Transcript Jan21

Astro 300B: Jan. 21, 2011
Equation of Radiative Transfer
Sign Attendance Sheet
Pick up HW #2, due Friday
Turn in HW #1
Chromey,Gaches,Patel: Do Doodle poll
First Talks: Donnerstein, Burleigh,
Sukhbold Fri., Feb 4
Radiation Energy Density
specific energy density
u
uν = energy per volume, per frequency range
Consider a cylinder with length
ds = c dt
c = speed of light
dA
uν(Ω) = specific energy density per solid angle
Then
ds
dE = uν(Ω) dV dΩ dν
energy
Hz
steradian
volume
energy
vo
steradian
l
Hz
But dV = dA c dt for the cylinder, so dE = uν(Ω) dA c dt dΩ dν
Recall that dE = Iν dA dΩ dt dν
so….
I
u () 
c
Integrate
uν(Ω) over all solid angle,
to get the energy density
uν


u
u

d




1
 I d

c
Recall
so

1
J 
4
u

4
0
4

J
c
I d
ergs cm-3 Hz-1
Radiation Pressure of an isotropic radiation field
inside an enclosure
What is the pressure exerted by each
photon when it reflects off the
wall?
Each photon transfers 2x its normal
component of momentum
photon
+ p┴
in
out
- p┴

2
2
p
I
d


cos
c

2
2
p

I
cos
d




c
Since the radiation field is isotropic, Jν = Iν



2
2
p

J
cos
d
d
c
 sin
 
2
Integrate over
2
J
d
cos
sin
d
2π steradians

c 0 0
only
 
Not 2π
2
 J

c

cos



 3 0
3
4
 1

J
c
3
But, recall that the energy density
So….

4

J
c
u
p
1

u
3
Radiation pressure of an isotropic radiation field =
1/3 of its energy density

Example: Flux from a uniformly bright sphere
(e.g. HII region)
θ
P
R
θc
r
At point P, Iν from the sphere is a constant (= B) if the ray intersects the
sphere, and Iν = 0 otherwise.
F

 I
2


c
0
0
cos d 
B
d

co

sin

s
d


R


1
where


sin

c
r

And…looking towards the
sphere from point P, in the
plane of the paper
dφ
So we integrate dφ
from 2π to 0
Fν
Or….
= π B ( 1 – cos2 θc )
= π B sin2 θc
2
R

F
 B 
r
Note: at r = R
F   B
Equation of Radiative Transfer
When photons pass through material, Iν changes due to
(a) absorption
(b) emission
(c) scattering
ds
Iν + dIν
Iν
+
dIν = dIν - dIν
Iν added by emission
-
- dIν
sc
Iν subtracted by
scattering
Iν subtracted by absorption
EMISSION: dIν+
DEFINE jν = volume emission coefficient

dI
j

ds
jν = energy emitted in direction Ĩ
per volume dV
per time dt
per frequency interval dν
per solid angle dΩ
Units: ergs cm-3 sec-1 Hz-1 steradians-1
Sometimes people write
emissivity
εν = energy emitted
per mass
per frequency
per time
integrated over all solid angle
So you can write:
or

d

j
E
d
d

d
V
d
t



d

dE

dt
d
 dV
4
Mass density
Fraction of energy
radiated into solid
angle dΩ

dI
ABSORPTION
Experimental fact:

dI
I

ds
Define ABSORPTION COEFFICIENT

such that
=


d
I
I

d s

has units of cm-1
Microscopic Picture
N absorbers / cm3
Each absorber has cross-section for absorption


has units cm2; is a function of frequency
I dI

I


ASSUME:
(1) Randomly distributed, independent absorbers
(2) No shadowing:

1
/
3

m
inte
cl
dis
ea

N
e
1
/
2
Then
N
dA
ds
Total area presented by the absorbers = N
dA
ds


Total # absorbers in the volume =
 
So, the energy absorbed when light passes through the volume is

dE

I
N
dA
ds
d

dt
d
In other words,

d
I

I


d s

d

I
N
I
ds




  N  

is often derivable from first principles
Can also define the mass absorption coefficient

 

Where ρ = mass density, ( g cm-3 )

Sometimes
has units cm2 g-1

is denoted

So… the Equation of Radiative Transfer is


d

I
d

I
d
  I


j
d

s
I
ds



OR
d
I

j


I

ds
emission
absorption
Amount of Iν removed by absorption is proportional to Iν
Amount of Iν added by emission is independent of Iν
TASK: find αν and jν for appropriate physical processes
Solutions to the Equation of
Radiative Transfer
dI
j


I

ds
(1) Pure Emission
(2) Pure absorption
(3) Emission + Absorption
d
I

j


I

ds
(1) Pure Emission Only
 0
So,
Absorption coefficient = 0
dI
 j
ds
s
)

I
s
)
I
s
)j
s
d
s
(
(
o
(
s
o
Incident specific intensity
Increase in brightness =
The emission coefficient
integrated along the line of
sight.
(2) Pure Absorption Only
j 0
Emission coefficient = 0
dI


I

ds
s



I
(
s
)

I
(
s
)
exp

(
s
)
d
s




o




o
s


incident
Factor by which Iν decreases = exp of the absorption coefficient integrated
along the line of sight
General Solution
l
I(L
2)
L2
I(L
1)
L1
dI
j


I

dl
Multiply Eqn. 1 by
Eqn. 1
l

expdl And rearrange


L2

 l

 l

dI   I dl exp  dl  j exp  dldl
L 2

L 2


 l

 l

d
  dl

  dl
dl
I exp
 j exp
L2

L2


Eqn. 2
Now integrate Eqn. 2 from L1 to L2
LHS=
L
2
 










d
I
exp
dl

I
exp
dl
 

 






L
L
L
1
2
2





L
1
L
2

l

l
L
1





I
(
L
)

I
(
L
)
exp
dl

2

1




L
2

2
L


l

I
(
L
)

I
(
L
)
ex


p
d

2

1

 
L
1

So…


L
L
L
2
2
2








I
(
L
)

I
(
L
)
exp

d

j
l
exp

d
d
ll

2

1






 


L
L
l


1
1




L
L
L
2
2
2









I
(
L
)

I
(
L
)
exp

d

j
l
exp

d
l
dl

2

1










L
L
l


1
1


I
at L2 is equal to
the incident specific intensity,
decreased by a factor of
I(L
1)
L2


exp dl
L

1

plus the integral of jν along the line of sight,
decreased by a factor of
L2


expdl


l


= the integral of αν from l to L2