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 gravita- tional 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 den- sity 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 momen- tum 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 .