Transcript Chapter 13 – Behavior of Spectral Lines

Chapter 13 – Behavior of Spectral Lines
We began with line absorption coefficients which give
the shapes of spectral lines. Now we move into the
calculation of line strength from a stellar atmosphere.
• Formalism of radiative transfer in spectral
lines
– Transfer equation for lines
– The line source function
• Computing the line profile in LTE
• Depth of formation
• Temperature and pressure dependence of
line strength
• The curve of growth
The Line Transfer Equation
• We can add the continuous absorption coefficient and the line
absorption coefficient to get the total absorption coefficient:
dtn = (ln+kn)rdx
• And the source function as the sum of the line and continuous
emission coefficients divided by the sum of the line and
continuous emission coefficients.
j  jn
Sn  n
ln  k n
l
c
• Or define the line and continuum source functions separately:
– Sl=jnl/ln
– Sc=jnc/kn
Sn 
( ln / k n ) S l  S c
1 ln / k n
• In either case, we still have the basic transfer equation:
dIn
  In  Sn
dt n

In (0)   Sn etn sec secdtn
0
The Line Source Function
• The basic problem is still how to obtain the source function to
solve the transfer equation.
• But the line source function depends on the atomic level
populations, which themselves depend on the continuum
intensity and the continuum source function. This coupling
complicates the solution of the transfer equation for lines.
• Recall that in the case of LTE the continuum source function is
just the Bn(T), the Planck Function.
• The assumption of LTE simplifies the line case in the same way,
and allows us to describe the energy level populations strictly
by the temperature without coupling to the radiation field.
• This approximation works when the excitation states of the
gas are defined primarily by collisions and not radiative
excitation or de-excitation.
Mapping the Line Source Function
• The line source function with depth maps into the
line profile
• The center of the line is formed at shallower
optical depth, and maps to the source function at
smaller t
• The wings of the line are formed in progressively
deeper layers
Computing the Line Profile
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The line profile results from the solution of the transfer equation
at each Dl through the line.
The line profile will depend on the number of absorbers at each
depth in the atmosphere
The simplifying assumptions are
– LTE
– Pure absorption (no scattering)
How well does this work?
To know for sure we must compute the line profile in the general
case and compare it to what we get with simplifying assumptions
Generally, it’s pretty good
Start with the assumed T(t) relation and model atmosphere
Recompute the flux using the line+continuous opacity at each
wavelength around the line
For blended lines, just add the line absorption coefficients
appropriate at each wavelength
Depth of Formation
• It’s straightforward to determine
approximately where in the atmosphere (in
terms of the optical depth of the
continuum) each part of the line profile is
formed
• But even at a specific Dl, a range of optical
depths contributes to the absorption at
that wavelength
• It’s not straightforward to characterize
the depth of formation of an entire line
• The cores of strong lines are formed at
very shallow optical depths.
The Ca II K Line @ 3933A
• Why does the line have an emission
peak (a reversal) in the center?
• In the Sun, the central emission peak
is self-absorbed. Why?
The Behavior of Line Strength
• The strengths of spectral lines depend on
– The width of the absorption coefficient
– Thermal and microturbulent velocities
– The number of absorbers
• Temperature
• Electron pressure or luminosity
• Atomic constants
– In strong lines – collisional line broadening
affected by the gas and electron pressures
The Effect of Temperature
• Temperature is
the main factor
affecting line
strength
• Exponential and
power of T in
excitation and
ionization
• Increase with T
due to increase in
excitation
• Decrease beyond
maximum an
increase in the
opacity or from
ionization
H-g Profiles
• H lines are sensitive
to temperature
because of the Stark
effect
The high excitation of the Balmer
series (10.2 eV) means excitation
continues to increase to high
temperature (max at ~ 9000K).
Most metal lines have disappeared
by this temperature. Why?
How do different kinds of lines
behave with temperature?
– Lines from a neutral species of a mostly neutral
element
– Lines from a neutral species of a mostly ionized
element
– Lines from an ion of a mostly neutral element
– Lines from an ion of a mostly ionized element
• Consider gas with H- as the dominant
opacity
5

2
kn  T Pe e
0.75 / kT
Neutral lines from a neutral species
• Number of absorbers proportional to exp(-c/kT)
• Number of neutrals independent of temperature
(why?)
• Ratio of line to continuous absorption coefficient
T5 2 
R

e
kn
Pe
ln
( c  0.75)
kT
• But Pe is ~ proportional to exp(T/1000), so…
1 dR 2.5 c  0.75


 0.001
2
R dT
T
T
DEQW
c  0.75


 DT  2T.5 

0
.
001

2
EQW
T

