astrophysics [part i] · • radiative processes in astrophysics (george rybicki & alan...

Astrophysics [Part I]

Lecture 1 Mar 7 (Thurs), 2019

Kwang-il SeonKASI / UST

Overview

Radiative processes• link astrophysical systems with astronomical observables• cover many areas of physics and astrophysics (electrodynamics,

quantum mechanics, statistical mechanics, relativity...)

Textbooks• Radiative Processes in Astrophysics (George Rybicki & Alan

Lightman)• 천체물리학: 복사와 기체역학 (구본철, 김웅태)• The Physics of Interstellar and Intergalactic Medium (Bruce T.

Draine)• The Physics of Astrophysics, Volume 1 Radiation (Frank H. Shu)

Chapter 1.Fundamentals of Radiative Transfer

Electromagnetic Radiation

Particle/wave duality

classically: electromagnetic waves• speed of light: • wavelength and frequency:

quantum mechanically: photons• quanta: massless, spin-1 particles (boson)• Plank:• Einstein:

= c/

E2 = (mc2)2 + (pc)2

p = E/c

E = h = hc/ (h = 6.625 1027 ergs)

c = 3 1010cm s1

Definition• Energy flux, , is defined as the energy passing through an

element of area in time interval

• depends on the orientation of elements of and the frequency (or wavelength).

• Unit:

Energy flux (or energy flux density)

2–7

In order to derive Poynting’s theorem, let’s look at the mechanical work done on a particle i by the electric and the magnetic fields. This work is given by

vi · Fi = qivi ·!

E +vi

c× B

"

= qivi · E (2.34)

since vi ⊥ (vi × B).

Summing over all particles at a certain position then givesP =

#

i

δ(x − xi(t))qivi · E = j · E (2.35)

Because of Ampère’s law,

∇× B =4πc

j +1c

∂E

∂t⇒ j =

c

4π(∇× B) −

14π

∂E

∂t(2.4)

we find

j · E =c

4π

$

E · (∇× B) −1cE ·

∂E

∂t

%

(2.36)

=c

4πE · (∇× B) −

18π

∂(E2)

∂t(2.37)

Analogously to (a × b) · c = (b × c) · a one hasE · (∇× B) = ∇B · (B × E) = −∇B · (E × B) (2.38)

where ∇B operates only on B. Therefore

j · E = −c

4π∇B · (E × B) −

18π

∂E2

∂t(2.39)

Note that∇ · (E × B) = ∇B · (E × B) + ∇E · (E × B) (2.40)

Therefore, add (c/4π)∇E(E × B) to Eq. 2.39:

j · E = −c

4π

!

∇B · (E × B) + ∇E(E × B) −∇E(E × B)"

−18π

∂E2

∂t(2.41)

= −c

4π∇ · (E × B) +

c

4π∇E(E × B) −

18π

∂E2

∂t(2.42)

2–7

but ∇E(E × B) = (∇× E) · B (similar to Eq. 2.38)

= −c

4π∇ · (E × B) +

c

4π(∇× E) · B −

18π

∂E2

∂t(2.43)

Calculate ∇× E using Faraday’s law (Eq. 2.2), i.e.,

c

4π(∇× E) · B = −

14π

∂B

∂t· B = −

18π

∂

∂t(B · B) = −

18π

∂B2

∂t(2.44)

such that we finally obtain Poynting’s theorem

−j · E =c

4π∇ · (E × B) +

∂

∂t

$

E2

8π+

B2

8π

%

(2.32)

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2–8

Radiation Quantities 1

Flux Density, I

dAWe now relate the electrodynamicquantities such as E, B, or S tomeasurables. Before we can do this,need to introduce some definitions.

Definition. Energy flux, F , is definedas the energy dE passing througharea dA in time interval dt:

dE = F dA dt (2.45)

Units of F are erg cm−2 s−1.F depends on the orientation of dA, and can also depend on the frequency.

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2–9


Flux Density, II

Flux from an isotropic radiation source, i.e., a source emitting equal amounts ofenergy in all directions.Spherically symmetric stars are isotropic radiation sources, other astronomical objects such as, e.g., ActiveGalactic Nuclei, are not.

r 1

r2

Because of energy conservation, fluxthrough two shells around source isidentical:

4πr21 F (r1) = 4πr22F (r2) (2.46)

and therefore we obtain the inversesquare law,

F (r) =const.

r2(2.47)

F dEdA dt

F dA

erg cm2 s1

dE = FdAdt

Inverse Square Law

Flux from an isotropic radiation source, i.e., a source emitting equal amounts of energy in all directions.

• Because of energy conservation, flux through two shells around the source must be the same.

• Therefore, we obtain the inverse square law.

2–7

In order to derive Poynting’s theorem, let’s look at the mechanical work done on a particle i by the electric and the magnetic fields. This work is given by

vi · Fi = qivi ·!

E +vi

c× B

"

= qivi · E (2.34)

since vi ⊥ (vi × B).

Summing over all particles at a certain position then givesP =

#

i

δ(x − xi(t))qivi · E = j · E (2.35)

Because of Ampère’s law,

∇× B =4πc

j +1c

∂E

∂t⇒ j =

c

4π(∇× B) −

14π

∂E

∂t(2.4)

we find

j · E =c

4π

$

E · (∇× B) −1cE ·

∂E

∂t

%

(2.36)

=c

4πE · (∇× B) −

18π

∂(E2)

∂t(2.37)

Analogously to (a × b) · c = (b × c) · a one hasE · (∇× B) = ∇B · (B × E) = −∇B · (E × B) (2.38)

where ∇B operates only on B. Therefore

j · E = −c

4π∇B · (E × B) −

18π

∂E2

∂t(2.39)

Note that∇ · (E × B) = ∇B · (E × B) + ∇E · (E × B) (2.40)

Therefore, add (c/4π)∇E(E × B) to Eq. 2.39:

j · E = −c

4π

!

∇B · (E × B) + ∇E(E × B) −∇E(E × B)"

−18π

∂E2

∂t(2.41)

= −c

4π∇ · (E × B) +

c

4π∇E(E × B) −

18π

∂E2

∂t(2.42)

2–7

but ∇E(E × B) = (∇× E) · B (similar to Eq. 2.38)

= −c

4π∇ · (E × B) +

c

4π(∇× E) · B −

18π

∂E2

∂t(2.43)

Calculate ∇× E using Faraday’s law (Eq. 2.2), i.e.,

c

4π(∇× E) · B = −

14π

∂B

∂t· B = −

18π

∂

∂t(B · B) = −

18π

∂B2

∂t(2.44)

such that we finally obtain Poynting’s theorem

−j · E =c

4π∇ · (E × B) +

∂

∂t

$

E2

8π+

B2

8π

%

(2.32)

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2–8


Flux Density, I

dAWe now relate the electrodynamicquantities such as E, B, or S tomeasurables. Before we can do this,need to introduce some definitions.

Definition. Energy flux, F , is definedas the energy dE passing througharea dA in time interval dt:

dE = F dA dt (2.45)

Units of F are erg cm−2 s−1.F depends on the orientation of dA, and can also depend on the frequency.

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2–9


Flux Density, II

Flux from an isotropic radiation source, i.e., a source emitting equal amounts ofenergy in all directions.Spherically symmetric stars are isotropic radiation sources, other astronomical objects such as, e.g., ActiveGalactic Nuclei, are not.

r 1

r2

Because of energy conservation, fluxthrough two shells around source isidentical:

4πr21 F (r1) = 4πr22F (r2) (2.46)

and therefore we obtain the inversesquare law,

F (r) =const.

r2(2.47)

4r21F (r1) = 4r22F (r2)

F =const.

r2

Flux = a measure of the energy carried by all rays passing through a given areaIntensity = the energy carried along by individual rays.

• Let be the amount of radiant energy which crosses in time the area with unit normal in a direction within solid angle centered about with photon frequency between .

• The monochromatic specific intensity is then defined by the equation.

• area normal to :• Unit:

(Specific) Intensity or (Surface) Brightness

dAdtdE

nk

d + d

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2–10


Specific Intensity, I

Better description of radiation: energy carried along individual ray.

dΩ

n

Problem: Rays are infinitely thin =⇒No energy carried by them. . .=⇒Look at energy passing through

area dA (with normal n) in allrays going into spatial directiondΩ.

The specific intensity, Iν, in the bandν, . . . , ν + dν is defined via

dE = Iν dA dt dΩ dν (2.48)

Iν is measured in units of erg s−1 cm−2 sr−1Hz−1 and depends on location,direction, and frequency.In an isotropic radiation field, Iν = const. for all directions.

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2–11


Specific Intensity, II

Using the definition of I , we can calculate the net flux in a direction n.n

dΩ

θ

Contribution to flux in direction n fromflux into direction dΩ:

dFν = Iν cos θ dΩ (2.49)Integrate over all angles to obtain the total flux:

Fν =

!

4π srIν cos θ dΩ =

! π

θ=0

! 2π

0Iν(θ, φ) cos θ sin θ dθ dφ (2.50)

2–11

θ

Especially in the theory of stellar atmospheres, where often deals with the radiative transportthrough a slab of material and can assume cylindrical symmetry, one writes

µ = cos θ (2.51)

where θ is the angle between the z-direction (“up”-“down” direction) and the direction of thelight ray. Then

dµ

dθ= − sin θ (2.52)

and

Fν =

!

−1

µ=+1

! 2π

φ=0Iν(µ, θ)(−µ) sin θ

dµ

sin θdφ (2.53)

=

! +1

µ=−1

! 2π

φ=0Iµ(µ, θ)µ dµ dφ (2.54)

taking into account the cylindrical symmetry,

= 2π! +1

−1Iν(µ)µ dµ (2.55)

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2–12


Specific Intensity, III

The specific intensity is constant along the line of sight.

Proof: Consider the following figure:

R

dAdA1 2

Energy carried through area dA1 anddA2 is given by

dE1 = I1 dA1 dt dΩ1 dν (2.56)dE2 = I2 dA2 dt dΩ2 dν (2.57)

where dΩ1: solid angle subtended bydA2 at dA1, and vice versa:

dΩ1 = dA2/R2 (2.58)

dΩ2 = dA1/R2 (2.59)

Energy conservation implies dE1 = dE2, i.e.,

I1 dA1 dtdA2

R2 dν = I2 dA2 dtdA1

R2 dν (2.60)

that is: I1 = I2 = const.. QED.

k

dAx

erg s1

cm2

sr1

Hz1

dAk = dA cos k

dE = I(k,x, t)dAkdddt

= I(k,x, t) cos dAdddt

Note

Moments of intensity• intensity : scalar (amplitude of the differential flux)• differential flux : vector• momentum flux (radiation pressure) : tensor

Intensity can be defined as per wavelength interval.

Integrated intensity is defined as the intensity over all frequencies.

I =

Z 1

0Id =

Z 1

0Id

I |d| = I|d|I = I

d

d

=c

2=

Constancy of Specific Intensity in Free Space

Consider a bundle of rays and any two points along the rays. Construct areas and normal to the rays at these points.Consider the energy carried by the rays passing through both areas. Because energy is conserved,

Here, is the solid angle subtended by at and so forth.

radiative transfer equation in free space:

dA1 dA2

dE1 = I1dA1dtd1d1 = dE2 = I2dA2dtd2d2

d1 dA2 dA1

d1 = dA2/R2

d2 = dA1/R2 ! I1 = I2

d1 = d2

dIds

= 0

dA1 dA2

R

Inverse Square Law for a Uniformly Bright Sphere

Let’s calculate the flux at from a sphere of uniform brightness

Therefore, there is no conflict between the constancy of intensity and the inverse square law.Note• The flux at a surface of uniform brightness is .• For stellar atmosphere, the astrophysical flux is defined by .

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2–13


Specific Intensity, IV

I = const. along a ray does not contradict the inverse square law!

θR c

θP

r

Proof: Assume sphere of uniform brightness,B, the flux measured at point P is the fluxfrom all visible points of sphere:

F =!

I cos θ dΩ (2.61)

= B ·! π

0

! θc

0sin θ cos θ dθ dφ (2.62)

where sin θc = R/r. Therefore

F = 2πB

! arcsin(R/r)

0sin θ cos θ dθ (2.63)

but because" α0 sin x cos x dx = 1

2 sin2 α,

= 2πB ·12

#

R

r

$2= πB

#

R

r

$2∝

1r2

(2.64)

=⇒ inverse square law is consequence of decreasing solid angle of objects!

2–13

One consequence of Eq. 2.64 is that the flux on the surface of a source of uniform brightness is given by

F = πB (2.65)

Therefore, especially for stellar atmospheres, one sometimes defines the astrophysical flux,

F :=F

π(2.66)

such that for stars (which roughly have uniform surface brightness),F = B (2.67)

Be aware of this source of possible confusion!

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2–14


Energy Density and Mean Intensity

The last important radiation quantity (for the moment) is the energy density, uν.For a certain direction, Ω, and volume element dV , uν is defined via

dE = uν(Ω) dV dΩ dν (2.68)

But for light, the volume element can be written as

dV = c dt · dA (2.69)

such that Eq. 2.68 becomes

dE = cuν(Ω) dΩ dt dν (2.70)

Compare this to the definition of the intensity:

dE = Iν dA dt dΩ dν (2.48)

Therefore,uν(Ω) = Iν/c (2.71)

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2–15



The total energy density at frequency ν is then

uν =

!

4π sruν(Ω) dΩ =

1c

!

4π srIν(Ω) dΩ =:

4πc

Jν (2.72)

where the mean intensity is defined by

Jν =14π

!

4π srIν(Ω) dΩ (2.73)

Note that for an isotropic radiation field, Iν(Ω) = Iν, Iν = Jν.

The total radiation density is obtained by integration:

u =

!

uν dν =4πc

! ∞

0Jν dν (2.74)

I = B

(Specific) Energy Density

Consider a bundle of rays passing through a volume element in a direction .Then, the energy density per unit solid angle is defined by

Since radiation travels at velocity c,

the definition of the intensity

Therefore,

Energy Density

consider a bundle of rays passing through

a small volume dV

energy density uν(Ω) for bundle

defined by dE = uν(Ω) dΩ dV

but dV = dAdh, and flux thru height dh

in time dt = dh/c, so

dV = c dA dt

dΩ

dAdh = c dt

volumedV = c dA dt

thus we have

dE = c uν(Ω) dA dt dΩ (6)

but by definition dE = Iν dA dt dΩ, so

uν(Ω) =Iνc

(7)

5


Integrating over all solid angle, we obtain

Mean intensity is defined by

Then, the energy density is

Total energy density is obtained by integrating over all frequencies.

- momentum of a photon:

- force:

- radiation pressure = force per unit area

- radiation pressure due to energy flux

propagating within solid angle and with

frequency between :

Integrating over solid angle,

Momentum Flux: Radiation Pressure (due to absorption)

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The first cosine factor is due to the area normal to and the second one is due to the projection of the differential flux vector to the normal vector .

k

n

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Radiation Pressure (due to reflection)

Consider a reflecting enclosure containing an isotropic radiation field.Each photon transfers twice its normal component of momentum on reflection. Thus, we have

The angular integration yields

Radiative Transfer Equation

As a ray passes through matter, energy may be added, subtracted, or scattered from it by emission, absorption, or scattering.

The intensity will not in general remain constant.

We need to derive the radiative transfer equation.

Emission coefficient and Emissivity

• (monochromatic) spontaneous emission coefficient = the energy emitted per unit time per unit solid angle and per unit volume

• (angle integrated) emissivity = the energy emitted spontaneously per unit frequency per unit time per unit mass. For isotropic emission,

• Then, we obtain

• In going a distance , a beam of cross section travels through a volume . Thus the intensity added to the beam is by spontaneous emission is:

Absorption Coefficient

Consider the medium with particle number density , each having effective absorbing area (cross section)• number of absorbers = • total absorbing area =energy taken out of beam

Absorption coefficient is defined by

where is the mass density and is called the mass absorption coefficient or the opacity coefficient.

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The Radiative Transfer Equation

Without scattering term,

Including scattering term, we obtain an integrodifferential equation.

• scattering coefficient• scattering phase function

• for isotropic scattering

Stimulated emission:• We consider “absorption” to include both “true absorption” and

stimulated emission, because both are proportional to the intensity of the incoming beam (unlike spontaneous emission).

Z(,0)d = 1

Emission Only & Absorption Only

For pure emission,

• The brightness increase is equal to the emission coefficient integrated along the line of sight.

For pure absorption,

• The brightness decreases along the ray by the exponential of the absorption coefficient integrated along the line of sight.

Optical Depth & Source Function

Optical depth:

• Then, for pure absorption, • A medium is said to be optically thick if• A medium is said to be optically thin if

Source function:

• The radiative transfer equation can now be written

Mean Free Path

• From the exponential absorption law, the probability of a photon absorbed between optical depths and :

• The mean optical depth traveled is thus equal to unity:

• The mean free path is defined as the average distance a photon can travel through an absorbing material without being absorbed. In a homogeneous medium, the mean free path is determined by

• A local mean path at a point in an inhomogeneous material can be also defined.

|dI | =dId

d & |dI | / P ()d ! P () = e

hi =Z 1

0P ()d =

Z 1

0nue

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hi = ↵` = 1 ! ` =1

↵=

1

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Formal Solution of the RT equation

The solution is easily interpreted as the sum of two terms:• the initial intensity diminished by absorption• the integrated source diminished by absorption.

Relaxation

For a constant source function, the solution becomes

“Relaxation”•

• The source function is the quantity that the specific intensity tries to approach, and does approach if given sufficient optical depth.

I > S ! dId

< 0, then I tends to decrease along the ray

I < S ! dId

> 0, then I tends to increase along the ray

Radiation Force

• radiation flux vector in direction :

• the vector momentum per unit area per unit time per unit path length absorbed by the medium is

• This is the force per unit volume imparted onto the medium by the radiation field. The force per unit mass of material is given by

• Home work: derive the Eddington luminosity (problem 1.4)

Thermal equilibrium• Equilibrium means a state of balance.• Thermal equilibrium refers to steady states of temperature, which may

be spatial or temporal.• In a state of (complete) thermodynamic equilibrium, no net flows of

matter or of energy, no phase changes, and no unbalanced potentials (or driving forces), within the system.

• When the material is in thermodynamic equilibrium, and only the radiation field is allowed to depart from its TE, we refer to the state of the system as being in local thermodynamic equilibrium (LTE).

• A blackbody is an idealized physical body that absorbs all incident radiation regardless of frequency or angle of incidence (i.e., perfect absorber).

• A blackbody in thermal equilibrium emits the blackbody radiation, which is itself in thermal equilibrium.

• Thermal radiation is radiation emitted by “matter” in thermal equilibrium.

Universal function

• In equilibrium, radiation field in box doesn’t change.

• Now, consider another enclosure (box 2), also at the same temperature, but made of different material or shape. If their escaping intensities are different, energy will flow spontaneously between the two boxes. This violets the second law of thermodynamics. Therefore, the escaping intensity should be a universal function of T and should be isotropic. The universal function is called the Planck function .

Radiation and Thermodynamics

consider an enclosure (“box 1”)

in thermodynamic equilibrium at temperature T

the matter in box 1

• is in random thermal motion

• will absorb and emit radiation

details of which depends on

the details of box material and geometry

• but equilibrium

→ radiation field in box doesn’t change

Iν,1T

box 1

open little hole: escaping radiation has intensity Iν,14

now add another enclosure (“box 2”), also at temperature Tbut made of different material

IνIν

filter

T T

box 1 box 2

,2 ,1

separate boxes by filter passing only frequency νradiation from each box incident on screenQ: imagine Iν,1 > Iν,2; what happens?Q: lesson?Q: how would Iν,1 change if we increased the box volume

but kept it at T?5

Kirchhoff Law (천문학백과사전)

Kirchhoff’s Law

• Consider an element of some thermally emitting material at temperature .

• Put this into a blackbody enclosure at the same temperature.• Let the source function of the material be .

• But, the presence of the material cannot alter the radiation, since the new configuration is also a blackbody enclosure at .

• Kirchhoff’s Law: in LTE, the ratio of the emission coefficient to the absorption coefficient is a function of temperature and frequency only.

ThermaiRIIciihtion 17

Fipm 1.9 Thermal emitter plnced in the opening of a blackbody enchum.

Relation (1.37), called Kirchhoff s law, is an expression between 4. and j,, and the temperature of the matter T. The transfer equation for thermal radiation is, then, [cf. Eq. (1.23)],

dI” - = - gr, + cu,B,( T ) , ds

or

-= - dru I, + B,( T ) . d.,

(1.38)

Since S, = B,, throughout a blackbody enclosure, we have that I,, = B,, throughout. Blackbody radiation is homogeneous and isotropic, so that p = j u .

At this point it is well to draw the distinction between blackbody radiation, where I , = B,, and thermal radiation, where S,, = B,. Thermal radiation becomes blackbody radiation only for optically thick media.

I

Thermodynamics of Blackbody Radiation

Blackbody radiation, ltke any system in the thermodynamic equilibrium, can be treated by thermodynamic methods. Let us make a blackbody enclosure with a piston, so that work may be done on or extracted from the radiation (Fig. 1.10). Now by the first law of thermodynamics, we have

dQ= dU+pdV, (1.39)

where Q is heat and U is total energy. By the second law of thermodynamics,

j = ↵B(T ) ! Kirchho↵ 0s Law

Note : j = B(T ) if ↵ = 1 (perfect absorber, i.e., blackbody)

Implications of Kirchhoff’s Law

• A good absorber is a good emitter, and a poor absorber is a poor emitter. A good reflector must be a poor absorber, and thus a poor emitter.

• It is not possible to thermally radiate more energy than a blackbody, at equilibrium.

• The radiative transfer equation in LTE:

• Note:blackbody radiation meansthermal radiation meansThermal radiation becomes blackbody radiation only for optically thick media.

Thermodynamics

• First Law of Thermodynamics: heat is energy in transit.

where Q is heat and U is total (internal) energy.

• Second Law of Thermodynamics: heat is entropy.

where S is entropy.

See “Fundamentals of Statistical and Thermal Physics” (Federick Reif)

Thermodynamics of Blackbody Radiation

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• Stefan-Boltzmann law:

• total energy density:

• the integrated Planck function

• emergent flux (another form of the Stefan-Boltzmann law)

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Stefan-Boltzmann constant

Entropy of Blackbody Radiation

• Entropy:

• Entropy density:

• The law of adiabatic expansion for blackbody radiation:

Thus, we have the adiabatic index for blackbody radiation:

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The Planck Spectrum (Quantum Mechanics)

How to calculate the Blackbody spectrum?- Intensity spectrum is related to the energy density:

- Energy density = Number density of photon states x Average energy of each state

- Density of photon states = number of states per solid angle per volume per frequency

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The Planck Spectrum (Quantum Mechanics)

(1) Number density of photon state:• Consider a photon propagating in direction inside a box with

dimensions in x, y, z directions.• wave vector: • If each dimension of the box is much longer than a wavelength,

the photon can be represented by standing wave in the box.• number of nodes in each direction:• number of node changes in a wave number interval (if ):

• number of states in 3D wave vector element :

two independent polarizations

• the density of states (number of states per solid angle per volume per frequency):

(2) Average energy of each state:• Each state may contain photons of energy . The energy of the

state is .• The probability of a state of energy is proportional to ,

where and is the Boltzmann’s constant. (from statistical mechanics)

• Therefore, the average energy is:

Average number of photons (occupation number):

Energy density:

Planck Law:

Stefan-Boltzmann constant & Riemann zeta function

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Rayleigh-Jeans Law & Wien Law

Rayleigh-Jeans Law (low-energy limit)

• Originally derived by assuming the classical equipartition energy

• ultraviolet catastrophe: if the equation is applied to all frequencies, the total amount of energy would diverge.

Wien Law (high-energy limit)

Monotonicity with Temperature

3–9

To get all possible photons we need to count the number of al distinguishable photons at the same frequency, i.e., photons with different spin or with a different number ofnodes (=different n).

Spin is the easy one: there are only 2 polarization states.

To calculate the number of nodes, look the number in the x-, y-, or z-direction. In either direction

nx =Lx

λ=

kxLx

2π⇐⇒ dnx =

Lx

λ=

Lx dkx

2π(3.41)

For n ≫ 1, we can go to a “continuum of states”. Eq. 3.41 then implies

dN = dnx dny dnz =LxLyLz d3k

(2π)3 =V d3k

(2π)3 (3.42)

Therefore, the total number of states per unit volume and per wave number is

nk

d3k= 2 ·

dN

V

1d3k

=2

(2π)3 (3.43)

where the factor 2 is due to spin.

Because of Eq. (3.40),

d3k = k2 dk dΩ =(2π)3

c3 ν2 dν dΩ (3.44)

such that the density of states, i.e., the number of states per solid angle, volume, and frequency is given by

ρs =nν

dν dΩ=

2(2π)3 ·

(2π)3

c3 ν2 =2ν2

c3 (3.45)

Third step: Black Body Spectrum

To summarize, we had

• the mean energy of the state:

⟨E⟩ =hν

ehν/kT − 1(3.39)

3–9

• the state density:

ρs =2ν2

c3 (3.45)

The total energy density is then

uν(Ω) = ⟨E⟩ · ρs =2hν3

c31

exp(hν/kT ) − 1(3.46)

(energy per volume per frequency per solid angle)

Because of Eq. (2.71) (uν = Iν/c), the intensity is given by

Iν =2hν3

c21

exp(hν/kT ) − 1=: Bν (3.18)

This is the spectrum of a black body.

I

EF

CO

DRI

L

A I

N

R

DN

XA

EA

ESII

C

M

LMVA

AI

AD

R

E

LG

E

3–10

Blackbody Radiation: Properties 1

Spectrum, I

Max Planck (1858–1947)

In wavelength space, the spectrum of ablack body is blackbody radiation:

dE

dA dt dΩ dλ=

Bλ =2hc2/λ5

exp(hc/λkT ) − 1(3.47)

Bλ: Energy emitted per second and wavelength interval

• h = 6.63 × 10−27 erg s: Planck’s constant• k = 1.38 × 10−16 erg K−1: Boltzmann constant

105 1010 1015 1020Frequency ν [Hz]

10-20

10-15

10-10

10-5

100

105

1010

I ν(T)

[erg

s-1 cm

-2 H

z-1 sr

-1]

3K

102 K

103 K

104 K

105 K

106 K

107 K

108 K

109 K

Rayleigh-Jeans Law

Wein Law

Monotonicity:Of two blackbody curves, the one with higher temperature lies entirely above the other.

Wien Displacement Law

Frequency at which the peak occurs:

Wavelength at which the peak occurs:

Be aware

Characteristic Temperatures

Brightness Temperature:

• The definition is used especially in radio astronomy, where the RJ law is usually applicable. In the RJ limit,

• radiative transfer equation in the RJ limit:

• In the Wien region, the concept is not so useful.

Color Temperature:• By fitting the spectrum to a blackbody curve without regarding to

vertical scale (absolute intensity scale), a color temperature is obtained.

• The color temperature correctly gives the temperature of a blackbody source of unknown absolute scale.

Effective Temperature:• The effective temperature of a source is obtained by equating the

actual flux to the flux of a blackbody at temperature .

(Excitation Temperature)<latexit sha1_base64="397KoSJA5RZ/ode1PVseVFrDEwc=">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</latexit>

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Stefan-Boltzmann law

Ludwig Boltzmann(1844–1906)

Jožef Stefan(1835–1893)

As shown on worksheet 2, the totalenergy density of black bodyradiation is given by theStefan-Boltzmann law

uBB(T ) =! ∞

0

4π

cBν dν

=8π5

15

"

kT

hc

#3kT

=: aT 4

="

T

3400 K

#4erg cm−3

(3.57)

where the radiation density constant,

a :=8π5k4

15c3h3 = 7.566 × 10−15 erg cm−3 K−4 (3.58)

Note: units of uBB(T ) are erg cm−3.

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Stefan-Boltzmann law

φ

Often we are interested in the radiation diffusing out of amedium in thermal equilibrium, such as the flux ofradiation at the surface of a star.

F =! 2π

0

! π/2

0

! ∞

0Bν dν dφ dθ (3.59)

=! 2π

0

! π/2

0

cuBB

4πcos φ sin φ dφ dθ (3.60)

=cuBB

2

! π/2

0

12

cos(2φ) dφ =cuBB

4(3.61)

where cos φ is the projection factor between the direction of Bν and the area (Lambert’s law).The total flux emitted by the surface is therefore given by

F =dE

dt dA=

c

4uBB =

ac

4T 4 =: σSBT

4 (3.62)

with the Stefan-Boltzmann constant

σSB :=2π5k4

15c2h3 = 5.671 × 10−5 erg cm−2 K−4 s (3.63)

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Applications 1

Effective Temperature

5000 10000 15000Wavelength [A]

0.0

0.2

0.4

0.6

0.8

1.0

1.2

I λ/I 55

56G0 V

Teff (λ>1000A)=6485KTeff (λ>5000A)=6094K

The effective temperature, Teff, ofan object with spectral shape Iν isthe temperature for which

F =!

Iν cos θ dν dΩ

= σT 4eff

(3.64)

Sometimes, Iν is only known over acertain wavelength range, anddepending on the spectrum themeasured Teff will depend on thisrange (see figure).

G0 V spectrum after Pickles (1998),PASP 110, 863

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Applications 2

Planetary Surface Temperatures

The temperature of an irradiated body is given from energy equilibrium:L⊙

4πa2πr2 = σSBT44πr2 (3.65)

where a: distance to Sun, r: planetary radius.Therefore

T ="

L⊙

16πσSBr2

#1/4=

281 K(a/1 AU)1/2 (3.66)

Last step used L⊙ = 4 × 1033 erg s−1 and 1 AU = 1.496 × 1013 cm.If the planet reflects part of the radiation and if the IR emissivity is only roughly a BB, thenEq. (3.65) is modified,

(1 − B)L⊙

4πa2πr2 = ϵσSBT44πr2 =⇒ T =

281 K(a/1 AU)1/2

"1 − B

ϵ

#1/4(3.67)

where B: Bond albedo, and ϵ: effective emissivityFor the Earth, B = 0.39, for Venus, B = 0.72. Thus, since TEarth ∼ 288 K, ϵEarth = 0.55 < 1(greenhouse effect).If the planet is not a fast rotator, replace 4πr2 by 2πr2.

G0V spectrum (Pickles 1998, PASP, 110, 863)

(Note that the solar spectral type is G2V.)

The Einstein Coefficients

• Consider a system with two discrete energy levels and degeneracies . Let be the number densities of atoms in levels (1, 2).

Spontaneous Emission from level 2 to level 1:The Einstein A-coefficient is the transition probability per unit time for spontaneous emission.

Two-Level Systems in Radiative Equilibrium

consider an ensemble of systems (“atoms”) with• two discrete energy levels E1, E2• and degeneracies g1, g2i.e., a number g1 of distinct states have energy E1

,

E=hΔ ν0emissionabsorption

2

1 g 1

E

E , level 1:

level 2: g 2

in thermodynamic equilibrium at T , emission and absorptionexchange energy with photon field

Q: when & why emit? absorb?Q: connection between emission, absorption rates in ensemble?

3

, n2

, n1

(n1, n2)

Absorption from level 1 to level 2 occurs in the presence of photons of energy .• The absorption probability per unit time is proportional to the

density of photons (or to the mean intensity) at frequency .• In general, the energy difference between the two levels have

finite width which can be described by a line profile function .

• The Einstein B-coefficient is defined by transition probability per unit time for absorptionwhere

Absorption

atoms in lower state E1 only promoted to state E2by absorbing a photon of energy ∆Ehν +X1 → X2

if levels were perfectly sharp, absorb only at ∆E = hν0but in general, energy levels have finite widthi.e., line energies “smeared out” by some amount h∆nuso transitions can be made by photons with frequenciesν0 −∆ν <∼ ν <∼ ν0 +∆ν

useful to define line profile function φ(ν)

with normalization!

φ(ν) dν = 1

e.g., Gaussian, Lorentzian, Voight functions

limiting case of sharp levels ∆ν → 0:

φ(ν) → δ(ν − ν0)0ν

Δν

φ(ν)

5

Stimulated emission from level 2 to level 1:• Another Einstein B-coefficient is defined by

transition probability per unit time for stimulated emission.• Einstein found that to derive Planck’s law another process was

required that was proportional to radiation field and caused emission of a photon.

• The stimulated emission is precisely coherent (same direction and frequency, etc) with the photon that induced the emission.

Note:• Be aware that the energy density is often used instead of intensity

to define the Einstein B-coefficients.

Relations between Einstein Coefficients

• In TE, total absorption rate = total emission rate:

• Populations of the atomic states follow the Boltzmann distribution.

• Therefore,

• In TE, for all temperatures. We must have the following Einstein relations:

n1B12J = n2A21 + n2B21J

! J =A21/B21

(n1/n2)(B12/B21) 1

J =A21/B21

(g1B12/g2B21) exp(h0/kBT ) 1

J = B

g1B12 = g2B21

A21 =2h3

c2B21

n1

n2=

g1 exp(E1/kBT )

g2 exp(E2/kBT )=

g1g2

exp(h0/kBT ).

Einstein relations:

• If we can determine any one of the coefficients, these relations allow us to determine the other two.

• These connect atomic properties and have no reference to the temperature. Thus, the relations must hold whether or not the atoms are in TE.

If the relations were only for TE, the relations would contain the dependence on T.

• Without stimulated emission, Einstein could not get Planck’s law, but only Wien’s law.

When (Wien’s limit), level 2 is very sparsely populated relative to level 1. Then, stimulated emission is unimportant compared to absorption.

g1B12 = g2B21

A21 =2h3

c2B21

(A21, B21, B12)

h kBT

Radiative Transfer Equation in terms of Einstein Coefficients

Emission coefficient:• assumption: the line profile function of the emitted radiation is the

same profile as for the absorption .• energy emitted in volume , solid angle , frequency range ,

and time :

Here, note that each atom emits an energy distributed over solid angle .

• Then, the emission coefficient is given by

()

dV d d

dt

4h

jdV dddt = (h/4)n2A21dV d()ddt

j =h

4n2A21()

Absorption coefficient:• energy absorbed out of a beam in frequency range ,

solid angle , time , and volume

• Then, the absorption coefficient (uncorrected for stimulated emission) is given by

• What about the stimulated emission? It is proportional to the intensity, in close analogy to the absorption process. Thus, the stimulated emission can be treated as negative absorption. The absorption coefficient, corrected for stimulate emission, is

dd dt dV

↵ =h

4n1B12()

↵ =h

4() (n1B12 n2B21)

↵IdV dtdd = (h/4)n1B12IdV dtd()d

Source function:

• Using the Einstein relations, the absorption coefficient and source function can be written

S =j↵

=n2A21

n1B12 n2B21

↵ =h

4n1B12

1 g1n2

g2n1

S =2h3

c2

g2n1

g1n2 1

1

! generalized Kirchho↵ 0s law

Thermal Emission (LTE)

• If the matter is in TE with itself (but not necessarily with the radiation), we have the Boltzmann distribution. The matter is said to be in LTE.

• In LTE, we obtain the absorption coefficient and the Kirchhoff’s law:

• The Kirchhoff’s law holds even in LTE condition.

n1

n2=

g1g2

exp (h/kBT )

↵ =h

4n1B12

1 exp

h

kBT

()

S = B(T ) ! Kirchho↵ 0s law in LTE

Normal & Inverted Populations

Normal populations:• In LTE,

• The normal populations is usually satisfied even when the material is out of thermal equilibrium. This is called “normal populations.”

Inverted populations:

• In this case, the absorption coefficient is negative and the intensity increases along a ray.

• Such a system is said to be a maser (microwave amplification by stimulated emission of radiation; also laser for light...).

• The amplification can be very large. A negative optical depth of -100 leads to an amplification by a factor of .

n1

g1<

n2

g2

n2g1n1g2

= exp

h

kBT

< 1 ! n1

g1>

n2

g2

e100 = 1043

Scattering Effects: Pure Scattering

• Assumptionsisotropic scattering: scattered equally into equal solid anglescoherent scattering (elastic or monochromatic scattering): the total amount of radiation scattered per unit frequency is equal to the total amount absorbed in the same frequency range.Thompson scattering (scattering from non-relativistic electrons) is nearly coherent.

• scattering coefficient

• source function

• radiative transfer equation

This is an integro-differential equation, and cannot be solved by the formal solution. Rosseland approximation, Eddington approximation, or random walks

S = J =1

4

ZId

dIds

= ↵sca (I J)

In the textbook,the scattering coefficient is denoted by .

j = ↵sca

Z(,0)I(

0)d0

= ↵sca

1

4

ZId = ↵sca

J

!

• Random walks: let’s consider a single photon rather than a beam of photons (i.e., ray).

• In an infinite, homogeneous medium, net displacement of the photon after N free paths is zero, because the average displacement, being a vector, must be zero.

• root mean square net displacement:

The cross terms involve averaging the cosine of the angle between the directions before and after scattering, and this vanishes for isotropic scattering and for any scattering with front-back symmetry (Thompson or Rayleigh scattering)

Random Walks (in infinite medium)

R = r1 + r2 + r3 + · · ·+ rN ! hRi = 0

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• In a finite medium, a photon generated somewhere within the medium will scatter until it escapes completely.

• For regions of large optical depth, the mean number of scatterings to escape is roughly determined by (the typical size of the medium).

• For regions of small optical depth, the probability of scatterings within is .

• For any optical thickness, the mean number of scatterings is

• However, for Ly scattering, as .

Random Walks (in finite medium)

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Combined Scattering and Absorption• The RT equation to the case of combined absorption and scattering, in

which scattering is coherent isotropic.

• Source function is an weighted average of the two source functions.• extinction coefficient:• optical depth:• If a matter element is deep inside a medium (i.e., in TE),

• If the element is isolate in free space,

• generalized mean free path:

• probability of a (random walk) step ending in absorption:

• probability for scattering (known as the single-scattering albedo)

• source function (for coherent isotopic scattering case) :

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• In an infinite medium, every photon is eventually absorbed.• Since a random walk can be terminated with probability

at the end of each free path, the mean number of free paths before absorption is given by

• diffusion length (thermalization length, effective mean path, or effective free path): a measure of the net displacement between the points of creation and destruction of a typical photon.

Random Walks with Scattering and Absorption

mean number of free paths x probability of termination = 1

(= ↵abs/↵ext)

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• In a finite medium:• The behavior depends on whether its size L is larger or smaller

than the effective free path .• effective optical thickness:

• If effectively thin or translucent , most photons will escape the medium before being destroyed.luminosity of thermal source with volume V is

• If effectively thick, we expect , and only the photons emitted within an effective path length of the boundary will have a reasonable chance of escaping before being absorbed.

(F = B at surface of the source)

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Approximate Solutions

How to solve the radiative transfer equation:

We will learn two approximations to solve the equation.• Rosseland approximation• Eddington approximation

dIds

= ↵ext (I S)

S = (1 )J + B and = ↵abs /↵ext

Radiative Diffusion: (1) Rosseland Approximation

• Imagine a plane-parallel medium (in which depend only on depth z ).

• “zeroth” approximation: when the point in question is deep in the material, all quantities changes slowly on the scale of a mean free path and the derivative term above is very small.

This is independent of the angle.• “first” approximation:

, T

dz

ds = dz/ cos

ds =dz

cos =

dz

µ! µ

@I(z, µ)

@z= ↵ext

(I S)

I(z, µ) = S µ

↵ext

@I@z

(l = 1/↵ext )

I(0) (z, µ) S(0) (T )

) J (0) = S(0)

and I(0) = S(0) = B

I(1) (z, µ) S(0) µ

↵ext

@I(0)

@z= B(T )

µ

↵ext

@B(T )

@z! linear in µ

• net specific flux along z : the angle-independent part of the intensity does not contribute to the flux.

• total integrated flux:

let’s define the Rosseland mean absorption coefficient

F(z) =

ZI(1) (z, µ) cos d = 2

Z +1

1I(1) (z, µ)µdµ

= 2

↵ext

@B

@z

Z +1

1µ2dµ

= 4

3↵ext

@B(T )

@T

@T

@z

F (z) =

Z 1

0F(z)d = 4

3

@T

@z

Z 1

0

1

↵ext

@B

@Td

1

↵R

R10

1↵ext

@B@T d

R10

@B@T d

use

Then, we obtain the Rosseland approximation to radiative flux

which is also called the equation of radiative diffusion.• The flux equation can be interpreted as a heat conduction with an

“effective heat conductivity,” .• At which frequencies the Rosseland mean becomes important?

The mean involves a weighted average of so that frequencies at which the extinction coefficient is small (transparent) tend to dominate.The weighting function has a shape similar to that of the Planck function, but it peaks at , instead of .

Z 1

0

@B

@Td =

@

@T

Z 1

0Bd =

@T 4/

@T=

4T 3

F (z) = 16T 3

3↵R

@T

@z! rT

= 16T 3/3↵R

@B/@T

1/↵ext

hmax = 2.8kBThmax = 3.8kBT

Radiative Diffusion: (2) Eddition Approximation

• In Eddington approximation, the intensities are assumed to approach isotropy, and not necessarily their thermal values.In the Rosseland approximation, the intensities approach the Planck function at large effective depths.

• Near isotropy can be introduced by assuming that the intensity is linear in . (frequency is suppressed for convenience)

• Let us take the first three moments.

Compare with the following equations for the isotropic radiation.

µ

I(, µ) = a() + b()µ

mean intensity:

flux:

radiation pressure:

J 1

2

Z +1

1Idµ = a

H 1

2

Z +1

1µIdµ =

b

3! K =

1

3J

K 1

2

Z +1

1µ2Idµ =

a

3

Eddington approximation

p =1

3u

p 1

c

ZI cos2 d, u() =

1

cI

optical depth and the transfer equation:

Note: source function is independent to (because ).Integrate the above equation and obtain the following equations.

The two equations can be combined to yield:

Let us define a new optical depthThe radiative equation is then

This equation is sometimes called the radiative diffusion equation. Given the temperature structure of the medium, , the equation can be solved for J.

d(z) ↵extdz, µ@I

@= I S

µ S = (1 )J + B

1

2

Z +1

1dµµ

µ@I

@= I S

! @K

@= H ! 1

3

@J

@= H

1

2

Z +1

1dµ

µ@I

@= I S

! @H

@= J S

p3 =

p3abs(abs + sca)

@2J

@2= J B

1

3

@2J

@2= J S ! 1

3

@2J

@2= (J B)

B()

To solve the second order differential equation, we need two boundary conditions. The boundary conditions can be provided in several ways. One way to do is to use two-stream approximation, in which the entire radiation field is represented by radiation at just two angles, i.e., :

The two terms denote the outward and inward intensities. Then, the three moments are

Using , we obtain .

µ = ±µ0

I(, µ) = I+()(µ µ0) + I()(µ+ µ0)

J =1

2

I+ + I

H =1

2µ0

I+ I

K =1

2µ20

I+ + I

! we obtain µ0 =

1p3

in order to satisty K =1

3J

H =1

3

@J

@I+ = J +

1

3

@J

@, I = J 1

3

@J

@

= 0 = 0

1p3

@J

@= J at = 0

1p3

@J

@= J at = 0

(0 = cos1 µ0 = 54.74)

I+(0) = 0 and I(0) = 0 !

Suppose the medium extends from to and there is no incident radiation. Then, we obtain two boundary conditions:

= 0 = 0

I+(0) = 0

I(0) = 0

z

0

Iteration Method• Recall

• Let be the intensity of photons that come directly from the source, the intensity of photons that have been scattered once by dust, and the intensity after n scatterings. Then,

• The intensities satisfy the equations.

• Then, the formal solutions are:

I0 I1In

I(s) =1X

n=0

In(s)

In

!I0() = eI0(0) +

Z

0e( 0)S( 0)d 0

In() = eIn(0) +

Z

0e( 0)Sn1(

0)d 0

dI0()

d= I0() + S()

dIn()

d= In() + a

Z(,0)In1(,

0)d0

In() + Sn1()

dI(s)

ds= ↵extI(s) + ↵sca

Z(,0)I(s,0)d0 + j(s)

ordI()

d= I() + a

Z(,0)I(,0)d0 + S()

d ↵extds, S() j()

↵ext

Sn1() a

Z(,0)In1(,

0)d0

Approximation: (1) application to the edge-on galaxiesThe solution can be further simplified by assuming that

Then, the infinite series becomes

Kylafis & Bahcall (1987) and Xilouris et al. (1997, 1998, 1999) applied this approximation to model the dust radiative transfer process in the edge-on galaxies.872 E.M. Xilouris et al.: Are spiral galaxies optically thin or thick?

Fig. 1. I-band observations of the galaxies NGC 4013, IC 2531,UGC 1082, NGC 5529 and NGC 5907 (top to bottom).

cylinder with radius Rmax = 3 max(hs, hd) and half height of6 max(zs, Re), so that practically all the galactic light as wellas the dust are included. A Henyey-Greenstein phase functionhas been used for the scattering of the dust (Henyey & Green-stein 1941). The values for the anisotropy parameter g and thealbedo ! have been taken from Bruzual et al. (1988). Our taskhas been to find those values of the parameters in Eqs. (1) - (6)which create images of the model galaxies as close as possibleto the images of the observed galaxies. For the model fittingtechniques that we use, the reader is referred to Papers I and II.

4. Results

In Tables 2 through 6, we give the parameters of the best fitmodels to the observed data along with their 95% confidenceintervals for NGC 4013, IC 2531, UGC 1082, NGC 5529 andNGC 5907 respectively (for Table 7 describing NGC 891 seeSect. 5). Using the values of the parameters that describe thestellar and dust distributions of the galaxies, given in the abovetables, we have created model images and have compared themwith the observed images (see Figs. 1 and 2). In Fig. 1 the I-band observations of all five galaxies are presented (NGC 4013,IC 2531, UGC 1082, NGC 5529 andNGC 5907 from top to bot-tom). In Fig. 2 we show the model images of these galaxies inthe I-band with the same scale and sequence as in Fig. 1 so thata direct comparison can be made between the two figures. InFigs. 3–7 we give a more detailed comparison between modeland observation for each galaxy and each filter by showing ver-tical cuts along the minor axis. For this demonstration we use

Fig. 2. I-band models of the galaxies NGC 4013, IC 2531, UGC 1082,NGC 5529 and NGC 5907 (top to bottom).

the ‘folded’ (around the minor axis) and photometrically aver-aged images of the galaxies, which are also the images that wereused for the model fit (see Paper I). In each plot, the horizontalaxis represents the offset (in kpc) along the vertical direction,with zero lying on the major axis of the disk. The vertical axisgives the surface brightness (inmag/arcsec2). Real data are indi-cated by stars, while the model is shown as a solid line for eachprofile. The six profiles in each plot are vertical cuts at six dif-ferent distances along the major axis. The cuts are at distances0, 0.5h

I

s, 1h

I

s, 1.5h

I

s, 2h

I

sand 2.5h

I

sand are plotted from bot-

tom to top, with hI

sbeing the scalelength of the stars as derived

from the I-band modelling. The magnitude scale correspondsto the lower profile of each set, the other profiles are shiftedupwards (in brightness) by 2, 4, 6, 8 and 10 magnitudes respec-tively. Profiles between 0 and 1.5h

I

scorrespond to brightness

averaged over 400 (7.500 for NGC 5907) parallel to the majoraxis, while profiles at 2h

I

sand 2.5h

I

sare averaged over 800 (1500

for NGC 5907) along the same direction. This was done to re-duce as much as possible the local clumpiness that may exist inparticular areas of the galaxy. In these plots, only the data abovethe limiting sigma level ( 3 sigma of the local sky) that wereused for the fit are plotted, while the foreground stars have beenremoved.

One can clearly see a good agreement between model andobservation despite the small deviations in some places. It isworth remembering here that a global fit was done to the ob-served galaxy’s image and the derived set of parameters (Ta-bles 2–6) gives the best galaxy image as a whole.

InIn1

I1I0

(n 2)

Xilouris et al. (1999)

In I0

1X

n=0

I1I0

n

=I0

1 I1/I0

Approximation: (2) solution for the forward scattering• Assume the very strong forward-scattering

• The iterative solutions are:

I(0)

0

S = 0

I0() = eI0(0)

! S0() = aI0() = aeI0(0)

I1() = e

Z

0e

0S0(

0)d 0 = (a)eI0(0)

! S1() = aI1() = (a2)eI0(0)

I2() = e

Z

0e

0S1(

0)d 0 =(a)2

2I0(0)

! S2() = aI2() = (a32)eI0(0)

I3() = e

Z

0e

0S2(

0)d 0 =(a)3

3 2I0(0)

...

! Sn1() = aIn1() = (ann1)eI0(0)

In() = e

Z

0e

0Sn1(

0)d 0 =(a)n

n!I0(0)

(,0) = (0 )

! Sn1() = aIn1()

The final solutions are:Idirec() = eI(0)

Iscatt() =1X

n=1

In() =1X

n=1

(a)n

n!eI(0)

= (ea 1)eI(0)

aeI(0) if a 1

aI(0) if 1

Itot() = Idirec() + Iscatt()

= e(1a)I(0)

= eabsI(0)

astrophysics [part i] · • radiative processes in astrophysics (george rybicki & alan...

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