Cluster modelling. Collisions. Stellar Dynamics & Structure of Galaxies handout #2. Just as a self-gravitating collection of objects.

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1 Stear Dynamics & Structure of Gaaxies handout # Custer modeing Just as a sef-gravitating coection of objects. Coisions Do we have to worry about coisions? Gobuar custers ook densest, so obtain a rough estimate of coision timescae for them. ρ M pc 3. M 0.8 M. n pc 3 is the star number density. We have σ r 7 km s 1 as the typica 1D speed of a star, so the 3D speed is 3 σ r (= σ x + σ y + σ z) 10 km s 1. Since M R (see Fuids, or Stars, course notes), have R 0.8R. For a coision, need the voume π(r ) σt co to contain one star, i.e. n 0 = 1/ ( π(r ) σt co ) (1) or t co = 1/ ( 4πR σn 0 ) () Putting in the numbers gives t co 5 10 s yr. So direct coisions between stars are rare, but if you have 10 6 stars then there is a coision every 10 9 years, so they do happen. 1

2 So, for now, ignore coisions, and we are eft with stars orbiting in the potentia from a the other stars in the system. Mode requirements Have a gravitationa potentia we Φ(r), approximatey smooth if the number of partices >> 1. Conventionay take Φ( ) = 0. Stars orbit in the potentia we, with time per orbit (for a gobuar custer) R h /σ 10 6 years << age. Stars give rise to Φ(r) by their mass, so for this potentia in a steady state coud average each star over its orbit to get ρ(r). The key probem is therefore sef-consistenty buiding a mode which fis in the terms: Φ(r) stear orbits ρ(r) Φ(r) (3) Note that in most observed cases we ony have v ine of sight (R), so it is even harder to mode rea systems. Sef-consistent = orbits & stear mass give ρ, which eads to Φ, which supports the orbits used to construct ρ. Basics Use: Newtons aws of motion Newtonian gravity

3 [Genera Reativity not needed, since v km s 1 is << c = km s 1 and GM 10 9 (gobuar custer) << 1.] rc The gravitationa force per unit mass acting on a body due to a mass M at the origin is f = GM ˆr = GM r r r (4) 3 We can write this in terms of a potentia Φ, using ( 1 r) = ˆr ( )1 1 (+0 r r.r ˆθ + 0 ˆφ) = 1 ( )3 1 ˆr r.ˆr r.r = 1 r ˆr (5) So f = Φ (6) where Φ is a scaar, Φ = Φ(r) = GM r (7) Hence the potentia due to a point mass M at r = r 1 is Φ(r) = GM r r 1 (8) 3

4 Orbits Partice of constant mass m at position r subject to a force F. Newton s aw: i.e. If F is due to a gravitationa potentia Φ(r), then d (mṙ) = F (9) dt m r = F (10) F = mf = m Φ (11) The anguar momentum about the origin is H = r (mṙ).then dh dt = r (m r) + mṙ ṙ = r F G (1) where G is the torque about the origin. The kinetic energy T = 1 mṙ.ṙ (13) dt dt = mṙ. r = F.ṙ (14) If F = m Φ, then dt dt = mṙ. Φ(r) (15) But if Φ is independent of t, the rate of change of Φ aong an orbit is (from the chain rue). Hence is constant for a given orbit. d dt Φ(r) = Φ.ṙ) (16) dt dt = m d Φ(r) (17) dt m d ( ) 1 dt ṙ.ṙ + Φ(r) = 0 (18) E = 1 ṙ.ṙ + Φ(r) (19) 4

5 Orbits in spherica potentias Φ(r) = Φ( r ) = Φ(r), so f = Φ = ˆr dφ dr. The orbita anguar momentum H = mr ṙ, and dh dt = r mf = mdφr ˆr = 0. (0) dr So the anguar momentum per unit mass h = H/m = r ṙ is a constant vector, and is perpendicuar to r and ṙ the partice stays in a pane through the origin which is perpendicuar to h. [If you want this in more detai - r h, r + δr = r + ṙδt h since both r and ṙ h, so partice remains in the pane.] Thus the probem becomes a two-dimensiona one to cacuate the orbit use -D cyindrica coordinates (R,φ,z) at z = 0, or spherica poars (r,θ,φ) with θ = π. So, in D, use (R,φ) and (r,φ) interchangeaby.. Equation of motion in two dimensions The equation of motion in two dimensions can be written in radia anguar terms, using r = rˆr = rê r + 0ê φ, so r =(r, 0). We know that and d dtêr = φê φ (1) d dtêφ = φê r () 5

6 [To see this: In time δt ê r ê r + δφê φ, and hence the first resut above, and in the same time interva ê φ ê φ δφê r, which gives the second.] Hence [or ṙ = v =(ṙ,r φ)] and so ṙ = ṙê r + r φê φ (3) r = rê r + ṙ φê φ + ṙ φê φ + r φê φ r φ ê r = ( r r φ )ê r + 1 d ( r φ)êφ r dt = a = [ r r φ, 1 d ( r φ) ] r dt (4) In genera f =(f r,f φ ), and then f r = r r φ, where the second term is the centrifuga ( force, since we are in a rotating frame, and the torque rf φ = r φ) (= r f). d dt In a spherica potentia f φ = 0, so r φ is constant. Path of the orbit To determine the shape of the orbit we need to remove t from the equations and find r(φ). It is simpest to set u = 1/r, and then from r φ = h obtain φ = hu (5) Then ṙ = 1 u u = 1 u du dφ φ = h du dφ 6 (6)

7 and So the radia equation of motion becomes r = h d u dφ φ = h u d u dφ. (7) r r φ = f r h u d u dφ 1 u h u 4 = f r (8) d u dφ + u = f r h u (9) Since f r is just a function of r (or u) this is an equation for u(φ), i.e. r(φ) - the path of the orbit. Note that it does not give r(t), or φ(t) - you need one of the other equations for those. If we take f r = GM r = GMu, then d u dφ + u = GM/h (30) (which is something you wi have seen in the Reativity course). The soution to this equation is r = u = 1 + e cos(φ φ 0) (31) which you can verify simpy by putting it in the differentia equation. Then e cos(φ φ 0) e cos(φ φ 0) = GM h so = h /GM and e and φ 0 are constants of integration. Note that if e < 1 then 1/r is never zero, so r is bounded in the range < r <. Aso, in a cases the orbit is symmetric about φ = φ 1+e 1 e 0, so we take φ 0 = 0 as defining the reference ine for the ange φ. is the distance 7

8 from the origin for φ = ± π (with φ measured reative to φ 0). We can use different parameters. Knowing that the point of cosest approach (periheion for a panet in orbit around the sun, periastron for something about a star) is at /(1 + e) when φ = 0 and the apheion (or whatever) is at /(1 e) when φ = π, we can set the distance between these two points (= major axis of the orbit)=a. Then 1 + e + 1 e = a (1 e) + (1 + e) = a(1 e ) (3) = a(1 e ) (33) r P = a(1 e) is the periheion distance from the gravitating mass at the origin, and r a = a(1+e) is the apheion distance. The distance of the sun from the midpoint is ae, and the anguar momentum h = GM = GMa(1 e ). We can transform to Cartesian (x,y) setting x = r cosφ+ae and y = r sin φ so the origin is at the midpoint of the major axis. Then fairy uninstructive agebra gives the standard Cartesian equation for an eipse x a + y b = 1 (34) where b is the ength of the minor axis, b = a (1 e ). 8

9 Energy per unit mass The energy per unit mass E = 1 ṙ.ṙ + Φ(r) = 1 ṙ + 1 r φ GM r (35) This is constant aong the orbit, so we can evauate it anywhere convenient - e.g. at periheion where ṙ = 0. Then φ = h and so rp E = 1 GMa(1 e ) GM a (1 e) a(1 e) = GM [ ( ) e 1 ] a 1 e 1 e = GM a (36) This is < 0 for a bound orbit, and is depends ony on the semi-major axis a (and not e). Keper s Laws... deduced from observations, and expained by Newtonian theory of gravity. 1 Orbits are eipses with the sun at a focus. Panets sweep out equa areas in equa time δa = 1 r δφ [= 1 r(rδφ)] (37) da dt = 1 h r φ = = constant (38) Keper s second aw is a consequence of a centra force, since this is why h is a constant. 3 (Period) (size of orbit) 3 9

10 In one period T, the area swept out is A = 1 ht = ( 0 But A = area of eipse = πab = πa 1 e [ Have so π 0 A = = = π 0 π π 0 0 r dφ rdr 0 1 r dφ dφ (1 + e cosφ) dx (a + b cosx) = π a a b a b A = Since = a(1 e ) this impies π 1 e 1 1 e A = πa 1 e da dt dt) and since b = a 1 e, ] Therefore A = πab T = πa 1 e h = πa 1 e GMa(1 e ) [since h = GMa(1 e ) ] a 3 T = π GM T a 3 (39) where in this case M is the mass of the sun. Note: Since E = GM πgm, the period T =. a ( E) 3 10

11 Unbound orbits r = 1 + e cos φ with e 1. If e > 1 then 1 + e cos φ = 0 has soutions φ where r =. cosφ = 1/e Then φ φ φ, and, since cosφ is negative, π < φ < π. The orbit is a hyperboa. If e = 1 then the partice just gets to infinity at φ = ±π - it is a paraboa. Energies for these unbound orbits: as r E 1 ṙ. Reca d dt of this and since h = r φ E = ṙ h r GM r r = 1 + e cosφ rṙ = e sin φ φ ṙ = eh sin φ As r cosφ 1/e E 1 ṙ = 1 e h (1 1 ) = GM e (e 1) (recaing that h = GM) Thus E > 0 if e > 1 and for paraboic orbits (e = 1) E = 0. 11

12 Escape veocity We have seen that in a fixed potentia Φ(r) a partice has constant energy E = 1 + Φ(r) aong an orbit. If we adopt the usua convention and take Φ(r) ṙ 0 as r, then if at some point r 0 the partice has veocity v 0 such that 1 v 0 + Φ(r 0 ) > 0 then it is abe to reach infinity. So at each point r 0 we can define an escape veocity v esc such that v esc = Φ(r 0 ) The escape veocity from the sun ( GM v esc = r 0 ) 1 ( ) r0 1 = 4. a.u. km s 1 Note: The circuar veocity v circ is such that r φ = GM r r φ = v circ = ( ) GM r0 1 = 9.8 r 0 a.u. km s 1 (= π a.u./yr). v esc = v circ for a point mass source of the gravitationa potentia. 1