Report

The Classification of Stellar Spectra 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. 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. 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 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 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 ~1016 cm-3). 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. Maxwell-Boltzmann Velocity Distribution For a gas in thermal equilibrium, the velocity component of the particles in a single direction of space (e.g., vx) has a Gaussian distribution. f(x) x Maxwell-Boltzmann Velocity Distribution For a gas in thermal equilibrium, the velocity component of the particles in a single direction of space (e.g., vx) has a Gaussian distribution. f(x) Note that: - the average speed in a given direction is 0 - the wings of a Gaussian function extends to ±∞ x Maxwell-Boltzmann Velocity Distribution For a gas in thermal equilibrium, the velocity component of the particles in a single direction of space (e.g., vx) has a Gaussian distribution. Compare with equation for thermal (Doppler) profile of spectral line: Spectral Line Profiles Three components contribute to the profile of stellar absorption lines: - natural broadening due to Heisenberg’s 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 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. 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 Maxwell-Boltzmann Velocity Distribution The velocity of a particle is the vector addition of all three orthogonal velocity components in space, so that v = (vx2 + vy2 + vz2)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 particle mass total number density of particles gas temperature Notice that the exponent of the distribution function is the ratio of the particle’s kinetic energy, ½ mv2, to its characteristic thermal energy, kT. The velocity distribution depends on the particle’s mass and the gas temperature. Note that there is no dependence on the gas density. Maxwell-Boltzmann Velocity Distribution The Maxwell-Boltzmann velocity distribution function Maxwell-Boltzmann Velocity Distribution The Maxwell-Boltzmann velocity distribution function Maxwell-Boltzmann Velocity Distribution The Maxwell-Boltzmann velocity distribution function 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. 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 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. 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. 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: - vmp = 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 - vrms = 11950.4 m/s, corresponding to an average kinetic energy of 0.75 eV An energy of at least 10.2 eV (3.7 × vrms = 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 vrms = 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? 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: - vmp = 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 - vrms = 11950.4 m/s, corresponding to an average kinetic energy of 0.75 eV An energy of at least 10.2 eV (3.7 × vrms = 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 vrms = 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 MaxwellBoltzmann velocity distribution. Collisional Excitation of Atoms Velocity distribution of hydrogen atoms in the solar photosphere. Collisional Excitation of Atoms Velocity distribution of hydrogen atoms in the solar photosphere. Collisional Excitation of Atoms Velocity distribution of hydrogen atoms in the solar photosphere. 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 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.) 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? 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 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 MaxwellBoltzmann 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. Distribution of Excited States Let sa stand for the specific set of quantum numbers that identifies a state of energy Ea for a system of particles. Similarly, let sb stand for the set of quantum numbers that identifies a state of energy Eb. For example, sa = {n = 1, l = 0, ml = 0, ms = +1/2}is the set of quantum numbers that identifies a ground state of the sa ➙ hydrogen atom with energy of -13.6 eV. sb = {n = 2, l = 0, ml = 0, ms = +1/2} is the set of quantum numbers that identifies a first excited state of the hydrogen atom with energy of -3.4 eV. s ➙ b Distribution of Excited States Let sa stand for the specific set of quantum numbers that identifies a state of energy Ea for a system of particles. Similarly, let sb stand for the set of quantum numbers that identifies a state of energy Eb. Let P(sa) stand for the probability that the system in the state sa. Let P(sb) stand for the probability that the system in the state sb. For example, let P(sa) stand for the probability of finding hydrogen atoms in the ground state sa, and let P(sb) stand for the probability of finding hydrogen atoms in the first excited state sb. The ratio of the probability P(sb) that the system is in the state sb to the probability that the system is in the state sa is given by the Boltzmann equation The term e-E/kT is called the Boltzmann factor. Distribution of Excited States The ratio of the probability P(sb) that the system is in the state sb to the probability that the system is in the state sa 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(sb) of finding hydrogen atoms in the first excited (sb) state to the probability P(sa) of finding hydrogen atoms in the ground (sa) state? Distribution of Excited States The ratio of the probability P(sb) that the system is in the state sb to the probability that the system is in the state sa 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(sb) of finding hydrogen atoms in the first excited (sb) state to the probability P(sa) of finding hydrogen atoms in the ground (sa) state? As T → 0 K, -(Eb-Ea)/kT → -∞; e-(Eb-Ea)/kT → e-∞ = 0 and so P(sb)/P(sa) → 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(sb) → 0. Distribution of Excited States The ratio of the probability P(sb) that the system is in the state sb to the probability that the system is in the state sa 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(sb) of finding hydrogen atoms in the first excited (sb) state to the probability P(sa) of finding hydrogen atoms in the ground (sa) state? Distribution of Excited States The ratio of the probability P(sb) that the system is in the state sb to the probability that the system is in the state sa 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(sb) of finding hydrogen atoms in the first excited (sb) state to the probability P(sa) of finding hydrogen atoms in the ground (sa) state? As T → ∞ K, -(Eb-Ea)/kT → 0; e-(Eb-Ea)/kT → e-0 = 1 and so P(sb)/P(sa) → 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(sb)/P(sa) > 0. If there is an unlimited reservoir of thermal energy available → ∞), all energy levels of the atom are accessible with equal probability. (T Distribution of Excited States Suppose that there are ga number of states with energy Ea, and gb number of states with energy Eb. (The energy levels Ea and Eb are both said to be degenerate.) ga and gb are called the statistical weights of the energy levels. For example, ga = 2 for the ground state of the hydrogen atom with energy 13.6 eV, and gb = 8 for the first excited sa ➙ state of the hydrogen atom with energy 3.4 eV. sb ➙ Distribution of Excited States Suppose that there are ga number of states with energy Ea, and gb number of states with energy Eb. (The energy levels Ea and Eb are both said to be degenerate.) ga and gb are called the statistical weights of the energy levels. The ratio of the probability P(Eb) that the system will be found in any of the gb degenerate states with energy Eb to the probability P(Ea) that the system is in any of the ga degenerate states with energy Ea is given by The ratio of the number of atoms Nb with energy Eb to the number of atoms Na with energy Ea in different states of excitation is given by the Boltzmann equation for a given element in a specified state of ionization (including neutral atoms). Distribution of Excited States Distribution of Excited States For hydrogen gas in the solar photosphere with Teff = 5778 K - N2/N1 = 5.0 × 10-9 For hydrogen gas in the atmosphere of an A0 star with Teff = 7500 K - N2/N1 = 5.6 × 10-7 For hydrogen gas in the atmosphere of a B star with Teff = 25,000 K - N2/N1 = 0.035 For hydrogen gas in the atmosphere of an O star with Teff = 50,000 K - N2/N1 = 0.37 In stellar atmospheres, most of the hydrogen atoms are in the ground state! 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. 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 Hδ Hγ Hβ increase with stellar effective temperatures. Hα Distribution of Excited States Why then do the strength of Balmer absorption lines in stellar spectra reach a maximum at spectral type A0? Hδ Hγ Hβ Hα Distribution of Excited States Why then do the strength of Balmer absorption lines in stellar spectra reach a maximum at spectral type A0? An increasing Hδ fraction of hydorgen atoms are ionized, Hγ leaving fewer available for line absorption. Hβ Hα