Transcript Note

Ch. 5 - Basic Definitions
Specific intensity/mean intensity
Flux
The K integral and radiation pressure
Absorption coefficient & optical depth
Emission coefficient & the source function
Scattering and absorption
Einstein coefficients
Specific
Intensity
We want to characterize the
radiation from
Normal
• Area DA
• at an angle of view
q from the normal
to the surface
• through an
increment of solid
angle Dw
q
to observer
DA
Dw
Assume no azimuthal dependence
Specific Intensity
• Average Energy (Eldl) is the amount of energy
carried into a cone in a time interval dt
• Specific Intensity in cgs (ergs s-1 cm-2 sr-1 Å-1)
• Intensity is a measure of brightness – the amount
of energy coming from a point on the surface
towards a particular direction at a given time, at a
frequency n
• For a black body radiator, the Planck function
gives the specific intensity (and it’s isotropic)
• Normally, specific intensity varies with direction
dEn
In 
cos qdAdwdtdn
dEl
Il 
cos qdAdwdtdl
In vs Il
• The shapes of In and Il are different
because dn and dl are different sizes
at the same energy of light:
dn = -(c/l2) dl
• For example, in the Sun, Il peaks at
~4500Å while In peaks at ~8000 Å
Mean Intensity
• Average of specific intensity over all
directions
Jn  In
1

I
d
w
n

4
• If the radiation field is isotropic (same
intensity in all directions), then <In>=In
• Black body radiation is isotropic and <In>=Bn
Flux
• The flux Fn is the net energy flow across
an area DA over time Dt, in the spectral
range Dn, integrating over all directions
Fn   In cos qdw
• energy per second at a given wavelength
flowing through a unit surface area (ergs
cm-2 s-1 Hz-1)
• for isotropic radiation, there is no net
transport of energy, so Fn=0
On the physical boundary of a
radiating sphere…
• if we define Fn =Fnout + Fnin
2

2
0
0
0
 /2
2

0
 /2
Fn   d  In sin q cos qdq   d  In sin q cos qdq   d  In sin q cos qdq
0
• then, at the surface, Fnin is zero
• we also assumed no azimuthal dependence, so
 /2
Fn  2 
0
In sin q cos qdq
• which gives the theoretical spectrum of a star
One more assumption:
• If In is independent of q, then
 /2
Fn  2 
0
In sin q cosqdq  In
• This is known as the Eddington
Approximation (we’ll see it again)
Specific Intensity vs. Flux
• Use specific intensity when the surface is
resolved (e.g. a point on the surface of the
Sun). The specific intensity is independent
of distance (so long as we can resolve the
object). For example, the surface
brightness of a planetary nebula or a
galaxy is independent of distance.
• Use radiative flux when the source isn’t
resolved, and we're seeing light from the
whole surface (integrating the specific
intensity over all directions). The radiative
flux declines with distance (1/r2).
1
Jn 
In dw

4
Fn   In cos qdw
The K Integral
1
2
Kn 
In cos qdw

4
• The K integral is useful because the radiation exerts
pressure on the gas. The radiation pressure can be
described as
1 
PR  
c 0
I
cos
q
d
w
d
n

n

2
4
c


0
Kn dn
Radiation Pressure
• Again, if In is independent of
direction, then
4
PR 
3c


0
In dn
• Using the definition of the black body
temperature, the radiation pressure
becomes
4 4
PR 
3c
T
Luminosity
• Luminosity is the total energy
radiated from a star, at all
wavelengths, integrated over a full
sphere.
Class Problem
• From the luminosity and radius of the
Sun, compute the bolometric flux, the
specific intensity, and the mean
intensity at the Sun’s surface.
• L = 3.91 x 1033 ergs sec
• R = 6.96 x 1010 cm
-1
Solution
• F= T4
• L = 4R2T4 or L = 4R2 F, F = L/4R2
• Eddington Approximation – Assume In
is independent of direction within the
outgoing hemisphere. Then…
• Fn = I n
• Jn = ½ In (radiation flows out, but not in)
The Numbers
• F = L/4R2 = 6.3 x 1010 ergs s-1 cm-2
• I = F/ = 2 x 1010 ergs s-1 cm-2 steradian-1
• J = ½I= 1 x 1010 ergs s-1 cm-2 steradian-1
(note – these are BOLOMETRIC – integrated
over wavelength!)
The K Integral and Radiation Pressure
Kn   In cos qdw
2
4 4
PR 
T
3c
• Thought Problem: Compare the
contribution of radiation pressure to
total pressure in the Sun and in other
stars. For which kinds of stars is
radiation pressure important in a
stellar atmosphere?
Absorption Coefficient and Optical Depth
• Gas absorbs photons passing through it
– Photons are converted to thermal energy or
– Re-radiated isotropically
• Radiation lost is proportional to
–
–
–
–
absorption coefficient (per gram)
density
dIn   n In dx
intensity
pathlength
 
 In d
• Optical depth is the integral of the
absorption coefficient times the density
along the path (if no emission…)
L
n   n dx
0
In ( n )  In (0)e
n
Class Problem
• Consider radiation with intensity In(0)
passing through a layer with optical
depth n = 2. What is the intensity of
the radiation that emerges?
Class Problem
• A star has magnitude +12 measured
above the Earth’s atmosphere and
magnitude +13 measured from the
surface of the Earth. What is the
optical depth of the Earth’s
atmosphere at the wavelength
corresponding to the measured
magnitudes?
Emission Coefficient
• There are two sources of radiation within a
volume of gas – real emission, as in the
creation of new photons from collisionally
excited gas, and scattering of photons into
the direction being considered.
• We can define an emission coefficient for
which the change in the intensity of the
radiation is just the product of the
emission coefficient times the density
times the distance considered.
dIn  jn dx
Note that dI does NOT depend on I!
The Source Function
• The “source function” is just the
ratio of the absorption coefficient
to the emission coefficient:
Sn 
jn
n
Sounds simple, but just wait….
Pure Isotropic Scattering
• The gas itself is not radiating – photons only arise
from absorption and isotropic re-radiation
• Contribution of photons proportional to solid angle
and energy absorbed:
n In dxd w
djn dx 
4
n
jn    n In dw / 4 
In dw   n Jn

4
• Jn is the mean intensity: dI/dn = -In + Jv
• The source function depends only on the radiation field
For pure isotropic scattering
n
jn    n In dw / 4 
In dw   n Jn

4
• Remember the definition of Jn
• So Jn = jn/n
• Hey! Then Jn = Sn for pure isotropic
scattering
Pure Absorption
• No scattering – all incoming photons are
destroyed and all emitted photons are
newly created with a distribution set by
the physical state of the gas.
• Source function given by Planck radiation
law
• Generally, use Bn rather than Sn if the
source function is the Planck function
Einstein Coefficients
• For spectral lines or bound-bound
transitions, assumed isotropic
• Spontaneous emission is proportional to Nu
x Einstein probability coefficient, Aul
jn = NuAulhn
• (Nu is the number of excited atoms per
unit volume)
• Induced emission proportional to intensity
n = NlBluhn – NuBulhn
Induced (Stimulated) Emission
• Induced emission in the same
direction as the inducing photon
• Induced emission proportional to
intensity
nIn = NlBluInhn – NuBulInhn
True absorption
Induced emission
Radiative Energy in a Gas
• As light passes through a gas, it is both
emitted and absorbed. The total change of
intensity with distance is just
dIn  n In dx  jn dx
• dividing both sides by -kndx gives
jn
1 dIn

 In 
n  dx
n
The Source Function
• The source function Sn is defined as
the ratio of the emission coefficient
to the absorption coefficient
• The source function is useful in
computing the changes to radiation
passing through a gas
Sn  jn / n
1 dIn
The Transfer Equation    dx  In  Sn
n
• We can then write the basic equation of
transfer for radiation passing through gas,
the change in intensity In is equal to:
dIn = intensity emitted – intensity absorbed
dIn = jndx – nIn dx
dIn /dn = -In + jn/n = -In + Sn
• This is the basic equation which must be
solved to compute the spectrum emerging
from or passing through a gas.
Special Cases
• If the intensity of light DOES NOT
VARY, then Il=Sl (the intensity is
equal to the source function)
• When we assume LTE, we are assuming
that Sl=Bl
dI l
 I l  Bl
d l
Thermodynamic Equilibrium
• Every process of absorption is balanced by a
process of emission; no energy is added or
subtracted from the radiation
• Then the total flux is constant with depth
Frad  Fsurface  T
4
e
• If the total flux is constant, then the mean
intensity must be equal to the source function:
<I>=S
Simplifying Assumptions
• Plane parallel atmospheres (the depth of a
star’s atmosphere is thin compared to its radius,
and the MFP of a photon is short compared to
the depth of the atmosphere
• Opacity is independent of wavelength (a gray
atmosphere)

I   In dn
0

S   Sn dn
0
Eddington Approximation
• Assume that the intensity of the radiation
(Il) has one value in all directions toward
the outward facing hemisphere and
another value in all directions toward the
inward facing hemisphere.
• These assumptions combined lead to a
simple physical description of a gray
atmosphere