Checklist Flashcards

Question

What sort of galaxy are we in?

Answer 1

Disk galaxy (spiral) Small-medium bulge Bar 2 major spiral arm (+ minor arm fragments) Likely Hubble type SBbc

Answer 2

Atomic Hydrogen (HI - radio wavelengths) Molecules, eg carbon monoxide (at mm wavelengths) Dust (Far infrared ~60 μm ) Old stars (Near infrared ~2 μm Original observations were affected by dust and confusion with Gould's belt (local feature, inclined ~20° to plane, likely part of spiral arm)

Answer 3

Bulge: R ~ 3kpc ~ 2kpc scale height ~10¹⁰ M☉︎ Vcirc ~ 100 km/s Thin Disk: Scale height ~100-300 pc 6 x10¹⁰ M☉︎ Young stars, 0-9 Gyrs Thick disk: Scale height ~1000 - 1300 pc Older stars, 10-12 Gyrs Stellar halo: ~ 15-20 kpc radius ~10kpc scale height ~2 x10⁹ M☉︎ 13 Gyrs old Pop II stars ~150 known globular clusters

Answer 4

Sgr A * ~10⁸ km ~ 1AU Sgr A * is stationary, occupies dynamical centre of Galaxy and lack of dynamics suggest it is massive. Sgr A contains circum-nuclear rotating molecular disk, density 10⁵ cm⁻³, mass 3 x10⁴ M☉︎, Vrot ~100km/s, radius 2pc Stars near central source Sgr A * observed over several years to determine orbits → Keplerian motion about Sgr A * . Velocity dispersion within 0.1pc is consistent with Keplerian V ∝ R^-0.5 Implies most mass of Sgr A is confined to centre (Sgr A * ) Orbits used to calculate mass of Sgr A * , current value M = 4.297 x10⁶ M☉︎

Answer 5

Extreme Pop I thin disk region: * O and B stars * Very young stars: ~ 10⁷ years * metal rich: Z ~ 2-4% | [Fe/H] ~ 0 to 0.5 Thin disk: * G, K and M type stars * Young(ish) stars: 0-9 Gyrs * metal rich: Z ~ 0.5-4% | [Fe/H] ~ -0.5 to 0.5 Thick disk: * Older stars: 10-12 Gyrs * less metal rich: [Fe/H] ≃ -0.6, tail down to ≃ -1.5 Bulge: * similar ages as thick disk * [Fe/H] ~ -1 to 0 Halo: * age 12-14 Gyrs * [Fe/H] ≃ -1 to -5

Answer 6

Random: Gaussian distribution with width σ. This is the velocity dispersion. Probability of star having random velocity v is P(v)dv = 1/σ√2π' exp(-v²/2σ²) dv Rotational component: orbital speed Vcirc of stars around a galaxy

Answer 7

Thin disk: σ ~ 20 km/s | Vcirc ~ 220 km/s Thick disk: σ ~ 40-60 km/s | Vcirc ~ 180 km/s Bulge: σ ~ 60-100 km/s | Vcirc ~ 100 km/s Stellar halo: σ ~ 100 km/s | Vcirc ~ 40 km/s

Answer 8

Pop I - ordered motion, circular orbits in the disk plane. Pop II - random motion, eccentric orbits passing through disk plane

Answer 9

Initial Mass Function Ψ(m) specifies the distribution in mass of a newly formed stellar population and it is frequently assumed to be a simple power law. * Mass range from ~0.1 – 100 M⊙ * Number of stars of a certain mass range can be calculated by ∫Ψ(m)dm in desired mass range * Total mass, ∫mΨ(m)dm * Salpeter IMF, for example, takes form Ψ(m) = Cmˣ, x = -2.35 * Salpeter breaks down at low masses, and x = -1.3 for 0.08 M⊙ < m < 0.5 M⊙

Answer 10

Stellar number density follows an exponential law n(z) = n₀exp(-|z|/hᶻ) hᶻ is disk scale height and n₀ is stellar number desity in disk plane.

Answer 11

O and B stars in extreme Pop I thin disk region hᶻ ~ 60 pc G, K, M stars in thin disk hᶻ ~ 200-300 pc

Answer 12

Difference between thick and thin disk rotation, ~ -40 km/s, thick slower than thin.

Answer 13

Average velocity of disk stars within ~50 pc Sun's velocity relative to LSR is ~17km/s towards l ~+56°, b ~+23°

Answer 14

m d²r/dt² = -m∇ϕ, r in cylindrical polar. ϕ is gravitational potential determined from Poisson's equation: ∇²ϕ = -4πGρ

Answer 15

Studying motions of stars gives a value for density of matter in the solar vicinity. Comparing this to the mass of all observable matter shows discrepencies, which can be explained using dark matter. The Oort limit, the dynamical mean density of matter in solar vicinity, has been determined to be ρ₀ = 0.076±0.015 M☉︎ pc⁻³. Observable matter estimated mass ρ ≃ 0.08 M☉︎ pc⁻³. Observable matter accounts for all dynamical mass, so there is no evidence for dark matter in the disk.

Answer 16

Gaseous disk: * Flares in outer regions of galaxy. * HI: solar vicinity hᶻ ~ 150pc | outer galaxy hᶻ ~ 1kpc * CO: solar vicinity hᶻ ~ 60pc | outer galaxy (16kpc) hᶻ ~ 200pc * CO: Highly concentrated towards plane, mainly within inner 5kpc of galaxy. Used to trace H₂ * Dust: far-infrared emission, similar distribution to HI and CO

Answer 17

mᵥ - Mᵥ = 5log₁₀(d/10pc) + Aᵥ Aᵥ is optical extinction. Aᵥ ~ 1 mag/kpc (clumpy, not smooth) Aᵥ ∝ N(HI) (hydrogen column density)

Answer 18

HI ~ 150pc H₂ ~ 60pc

Answer 19

Origin: Gas ionised from UV radiation emitted by massive O and B type stars (Photons E>13.6). HII regions surround these stars, known as Stromgren spheres. O & B stars typically located 0-60pc from central plane, so HII originates there. HII T~10^4 K, surrounding neutral gas T~100 K Ionised pressure > neutral gas pressure, so HII region expands to form diffuse ionised medium along plane.

Answer 20

Oort’s constant A expresses local shear, velocity relative to LSR. A = ½(V₀/R₀ - (dV/dr)|R₀) = ½r(dω/dr)|R₀ Oort’s constant B expresses local vorticity, tangential velocities B = -½(V₀/R₀ + (dV/dr)|R₀) = -½r(dω/dr + 2ω/r)|R₀ They can be determined by plotting Vᵣ/d and Vₜ/d against galactic longitude. Vᵣ/d = Asin(2l) gives A = 14.8 ± 0.8 km s⁻¹ kpc⁻¹ Vₜ/d = Acos(2l) gives B = -12.4 ± 0.6 km s⁻¹ kpc⁻¹ They rule out solid body rotation, as A is not equal to 0. Used to estimate local differntial rotation, A+B = -(dV/dr)|R₀ Flat rotation curve = V constant, so (dV/dr) = 0, aka A=-B This is consisten with observed values, so rotation curve approximately flat in vicinity of Sun. Also used to calculate local circular speed A-B = V₀/R₀ Vcirc ~ 210 km s⁻¹ | |R₀ means evaluated at R₀

Answer 21

Atomic hydrogen - broad distribution from 4 - 15 kpc Molecular, ionised gas, and supernova remnants - peak between 4 - 7 kpc

Answer 22

We know how to model galactic rotation, so can calculate distance by comparing radial velocity to our model. Limitations: * assumes circular velocity, invalid for r < 3kpc * requires small intrinsic (random) motion * inaccurate near l = 0° and 180° as Vᵣ ~ 0 km/s * potential distance ambiguity for r < R₀ * doesnt work for r ~ R₀ as Vᵣ ~ 0 km/s

Answer 23

if r > R₀: Vᵣ = (ω(r) - ω₀)R₀ sin(l) ω(r) = V(r)/r combine to get V(r)/r = Vᵣ/R₀sin(l) sub in V₀ = 220 km/s | R₀ = 8 kpc | V(r) = V₀ (flat rotation curve) if r < R₀: ω(r) could have one of two values Vᵣ = (ω(r) - ω₀)R₀ sin(l) rearrange to get V(r)/r = Vᵣ/R₀sin(l) + V₀/R₀ sub in V₀ = 220 km/s | R₀ = 8 kpc | V(r) = V₀ (flat rotation curve) both: galactocentric distance r = V(r)/ω(r) can then use cosine rule to get two possibilities for d. can rule out a possibility if d is -ve

Answer 24

Line spectra are shifted differntly depending on longitude. for 0° < l < 90°, redshift corresponds to r < R₀, blueshift to r > R₀. Opposite way round for 90° < l < 180°. Shift corresponds to radial velocities, and plotting these velocities against longitudes allows for a sort of "map" of gas. Spiral arm and bar like structures can be seen . | Note that central 3kpc affected by non-circular motions.

Answer 25

l - v plot: central '3 kpc arm' structure implies bar motion infrared: asymmetry in star counts about l = 0° indicates a 'near' and 'far' side, consistent with bar-like geometry

Answer 26

Sgr A radio source - supernova remnant Sgr A East, spiral structure Sgr A West, central source Sgr A * , likely a massive black hole

Answer 27

3 rotation candidates are solid body (Vcirc ∝ r), Keplerian (Vcirc ∝ r^⁻½) and flat (Vcirc = constant) Not Keplerian, not solid body, but flat-ish Rotational velocity is much larger than that expected based on visible matter only, and this additional velocity is attributed to the dark matter halo.

Answer 28

Disk: exponential law I(r) = I₀ exp(-r/r₀) where r₀ = characteristic length and I₀ = central surface brightness Bulge: follows r^1/4 law, ellipticals only have this part I(r) = Iₑ exp{ -7.67( {r/Rₑ}^1/4 - 1)} where Rₑ = radius containing half of total projected luminosity and Iₑ = intensity at Rₑ

Answer 29

HI mass can be measured if telescope beam > galaxy extent. Mᴴᴵ/M☉︎ = 2.36 x 10¹⁹D² ±∞∫ Sᵥ dVᵣ where Sᵥ is flux density in W/m^2/Hz and Vᵣ is radial velocity in km/s and D is distance in pc Total HI mass % (wrt visible matter) increases from Sa to Sc (~2-7%). Irr galaxies have the most (10-15%) Ellipticals have the least (≪ 1%)

Answer 30

More mass, faster rotation, broader HI line width (doppler shift). Consider outermost HI gas cloud, assume all mass contained within its orbit, equate centripetal and gravitational forces. mV²/R ~ GMm/R² rearrange, M ~ RV²/G using ΔV ~ 2V/sin(i) (i is inclination angle. i=0° face on, i=90° edge on, estimated assuming circular disk, i = cos⁻¹(b/a)) M ~ R(ΔVsin(i))²/4G

Answer 31

All empirically derived, all useful for distance calculations. Tully-Fisher: * luminosity of a galaxy is directly proportional to the fourth power of its rotational velocity * L ∝ Vcirc⁴ ∝ (ΔVsin(i))⁴ * Intrinsic luminosity gives absolute magnitude * Comparison with apparent magnitude gives distance * Scatter ±0.2 magnitudes Luminosity-Dispersion: * Relation between Luminosity L and the Stellar dispersion σ of elliptical galaxies. * Scatter ±2 magnitudes Fundamental Plane: * Rₑ, Σₑ and σ₀ are related by Rₑ ∝ σ₀^1.4 Σₑ^-0.85 * surface in 3D ⇒ ‘The Fundamental Plane’ * Rₑ is the half light radius * Σₑ is the mean surface brightness within Rₑ: Σₑ = 1/2 L/πRₑ² * σ₀ is the central velocity dispersion

Answer 32

Flat rotation curves mean that inner parts of galaxy have higher angular velocity than outer parts, hence spiral arms shoud 'wind up'. Rotation period ~ 100-400 million years, spiral arms should wind up in ~10^8 years. Implies Sa may be evolution of Sc, however not the case as timescales unrealistic. Sa much older than 10^8 years, Sa more massive than Sc, and Sa have significantly lower gas content (10^8 years too short for such a change)

Answer 33

Spiral arms are waves of stellar density. Arm pattern moves around the galaxy at constant angular velocity (pattern speed) but different circular speed Stars and gas move at approx. constant circular speed vcirc (km/s) and variable angular velocity w(r). Perturbations cause radial oscillations in orbits, so they are slightly elliptical. Conservation of angular momentum means matter moves slightly faster and slower at differnt points, 'slow' regions have higher stellar density. A series of tilted ellipticals can create a spiral of density maxima.

Answer 34

* Spiral arms are over-dense regions of the disk which move round at a different speed to the stars themselves * Stars move in and out of the spiral arm, alleviating winding dilemma * Existence of rings in centres of spirals (ILR) * Prevalence of 2-armed spirals

Answer 35

Stochastic self-propagative star formation. Star formation triggers more star formation nearby, then galactic rotation creates appearence of spiral. "Stochastic" because there is a small probability of random star formation in all areas in the disk

Answer 36

Relaxation time for stellar interactions in galaxies is shorter than galactic lifetimes, but longer within globular clusters.

Answer 37

Stellar orbits are superposition of circular orbits (rad R, ang freq ω) with small retrograde epicycles (ang freq κ) κ(r) is epicyclic frequency required for long-lived spiral structure κ²(r) = (2ω)² (1+ r/2ω ∂ω/∂r) For a circular orbit, R, θ = ωt, z = 0 For an epicyclic perturbation, x = small radial deviation, ε = small azimuth deviation, Φ(R,z) is gravitational potential Epicyclic motion: r = R + x θ = ωt + ε

Answer 38

Oorts constants A = ½(V₀/R₀ - (dV/dR)|R₀) B = -½(V₀/R₀ + (dV/dR)|R₀) A - B = V₀/R₀, A+B = -(dV/dR)|R₀ ω = v/r dv/dr = d/dr(rω) = r dω/dr + ω substituting gives κ₀²(R₀) = -4B (A - B)

Answer 39

Φ₁(r, θ, t) = Φ₁(r)exp( -i(mΩₚt - mθ + ψ(r)) where m = number of arms (usually m=2) Ωₚ = pattern speed ψ(r) = phase term determining pitch of arms Total potential is symmetric potential + spiral potential (Φ₁) Stars will experience gravitational potential varying with frequency m( ω(r) - Ωₚ )

Answer 40

Stars and gas radially oscillate with epicyclic frequency κ. Density wave produces periodic variation in global galactic potential Φₜₒₜ. When frequencies are equal, κ = ±m(ω(r) - Ωₚ), (m normally 2) Ωₚ = ω(r) ± κ/2 Inner Lindblad Resonance (ILR) Ωₚ = ω(r) - κ/2, Resonance is in inner part of galaxy Stars and gas have higher velocity than the density wave and so ‘catch up’ with the density wave. Often seen as a ring around the centre of a galaxy, spiral arms start from this ring Outer Lindblad Resonance (OLR) Ωₚ = ω(r) + κ/2, Resonance is in outer part of galaxy Stars and gas have a lower velocity than the density wave so the density wave ‘catches up’ the stars and gas. Rarely observed, not as important as ILR. Density waves can only propagate in the region between the ILR and OLR, and this region is largest for 2 armed spirals.

Answer 41

When Ωₚ = ω(r) stars and gas have the same angular velocity as the density wave. This occurs at the corotation radius

Answer 42

Formed in early universe, mostly in first billion years. Disk galaxies instead formed more uniformly over many billions (~10)

Answer 43

Eₙ: E0 - E7 where n = 10(a-b)/a

Answer 44

Ellipticals are isophotes, approx. elliptical in shape. Spherical: a=b=c Oblate: a=b > c, flat-ish Prolate: a=b < c, pulled-ish in direction of axis Triaxial: a ≠ b ≠ c

Answer 45

log I(r) ∝ r¼ I(r) = Iₑ exp( -7.67( (r/Rₑ)^1/4 -1) ) Rₑ = radius within which half of the total light is contained Iₑ = surface brightness at Rₑ | I is capital i

Answer 46

* Stellar absorption spectra * Stars are old, with low mass, so have low luminosity. * Strong absorbtion lines due to metals in stellar population. * Average spectrum of many stars in line of sight * Line width Δλ is measure of stellar velocity dispersion σ * Variation of λ as a function of distance from the centre of the galaxy can tell us about net rotation * Massive ellipticals show – Low rotation velocities ~ 50-100 km/s – High dispersion velocities ~200-300 km/s * Velocity/dispersion ratio v/σ typically ranges from 0 to 1 * To be supported by rotation, v/σ ~ (ε/1-ε) where ε = 1 - b/a

Answer 47

* Relates kinetic energy to its gravitational poteltial energy * 2T + Ω = 0 * assumes system in a stable statistical equilibrium under its own gravitation * assumes stellar interactions are negligible

Answer 48

T = ½ Σmᵢvᵢ² = ½M⟨v²⟩ M is total mass of stellar system and ⟨v²⟩ is mean value of vᵢ² Ω = -Σᵢ Σᵢ≠ⱼ Gmᵢmⱼ/rᵢⱼ = -αGM²/Rₑ α is a factor of order unity Combined, M = Rₑ⟨v²⟩/αG ~ Rₑ⟨v²⟩/G (α ~ 1)

Answer 49

* Large stellar halo * Luminosities typically 10x Lstar (recall Schechter luminosity function: Lstar = 2x1010 L⊙) * Halo extends 50-100kpc * Origin – likely formed from mergers of galaxies in the cluster core

Answer 50

r = (m/2M)^1/3 R R is critical distance of approach, corresponds to the ‘Roche limit’ for M to disrupt m.

Answer 51

Groups: * A few to 10 bright (Lstar = 2 x10¹⁰ L⊙) members * ~100 dwarf galaxies * Mass ~5 x 10¹² to 1 x 10¹⁴ M⊙ * Sizes ~ 1Mpc Clusters: * Around 100 bright (Lstar = 2x10¹⁰ L⊙) members * Thousands of dwarfs * Mass around 10¹⁴ to a few x 10¹⁵ M⊙ * Size typically around 3 Mpc * Separation between clusters ~ 50 Mpc

Answer 52

UV continuum - works with little dust, but high star formation often corresponds with high extinction Far infrared - reemission of UV

Answer 53

* Age correlates with vertical velocity * Young stars formed close to the disk plane * Old stars formed at range of heights above the plane * must be fast collapse to produce eccentric orbits for old stars

Answer 54

Fg = GMm/r^2

Answer 55

Fc = mv^2/r

Answer 56

Ug = −GMm/r

Answer 57

K = 1/2 mv^2

Checklist Flashcards

(87 cards)