Feb. 2, 2011Rosseland Mean Absorption

Poynting VectorPlane EM Waves

The Radiation Spectrum: Fourier Transforms

Rosseland Mean Opacity

Recall that for large optical depth

€

τ

€

Sν →Bν

In a star,

€

τ is large, but there is a temperature gradient

How does F(r) relate to T(r)?

Plane parallel atmosphere

€

ds =dz

cosθ

dz ds

z

€

θ

€

μ ≡cosθ

so

€

ds =dz

μ

Equation of radiative transfer:

€

dIν (z,μ)

ds= jν − αν Iν −σ ν (Iν − Jν )

emissionabsorption

scattering

€

dIν (z,μ)

ds= jν − αν Iν −σ ν (Iν − Jν )

For thermal emission:

€

jν = αν Bν (T)

€

Bν = Planck function

so

€

dIνds

= αν Bν − αν Iν −σ ν (Iν − Jν )

€

μdIνdz

= −(αν +σ ν )(Iν − Sν )

€

ds =dz

μ

Where the source function

€

Sν =αν Bν +σ ν Jν

αν +σ ν

€

μdIνdz

= −(αν +σ ν )(Iν − Sν )

€

Sν =αν Bν +σ ν Jν

αν +σ ν

Rewrite:

€

Iν = Sν −μ

αν +σ ν

⎛

⎝ ⎜

⎞

⎠ ⎟dIνdz

(1)

“Zeroth” order approximation:

€

Iν0 ≈ Sν

0(T)

= Bν (T) Independent of μ

1st order approximation:

€

dIνdz

=dBν

dz

Equation (1)

€

Iν(1)(z,μ) = Bν (T) −

μ

αν +σ ν

dBν

dzDepends on μ

€

Fν = Iν(1)∫ cosθ dΩ

= 2π Iν μ dμ-1

+1

∫

= 2π Bν −μ

αν +σ ν

dBν

dz

⎛

⎝ ⎜

⎞

⎠ ⎟μ dμ

-1

+1

∫

= 2πBν μ dμ-1

+1

∫ − 2π

αν +σ ν

dBν

dzμ 2dμ

−1

+1

∫

= 0

€

=−4π

3 αν +σ ν( )

dBν

dz

= −4π

3(αν +σ ν )

dBν

dT

dT

dz

Integrate over all frequencies:

€

F = Fν dν0

∞

∫

= −4π

3

dT

dz

1

αν +σ ν

dBν

dT dν

0

∞

∫

Now recall:

€

dBν

dTdν

0

∞

∫ = d

dTBν dν

0

∞

∫

=d

dTB(T)

=d

dT

σ

πT 4

= 4σT 3

π

= Stefan-Boltzman Constant

Define Rosseland mean absorption coefficient:

€

αR ≡

1

αν +σ ν

dBν

dTdν

0

∞

∫dBν

dTdν

0

∞

∫

Combining,

€

F(z) = −16σT 3

3α R

dT

dzEquation of radiative diffusionIn the Rosseland Approximation

• Flux flows in the direction opposite the temperature gradient• Energy flux depends on the Rosseland mean absorption coefficient, which is the weighted mean of

Transparent regions dominate the mean

€

1

α R +σ ν

€

αR

Conservation of Charge

t

D

cj

cH

14

€

∇⋅ ∇×H( ) =4π

c∇⋅ j +

1

c∇⋅

∂D

∂t

€

0 =∇⋅ j +1

4π∇⋅

∂D

∂t

Follows from Maxwell’s Equations

4 D

0

t

j

Take

But so

current density

charge density

Poynting Vector

HEc

S

4

One of the most important properties of EM waves is that they transportenergy— e.g. light from the Sun has traveled 93 million miles and still has enough energy to do work on the electrons in your eye!

Poynting vector, S: energy/sec/area crossing a surface whose normal is parallel to S

Poynting’s theorem: relates mechanical energy performed by the E, Bfields to S and the field energy density, U

€

U field =1

8πεE 2 +

B2

μ

⎛

⎝ ⎜

⎞

⎠ ⎟

Mechanical energy:

B

cEqF

v

dF

vF

c

BqEq

BEq

vvv

c

vv

0vv B

Lorentz force:

work doneby force

rate of work

=0 since

dt

dmF

v

but also

€

rv ⋅

r F = m

r v ⋅

dr v

dt

=d

dt12 m

r v 2( )

€

qr v ⋅

r E =

d

dt12 m

r v 2( )so...

More generally,

i

i0V

vV

1lim

densitycurrent

iq

j

dt

dUEj mech

€

rj ⋅

r E =

d

dt12 mv2 /volume( )

where U(mech) = mechanical energy / volume

Back to Poynting’s Theorem

Maxwell’s Equation

t

D

cj

cH

14

rearrange and dot E

€

rj ⋅

r E =

1

4πc ∇ ×

r H ( ) ⋅

r E −

r E ⋅

∂r D

∂t

⎡

⎣ ⎢

⎤

⎦ ⎥

use the vector identity

€

rE ⋅ ∇ ×

r H ( ) =

r H ⋅ ∇ ×

r E ( ) −∇ ⋅

r E ×

r H ( )

€

rj ⋅

r E =

1

4πc

r H ⋅ ∇ ×

r E ( ) − c∇ ⋅

r E ×

r H ( ) −

v E ⋅

∂r D

∂t

⎡

⎣ ⎢

⎤

⎦ ⎥

But t

B

cE

1and HBED

μ ,

€

rj ⋅

r E +

1

8π

∂

∂tεE 2 +

B2

μ

⎛

⎝ ⎜

⎞

⎠ ⎟= −

c

4π∇⋅

r E ×

r H ( )

Now

€

U field =1

8πεE 2 +

B2

μ

⎛

⎝ ⎜

⎞

⎠ ⎟ = field energy / volume

€

c

4π

r E ×

r H ( ) =

r S the Poynting vector

(1)

per volumeenergy mechanical of change of rateEj

S

So (1) says

rate of change ofmechanical energyper volume

+rate of changeof field energyper volume

Plane Electromagnetic Waves

Maxwell’s Equations in a vacuum: 0 0; ;1 j

μ

0E

0 B

t

B

cE

1

t

E

cB

1

(1)

(2)

(3)

(4)

These equations predict the existence of WAVES for E and B which carry energy

Curl of (3)

€

∇× ∇× r

E ( ) = −1

c

∂

∂t∇ ×

r B ( )

use (4) 2

2

2

1

t

E

c

€

∇× ∇× r

E ( ) =∇ ∇⋅r E ( ) −∇2

r E Use vector identity:

=0 from (1)

01

2

2

22

t

E

cE

01

2

2

22

t

B

cB

SOVectorWave

Equation

Similarly, for B:

Note: Laplacian2

zyx EkEjEiE 2222 ˆˆˆ

2

2

2

2

2

2

2

2 1

t

E

cz

E

y

E

x

E xxxx

operates on each component of BE

and

so these 2 vector wave equations are actually 6 scalar equations

So, for example, one of the equations is

Similarly for Ey, Ez, Bx, By and Bz

What are the solutions to the wave equations?

First, consider the simple 1-D case --2

2

22

2

v

1

tx

Wave equation

A solution is )v(sin),( txkAtx

A = constant (amplitude)[kx] = radians [kvt] = radians

)v(sin

)v(cos

)v(sin

2

22

2

txkAk

txkx

kA

txkAxx

)v(sinv222

2

txkAkt

So the wave equation is satisfied

Ψ is periodic in space (x) and time (t)

)v(sin),( txkAtx

The wavelength λ corresponds to a change in the argument of the sine by 2π

)v(sin),( txkAtx

)2)v(sin()v(sin txktxk period spatial 2kττ period l tempora2v k

frequencyτ

v1 v

angular frequency radians/s 22 τ

)sin(),( tkxAtx