Astro 300B: Jan. 21, 2011 Equation of Radiative Transfer Sign Attendance Sheet Pick up HW #2, due...

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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
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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

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

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

ν F ≡ Iν∫ cosθ dΩ

Fν = π B ( 1 – cos2 θc ) = π B sin2 θc

Or…. 2

r

R

BF ν

Note: at r = R

Fν = π B

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

dlldjdlLILIL

l

L

L

L

L

22

1

2

1

expexp)()( 12 ννννν

at L2 is equal to

the incident specific intensity,

decreased by a factor of

plus the integral of jν along the line of sight, decreased by a factor of

νI

)( 1LIν

2

1

expL

L

dlν

2

expL

l

ldν = the integral of αν from l to L2