav = fradiation?
Force per unit mass = force per unit massdue to gravity due to absorption of radiation
r
M,L
fradiation = c
F F = radiation flux, integrated over frequency
2 4 r
L
c
L = luminosity of radiation, ergs/sec
r = distance between blob and the source
Κ = mass absorption coefficent
fgravity = 2r
GMM = mass of the source
So…
cr
L
r
GM 22 4
GcL
M
4
Define Eddington Luminosity = the L at which f(gravity) = f(radiation)
4
GcMLedd
4
GcMLedd
Note: independent of r
A “minimal” value for κ is the Thomson cross-section,
For Thomson scattering of photons off of free electrons,assuming the gas is completely ionized and pure hydrogen
Other sources of absorption opacity, if present, will contribute to
larger κ, and therefore smaller L
Maximum luminosity of a source of mass M
Thomson cross-section
σT = 6.65 x 10-25 cm2
Independent of frequency (except at very high frequencies)
H
TT m
Where mH= mass of hydrogen atom
T
Hedd
GMcmL
4
T
Hedd
GMcmL
4
If M = M(Sun), then Ledd = 1.25 x 1038 erg/sec
Compared to L(sun) = 3.9 x 1033 erg/sec
Another example of a cross-section for absorption:
Photoionization of Hydrogen from the ground state
H atom + hν p + e-
Only photons more energetic than threshold χ can ionize hydrogen, where χ = 13.6 eV 912 Å 1 Rydberg Lyman limit ν1 = 3.3x1015 sec-1
The cross-section for absorption is a function of frequency,2
3
118 1063.6 cm
where ν1=3.3x1015 sec-1
More energetic photons are less likely to ionize hydrogen than photons at energies near the Lyman Limit
Note: αν : photon-particle cross-sections; σν: particle-particle cross-sections
Similarly, one can consider the ionization of He I He II He II He III
Thresholds: Hydrogen hν = 13.6 eV 912 Å Helium I 24.6 eV 504 Å Helium II 54.4 eV 228 Å
For HeI α(504 Å) = 7.4 x 10-18 cm2 declines like ν2
For HeII α(228 Å) = 1.7 x 10-18 cm2
declines like ν3
Thermal Radiation, and Thermodynamic Equilibrium
Thermal radiation is radiation emitted by matter in thermodynamic equilibrium.
When radiation is in thermal equilibrium, Iν is a universal function of frequency ν and temperature T – the Planck function, Bν.
Blackbody Radiation: BI
In a very optically thick media, recall the SOURCE FUNCTION
Ij
S
So thermal radiation has BjBS and
And the equation of radiative transfer becomes
)(or TBId
dIBI
dl
dI
THERMODYNAMIC EQUILIBRIUM
When astronomers speak of thermodynamic equilibrium, they mean a lot
more than dT/dt = 0, i.e. temperature is a constant.DETAILED BALANCE: rate of every reaction = rate of inverse reaction on a microprocess level
If DETAILED BALANCE holds, then one can describe
(1) The radiation field by the Planck function(2) The ionization of atoms by the SAHA equation(3) The excitation of electroms in atoms by the Boltzman distribution(4) The velocity distribution of particles by the Maxwell-Boltzman distribution
ALL WITH THE SAME TEMPERATURE, T
When (1)-(4) are described with a single temperature, T, then the system is said to be in THERMODYNAMIC EQUILIBRIUM.
In thermodynamic equilibrium, the radiation and matter have the same temperature, i.e. there is a very high level of coupling between matter and radiation Very high optical depth
By contrast, a system can be in statistical equilibrium, or in a steady state, but not be in thermodynamic equilibrium.
So it could be that measurable quantities are constant with time, but there are 4 different temperatures:
T(ionization) given by the Saha equationT(excitation) given by the Boltzman equationT(radiation) given by the Planck FunctionT(kinetic) given by the Maxwell-Boltzmann distribution
WhereT(ionization) ≠ T(excitation) ≠ T(radiation) ≠ T(kinetic)
If locally, T(ion) = T(exc) = T(rad) = T(kinetic)
Then the system is in LOCAL THERMODYNAMIC EQUILIBRIUM, or LTE
This can be a good approximation if the mean free path for particle-photon interactions << scale upon which T changes
LOCAL THERMODYNAMIC EQUILIBRIUM (LTE)
Example: H II Region (e.g. Orion Nebula, Eagle Nebula, etc)
Ionized region of interstellar gas around a very hot star
Radiation field is essentially a black-body at the temperature of the centralStar, T~50,000 – 100,000 K
However, the gas cools to Te ~ 10,000 K (Te = kinetic temperature of electrons)
H IH II
O star
Q.: Is this room in thermodynamic equilibrium?
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