Astro 300B: Jan. 24, 2011

Optical Depth Eddington Luminosity Thermal radiation and Thermal Equilibrium



Radiation pressure: why cos

2

?

Each photon of energy E=h n momentum h n /c has Want the component of the momentum normal To direction defined by dA, this will be

h

n cos 

c

Pressure = net momentum normal to dA/time/area or the net energy x cos   The net energy = flux in direction n, i.e.

F

 

I

n cos 

d



OPTICAL DEPTH

 n It’s useful to rewrite the transfer equation in terms of the optical depth:   or n ) 2 n 1  

l

) 

l



l o



l



l

     n emergent  n >1: optically thick incident opaque, typical photon will be absorbed  n <1: optically thin transparent, typical photon will traverse the medium without being absorbed

Source Function

S

n Define

S

n 

j

n  n so that the Equation of Radiative Transfer is Which has solution  n   

dI

      0 If 

S

then

dI

n

d

 n  0 Specific intensity decreases along path If then

dI

n

d

 n  0 Specific intensity increases along path 

S

so

Mean Free Path

The mean free path,  n  n is the average distance a photon travels before being absorbed also Or in other words, the distance through the absorbing material corresponding to optical depth = 1 Recall so   n  1   where   or  n  n = absorption coefficient  1  n   hence  n  1  n #absorbers/Vol Cross-section for absorption

 n  1  n Makes sense: If N increases,  n

decreases

If σ ν increases,  n

decreases

Radiation Force: The Eddington Limit see 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 Consider a source of radiation, with luminosity L And a piece of material a distance r from the source (ergs/sec) Each photon absorbed imparts momentum = E / c Specific flux = F

ν ergs s -1 cm -2 Hz -1 Momentum flux = F ν / c momentum s -1 Momentum imparted by absorbed photons =

 n 

-1

c

cm -2

F

n

Hz -1

 n

F

n

c

Momentum /area /time /Hz /pathlength

through absorber Now, area x pathlength = volume so  n

F

n

c

Is momentum/time /Hz /volume But momentum /time = Force So, integrating over frequency, the Force/volume imparted by the absorbed photons is  

F c

n

Likewise, in terms of the mass absorption coefficient, κ

f c

  n n An important application of this concept is the Eddington Luminosity, or Eddington Limit This is the maximum luminosity an object can have before it ejects

hydrogen by radiation pressure

Eddington Luminosity c.f. Accretion onto a black hole

When does f

grav

= f

radiation

?

r M,L Force per unit mass = force per unit mass due to gravity due to absorption of radiation f radiation 

F

=  

c c L

4 

r

2 F = radiation flux, integrated over frequency L = luminosity of radiation, ergs/sec r = distance between blob and the source Κ = mass absorption coefficent

f gravity = So…

GM r

2 M = mass of the source 2

r



L

 2 

c M L

 4   Define Eddington Luminosity = the L at which f(gravity) = f(radiation)

L ed d

 4  

L ed d

 4   Maximum luminosity of a source of mass M 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

Thomson cross-section

σ T = 6.65 x 10 -25 cm 2 Independent of frequency (except at very high frequencies) 

T

 

T m H

Where m

H

= mass of hydrogen atom

L



edd

 

T

L



edd

 

T

If M = M(Sun), then L edd = 1.25 x 10 38 erg/sec Compared to L(sun) = 3.9 x 10 33 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 ν

1 = 3.3x10

15 sec -1

The cross-section for absorption is a function of frequency,   n n 2

cm

1 =3.3x10

15

sec -1

More energetic photons are less likely to ionize hydrogen than photons at energies near the Lyman Limit Note: α ν : photon-particle cross-sections; σ ν : 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 cm

2 declines like ν 2

For HeII α(228 Å) = 1.7 x 10-18 cm

2 declines like ν 3

Thermal Radiation, and Thermodynamic Equilibrium

Thermal radiation is radiation emitted by matter in thermodynamic equilibrium.

When radiation is in thermal equilibrium,

I ν

frequency

ν

and temperature

T

is a universal function of – the Planck function,

B ν

. Blackbody Radiation:

I

n  In a very optically thick media, recall the SOURCE FUNCTION So thermal radiation has 

j

 n n n  And the equation of radiative transfer becomes n

dl B

 n n

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

(3) The

excitation

of atoms by the SAHA equation 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 equation T(excitation) given by the Boltzman equation T(radiation) given by the Planck Function T(kinetic) given by the Maxwell-Boltzmann distribution Where T(ionization) ≠ T(excitation) ≠ T(radiation) ≠ T(kinetic)

LOCAL THERMODYNAMIC EQUILIBRIUM (LTE)

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

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 central Star, T~50,000 – 100,000 K However, the gas cools to T e ~ 10,000 K (T e = kinetic temperature of electrons) O star H I H II

Q.: Is this room in thermodynamic equilibrium?