Astro 300B: Jan. 21, 2011 Equation of Radiative Transfer Sign Attendance Sheet Pick up HW #2, due...
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Transcript of Astro 300B: Jan. 21, 2011 Equation of Radiative Transfer Sign Attendance Sheet Pick up HW #2, due...
Astro 300B: Jan. 21, 2011
Equation of Radiative Transfer
Sign Attendance SheetPick up HW #2, due Friday
Turn in HW #1Chromey,Gaches,Patel: Do Doodle poll
First Talks: Donnerstein, Burleigh, Sukhbold Fri., Feb 4
Radiation Energy Density
specific energy density uν = energy per volume, per frequency range
Consider a cylinder with length ds = c dt c = speed of lightdA
ds
uν(Ω) = specific energy density per solid angle
Then dE = uν(Ω) dV dΩ dν
uν
energy
Hz steradian vol
energy volumesteradian
Hz
But dV = dA c dt for the cylinder, so dE = uν(Ω) dA c dt dΩ dνRecall that dE = Iν dA dΩ dt dν so….
€
uν (Ω) = Iνc
Integrate uν(Ω) over all solid angle,
to get the energy density uν
duu νν
dI c
1 ν
Recall
soergs cm-3 Hz-1
€
Jν = 1
4π Iν dΩ0
4π
∫
€
uν = 4π
c Jν
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
in
out
+ p┴
- p┴ dI
cp cos
2 2νν
dIc
p cos 2
2νν
Since the radiation field is isotropic, Jν = Iν
νν ddJc
p sin cos 2
2
sin cos 2
0
2
0
ν ddJc
3cos
3
0
2
νJc
Integrate over2π steradians
only
Not 2π
3
1
4 ν
J
c
But, recall that the energy density
So….
Radiation pressure of an isotropic radiation field = 1/3 of its energy density€
pν = 1
3 uν
€
uν = 4π
c Jν
Example: Flux from a uniformly bright sphere (e.g. HII region)
At point P, Iν from the sphere is a constant (= B) if the ray intersects the sphere, and Iν = 0 otherwise.
RP
r
θ
θc
sin cos B 0
2
0
ddc
r
R 1-sin where c
And…looking towards the sphere from point P, in the
plane of the paper
So we integrate dφ from 2π to 0
dφ
€
ν F ≡ Iν∫ cosθ dΩ
Equation of Radiative Transfer
When photons pass through material, Iν changes due to (a) absorption (b) emission
(c) scatteringds
IνIν + dIν
dIν = dIν+ - dIν
- - dIνsc
Iν added by emission Iν subtracted by absorption
Iν subtracted by scattering
EMISSION: dIν+
DEFINE jν = volume emission coefficient
dsjdI νν
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:
νν ddtddVjdE or
νν 4
d
ddtdVdE
Mass density Fraction of energy radiated into solid
angle dΩ
ABSORPTIONνdI
Experimental fact: dsIdI νν
Define ABSORPTION COEFFICIENT =
such that dsIdI ννν
ν
ν has units of cm-1
Microscopic Picture
N absorbers / cm3
Each absorber has cross-section for absorption ν
has units cm2; is a function of frequencyν
νI νν dII
ASSUME:(1) Randomly distributed, independent absorbers(2) No shadowing:
3/12/1 distance cleinterpartimean Nν
Then
dsdAN
ν dsdANTotal area presented by the absorbers =
So, the energy absorbed when light passes through the volume is
Total # absorbers in the volume =
νννν ddtddsdANIdE
dsINdI ννν
dsIdI ννν
In other words,
€
α ν = N σ ν
ν is often derivable from first principles
Can also define the mass absorption coefficient
ν
ν
Where ρ = mass density, ( g cm-3 )
ν has units cm2 g-1
Sometimes is denoted ν ν
So… the Equation of Radiative Transfer is
ννν dIdIdI
dsIdsj ννν
OR νννν Ijds
dI
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
νννν Ijds
dI
(1) Pure Emission(2) Pure absorption(3) Emission + Absorption
νννν Ijds
dI
(1)Pure Emission Only
0ν Absorption coefficient = 0
So, νν jds
dI
s
s
sdsjsIsI
o
o )( )()( ννν
Increase in brightness = The emission coefficient
integrated along the line ofsight.
Incident specific intensity
(2) Pure Absorption Only
0νj Emission coefficient = 0
ννν Ids
dI
s
s
o
o
sdssIsI )(exp)()( ννν
Factor by which Iν decreases = exp of the absorption coefficient integrated along the line of sight
incident
General Solution
)( 1LIν
)( 2LIν
νννν Ijdl
dI
L1
L2
Eqn. 1
Multiply Eqn. 1 by
l
L
dl2
exp ν And rearrange
€
dIν +αν Iν dl[ ]exp αν dlL2
l
∫ ⎛
⎝ ⎜ ⎜
⎞
⎠ ⎟ ⎟= jν exp αν dl
L2
l
∫ ⎛
⎝ ⎜ ⎜
⎞
⎠ ⎟ ⎟dl
Eqn. 2
l
€
d Iν exp αν dlL2
l
∫ ⎛
⎝ ⎜ ⎜
⎞
⎠ ⎟ ⎟
⎛
⎝ ⎜ ⎜
⎞
⎠ ⎟ ⎟= jν exp αν dl
L2
l
∫ ⎛
⎝ ⎜ ⎜
⎞
⎠ ⎟ ⎟dl
Now integrate Eqn. 2 from L1 to L2
2
12
2
1 2
expexp
L
L
l
L
L
L
l
L
dlIdlId
νννν LHS=
1
2
exp)()( 12
L
L
dlLILI ννν
2
1
exp)()( 12
L
L
dlLILI ννν
So… dldljdlLILIL
l
L
L
L
L
22
1
2
1
expexp)()( 12 ννννν