–1– (AST 461) LECTURE 1: Basics of Radiation Transfer Almost all

121 Pages·2011·0.42 MB·English

(AST 461) LECTURE 1: Basics of Radiation Transfer
Almost all we know about the astronomical universe comes from from
radiation emanating from faint sources. We need to know how to interpret
this radiation.
Electromagnetic spectrum, wavelength and frequency:
νλ = c, (1)
ν is frequency, λ is wavelength. The speed of light depends on the index of
refraction.
Justification for Macroscopic Treatment of Radiation
Scale of system >> λ of the radiation, the radiation can been considered
to travel in “straight lines” =rays. dA >> λ2
Amount of energy passing through source dA is in time dt is FdAdt where
F is the flux. Units are (erg/cm.s2) in CGS.
Flux from an isotropic source
Isotropic means energy emitted in all directions. Consider 2 different
spherical surfaces S and S . (fig 1.)
1 2
F dA dt = F dA dt (2)
1 1 1 2 2 2
but dt = dt so that
1 2
F /F = dA /dA = r2/r2. (3)
1 2 2 1 2 1
This is the inverse square law.
Intensity
Flux measures energy from all rays in given area. A more detailed ap-
proach is to consider energy carried from individual rays, or sets of rays
differing infinitesimally from the initial ray. (fig 2.)
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dE = (kˆ·nˆ)I dAdtdΩdν, (4)
ν
where I is the Specific Intensity and represents Energy /(time × area × sold
ν
angle × frequency).
Now suppose we have an isotropic radiation field. The differential flux at
a given frequency is
dF = (kˆ·nˆ)I dΩ = I cosθdΩ. (5)
ν ν ν
But if I is isotropic, then F = 0.
ν ν
Constancy of Intensity in Free Space:
Specific intensity is constant along a ray in free space.
Here’s why: Consider all rays passing through both dA and dA and use
1 2
conservation of energy: (fig 3)
dE = I dA dt dΩ dν = dE = I dA dt dΩ dν (6)
1 ν,1 1 1 1 1 2 ν,2 2 2 2 2
Note that dt = dt , dν = dν . Note also that dΩ is the solid angle
1 2 1 2 1
subtended by dA at dA and dΩ is the solid angle subtended by dA at
2 1 2 1
dA . Thus we have
2
dΩ = dA /r2 (7)
1 2
and
dΩ = dA /r2. (8)
2 1
Thus
I = I . (9)
ν,1 ν,2
The intensity is constant along free space.
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Application to a Telescope (fig 3.)
Assume large distance between object and lens r >> f, where f is the
focallength. DetermineimageintensityI givenobjectintensityI . Infinites-
i 0
imal amount of surface area of object dA has intensity I . We have
0 0
I dA dΩ = I A dA /r2. (10)
0 0 T,0 0 T 0
where dΩ is the solid angle of the telescope as measured from the dA of
T,0 0
the object. All photons from dA must strike dA on focal plane.
0 1
Thus
I dA dΩ = I dA A /r2 = I dA dΩ = I dA A /f2, (11)
0 0 T,0 0 0 T i i T,i i i T
but solid angle sub-tending image from telescope equals solid angle sub-
tending object from telescope so
dA /r2 = dA /f2, (12)
0 i
and as expected, I = I .
i o
Do telescopes measure flux or intensity?
If resolution of telescope is crude, and we cannot resolve object, then we
measure flux. If telescope can resolve object, then we measure intensity.
Why? Consider the case when the source is unresolved. Now imagine
pushing the source to farther distance. As the distance increases, the number
of photons falls as r2. The flux is measured.
If instead the the source is resolved, then as the source is pushed far-
ther away, more area of the source would be included with the solid angle,
which compensates for the the increased distance and the collected number
of photons remain the same.
Flux from a uniformly bright sphere: read in text.
Relationship of Intensity to Stat Mech Quantities:
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Consider photon distribution function f such that f (x,p,t)d3xd3p is the
α
number of photons in x,p space, with spin index α.
p = hk = (hν/c)kˆ (13)
so
2
dE = (cid:88)hνf (x,p,t)d3xd3p. (14)
α
α=1
But photons traveling in direction kˆ for time dt, through an element of
area dA whose normal = nˆ, occupy
d3x = cdt(kˆ·nˆ)dA (15)
and
d3p = p2dΩdp = (E2/c2)dΩdE/c = (h3ν2/c3)dΩdν, (16)
where we have used E = pc and E = hν. Thus,
2
dE = (kˆ·nˆ)(cid:88)(h4ν3/c3)f (x,p,t)dAdtdΩdν (17)
α
α=1
so
2
I = (cid:88)(h4ν3/c3)f (x,p,t). (18)
ν α
α=1
In stat mech, the occupation number for each photon is
N = h3f , (19)
α α
which is dimensionless so
2
I = (cid:88)(hν3/c2)N (x,p,t). (20)
ν α
α=1
Later we will see for example that I corresponds to B , the Planck func-
ν ν
tion for b-body radiation. The occupation number corresponds to N =
α
(ehνα/kT−µα −1)−1 for Bose-Einstein stats.
Radiation Pressure
photon carries momentum p = E /c. (fig. 4)
ν ν
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Can compute pressure by considering incident photons reflecting off of a
wall: The angle of incidence equals the angle of reflection. The change in
the z component of momentum of photon between frequency ν and ν +dν
reflected in time dt from area dA is
dp dν = [(p ) −(p ) ]dν = (1/c)(E cosθ−(−E cosθ))dν = (2/c)E cosθdν
ν ν f ν i ν ν ν
(cid:16) (cid:17)
= 2 I dνdtdAcos2θdΩ,
c ν
(21)
where we used (4). Now the change in momentum per unit area per unit
time is a differential force/area= differential pressure. Integrating over solid
angle gives the total pressure:
dp (cid:90) (cid:90) 2π (cid:90) π/2
P = ν = (2/c) I cos2θdΩ = (2/c) I cos2θsinθdθdφ.
rad,ν ν hem ν
dtdA
φ=0 θ=0
(22)
Now imagine removing the reflecting surface. Thus instead of the factor of 2
in momentum, we would have photons coming in from the other side. Thus
we can remove the factor of 2, and integrate over the full sphere.
(cid:90) (cid:90) 2π (cid:90) π
P = dp /dtdA = (1/c) I cos2θdΩ = (1/c) I cos2θsinθdθdφ.
rad,ν ν ν ν
φ=0 θ=0
(23)
Energy Density
Consider a cylinder of cross section dA. (fig 5)
dE = u (Ω)dVdνdΩ = u (Ω)cdtdνdAdΩ = I dAdΩdtdν (24)
ν ν ν
so the energy density is
u (Ω) = (I /c). (25)
ν ν
Then
(cid:90) (cid:90)
u = u (Ω)dΩ = (1/c) I dΩ = 4πI /c (26)
ν ν ν ν
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For isotropic radiation field
I = I (27)
ν ν
Note also that the pressure for an isotropic radiation field from (23) and (26)
is
(cid:90) (cid:90) 1 (cid:90)
P = (1/c) I cos2θdΩdν = (I /c) cos2θdΩdν = u dν. (28)
ν ν ν
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LECTURE 2
Equation of Radiative Transfer
Condition that I is constant along rays means that
ν
dI /dt = 0 = ∂ I +ck·∇I , (29)
ν t ν ν
where k · ∇ = dI /ds is the ray-path derivative. This is equation is the
ν
statement that there are no sources or sinks. But real systems may have
sources and sinks: emission (source) and absorption (sink).
If light travels past a system much faster than the time scale for which
the system evolves then ∂ I (cid:39) 0 and we have then
t ν
dI /dt = 0 = ck·∇I = dI /ds = sources − sinks, (30)
ν ν ν
where ds is the path. We now construct the source and sink terms in this
radiative transfer equation.
Emission coefficient (source)
Spontaneous emission coefficient j is defined as the energy emitted per
ν
unit time per unit solid angle, per unit volume per unit frequency
dE = j dVdΩdtdν, (31)
ν
where j has units erg/cm3 · steradian · sec · hz. The emissivity is defined
ν
as (cid:15) = 4πj /ρ, where ρ is the mass density. Since dV = dAds, the “source”
ν ν
contribution to the radiative transfer equation is given by
dI = j ds. (32)
ν ν
Absorption coefficient (sink)
Consider the propagation of a beam through a cylinder (again) of cross
section dA and length ds, populated with a number density n of absorbers
(particles), with absorption cross section σ . (fig 6)
ν
The loss of intensity, or the energy absorbed out of the beam is given by
dE = (I −I )dAdΩdtdν = I (nσ dAds)dΩdtdν. (33)
ν i,ν f,ν ν ν
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Thus the “sink” contribution to the radiative transfer equation is
dI = −nσ I ds. (34)
ν ν ν
The absorption coefficient α is defined such that
ν
dI = −α I ds, (35)
ν ν ν
so α = nσ and α is units of cm−1. Sometimes it’s written α = ρκ , where
ν ν ν ν ν
ρ is the mass density and κ is the mass absorption or “opacity” coefficient.
ν
Note that the emission and absorption coefficients have different units. Also,
note that absorbing particles are assumed to be randomly distributed and
much smaller than inter-particle spacing.
The term “absorption” is used somewhat loosely: Stimulated emission is
also included in the absorption coefficient because it is proportional to the
incident intensity.
We have thus derived the radiative transfer equation which we will soon
solve:
dI /ds = −α I +j . (36)
ν ν ν ν
Solutions of Equation of Radiative Transfer
A primary goal is to see what the values of α and j are for different
ν ν
absorption and emission processes. Scattering complicates things since radi-
ation from the initial solid angle is scattered into a different solid angle for
which the scatter depends on the initial solid angle, so we need numerics.
Consider some simple cases here.
Emission Only:
dI /ds = j , (37)
ν ν
where the solution is
(cid:90) s
I (s) = I (s )+ j (s(cid:48))ds(cid:48). (38)
ν ν 0 ν
s0
Intensity increase is equal to the integrated emission coefficient along the
propagation path.
Absorbtion Only: Then j = 0. In this case
ν
(cid:90) s
I (s) = I (s )Exp[− α(s(cid:48))ds(cid:48)]. (39)
ν ν 0
s0
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The intensity decreases exponentially along the ray path.
Optical Depth
Define
dτ = α ds, (40)
ν ν
or
(cid:90) s
τ (s) = α(s(cid:48))ds(cid:48). (41)
ν
s0
The optical depth is measure of how transparent a medium is to radiation.
For τ ≥ 1 the medium is optically thick, and for τ < 1 the medium is
ν ν
optically thin. We rewrite the transfer equation (36) as
dI /dτ = −I +S , (42)
ν ν ν ν
wherewedefinethesourcefunctionS ≡ j /α astheratiooftheemissionto
ν ν ν
the absorption coefficient. Optical depth is a physically meaningful variable
because it contains the information about transparency along the path.
Now solve (42) by integrating factor: Multiplying by eτν, we obtain:
d(I eτν)/dτ = eτνS , (43)
ν ν ν
so the solution is
I eτν = I (0)+(cid:90) τν S (τ(cid:48))eτν(cid:48)dτ(cid:48) (44)
ν ν ν ν ν
0
or
I = I (0)e−τν +(cid:90) τν S (τ(cid:48))e−(τν−τν(cid:48))dτ(cid:48). (45)
ν ν ν ν ν
0
The right side is the sum of the initial intensity attenuated by absorption,
and the emission source attenuated by absorption. For a constant source
function we have
I = I (0)e−τν +S (1−e−τν) = S +e−τν(I (0)−S ). (46)
ν ν ν ν ν ν
At large optical depth, I (cid:39) S . Why? because the source function is the
ν ν
local input of radiation–little is contributed to the to I from matter far from
ν
the location of interest when τ >> 1.
ν
Note also that there is a relaxation process (evidenced by (42) where I
ν
tends toward the source function: If I > S , then dI /dτ < 0. If I < S ,
ν ν ν ν ν ν
then dI /dτ > 0.
ν ν
Mean Free Path
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Interpreting the exponential in(39) (which equals e−τν) as the probability
of traveling an optical depth τ , the mean optical depth traveled
ν
(cid:90) ∞
(cid:104)τ (cid:105) = τ e−τνdτ = 1. (47)
ν ν ν
0
The associated mean free path (MFP) is
l = (cid:104)τ (cid:105)/α = 1/nσ . (48)
ν ν ν ν
MFP is important concept for particles as well as radiation.
Thermal Radiation and Blackbody Radiation
Thermal radiation is radiation emitted from matter in thermal equilib-
rium.
A blackbody absorbs all incident radiation.
Blackbody radiation is radiation which itself is in thermal equilibrium
and in thermal equilibrium with surrounding matter.
Consider a thermally insulated box kept at temp T. I = I (T) = B , the
ν ν ν
Planck function for such a black body box. Dependence only on temperature
can be seen by considering two connected boxes at same T. Energy cannot
flow without violating second law. Thus I is same in both boxes.
ν
I is isotropic because otherwise there would be a net energy flow from
ν
one part of the box to another which is not possible when entire system is at
uniform T.
For a blackbody,
2hν3/c2
I = B (T) = , (49)
ν ν
Exp[hν/kT]−1
the Planck function, and is only a function of the temperature and frequency.
Kirchoff’s Law
Put thermally emitting material at temperature T with source function
S (T) in the box.
ν
If S > B then the intensity I will tend to increase such that I > B
ν ν ν ν ν
and if S < B then the intensity will decrease to make I < B . But in
ν ν ν ν
equilibrium, the new box is still a blackbody, and so S = B or
ν ν
j = α B , (50)
ν ν ν
whichisKirchoff’slawforathermalemitter. Ingeneral, Kirchoffslawrelates
emission to absorption.
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