8. Galactic Dynamics Aim: Understand equilibrium of galaxies 1. What are the dominant forces? 2. Can we define some kind of equilibrium? 3. What are the relevant timescales? 4. Do galaxies evolve along a sequence of equilibria? [like stars]? Equilibrium of stars as comparison: gas pressure versus gravity hydrostatic equilibrium, thermal equilibrium timescales corresponding to the above, plus nuclear burning structure completely determined by mass plus age (and metallicity) very little influence from initial conditions Galaxies: Much harder! 1
Galaxy dynamics: assume ensemble of point masses, moving under influence of their own gravity Point masses can be: stars, mini BHs, planets, very light elementary particles... Overview 8.1 What are the equations of motion? 8.2 Do stars collide? (BT p. 3+4) 8.3 Virial Theorem 8.4 Application: Galaxy masses (BT p. 211-214) 8.5 Binding Energy and Formation of Galaxies (BT p. 214) 8.6 Scaling Relations 8.7 Relevant Time scales (BT p. 187-190) (Dynamical timescale, particle interaction timescale, Relaxation time for large systems) 2
8.1 Equations of motion Assume a collection of masses m i at location x i, and assume gravitational interaction. Hence the force F i on particle i is given by: F i = m i d 2 dt 2 x i = j i x j x i x j x i 3Gm im j Only analytic solution for 2 point masses Easy to solve numerically (brute force) - but slow for 10 11 particles. See for example: Analytic approximations necessary for a better understanding of solution In the following we investigate properties of gravitational systems without explicitly solving the equations of motion. 3
8.2 Do stars collide? Is it safe to ignore non-gravitational interactions? Example: our galaxy 10 11 stars with a total mass of 5 10 10 M A disk of 20 kpc in size, and thickness of 1 kpc The sun is 8 kpc from the center Cross section of the sun σ = π (2R ) 2 with R = 7 10 10 cm Typical orbital speed 200 km/s. 1 km/s 1 pc in 1 Myr. One orbit: 250 Myr. Age of the galaxy: 10 1 0 yr Sun has made 30 revolutions and disk is in quasi steady state 4
Calculate number of collisions between t 1 and t 1 + dt of stars coming in from the left, in a galaxy with homogeneous number density n. incoming star has velocity v suppose all stars have radius r incoming star will collide with stars in cylinder with volume V 1 : > V 1 = π(2r ) 2 v dt number of stars in V 1 : N 1 = nv 1 number of collisions per unit time: N 1 dt = nv 1 dt = 4πr2 nv 5
This gives a collision rate of 1.6 10 26 sec 1 = 5 10 19 yr 1. Note that the age of the Universe is 13.75 ± 0.11 Gyr So collisions are very rare indeed. Hence we can ignore these collisions without too much trouble. Komatsu, E., et.al., http://arxiv.org/abs/1001.4538 6
8.3 Virial Theorem: Relation for global properties: Kinetic energy and Potential energy. Again consider our system of point masses m i with positions x i. Construct i p i x i and differentiate w.r.t. time: d dt i p i x i = d dt i m i d x i dt x i = d dt i 1 d 2dt (m ix 2 i ) where I = i m i x 2 i inertia. = 1 2 d 2 I dt 2, which is the moment of However, using the chain rule, we can also write: d dt i p i x i = i d p i dt x i + i p i d x i dt 7
Then i p i d x i dt = i m i v 2 i = 2K with K the kinetic energy. Since d p i dt = F i we have 1d 2 I 2 dt 2 = i F i x i + 2K. Now assume the galaxy is quasi-static, i.e. its properties change only slowly so that d 2 I dt2 = 0. Then the equation above implies. K = 1 2 i F i x i 8
Then assume that the force F i can be written as a summation over pairwise forces F ij F i = j,j i F ij Now realize that F i x i = i F ij x i i j i This summation can be rewritten. We sum the terms over the full area 0 < i N, 0 < j N, i j. However, we can also limit the summation over just half this area: 0 < i N, i < j N, and add the (j, i) term explicitly to the (i, j) term within the summation. 9
Hence instead of summing over F ij x i, we sum over F ij x i + F ji x j Hence i j i F ij x i = i j>i ( F ij x i + F ji x j ) This is simply a change in how the summation is done, it does not use any special property of the force field. Because F ij = F ji (forces are equal and opposite for pairwise forces) the last term can be rewritten F i x i = i F ij ( x i x j ) i j>i For gravitational force F ij = Gm im j x i x j 2 x i x j x i x j 10
i F i x i = i j>i Gm i m j x i x j 3( x i x j )( x i x j ) which equals = i j>i Gm i m j x i x j = 1 2 i j i Gm i m j x i x j = W eq.(1) with W the total potential energy. Therefore for a galaxy in quasi-static equilibrium: K = 1 2 W, which is the virial theorem for quasi-static systems. The more general expression for non-static systems is: 1d 2 I 2 dt 2 = W + 2K. 11
8.4 Application: Galaxy masses Consider a system with total mass M Kinetic energy K = 1 2 M v2 with v 2 = mean square speed of stars (assumption: speed of star not correlated with mass of star) Define gravitational radius r g W = GM 2 Spitzer found for many systems that r g = 2.5r h, where r h is the radius which contains half the mass r g Virial theorem implies: M v 2 = GM 2 r g 12
or M = v 2 r g G 1 Hence, we can estimate the mass of galaxies if we know the typical velocities in the galaxy, and its size! 13
8.5 Binding Energy and Formation of Galaxies The total energy E of a galaxy is E = K + W = K = 1/2W Bound galaxies have negative energy - cannot fall apart and dissolve into a very large homogeneous distribution A galaxy cannot just form from an unbound, extended smooth distribution > E total = E start 0, E gal = K, so energy must be lost or the structure keeps oscillating: Possible energy losses through Ejection of stars Radiation (before stars would form) 14
8.6. Scaling Relations Consider a steady state galaxy with particles m i at location x i (t). Can the galaxy be rescaled to other physical galaxies? scaled particle mass ˆm i = a m m i scaled particle location ˆx i = a x x i (a t t) a m, a x, a t are scaling parameters In the rescaled galaxy we have ˆ F i = a m m i d 2 dt 2a x x(a t t) = a m a x a 2 t F i,orig The gravitational force is equal to ˆ F G = j i a 2 m a 2 x FG,orig a x x j a x x i a x x j a x x i 3Ga mm i a m m j = 15
Equilibrium is satisfied if the two terms above are equal ˆ F i = ˆ FG Since we have equilibrium when all scaling parameters are equal to 1, we obtain or a m a x a 2 t = a2 m a 2, x a m = a 3 xa 2 t. Now how do velocities scale? ˆv = d dtˆx = d dt a xx(a t t) = a x a t v orig Hence, the scaling of the velocities satisfies: a v = a x a t. We can write the new scaling relation as a m = a x a 2 v Hence ANY galaxy can be scaled up like this! 16
Notice that it is trivial to derive this from the virial theorem - the expression for the mass has exactly the same form. As a consequence, if we have a model for a galaxy with a certain mass and size, we can make many more models, with arbitrary mass, and arbitrary size. 17
8.7 Time scales Dynamical timescale, particle interaction timescale Is gravitational force dominated by short or long range encounters? (N.B. in a gas, only short range forces are relevant). In a galaxy, the situation is different. Consider force with which stars in cone attract star in apex of cone. 18
Force 1/r 2, with r the distance from apex. If ρ is almost constant, then the mass in a shell with width dr increases as r 2 dr. Hence differential force is constant at each r, and we have to integrate all the way out to obtain the total force. Realistic densities decrease after some radius, so that the force will be determined by the density distribution on a galactic scale (characterized by the half mass radius). 19
Relaxation time Short range encounters do not dominate Approximate force field with a smooth density ρ(x) instead of point masses. Contrary of situation in gas: only consider long range encounters (long range scale of the galaxy) Assume all stars have mass m. Analyze perturbations due to the fact that density is not smooth, but consists of individual stars. Simplify, and look first at single star-star encounter. What is the amount δv induced by one star crossing another? Estimate: straight line trajectory past stationary perturber 20
~ gives The perpendicular force F perturbation δ~v : 2 cos θ 2b 2 cos θ Gm Gm Gm ~ = F = = 2 2 2 r (b +x ) (b2 +x2)3/2 = Gm2 b2[1+(vt/b)2]3/2 ~ d F Newton: δ~v = dt m Z ~ Z F Gm δ~v = dt = dt = 3/2 2 2 m b [1 + (vt/b) ] Gm bv Z ds Gm s 2Gm q = = bv 1 + s2 bv (1 + s2)3/2 In words: δv is roughly equal to acceleration at closest approach, Gm/b2, times the duration 2b/v. 21
Note: approximation fails when δ v > v b < Gm/v 2 b min Galaxy has characteristic radius R. Define crossing time t c as the time it takes a star to move through the galaxy t c = R/v Calculate number of perturbing encounters per crossing time t c In a crossing time, the star has 1 encounter with each other star in the galaxy The impact parameter of each encounter can be derived by projecting each star onto a plane perpendicular to the unperturbed motion of the star Hence flatten the galaxy in the plane perpendicular to the motion of the star, and 22
assume that the stars are homogeneously distributed in that plane, out to a radius R, and no stars outside R. This is obviously a simplifying assumption, but it is reasonably accurate. This can be used to derive the distribution of impact parameters: N stars in total in Galaxy, distributed over total surface πr 2 stellar surface density / # per unit area: N πr 2 In a crossing time, the star has δn encounters with impact parameter between b and b + db. δn is given by the area of the annulus 2πbdb times the density of stars on the surface, which is N/(πR 2 ): δn = N πr 22πbdb = 2N R 2 b db 23
Result: δv = 0 as the perturbations are randomly distributed, and will not change the average velocity δv 2 = δv 2 δn = ( 2Gm bv ) 2 2Nb R 2 db = 8N ( ) Gm 2 db Rv b as each perturbation adds to δv by an equal amount (2Gm/bv) 2. The encounters do not produce an average perpendicular velocity, but they do produce an average (perpendicular velocity) 2. Hence, on average, the stars still follow their average path, but they tend to diffuse around it. 24
The total increase in rms perpendicular velocity can be calculated by integrating over all impact parameters from b min to infinity: Total rms increase: v 2 = R b min δv 2 db = R b min 8N ( ) Gm 2 db/b = Rv =8N( Gm Rv )2 ln Λ with ln Λ = Coulomb logarithm = ln R b min We can rewrite this equation. Recall: b min Gm/v 2 From virial theorem: v 2 = GM/R = GNm/R 25
Hence: b min = Gm/(GNm/R) = R/N ln Λ = ln R/b min = ln R R/N = ln N Furthermore from virial theorem: GM 2 R = Mv2 GM R = v2 GNm R = v2 N = v2 R Gm so that: v2 v 2 = 8 ln N N This last number is the fractional change in energy per crossing time. Hence we need the inverse number of crossings N/(8 ln N) to get v 2 v2 26
The timescale t relax is defined as the time it takes to deflect each star significantly by two body encounters, and it is therefore equal to t relax = N 8 ln N t c 0.1N ln N t c with crossing time: t c R/v 27
Conclusions effect of point mass perturbations decreases as N increases even for low N=50, v 2 /v2 = 0.6, hence deflections play a moderate role. for larger systems the effect of encounters become even less important Notice: one derives the same equation when the exact formulas for the encounters are used. Put in another way, the encounters with b < b min do not dominate. 28
Relaxation time for large systems System N t c (yr) t relax (yr) globular cluster 10 5 10 5 2 10 8 galaxy 10 11 10 8 10 17 galaxy cluster 10 3 10 9 3 10 10 Age of Universe 13.7 Gyr. Galaxies are collisionless systems motion of a star accurately described by single particle orbit in smooth gravitational field of galaxy no need to solve N-body problem with N = 10 11 (!) 29
Some more typical questions that can be asked at exam: (not homework assignments) What is the virial theorem? How can we use it to scale galaxies? Do galaxies have a main sequence like stars? How is relaxation time defined? What is the approximate expression? What is the typical relaxation time for a galaxy? How does the long relaxation time make it easier to produce models for galaxies? 30
8. Homework Assignments: 1) How empty are galaxies? Calculate the mean distance between stars. Take the radius of the sun as typical radius for a star. Take the ratio of the two. Compare this to the same ratio of galaxies: take the average distance between galaxies, and the radius of galaxies. 2) calculate the mass of our galaxy for two assumed radii: a.) 10 kpc (roughly the distance to the sun) b.) 350 kpc (halfway out to Andromeda) Take as a typical velocity the solar value of 200 km/s. 3) Show explicitly that the virial theorem produces the same result for the scaling relations. 4) Take typical sizes, velocities, number of stars for globular clusters, galaxies, and 31
galaxy clusters. Calculate for each of the systems, the average time between stellar collisions, crossing times and relaxation times.