feb. 2, 2011 rosseland mean absorption poynting vector plane em waves the radiation spectrum:...
Post on 21-Dec-2015
216 views
TRANSCRIPT
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
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)