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?