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Astronomy 200Spring 1998

Lecture Notes - I

Lecture 1. Goals; Introducing Solar System Orbits 21 Jan 1998

- Hand out syllabus & discuss- Get info cards filled out

Themes

• Gravity: why things collapse (or don't)• Energy: what keeps things going (for a while)

Organization: inside-out

Gravity and Orbits in the Solar System

Naked-Eye Sky. We see only angular motions on the sky.

Naked-eye motions can be represented as occurring as lines on a (very large) sphere called"Celestial Sphere". The "fixed stars" appear to be (as far as human eyes can detect over a humanlifetime) fixed on this sphere.

Geometric models attempt to explain the motions of solar system objects on the Celestial Sphereas 2-dimensional projections of motions in a three-dimensional universe. The HeliocentricModel has now been verified (best way: sending probes to the objects); this is what we will use.

Observed angular motions are result of nested three-D motions, some real, some apparent (due tomotion of the observer)

1) Rotation of the Earth counterclockwise (as observed from above the North pole) or West toEast, causes entire Celestial Sphere to appear to rotate East to West about the North and SouthCelestial Poles. (an apparent motion)

The NCP and SCP, together with the Equatorial Plane (projection of Earth's equator), server todefine a celestial latitude("declination") and longitude ("Right Ascension") system which isfixed relative to the position of the stars.---

All Naked-Eye solar system objects have paths within a band around the Celestial Sphere called

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the Zodiac. (this is because the orbits are pretty much coplanar)

2) The orbit of the Earth counterclockwise around the Sun in one year causes the Sun to appear tomove West to East (counter to the apparent rotation of the Celestial Sphere) rather steadily at1E/day along the Ecliptic (at the center of the Zodiac). (another apparent motion).

W->E is the predominant apparent motion of Solar System objects and is called "Direct"; E->Wis called "Retrograde". (the predominance of one direction is because all the orbits have thesame sense: counterclockwise).---

The apparent motions of the planets are different depending on whether they orbit the Sun closerthan he Earth ("inferior" planets) or farther from the Sun ("superior" planets). The planetarypositions are described by their "elongation": the angular distance from the Sun (the Earth-Sun-Planet angle in the heliocentric model)

3) Inferior Planets (Mercury, Venus)

Alternate between direct and retrograde motion in such a way as to stay near the sun, within amaximum elongation of 28E (Mercury) and 48E (Venus). When they are East of the Sun, theymay be seen in the early evening, and when they are West, in the dawn.

4) Superior Planets (Mars, Jupiter, Saturn, Uranus, Neptune, Pluto)

These are usually in direct motion, but when the elongation approaches 180E (opposite the Sun inthe sky, or "opposition"), they reverse direction for a few months, making a "retrograde loop". This is because the planetary orbital periods get longer as one goes out in the solar system, andthe Earth passes the outer planets on the inside at opposition, causing the outer planets to appearto fall back.

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Lecture 2. Keplers's Empirical Laws 23 Jan 1998

• Class web site tour• Sky & Tel Page• Mercury- Mars- Jupiter exercise

So from simple naked-eye observations plus a 3-D model, we have already deduced thefollowing about orbits in the Solar System

• They are all nearly coplanar• They all have the same sense (counterclockwise)• They get slower as you go out from the Sun

For the latter, we can be more quantitative: We can easily measure the time it takes for the angle between the Earth and the planet as seenfrom to Sun to go through 360E: this is the "synodic period", or the time between conjunctions.

If the time it takes for the Sun- Planet vector to rotate through 360E relative to the stars (the"sidereal period") is P, and the time it takes for the Sun- Earth vector to revolve similarly is E (=1 year), then the mean rate of change of the Earth- Sun- Planet angle (over one synodic period) is

360E/S = ±(360E/P - 360E/E ) (+: inferior; -: superior)

so that 1/P = ± ( 1/S - 1/E )

eg, for Mars S = 780 days = 2.14 yrP = 1 / ( 1 - 1/2.14 ) = 1.88 yr = 686

days

Kepler Surveys the Solar System

Copernicus assumed circular uniform motion for hisplanets and got pretty bad agreement. Kepler realizedthat the heliocentric model actually allows one tomeasure the 3-D paths of the orbits once and for all bytriangulation from the moving Earth (this works onlyif the orbits are closed paths, which fortunately they are). Newton managed to extract thefollowing three useful empirical results from Kepler's rather obscure writings:

I. The orbits are not circles, but ellipses, with the Sun being at one focus of the ellipse (see textfor properties of ellipses)

II. The planet do not move uniformly in their orbits: "a line from the sun to the planet sweeps

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out equal areas in equal times". Basically this means that the tangential velocity vt = const / r,and it moves most rapidly when nearest the Sun.

III. The periods of different orbits around the sun = const × size3/2. The size of an ellipse isquantified by 1/2 of its largest diameter, the "semi-major axis" a. Formally a3 = const × P2.

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Lecture 3. Deriving Gravity 26 Jan 1998

Newton Derives Gravity

As one application of his theory of how things move ("Newtonian" or "classical" mechanics),Newton used Kepler Laws to infer the form of the force of gravity.

Classical Mechanics can be summarized by (bold = vector)

dp / dt = F, where p = m v

dL / dt = r × F, where L = m r × v

m = massp = linear momentumF = forcev = velocityL = angular momentumr = radius vector

The properties of F(gravity) are deduced from Kepler's Laws:

K's IInd: since vt % 1/r, L = const, and dL/ dt = 0=> F(gravity) is parallel to r. Gravity is a central force

K's IIrd: F(gravity) % 1/r2

This is demonstrated in the text for circular orbits, a special case of the ellipse: Basically dp/dt % v2/r = (2Ba/P)2 / a % a/P2 % a/a3 = 1/a2

K's Ist: follows from the 1/r2 falloff, with a fair amount of work.

Galileo's observations on the Earth that things of different masses have the same acceleration dueto gravity implies that F(gravity) % m.

Newton decided that symmetry demands that it depends on both attracting masses m and M, andis a universal attractive force between any two masses. (since verified in all experiments).

Finally F(gravity) = -er G m1 m2 / r2

Where G = 6.67 × 10-11 m3 / kg s2, the Universal Constant of Gravity.

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Orbits from a Physical Viewpoint

With this physicalization of orbits, a number of useful results fall out:

1) Mutual orbits. Conservation of the total linear momentum of orbiting bodies requires that theforce of gravity of object 1 on object 2 is equal and opposite to the force of object 2 on object 1, That means that neither object is actually fixed, but both orbit a common center of mass, aposition between the two objects where

m1 r1 = m2 r2

This was not noticed by Kepler since heassumed the Sun to be fixed, but accuratemeasurements of the position of the solarsystem show that in fact the Sun movesaround the solar system "barycenter" (the center of mass of the whole solar system), which ishowever inside the Sun itself (since m(Sun) >> m(planets), r(Sun) << r(planets)).

This is more important for binary stars, whose masses are similar.

2) Putting in the proper constants in the force of gravity, can rederive K's IIIrd Law:

P2 = 4 B2 a3 / G (m1 + m2)

Called "Newton's form of Kepler's IIIrd Law". If you can measure the period and size of the orbitof two bodies, you have the sum of their masses. This is basically the only way to get masses inAstronomy!

3) Kepler's laws work exactly only if there are only two objects (Sun and planet) gravitating. Inthe solar system , the planets must also be pulling on each other (if gravity is truly universal). These are small forces, which cause "perturbations" in the orbits away from pure stationaryellipses. The folks at NASA JPL have a giant program which keeps track of all these effects.Called the "JPL ephemeris" (ephemeris = table of positions of things).

Other flies in the ointment:

4) K's laws work exactly for two point masses, or for objects that are spherically symmetric (N.proved that these behave like point masses as far as objects on the outside are concerned.

5) When the gravity gets large, General Relativistic effects become important. More on thislater

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Lecture 4. Gravitational Mechanics in the Solar System 28 Jan 1998 (Chap 2-3)

Define orbits in terms ofSemi-major axis a,

measured in units of the Earth's orbit a(Earth) = 1.5×108 km = "Astronomical Unit" A.U.Period P

measured in " " " P(Earth) = 365.2422 d = 1 yrInclination i of planets orbit to the at of the Earth (the ecliptic plane)Eccentricity e of ellipse: ae = center to focus distance

minor axis / major axis = b/a = (1 - e2)1/2

Note that in these units, Newton's form of Kepler's Third Law isa(AU)3 / P(yr)2 = 1 + (mp / Mu ) . 1

In the Solar system we have:

1) One very large mass, the Sun: Mu = 2× 1030 kg

2) Four medium size masses, the Jovian planets

3) Four quite small masses, the Terrestrial planets (Earth, Venus, Mars, Mercury) < 6×1024 kg

Planet Mass/ 1027 kg a (A.U.) P (yrs) Inclin Eccen

Jupiter 1.9 5.2 11.9 1.3E 0.05

Saturn 0.57 9.5 29.5 2.5E 0.06

Neptune 0.09 19.2 84 0.8E 0.05

Uranus 0.10 30 165 1.8E 0.01

Planet Mass/ 1024 kg a (A.U.) P (yrs) Inclin Eccen

Mercury 0.3 0.39 0.24 7.0E 0.21

Venus 4.9 0.72 0.62 3.4E 0.01

Earth 6.0 1 1 0 0.02

Mars 0.6 1.52 1.88 1.9E 0.09

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4) A lot of icy stuff beyond Neptune (the Kuiper Belt): Pluto, Charon, etcPluto is the largest known:

5) A lot of small rocky stuff more or less in the plane of the ecliptic between Mars and Jupiter:(asteroid belt)

eg most massive: Ceres, M = 1021 kg (poorly known: from perturbations of other asteroids)

Includes:Trojan Asteroids: same orbit as Jupiter, but 60E in front or behind. These are the "Trojan

points" - stable 3-body orbits with m1 > m2 >> m3

Earth- crossing Asteroids: small number have eccentric orbits that cross Earth's orbit. Biggest collision hazard

6) Comets: very small icy bodies in very eccentric orbits

eg Halley, diameter 8×8×16 km => mass . 8×1014 kg

New comets: a > 104 AU, e . 1, all inclinations (including backwards orbits)from "Oort Cloud", spherical distribution of material left over from Solar System formation

Old comets: a as small as a few AU, tend to be direct orbitsfrom Kuiper Belt, perturbed into inner Solar System by outer planets

Planet Mass/ 1024 kg a (A.U.) P (yrs) Inclin Eccen

Pluto 0.01 39 250 17E 0.25

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Lecture 5. Gravity in the Solar System (Con't) 30 Jan 1998

7) Satellites around all objects Earth and beyond

direct orbits: formed with planet (Moon, Jovian Galilean satellites)retrograde orbits: captured via 3-body interaction

8) Rings around the Jovian planets (chapter 3-4)

Unconsolidated material in orbit near planet. These are foundinside the point (the "Roche Limit") at which differential gravity(the tides) across a larger body is larger than self-gravitation

Tidal acceleration: At = GM/d2 - GM/(d + r)2 (planet: M; satellite: m, orbit d, size r). 2GMr/d3

Self-Gravitation: As = Gm/r2

Set equal: d(Roche) = r(2M/m)1/3 (2 -> 3, if the satellite is locked: rotation =orbital period)

Define mean density of planet DM = M/ 4/3 B R3; satellite Dm = m/ 4/3 B r3

d(Roche) = R (2 or 3 DM / Dm )1/3 = (1.25 or 1.44) R (DM / Dm)1/3

Dramatic illustration: Comet Shoemaker-Levy 9 and Jupiter. In July, 1992 SL9 passed within1.3 RJ and broke into 20 pieces, which crashed onto Jupiter in July 1994 (next pass).

Orbital Velocity and Astrogation in the Solar System

(Chap 1-5C). The most useful relation for understanding astrogation is "vis-viva" equation,which follows from energy conservation. Gives the total orbital speed v(r):

v2 = G (m1 + m2) [ (2/r) - (1/a) ]

Escape speed from surface of an object of mass M, radius r = R. (Set a = infinity:)

v(esc) = (2GM/R)1/2 (eg from Earth surface v(esc) = 11.2 km/sec)

Circular orbital speed at r (set r = a)

v(circ) = (GM/r)1/2 = v(esc, R=r) / 21/2 (eg circ orb speed at Earth surface = 7.9 km/sec)Note: v(circ) % r-1/2, is called "Keplerian" velocity dependence

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Minimum energy transfer orbit. To get from circular orbit r1 to circular orbit r2

1) initial speed v1 = (GMu/r1)1/2

Transfer orbit has its extreme points rmin ("perihelion") andrmax ("aphelion") at r1 and r2, so

a(transfer) = (r1 + r2)/22) v(transfer, r=r1) = (GMu [(2/r1) - (2/r1+r2)])

1/2

3) v(transfer, r=r2) = (GMu [(2/r2) - (2/r1+r2)])1/2

4) final speed v2 = (GMu/r2)1/2

Procedure:1-2) insertion into transfer orbit; wait 1/2 period of transferorbit; 3-4) enter new orbit

"Venus delta-v assist" (Venus is best, since it has a large orbital velocity)1) transfer orbit to Venus, aim to just trail planet (do not enter Venus orbit)2) torque from Venus increases transfer orbit velocity, a, and aphelion to outer planet

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Lecture 6. Rotation and Gravity; Light 2 Feb 1998

For a non-point object, rotation causes non-spherical form, which then couples with orbits. Illustrate using Earth.

Earth Sidereal Rotation: 23h 56m 04s = 86164 sec (solar secs, giving 24h per mean solar day)Earth Polar axis: Near star " UMi ("Polaris")

inclination to the ecliptic pole (obliquity) = 23.5Eintersection of Eq plane with Ecl plane = "equinox" in constell Aries sense: W to E (same as orbit)

The rotation is slowing down due to tides from the Sun and Moon by about 2 millisec/ century

The obliquity of the axis is the major cause of the seasons: when N pole is pointed towards Sun,have summer in the North and winter in the South. Currently the equinox is pointed so that theEarth is a aphelion in Northern summer and perihelion in Northern winter, so the northernseasons are somewhat less severe than the southern seasons.

Oblateness: Rotation causes Earth to be oblate spheroid (equatorial diameter 0.3% larger thanpolar diameter). This plus non-zero-obliquity causes sun to apply torque, making Earth rotationwobble like a top:

Precession: The direction of the Pole and of the equinox relative to the stars rotates about theecliptic pole in 26,000 years. (practical problem: RA- Dec coordinate system then moves, andstellar positions must give the date of the equinox which defines the coordinates)

Astronomical climate change

There are a number of slow changes in the Earth's orbit and rotation that can cause climatechange:

1) Precession (26,000 yr cycle) causes the relative timing of perihelion and Northern winter tochange. When this is adverse, ice builds up in the far North

2) Three-body (Sun- Earth- Moon) effects change the orientation of the Moon's orbit, whichchanges the obliquity between 22 and 24E with a period of 41,000 yrs.

3) Three-body (Sun- Earth- Jupiter) effects change the eccentricity of the Earth's orbit between 0and ??? with a period of 93,000 yrs.

Milankovich Hypothesis suggests these explain the Ice Age timings in the last 106 yrs. Maybe.

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Light and Matter

The physics of light is of great importance in Astronomy because all astronomical objects outsideof the solar system can only be studied remotely by the energy and particles they emit, and lightis the main channel:

Channels for remote sensing:Light (electromagnetic radiation)Particles "Cosmic Rays": (atomic nuclei, subatomic particles like the neutrino have been

detected)Gravitational radiation (proposed by Einstein, not yet detected)

Light

According to Maxwell's equations, when you accelerate a charge, it changes the surroundingelectric and magnetic fields, and this disturbance propagates to infinity. This disturbance can bedescribed as waves (classical) or as particles (photons, quantum mechanical):

Wave description. The light wave hasSpeed c = 3×105 km/secWavelength 8Frequency < = c / 8

Wavelength and frequency have a huge range corresponding to different parts of theelectromagnetic spectrum, which are given names and arbitrary boundaries

8 < E

GammaRays

<10-3 nm > 1.2 MeV

X Rays <10 nm (100 D) > 120 eV

Ultraviolet 10 - 300 nm (100 - 3000 D) 1015 - 3×1016 Hz 4 - 120 eV

Visible 300 - 1000 nm (3000 D - 1µm)

3×1014 - 1015 Hz 1.2 - 4 eV

Infrared 1 µm - 1000 µm 3×1011 - 3×1014 Hz .0012 - 1.2 eV

Radio > 1 mm < 3×1011 Hz

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Lecture 7. Light and Atomic Structure 4 Feb 1998

In the EM wave, the electric field oscillates perpendicular to the direction of propagation, and themagnetic field oscillates perpendicular to both.

In a single coherent wave, the E vector oscillates in only one plane, say x, and is said to be 100%polarized. We detect the square of the oscillation amplitude in two perpendicular polarizations:

Intensity = <Ex>2 + <Ey>

2 = Ix + Iy

Astronomical light consists of many waves, each with an uncorrelated phase and possibly manydirections of polarization. This results in partially polarized light. You can measure the degreeof polarization by measuring the intensity through a polaroid filter, finding the angle at which theintensity is greatest Imax and 90E from that Imin

Polarization = (Imax + Imin) / (Imax + Imin)

"Natural" light (Sun, tungsten bulb) is unpolarized; scattering polarizes it.

Particle description

Comes in quantized energy packets called "photons" which can be thought of "wave packets"E = h< (h = 6.6×10-34 J-s, Planck's constant)

Photon energies customarily measured in electron volts eV = 1.6×10-19 J

Matter

Matter interacts with light on an atomic scale (or below), constructed from subatomic particles.

electron: m = me = 9.1×10-31 kg charge = - e = 1.6×10-19 coulproton: mp = 1836 me charge = +eneutron: mn = 1838 me charge = 0

As far as atoms are concerned, all electrons are alike, etc, and the only differences among atomsis the number of e's, p's, and n's, and the energy of the atom. The positively charged p's and n'sare in a very small nucleus, kept there by the nuclear Strong Force; negatively charge e'ssurround the nucleus, kept there by electromagnetic force.

Atomic number = Z = # of protons => chemical element, hydrogen - uranium Z = 1 to 92Atomic weight = A = Z + N = # protons+neutrons => isotope of elementIon stage = Z - E = #protons - #electrons => net charge

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Isotope Notation: AX, where X is symbol forchemical element of atomic number Z:

(See Appendix A5 for periodic table giving thesymbols)

Ion Stage Notation: Xcharge or X{Roman Num}

H0 = HI Neutral, one eH+ = HII charge +1, no e'sC++ = CII charge +2, 4 e's

Name Symbol Z N

Hydrogen 1H 1 0

Deuterium 2H 1 1

Helium 4He 2 2

Carbon-13 13C 6 7

Uranium 238U 92 146

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Lecture 8. Energy Levels and Spectra 6 Feb 1998

Atomic Energy Levels

Light interacts with an atom depending on its internal energy, which can be different dependingon the "orbital" motion of the most loosely bound electron. By quantum mechanics, the motionof this electron must have an angular momentum which is an integral multiple of h/2B:

me ve r = L h / 2B = L S

• Each allowed orbit L corresponds to a discrete energy En

• For each ion there is a maximum possible energy beyond which the least bound electronbecomes unbound, the ionization energy E(ion)

The set of possible energy levels is unique to each ion of each element

(they are even slightly different among different isotopes of the same element)

Atomic interaction

An atom may change its energy level by gaining or losing just the right amount of energy to getto a different level or change its ionization state

excitation/ de-excitation:X(na) + )Eab <-> X(nb) ()Eab = Eb - Ea)

ionization/ recombinationX(na) + )Ea(ion) <-> X+(n=1) + e- ()Ea(ion) > E(ion) - Ea)

This energy may come from:

• Collision with another atom or electron

• Absorption of a photon of the right frequency (excitation, ionization)h<ab = )Eab or h< > h<a(ion) = )Ea(ion)

X(na) + h<ab -> X(nb) (the frequency <ab is removed from a continuousspectrum, giving an absorption line)

X(na) + h<(ion) -> X+(n=1) + e- (any frequency > <a(ion) is removed from acontinuous spectrum, giving an absorption edge)

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• Emission of a photon of the right frequency (de-excitation, recombination)

X(na) -> X(nb) + h<ab (an emission line)

X+(n=1) + e- -> X(na) + h< (a continuous spectrum ending at <a(ion))

Example: Hydrogenic energy levelsThe energy levels of ions which have the same number of electrons (called isoelectronic series)are similar:

HI, HeII, LiIII, ... has one electron Hydrogenic ionsHeI, LiII, ... 2 electrons "Helium-like" ions

The hydrogenic energy levels are easy to calculate, since it is a 2-body problem: 1 electron +nucleus of charge Z. This the Bohr model of Hydrogen.

Turns out:E(n) = -(2B2 me e

4 k2 / h2) Z2 (1- 1/ n2) = -R' Z2 (1 - 1/ n2) (R' = "Rydberg" = 2.2×10-18 J = 13.6 eV)

(Note: this is relative to the ground state n=1. The absolute energy is not significant.

Spectrum of Hydrogen

h<ab = hc/8ab = E(nb) - E(na) = R' (1/na2 - 1/nb

2 )

na = 1 (lines out of the ground state)nb 2 3 4 .. ionization 8ab(nm) 122 103 97.2 < 91.2 (Lyman; UV)

Ly" Ly$ Ly( Ly continuum

na = 2 (lines out of the ground state)nb 3 4 5 .. ionization 8ab(nm) 656 486 434 < 365 (Balmer; visible)

H" H$ H( Balmer continuum

na = 3, 4 ... Paschen (near IR), Brackett (IR), etc

These lines and continua are very strong in the hotter stars

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Lecture 9. Spectra of real gases 9 Feb 1998

Hydrogenic spectra: He II (Z=2). Energies Z2 = 4× bigger, Wavelengths 4× smaller. eg "HeIILy"" is at 122 / 4 = 30.4 nm in the extreme UV.

Other isoelectronic spectra. These can get pretty complicated, since multiple electrons caninteract with each other both via electric field and via magnetic field due to electron's spin Sinteracting with the orbital angular momentum L, giving total J (all quantized!).

Before we consider the spectra from ensembles of atoms, we need to put in the effect of theatoms motion on the light:

Doppler Effect

8(observed) = 8(emitted) × [ (1 + vr/c) / (1 - vr/c)]1/2 (relativistic exact formula)-> 8(emitted) × (1 + vr/c) (vr << c; use this from now on)

vr is the radial component of the atom's velocity

We can express this as a Doppler Shift )8 = 8(obs) - 8(em))8/8 = vr/c

The doppler shift is the same percentage as v/c

vr > 0: receding: )8 > 0 : wavelengths increase ("redshift")vr < 0; approaching: )8 < 0 : " decrease ("blueshift")

Notice that all wavelengths from an atom are changed by the same factor. This means thatalthough the lines are no longer at the "laboratory" wavelengths, the ratio of the line wavelengthsremains the same:

Identification procedure:• identify atom from pattern of line wavelength ratios• calculate vr = c )8 / 8 from the difference with lab wavelength

Line Profiles

Observed absorption or emission lines are not infinitely sharp, even if we had an infinitely goodspectrometer to observe them with:

1) Natural width. The energy of atomic levels is not perfectly determinable. By QM, there is anenergy uncertainty )E = h/ / )t, where )t is the time spent in the level. This is largest for thestrongest lines with )t (upper level ) as small as 10-8 sec.

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h )< = )E = h / 2B )t)v = 1 / 2B)t - 107 Hz)< / < = )8 / 8 - 10-8 (a pretty small effect!)

2) Thermal (doppler) broadening. In a gas, the atoms are all in motion with random velocitieswhose mean kinetic energy is proportional to the temperature:

1/2 m <v2> = kT (k = 1.38×10-23 J/EK, Boltzmann's Constant)

so the rms thermal velocity is<v2>1/2 = vth = (2kT / m)1/2

the random doppler shift associated with these atoms is)8th/8 = vth / c

eg, for H at T = 10,000 EKvth = (2×104 × 1.38×10-23 / 1.7×10-27 kg)1/2 = 12.8 km/s)8th/8 (H, 104 EK) = 4×10-5

Note that heavier atoms at the same temperature have smaller thermal broadening)8th(H) / )8th(Fe) = (m(H) / m(Fe) )-1/2 = 561/2 = 7.5

3) Collisional ("Stark", "pressure") broadening. Having nearby atoms perturbs the atomic levelsin a variety of ways, different for each atom and each perturber. In general:

)8(coll)/8 % pressure

This can be larger than thermal broadening for H and for high-gravity stars.

4) Turbulent broadening This is doppler broadening, just like thermal, but all atoms have thesame turbulent broadening regardless of atomic mass.

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Lecture 10. Line Strength; Blackbody Continuum 11 Feb 1998

Line Strength

The strength of an emission lineI = I I(8) d8

or of an absorption line in a continuum of intensity Ic

EW = I (Ic - I(8)) d8 / Ic (this is called the equivalent width of the line)

depends on the number of atoms along the line of sight which are capable of emitting orabsorbing it:

EW(8ab) = fn ( I N(a) dz )

To calculate the relative line strengths of all the possible line strengths in a gas we need

• The relative abundance of each element, H, He, Li ...• The relative number of each ion stage for each element• The relative number of ions in each energy level (the level populations)

Thermal Equilibrium

In a general gas the ionization levels and energy levels may have all sorts of relative populations,and the problem of predicting the spectrum is very difficult.

In a gas in thermal equilibrium ("TE") the ionization and level populations depend only on thecollision rates, and thus only on the temperature T and electron density Ne

Conversely, by observing the relative line strengths in a gas in TE, we can deduce the elementabundances, the temperature, and the electron density (pressure) in the gas.

Ionization: the Saha EquationNi+1 / Ni = [( 2BmekT/h2 )3/2 / Ne] exp (-P/kT) (P is ionization energy from i to i+1)

high T and low Ne (pressure) tends to increase the ionization stage

Levels: the Boltzmann Equation:Nb / Na = (gb / ga) exp (-)Eab / kT) ()Eab is energy difference state a to state b)gn is the statistical weight of level n from QM; for H, gn = n2

high T tends to increase the excitation level

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Useful conversion: kT = 1 eV at 11,600EK

eg: in stars, even though H is the chief element by far, the Balmer lines are strong only for T =6000 - 20000 EK:

• For T < 6000 EK , the second level of H (from which Balmer lines originate) is not populatedsince )Eab = 10.2 eV >> kT = 0.5 eV

• For T > 20000 EK, most of the H is ionized since kT = 2 eV is enough and Ne is smallenough the N+ / N0 >> 1

Thermal "Blackbody" Continuum Radiation

In a gas in complete thermal equilibrium, even the number of photons of each energy is governedby a statistical equilibrium equation:

Photons: the Planck Radiation LawI(<) = B<(T) = (2h<3 / c2) / (exp(h</kT) - 1)

I(<)d< is the specific intensity, units W/m2 ster Hz:Power (W) passing through an area of 1 m2 from a solid angle of one steradian per 1 Hz

frequency intervalA steradian is the area analog of radian: there are 2B radians in a circle, and 4B steradians in

a sphere.

Can also be stated in wavelength units. Since < = c/8, |d<| = c/82 d8:

I(8) = B8(T) = (2hc2/85) / (exp(hc/8kT) - 1) (W/ m2 ster m)

• B8(T) has a maximum at wavelength (Wien's Law) 8max = 2898 µ / T(EK)

Hotter objects radiate bluer light

• the total intensity integrated over all wavelengths is (Stefan-Boltzmann Law)B(T) = IB8(T) d8 = ( F/B ) T4 (F = 5.67×10-8 W/m2 EK4)

Hotter objects radiate more light

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Lecture 11. Radiation Transfer, Flux, and Luminosity 13 Feb 1998

The spectrum from a gas cloud depends on whether it is opaque:

At every frequency there is an opacity 6< (units inverse length) due to all absorption andscattering processes, and there is an emissivity j< (units W/m3 ster Hz). The intensity along a lineof sight decreases with opacity and increases with emissivity:

dI</ds = j< - 6< I< (Equation of Radiative Transfer)

If we define a dimensionless length variable the optical depthdJ< = 6< ds

Then we getdI< / dJ< = S< - I<S< = j< / 6< is called the source function

For instance, for a uniform gas:If the gas is optically thin J< = I6< (s) ds << 1

I< = J< S< = J< (j< / 6<)

If the gas is optically thick J< >> 1

I< = S<

In thermal equilibrium, S< = B<

• If a gas is optically thick at all frequencies, we get a continuum I< = B< (Planck Curve)• If it is optically thin at some frequencies, we get emission lines % J< where J< is large, with

maximum I< = B<.

and eg in stellar atmospheres:• If a partially optically thin slab at temp T1 is in front of an optically thick slab at T2, we get

I< = B<(T1) at frequencies where it is thick, and = Bn(T2) where it is thin

=> continuum plus (T1 < T2) absorption lines; (T1 > T2) emission lines

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Flux and Luminosity

Intensity I<(2,N) is the amount of energy going in a particular direction at some point in space.

Flux F< is the net amount of energy passing through a fixed areafrom all directions:

F< = I I< (2,N) cos 2 dS = I I< (2,N) cos 2 sin 2 d2 dN(units W/m2 Hz)

(S is the solid angle; 2 is measured relative to the normal to thearea, N is the azimuthal angle around the axis)

(eg, at the surface of a blackbody, I< = B<(T) going out, and = 0coming in, so

F< = B B<(T)F = F T4 This is the total W/m2 escaping

from the blackbody)

The luminosity of the star is the flux integrated over its surfaceL< = 4BR2 F<; L = 4BR2 F This is the total

power in W of the star

( for a blackbody, L = 4BR2 FT4 Blackbody luminosity is a common stellar estimate)

Note that in free space a distance d outside of an emitting/ absorbing body like a star

I is constant along a line of sight: = I* in the direction of the star, = 0 elsewhere.

F = I I cos 2 dS % I S(star) = I BR2 / d2 drops off as 1 / d2

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Lecture 12. The Sun; Stellar Atmospheres 16 Feb 1998

The Sun

So, now apply all this to our star, the Sun. We know more about the Sun than any other starbecause• It is so close that very sensitive observations can be done• We can resolve features on its surface• We know its age (assumed = age of oldest solar system rocks, 4.5×109 yr, or 4.5 Gyr)

Gravity- related:Mu = 2×1030 kg Ru = 6.96×108 m

=> au = GMu / Ru2 = 274 m/s2 = 28 ar (28 "G's")

v(esc) = (2GMu / Ru) = 618 km/s<Du> = 1.4 gm/cm3 (1.4 × water)

Energy- relatedLu = 3.86×1026 W Teff u = 5770EK (Teff = blackbody temp that gives L)

AtmosphereFrom these data, we can estimate how thick the atmosphere is

Assume: )z(atm) << R (called "plane parallel" atmosphere)T(atm) = const (called "isothermal" atmosphere)The atmosphere is static (not being accelerated)The pressure is given by the Perfect Gas Law

Pg = NkT = kT (D/µmH) (N = #/vol; µ = mean atomic wt)Balance the weight of a slab of gas of thickness dz with the difference in pressure between its topand bottom:

dPg / dz = -g D(z) = -g µ mH Pg / kT (g = acceleration due to gravity)

If µ, g, and T are constant with z, then

Pg(z) / Pg(0) = exp (-z g µ mH / kT) (z = 0 is defined arbitrarily)= exp (-z / H)= D(z) / D(0)

H = kT / g µ mH is the isothermal scale height: Pg and D drop by 1/eeg Hr = 8 km (26,000 ft) (g = 9.8m/s2, µ = 29, T = 273 EK)

Hu = 110 km (g = 274 m/s; µ = 1; T = 5800 EK)

Note: Hu << Ru => plane parallel approximation is OK

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Structure:Interior. Outer 20-30% is convective

Atmosphere:Photosphere. Visible layer ~ 300 km (3H) thick,

J(continuum) 0.005 - 1T = 4500 - 6500 KSpectrum: blackbody at 5800EK + absorption lines of neutral and once ionized

metals: Na, CaII, Mg, Fe. Weak H Balmer lines.

Chromosphere. Upper atmosphere visible in eclipses ~ 2000 km thickT = 4200 - 20,000 KSpectrum: additional absorption + emission lines of H, CaII

Corona. Extreme upper atmosphere seen in eclipses, X-Rays ~ 2 Ru

T -> 106 K (except mostly not in TE, so temperature becomes meaningless)Spectrum: emission lines of highly ionized ions, eg FeXIV

Solar Wind. Ionized outward flow at ~ 300 km/sec bathes whole solar system

Elemental Abundances (from photospheric spectrum)H 0.9 C 5×10-4 Ne 1×10-4 S 0.2×10-4

He 0.1 N 1 " Mg 0.4 " Fe 0.5 "O 8 " Si 0.4 "

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Lecture 13. Stellar Distances and Luminosities 18 Feb 1998

Other Stars

Distance

The biggest problem in learning about other stars is measuring their distance. The only directway is by triangulation using the Earth's orbit (the same way Kepler derived the orbit of Mars).

Parallax B = semimajor axis of angular shift with respect to distant stars.

tan B – B = 1 AU / d

d (AU) = 1/B (rad)

This leads to a standard set of distance units

d (parsec) = 1/B (arcsec)

1 parsec (pc) = 206265 AU = 3.086×1016 m = 3.26 light year (ly)

Also: kiloparsec = kpc; megaparsec = Mpc; gigaparsec = Gpc

Modern parallaxes. Satellite Hipparcos just recently measured 105 stars to < 10-3 arcsec, about10× better than before.

Nearest star " Cen. B = 0.76 arcsec. d = 1.3 pc = 4 lyBetelgeuse " Ori. B = 0.0763 ± 0.0016 arcsec. d = 131 ± 27 pc

Flux, Luminosity, and magnitudes

If we measure distance d and the flux F from star at the earth, can get the luminosity L:

L = 4Bd2 F

Astronomers employ an awkward, obsolete unit system for flux and luminosity

Flux ratios can be stated as apparent magnitude differences:m - m0 = -2.5 log10 (F / F0)

where F0 is the flux from some standard object. The star Vega (" Lyr) is often used as a standard:m(Vega) = 0

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m(faintest naked-eye star) - 5 (F/F(Vega) = 1/100)m(faintest objects seen by HST - 25 (F/F(Vega) = 10-10)m(Sun) = -26.5 (Fu/ F(Vega) = 1026.5/2.5 = 4×1010)

The flux (and magnitude) can be measured for• monochromatic m8 - m80 = -2.5 log10 (F8 / F80 )• "filter" mV - mV0 = -2.5 log10 (FV / FV0)

where FV = I F8 SV(8) d8

SV(8) is the response of the "V" filter "UBV": a standard set of broadband filters:

"mV" is often called "V" for short, etc

• integrated over wavelength or "bolometric"

mbol - mbol(0) = -2.5 log10 (F / F0)

Luminosity Ratios can be stated as absolutemagnitude differences, denote by capital M:

M - M0 = -2.5 log10 (L / L0)

These can be quoted for all the types of magnitudes above.

A common reference for absolute magnitudes is to state the apparent magnitude an object wouldhave at a standard distance, usually 10 pc:

m - M = -2.5 log10 ( (10/d(pc))2 ) = 5 log10 d(pc) -5 called the "distance modulus"

eg (m -M)(u) = 5 log 1/206265 - 5 = -31.57 = -26.74 - 4.83 =-26.82 - 4.75(m-M) (" Cen) = 5 log 1.3 - 5 = -4.4 = 0.01 - 4.4(m-M) (" Lyr) = 5 log 8.1 - 5 = -0.5 = 0.00 - 0.5(m-M) (" Ori) = 5 log 131 - 5 = 5.6 = 0.8 - (-4.8)

Name 8(effective)

)8 m(u) M(u)

U 365 70 -25.96 5.61

B 440 100 -26.09 5.48

V 550 90 -26.74 4.83

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Lecture 14. Stellar Color and Classes; Binary Stars 20 Feb 1998

The difference of different filter magnitudes is called the color index

B - V = -2.5 log FB / FV = MB - MV = -2.5 log LB / LV

The color index is an intrinsic quantity, depends on the continuum slope at B and V, so is a crudemeasure of temperature. Has been measured for > 105 stars, often used in statistical studies andfor very faint objects.

Fly in the ointment: Interstellar extinction. Fine solid particles between the stars cause"extinction" = scattering + absorption.

• Flux is lower than it should be:F(obs) = F(intrinsic) exp -J(ISM)

=> m(obs) - m0 = -2.5 log F/F0 + AA = -2.5 log10 (exp -J) = 1.085 J

• The extinction is different at different wavelengths (increases into the blue, so is called"reddening"), so the observed color index is also affected:

(B - V) (obs) = (B - V) (intr) + (AB - AV)= (B - V) (intr) + EB-V

A (J) is larger for larger distances, but different in different directions. Correction methodsinvolve deducing what B-V should be (from spectra), thus getting EB-V. Since R = AV / (AB - AV) (the color of the extinction) is roughly the same in all directions (about 3.2), this gives AV and theintrinsic apparent magnitude and color.

Stellar Classes and the Hertzsprung-Russell (H-R) diagram (chap 13)

We now have enough information to separate stars into classes, using a classification diagram: plot a sample of objects with one intrinsic independent observable vs another. Look to seewhether there are groupings in this diagram.

Observable 1) (abscissa): Surface temperature or color (from continuous spectrum)Observable 2) (ordinate): Luminosity or absolute magnitude

These are independent of each other since L = 4BR2 T4

and R might (in principle) take on any value

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We find that there are three main groupings

1) The Main Sequence. Runs from high luminosity hotstars to low luminosity cool stars. The great majority ofstars are "on the main sequence".

Example: the Sun, Sirius A, Vega

2) Red Giants. High luminosity cool stars (They are"giants" since R2 = L/4B T4 is large)

Example: Mira (o Cet), Betelgeuse (" Ori)

3)White Dwarfs. Low luminosity hot stars. ("dwarfs" since L/T4 is very small)

Example: Sirius B

It is best to consider these groupings as grouped by size. These are the three different structures,with three different balances between energy and gravity. Much more on this later, after weconsider one more intrinsic observable property:

Stellar Mass and Binary Stars

The only information we have on stellar masses comes from observing the mutual orbits ofmultiple stars. Maybe 50% of stars are in binaries (the Sun is not), but most of the binaries donot have observable orbits. Two classes which are observable:

Visual Binaries. Directly measure a and P of mutual orbits• The stars have a large enough angular separation that the stellar images are separate (2 = a/d

> 1 arcsec)• the orbital period is short enough that they can be seen to move on a human timescale (P <

200 years)

These are all relatively nearby, to give the largest 2 for the smallest a and P. eg Sirius A and B.

Spectroscopic binaries. Measure orbital velocities and P, infer a.• The stars have very small a, giving large orbital velocity doppler shift.

These tend to be unresolved binaries with very short periods, and can be seen at all distances.

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Lecture 15. Binary Star Analysis 23 Feb 1998

Visual Binary analysis

Terminology: relative orbit is orbit of one star relative to the other;absolute orbit is orbit of both with respect to the center of mass (to due this, use the surrounding

starfield as a fixed reference).

inclination is the angle between the axis of the orbit and the lineof sight

line of nodes is the line of intersection between the orbit planeand the plane of the sky

apparent orbit is the orbit as seen from the Earth; true orbit is the orbit as seen from along the axis (corrected forthe inclination)

Procedure:1) (after collecting data over a large fraction of the period) plot the

relative apparent orbit2) deduce the inclination and orientation of the line of nodes by

applying Kepler's first law. The relative apparent orbit is anellipse, but inclination makes the reference star appear neither at the focus of the apparentellipse nor on its major axis.

3) plot the true, absolute orbits4) measure a1(arcsec), a2(arcsec) and the distance d, giving

a(AU) =( a1(arcsec) + a2(arcsec) )×d(pc)5) measure P from the time it takes to go around

(M1 + M2) / Mu = a3(AU) / P2(yr)M1 / M2 = a2 / a1

Main problem: we need lots of time, and we need the distance

Spectroscopic Binary AnalysisObserve doppler shift of spectral lines to vary periodically => radial component of orbitalvelocity vr/c = )8/8

terminology:double-line spectroscopic binary: see sum of two spectra; absorption lines doppler shifted apartsingle-line: see only one spectrum, but periodic variation of doppler shift implies orbit around

unseen companion

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Simplest case: circular orbits2Br1 = 2Ba1 = v1 P2Br2 = 2Ba2 = v2 P

We observe the velocity curvev1r(t) = K1 sin (2Bt / P); K1 = v1 sin iv2r(t) = -K2 sin (2Bt / P); K2 = v2 sin i

=> K1, K2, P

We inferM1 / M2 = r2 / r1 = v2 / v1 = K2 / K1

a = a1 + a2 = (v1 + v2)P/2B = (K1 + K2) P / (2B sin i)

Severe problem: there is often no way of knowing the inclination, so all we can measure is a sin i

=> (M1 + M2) sin3 i = (a sin i)3 / P2

Complications:• elliptical orbit. Velocity curve is no longer

sinusoidal, but from its form one can deduceeccentricity e and orientation S of the orbit,and M1 / M2 = K2 / K1 still holds

• single line binary. Now we can't get the massratio M1 / M2, and can only measurea1 sin i and P. This is of some use if we haveother information (cf black hole candidateslater)

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Lecture 16. Mass-Luminosity Relation; Spectral Type 25 Feb 1998

Extra Goodies:• eclipsing binaries. For i - 90E, the stars occult

each other twice an orbit, giving a "primaryeclipse" (the deepest one) and a "secondaryeclipse". Besides giving i and thus the masses,one can also get the radii of the individual starsfrom the shape of the "light curve"

eg, circular orbits:the relative sideways velocity at eclipse is v = v1 + v2 = 2Ba/PR(large) = 1/2 v (t4 - t2) = Ba (t4 - t2) / PR(small) = Ba (t2 - t1) / P

Mass-Luminosity RelationBottom-line results:• For Main Sequence stars, there is a tight relation between mass and luminosity. Empirically:

L/Lu = 1.0 (M/Mu)4.0 (M/Mu > 0.43)

= 2.3 (M/Mu)2.3 (M/Mu < 0.43)

Notice the large exponent! More massive MS stars are much more luminous (important fact)

MS stars mass range: 0.08 < M/Mu < 100

• For Giants, there is no good correlation between M and L, although the most massive onestend to be luminous. Masses range from -1 Mu to the upper limit for MS stars

• For White dwarfs, they are all in a narrow mass range 0.5 - 1.3 Mu, and there is no relation toluminosity.

Spectral Classification (13-2)

Mbol, T(surf) (or B-V color), and Mass derived from continuum light define the basic stellarproperties. But many of these things are also observable by looking at the details of the spectrallines: spectral classification. Advantages:

• does not require the distance• does not require detailed calibration of flux• does not depend on interstellar extinction and reddening• flags peculiar stars

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Modern classification scheme is two-dimensional + additional "peculiarity" indicators:

1) Temperature (Harvard) sequence based on relative strength of lines in the blue part of thespectrum. In decreasing temperature order, indicated by

• letter, the spectral type OBAFGKM, • number, the spectral subtype 0-9

eg Vega A0Sun G2Betelgeuse M1.5

This correlates with temperature because of excitation/ionization effects like in the Balmer lines.

• Calibrated to temperature using stars of known Teff.• Calibrated to B-V using stars with B-Vcorrected

for interstellar

Old, bad terminology: O stars are "early type: and Mstars are "late type"

Type DominantLines

T/1000K

O HeII >25

B HeI 10-25

A H Balmer 8-10

F H + metals 5-6

G CaII, FeII 4-5

K neutral metals 3-4

M TiO molecule <3