Transcript Feb. 2, 2011 Rosseland Mean Absorption Plane EM Waves The

Feb. 2, 2011
Rosseland Mean Absorption
Poynting Vector
Plane EM Waves
The Radiation Spectrum:
Fourier Transforms
Rosseland Mean Opacity
Recall that for large optical depth

S B
In a star,


is large, but there is a temperature gradient

How does F(r) relate to T(r)?
Plane parallel atmosphere
z
dz
dz
ds 
cos

  cos 
ds



Equation of radiative transfer:
so
dz

dI (z,
)

 j   I   (I  J )
ds
emission

ds 
absorption
scattering
dI (z, )
 j   I   (I  J )
ds
For thermal emission:
so
j   B (T)
B = Planck function
dI
 
B   I   (I  J ) 
ds
dI

 (   )(I  S )
dz
Where the source function
S 
 B   J

  
ds 
dz

dI

 (   )(I  S )
dz
Rewrite:
  dI
I  S  




 
  dz
“Zeroth” order approximation:

1st order approximation:

S 
(1)

I0  S0 (T)
 B (T)
Independent of 
dI dB

dz
dz

dB
I (z, )  B (T) 
Equation (1) 
   dz

(1)
 B   J
  
Depends on 
F 
(1)
I
  cos d
 2
+1
 I  d
-1

 dB 
 2  B 
 d
   dz 
-1 
+1
2 dB
 2B   d 
   dz
-1
+1
=0
4
dB

3    dz
4
dB dT

3(   ) dT dz
1
2

 d
1
Integrate over all frequencies:

 F d
F =
0
4 dT

3 dz
Now recall:



0


0
1
dB
d
   dT
dB
d
d =
dT
dT

 B d
0
d
=
B(T)
dT
d  4
=
T
dT 
4T 3
=

 = Stefan-Boltzman Constant
Define Rosseland mean absorption coefficient:

R 
Combining,


0
1
dB
d
   dT

dB
 dT d
0
16T dT
F(z)  
3 R dz
3
Equation of radiative diffusion
In 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

1
 R  
Transparent regions dominate the mean
R
Conservation of Charge
Follows from Maxwell’s Equations
4

1

D


H
 j
c c
t
Take

4
1 D
   H 
 j  
c
c
t


1
D
0   j 

4
t
But

so
D

4


charge density



j 

0

t
current density
Poynting Vector
One of the most important properties of EM waves is that they transport
energy—
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
 c 
S E

H
4

Poynting’s theorem: relates mechanical energy performed by the E, B
fields to S and the field energy density, U
1  2 B 2 
U field  E  
8 
 
Mechanical energy:
Lorentz force:
work done
by force
rate of
work
 
  v

F

q
E


B


 c 

Fd
 
    v

Fv qvE B
c




   vB
qv Eqv
c
=0 since
 
 
v
v

B

0
but also

so...


dv
Fm
dt
r
r r
r dv
v F  mv
dt
d 1 r2
 2 mv 
dt
r r d 1 r2
qv E  2 mv 
dt
More generally,

j current
density

1
lim 
qivi

V

0
Vi
r r d 1 2
j  E  2 mv /volume
dt

  dU
mech
jE

dt
where U(mech) = mechanical energy / volume
Back to Poynting’s Theorem 
Maxwell’s Equation
4

1

D


H
 j
c c
t

r
dot
E
and
rearrange


r
r r r
r
1
D
jE
c   H  E  E  
4 
t 

use the vector identity


r
r
r
r
r r
E    H  H    E   E  H






r
r
r r
r r
v D
1  r
jE
cH    E  c  E  H  E  
4 
t 

But


 1
B


E
c
t

and


 

 
D

E
,B

H
2 

r
r
r r
c
B
1 
2
 E  H
j E 
 
E 
4
 
8 t 

1  2 B2 
U field 
E  
8 
 
Now
r
c r r
EH S
4




(1)
= field energy / volume
the Poynting vector


j

E

rate
of
cha
of
me
en
pe
n
So (1) says

rate of change of
mechanical energy
per volume
+
rate of change
of field energy
per volume



S
Plane Electromagnetic Waves





1
;
0;
j

0
Maxwell’s Equations in a vacuum:
(1)


E

0
(2)


B

0

 1
B


E
c
t

 1
E

B
c
t
These equations predict the existence of WAVES
for E and B which carry energy
Curl of (3) 
r
r
1
  E 
B
c t




1 E
 2 2
c t
2
use (4) 


(3)
(4)
Use vector identity:
r
r
r
2
    E    E   E

 

=0 from (1)

SO

2

1E
2

E
2 2 
0
c
t
Similarly, for B:

2

1B
2

B
2 2 
0
c
t
Vector
Wave
Equation
Note:
2


La pla cia
operates on each component of


E
and
B
so these 2 vector wave equations are actually 6 scalar equations
2 2 2
ˆ
ˆ
ˆ

E

i

E

j

E

k
E
x
y
z
2
So, for example, one of the equations is
2
2
2
2

E

E

E

E
1
x
x
x
x



2
2
2
2

x

y

z
c

t
Similarly for Ey, Ez, Bx, By and Bz
What are the solutions to the wave equations?
First, consider the simple 1-D case -A solution is
Wave equation
2
2

 1

2 2
2

x v
t

(
x
,
t
)

A
sin
k
(
x

v
t
)
A = constant (amplitude)
[kx] = radians
[kvt] = radians
2 
 2 Asink(xvt)
2
x
x

k A cos
k(xvt)
x
k2Asink(xvt)
2

 22

Ak
v
sin
k
(
x

v
t
)
2

t
So the wave equation is satisfied
Ψ is periodic in space (x) and time (t)

(
x
,
t
)

A
sin
k
(
x

v
t
)

(
x
,
t
)

A
sin
k
(
x

v
t
)


The wavelength λ corresponds to a change in the argument of
the sine by 2π




sin
k
(
x

v
t
)

sin
k
(
x

v
t
)

2
) (

k

2
spati
peria
o
k
v

2 l
per
tem
frequency
angular frequency
1 v
 
 


v

 

(
x
,
t
)

A
sin(
kx

t
)
2

2 radians/s
