Atoms in stellar atmospheres are excited and ionized primarily
by collisions between atoms/ions/electrons (along with a small
contribution from the absorption of photons). The temperature
dependence of stellar absorption lines reflect how collisions
between atoms/ions/electrons (due to their thermal kinetic energy)
affect the excitation and ionization of atoms/ions. The
Classification of Stellar Spectra Slide 2 Excitation of Atoms in
Stellar Atmospheres Atoms in stellar atmospheres are excited
primarily by collisions between atoms/ions/electrons. To excite an
atom from a lower to higher excitation state, the colliding
atom/ion/electron must have a kinetic energy equal to or greater
than the corresponding excitation energy of the other atom. To
understand collisional excitation in stars, we first have to
understand the distribution in speeds (and therefore kinetic
energies) of atoms in stars. Slide 3 Learning Objectives Gas
Velocity Distribution in Stars Maxwell-Boltzmann velocity
distribution in 1-dimension Spectral line profile Maxwell-Boltzmann
velocity distribution in 3-dimensions Distribution of Electronic
Excited States Boltzmann equation Slide 4 Learning Objectives Gas
Velocity Distribution in Stars Maxwell-Boltzmann velocity
distribution in 1-dimension Spectral line profile Maxwell-Boltzmann
velocity distribution in 3-dimensions Distribution of Electronic
Excited States Boltzmann equation Slide 5 Maxwell-Boltzmann
Velocity Distribution Consider a gas comprising a huge number of
particles interacting through their mutual electromagnetic forces.
Computing the velocity of any one particle, and how the velocity of
this particle changes with time, is a hugely complex problem (i.e.,
if the gas comprises N particles, computing the velocity of any one
particle is an N-body problem; density of solar photosphere is ~10
16 cm -3 ). Slide 6 Maxwell-Boltzmann Velocity Distribution
Computing the distribution in speeds (velocity distribution) of the
ensemble of particles, however, is a much simpler problem to solve.
Study of the behavior of systems comprising a large number of
particles is a branch of physics known as statistical mechanics.
This problem was first solved by James Clerk Maxwell (arguing, by
symmetry, that the speed distribution is the same in all
directions) and extended/generalized by Ludwig Eduard Boltzmann.
The distribution in speeds of an ensemble of particles in thermal
equilibrium (not changing in temperature, or changing in
temperature much more slowly that the interval over which the
measurement is made) is now known as the Maxwell-Boltzmann velocity
distribution. Although the speed of each particle is different and
changes with time, the overall distribution in particle speeds
remain the same over time. Slide 7 Maxwell-Boltzmann Velocity
Distribution For a gas in thermal equilibrium, the velocity
component of the particles in a single direction of space (e.g., v
x ) has a Gaussian distribution. x f(x) Slide 8 Maxwell-Boltzmann
Velocity Distribution For a gas in thermal equilibrium, the
velocity component of the particles in a single direction of space
(e.g., v x ) has a Gaussian distribution. Note that: -the average
speed in a given direction is 0 -the wings of a Gaussian function
extends to x f(x) Slide 9 Maxwell-Boltzmann Velocity Distribution
For a gas in thermal equilibrium, the velocity component of the
particles in a single direction of space (e.g., v x ) has a
Gaussian distribution. Compare with equation for thermal (Doppler)
profile of spectral line: Slide 10 Spectral Line Profiles Three
components contribute to the profile of stellar absorption lines:
-natural broadening due to Heisenbergs uncertainty principle
(Lorentz profile) -Doppler broadening due to the random motion of
hot gas (Gaussian profile) -pressure broadening due to the
perturbation of atomic orbitals via collisions with neutral atoms
or the electric fields of ions (Lorentz profile) Lorentz Gaussian
Slide 11 Spectral Line Profiles The combination (convolution) of a
Lorentzian and Gaussian profile is known as a Voigt profile. The
profile of spectral lines is therefore resembles a Gaussian profile
in its core and a Lorentzian profile in its wings. In virtually all
astrophysical situations, Doppler broadening dominates the widths
of spectral lines in their cores. Slide 12 Spectral Line Profiles
The width of spectral lines therefore depends on (at least) four
factors: -the atom/ion involved -gas temperature -number of
atoms/ions emitting/absorbing in that line -gas pressure Slide 13
Maxwell-Boltzmann Velocity Distribution The velocity of a particle
is the vector addition of all three orthogonal velocity components
in space, so that v = (v x 2 + v y 2 + v z 2 ) 1/2. For a gas in
thermal equilibrium, the number of gas particles having velocities
between v and v + dv is given by the Maxwell-Boltzmann velocity
distribution function Notice that the exponent of the distribution
function is the ratio of the particles kinetic energy, mv 2, to its
characteristic thermal energy, kT. The velocity distribution
depends on the particles mass and the gas temperature. Note that
there is no dependence on the gas density. total number density of
particles particle mass gas temperature Slide 14 Maxwell-Boltzmann
Velocity Distribution The Maxwell-Boltzmann velocity distribution
function Slide 15 Maxwell-Boltzmann Velocity Distribution The
Maxwell-Boltzmann velocity distribution function Slide 16
Maxwell-Boltzmann Velocity Distribution The Maxwell-Boltzmann
velocity distribution function Slide 17 Maxwell-Boltzmann Velocity
Distribution The Maxwell-Boltzmann velocity distribution function
The Maxwell-Boltzmann velocity distribution function peaks whether
the particle kinetic energy is equal to its characteristic thermal
energy, at a most probable speed of Much fewer particles have
kinetic energies much less or much greater than the characteristic
thermal energy. Slide 18 Maxwell-Boltzmann Velocity Distribution
The Maxwell-Boltzmann velocity distribution function Because of the
exponential tail in the distribution, the average speed is higher
than the most probable speed The root-mean-square (rms) speed Slide
19 Gas Velocity Distribution in Stars In the previous discussion,
we have ignored the possibility that collisions between particles
also can: -excite atoms -ionize atoms Understanding how atoms are
excited and ionized in stellar atmospheres is the key to
understanding stellar absorption lines. In the previous discussion,
we also have ignored fact that gas loses energy through radiation.
All three abovementioned processes remove thermal energy from the
gas. In stellar atmospheres, the thermal energy transferred to
excitation/ionization and lost to radiation is replaced by
(collisional de-excitation and) thermal energy from the stellar
interior (generated ultimately by nuclear fusion at the center of
the star), so that stellar atmospheres are in thermal equilibrium.
The velocity distribution of gas particles in stars therefore obey
the Maxwell-Boltzmann velocity distribution. Slide 20 Collisional
Excitation of Atoms Consider two hydrogen atoms in their ground
state; i.e., with their individual electrons in the n = 1 quantum
state. In a collision between these two atoms, a part of their
kinetic energy can be transferred into exciting one or both atoms;
i.e., cause their individual electrons to make a transition from
the n = 1 to the n = 2 level or higher. Slide 21 Collisional
Excitation of Atoms Consider two hydrogen atoms in their ground
state; i.e., with their individual electrons in the n = 1 quantum
state. In a collision between these two atoms, a part of their
kinetic energy can be transferred into exciting one or both atoms;
i.e., cause their individual electrons to make a transition from
the n = 1 to the n = 2 level or higher. Consider the solar
photosphere, which is at a temperature of 5778 K: -v mp = 9779.1
m/s, corresponding to a kinetic energy of 0.50 eV - v = 11026.4
m/s, corresponding to a kinetic energy of 0.64 eV -v rms = 11950.4
m/s, corresponding to an average kinetic energy of 0.75 eV An
energy of at least 10.2 eV (3.7 v rms = 44070.9 m/s) is required to
excite a hydrogen atom from the ground state, much higher than the
characteristic energy of hydrogen atoms in the solar photosphere.
(For v rms = 10.2 eV, T = 78581 K.) Yet, the presence of Balmer
absorption lines in the solar photosphere indicate that at least
some hydrogen atoms are in the first excited state. So, how is it
that at least some hydrogen atoms are excited to n = 2 by
collisions in the solar photosphere? Slide 22 Collisional
Excitation of Atoms Consider two hydrogen atoms in their ground
state; i.e., with their individual electrons in the n = 1 quantum
state. In a collision between these two atoms, a part of their
kinetic energy can be transferred into exciting one or both atoms;
i.e., cause their individual electrons to make a transition from
the n = 1 to the n = 2 level or higher. Consider the solar
photosphere, which is at a temperature of 5778 K: -v mp = 9779.1
m/s, corresponding to a kinetic energy of 0.50 eV - v = 11026.4
m/s, corresponding to a kinetic energy of 0.64 eV -v rms = 11950.4
m/s, corresponding to an average kinetic energy of 0.75 eV An
energy of at least 10.2 eV (3.7 v rms = 44070.9 m/s) is required to
excite a hydrogen atom from the ground state, much higher than the
characteristic energy of hydrogen atoms in the solar photosphere.
(For v rms = 10.2 eV, T = 78581 K.) Yet, the presence of Balmer
absorption lines in the solar photosphere indicate that at least
some hydrogen atoms are in the first excited state. So, how is it
that at least some hydrogen atoms are excited to n = 2 by
collisions in the solar photosphere? By collisions between hydrogen
atoms at the tail of the Maxwell- Boltzmann velocity distribution.
Slide 23 Collisional Excitation of Atoms Velocity distribution of
hydrogen atoms in the solar photosphere. Slide 24 Collisional
Excitation of Atoms Velocity distribution of hydrogen atoms in the
solar photosphere. Slide 25 Collisional Excitation of Atoms
Velocity distribution of hydrogen atoms in the solar photosphere.
Slide 26 Learning Objectives Gas Velocity Distribution in Stars
Maxwell-Boltzmann velocity distribution in 1-dimension Spectral
line profile Maxwell-Boltzmann velocity distribution in
3-dimensions Distribution of Electronic Excited States Boltzmann
equation Slide 27 Distribution of Excited States Atoms of a gas can
gain energy during a collision in the form of -kinetic energy
-excitation of an electron to a higher energy level Atoms of a gas
also can lose energy during a collision in the form of -kinetic
energy -de-excitation of an electron to a lower energy level
(energy transferred as kinetic energy to another atom) (Atoms in
stars also can gain/lose energy due to absorption/emission of
photons.) Slide 28 Distribution of Excited States Consider hydrogen
atoms subjected to excitation (and de-excitation) by mutual
collisions in thermalized gas. In this situation, would you expect
fewer or more atoms excited to higher energies? What would you
expect the distribution of atoms in different excited states to
depend upon? Slide 29 Distribution of Excited States Consider
hydrogen atoms subjected to excitation (and de-excitation) by
mutual collisions in thermalized gas. In this situation, would you
expect fewer or more atoms excited to higher energies? Fewer, as
there are fewer particles with increasing kinetic energies. What
would you expect the distribution of atoms in different excited
states to depend upon? Slide 30 Distribution of Excited States
Consider hydrogen atoms subjected to excitation (and de-excitation)
by mutual collisions in thermalized gas. In this situation, would
you expect fewer or more atoms excited to higher energies? Fewer,
as there are fewer particles with increasing kinetic energies. What
would you expect the distribution of atoms in different excited
states to depend upon? Distribution of kinetic energies as given by
the Maxwell- Boltzmann velocity distribution for a given particle
(hydrogen in stellar atmospheres). In a collection of atoms excited
and de-excited by collisions, the distribution of excited states
depends on the distribution in kinetic energies of the atoms. In
stellar atmospheres, the distribution in speeds of the impacting
atoms given by the Maxwell-Boltzmann velocity distribution produces
a definite distribution of excited states. The distribution of
excited states is governed by a fundamental result of statistical
mechanics: orbitals of a higher energy are less likely to be
occupied by electrons. Slide 31 Distribution of Excited States Let
s a stand for the specific set of quantum numbers that identifies a
state of energy E a for a system of particles. Similarly, let s b
stand for the set of quantum numbers that identifies a state of
energy E b. For example, s a = {n = 1, l = 0, m l = 0, m s =
+1/2}is the set of quantum numbers that identifies a ground state
of the hydrogen atom with energy of -13.6 eV. s b = {n = 2, l = 0,
m l = 0, m s = +1/2} is the set of quantum numbers that identifies
a first excited state of the hydrogen atom with energy of -3.4 eV.
sa sa sb sb Slide 32 Distribution of Excited States Let s a stand
for the specific set of quantum numbers that identifies a state of
energy E a for a system of particles. Similarly, let s b stand for
the set of quantum numbers that identifies a state of energy E b.
Let P(s a ) stand for the probability that the system in the state
s a. Let P(s b ) stand for the probability that the system in the
state s b. For example, let P(s a ) stand for the probability of
finding hydrogen atoms in the ground state s a, and let P(s b )
stand for the probability of finding hydrogen atoms in the first
excited state s b. The ratio of the probability P(s b ) that the
system is in the state s b to the probability that the system is in
the state s a is given by the Boltzmann equation The term e -E/kT
is called the Boltzmann factor. Slide 33 Distribution of Excited
States The ratio of the probability P(s b ) that the system is in
the state s b to the probability that the system is in the state s
a is given by the Boltzmann equation Imagine that we cool down
hydrogen gas so that T 0 K. At this low temperature, what is the
ratio of the probability P(s b ) of finding hydrogen atoms in the
first excited (s b ) state to the probability P(s a ) of finding
hydrogen atoms in the ground (s a ) state? Slide 34 Distribution of
Excited States The ratio of the probability P(s b ) that the system
is in the state s b to the probability that the system is in the
state s a is given by the Boltzmann equation Imagine that we cool
down hydrogen gas so that T 0 K. At this low temperature, what is
the ratio of the probability P(s b ) of finding hydrogen atoms in
the first excited (s b ) state to the probability P(s a ) of
finding hydrogen atoms in the ground (s a ) state? As T 0 K, -(E b
-E a )/kT -; e -(E b -E a )/kT e - = 0 and so P(s b )/P(s a ) 0. As
the temperature approaches 0 K, there are progressively fewer
hydrogen atoms with sufficient kinetic energies to be able to
excite other hydrogen atoms to the first (let alone higher) excited
state, so that P(s b ) 0. Slide 35 Distribution of Excited States
The ratio of the probability P(s b ) that the system is in the
state s b to the probability that the system is in the state s a is
given by the Boltzmann equation Imagine that we heat hydrogen gas
so that T K. At this high temperature, what is the ratio of the
probability P(s b ) of finding hydrogen atoms in the first excited
(s b ) state to the probability P(s a ) of finding hydrogen atoms
in the ground (s a ) state? Slide 36 Distribution of Excited States
The ratio of the probability P(s b ) that the system is in the
state s b to the probability that the system is in the state s a is
given by the Boltzmann equation Imagine that we heat hydrogen gas
so that T K. At this high temperature, what is the ratio of the
probability P(s b ) of finding hydrogen atoms in the first excited
(s b ) state to the probability P(s a ) of finding hydrogen atoms
in the ground (s a ) state? As T K, -(E b -E a )/kT 0; e -(E b -E a
)/kT e -0 = 1 and so P(s b )/P(s a ) 1. As the temperature
increases from almost 0 K, there are progressively more hydrogen
atoms with sufficient kinetic energies to be able to excite other
hydrogen atoms to the first (as well as higher) excited states, and
so P(s b )/P(s a ) > 0. If there is an unlimited reservoir of
thermal energy available (T ), all energy levels of the atom are
accessible with equal probability. Slide 37 Distribution of Excited
States Suppose that there are g a number of states with energy E a,
and g b number of states with energy E b. (The energy levels E a
and E b are both said to be degenerate.) g a and g b are called the
statistical weights of the energy levels. sa sa sb sb For example,
g a = 2 for the ground state of the hydrogen atom with energy -
13.6 eV, and g b = 8 for the first excited state of the hydrogen
atom with energy - 3.4 eV. Slide 38 Distribution of Excited States
Suppose that there are g a number of states with energy E a, and g
b number of states with energy E b. (The energy levels E a and E b
are both said to be degenerate.) g a and g b are called the
statistical weights of the energy levels. The ratio of the
probability P(E b ) that the system will be found in any of the g b
degenerate states with energy E b to the probability P(E a ) that
the system is in any of the g a degenerate states with energy E a
is given by The ratio of the number of atoms N b with energy E b to
the number of atoms N a with energy E a in different states of
excitation is given by the Boltzmann equation for a given element
in a specified state of ionization (including neutral atoms). Slide
39 Distribution of Excited States Slide 40 For hydrogen gas in the
solar photosphere with T eff = 5778 K -N 2 /N 1 = 5.0 10 -9 For
hydrogen gas in the atmosphere of an A0 star with T eff = 7500 K -N
2 /N 1 = 5.6 10 -7 For hydrogen gas in the atmosphere of a B star
with T eff = 25,000 K -N 2 /N 1 = 0.035 For hydrogen gas in the
atmosphere of an O star with T eff = 50,000 K -N 2 /N 1 = 0.37 In
stellar atmospheres, most of the hydrogen atoms are in the ground
state! Slide 41 Distribution of Excited States Relative number of
hydrogen atoms in the first exited state to the total number of
hydrogen atoms in the ground and first excited states. Slide 42
Distribution of Excited States Because the fraction of hydrogen
atoms in the first excited state to ground state increases with
increasing stellar effective temperatures, the strength of Balmer
absorption lines in stellar spectra should also increase with
stellar effective temperatures. H H HH Slide 43 Distribution of
Excited States Why then do the strength of Balmer absorption lines
in stellar spectra reach a maximum at spectral type A0? H H H H
Slide 44 Distribution of Excited States H H H H Why then do the
strength of Balmer absorption lines in stellar spectra reach a
maximum at spectral type A0? An increasing fraction of hydrogen
atoms are ionized, leaving fewer available for line absorption.
