Spectroscopy – Lecture 2

I. Atomic excitation and ionization

II. Radiation Terms

III. Absorption and emission coefficients

IV. Einstein coefficients

V. Black Body radiation

I. Atomic excitation and ionization

· · · · · ·

1

2

3

n

∞

I

qualitative energy level diagram

Mechanisms for populating and depopulating the levels in stellar atmospheres:

• radiative

• collisional

• spontaneous transitions

E=–I

E=0

E>0

H, He relatively hard to ionize → hot stars you see absorption lines of Hydrogen

Metals relatively easy for first ionization

I. Atomic excitation and ionization

The fraction of atoms (or ions) excited to the nth level is:

Nn = constant gn exp(–n/kT)

Boltzmann factorstatistical weight

Statistical weight is 2J+1 where J is the inner quantum number (Moore 1945)1. For hydrogen gn=2n2

1 Moore, C.E. 1945, A Multiplet Table of Astrophysical Interest, National Bureau of Standards

I. Atomic excitation and ionization

Ratio of populations in two levels m and n :

Nn

Nm

=gn

gm( kT ) –exp

= n – m

I. Atomic excitation and ionization

The number of atoms in level n as fraction of all atoms of the same species:

Nn

N=

g1

gn ( kT ) –expn

+ g2 ( kT ) –exp2 + g3 ( kT ) –exp

3 ...+

=gn

u(T) ( kT ) –expn

u(T) = ( kT ) –expigi Partition Function

Nn

N=

gn

u(T)10 –

n = log e/kT = 5040/T

From Allen‘s Astrophysical Quantities

= 5040/T

Y = stage of ionization. Y = 1 is neutral, Y = 2 is first ion.

I. Atomic excitation and ionization

If we are comparing the population of the rth level with the ground level:

Nr

N1

=T

gr

g1log

–5040 + log

I. Atomic excitation and ionization

Example: Compare relative populations between ground state and n=2 for Hydrogen

g1 = 2, g2=2n2 =8

Temp. (K) =5040/T N2/N1

6000 0.840 0.000000018000 0.630 0.000001610000 0.504 0.0003115000 0.336 0.0015520000 0.252 0.0110040000 0.126 0.209

I. Atomic excitation and ionization

2000010000 40000 60000

N2/N1

I. Atomic excitation and ionization : Saha Eq.

For collisionally dominated gas:

=N1

NPe

h3

( 2m ) 23

( kT ) 25

2u1(T) u0(T) ( kT ) –exp

I

N1

N= Ratio of ions to neutrals

u1

u0

= Ratio of ionic to neutral partition function

m = mass of electron, h = Planck´s constant, Pe = electron pressure

I. Saha Equation

Numerically:

N1

N=

TPe

u1 u0

log–5040

I + 2.5 log T + log – 0.1762

orN1

N=

(T)Pe

(T) = 0.65 u1 u0

T 25

10–5040I/kT

I. Saha Equation

Example: What fraction of calcium atoms are singly ionized in Sirius?

log N1/N0 = 4.14 no neutral Ca

T = 10000 KPe = 300 dynes cm–2

Stellar Parameters:

Ca I = 6.11 evlog 2u1/uo = 0.18

Atomic Parameters:

I. Saha Equation

Maybe it is doubly ionized:

Second ionization potential for Ca = 11.87 ev

u1 = 1.0 log 2u2/u1 = –0.25

log N2/N1 = 0.82N2/N1 = 6.6

N1/(N1+N2) = 0.13

In Sirius 13% of the Ca is singly ionized and the remainder is doubly ionized because of the low ionization potential of Ca.

25000 10000 6300 4200T

The number of hydrogen atoms in the second level capable of producing Balmer lines reaches its maximum at Teff ≈ 10000 K

From Lawrence Allen‘s The Atmospheres of the Sun and Stars

Behavior of the Balmer lines (H)

Ionization theory thus explains the behavior of the Balmer lines along the spectral sequence.

How can a T=40000 star ionize Hydrogen?

I = 13.6 ev = 2.2 x 10–11 ergs E = kT → T = 160.000 K

So a star has to have an effective temperature to of 160.000 K to ionize hydrogen? Answer later.

Predicted behavior according to Ionization Theory

Observed behavior according to Ionization Theory

Ionization theory‘s achievement was the intepretation of the spectral sequence as a temperature sequence

II. Radiation Terms: Specific intensity

Normal

Observer

A

I =

E

cos A t lim

I = dE

cos dA d dt d

Consider a radiating surface:

II. Specific intensity

Can also use wavelength interval:

Id = Id

Note: the two spectral distributions (,) have different shape for the same spectrum

For solar spectrum:

I = max at 4500 Ang

I = max at 8000 Angc= d = –(c/2) d

Equal intervals in correspond to different intervals in . With increasing , a constant d corresponds to a smaller and smaller d

Circle indicates the integration is done over whole unit sphere on the point of interest

II. Radiation Terms: Mean intensity

I. The mean intensity is the directional average of the specific intensity:

J = 14 ∫ Id

II. Radiation Terms: Flux

Flux is a measure of the net energy across an area A, in time t and in spectral range

Flux has directional information:

-F

+F

A

II. Radiation Terms: Flux

F= limE

A t

dE= ∫dA dt d

∫ Icosd

F=

= ∫ Icosd

I = dE

cos dA d dt dRecall:

II. Radiation Terms: Flux

F = d

Looking at a point on the boundary of a radiating sphere

∫0

2

∫0

Icos sin d

= d∫0

2

∫0

Icos sin d

+ d∫0

2

∫/2

Icos sin d= 0

For stars flux is positive

Outgoing flux

Incoming flux

II. Radiation

Astronomical Example of Negative Flux: Close Binary system:

Hot star (DAQ3)

Cool star (K0IV)Hot Spot

II. Radiation Terms: Flux

F = 2

If there is no azimuthal () dependence

∫0

Isin cos d

Simple case: if I is independent of direction:

F = I (∫ sin cos d= 1/2 )

Note: I is independent of distance, but F obeys the standard inverse square law

L = 4R2I (=T4)

d

Flux radiating through a sphere of radius d is just F = L/4d2

r

F

Energy received ~ IA1/r2

Source image

SourceDetector element

A1

2r

F

Energy received ~ IA2/4r2 but A2 =4A1

= IA/r2

A2

10r

F

Energy received ~ IA3/100r2 but A3 = area of source

Since the image source size is smaller than our detector element, we are now measuring the flux

The Sun is the only star for which we measure the specific intensity

Detector element

Source image

A3

II. K-integral and radiation pressure

= d = ∫0

∫ ∫0

sin d d ∫–1

2 d

= cos

K = ∫ I cosd14

K = 12

d∫–1

II. K-integral and radiation pressure

This intergral is related to the radiation pressure.

Radiation has momentum = energy/c. Consider photons hitting a solid wall

Pressure= 2c

d E cosdt dA

component of momentum normal to wall per unit area per time = pressure

II. K-integral and radiation pressure

P d d = cos2 d d 2Ic

Pd = ∫ I cos2 d d/c

P = 4 ∫0

I () 2d = 2 ∫-1

I () 2dc c

4c

KP =

II. K-integral and radiation pressure

Special Case: I is indepedent of direction

3cP = 4 I

Total radiation pressure:

P = ∫0

Id3c4

=3c4

T4For Blackbody radiation

P = 2 ∫-1

I () 2dc

Radiation pressure is a significant contribution to the total pressure only in very hot stars.

II. Moments of radiation

J = 14 ∫ Id

J = 1 ∫–1

I () d2

K = 12

Id∫–1

H = 12

Id

∫–1

Mean intensity

Flux = 4H

Radiation pressure

= cos

III. The absorption coefficient

II+ dI

dx

is the absorption coefficient/unit mass [ ] = cm2/gm.

comes from true absorption (photon destroyed) or from scattering (removed from solid angle)

dI= – I dx

III. Optical depth

I I+ dI

The radiation sees neither or dx, but a the combination of the two over some path length L.

=∫o

L

dx Optical depth

L

Units: cm2

gmgmcm3

cm

III. Optical depth

Optically thick case:

>> 1 => a photon does not travel far before it gets absorbed

Optically thin case:

<< 1 => a photon can travel a long distance before it gets absorbed

Luca Sebben

Optically thin < 1 ≈1

Optically thick > 1

III. Simple solution to radiative transfer equation

II+ dI

dx

dI= – I d

I= I e–

Optically thin e– = 1- I = Io(1-)

dI= – I dx

III. The emission coefficient

II+ dI

dxj

dI= j I dx

jis the emission coefficient/unit mass

[ ] = erg/(s rad2 Hz gm)

j comes from real emission (photon created) or from scattering of photons into the direction considered.

III. The Source Function

The ratio of the emission to absorption coefficients have units of I. This is commonly referred to as the source function:

S = j/

The physics of calculating the source function S can be complicated. Let´s consider the simple cases of scattering and absorption

III. The Source Function: Pure isotropic scattering

•

isotropic scattering

d

dj to observer

The scattered radiation to the observer is the sum of all contributions from all increments of the solid angle like dRadiation is scattered in all directions, but only a fraction of the photons reach the observer

III. The Source Function: Pure isotropic scattering

The contribution to the emission from the solid angle d is proportional to d and the absorbed energy I. This is isotropically re-radiated:

dj = I d/4

∫j = I d/4

S = j

= ∫ I d/4 = J

The source function is the mean intensity

III. The Source Function: Pure absorption

All photons are destroyed and new ones created with a distribution governed by the physical state of the material.

Emission of a gas in thermodynamic equilibrium is governed by a black body radiator:

S =2h3

c2

1exp(h/kT) – 1

III. The Source Function: Scattering + Pure absorption

j = SI +

AB(T)

S = j/ where = S+

A

S =

S

S+

A J

A

S+

A B+

Sum of two source functions weighted according to the relative strength of the absorption and scattering

IV. Einstein Coefficients

When dealing with spectral lines the probabilities for spontaneous emission can be described in terms of atomic constants

Consider the spontaneous transition between an upper level u and lower level l, separated by energy h.

The probability that an atom will emit its quantum energy in a time dt, solid angle d is Aul. Aul is the Einstein probability coefficient for spontaneous emission.

IV. Einstein Coefficients

If there are Nu excited atoms per unit volume the contribution to the spontaneous emission is:

j = Nu Aul h

If a radiation field is present that has photons corresponding to the energy difference between levels l and u, then additional emission is induced. Each new photon shows phase coherence and a direction of propagation that is the same as the inducing photon.

This process of stimulated emission is often called negative absorption.

IV. Einstein Coefficients

The probability for stimulated emission producing a quantum in a time dt, solid angle d is Bul I dt d Bul is the Einstein probability coefficient for stimulated emission.

True absorption is defined in the same way and the proportionality constant denoted Blu.

I = NlBluIh – NuBulIh

The amount of reduction in absorption due to the second term is only a few percent in the visible spectrum.

IV. Einstein Coefficients

Nu

Nl

BluITrue absorption, dependent on I

Principle of detailed balance:

Nu[Alu + BulI] = NlBulI

Aul

Spontaneous emission, independent of I

Negative absorption, dependent on I

+BulI h

V. Black body radiation

Light enters a box that is a perfect absorber. If the container is heated walls will emit photons that are reabsorbed (thermodynamic equilibrium). A small fraction of the photons will escape through the hole.

Detector

V. Black body radiation: observed quantities

I =c4

5F(c/T)

I = 3 F(T)

F is a function that is tabulated by measurements. This scaling relation was discovered by Wien in 1893

I = 2kT2

c2

2ckT4

Rayleigh-Jeans approximation for low frequenciesI =

V. Black body radiation: The Classical (Wrong) approach

Lord Rayleigh and Jeans suggested that one could calculate the number of degrees of freedom of electromagnetic waves in a box at temperature T assuming each degree of freedom had a kinetic energy kT and potential energy

2kT2

c2I =

but as ∞, I

This is the „ultraviolet“ catastrophe of classical physics

Radiation energy density = number of degrees of freedom×energy per degree of freedom per unit volume.

V. Black body radiation: Planck´s Radiation Law

Derive using a two level atom:

Nn

Nm

=gn

gm( kT ) –exp

h

Number of spontaneous emissions: NuAul

Rate of stimulated emission: NuBulI

Absorption: NlBulI

V. Black body radiation: Planck´s Radiation Law

In radiative equilibrium collisionally induced transitions cancel (as many up as down)

NuAul + NuBulI = NlBluI

I = Aul

Blu(Nl/Nu) – Bul

I = Aul

(gl/gu)Bluexp(h/kT) – Bul

V. Black body radiation: Planck´s Radiation Law

This must revert to Raleigh-Jeans relation for small

I ≈ Aul

(gl/gu)Blu – Bul + (gl/gu)Bluh/kT

Expand the exponential for small values ex = 1+x)

h/kT << 1 this can only equal 2kT2/c2 if

Bul = Blugl/gu Aul = 2h3

c2Bul

Note: if you know one Einstein coeffiecient you know them all

V. Black body radiation: Planck´s Radiation Law

I = 1

(exp(h/kT) – 1)

2h3

c2

I = 1

(exp(hc/kT) – 1)

2hc2

5

Maximum I = T = 0.5099 cm K

Maximum I = 5.8789×1010 Hz K

V. Black body radiation: Planck´s Radiation Law

I = 2kT2

c2

Rayleigh-Jeans approximation → 0

2ckT4I =

I = 2h3

c2e–h/kT

I = 2hc2

5e–hc/kT

Wien approximation→∞

V. Black body radiation: Stefan Boltzman Law

In our black body chamber escaping radiation is isotropic and no significant radiation is entering the hole, therefore F = I

F d = ∫0

∞

∫0

∞ 1

(exp(h/kT) – 1)

2h3

c2d

Integral = 4/15

4 x3

= 2 c2 ( kT

h) ∫

0

∞

ex–1dx

∞

0

25k4

F∫ d =15h3c2

T4 = T4

x=hkT

V. Note on Einstein Coefficients and BB radiation

In the spectral region where h/kT >> 1 spontaneous emissions are more important than induced emissions

In the ultraviolet region of the spectrum replace I by Wien´s law:

BulI = Bul

2h3

c2 e–h/kT = Aul e–h/kT << Aul

Induced emissions can be neglected in comparison to spontaneous emissions

V. Note on Einstein Coefficients and BB radiation

In the spectral region where h/kT << 1 negative absorption (induced emissions) are more important than spontaneous emissions

In the far infrared region of the spectrum replace Iby Rayleigh-Jeans law:

BulI = Bul

2kTc2

= Aulc2

2h3

2kTc2 = Aul

kTh

>>Aul

The number of negative absorptions is greater than the spontaneous emissions

IU B J K

log I ~ –4 log

log I ~ –5 log – 1/

40000 20000 10000

50003000 15001000

500750

T (K)

Fra

ctio

n of

tota

l par

ticle

s

T=1000 K

T=6000 K

T=40000 K

1. Blackbody has a distribution of energies and some photons have the energy to ionize hydrogen

2. The thermal velocities have a Maxwell Boltzmann distribution and some particles have the thermal energy to ionize Hydrogen.

How can a T=40000 star ionize Hydrogen?

T = 6000 KI

I

V. Black body radiation: Photon Distribution Law

N = 1

(exp(h/kT) – 1)

2h2

c2

N = 1

(exp(hc/kT) – 1)

2hc2

4

Detectors detect N, not I !

Steven Spangler, Univ. of Iowa

Temperature of the Sun is almost a black body

But the corona has a much higher temperature

• The 1905 book “The Sun” by Abbott commented on the unidentified green and red lines in eclipse spectra

• Red and green lines are FeX and FeXIV, indicating temperatures of 1 - 2 million K

kT ≈ 262 ev

→ T ≈ 3 x 106 K