Chapter 2 Stellar Dynamics in Galaxies

Chapter 2 Stellar Dynamics in Galaxies 2.1 Introduction A system of stars behaves like a fluid, but one with unusual properties. In a normal fluid two-body interactions are crucial in the dynamics, but in contrast star-star encounters are very rare. Instead stellar dynamics is mostly governed by the interaction of individual stars with the mean gravitational field of all the other stars combined. This has profound consequences for how the dynamics of the stars within galaxies are described mathematically, allowing for some considerable simplifications. This chapter establishes some basic results relating to the motions of stars within galaxies. The virial theorem provides a very simple relation between the total potential and kinetic energies of stars within a galaxy, or other system of stars, that has settled down into a steady state. The virial theorem is derived formally here. The timescale for stars to cross a system of stars, known as the crossing time, is a simple but important measure of the motions of stars. The relaxation time measures how long it takes for two-body encounters to influence the dynamics of a galaxy, or other system of stars. An expression for the relaxation time is derived here, which is then used to show that encounters between stars are so rare within galaxies that they have had little effect over the lifetime of the Universe. The motions of stars within galaxies can be described by the collisionless Boltzmann equation, which allows the numbers of stars to be calculated as a function of position and velocity in the galaxy. The equation is derived from first principles here. Similarly, the Jeans Equations relate the densities of stars to position, velocity, velocity dispersion and gravitational potential. 2.2 The Virial Theorem 2.2.1 The basic result Before going into the main material on stellar dynamics, it is worth stating { and deriving { a basic principle known as the virial theorem. It states that for any system of particles bound by an inverse-square force law, the time-averaged kinetic energy T and the time-averaged potential energy U satisfy h i h i 2 T + U = 0 ; (2.1) h i h i 15 for a steady equilibrium state. T will be a very large positive quantity and U a very large negative quantity. Of hcourse,i for a galaxy to hold together, the total energyh i T + U < 0 ; the virial theorem provides a much tighter constraint than this alone. Th ypicallyi h i, T and U 1050 to 1054 J for galaxies. In practice,h i man−hy systemsi ∼ of stars are not in a perfect final steady state and the virial theorem does not apply exactly. Despite this, it does give important, approxi- mate results for many astronomical systems. The virial theorem was first devised by Rudolf Clausius (1822{1888) to describe motions of particles in thermodynamics. The term virial comes from the Latin word for force, vis. 2.2.2 Deriving the virial theorem from first principles To prove the virial theorem, consider a system of N stars. Let the ith star have a mass mi and a position vector xi. The position vector will be measured from the centre of mass of the system, and we shall assume that this centre of mass moves with uniform motion. The velocity of the ith star is x_ dx =dt, where t is the time. i ≡ i Consider the moment of inertia I of this system of stars defined here as N N I m x .x = m x2 : (2.2) ≡ i i i i i i=1 i=1 X X (Note that this is a different definition of moment of inertia to the moment of inertia about a particular axis that is used to study the rotation of bodies about an axis.) Differentiating with respect to time t, dI d d = m x .x = m x .x dt dt i i i dt i i i i ! i X X d = m x .x assuming that the masses m do not change i dt i i i i X . . = mi x_ i xi + xi x_ i from the product rule i X . = 2 mi x_ i xi i X Differentiating again, d2I d d d = 2 m x_ .x = 2 m x_ .x = 2 m x_ .x dt2 dt i i i dt i i i i dt i i i i i X . X X = 2 mi xï xi + x_ i x_ i i X . 2 = 2 mi xï xi + 2 mi x_i (2.3) i i X X 1 2 The kinetic energy of the ith particle is 2 mix_i . Therefore the total kinetic energy of the entire system of stars is 1 2 2 T = 2 mi x_ i ) mi x_ i = 2 T i i X X 16 Substituting this into Equation 2.3, d2I = 4 T + 2 m x¨ .x ; (2.4) dt2 i i i i X at any time t. We now need to remember that the average of any parameter y(t) over time t = 0 to τ is 1 τ y = y(t) dt h i τ Z0 Consider the average value of d2I=dt2 over a time interval t = 0 to τ. d2I 1 τ = 4 T + 2 m x¨ .x dt dt2 τ i i i 0 i ! Z X 4 τ 2 τ = T dt + m x¨ .x dt τ τ i i i 0 0 i Z Z X m τ = 4 T + 2 i x¨ .x dt assuming m is constant over time h i τ i i i i 0 X Z = 4 T + 2 m x¨ .x (2.5) h i i h i ii i X When the system of stars eventually reaches equilibrium, the moment of inertia I will be constant. So, d2I=dt2 = 0. An alternative way of visualising this is by considering that I willh be boundedi in any physical system and d2I/dt2 will also d2I be finite. Therefore the long-time average dt2 will vanish as τ becomes large, i.e. 2 2 h i lim d I = lim ( 1 τ d I dt) 0 because d2I=dt2 remains finite. τ!1h dt2 i τ!1 τ 0 dt2 ! Substituting for d2I=dt2 = 0 into Equation 2.5, h iR 4 T + 2 m x¨ .x = 0 : h i i h i ii i X ) 2 T + m x¨ .x = 0 : (2.6) h i i h i ii i X The term m x¨ .x is related to the gravitational potential. We next need to show i i h i ii how. P Newton's Second Law of Motion gives for the ith star, mi xï = Fij j Xj6=i where Fij is the force exerted on the ith star by the jth star. Using the law of universal gravitation, G mi mj mi xï = 3 (xi xj) : − xi xj − j j − j Xj6=i Taking the scalar product (dot product) with xi, . G mi mj . mi xï xi = 3 (xi xj) xi − xi xj − j j − j Xj6=i 17 Summing over all i, . G mi mj . G mi mj . mixï xi = 3 (xi xj) xi = 3 (xi xj) xi − xi xj − − xi xj − i i j j − j i;j j − j X X Xj6=i Xi6=j (2.7) Switching i and j, we have . G mj mi . mj x¨j xj = 3 (xj xi) xj (2.8) − xj xi − j j;i j − j X Xi6=j Adding Equations 2.7 and 2.8, . G mi mj . G mj mi . mixï xi+ mjx¨j xj = 3 (xi xj ) xi 3 (xj xi) xj − xi xj − − xj xi − i j i;j j − j i;j j − j X X Xi6=j Xi6=j . G mi mj . ) 2 mi xï xi = 3 (xi xj) xi + (xj xi) xj − xi xj − − i i;j j − j X Xi6=j But (x x ).x + (x x ).x = (x x ).x (x x ).x i − j i j − i j i − j i − i − j j = (x x ).(x x ) (factorising) i − j i − j = x x 2 j i − jj . G mi mj 2 ) 2 mi xï xi = 3 xi xj − xi xj j − j i i;j j − j X Xi6=j . 1 G mi mj ) mi xï xi = (2.9) − 2 xi xj i i;j j − j X Xi6=j We now need to find the total potential energy of the system. The gravitational potential at star i due to star j is G m Φ = j i j − x x j i − jj Therefore the gravitational potential at star i due to all other stars is G mj Φi = Φi j = − xi xj j j j − j Xj6=i Xj6=i Therefore the gravitational potential energy of star i due to all the other stars is G mj Ui = mi Φi = mi − xi xj j j − j Xj6=i 18 The total potential energy of the system is therefore 1 G mj U = Ui = 0 mi 1 2 − xi xj i i j j − j X X B Xj6=i C B C @ A 1 The factor 2 ensures that we only count each pair of stars once (otherwise we would count each pair twice and would get a result twice as large as we should). Therefore, 1 G m m U = i j − 2 xi xj i;j j − j Xi6=j Substituting for the total potential energy into Equation 2.9, .

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