Galaxies: Structure, Dynamics, and Evolution
Spiral Galaxies/Disk Galaxies (III): Spiral Structure
Review: Collisionless Boltzmann Equation and Collisionless Dynamics Layout of the Course Feb 5: Review: Galaxies and Cosmology Feb 12: Review: Disk Galaxies and Galaxy Formation Basics Feb 19: Disk Galaxies (I) Feb 26: Disk Galaxies (II) Mar 5: Disk Galaxies (III) / Review: Vlasov Equations this lecture Mar 12: Elliptical Galaxies (I) Mar 19: Elliptical Galaxies (II) Mar 26: Elliptical Galaxies (III) Apr 2: (No Class) Apr 9: Dark Matter Halos Apr 16: Large Scale Structure Apr 23: (No Class) Apr 30: Analysis of Galaxy Stellar Populations May 7: Lessons from Large Galaxy Samples at z<0.2 May 14: (No Class) May 21: Evolution of Galaxies with Redshift May 28: Galaxy Evolution at z>1.5 / Review for Final Exam June 4: Final Exam You have a homework assignment that is due on Monday, Mar 9, before noon
There will be a new homework assignment that will be due on Monday, Mar 16, before noon First, let’s review the important material from last week Multiple arm spiral
Grand design spiral
How doNGC 6946the arms in spiral galaxies evolve with time?
3-4-12see http://www.strw.leidenuniv.nl/˜ franx/college/galaxies12 12-c02-3 Most spiral3-4-12see arms http://www.strw.leidenuniv.nl/˜are This could franx/college/galaxbe determinedies12 by 12-c02-4 looking at Flocculent spiral Most spiralfound arms toare trailingbe trailing. reddening in globular clusters / novae
globular clusters seen around disk galaxy. amount of reddening indicated by whether circles are solid or open
allows us to determine which way a spiral galaxy is 3-4-12see http://www.strw.leidenuniv.nl/˜ franx/college/galaxies12 12-c02-3 3-4-12see http://www.strw.leidenuniv.nl/˜tilted. franx/college/galaxies12 12-c02-4
--2e tp/wwsr.ednnvn/ franx/college/galax http://www.strw.leidenuniv.nl/˜ 3-4-12see franx/college/galax 12-c02-3 ies12 http://www.strw.leidenuniv.nl/˜ 3-4-12see 12-c02-4 ies12 What is nature of the arms in spiral arms?
Do the spiral arms travelWinding at the same problem speed as the stars? If spiral arms did,one would predict that the spiral arms in a galaxy would wind up very quickly.
Winding problem
12 The predicted outcome is in contrast to what is observed! The problem: most spiral galaxies would be tightly wound by now, which is inconsistent with observations.
Spiral arms cannot be a static structure (i.e. at di↵erent times, arms must be made of di↵erent stars)
13 How can we solve the winding problem?
Density Wave Theory Lin & Shu (1964-1966)
The spiral arms in disk galaxies are not fixed structures that rotate around the center of disk galaxies, but rather density waves.
These density waves can move at a different speed than the stars within the galaxy itself.
The speed at which the spiral density waves propagate around the disk of a spiral galaxy is called the pattern speed Ωp. 3-4-12see http://www.strw.leidenuniv.nl/˜ franx/college/galaxies12 12-c02-7 3-4-12see http://www.strw.leidenuniv.nl/˜ franx/college/galaxies12 12-c02-8
3-4-12see http://www.strw.leidenuniv.nl/˜ franx/college/galaxies12 12-c02-7 3-4-12see http://www.strw.leidenuniv.nl/˜ franx/college/galaxies12 12-c02-8
Epicycle Approximation IV
An important question is: “When is the epicycle approximation valid?” First consider the z-motion: The equation of motion, z¨ = ν2z implies a − constant density in the z-direction. Hence, the epicycle approximation is valid as long as ρ(z) is roughly constant. This is only approximately true very close to equatorial plane. In general, however, epicycle approx. is poor Starsfor motion inin zSpiral-direction. Galaxies are on Epicyclic orbits In the radial direction, we have to realize that the Taylor expansion is only accurate sufficiently close to R = Rg. Hence, the epicycle approximation is Theonly motionvalid for small canlibrations be approximatelyaround the guiding describedcenter; i.e., for asorbits thewith combination of an angular momentum that is close to that of the corresponding circular orbitalorbit. motion around a disk galaxy and an epicyclic motion in radius:
Epicyclic Motion Orbital Frequency Frequency of epicyclic around Disk Galaxy = motion = 2π/κ 2π/Ω
The orbital frequency of a star Ω(R) can be written as follows
Meanwhile, the frequency of epicyclic motion κ(R) can be written as follows:
The frequency of epicycle motion is very similar to the orbital frequency: In general, Ω < κ < 2 Ω Near the solar system, the epicycle frequency κ ~ 1.3 Ω NOW new material for this week Epicyclic orbits
For the case of a point mass (Ω ∝ R-3/2), e.g., solar system, the epicyclical time perfectly matches the rotation time around the central body so that orbits close on each other.
There can be no more than 6 integrals of motion. Typically there is 12-10-07see http://www.strw.leidenuniv.nl/˜ franx/college/ mf-sts-07-c4-17 12-10-07see http://www.strw.leidenuniv.nl/˜ franx/college/ mf-sts-07-c4-18 at least one integral of motion (energy). A general 3-dimensional potential A Simple recipe to build galaxies Here are some examples of orbits where the phase space is only Schwarzschild’s method: • Define density ρ 12-10-07 see http://www.strw.leidenuniv.nl/˜incompletely franx/college/ filled: mf-sts-07-c4-3 2 2 12-10-07 see http://www.strw.leidenuniv.nl/˜ franx/college/ mf-sts-07-c4-4 St¨ackel potential( ρ =1/(1 + m ) ) • Calculate potential, forces Stacker potential: triaxial potential In general,typical orbit inthis spherical is not potential true, forms however. a Orbits regress and one finds a • Integrate orbits, find orbital densities ρi Typical orbitplanar in arosette spherical potential is a planarthat admits rosette three integrals of motion • Calculate weights wi > 0 such that planar rosette. ⃗¨r = −Ω2⃗r ρ = ρiwi or in cartesian coordinates x, y !
x¨ = −Ω2x y¨ = −Ω2y Examples: build a 2D galaxy in a logarithmic potential 2 2 Hence solutions are Φ =ln(1+x + y /a). • As we saw, box orbits void the outer x-axis x = X cos(Ωt + c•xAs) we saw,y = loopY orbitscos( voidΩt + thec innery) x-axis → both box and loop orbits are needed. where X, Y, cx and cy are arbitracy constants. Hence, BT 3.1: page 106 Suppose we have constructed a model. Angle ∆ψ between successive apocenter passages de- even though energy and angular momentum restrict orbit to a “rosetta”, these• What orbits kind of are rotation even can more we expect special: ? pendsin most ongeneral mass case, distribution: the orbit is not closed box orbits: no net rotation they do not fill the area betweenloop orbits: the can rotate minimum either way: and positive, nega- and will fill entire area between rmin and rmax maximum radius, but are always closed ! π < ∆ψ < 2π tive, or “neutral”. BT 3.4: page 155 The same holds for Kepler potential. But beware, for Hence: a maximum rotation is defined if all loop the homogeneous sphereorbits the particlerotate the same does way. two The radial rotation can vary homogeneous sphere point mass excursions per cycle aroundbetween the zero, center, and this for maximum the Kepler rotation potential, it does one radial excursion per angular cy- cle. Special cases We now wish to “classify” orbits and their density dis- 2 tribution in a systematic way. For that we use Integrals v⊥ dΦ GM(r) rmin = rmax circular orbit = = of motion. r dr r2
1 2 L =0 ⇒ radial orbit 2 r˙ = E − Φ(R)
Homogeneous sphere
1 2 2 Φ(r)= 2 Ω r + Constant
In radial coordinates Epicyclic orbits
Using the measured values for κ and Ω at the radial position of the sun in our galaxy is as follows:
κ = 1.3 Ω similar to the case for an isothermal sphere...
Period for orbit around galaxy = 2π/Ω
Period for epicyclic orbit = 2π/κ Which resonances drive spiral density wave growth?
Now let us now consider a possible spiral density wave in the disk of a galaxy:
Rotational Frequency of Spiral Density Wave = Ωp
In these illustrations, let’s adopt the most common type of “grand design” spiral galaxy where we just have 2 arms (rotational symmetry = 180 degrees) Which resonances drive spiral density wave growth? When might we expect growth of a spiral density wave?
Rotational Frequency of Spiral Density Wave = Ωp
We might expect such if a star completes one period of epicyclic motion every time it encounters the spiral density wave in its orbit around the galaxy. Which resonances drive spiral density wave growth? Let us consider a few examples of the orbit of stars that would finish a complete epicyclic orbit in the spiral density wave itself:
Example #1: The star is moving at the same speed as the spiral density in orbiting around the center of a galaxy.
Let’s consider snapshots in time where the star completes an entire epicyclic orbit. Typically a star must complete 70% of a revolution around a galaxy before this happens. spiral arm
spiral star arm
spiral spiral arm arm one epicyclic orbit two epicyclic orbits time = 2π/κ time = 4π/κ three epicyclic orbits spiral pattern ~ 250 deg spiral pattern ~ 500 deg time = 6π/κ star orbitted ~ 250 deg star orbitted ~ 500 deg spiral pattern ~ 750 deg star orbitted ~ 750 deg Which resonances drive spiral density wave growth?
Let us consider a few examples of the orbit of stars that would finish a complete epicyclic orbit in the spiral density wave itself:
Example #2: The star is traveling much faster than the speed of the spiral density wave.
Let’s consider snapshots in time where the star completes an entire epicyclic orbit. In this case, the star again completes 70% of an orbit, but the spiral arm orbits 0.2 times
spiral arm spiral arm
spiral spiral arm arm
one epicyclic orbit two epicyclic orbits three epicyclic orbits π κ time = 0 time = 2π/κ time = 4 / time = 6π/κ spiral pattern ~ 70 deg spiral pattern ~ 140 deg spiral pattern ~ 210 deg star orbitted ~ 250 deg star orbitted ~ 500 deg star orbitted ~ 750 deg Which resonances drive spiral density wave growth?
Let us consider a few examples of the orbit of stars that would finish a complete epicyclic orbit in the spiral density wave itself:
Example #3: The star is traveling much slower than the speed of the spiral density wave.
Let’s consider snapshots in time where the star completes an entire epicyclic orbit. In this case, the star again completes 70% of an orbit, but the spiral arm orbits 1.2 times (instead of just 0.2 times)
spiral arm spiral arm
spiral spiral arm arm
one epicyclic orbit two epicyclic orbits three epicyclic orbits π κ time = 0 time = 2π/κ time = 4 / time = 6π/κ spiral pattern ~ 430 deg spiral pattern ~ 860 deg spiral pattern ~ 1290 deg star orbitted ~ 250 deg star orbitted ~ 500 deg star orbitted ~ 750 deg Which resonances drive spiral density wave growth?
Let us look at a few movies that illustrate these concepts rather directly: Corotation
http://cosmo.nyu.edu/~jb2777/resonance.html Credit: Jo Bovy Which resonances drive spiral density wave growth?
Let us look at a few movies that illustrate these concepts rather directly: Inner Lindblad Resonance
Credit: Jo Bovy Which resonances drive spiral density wave growth?
Let us look at a few movies that illustrate these concepts rather directly: Outer Lindblad Resonance
Credit: Jo Bovy Which resonances drive spiral density wave growth?
To ensure that some arbitrary star can complete an epicyclic orbit in the same time it takes to move from one region in the spiral arm to another, the following condition must be satisfied:
some integer
m(Ωp - Ω) = nκ
Epicyclic (or radial) # of Spiral Arms Frequency
Orbital Frequency of Orbital (or Azimuthal) Spiral Arms Frequency of Stars on Circular Orbits
The only integers n for this relation that are interesting are 0, +1, -1. Which resonances drive spiral density wave growth?
This results in a number of well known resonances:
Inner Lindblad resonance: Most relevant cases:
Ωp = Ω − κ/m Ωp = Ω − κ/2
Outer Lindblad resonance:
Ωp = Ω + κ/m Ωp = Ω + κ/2 Corotational radius:
Ωp = Ω Ωp = Ω
In most cases, the only relevant case is that of two spiral arms, i.e., m = 2 At what orbital frequencies for the spiral arms are these resonances relevant?
Compute Ωp = Ω − κ/2, Ω, Ω + κ/2 3-4-12see http://www.strw.leidenuniv.nl/˜ franx/college/galaxies12 12-c02-9 3-4-12see http://www.strw.leidenuniv.nl/˜ franx/college/galaxies12 12-c02-10 How does the resonant frequencies vary by radius?
isochrone
Ωp potential Orbital frequency for spiral arms at which these resonances become important One thing you should note is the extended range in radius where the rotational frequency for one of these resonances, i.e., Ω − κ/2 is approximately constant. Radius 3-4-12see http://www.strw.leidenuniv.nl/˜ franx/college/galaxies12 12-c02-9 3-4-12see http://www.strw.leidenuniv.nl/˜ franx/college/galaxies12 12-c02-10
At what orbital frequencies for the spiral arms are these resonances relevant?
Compute Ωp = Ω − κ/2, Ω, Ω + κ/2 How does the resonant frequencies vary by radius?
model 1 for Ωp our Galaxy Rotational from BT 2.7 frequency for spiral arms at which these resonances become important Again Ω − κ/2 is almost independent of radius!
Radius What are the typical physical radii where these resonances Lindbladapply?Resonances III
Let’s zoom in on this plot and look at it more closely at the highly relevant m = 2 case Frequency
Rotational frequency for (typically ~20 kpc in spiral arms (typically a spiral galaxy like ~14 kpc) our own) (typical pattern speed for spiral (typically Corotation Outer Lindblad arms Ω ~3 kpc) Radius Resonance p p ~15 km/s / kpc) Ω Ω+κ/2 Ω Ω−κ/2 IILR OILR CR OLR Radius LindbladInnerResonances Lindbladpla y(Canimpor betant morerole for orbits in barred potentials. Resonances than one) How might this growth occur in practice?
Let’s say you have a strong bar in the galaxy (acts like a spiral arm with m=2) and you have particles leading or lagging the bar. What happens?
Diagram from Yanqin Wu What do these orbits look like in a rotating frame?
3-4-12see http://www.strw.leidenuniv.nl/˜ franx/college/galaxies12 12-c02-7 3-4-12see http://www.strw.leidenuniv.nl/˜Let’s choose a frame moving franx/college/galax at Ω − nκ/m ies12 12-c02-8 Outline Density wave theory Spiral structure Stochastic star formation model Stellar kinematics
In a disk where this propertyLet’s is look constant at a overrotating most frame radii wefor can a set get of orbits... the following situation, where the stars are forced in orbits that line up as a spiral pattern. Let’s choose a frame rotating at a fixed rotation speed of Ω − κ/2
In a coordinate frame, rotating with the pattern speed ⌦ = ⌦ /2, the spiral p pattern remains unchanged.
We can easily set up a series of epicyclic orbits for stars such that they look very much like a spiral shape: Piet van der Kruit, Kapteyn Astronomical Institute Stellar kinematics and spiral structure Additional Properties of Spiral Density Waves
Spiral density waves can only survive and grow between the inner Lindblad resonance and outer Lindblad resonance.
These waves cannot pass through the inner Lindblad resonance (they are damped inside this radius) Why are there Spiral Arms? Density wave theory: It only takes a few orbits for arms to become completely wound up and lose spiral structure! This is the “winding problem”. •! We think that spiral arms are caused by a density perturbation that moves along at a speed different from the speed of the objects within it. The density wave resists the spiral’s tendency to wind up and causes a rigidly rotating spiral pattern •! Think about what happens when there is a slow-moving car on a freeway … •! The spiral pattern is a density wave rotating through the galaxy at a fixed angular speed, called the pattern speed
What astrophysical processes drive these spiral density waves as they rotate around a spiral disk?
Spiral Density Waves Spiral Density Waves
•! There are initial “seed” perturbations in the When the gas in the spiral•! When density infalling wave gas collides is compressed, with gas in the spiral disk. These come from either initial density wave, stars formed, either due to asymmetries in the disk and/or halo (galaxy it results in the formation of stars (due to the high gas formation processes), or induced via galaxy simple Jeans criteria collapse, or induced encounters (like the M51 system) densities induced bythrough these shocks.compression waves)
•! Thus there are regions of slightly higher After the stars form,• !theyMaterial will willapproximately continue to drift move through at the density than their surroundings. The the circular velocity of thedensity spiral wave, galaxy though -- whichthe local is gravityoften higher density accelerates matter into the will cause a slight deceleration to the wave. faster than the patternmotion. speed of the spiral arm •! In the inner disk, stars move faster than the pattern speed and overtake the density The high mass stars•! Theformed high-mass in the stars spiral don’t density go far before wave; in the outer disk, the density wave compression waves die they(SNe go explosionssupernova or or otherwise otherwise) die. This overtakes the stars. Either way, material shortly after leaving theenhances spiral arm the visibilitycompression of the density wave, wave at bluer wavelengths. will encounter the wave. but the lower mass (redder) stars continue to rotate around the disk. Outline Density wave theory Spiral structure Stochastic star formation model Stellar kinematics
The strongest confirmationWhat astrophysical came from processes studies drive of thethese interstellar spiral density medium. waves as they rotate around a spiral disk? at inner radii in spiral galaxies, stars travel faster than the spiral density wave.
gas and dust lanes (formed from the metal output of the supernovae explosions) indicate the position of the high density spiral density wave
hot (massive) stars do not travel much beyond the spiral density wave in which they are formed
it is only the old (low Credit: van der Kruit the hot stars are somewhat ahead of the gas/ mass) stars that can dust lane, since there is some time lag between travel far enough to get when gravitational collapse begins and when the ahead of the spiral wave stars finally form (i.e., are on the main sequence) Piet van der Kruit, Kapteyn Astronomical Institute Stellar kinematics and spiral structure What astrophysical processes drive these spiral density waves as they rotate around a spiral disk? Density Wave theory C.C. Lin & F. Shu (1964-66) This is the preferred model • A Similar Schematic: for grand design spirals. The spiral arms are over- • dense regions which move around at a di↵erent speed than star: stars thus move in and out of the spiral arm How these density waves • are set up is unclear, but it may have to do with interactions. Once they are set up, they must last for a long enough time to be consistent with the observed number of spiral Credit:galaxies Ivezic
14 Where is theSpiral spiral Density structure Waves most evident? Grand design spiral – M51
Since the brightest (bluest) stars die beforeBecause leaving of the the hot spiral blue starsarm, being the spiralpredominantly formed in the densityspiral waves density must waves show in disk up galaxies better and at living for a very short time, we would expect the spiral structure to be much clearer at ultraviolet wavelengths. bluer or ultraviolet wavelengths where we just see the hot blue stars.
Density wave theory: NGC 3351, inner ring
•! Spiral arm pattern is amplified by resonances between the epicyclic frequencies of the stars (deviations from circular orbits) and the angular frequency of the spiral pattern –! Spiral waves can only grow between the inner and outer
Lindblad resonances ((p = ( - )/m ; (p = ( + )/m ) where ) is the epicyclic frequency and m is an integer (the # of spiral arms) –! Stars outside this region find that the periodic pull of the spiral is faster than their epicyclic frequency, they don’t respond to the spiral and the wave dies out
–! Resonance can explain why 2 arm spirals are more prominent •! We observe resonance patterns in spirals What is the Origin of these Spiral Waves?
How are these spiral density waves set up in disk galaxies in the first place?
Either from asymmetries in the dark and/or halo (galaxy formation processes)
Or from interactions with a nearby neighbor (as in the case of spiral galaxy M51) Another important concept in spiral galaxies is the idea of disk stability / instability Jeans Instability
Consider homogeneous fluid that is in equilibrium, with density ρ, pressure p, with no internal motion. Assume that the fluid is spherically symmetric. We shall consider the fluid is the matter inside some sphere with radius r.
Suppose that we compress the fluid element so that it now has a radius r(1-α/3) where α is much smaller than 1.
What will be the force acting on the surface of the sphere after this small compression?
To first order, the density perturbation is ρ1 = αρ
To first order, the pressure perturbation is 2 p1 = (dp/dρ)αρ = αρvs where vs is the sound speed. Jeans Instability
The pressure force per unit mass is
Fp = ∇p / ρ The gravitational force per unit mass is
The additional pressure force per unit Fg = ρG(4π/3)r mass is The additional gravitational force per 2 unit mass is dFp = ∇(αρvs )/ρ
2 2 dFp = αvs /r dFg = αGM/r
If the additional force on the surface of the sphere from the pressure of the fluid, i.e., dFp, is greater than the additional force on the surface of the sphere from gravity, i.e., dFg, then the pressure force resists the radial perturbation.
However, if the additional force from the gas pressure dFp is less than the additional force from gravity dFg, then the force of gravity will only accelerate the collapse. Jeans Instability
In summary, for dFp > dFg ==> fluid pressure resists gravitational collapse
However, for dFp < dFg ==> system undergoes gravitational collapse
dFp = dFg represents a specific physical scale.
2 2 dFp = αvs /r = αGM/r = dFg Using M = ρ 4/3 π r3,
2 vs /r = Gρ 4/3 π r We find: 2 1/2 rJ~ (3 vs / (4π Gρ))
Perturbations on a larger scale than the Jeans scale rJ will result in a gravitational collapse. 3-4-12see http://www.strw.leidenuniv.nl/˜ franx/college/galaxies12 12-c02-13 3-4-12seeToomre http://www.strw.leidenuniv.nl/˜ Instability franx/college/galax Criterionies12 12-c02-14
3-4-12see http://www.strw.leidenuniv.nl/˜ franx/college/galaxies12 12-c02-13 3-4-12see http://www.strw.leidenuniv.nl/˜ franx/college/galaxies12 12-c02-14 In spiral/diskσ to derive galaxies, the Jeans the stability mass. Hence, criterion if the systemis more is complex, due classic instability: Jeans mass sufficiently large, it will become unstable and fragment σ to deriveto the the Jeans shearing mass. Hence, type if motion the system is classic instability: Jeans mass into smallersufficiently pieces. large, it will become unstable and fragment Similar,into for smaller a disk pieces. one can obtain stability criteria. SimilarlyFor for aSimilar, cold a disk (fluid) for galaxies, a disk disk, one any canthere perturbations obtain is stabilityalso a withcriteria.stability wave- criterion. Any lengthFor larger a cold than (fluid)λcrit disk,will any be perturbations unstable with wave- perturbationslength larger with than wavelengthλcrit will be unstablelarger than λcrit are unstable. 2 2 λcrit =2π/kcrit =4π2 GΣ/κ2 λcrit =2π/kcrit =4π GΣ/κ
(BT6.65(BT6.65 page page 495) 495) RandomHowever, motionsHowever, random by random stars stellar stellar tend motions motions to stabilize will will tend tend todisks. to stabilize stabilize Disks with the diskthe disk (juststellar (just like like motions in in the the case case are of of theunstable the Jeans Jeans equation). equation).if DisksDisks with with stellar stellar motions motions are are unstable unstable if if σRκ Q = σRκ < 1 Q = 3.36 G Σ < 1 This is the Toomre criterion.3.36 G ItΣ assumes that the disk Thisis is very the thin, Toomre and that criterion. the unstable It assumes modes havethat wave- the disk is verylength thin, substantially and that the smaller unstable than the modes size of have the disk. wave- This is thelength ToomreAgain, substantially if thecriterion. dispersion smaller Itσ Rassume thanis high the enough, that size ofthe it thewill disk stabi- disk. is thin and lize the disk. unstable Again,modesFor if ourare the galaxy, dispersionmuchκ smaller= (37σR±is3) than highkm/sec/kpc enough,the size it of will the stabi- sky. Again, lize the disk. significant dispersion in the velocities± of stars will stabilize the For ourold galaxy, stars: σr κ== 38 (37± 2km/sec3)km/sec/kpc disk. −2 old stars:Σ∗σ=r 36=± 385M±⊙2/pckm/sec
Hence Q∗ =2.7 ± 0.4−2 Σ∗ = 36 ± 5M⊙/pc
± 2 HenceIncludingQ∗ =2 interstellar.7 0.4 gas: 13M⊙/pc , 7 km/sec − > Q=1.5 2 − Includingcombined interstellar effect of gas: gas and13M stars⊙/pc is worse, 7 km/sec> nearly− > For a stellar system, we simply need to replace vs with Q=1.5unstable combined effect of gas and stars is worse − > nearly For a stellar system, we simply need to replace vs with unstable What happens if Q < 1?
14.3. DISPERSION RELATIONS & THE Q PARAMETER 103
This is an illustration of how the disk becomes clumpy because of a Toomre-like instability.
Credit: Barnes
Figure 14.1: Unstable disk simulation. Initially, this disk has Q 0 63, producing the violent, essentially local instability seen here. The galaxy model includes a bulge and a halo (not shown); What is Qthe diskfor is 15% of the totalour mass, and a version galaxy? with a higher Q value is stable (Fig. 15.2). Each frame is 15 disk scale lengths on a side; times are given in units of the rotation period at 3disk scale lengths.
for a stable gas disk, and κσR For our disk where κ = 37±3 km/s/kpc,Qstars σR =1 38±2 km/s,(14.25) Σ = 3 36GΣ for a stable stellar2 disk (Toomre 1964). 36±5 M⦿ / It’spc worth comparing, Q* (14.25), = the2.7±0.4. product of a WKB analysis, with (14.13), which was derived by simple physical considerations. The condition Qstars 1mayberearrangedtogive GΣ σR 3 36 2 (14.26) Including interstellar gas 13 M⦿ / pcκ , Q = 1.5. This is very similar to (14.13); apart from the numerical factor, the only difference is that the epicyclic frequency κ replaces the Oort constant B in the denominator. What about the presence of bars in spiral galaxies?
Bars likely act very similar to spiral structure in disk-like galaxies.
For bars also, one has m=2 symmetry, much like most grand- design spirals. Next topic is elliptical galaxies... Elliptical galaxies consist of large numbers of stars on diverse orbits.
While spiral galaxies are rotation supported, elliptical galaxies are supported by the random motions of stars they contain
Their behavior can largely be described using collisionless dynamics. We will therefore be reviewing some concepts from collisionless dynamics from the Leiden Bachelor course
(this will take ~1.3 lectures) 8. Stars dont collide calculate from typical density, velocity, and cross- section of stars. To understand equilibrium, one can ignore stellar collisions. The main challenge is to understand a system with > 1011 particles moving in their common gravitational potential.
9. Individual interactions between stars can be ignored. Hubble sequence: ellipticals, S0s, spirals (barred and unbarred), Sa-Sd, irregulars, mergers The relaxation time is the time needed to a↵ect the motion of the star by indi- vidual star-star interactions. This timescale is very high (1016 years) - too high to be relevant for galaxies. As a re- sult, we can describe the potential as a smooth, 3D, potential ⇢(~x)), instead of the potential produced by 1011 particles. Similarly, we can derive the phase-space density (a 6D function f(~x,~v)), by calculating the average density in space and velocity of the 1011 Keyor concepts more particles. from bachelor course
10. Galaxy formation is driven by the formation of the halo At very high redshift, the density of the universe is close to critical. Any perturbation will drive the density to above critical. Regions with density above critical density will collapse and become self-gravitating.
11. Galaxies form inside the collapsed halos, as the gas can cool radia- tively, become self-gravitating, and form stars.
12. The equations of motion for a self-gravitating system are simple. They imply that any system can be scaled in mass, size, and velocity if the scaling factors satisfy one single relaiton This implies that any equilibrium model can be scaled in mass and size !
13. In practice, the halos grow in mass due to merging with smaller halos. Merging is very e↵ective in the universe, due to the fact that the halos are extended and not point-sources.
14. The time-evolution of the distribution function is defined by the dis- tribution function at that time, spatial derivatives, and the gradients in the potential (Vlasov-Equation). This follows directly from the conservation of stars. 15. Integrals of motion are functions of x and vvwhichareconstantalong an orbit. They are not explicit functions of time. Examples: Energy; angular momentum in a spherical potential. Most 3D densities allow for 3 integrals of motion, 2 of which are non-classical.2
16. Equilibrium models can easily made by requiring that the distribution function is a function of integrals of motion. By definition, the distribution function is an integral of motion itself.
17. Another way to construct equilibrium models is to make a library of orbits, calculate their spatial densities, and calculate the weight function which reproduces the density distribution for which the orbits were calcu- lated. This is Schwarzschilds method.
18. Using modern telescopes, we can trace galaxies to z =6andbeyond, and study galaxy formation and evolution directly.
3 15. Integrals of motion are functions of x and vvwhichareconstantalong an orbit. They are not explicit functions of time. Examples: Energy; angular momentum in a spherical potential. Most 3D densities allow for 3 integrals Key ofconcepts motion, 2 of from which are bachelor non-classical. course
16. Equilibrium models can easily made by requiring that the distribution function is a function of integrals of motion. By definition, the distribution function is an integral of motion itself.
17. Another way to construct equilibrium models is to make a library of orbits, calculate their spatial densities, and calculate the weight function which reproduces the density distribution for which the orbits were calcu- lated. This is Schwarzschilds method.
18. Using modern telescopes, we can trace galaxies to z =6andbeyond, and study galaxy formation and evolution directly.
3 MOTIVATION: We have this very complicated situation: how can we model the orbits of all the stars in a galaxy simultaneously
Seems difficult!
Before even thinking about how to solve it, how shall we even try to model it?
Ignore the fact that stars are discrete sources and assume that we treat them as a fluid with each star individually having an infinitesimal mass.
We will use a 7-dimension distribution function to describe where they are in 6-dimensional phase space at some time t. 8. Stars dont collide calculate from typical density, velocity, and cross- section of stars. To understand equilibrium, one can ignore stellar collisions. The main challenge is to understand a system with > 1011 particles moving in their common gravitational potential.
9. Individual interactions between stars can be ignored. Therelaxation time is the time needed to affect the motion of the star by individual star-star interactions. This timescale is very high (1016 years) - too high to be relevant for galaxies. As a result, we can describe the potential as a smooth, 3D, po- tential ρ(⃗x)), instead of the potential produced by 1011 particles. Similarly, we can derive the phase-space density (a 6D function f(⃗x,⃗v)), by calculating the average density in space and velocity of the 1011 or more particles.
10. Galaxy formation is driven by the formation of the halo At very high redshift, the density of the universe is close to critical. Any perturbation will drive the density to above critical. Regions with density above critical density will collapse and become self-gravitating.
11. Galaxies form inside the collapsed halos, as the gas can cool radia- tively, become self-gravitating, and form stars.
12. The equations of motion for a self-gravitating system aresimple. They imply that any system can be scaled in mass, size, and velocity if the scaling factors satisfy one single relaiton This implies that any equilibrium model can be scaled in mass and size !
13. In practice, the halos grow in mass due to merging with smaller halos. Merging is very effective in the universe, due to the fact that the halos are extended and notREVIEW point-sources. (from Bachelor course) 14. The time-evolution of the distribution function is defined by the dis- tribution function at that time, spatial derivatives, and the gradients in the potential (Vlasov-Equation). This follows directly from theconservationof stars.
At any time t, one can describe the collective positions and velocities for stars in a dynamical system by a distribution function f(x,v,t)
To describe the time evolution, we2 define a six dimensional vector w = (x,v)
The flow of stars in the six dimensional phase can be described as dw/dt = (v,-∇Φ)
The flow dw/dt conserves stars...
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Consider a system with a large nunberConsider of stars a systemf/www.strw.leidenuniv.nl/" with a large nunber franx/coliege/ of stars mf-sts-07-c5-l W15-10-07 e L see ^ http://www.strw.leidenuniv.nl/" r v ^ a ^ ^ franx/college/ mf-sts-07-c5-2 Collisionless BoltzmannCollisionless Equation Boltzmann f(j}\ci Equation ^Oy^ < f(j}\ci ^Oy^star < is given by star is given by At any t define the distributionAt function any t f(x,v,t)dxdvdefine the distribution functionHence we f(x,v,t)dxdv have the continuity equation:Hence we have the continuity equation: = # of stars in volume dx with velocities= # of stars in range in volume dv dx with velocities in range dv Collisionless BoltzmannBT 4.1 p. 190-193 EquationBT 4.1 f(j}\ci p. 190-193 ^Oy^ < star is given by (centered on x, v). 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The in second range dv term Atis any equal time t, w to:one ume.= can(2?, describe The v) =second the (v, collective — V$)term positions is equal and velocities to: We now have reduced QN functionsWe Xi,Vi now have into one reduced 7- QN functions Xi,Vi into one 7- for stars in__/ a dynamical systemd(fWg) by a distribution= 0 functiond(fWg) f(x,v,t) BT 4.1 p. 190-193 + £ __/w =+ (2?, £ v) = (v, —= V$) 0 DIMENSIONAL/(xjV^t) FUNCTION. is called /(xjV^t) theDIMENSIONAL(centered distribution is called on x, function the FUNCTION. v). distribution or the phase function or the phase at dwa at dwa W eThe L ^ flow r w conserves v ^ a ^ stars.a=l ^ __/ a=l d(fWg) = 0 spaceConsider densityf/www.strw.leidenuniv.nl/" a systemspace with density a large franx/coliege/ nunber mf-sts-07-c5-lof stars 15-10-07/ v-(/ + =0 “Vlasov Equation” + ⃗v. f Φ. =0 bution functions, they ∂ aret much simplerwα than the 6N Hence the CBE∂t can∇ be written− ∇ ∂⃗ asv ignore individual particles ! α!=1 This is theCollisionless Boltzmann surface integraldw over the(dvi flow . out dvi of the 1-dimensional functions.) volume. Hence the equation guarantees that+ stars are These equations are the Collisionlessdf Boltzmanndvi Equa- (Notice thator we can always rewrite f(x,v,t) as a sum conservedtion (the (CBE). densityalso cancan write only as increase=0 if stars move mationIn of order S functions. to derive We the would 6 time evolution, then get back first define our a new dt ∂f ∂f ∂w˙ α into theNoticeThe volume). CBE that is by su ffi definitioncient to calculate dvi/dxi = the 0 because evolution x* of and any coordinate w: + .w˙ α + f =0 "original" particles. 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The flow of the flow-line). cc=l 6 dvi In order to derive the time ∂ evolution,f first ∂ definef a new + w˙ α =0 Notice that by definition dvi/dxi = 0 because x* and coordinate w: ∂t ∂wα α!=1 Vi are independent coordinates. The second term is OrW we = write (x,v) this = as (W1,«72,...W6) equal to A d / d $ \ where w\ = x\, W2 = #2,∂f ■•■ 3 w* =∂ vi,f etc.∂ Φ Hence∂f the star has coordinate w in phase-space.+ v The flow of=0 the dvi \ dx. ∂t i ∂x − ∂x ∂v cc=l !i=1 i i i or ∂f ∂f + ⃗v.⃗ f ⃗ Φ. =0 ∂t ∇ − ∇ ∂⃗v These equations are the Collisionless Boltzmann Equa- tion (CBE). The CBE is sufficient to calculate the evolution of any f with time. Adifferent description of the same equation: consider the evolution of f if one moves along with a particle (this is the Lagrangian derivative):