Basic Galactic Dynamics

5 Basic galactic dynamics

5.1 Basic laws govern galaxy structure The basic force on galaxy scale is gravitational • Why? The other long-range force is electro-magnetic force due to electric • charges: positive and negative charges are diﬃcult to separate on galaxy scales.

5.1.1 Gravitational forces

Gravitational force on a point mass M1 located at x1 from a point mass M2 located at x2 is GM M (x x ) F = 1 2 1 − 2 . 1 − x x 3 | 1 − 2| Note that force F1 and positions, x1 and x2 are vectors. Note also F2 = F1 consistent with Newton’s Third Law. Consider a stellar− system of N stars. The gravitational force on ith star is the sum of the force due to all other stars

N GM M (x x ) F i j i j i = −3 . − xi xj Xj=6 i | − | The equation of motion of the ith star under mutual gravitational force is given by Newton’s second law, dv F = m a = m i i i i i dt and so N dvi GMj(xi xj) = − 3 . dt xi xj Xj=6 i | − |

5.1.2 Energy conservation

Assuming M2 is ﬁxed at the origin, and M1 on the x-axis at a distance x1 from the origin, we have

m1dv1 GM1M2 = 2 dt − x1

1 2

Using 2 dv1 dx1 dv1 dv1 1 dv1 = = v1 = dt dt dx1 dx1 2 dx1 we have 2 1 dv1 GM1M2 m1 = 2 , 2 dx1 − x1 which can be integrated to give

1 2 GM1M2 m1v1 = E , 2 − x1 The ﬁrst term on the lhs is the kinetic energy • The second term on the lhs is the potential energy • E is the total energy and is conserved •

5.1.3 Gravitational potential

The potential energy of M1 in the potential of M2 is M Φ , − 1 × 2 where GM2 Φ2 = − x1 is the potential of M2 at a distance x1 from it. The gravitational acceleration at x1 is

GM2 dΦ2 a1 = 2 = − x1 − dx1 For more than one mass points: GM Φ(x)= i , − x xi Xi | − | and the acceleration at x is ∂Φ ∂Φ ∂Φ a = Φ(x)= e , e , e x ∂x y ∂y z ∂z −∇ − For continuous mass distribution: Gρ(x0)d3x0 Φ(x)= . − x x0 Z | − | It can then be shown that Φ obeys the Poisson equation: 2Φ(x) = 4πGρ(x) , ∇ where ∂2Φ ∂2Φ ∂2Φ 2Φ + + . ∇ ≡ ∂x2 ∂y2 ∂r2 Basic galactic dynamics 3

Fig. 5.1. The gravitational force within a spherical shell of uniform density.

Potential energy and escaping velocity: 1 E = mv2 + mΦ(x) . 2 A star originally at x can escape the potential well if v > 0 at x where Φ = 0. This deﬁnes an escaping velocity: ∼ | |→∞ v2 = 2Φ(x) . esc − Question: calculate the escaping velocity from the surface of the earth. The radius and mass of the earth are 6380km, 6 1024 kg, and the gravitational constant is G = 6.67 11m3 kg−1s−2. × −

5.1.4 Newtonian gravity of spherically symmetric mass distribution Theorem I The gravitational force inside a spgerical shell of uniform density is zero. Proof: The force on ‘P’ due to mass elements M1 and M2 on the shell is

GmP M1 GmP M2 F = 2 2 . r1 − r2

M1 = A1Σ M2 = A2Σ , 4

Fig. 5.2. The gravitational force outside a spherical shell of uniform density.

where A1 and A2 are the areas of the two caps, and Σ is the surface density. θ 2 1 θ 2 1 A = π r sin A = π r sin 1 1 2 sin α 2 2 2 sin α Thus, 2 2 2 GmP Σπ sin (θ/2) r1 r2 F = 2 2 = 0. sin α "r1 − r2 # Since this is true for all lines through P , we have F = 0. Clearly the force is also zero within a spherical hole in the inner part of a spherically symmetric mass distribution.

Theorem II Outside any spherically symmetric mass distribution, the gravitational force is the same as if all its mass was concentrated at the center. Here it is easier to work with the potential instead of the force. The law of cosine: 2 2 2 r = R + x 2xP R cos θ . P − The mass of a thin ring on the shell: ∆M =∆AΣ , ∆A = R∆θ2πy = 2πR2∆θ sin θ. Basic galactic dynamics 5

Thus, the potential due to the ring is GΣ2πR2∆θ sin θ ∆Φ = . − 2 2 R + x 2xP R cos θ P − q Summing over all ‘rings’ on the shell: π sin θdθ Φ= 2πGΣR2 − 2 2 Z0 R + x 2xP R cos θ P − q π d cos θ = 2πGΣR2 2 2 Z0 R + x 2xP R cos θ P − q 2 2 2πGΣR2 R +xP −2xP R dw = 2 2 − 2x R P √w P ZR +xP −2x R

2 2 2πGΣR (R+xP ) 4πGΣR GM √ 2 = w (R−xP ) = − = , − xP | xP − xP where M is the mass of the spherical shell, and potential is the same as a point mass of mass M at the origin. This also holds for any spherical mass distribution! Poisson equation for spherically symmetric system: 1 d d r2 Φ(r) = 4πGρ(r) . r2 dr dr

5.1.5 Application of these two theorems For a galaxy with spherical mass distribution, with density proﬁle ρ(r), the mass within R is R M(< R)= 4πr2ρ(r)dr , Z0 and the gravitational force on a particle of mass m at R is GM(< R)m Fg = . − R2

Suppose the particle is in circular orbit with a speed Vc. The centrifugal force is mV 2 F = c . c R

In balance, we have Fc + Fg = 0, which gives: V 2 GM(< R) c = . R R Thus: V 2R M(< R)= c . G 6

Thus we can measure the mass proﬁle M(< R) by measuring Vc as a function of R (the rotation curve). For a star moving in a spherically symmetric galaxy, its angular momentum is conserved, since the torque T = r F = 0 . × Thus, the angular momentum is conserved: L = mr v = const. ×

5.1.6 How stars move in a galaxy? Strong encounters of stars are very rare: Strong encounter: at the closest approach the potential energy is larger than the initial kinetic energy so that the velocity of the star can change signiﬁcantly after the encounter: GM 2 1 > Mv2 , r 2 0 which requires 2GM r < rs 2 . ≡ v0 How often does this happen? Within a time t, a star will move a distance v0t. Since all other stars within a distance rs of the path of the star will produce a strong encounter, the total number of strong encounter within time t is 2 Ns = πrs v0tn , where n is the average number density of stars. Thus, the time for a strong encounter to occur is given by setting Ns = 1, which gives the average time a star experiences a strong encounter: 1 v3 v 3 M −2 n −1 t = = 0 4 1012 0 yr , s nπr2v 4πG2M 2n 10kms−1 M 1pc−3 s 0 ∼ × which is much longer than the age of the universe. Conclusions: Strong encounters are extremely rare in galaxies • The motion of a star may be considered that in a smooth potential •

5.2 Virial theorem Consider the moment of inertia of a system of point masses: 2 I = mir = miri ri . i · Xi Xi Thus, we have dI I˙ = 2 miri r˙ i ≡ dt · Xi Basic galactic dynamics 7

dI˙ I¨ = = 2 mir˙ i r˙ i + 2 miri ¨ri dt · · Xi Xi 1 I¨ = 2K + W, 2 where 1 1 2 K = mir˙ i r˙ i = miv 2 · 2 i Xi Xi is the total kinetic energy of the system, and

W = miri ¨ri . · Xi Using Newtonian gravity: r r r j i ¨i = Gmj − 3 , rj ri Xj=6 i | − | we have r r r j i W = Gmimj i − 3 · rj ri Xi Xj=6 i | − |

1 r r r r r j i r j i = Gmimj i − 3 Gmimj j − 3 2  · rj ri − · rj ri  Xi Xj=6 i | − | Xi Xj=6 i | − |   1 Gmimj(rjri) (rjri) = ·3 −2 rj ri Xi Xj=6 i | − | 1 Gm m = i j , −2 rj ri Xi Xj=6 i | − | which is the total gravitational potential energy of the system. For a static system I¨ = 0, we have 2K + W = 0 , which is the virial theorem, holds for all static system.

5.2.1 Application of virial theorem For an elliptical galaxy with size R and velocity dispersion σ: 1 GM 2 K Mσ2 , W = ∼ 2 − R Virial theorem gives: GM 2 Mσ2. R ∼ 8

Fig. 5.3. An illustration of conservation of phase space density.

Thus, the mass can be written as σ2R M . ∼ G Since R can be measured from photometry and distance, σ can be measured through spectroscopy, the above relation provides a way of measuring mass!

5.3 The collisionless Boltzmann equation At any given time t, the state of a star is speciﬁed by its position in space and its velocity. For a system containing many stars, it is described by a distribution function deﬁned through

dN = f(x, v,t)dxdydzdvxdvydvz . Thus, at time t the number density of stars near x is

n(x,t)= f(x, v,t)dvxdvydvz . Z The average velocity at a given point can be obained from

n(x,t) v(x,t) = vf(x, v,t)dvxdvydvz . h i Z Basic galactic dynamics 9

For a collisionless system, such as a galaxy, f is conserved: df = 0 . dt This can be understood using a one-dimensional example, as shown in the ﬁgure. In a 1-dimensional problem f = f(x, vx,t), and the collisionless Boltz- mann equation can be written as

df(x, vx,t) ∂f ∂f ∂f = + vx +v ˙x = 0 . dt ∂t ∂x ∂vx

Note thatv ˙x = ax = dΦ/dx. −

5.3.1 Jeans equations The collisional Boltzmann equation can be solved by looking at the velocity moments of f(x, vx,t). Integrating the Boltzman equation over vx, we have

∂n ∂(n vx ) + h i = 0 . ∂t ∂x This is called Jeans ﬁrst equation; it just says the number of particles is conserved. Multiplying both sides of the Boltzmann equation by vx and then inter- grate over vx, we have

∂ vx ∂ vx 1 ∂ 2 ∂Φ h i + vx h i + [nσ (x,t)] + = 0 , ∂t h i ∂x n ∂x x ∂x 2 2 2 where σx = vx(x,t) v(x,t) is the velocity dispersion. The above equation is calledh Jeansi − second h equation.i In three-dimensional case, the above two equations can be written as ∂n ∂ + [n vi ]=0; ∂t ∂xi h i Xi ∂ ∂ ∂Φ [n vj ]+ [n vivj ]+ n = 0 . ∂t h i ∂xi h i ∂xj Xi Note that we have used the notation: x1 = x, x2 = y and x3 = z.

5.3.2 Application I: Mass from velocity dispersion Consider a steady state system, the Jeans second equation becomes ∂ ∂Φ [n vivj ]+ n = 0 . ∂xi h i ∂xj Xi Written in spherical coordinates, we have

d n v2 n dΦ h r i + 2 v2 v2 v2 = n , dr r h r i − h ϑi − h ϕi − dr h i 10

2 2 2 where vϑ = vϕ because of spherical symmetry. Since dΦ/dr = GMtot(r)/r , we geth i h i 2 2 vr r d ln n d ln vr Mtot(r)= h i + h i + 2β , − G " d ln r d ln r #

2 2 where β(r) 1 vϑ / vr describes the anisotropy of the velocity ﬁeld at radius r. ≡ − h i h i If we can infer the density proﬁle n(r) of a population of objects (e.g. stars 2 in a galaxy ) and measure their velocity dispersion, vr and β as functions of r, the above equation can be used to infer the totalh imass distribution in a spherical system. Zwicky’s inference: For clusters, Zwicky found that the typical value of v2 1000km s−1 with a radius of about 1Mpc. What is the total massh ini a ∼ cluster? The 12 observed total mass in visible component is about 10 M . What can you infer?

5.3.3 Application II: mass from X-ray The above derivation can also be applied to gas particles in hydrostatic equilibrium within a gravitational potential well. For an ideal gas, v2 = v2 and v2 = k T (r)/µm , where T (r) is the h ϑi h r i h r i B p temperature proﬁle and µmp is the mean molecular weight. We then have k T (r)r d ln ρ d ln T M (r)= B gas + , tot Gµm d ln r d ln r − p where ρgas is the mass density of gas. This equation can be used to infer the total mass distribution in galaxy clusters where both the density and temperature proﬁles can be inferred from X-ray data.

5.3.4 Application III: Disk surface density and disk scale-height In cylindrical coordinates, the z-component of steady state Jeans second equation is

2 ∂ n v ∂ (n vRvz ) n vRvz ∂Φ h z i + h i + h i + n = 0 . ∂z ∂R R ∂z 2 For a highly ﬂattened system, z /R 1 and vRvz v . Thus | | h i h z i 1 ∂(n v2 ) ∂Φ h z i = . n ∂z − ∂z For such systems, Poisson’s equation can approximately written as ∂2Φ = 4πGρ , ∂z2 tot Basic galactic dynamics 11 where ρtot is the total mass density. It then follows that 2 2 1 ∂ vz ∂ ln n ∂ ln vz ρtot = h i + h i . −4πG ∂z " z ∂ ln z ∂ ln z !# The mass surface density is z v2 ∂ ln n ∂ ln v2 Σ (z)= ρ (z0)dz0 = h z i + h z i . (5.1) tot tot −2πGz ∂ ln z ∂ ln z Z−z ! If we can infer the vertical distribution of a population of objects in a ﬂattened system and measure their vertical velocity dispersion, we can estimate the vertical mass distribution. This method has been used to estimate the surface mass densities in the solar neighborhood. The inferred mass is similar to that in stars: not much dark matter in the disk! The above equation can also be used to understand the thickness of a disk. The larger v2 the thicker the disk. h z i 12

5.4 How about the gas component? For stars, collisions are very rare: mean-free time is about 1012yr. The mean-free time for gas particles in a galaxy 1 t = , ngasσv where n is the number density of gas particles; σ is the collision cross section, and v is the typical velocity of gas particles (related to the temperature by v2 kT/m). Inserting∼ typical values of quantity, we have n −1 σ −1 v −1 t 30, 000yr 3 −3 −19 2 −1 . ∼ 10 m 10 m 10kms Thus, gas particales collide with each other very frequently. When gas particles collide, atoms and molecules can be excited. Gas can then cool due to radiation. Gas can then lose internal energy (or pressure) and collapse further, en- hancing cooling. This is an run-away process. The end will be a blackhole? This process can be truncated by Star formation • Angular momentum of gas • If star formation is not effective, gas will first settle into a thin gaseous disk: why? Cooling can get rid of energy but not angular momentum. Minimize random motion but conserve orderly motion. If stars form very quickly during the collapse, random-motion energy due to the collapse cannot be effectively got rid of, we have a ‘hot’ system, an elliptical.