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Transcript PPT - University of Arizona
Astro 300B: Jan. 19, 2011
Radiative Transfer
Read:
Chapter 1, Rybicki & Lightman
c
Light
where
Energy
= wavelength
-1
= frequency units: Hz or sec
c = 3.00 x 1010 cm sec-1
velocity of light in a vacuum
E
h
where
h = 6.625 x10-27 erg sec
Planck’s constant
Region
Gamma Ray
Wavelength
λ< 0.1 Å
X-ray
0.1 Å < λ< 100 Å
Ultraviolet
100 Å < λ< 4000Å
Visible
4000 Å < λ< 7000Å
Infrared
7000Å < λ<1 mm
Microwave
1 mm < λ< 10 cm
Radio
10 cm < λ
Units and other important facts
-8
• Angstrom Å = 10
-10
cm = 10 m
-4
-6
• Micron
µm =10 cm = 10 m
-1
• Wave number = (2π)/ λ
cm
(number of wavelengths / distance)
1 eV = 1.6 x 10-12 ergs
Jansky = 1 Jy
= 10-26 W m-2 Hz -1
= 10-23 ergs s-1 cm-2 Hz-1
Radio
λ=few cm
VLA, GBT
Mm
λ = 1,2,3mm
ARO 12m, Interferometers – BIMA, OVRO
ALMA
Submm λ = 800 microns
JCMT, CSO, SMT, South Pole
230, 345, 492 GHz
Mid/far IR λ = 20-350 microns
Near IR
λ = 1-10 microns
Space only
IRAS, ISO, Spitzer
J: 1.25 microns
H: 1.60 microns
K: 2.22 microns
L: 3.4 microns
N: 10.6 microns
Sky emission lines very bright for λ= 8000 Å – 2 µm
Many atmospheric absorption features
λ > 2.2 µm: Thermal emission from telescope dominates
NICMOS, WFPC3 on HST
JWST
Shape arbitrary – see also UKIRT home page
Optical:
λ = 3200 Å -- 9000 Å
Earth’s atmosphere opaque for λ < 3200 Å
silicon (ccds) transparent for λ > 9000 Å
UV:
911 Å -- Milky Way is opaque for λ < 911 Å
1215 Å Lyman alpha, n=2 n=1 for Hydrogen
IUE, HST, GALEX: 1150 – 3200 Å
HUT, FUSE
911 -1200 Å (MgFl cutoff)
X-ray
The γ- rays
E = 0.2 – 10 keV
E > 10 keV
Einstein, ROSAT, Asca
Chandra, XMM, Astro-E
XTE, GRO, INTEGRAL
I
Definitions:
1. Specific Intensity
J F p
I
Î
Consider photons flowing in direction Î,
into solid angle dΩ, centered on Î
The vector n is the normal to dA
Energy
(ergs)
dE
dA
cos
dt
d
d
Projection of area dA perpendicular to Î
radiant energy flowing through dA,
in time dt, in solid angle dΩ, in direction Î
I
Is defined as the constant of proportionality
dE
I
dA
co
dt
d
d
s
n
θ
dΩ
Perhaps it’s easier to visualize photons falling from the sky
from all directions on a flat area, dA,
on the surface of the Earth
Recall polar coordinates
Solid angle: dΩ = sin θ dθ dφ
d
units=steradian
2
0
d
0
sin d
Specific
Intensity
I
dE
dA cos dt d d
Comments:
• units:
ergs cm-2 s-1 Hz-1 steradian -1
• depends
on location in space
on direction F I
on frequency
cos d
• in the absence of interactions
with matter, Iν is constant
• in “thermodynamic equilibrium”, Iν is the blackbody, or Planck
function – a universal function of temperature T only
32
2
h
c
I
B
T
)
(
h
exp
1
kT
Now consider various moments of the specific intensity:
i.e. multiply Iν by powers of cosθ and integrate over dΩ
2. Mean Intensity
J
Zeroth moment
J geometrical mean ofI
over solid angle
d
1 4
J
I
d
4 0
d
d
sin
d
2
where
0
0
Comments:
• Jν has the same units as Iν
ergs cm-2 sec-1 Hz-1 steradian-1
Even though you integrate over dΩ Iν to get Jν, you divide by 4π
• Some people define Jν without the 4 π,
so the units are
ergs cm-2 sec-1 Hz-1
• Jν is what you need to know to compute photoionization rates
3. Net Flux
F
F
I
cos d
• Fν is the thing you observe:
net energy crossing surface dA
in normal direction n,
from Iν integrated over all solid angle
reduced by the effective area cosθ dA
• If Iν is isotropic (not a function of angle), then
F
0
since
cos
d
0
There is as much energy flowing in the +n direction
as the –n direction
• Also, if Iν is isotropic, then
I J
4. Radiation pressure
p
Momentum of a photon = E/c
Pressure = momentum per unit time, per unit area
Momentum flux in direction θ is
dF
c
Component of momentum flux normal to dA is
dF
cos
c
Radiation pressure is therefore
p
1
c
I cos2 d
dynes cm-2 Hz-1
We have written everything as a function of frequency, ν
However, you can also integrate these quantities over ν,
either ν= 0
or ν over some passband.
F
(
ergs
s
cm
)
F
d
1
2
p
(
dcm
ynes
)
p
d
2
I
(
ergs
s
cm
ster
)
I
d
1
2
1
Iν=0 in all directions except
towards the point source
Fν vs. Iν
Specific
Intensity
Point
Source
Iν=0
I 0
Iν=0
Fν= 0 normal to the line
of sight to the point
source because cosθ = 0
F
0
Fν=0
In every other direction
F
0
FLUX
• Uniform, isotropic, homogeneous radiation field:
Fν =0
J ν = Iν
•
Iν
•
Fν
is independent of distance from the source
obeys the inverse square law