Transcript Spectroscopy – Lecture 1

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
E>0
∞ E=0
··
··
I
c
qualitative energy level
diagram
n
3
2
1
E=–I
Mechanisms for populating
and depopulating the levels
in stellar atmospheres:
• radiative
• collisional
• spontaneous transitions
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(–cn/kT)
statistical weight
Boltzmann factor
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
Dc
exp –
gm
kT
(
Dc = cn – cm
)
I. Atomic excitation and ionization
The number of atoms in level n as fraction of all atoms of
the same species:
cn
gn exp –
Nn
kT
=
c2
c3
N
g1 + g2 exp –
+ g3 exp –
+ ...
kT
kT
gn
cn
exp –
=
u(T)
kT
( )
( )
( )
c
u(T) = S gi exp – i
kT
(
Nn
N
gn
10 –qcn
=
u(T)
)
(
)
Partition Function
q= log e/kT = 5040/T
From Allen‘s
Astrophysical
Quantities
Q = 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:
log
Nr
N1
=
gr
–5040
c + log g
T
1
I. Atomic excitation and ionization
Example: Compare relative populations between
ground state and n=2 for Hydrogen
g1 = 2, g2=2n2 =8
Temp. (K)
6000
8000
10000
15000
20000
40000
q=5040/T
0.840
0.630
0.504
0.336
0.252
0.126
N2/N1
0.00000001
0.0000016
0.00031
0.00155
0.01100
0.209
I. Atomic excitation and ionization
N2/N1
10000
20000
40000
60000
I. Atomic excitation and ionization : Saha Eq.
For collisionally dominated gas:
3
2
N1
N
N1
N
u1
u0
Pe =
=
=
( 2pm) ( kT )
h3
5
2
2u1(T)
I
exp –
kT
u0(T)
(
)
Ratio of ions to neutrals
Ratio of ionic to neutral partition function
m = mass of electron, h = Planck´s constant,
Pe = electron pressure
I. Saha Equation
Numerically:
log
or
N1
N
Pe =
–5040
T
N1
N
=
u1
I + 2.5 log T + log
– 0.1762
u0
F(T)
Pe
F(T) = 0.65
u1
u0
T
5
2
10–5040I/kT
I. Saha Equation
Example: What fraction of calcium atoms are singly
ionized in Sirius?
Stellar Parameters:
T = 10000 K
Pe = 300 dynes cm–2
Atomic Parameters:
Ca I = 6.11 ev
log 2u1/uo = 0.18
log N1/N0 = 4.14
no neutral Ca
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 = 0.82
N2/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.
From Lawrence Allen‘s The Atmospheres of the Sun and Stars
T
25000
10000
6300
4200
The number of hydrogen atoms in the second level
capable of producing Balmer lines reaches its
maximum at Teff ≈ 10000 K
Behavior of the Balmer lines (Hb)
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
Ionization theory‘s
achievement was the
intepretation of the
spectral sequence as a
temperature sequence
Observed
behavior
according to
Ionization
Theory
II. Radiation Terms: Specific intensity
Consider a radiating surface:
Normal
q
Observer
DA
DEn
In = lim
cos q DA Dw Dt Dn
In =
dEn
cos q dA dw dt dn
Dw
II. Specific intensity
Can also use wavelength interval:
Indn = Ildl
Note: the two spectral distributions (n,l) have different
shape for the same spectrum
For solar spectrum:
Il = max at 4500 Ang
In = max at 8000 Ang
c=ln dn = –(c/l2) dl
Equal intervals in l correspond to different intervals in n. With
increasing l, a constant dl corresponds to a smaller and smaller dn
II. Radiation Terms: Mean intensity
I. The mean intensity is the directional average of
the specific intensity:
Jn = 1
4p
∫
In dw
Circle indicates the integration is done over whole unit
sphere on the point of interest
II. Radiation Terms: Flux
Flux is a measure of the net energy across an area
DA, in time Dt and in spectral range Dn
Flux has directional information:
+Fn
-Fn
DA
II. Radiation Terms: Flux
S DEn
Fn = lim
=
Recall:
DA Dt Dn
∫ dEn
=
dA dt dn
∫
In cos q dw
dEn
In =
cos q dA dw dt dn
Fn =
∫
In cos q dw
II. Radiation Terms: Flux
Looking at a point on the boundary of a radiating
sphere
p
2p
Fn
=
∫ df∫ I cos q sin q dq
n
0
2p
=
p/2
∫ df∫ I cos q sin q dq
df
∫ ∫ I cos q sin q dq
n
0
0
Outgoing flux
0
2p
+
0
p
p/2
=0
n
For stars flux is positive
Incoming flux
II. Radiation
Astronomical Example of Negative Flux: Close
Binary system:
Cool star (K0IV)
Hot Spot
Hot star (DAQ3)
II. Radiation Terms: Flux
If there is no azimuthal (f) dependence
p/2
Fn
=
2p
∫ I sin q cos q dq
n
0
Simple case: if In is independent of direction:
Fn = p In
(∫ sin q cos q dq = 1/2 )
Note: In is independent of distance, but Fn obeys the
standard inverse square law
Flux radiating through
a sphere of radius d is
just F = L/4pd2
L = 4pR2In (=sT4)
d
Energy received ~ InDA1/r2
Detector element
Source
F
r
DA1
Source image
Energy received ~ InDA2/4r2
but DA2 =4DA1
= In DA1/r2
F
2r
DA2
Energy received ~ InDA3/100r2
but DA3 = area of source
Source image
F
DA3
10r
Detector element
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
II. K-integral and radiation pressure
∫
Kn = 1
4p
2p
∫ dw = ∫ ∫
0
In cos2q dw
p
sin q dq df = 2p
0
∫
+1
dm
–1
m = cos q
Kn = 1
2
∫
+1
–1
m2 dm
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=
2
c
d En cosq
dt dA
q
component of momentum normal to wall per unit area per time =
pressure
II. K-integral and radiation pressure
2In 2
Pn dn dw = c cos q dn dw
Pndn =
Pn = 4p
c
∫
∫
+1
0
In (m)
In cos2q dn dw/c
m2
dm = 2p
c
Pn = 4p Kn
c
∫
+1
-1
In (m) m2 dm
II. K-integral and radiation pressure
Special Case: In is indepedent of direction
Pn = 2p
c
∫
+1
In (m) m2 dm
-1
Pn = 4p In
3c
Total radiation pressure:
Pn
4p
=
3c
∫
0

In dn
= 4s T4
3c
For Blackbody
radiation
Radiation pressure is a significant contribution to the total pressure only
in very hot stars.
II. Moments of radiation
∫
Jn = 1
4p
Jn = 1
2
Hn = 1
2
Kn = 1
2
∫
+1
–1
∫
In dw
In (m) dm
+1
Mean intensity
I(m) m dm
Flux = 4pH
I(m) m2 dm
Radiation pressure
–1
∫
+1
–1
m = cos q
III. The absorption coefficient
In + dIn
In
dx
kn
kn is the absorption coefficient/unit mass [ ] = cm2/gm.
kn comes from true absorption (photon destroyed) or
from scattering (removed from solid angle)
dIn = –knr In dx
III. Optical depth
L
In + dIn
In
kn
The radiation sees neither knr or dx, but a the combination
of the two over some path length L.
tn =
L
∫ k r dx
n
Optical depth
o
Units: cm2 gm cm
gm cm3
III. Optical depth
Optically thick case:
t >> 1 => a photon does not travel far
before it gets absorbed
Optically thin case:
t << 1 => a photon can travel a long
distance before it gets absorbed
Optically thin t < 1
t ≈1
Optically thick t > 1
Luca Sebben
III. Simple solution to radiative transfer equation
In + dIn
In
dIn = –knr In dx
dx
kn
dIn = – In dt
In = Ino e–t
Optically thin e–t = 1-t
In = Ino(1-t)
III. The emission coefficient
In + dIn
In
dx
jn
dIn = jnr In dx
jn is the emission coefficient/unit mass
[ ] = erg/(s rad2 Hz gm)
jn 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 In. This is commonly referred to as the
source function:
Sn = jn/kn
The physics of calculating the source function Sn can be
complicated. Let´s consider the simple cases of
scattering and absorption
III. The Source Function: Pure isotropic scattering
dw
djn to
observer
•
isotropic scattering
The scattered radiation to the observer is the sum of all
contributions from all increments of the solid angle like
dw. Radiation 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 dw is
proportional to dw and the absorbed energy knIn. This is
isotropically re-radiated:
djn = knIn dw/4p
jn =
∫
jn
Sn = k
n
knIn dw/4p
=
∫I
n
dw/4p
= Jn
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:
Sn =
1
2hn3
c2
exp(hn/kT) – 1
III. The Source Function: Scattering + Pure absorption
jn = knSIn +knABn(T)
Sn = jn/kn where kn = knS+ knA
knS
knA
Sn =
Jn +
Bn (T)
S
A
S
A
kn + kn
kn + kn
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 hn.
The probability that an atom will emit its quantum
energy in a time dt, solid angle dw 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:
jnr = Nu Aul hn
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 dw is Bul In dt dw.
Bul is the Einstein probability coefficient for stimulated
emission.
True absorption is defined in the same way and the
proportionality constant denoted Blu.
knrIn = NlBluInhn – NuBulInhn
The amount of reduction in absorption due to the second
term is only a few percent in the visible spectrum.
IV. Einstein Coefficients
Nu
True
absorption,
dependent
on In
BluIn
Aul +BulIn
Nl
Spontaneous emission,
independent of In
Negative absorption,
dependent on In
Principle of detailed balance:
Nu[Alu + BulIn] = NlBulIn
hn
V. Black body radiation
Detector
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.
V. Black body radiation: observed quantities
Il =
c4
l5
F(c/lT)
In = n3 F(n/T)
2
2kT
n
I n=
c2
2pckT
I l= l4
F is a function that
is tabulated by
measurements. This
scaling relation was
discovered by Wien
in 1893
Rayleigh-Jeans
approximation for
low frequencies
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
Radiation energy density = number of degrees of
freedom×energy per degree of freedom per unit volume.
2kTn2
I n = c2
but as n
∞, In

This is the „ultraviolet“ catastrophe of classical physics
V. Black body radiation: Planck´s Radiation Law
Derive using a two level atom:
Nn
Nm
=
gn
hn
exp –
gm
kT
(
)
Number of spontaneous emissions: NuAul
Rate of stimulated emission: NuBulIn
Absorption: NlBulIn
V. Black body radiation: Planck´s Radiation Law
In radiative equilibrium collisionally induced transitions
cancel (as many up as down)
NuAul + NuBulIn = NlBluIn
In =
In =
Aul
Blu(Nl/Nu) – Bul
Aul
(gl/gu)Bluexp(hn/kT) – Bul
V. Black body radiation: Planck´s Radiation Law
This must revert to Raleigh-Jeans relation for small n
Expand the exponential for small values (ex = 1+x)
In ≈
Aul
(gl/gu)Blu – Bul + (gl/gu)Bluhn/kT
hn/kT << 1 this can only equal 2kTn2/c2 if
Bul = Blugl/gu
Aul =
2hn3
c2
Bul
Note: if you know one Einstein coeffiecient you know them all
V. Black body radiation: Planck´s Radiation Law
In =
Il =
2hn3
1
c2 (exp(hn/kT) – 1)
2hc2
1
l5 (exp(hc/lkT) – 1)
V. Black body radiation: Planck´s Radiation Law
Maximum Il = lT = 0.5099 cm K
Maximum In = 5.8789×1010 Hz K
In =
I n=
I l=
2kTn 2
c2
I l=
2phn 3 e–hn/kT
c2
2phc2
l5
e–hc/klT
2pckT
l4
Rayleigh-Jeans
approximation n→ 0
Wien approximation
n→∞
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 Fn = pIn
∫
∞
Fn dn = p
0
∫
∞
1
2hn3
c2 (exp(hn/kT) – 1)
4
∞ x3
2p kT
dx
= 2( )
x
e –1
c
h
0
0
∫
Integral = p4/15
dn
2p5k4 4
T
Fn dn =
3c2
15h
0
∫
∞
x=hn/kT
= sT4
V. Note on Einstein Coefficients and BB radiation
In the spectral region where hn/kT >> 1
spontaneous emissions are more
important than induced emissions
In the ultraviolet region of the spectrum
replace In by Wien´s law:
2hn3 –hn/kT
= Aul e–hn/kT << Aul
BulIn = Bul 2 e
c
Induced emissions can be neglected in
comparison to spontaneous emissions
V. Note on Einstein Coefficients and BB radiation
In the spectral region where hn/kT << 1 negative
absorption (induced emissions) are more
important than spontaneous emissions
In the far infrared region of the spectrum
replace In by Rayleigh-Jeans law:
BulIn = Bul
2
2nkT
c
=
2
c
2hn3
Aul 2nkT = Aul kT
c2
hn
>>Aul
The number of negative absorptions is
greater than the spontaneous emissions
log I ~ –4 log l
U B I
J
K
T (K)
40000
20000
10000
5000
3000
1500
1000
750
500
log I ~ –5 log l – 1/l
How can a T=40000 star ionize Hydrogen?
1. Blackbody has a distribution of energies and some
photons have the energy to ionize hydrogen
Fraction of total particles
2. The thermal velocities have a Maxwell Boltzmann
distribution and some particles have the thermal energy to
ionize Hydrogen.
T=1000 K
T=6000 K
T=40000 K
Il
T = 6000 K
In
V. Black body radiation: Photon Distribution Law
Nn =
Nl =
2hn2
1
c2 (exp(hn/kT) – 1)
2hc2
1
l4 (exp(hc/lkT) – 1)
Detectors detect N, not I !
Temperature of the Sun is almost a black body
Steven Spangler, Univ. of Iowa
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