Gravitational Dynamics. Gravitational Dynamics can be applied to: Two body systems:binary stars Planetary Systems Stellar Clusters:open & globular Galactic.

Gravitational Dynamics

Gravitational Dynamics can be applied to: Two body systems:binary stars Planetary Systems Stellar Clusters:open & globular Galactic Structure:nuclei/bulge/disk/halo Clusters of Galaxies The universe:large scale structure

Syllabus Phase Space Fluid f(x,v) –Eq n of motion –Poisson’s equation Stellar Orbits –Integrals of motion (E,J) –Jeans Theorem Spherical Equilibrium –Virial Theorem –Jeans Equation Interacting Systems –Tides  Satellites  Streams –Relaxation  collisions

How to model motions of 10 10 stars in a galaxy? Direct N-body approach (as in simulations) –At time t particles have (m i,x i,y i,z i,vx i,vy i,vz i ), i=1,2,...,N (feasible for N<<10 6). Statistical or fluid approach (N very large) –At time t particles have a spatial density distribution n(x,y,z)*m, e.g., uniform, –at each point have a velocity distribution G(vx,vy,vz), e.g., a 3D Gaussian.

N-body Potential and Force In N-body system with mass m 1 …m N, the gravitational acceleration g(r) and potential φ(r) at position r is given by: r 12 RiRi r mimi

Eq. of Motion in N-body Newton’s law: a point mass m at position r moving with a velocity dr/dt with Potential energy Φ(r) =mφ(r) experiences a Force F=mg, accelerates with following Eq. of Motion :

Orbits defined by EoM & Gravity Solve for a complete prescription of history of a particle r(t) E.g., if G=0  F=0, Φ(r)=cst,  dx i /dt = vx i =c i  x i (t) =c i t +x 0, likewise for y i,z i (t) –E.g., relativistic neutrinos in universe go straight lines Repeat for all N particles.  N-body system fully described

Example: Force field of two-body system in Cartesian coordinates

Example: 4-body problem Four point masses Gm=1 at rest (x,y,z)=(0,1,0),(0,-1,0),(-1,0,0),(1,0,0). What is the initial total energy? Integrate EoM by brutal force with time step=1 to find the positions/velocities at time t=1. i.e., use straight-orbit V=V0+gt, R=R0+V0t+gt 2 /2. What is the new total energy?

Star clusters differ from air: Size doesn’t matter: –size of stars<<distance between them –  stars collide far less frequently than molecules in air. Inhomogeneous In a Gravitational Potential φ(r) Spectacularly rich in structure because φ(r) is non-linear function of r

Why Potential φ(r) ? More convenient to work with force, potential per unit mass. e.g. KE  ½v 2 Potential φ(r) is scaler, function of r only, –Easier to work with than force (vector, 3 components) –Simply relates to orbital energy E= φ(r) +½v 2

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Example: energy per unit mass The orbital energy of a star is given by: 0 since and 0 for static potential. So orbital Energy is Conserved in a static potential.

Example: Energy is conserved The orbital energy of a star is given by: 0 since and 0 for static potential. So orbital Energy is Conserved in a static potential.

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A fluid element: Potential & Gravity For large N or a continuous fluid, the gravity dg and potential dφ due to a small mass element dM is calculated by replacing m i with dM: r 12 R r dM d3Rd3R

Potential in a galaxy Replace a summation over all N-body particles with the integration: Remember dM=ρ(R)d 3 R for average density ρ(R) in small volume d 3 R So the equation for the gravitational force becomes: RRiRRi

Poisson’s Equation Relates potential with density Proof hints:

Poisson’s Equation Poissons equation relates the potential to the density of matter generating the potential. It is given by:

Gauss’s Theorem Gauss’s theorem is obtained by integrating poisson’s equation: i.e. the integral,over any closed surface, of the normal component of the gradient of the potential is equal to 4  G times the Mass enclosed within that surface.

Laplacian in various coordinates

4 th Lec Potential,density,orbits

From Potential to Density Use Poisson’s Equation The integrated form of Poisson’s equation is given by: From Gravitational Force to Potential

More on Spherical Systems Newton proved 2 results which enable us to calculate the potential of any spherical system very easily. NEWTONS 1 st THEOREM:A body that is inside a spherical shell of matter experiences no net gravitational force from that shell NEWTONS 2 nd THEOREM:The gravitational force on a body that lies outside a closed spherical shell of matter is the same as it would be if all the matter were concentrated at its centre.

From Spherical Density to Mass M(r) M(r+dr)

Poisson’s eq. in Spherical systems Poisson’s eq. in a spherical potential with no θ or Φ dependence is:

Proof of Poissons Equation Consider a spherical distribution of mass of density ρ(r). r g

Take d/dr and multiply r 2  Take d/dr and divide r 2 

Sun escapes if Galactic potential well is made shallower

Solar system accelerates weakly in MW 200km/s circulation g(R 0 =8kpc)~0.8a0, a0=1.2 10 -8 cm 2 s -1 Merely g n ~0.5 a0 from all stars/gas Obs. g(R=20 R 0 ) ~20 g n ~0.02 a0 g-g n ~ (0-1)a0 “GM” ~ R if weak! Motivates –M(R) dark particles –G(R) (MOND)

Circular Velocity CIRCULAR VELOCITY= the speed of a test particle in a circular orbit at radius r. For a point mass:For a homogeneous sphere

Escape Velocity ESCAPE VELOCITY= velocity required in order for an object to escape from a gravitational potential well and arrive at  with zero KE. It is the velocity for which the kinetic energy balances potential. -ve

Tutorial Question 1: Singular Isothermal Sphere Has Potential Beyond r o : And Inside r<r 0 Prove that the potential AND gravity is continuous at r=r o if Prove density drops sharply to 0 beyond r0, and inside r0 Integrate density to prove total mass=M0 What is circular and escape velocities at r=r0? Draw Log-log diagrams of M(r), Vesc(r), Vcir(r), Phi(r), rho(r), g(r) for V0=200km/s, r0=100kpc.

Tutorial Question 2: Isochrone Potential Prove G is approximately 4 x 10 -3 (km/s) 2 pc/Msun. Given an ISOCHRONE POTENTIAL For M=10 5 Msun, b=1pc, show the central escape velocity = (GM/b) 1/2 ~ 20km/s. Argue why M must be the total mass. What fraction of the total mass is inside radius r=b=1pc? Calculate the local Vcir(b) and Vesc(b) and acceleration g(b). What is your unit of g? Draw log-log diagram of Vcir(r). What is the central density in Msun pc -3 ? Compare with average density inside r=1pc. (Answer in BT, p38)

Example:Single Isothermal Sphere Model For a SINGLE ISOTHERMAL SPHERE (SIS) the line of sight velocity dispersion is constant. This also results in the circular velocity being constant (proof later). The potential and density are given by:

Proof: Density  r -2 n=-2 Log(  ) Log(r)

Proof: Potential We redefine the zero of potential  If the SIS extends to a radius r o then the mass and density distribution look like this: M r  r roro roro

Beyond r o : We choose the constant so that the potential is continuous at r=r o. r  logarithmic  r -1

Plummer Model PLUMMER MODEL=the special case of the gravitational potential of a galaxy. This is a spherically symmetric potential of the form: Corresponding to a density: which can be proved using poisson’s equation.

The potential of the plummer model looks like this:  r 

Since, the potential is spherically symmetric g is also given by:  The density can then be obtained from: dM is found from the equation for M above and dV=4  r 2 dr. This gives (as before from Poisson’s)

Isochrone Potential We might expect that a spherical galaxy has roughly constant  near its centre and it falls to 0 at sufficiently large radii. i.e. A potential of this form is the ISOCHRONE POTENTIAL.

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Stellar Orbits Once we have solved for the gravitational potential (Poisson’s eq.) of a system we want to know: How do stars move in gravitational potentials? Neglect stellar encounters use smoothed potential due to system or galaxy as a whole

Motions in spherical potential

Proof: Angular Momentum is Conserved  Sincethen the force is in the r direction.  both cross products on the RHS = 0. So Angular Momentum L is Conserved in Spherical Isotropic Self Gravitating Equilibrium Systems. Alternatively:  =r×F & F only has components in the r direction   =0 so

In static spherical potentials: star moves in a plane (r,  ) central force field angular momentum equations of motion are –radial acceleration: –tangential acceleration:

Orbits in Spherical Potentials The motion of a star in a centrally directed field of force is greatly simplified by the familiar law of conservation (WHY?) of angular momentum. Keplers 3 rd law pericentre apocentre

Energy Conservation (WHY?)  eff

Radial Oscillation in an Effective potential Argue: The total velocity of the star is slowest at apocentre due to the conservation of energy Argue: The azimuthal velocity is slowest at apocentre due to conservation of angular momentum.

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at the PERICENTRE and APOCENTRE There are two roots for One of them is the pericentre and the other is the apocentre. The RADIAL PERIOD T r is the time required for the star to travel from apocentre to pericentre and back. To determine T r we use:

The two possible signs arise because the star moves alternately in and out. In travelling from apocentre to pericentre and back, the azimuthal angle  increases by an amount:

The AZIMUTHAL PERIOD is In general will not be a rational number. Hence the orbit will not be closed. A typical orbit resembles a rosette and eventually passes through every point in the annulus between the circle of radius r p and r a. Orbits will only be closed if is an integer.

Examples: homogeneous sphere potential of the form using x=r cos  and y = r sin  equations of motion are then: –spherical harmonic oscillator Periods in x and y are the same so every orbit is closed ellipses centred on the centre of attraction.

homogeneous sphere cont. orbit is ellipse define t=0 with x=A, y=0 One complete radial oscillation: A to -A azimuth angle only increased by š A A B B t=0

Radial orbit in homogeneous sphere –equation for a harmonic oscillator angular frequency 2  /P

Altenative equations in spherical potential Let

Kepler potential Equation of motion becomes: –solution: u linear function of cos(theta): with and thus Galaxies are more centrally condensed than a uniform sphere, and more extended than a point mass, so

Tutorial Question 3: Show in Isochrone potential –radial period depends on E, not L Argue, but for –this occurs for large r, almost Kepler

1 st Tutorial g  (r)  (r)  2  (E) v esc M

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Tidal Stripping TIDAL RADIUS:Radius within which a particle is bound to the satellite rather than the host system. Consider a satellite of mass M s inside radius R is moving in a spherical potential  (r) made from a point mass M. r R M(r)

The condition for a particle to be bound to the satellite rather than the host system is: Differential (tidal) force on the particle due to the host galaxy Force on particle due to satellite

Generally, fudge factor k=1  4, bigger for radial orbits, bigger for point-like mass. Therefore Tidal Radius is (ambiguously defined as): The tidal radius is smallest at pericentre where r is smallest. Often tidal radius is only defined when r(t)=pericentre. As a satellite losses mass, its tidal radius shrinks.

The meaning of tidal radius The inequality can be written in terms of the mean densities. The less dense part of the satellite is torn out of the system, into tidal tails.

Size and Density of a BH A black hole has a finite (schwarzschild) radius R bh =2 G M bh /c 2 ~ 2au (M bh /10 8 M sun ) –verify this! What is the mass of 1cm BH? A BH has a density (3/4Pi) M bh /R bh 3, hence smallest holes are densest. –Compare density of 10 8 Msun BH with Sun (or water) and a giant star (10Rsun).

Growth of a BH by capturing objects in its Loss Cone A small BH on orbit with pericentre r p <R bh is lost (as a whole) in the bigger BH. –The final process is at relativistic speed. Newtonian theory is not adequate (Nearly radial) orbits with angular momentum J<J lc =2*c*R bh =4GM bh /c 2 enters `loss cone` (lc) When two BHs merger, the new BH has a mass somewhat less than the sum, due to gravitational radiation.

Tidal disruption near giant BH A giant star has low density than the giant BH, is tidally disrupted first. The disruption happens at radius r dis > R bh, M bh /r dis 3 ~ M*/R* 3 Giant star is shreded. Part of the tidal tail feeds into the BH, part goes out.

Adiabatic Compression due to growing BH A star circulating a BH at radius r has a velocity v=(GM bh /r) 1/2, an angular momentum J = r v =(GM bh r) 1/2, As BH grows, Potential and Orbital Energy E changes (t) But J conserved (no torque!), still circular! So J i = (GM i r i ) 1/2 =J f =(GM f r f ) 1/2 Shrink r f /r i = M i /M f < 1, orbit compressed!

Adiabatic Invariance Suppose we have a sequence of potentials  p (|r|) that depends continuously on the parameter P(t). P(t) varies slowly with time. For each fixed P we would assume that the orbits supported by  p (r) are regular and thus phase space is filled by arrays of nested tori on which phase points of individual stars move. Suppose

The orbit energy of a test particle will change. Suppose The angular momentum J is still conserved because r  F=0. Initial Final

In general, two stellar phase points that started out on the same torus will move onto two different tori. However, if potential is changed very slowly compared to all characteristic times associated with the motion on each torus, all phase points that are initially on a given torus will be equally affected by the variation in Potential Any two stars that are on a common orbit will still be on a common orbit after the variation in Pot is complete. This is ADIABATIC INVARIANCE.

Stellar interactions When are interactions important? Consider a system of N stars of mass m evaluate deflection of star as it crosses system consider en encounter with star of mass m at a distance b: b r  X=vt v F perp

Stellar interactions cont. the change in the velocity  v perp is then –using s = vt / b Or using impulse approximation: –where g perp is the force at closest approach and –the duration of the interaction can be estimated as :  t = 2 b / v

Number of encounters with impact parameter b - b +  b let system diameter be: 2R Star surface density ~ N/R 2 /Pi the number encountering –each encounter has effect  v perp but each one randomly oriented –sum is zero: b b+  b

change in kinetic energy but suming over squares (  v perp 2 ) is > 0 hence now consider encounters over all b –then –but @ b=0, 1/b is infinte! –need to replace lower limit with some b min –expected distance of closest R/N

Relaxation time hence v 2 changes by  v 2 each time it crosses the system where  v 2 is: Orbit deflected when v 2 ~  v 2 –after n relax times across the system and thus the relaxation time is:

Relaxation time cont. collisionless approx. only for t < t relax ! mass segregation occurs on relaxation timescale –also referred to as equipartition –where kinetic energy is mass independent –Hence the massive stars, with lower specific energy sink to the centre of the gravitational potential.

globular cluster, N=10 5, R=10 pc –t cross ~ 2 R / v ~ 10 5 years –t relax ~ 10 8 years << age of cluster: relaxed galaxy, N=10 11, R=15 kpc –t cross ~ 10 8 years –t relax ~ 10 15 years >> age of galaxy: collisionless cluster of galaxie: t relax ~ age

Dynamical Friction DYNAMICAL FRICTION slows a satellite on its orbit causing it to spiral towards the centre of the parent galaxy. As the satellite moves through a sea of stars I.e. the individual stars in the parent galaxy the satellites gravity alters the trajectory of the stars, building up a slight density enhancement of stars behind the satellite The gravity from the wake pulls backwards on the satellites motion, slowing it down a little

The satellite loses angular momentum and slowly spirals inwards. This effect is referred to as “dynamical friction” because it acts like a frictional or viscous force, but it’s pure gravity.

More massive satellites feel a greater friction since they can alter trajectories more and build up a more massive wake behind them. Dynamical friction is stronger in higher density regions since there are more stars to contribute to the wake so the wake is more massive. For low v the dynamical friction increases as v increases since the build up of a wake depends on the speed of the satellite being large enough so that it can scatter stars preferentially behind it (if it’s not moving, it scatters as many stars in front as it does behind). However, at high speeds the frictional force  v -2, since the ability to scatter drops as the velocity increases. Note: both stars and dark matter contribute to dynamical friction

The dynamical friction acting on a satellite of mass M moving at v s kms -1 in a sea of particles of density mXn(r) with gaussian velocity distribution Only stars moving slower than M contribute to the force. This is usually called the Chandrasekhar Dynamical Friction Formula.

For an isotropic distribution of stellar velocities this is: For a sufficiently large v M, the integral converges to a definite limit and the frictional force therefore falls like v M -2. For sufficiently small v M we may replace f(v M ) by f(0), define friction timescale by:

Friction & tide: effects on satellite orbit When there is dynamical friction there is a drag force which dissipates angular momentum. The decay is faster at pericentre resulting in the staircase-like decline of J(t). As the satellite moves inward the tidal force becomes greater so the tidal radius decreases and the mass will decay.

Collisionless Systems stars move under influence of a smooth gravitational potential –determined by overall structure of system Statistical treatment of motions –collisionless Boltzman equation –Jeans equations

provide link between theoretical models (potentials) and observable quantities. instead of following individual orbits study motions as a function of position in system Use CBE, Jeans eqs. to determine mass distributions and total masses

Collisionless Systems We showed collisions or deflections are rare Collisionless: stellar motions under influence of mean gravitational potential! Rational: Gravity is a long-distance force, decreases as r -2 –as opposed to the statistical mechanics of molecules in a box

Fluid approach:Phase Space Density PHASE SPACE DENSITY:Number of stars per unit volume per unit velocity volume f(x,v) (all called Distribution Function DF). The total number of particles per unit volume is given by:

E.g., air particles with Gaussian velocity (rms velocity = σ in x,y,z directions): The distribution function is defined by: mdN=f(x,v)d 3 xd 3 v where dN is the number of particles per unit volume with a given range of velocities. The mass distribution function is given by f(x,v).

The total mass is then given by the integral of the mass distribution function over space and velocity volume: Note:in spherical coordinates d 3 x=4πr 2 dr The total momentum is given by:

The mean velocity is given by:

Example:molecules in a room: These are gamma functions

Gamma Functions:

DF and its moments

Additive equations

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Liouvilles Theorem We previously introduced the concept of phase space density. The concept of phase space density is useful because it has the nice property that it is incompessible for collisionless systems. A COLLISIONLESS SYSTEM is one where there are no collisions. All the constituent particles move under the influence of the mean potential generated by all the other particles. INCOMPRESSIBLE means that the phase-space density doesn’t change with time.

Consider Nstar identical particles moving in a small bundle through spacetime on neighbouring paths. If you measure the bundles volume in phase space (Vol=Δ x Δ p ) as a function of a parameter λ (e.g., time t) along the central path of the bundle. It can be shown that: df/dt=0! It can be seen that the region of phase space occupied by the particle deforms but maintains its area. The same is true for y-p y and z-p z. This is equivalent to saying that the phase space density f=Nstars/Vol is constant.  df/dt=0! pxpx x p x x

motions in phase-space Flow of points in phase space corresponding to stars moving along their orbits. phase space coords: and the velocity of the flow is then: –where wdot is the 6-D vector related to w as the 3-D velocity vector v relates to x

stars are conserved in this flow, with no encounters, stars do not jump from one point to another in phase space. they drift slowly through phase space

fluid analogy regard stars as making up a fluid in phase space with a phase space density assume that f is a smooth function, continuous and differentiable –good for N 10 5

as in a fluid, we have a continuity equation fluid in box of volume V, density , and velocity v, the change in mass is then: –Used the divergence theorem

continuity equation must hold for any volume V, hence: in same manner, density of stars in phase space obeys a continuity equation: If we integrate over a volume of phase space V, then 1 st term is the rate of change of the stars in V, while 2 nd term is the rate of outflow/inflow of stars from/into V. 0

Collisionless Boltzmann Equation Hence, we can simplify the continuity equation to the CBE: Vector form

in the event of stellar encounters, no longer collisionless require additional terms to rhs of equation

CBE cont. can define a Lagrangian derivative Lagrangian flows are where the coordinates travel along with the motions (flow) –hence x= x 0 = constant for a given star then we have: and rate of change of phase space density seen by observer travelling with star the flow of stellar phase points through phase space is incompressible f around the phase point of a given star remains the same

incompressible flow example of incompressible flow idealised marathon race: each runner runs at constant speed At start: the number density of runners is large, but they travel at wide variety of speeds At finish: the number density is low, but at any given time the runners going past have nearly the same speed

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DF & Integrals of motion If some quantity I(x,v) is conserved i.e. We know that the phase space density is conserved i.e Therefore it is likely that f(x,v) depends on (x,v) through the function I(x,v), so f=f(I(x,v)).

Jeans theorem For most stellar systems the DF depends on (x,v) through generally three integrals of motion (conserved quantities), I i (x,v),i=1..3  f(x,v) = f(I 1 (x,v), I 2 (x,v), I 3 (x,v)) E.g., in Spherical Equilibrium, f is a function of energy E(x,v) and ang. mom. vector L(x,v).’s amplitude and z-component

Analogy DF(x,v) Analogous to density(x,y,z), DF(E,L,Lz) analogous to density(r,theta,phi), E(x,v) analogous to r(x,y,z) Integrals analogous to spherical coordinates Isotropic F(E) analogous to spherical density(r) Normalization dM=f(E)*dx^3*dv^3= density(r)*dv^3

Spherical Equilibrium System Described by potential φ(r) SPHERICAL: density ρ(r) depends on modulus of r. EQUILIBRIUM:Properties do not evolve with time.

Anisotropic DF f(E,L,Lz). Energy is conserved as: Angular Momentum Vector is conserved as: DF depends on Velocity Direction through L=r X v

e.g., F(E,L) is an incompressible fluid The total energy of an orbit is given by: 0 for static potential, 0 for spherical potential So F(E,L) constant along orbit or flow

Stress Tensor describes a pressure which is anisotropic –not the same in all directions and we can refer to a “pressure supported” system the tensor is symmetric. can chose a set of orthogonal axes such that the tensor is diagonal Velocity ellipsoid with semi-major axes given by

To prove the above Jeans eq.

Velocity dispersions and masses in spherical systems For a spherically symmetric system we have a non-rotating galaxy has –and the velocity ellipsoids are spheroids with their symmetry axes pointing towards the galactic centre Define anisotropy

Spherical mass profile from velocity dispersions. Get M(r) or Vcir from: rhs observations of dispersion and  as a function of radius r for a stellar population.

Total Mass of spherical systems E.g. Motions of globular clusters and satellite galaxies around 100kpc of MW –Need n(r), v r 2,  to find M(r), including dark halo Several attempts all suffer from problem of small numbers N ~ 15 For the isotropic case, Little and Tremaine TOTAL mass of 2.4 (+1.3, -0,7) 10 11 M sol 3 times the disc  need DM

Isotropic orbits: Radial orbits If we assume a power law for the density distribution –E.g. Flat rotation a=1, Self-grav gamma=2, Radial anis.  –E.g., Point mass a=0, Tracer gam=3.5, Isotro  

Mass of the Milky Way We find Drop first term, solutions

Scalar Virial Theorem The Scalar Virial Theorem states that the kinetic energy of a system with mass M is just where is the mean-squared speed of the system’s stars. Hence the virial theorem states that Virial

Equation of motion: This is Tensor Virial Theorem

E.g. So the time averaged value of v 2 is equal to the time averaged value of the circular velocity squared.

In a spherical potential So =0 since the average value of xy will be zero.  =0

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Spherical Isotropic Self Gravitating Equilibrium Systems ISOTROPIC:The distribution function only depends on the modulus of the velocity rather than the direction. Note:the tangential direction has  and  components

Isotropic DF f(E) In a static potential the energy of an particle is conserved. So,if we write f as a function of E then it will agree with the statement: Note:E=energy per unit mass For incompressible fluids

So: E=cst since For a bound equilibrium system f(E) is positive everywhere (can be zero) and is monotonically decreasing.

SELF GRAVITATING:The masses are kept together by their mutual gravity. In non self gravitating systems the density that creates the potential is not equal to the density of stars. e.g a black hole with stars orbiting about it is not self gravitating.

Eddington Formulae EDDINGTON FORMULAE:These can be used to get the density as a function of r from the energy density distribution function f(E).

Proof of the 1 st Eddington Formula

So, from a given distribution function we can compute the spherical density.

Relation Between Pressure Gradient and Gravitational Force Pressure is given by:

So, this gives: This relates the pressure gradient to the gravitational force. This is the JEANS EQUATION. Note:  2 =P Note:-ve sign has gone since we reversed the limits.

g  (r)  (r)  2  (E) v esc M So, gravity, potential, density and Mass are all related and can be calculated from each other by several different methods.

2 nd Tutorial g  (r)  (r)  2  (E) v esc M

Proof: Situation where V c 2 =const is a Singular Isothermal Sphere From Previously: Conservation of momentum gives:

 

 Since the circular velocity is independent of radius then so is the velocity dispersion  Isothermal.

Finding the normalising constant for the phase space density If we assume the phase space density is given by: 

We can then find the normalizing constant so that  (r) is reproduced. Note: you want to integrate f(E) over all energies that the star can have I.e. only energies above the potential  We are integrating over stars of different velocities ranging from 0 to .

One way is to stick the velocity into the distribution function: Using the substitution gives:

Now  can also be found from poissons equation. Substituting in  from before gives: Equating the r terms gives: as before.

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Flattened Disks Here the potential is of the form  (R,z). No longer spherically symmetric. Now it is Axisymmetric

Orbits in Axisymmetric Potentials (disk galaxies) cylindrical (R, ,z) symmetry z-axis stars in equatorial plane: same motions as in spherically symmetric potential –non-closed rosette orbits stars moving out of plane –can be reduced to 2-D problem in (R,z) –conservation of z-angular momentum, L z z R y x  R 2 =x 2 +y 2

Orbits in Axisymmetric Potentials We employ a cylindrical coordinate system (R, ,z) centred on the galactic nucleus and align the z axis with the galaxies axis of symmetry. Stars confined to the equatorial plane have no way of perceiving that the potential is not spherically symmetric.  Their orbits will be identical to those in spherical potentials. R of a star on such an orbit oscillates around some mean value as the star revolves around the centre forming a rosette figure.

Reducing the Study of Orbits to a 2D Problem This is done by exploiting the conservation of the z component of angular momentum. Let the potential which we assume to be symmetric about the plane z=0, be  (R,z). The general equation of motion of the star is then: The acceleration in cylindrical coordinates is: Motion in the meridional plane

The component of angular momentum about the z- axis is conserved. If  has no dependence on  then the azimuthal angular momentum is conserved (r  F=0). Eliminating in the energy equation using conservation of angular momentum gives: Specific energy density in 3D  eff

Motions in Meridional Plane EoM in  (R,z) –: –in cylindrical coords: –and Lz conserved

Thus, the 3D motion of a star in an axisymmetric potential  (R,z) can be reduced to the motion of a star in a plane. This (non uniformly) rotating plane with cartesian coordinates (R,z) is often called the MERIDIONAL PLANE.  eff (R,z) is called the EFFECTIVE POTENTIAL.

effective potential  eff (R,z) coupled equations for oscillations in R,z directions use L z to replace by reduced to motion in meridional plane (R,z)

So the minimum in  eff occurs at the radius at which a circular orbit has angular momentum L z. The value of  eff at the minimum is the energy of this circular orbit. R  eff E  R cir

The orbits are bound between two radii (where the effective potential equals the total energy) and oscillates in the z direction.

Example: Logarithmic potential oblate galaxy with V circ ~ V 0 =100km/s Draw contours of the corresponding Self- gravitating Density to show it is unphysical. Plot effective potential contours for Lz=100kpckm/s. orbits with E=Φ(1kpc,0), what is maximum z-height? What is R g of a circular orbit with E= Φ(1kpc,0)?

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Total Angular momentum almost conserved These orbits can be thought of as being planar with more or less fixed eccentricity. The approximate orbital planes have a fixed inclination to the z axis but they process about this axis. star picks up angular momentum as it goes towards the plane and returns it as it leaves.

Orbital energy Energy of orbit is (per unit mass) effective potential is the gravitational potential energy plus the specific kinetic energy associated with motion in  direction orbit bound within

The angular momentum barrier for an orbit of energy E is given by The effective potential cannot be greater than the energy of the orbit. The equations of motion in the 2D meridional plane then become:.

The effective potential is the sum of the gravitational potential energy of the orbiting star and the kinetic energy associated with its motion in the  direction (rotation). Any difference between  eff and E is simply kinetic energy of the motion in the (R,z) plane. Since the kinetic energy is non negative, the orbit is restricted to the area of the meridional plane satisfying. The curve bounding this area is called the ZERO VELOCITY CURVE since the orbit can only reach this curve if its velocity is instantaneously zero.

Minimum in  eff The minimum in  eff has a simple physical significance. It occurs where: (2) is satisfied anywhere in the equatorial plane z=0. (1) is satisfied at radius R g where This is the condition for a circular orbit of angular speed

conditions for a circular orbit at R g minimum in effective potential at R,z = R g,0 with angular speed circular orbit with angular momentum L z

If the energy of the orbit is reduced the two points between which the orbit is bound eventually become one. You then get no radial oscillation. You have circular orbits in the plane of the galaxy. This is one of the closed orbits in an axisymmetric potential and has the property that. (Minimum in effective potential.) Centrifugal Force Gravitational force in radial direction

Nearly circular orbits: epicycles In disk galaxies, many stars (disk stars) are on nearly-circular orbits EoM: x=R-R g –expand in Taylor series about (x,z)=(0,0) –then

When the star is close to z=0 the effective potential can be expanded to give Zero, changes sign above/below z=0 equatorial plane. 2 So, the orbit is oscillating in the z direction.

epicyclic approximation ignore all higher / cross terms: EoM: harmonic oscillators –epiclyclic frequency : –vertical frequency : –with

epicycles cont. using the circular frequency , given by –so that disk galaxy:  ~ constant near centre –so  ~ 2   ~ declines with R, »slower than Keplerian R -3/2 »lower limit is  ~  in general  <  2  R V rot

Example:Oort’s constants near Sun –where R 0 is the galacto-centric distance then  2 = -4A(A-B) + 4(A-B) 2 = -4B(A-B) = -4B  0 Obs. A = 14.5 km/s /kpc and B=-12 km/s /kpc

the sun makes 1.3 oscillations in the radial direction per azimuthal (2  ) orbit –epicyclic approximation not valid for z-motions when |z|>300 pc

16 th Lec orbits

JEANS EQUATION for oblate rotator : a steady-state axisymmetrical system in which  2 is isotropic and the only streaming motion is azimuthal rotation:

The velocity dispersions in this case are given by: If we know the forms of  (R,z) and  (R,z) then at any radius R we may integrate the Jeans equation in the z direction to obtain  2.

Obtaining  2 Inserting this into the jeans equation in the R direction gives:

Example:

17 th Lec orbits

Orbits in Planar Non-Axisymmetrical Potentials Here the angular momentum is not exactly conserved. There are two main types of orbit –BOX ORBITS –LOOP ORBITS

LOOP ORBITS Star rotates in a fixed sense about the centre of the potential while oscillating in radius Star orbits between allowed radii given by its energy. There are two periods associated with the orbit: –Period of the radial oscillation –Period of the star going around 2  The energy is generally conserved and determines the outer radius of the orbit. The inner radius is determined by the angular momentum. pericentre apocentre

Box Orbits Have no particular sense of circulation about the centre. They are the sum of independent harmonic motions parallel to the x and y axes. In a logarithmic potential of the form box orbits will occur when R >R c.

Orbits in 2D elliptical potentials bars in nulcear regions of disc galaxies: SBs E.g., non-rotating logarithmic potential –equipotential ellipses constant axial ratios, q –for small R << R c, expand: –potential of a 2-D harmonic oscillator (same as for an homogenous ellipsoid)

non-rotating potentials cont. Box Orbits –Within core, harmonic oscillators: –unless frequencies are commensurable: –independent oscillations, stars eventually pass close to every point inside a rectangle:

For large R, R >> R c, –numerical integrations are required if launched with a tangential velocity LOOP Orbits orbits rotate in fixed sense about centre of potential –oscillate in radius between R min and R max –never approach centre! –fills in annulus, of width determined by L z,

Loop and Box orbits motions in plane y=0 and with v Y > 0 –with given energy E –loop orbits are in annuli around “closed loop orbit” »equivalent to circular orbit in axisymmetric potential –box orbits outermost curve denotes y=0 and v Y =0; “closed box orbit” motion simply parallel to x-axis x vxvx loop orbits, anticlockwise box orbits loop orbits, clockwise 0 0

relative proportion of loop versus box orbits More box orbits when energy is lower –all orbits with E < Phi(R c ) are box orbits More box orbits when increasingly non- axisymmetric all orbits are loop orbits in axisymmetric potentials –always rotate in fixed sense (conserves L) –loop orbits become more elongated –less epicyclic motion necessary to fill in central hole becomes a box orbit

Orbits in 3-D Triaxial Potentials –E.g., Elliptical Galaxies For given Energy have Within core of potential: (box orbits) –3-D harmonic oscillator –adopt long-axis orbit as parent to family –all axial orbits stable Outside core : potentially 3 axial orbits and 3 loop orbits –one about each axis, stable? –short and middle-axis (box) orbits are unstable –middle-axis tube (loop) axis unstable

have one box orbit (long axis) and two loop orbits (long-axis and short-axis) closed orbits and hence parents of non-closed families disc stars in spirals are short-axis tube orbits

18 th Lec orbits

2-D rotating potentials non-axisymmetric potentials generally rotate! frame of reference (x,y,z) in which  L is static rotates at angular velocity  b =  b e z EoM in this rotating cordinate system are: centrifugal coriolis Effective potential: as used in the binary stars section

Five Lagrange points as in Roche-lobe potential –L 3, potential minimum at central stationary point –L 1 and L 2, saddle points : unstable –L 4 and L 5, potential maxima: stable for certain  –Annulus bounded by L 1,2,4,5 : stars appear to be stationary, called Co-Rotation Radius L5L5 L4L4 L2L2 L1L1 L3L3

Co-Rotation Radius Co-Rotation:  b –where angular velocity of a star is the angular velocity of the rotating potential stars stay in same place relative to potential For R>> R Co-Rotation, all closed orbits nearly circular Barred potential spins much faster than stars –so asymmetry is averaged out For R < R Co-Rotation, most heavily populated orbits parented by long-axis (box) orbits –aligned with potential

Lindblad Resonances barred potential in polar coordinates (R,  ) in frame rotating with potential where  =0 is long axis loop orbits as circular with small epicyclic oscillations for a barred potential: –m=2 ensures a barred pot. small motions R 1 (t) are harmonic oscillator, frequency  0, driven at frequency m(  0 -  b )

Lindblad Resonances cont. At Co-Rotation, no oscillations At Lindblad resonances, m(  0 -  b )= ±  0, and star encounters successive crests of the potential at a frequency that coincides with natural frequency of radial oscillations. + sign: star overtakes potential --- Inner LR - sign: potential overtakes star --- Outer LR e.g. MW: outer LR inner LR bar

Lagrange Points There is a point between two bodies where a particle can belong to either one of them. This point is known as the LAGRANGE POINT. A small body orbiting at this point would remain in the orbital plane of the two massive bodies. The Lagrange points mark positions where the gravitational pull of the two large masses precisely cancels the centripetal acceleration required to rotate with them. At the lagrange points:

Effective Force of Gravity A particle will experience gravity due to the galaxy and the satellite along with a centrifugal force and a coriolis term. The effective force of gravity is given by: The acceleration of the particle is given by: Coriolis term Centrifugal term

Jakobi’s Energy The JAKOBI’S ENERGY is given by: Jakobi’s energy is conserved because

For an orbit in the plane (r perpendicular to  )

E J is also known as the Jakobi Integral. Since v 2 is always positive a star whose Jakobi integral takes the value E J will never tresspass into a region where  eff (x)>E J. Consequently the surface  eff (x)>E J, which we call the zero velocity surface for stars of Jakobi Integral E J, forms an impenetrable wall for such stars.

By taking a Taylor Expansion r=r s, you end up with: Where we call the radius r J the JAKOBI LIMIT of the mass m. This provides a crude estimate of the tidal radius r t.

19 th Lec orbits

The Jeans Equations The DF (phase space density f) is a function of 7 variables and hence generally difficult to solve Can gain insights by taking moments of the equation. where integrate over all possible velocities –where the summation over subscripts is implicit

Jeans equations cont. –first term: velocity doesn’t depend on time, hence we can take partial w.r.t. time outside –second term, v i does not depend on x i, so we can take partial w.r.t x i outside –third term: apply divergence theorem so that but at very large velocities, f 0, hence last term is zero must be true for all bound systems

First Jeans equation define spatial density of stars n(x) and the mean stellar velocity v(x) then our first (zeroth) moment equation becomes this is the first Jeans equation analogous to the continuity equation for a fluid.

2 nd Jeans equation multiply the CBE by v j and then integrate over all velocities We get

3rd Jeans Equation similar to the Euler equation for a fluid flow: –last term of RHS represents some sort of pressure force

Jeans Equation Compact form, s=x, y, z, R, r, … e.g., oblate spheroid, s=[R,phi,z], –Isotropic rotator, a=[-Vrot^2/R, 0, 0], sigma=sigma_s –Tangential anisotropic (b<0), a=[b*sigma^2)/R, 0, 0], sigma=sigma_R=sigma_z=sigma_phi/(1-b),

e.g., non-rotating sphere, s=[r,th,phi], a=[-2*b*sigma_r^2/r, 0, 0], sigma_th=sigma_phi=(1-b)*sigma_r

20 th Lec orbits

Applications of the Jeans Equations I. The mass density in the solar neighbourhood Using velocity and density distribution perpendicular to the Galactic disc –cylindrical coordinates. –Ignore R dependence

Vertical Jeans equation Small z/R in the solar neighbourhood, R~8.5 kpc, |z|< 1kpc, R-dependence neglected. Hence, reduces to vertical hydrostatic eq.:

mass density in solar neighbourhood Drop R, theta in Poisson’s equation in cylindrical coordinates:

local mass density  =  0 Finally all quantities on the LHS are, in principle, determinable from observations. RHS Known as the Oort limit. Uncertain due to double differentiation!

local mass density Don’t need to calculate for all stars –just a well defined population (ie G stars, BDs etc) –test particles (don’t need all the mass to test potential) Procedure –determine the number density n, and the mean square vertical velocity, v z 2, the variance of the square of the velocity dispersion in the solar neighbourhood. –need a reliable “tracer population” of stars whose motions do not reflect formation hence old population that has orbited Galaxy many times ages > several x 10 9 years N.B. problems of double differentiation of the number density n derived from observations need a large sample of stars to obtain v z as f(z) –ensure that v z is constant in time –ie stars have forgotten initial motion

local mass density > 1000 stars required Oort :  0 = 0.15 M sol pc -3 K dwarf stars (Kuijken and Gilmore 1989) –MNRAS 239, 651 Dynamical mass density of  0 = 0.11 M sol pc -3 also done with F stars (Knude 1994)

Observed mass density of stars plus interstellar gas within a 20 pc radius is  0 = 0.10 M sol pc -3 can get better estimate of surface density out to 700 pc  ~ 90 M sol pc -2 from rotation curve  rot ~ 200 M sol pc -2

Helpful Math/Approximations (To be shown at AS4021 exam) Convenient Units Gravitational Constant Laplacian operator in various coordinates Phase Space Density f(x,v) relation with the mass in a small position cube and velocity cube

21 th Lec: MOND orbits

Gravitational Dynamics. Gravitational Dynamics can be applied to: Two body systems:binary stars Planetary Systems Stellar Clusters:open & globular Galactic.

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Gravitational Dynamics. Gravitational Dynamics can be applied to: Two body systems:binary stars Planetary Systems Stellar Clusters:open & globular Galactic.

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Presentation on theme: "Gravitational Dynamics. Gravitational Dynamics can be applied to: Two body systems:binary stars Planetary Systems Stellar Clusters:open & globular Galactic."— Presentation transcript:

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