STELLAR DYNAMICS Preliminary version, 2010 Hans Rickman Dept. of Astronomy & Space Physics, Uppsala Univ. PAS Space Research Center, Warsaw These notes are intended as an introduction to fundamental concepts of stellar dynamics, and while observed stellar systems will often be taken as examples, the aim is not to describe those systems in detail but rather to focus on the dynamical processes. 1 Fundamental equations 1.1 The N-body problem Consider a self-gravitating system of N objects that do not experience any external forces { i.e., the system is isolated. The objects in question will always be stars, so we will use the term stars instead of `objects'. Let us number the stars from 1 to N. Consider a cartesian coordinate system x,y,z. We denote the coordinates of star i by xi,yi,zi. These are functions of time, since the star is moving. The velocity has components dx i = x_ = u dt i i dy i = y_ = v dt i i dz i = z_ = w dt i i (the dots traditionally stand for time derivatives). For convenience we name them ui,vi,wi. Finally, let the mass of star i be mi. By ri we denote the coordinate vector (xi,yi,zi) that connects the origin to star i. We denote by rij the distance between stars i and j: r = r r = (x x )2 + (y y )2 + (z z )2 (1) ij j j − ij j − i j − i j − i q The force of attraction between stars i and j has the abso- 2 lute value Gmimj=rij (G is the gravitational constant). Its direction is along the line i{j. Thus, in vector form, the force exerted by j on i is: Gm m r r i j j − i r2 · r r ij j j − ij (the second factor is the unit vector from i toward j). Hence the total force acting on star i due to all the others is: N Gmimj(rj ri) mi¨ri = 3 − (i = 1; : : : ; N) (2) j=1 rij Xj6=i 1 According to Newton's 2nd law of motion, this force equals the product of the acceleration and the mass (thus the above equation). This is a vector equation, which represents three scalar equations; the first is obtained by isolating the x components: N Gmimj(xj xi) mix¨i = 3 − (i = 1; : : : ; N) j=1 rij Xj6=i and similarly for y and z. Equation (1) provides the foundation for the whole subject of stellar dynamics, as well as celestial mechanics. Such an equation can be written down for each star, i.e., for all values of i, from 1 to N. We have a system of 3N simultaneous second-order differential equations and thus a dynamical system of order 6N. If one knows the exact positions and velocities of all the stars at an initial moment, these equations in principle allow to compute the subsequent evolution of the stellar system for all times. However, such an exact solution is impossible to achieve by numerical integration owing to round-off errors and finite time steps of the integration. We hence wish to proceed as far as possible by analytic means. 1.2 The ten integrals One may obtain some important information about the system from the basic equations of motion without integrating them. Let us first seek integrals of the system, i.e., entities that do not change with time. 1.2.1 Motion of the center of mass Let us form the sum of the equations, from 1 to N: N N N Gmimj(rj ri) mi¨ri = 3 − (i = 1; : : : ; N) i=1 i=1 j=1 rij X X Xj6=i In the right-hand member we have all pairs (i; j) represented except those for which i = j. The sum of the two terms (i; j) and (j; i) is: Gmimj(rj ri) Gmjmi(ri rj) 3 − + 3 − rij rji But rji = rij, and hence this sum equals zero. In this way the terms cancel out in pairs, and we have: N mi¨ri = 0 (3) Xi=1 Physical Interpretation: Eq. (3) means that the sum of all forces within the system is zero, which is always true for an isolated system. If we integrate over time, we get: N mir_ i = a; Xi=1 where a is a constant vector, and then: N miri = at + b; Xi=1 2 where b is yet another constant vector. The center of mass of the system, , is defined by: G m r at + b r = i i = (4) G P mi M where M is the total mass of the system, whicP h is also a constant. This shows that the center of mass moves with constant velocity along a straight line, which is obvious, since no force is acting upon the system. In mechanics we can always replace one frame of reference by another one in uniform, rectilinear motion with respect to the first. Hence, in all that follows, we shall take a coordinate system anchored at the center of mass, i.e., which has its origin in and moves with . In this G G system we thus have: r 0, i.e.: a = 0 and b = 0. G The choice of the center≡ of mass frame simply means that we eliminate the common motion of the system, which does not interest us. We shall only investigate the internal motions. 1.2.2 Total angular momentum integral Take the cross product of both members of the basic equation by ri, and form the sum over i: N N N Gmimjri (rj ri) miri ¨ri = ×3 − (i = 1; : : : ; N) i=1 × i=1 j=1 rij X X Xj6=i All the terms r r in the right-hand member vanish. Let us again form pairs (i; j) and (j; i): i × i Gmimjrj ri Gmjmiri rj 3 × + 3 × = 0; rij rji since rj ri = ri rj. Hence: × − × N m r ¨r = 0 (5) i i × i Xi=1 Thus we have: d N N miri r_ i = miri ¨ri = 0 dt × ! × Xi=1 Xi=1 which yields: N m r r_ = c (6) i i × i Xi=1 where, again, c is a constant vector. Physical Interpretation: This quantity is the total angular momentum of the system, and it is conserved for an isolated system, on which no external torque is acting. It cannot be eliminated by a change of reference system, like we did with a and b. In other words, c is an invariant vector that characterizes the system. In particular, it defines a characteristic direction. If the system has an axis of symmetry, this axis should coincide with c. We shall also see below that the absolute value of c, which measures the amount of common rotation in the system, is linked to the flattening of the system. 3 1.2.3 Total energy integral Let us define a pairwise potential energy of stars i and j as the quantity: Gmimj Ωij = − rij and the total potential energy as the sum of all pairwise ener- gies, i.e.: N 1 N − Gm m Ω = i j (7) − rij Xi=1 j=Xi+1 Each pair should appear once and only once, which explains the definition of the sums. It can be shown that Ω is the work performed by the system, if all the stars are removed infinitely far from each other. Thus the name potential energy, i.e., work that potentially may be performed. In reality Ω is negative, i.e., work must be provided to the system in order to push the stars away from each other. This is obvious, since they are attracting each other. Ω is a function of all the coordinates: Ω(x1; y1; z1; x2; y2; z2; : : : ; xN ; yN ; zN ) { these appear in the expressions for rij. Let us differentiate with respect to one of them and form @Ω=@xi. Only the pairs formed by star i and another star yield a non-vanishing derivative. Thus: N N @Ω @ Gmimj Gmimj @rij = = 2 @xi − j=1 @xi rij ! j=1 rij @xi Xj6=i Xj6=i We have: r2 = (x x )2 + (y y )2 + (z z )2, and hence: ij j − i j − i j − i @r x x ij = j − i @xi − rij and: N @Ω Gmimj(xj xi) = 3 − @xi − j=1 rij Xj6=i But we saw before that this expression is exactly: m x¨ . Thus: − i i @Ω mix¨i = −@xi and similarly: @Ω miy¨i = −@yi @Ω miz¨i = −@zi which we merge into: m ¨r = r Ω (8) i i − i (a notation that is not strictly correct but convenient). So we have yet another quite general result: The force equals minus the gradient of the potential energy. 4 Let us take the scalar product of both members with r_ i and form the sum over i: N N m r_ ¨r + r_ r Ω = 0 (9) i i · i i · i Xi=1 Xi=1 or: d N m r_ 2 dΩ i i + = 0 (10) dt 2 dt Xi=1 since the second term of Eq. (9) can be written: @Ω @Ω @Ω dx1 @Ω dzN dΩ x_ 1 + : : : + z_N = + : : : + = @x1 @zN @x1 dt @zN dt dt The sum in the first term of Eq. (10) is called the total kinetic energy (T ) of the system, i.e., the sum of the individual kinetic energies of the stars. We integrate over time and obtain: T + Ω = E (11) where E is a constant called the total energy.