“The difference between science and magic is that magicians usually know what they're doing.” Ashleigh Brilliant HW2 is on-line now and due next Thursday Reading: Chapter 1

Last time: Parts of a stellar atmosphere: 1) Photosphere

Roughly a few thousand km thick. 2) Chromosphere. Roughly a few thousand km thick.

3) The corona.

Temperature and density of the Sun's atmosphere (photosphere, chromosphere and corona).

Activity increases with later spectral types and decreases with age (as does rotation)

Light Energy of a photon: E=hc/ Making or destroying photons: 1) bound-bound (atomic) transition 2) bound-free (ionization) 3) Free-free (Bremsstrahlung) absorption 4) Thomson (electron) scattering

Bound-Bound: Atomic Emission/Absorption This is when a bound electron changes orbitals.

DE=hc/l

hc l

Transitions for hydrogen for the Bohr model

IR Optical UV

Atomic Series n=1: Lyman n=2: Balmer n=3: Paschen n=4: Bracken Within each series, the lowest energy transition is labeled , the second lowest , and so on. So L is the n=2 to n=1 (emission) transition.

Atomic Series Because the Balmer series for hydrogen occurs in the optical, it is often written as H, H, etc. with it being understood that the H means the Balmer series for hydrogen.

Hd Hg Hb Ha

Small side note: degeneracy Only a certain number of electrons can fit into each orbital. Once that is full, they must go into the next higher excited state. This degeneracy is 2n2, so in the ground state (n=1), 2 electrons can fit, in the first excited state (n=2), 8 electrons can fit, and so on.

Bound-free (photoionization): Ionization energy An atom can absorb any photons that have energy higher than the ionization energy (13.6eV for H). The extra energy goes into the momentum of the electron. This is also called Compton scattering. So gas is very opaque to photons with wavelengths shorter than the ionization energy.

Ionization energy hc/l=13.6eV Since stars are mostly H (92% by number), they become mostly opaque to photons higher than the H ionization energy of 13.6eV. What wavelength does this correspond to? hc/l=13.6eV

Ionization energy hc/l=13.6eV, l=91.2nm. Since stars are mostly H (92% by number), they become completely opaque to photons higher than the H ionization energy of 13.6eV. What wavelength does this correspond to? hc/l=13.6eV, l=91.2nm.

This corresponds to a wavelength of 364.7nm The Balmer Jump In fact, opacity greatly increases for electrons being stripped from the second orbital of H (the first excited state), which only requires 3.4eV. This corresponds to a wavelength of 364.7nm (3647 angstroms), which is also called the Balmer jump, since blueward of this wavelength, hot stars do NOT fit a blackbody spectrum.

Bound-free(photoionization): Important feature: Balmer jump

Free-bound: Recombination When an atom captures an electron, it is called recombination. The amount of energy released is very nearly equal to the ionization energy.

Free-free: Bremsstrahlung A free electron interacts with an ion.

Electron scattering Thomson scattering: free electrons interact with photons, mostly changing their direction and less their energy.

Thompson Scattering A free electron can interact and exchange energy with a photon. Any wavelengths are allowed. BUT.. momentum must be conserved, so the photon survives, but with a different energy.

Some notes on notation: In chemistry ionized elements are notated as: A+ if singly ionized, A++ if doubly ionized, and then A3+ if triply ionized. In astronomy, neutral atoms are written as AI, so ionized ones are written as: AII, AIII, and so on. So iron missing 13 electrons would be written as FeXIV (and said as iron fourteen). Negative ions are noted as A-. Molecular forms written as: A2.

The Sun viewed in Fe XIV

Light Can you now describe these in one sentence? The processes are: 1) bound-bound (atomic) transition 2) bound-free (ionization) 3) Free-free (Bremsstrahlung) absorption 4) Thomson (electron) scattering

Molecular Lines Similar to atomic lines, these can be caused by the creation or destruction of molecules. Also, by changes in their rotation or vibration energy (which are also quantized). Only for very cool stars. In hot stars, molecules are broken by energetic photons.

Types of spectra reveal physical insight. Continuous spectra: from a high-pressure source. All solids and liquids and gas depending on pressure. This is the product of “thermal emission” which is really just a combination of Compton and Thompson scattering. Emission spectrum: low pressure gas hot compared to surroundings. Emits in all directions. Bound-bound transitions. Absorption spectrum: low pressure gas in front of a hot source.

Emission and Absorption Spectra These depend on the element. So we roughly get composition from these lines.

And an A0 star is made only of H? So does that mean our Sun is made mostly of Ca, Na, Mg, and Fe and not so much H? And an A0 star is made only of H?

NO! compositionally they are very nearly the same. The difference is how likely each line is to form.

The formation of spectral lines The basic premise behind spectral classification are these 2 questions: In what orbitals are the electrons most likely to be? What are the relative numbers of atoms in various stages of ionization? These are done statistically.

Velocity distribution For a gas in thermal equilibrium, the fraction of particles having a given speed remains stable and is described by the Maxwell-Boltzmann distribution function. The number of gas particles per unit volume having a speed between v and v+Dv is: Where n is the number density, m is the particle's mass, k is Boltzmann's constant.

T=298.15K (25oC) The area under the curve between 2 speeds is equal to the fraction of gas particles in that range of speeds.

The exponent in the distribution function is a ratio of the gas particle's kinetic energy to the characteristic thermal energy It is difficult for too many particles to deviate much from the thermal energy. Therefore the most probable speed is: But because of the long exponential tail, the root-mean-square speed is:

vmp vrms The most probable speed is the important one for the Boltzmann distribution (it sets the peak) and is what we use later for line widths.

vrms is important in pressure calculations and turbulent flow. vmp vrms vrms is important in pressure calculations and turbulent flow.

The exponent in the distribution function is a ratio of the gas particle's kinetic energy to the characteristic thermal energy This is the Boltzmann distribution for particle velocities for a given velocity range.

Remember this exponential tail, as it will be important for nuclear fusion later.

What is vrms for a hydrogen atom in the Sun's photosphere at 5,800K? Example What is vrms for a hydrogen atom in the Sun's photosphere at 5,800K? mH=1.67x10-27, and k= 1.38x10-23 m2kg.s-2K-1

What is vrms for a hydrogen atom in the Sun's photosphere at 5,800K? mH=1.67x10-27, and k= 1.38x10-23 m2kg.s-2K-1 11,991 m/s. About 12km/s.

The atoms gain or lose energy as they collide The atoms gain or lose energy as they collide. Because of the distribution of speeds of the impacting atoms, there is a distribution of electrons among the atomic orbitals. This distribution is statistical as well.

At the common temperature T. Let Sa stand for (the specific set of quantum numbers that identifies) a state of energy Ea for a system of particles. The same for Sb and Eb. From the Maxwell-Boltzmann distribution, the ratio of the probability P(Sb) that the system is in state Sb to the probability P(Sa) of the system being in state Sa is given by: At the common temperature T.

At the common temperature T. PLEASE NOTE: This equation is constructed such that the energies are binding energies- and therefore negative. So for the H ground state, E= -13.6 eV, then E2 ~ -13.6/n2 = -3.4 eV, and so on. At the common temperature T.

ANOTHER NOTE: in eV, Boltzmann's constant is k=8.6174x10-5eV/K. At the common temperature T.

But we have left out the degeneracies for each state But we have left out the degeneracies for each state. Since only so many electrons can inhabit a specific energy state (orbital), what cannot fit get distributed to other states. So we have to add in a degeneracy parameter gx.

The degeneracy parameters gx are looked up from tables The degeneracy parameters gx are looked up from tables. But usually (and for us) g = 2n2 where n is the orbital energy level.

The degeneracy parameters gx are looked up from tables The degeneracy parameters gx are looked up from tables. g = 2n2 where n is the orbital energy level. So for H, g1= 2(1)2=2, g2= 2(2)2=8, and so on.

Since stellar atmospheres contain vast amounts of atoms, the ratios of the probabilities is equal to the ratio of the number of atoms inhabiting each energy state and so the ratio of the number Nb of atoms with energy Eb to the number Na of atoms with energy Ea is the same function (called the Maxwell equation):

Example 1: For a H gas at 10,000 K, what is the ratio of atoms in the first excited state (n=2) to that of the ground state (n=1)? Recall that gn=2n2 and k=8.6174x10-5eV/K, En= -13.6/n2 eV

Example 1: For a H gas at 10,000 K, what is the ratio of atoms in the first excited state (n=2) to that of the ground state (n=1)? Recall that gn=2n2 and k=8.6174x10-5eV/K, En=-13.6/n2 eV For every atom in the first excited state, there are 34,600 atoms in the ground state.

Example 2: At what temperature will the number of atoms in the first excited state (n=2) equal that of the ground state (n=1)? Recall that gn=2n2 and k=8.6174x10-5eV/K, En=-13.6/n2 eV

Example 2: At what temperature will the number of atoms in the first excited state (n=2) equal that of the ground state (n=1)? Recall that gn=2n2 and k=8.6174x10-5eV/K, En=-13.6/n2 eV

But would there be any neutral atoms left? Example 2: At what temperature will the number of atoms in the first excited state (n=2) equal that of the ground state (n=1)? Recall that gn=2n2 and k=8.6174x10-5eV/K, En=-13.6/n2 eV At 85,400 K, the ground and first excited state would have equal numbers. But would there be any neutral atoms left?

e.g. to ionize H from the ground state (HI to HII) would be cI=13.6eV. Ionization states The Boltzmann equation also enters into the expression for the relative number of atoms in different stages of ionization. Let ci be the ionization energy needed to remove an electron from an atom in the ground state, going from ionization state i to (i+1). e.g. to ionize H from the ground state (HI to HII) would be cI=13.6eV.

To ionize H from the ground state (HI to HII) would be cI=13.6eV. Ionization states To ionize H from the ground state (HI to HII) would be cI=13.6eV. However, the atoms may not all be in the ground state. An average must be taken over the orbital energies to allow for the possible partitioning of the atom's electrons among its orbitals. The partition function, Z, is the weighted sum of the number of ways the atom can arrange its electrons with the same energy, with more energetic (and therefore less likely) configurations, receiving less weight from the Boltzmann factor.

Partition functions If Ej is the energy of the jth energy level and gj is the degeneracy of that level, then Usually this becomes a simplified expression (e.g. nearly all of the atoms are in 1 level, and so this only has to be calculated for 1 level).