spectroscopy – lecture 2
DESCRIPTION
Spectroscopy – Lecture 2. Atomic excitation and ionization Radiation Terms Absorption and emission coefficients Einstein coefficients Black Body radiation. ·. ·. ·. ·. ·. ·. I. c. I. Atomic excitation and ionization. E >0. E =0. ∞. qualitative energy level diagram. n. - PowerPoint PPT PresentationTRANSCRIPT
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Spectroscopy – Lecture 2
I. Atomic excitation and ionization
II. Radiation Terms
III. Absorption and emission coefficients
IV. Einstein coefficients
V. Black Body radiation
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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
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H, He relatively hard to ionize → hot stars you see absorption lines of Hydrogen
Metals relatively easy for first ionization
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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
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I. Atomic excitation and ionization
Ratio of populations in two levels m and n :
Nn
Nm
=gn
gm( kT ) –exp
= n – m
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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
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From Allen‘s Astrophysical Quantities
= 5040/T
Y = stage of ionization. Y = 1 is neutral, Y = 2 is first ion.
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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
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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
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I. Atomic excitation and ionization
2000010000 40000 60000
N2/N1
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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
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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
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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:
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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.
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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
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Behavior of the Balmer lines (H)
Ionization theory thus explains the behavior of the Balmer lines along the spectral sequence.
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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.
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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
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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:
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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
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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
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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
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II. Radiation Terms: Flux
F= limE
A t
dE= ∫dA dt d
∫ Icosd
F=
= ∫ Icosd
I = dE
cos dA d dt dRecall:
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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
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II. Radiation
Astronomical Example of Negative Flux: Close Binary system:
Hot star (DAQ3)
Cool star (K0IV)Hot Spot
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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
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L = 4R2I (=T4)
d
Flux radiating through a sphere of radius d is just F = L/4d2
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r
F
Energy received ~ IA1/r2
Source image
SourceDetector element
A1
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2r
F
Energy received ~ IA2/4r2 but A2 =4A1
= IA/r2
A2
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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
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II. K-integral and radiation pressure
= d = ∫0
∫ ∫0
sin d d ∫–1
2 d
= cos
K = ∫ I cosd14
K = 12
d∫–1
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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
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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 =
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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
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Radiation pressure is a significant contribution to the total pressure only in very hot stars.
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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
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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
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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
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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
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Luca Sebben
Optically thin < 1 ≈1
Optically thick > 1
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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
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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.
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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
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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
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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
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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
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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
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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.
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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.
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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.
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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
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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
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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 =
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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.
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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
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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
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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
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V. Black body radiation: Planck´s Radiation Law
I = 1
(exp(h/kT) – 1)
2h3
c2
I = 1
(exp(hc/kT) – 1)
2hc2
5
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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→∞
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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
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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
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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
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IU B J K
log I ~ –4 log
log I ~ –5 log – 1/
40000 20000 10000
50003000 15001000
500750
T (K)
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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?
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T = 6000 KI
I
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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 !
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Steven Spangler, Univ. of Iowa
Temperature of the Sun is almost a black body
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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
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kT ≈ 262 ev
→ T ≈ 3 x 106 K