9. Orbits in stationary Potentials We have seen how to calculate forces and potentials from the smoothed density ρ. Now we can analyse how stars move in this potential. Because two body interactions can be ignored, we can analyse each star by itself. We therefore speak of orbits. The main aim is to focus on the properties of orbits. Given a potential, what orbits are possible? Overview 9.1 Orbits in spherical potentials (BT p. 13-17) 9. Constants and Integrals of motion (BT p. 11-117) 9..1 Spherical potentials 9.. Integrals in dimensional flattened potentials 9..3 Axisymmetric potentials 9.3 A general 3-dimensional potential 9. Schwarzschild s method 1
9.1 Orbits in spherical potentials Appropriate for for example globular clusters. Potential function Φ = Φ(r), with r = r Equation of motion for a star with unit mass d r dt = F (r) e r = Φ Conservation of angular momentum Define angular momentum per unit mass: using that r r = for any r: d dt L = d dt ( r d r dt ) L r d r dt = d r dt d r dt + r d r dt = F (r) r e r =
Hence L = r r is constant with time. L is always perpendicular to the plane in which r and v lie. Since it is constant with time, these vectors always lie in the same plane. Hence the orbit is constrained to this orbital plane. Geometrically, L is equal to twice the rate at which the radius vector sweeps out area. Use polar coordinates (r, ψ) in orbital plane and rewrite equations of motion in polar coordinates Using BT (App. B.), the acceleration d r dt in cylindrical coordinates can be written as: d r dt = ( r r ψ ) e r + (ṙ ψ + r ψ) e ψ With the equation of motion d r dt = F (r) e r, this implies: r r ψ = F (r) ṙ ψ + r ψ = ( ) ( ) 3
Hence: ṙ ψ + r ψ = 1 r dr ψ dt = r ψ = rv = L = c st Using ψ = L r and equation (**): r r ψ = r L r 3 = dφ dr 3 where Φ is the potential. Integrate last equation to obtain: 1 ṙ = E Φ L r = E Φ eff(r) with E the energy. E is the integration constant obtained for r This equation governs radial motion through the effective potential Φ eff : Φ eff = Φ + L r
Veff(r) rmin rmax r E Motion possible only when ṙ r min r r max pericenter < r < apocenter Typical orbit in a spherical potential is a planar rosette 5
Angle ψ between successive apocenter passages depends on mass distribution: π (homogeneous sphere) < ψ < π (pointmass) 6
Special cases Circular orbit: r min = r max v r = dφ dr = GM(r) r Radial orbit: L = 1 ṙ = E Φ(R) Homogeneous sphere Φ(r) = 1 Ω r + Constant Equation of motion in in radial coordinates: r = Ω r or in cartesian coordinates x, y ẍ = Ω x ÿ = Ω y Hence solutions are x = X cos(ωt + c x ) y = Y cos(ωt + c y ) where X, Y, c x and c y are arbitracy constants. 7
Hence, even though energy and angular momentum restrict orbit to a rosetta, these orbits are even more special: they do not fill the area between the minimum and maximum radius, but are always closed! The same holds for Kepler potential. But beware, for the homogeneous sphere the particle does two radial excursions per cycle around the center, for the Kepler potential, it does one radial excursion per angular cycle. We now wish to classify orbits and their density distribution in a systematic way. For that we use integrals of motion. 8
9. Constants and Integrals of motion First, we define the 6 dimensional phase space coordinates ( x, v). They are conveniently used to describe the motions of stars. Now we introduce: Constant of motion: a function C( x, v, t) which is constant along any orbit: C( x(t 1 ), v(t 1 ), t 1 ) = C( x(t ), v(t ), t ) C is a function of x, v, and time t. Integral of motion: a function I( x, v) which is constant along any orbit: I[ x(t 1 ), v(t 1 )] = I[ x(t ), v(t )] 9
I is not a function of time! Thus: integrals of motion are constants of motion, but constants of motion are not always integrals of motion! E.g.: for a circular orbit ψ = Ω t + ψ o, so that C = t ψ/ω. C is constant of motion, but not an integral as it depends on t. Constants of motion 6 for any arbitrary orbit: Initial position ( x, v ) at time t = t. Can always be calculated back from x, v, t. Hence ( x, v ) can be regarded as six constants of motion. 1
Integrals of motion are often hard or impossible to define. Simple exceptions include For all static potentials: Energy E( x, v) = 1 v + Φ For axisymmetric potentials: L z For spherical potentials: the three components of L Integrals constrain geometry of orbits, lowering the number of dimensions in the 6 dimensional phase space, where the orbit can exist. Examples:..1 Spherical potentials.. Integrals in dimensional attened potentials..3 Axisymmetric potentials 11
9..1. Spherical potentials E, L x, L y, L z are integrals of motion, but also E, L and the direction of L (given by the unit vector n, which is defined by two independent numbers). n defines the plane in which x and v must lie. Define coordinate system with z axis along n x = (x 1, x, ) v = (v 1, v, ) x and v constrained to D region of the 6D phase space. In this dimensional space, L and E are conserved. This constrains the orbit to a dimensional space. Hence the velocity is uniquely defined for a given x (see page ). v r = ± (E Φ) L /r v ψ = ±L/r 1
9... Integrals in dimensional flattened potentials Examples: Circular potential V (x, y) = V ( r) Two integrals: E, L z. Flattened potential V (x, y) = ln(x + y a + 1) Only classic integral of motion: E Figures on the next page show the orbits that one gets by integrating the equations of motion this flattened potential 13
Box orbits no net angular momentum, (L z = x v y y v x ) avoid outer x-axis - - - - - - - - - - - - - - - - - x=.5 x=.5 y= vx= y= vx= vy=1 vy=1 - - - - - x=1 x=1 y= y= vx= vy=1 - - - - - - - - - - - - -x=1.1 - -y= vx= vy=1 x=1.1 x=1.1 y= y= vx= vx= vy=1 vy=1 - - - - - - x=1. - - y= - vx= vy=1 x=1. x=1. y= y= vx= vx= vy=1 vy=1 1
Loop orbits with - net angular momentum avoid inner x-axis - - x=1 y= vx= vy=1 can circulate in two directions. - =.5 y= vx= vy=1 - - - - =1.1 y= vx= vy=1 - - x=1. y= vx= vy=1 - - - - - - x= y= vx= vy=1 - - x= y= vx= vy=1 15
Clearly the orbits are regular and do not fill equipotential surface Furthermore, they do not traverse each point in a random direction, but generally only in directions Conclusion: the orbits do not occupy a 3 dimensional space in the -dimensional phase-space, but they occopy only a -dimensional space! This indicates that there is an additional integral of motion: a non-classical integral The non-classical integral, plus the regular Energy, constrain the orbit to lie on a dimensional surface in the dimensional phase-space. 16
A homogeneous ellipsoid The homogeneous ellipsoid helps us to understand how additional integrals of motions, and box orbits, exist. Consider a density distribution: ρ = ρ H(1 m ), with m = x a + y b + z c and H(x) = 1 for x, H(x) = for x < Potential inside the ellipsoid: Φ = A x x + A y y + A z z + C Forces are of the form F i = A i x i, i.e. 3 independent harmonic oscillators: x i = a i cos(ω i t + ψ,i ) 3 integrals of motion, E i 17
Orbits for general homogeneous ellipsoid All orbits are box orbits - - - - - - x=.5 y= vx= vy=1 - - x=1 y= vx= vy=1 - - - - - - x=1.1 y= vx= vy=1 - - x=1. y= vx= vy=1 18
- - - - - - x= y= vx= vy=1 - - x= y= vx= vy=1 19
Special case: a=b All orbits are loop orbits - - - - x=.5 x=1.1 x=1. x= x= y= vx= vy=1
9..3. Axisymmetric potentials Φ = Φ(R, z ), with R = x + y. For z=: orbits as if potential were circular. For the general case, the equation of motions of a star is: d~ r ~ = Φ(R, z). dt 1
With e r, e φ, e z unit vectors in r, φ and z direction, we can write: r = R e R + z e z, Φ = Φ R e R + Φ φ e φ + Φ z e z with Φ φ =. Using BT (App. B.), the acceleration d r dt in cylindrical coordinates can be written as: d r dt = ( R R φ ) e R + (Ṙ φ + R φ) e φ + z e z. Note that: d dt L z d dt R φ = Ṙ φ + R φ =. and conclude that the angular momentum about the z-axis is conserved. If we use this to eliminate φ, we obtain for the equation of motions in the e r and e z directions:
d R dt = Φ eff R ; d z dt = Φ eff z with Φ eff = Φ(R, z) + L z R. Hence 3D motion can be reduced to motion in (R,z) plane or meridional plane, under influence of the effective potential Φ eff. Note that this meridonial plane can rotate. Application to a logarithmic potential Recall: the axisymmetric logarithmic potential (R c taken to be zero): Φ(R, z) = 1 v ln(r + z q ). Then: Φ eff = 1 v ln(r + z q ) + L z R. 3
Total energy: E = 1 [Ṙ + (r φ) + ż ] + Φ = 1 (Ṙ + ż ) + ( Φ + L z R ) = 1 (Ṙ + ż ) + Φ eff. Allowed region in meridional plane Since the kinetic energy is non-negative, orbits are only allowed in areas where: Φ eff < E. E.g. Lines of constant Φ eff are shown in Figure 3.. Stars with energy E have zero velocity at curves of Φ eff = E 5
Equations of motions for this potential solved numerically for two stars with the same energy and angular momentum, but with different initial conditions. Not all orbits fill the space Φ eff < E fully! 6
Two integrals (E, L z ) reduce the dimensionality of the orbit from 6 to (e.g. R, z, ψ, v z ). Therefore another integral of motion must play a role dimensionality reduced to 3 (e.g. R, z, ψ). This integral is a non-classical integral of motion. 7
9.3 A general 3-dimensional potential Stäckel potential: ρ = 1/(1 + m ) with m = x a + y b + z c. 8
9
9. Schwarzschild s method A simple recipe to build galaxies Define density ρ. Calculate potential, forces. Integrate orbits, find orbital densities ρ i. Calculate weights w i > such that ρ = ρ i w i. Examples: build a D galaxy in a logarithmic potential Φ = ln(1 + x + y /a). 3
As we saw, box orbits void the outer x-axis. As we saw, loop orbits void the inner x-axis. both box and loop orbits are needed. Suppose we have constructed a model. What kind of rotation can we expect? box orbits: no net rotation. loop orbits: can rotate either way: positive, negative, or neutral. Hence: The rotation can vary between zero, and a maximum rotation A maximum rotation is obtained if all loop orbits rotate the same way. 31
Is the solution unique? box orbits are defined by integrals of motion, say the coordinates of the corner loop orbits have two integrals of motion Hence, we have two construct a dimensional function from the superposition of two -dimentional functions ρ( x, y) = w box (I 1, I )ρ box (I 1, I )+ w loop (I 1, I )ρ loop (I 1, I ) The unknown functions are w box (I 1, I ) and w loop (I 1, I ). The system is underdetermined. Hence, many solutions are possible. 3
9. Homework Assignments: 1) Box orbits are characterized by the fact they they go through the center, and have no net angular momentum. Explain how it comes that they don t have net angular momentum, even though a star in a box orbit has a non-zero angular momentum at most times. ) How is it possible that the box orbit touches the equipotential surface given by Energy = Φ? 3) Why does a loop orbit not touch that surface? ) Why do we expect virtually no loop orbits in a homogeneous ellipsoid? 5) Figure 3- (page ) shows contours plot for effective potentials related to an axisymmetric logarithmic potential given on 33
page 3. When does the minimum occur? What kind of orbit does this minimum represent? 6) Calculate at least 3 orbits in the -dimentional potential Φ = ln(1 + x + y /). Do this as follows: Start the star at a given location (x, ) with velocity (, v ). Calculate the shift in position at time dt by dt* v. Calculate the shift in velocity at time dt by dt* F. And keep integrating! Also plot the equipotential curve where Energy E = Φ.