KeyKey12-10-07 see PointPoint http://www.strw.leidenuniv.nl/˜ #15#10 fromfrom franx/college/ BachelorBachelor mf-sts-07-c4-5 Course:Course:12-10-07 see http://www.strw.leidenuniv.nl/˜ IntegralsIntegrals ofof franx/college/ MotionMotion mf-sts-07-c4-6

4.2 Constants and Integrals of motion Integrals (BT 3.1 p 110-113) Much harder to define. E.g.: 1 2 Energy (all static potentials): E(⃗x,⃗v)= 2 v + Φ First, we define the 6 dimensional “phase space” coor- dinates (⃗x,⃗v). They are conveniently used to describe Lz (axisymmetric potentials) the motions of stars. Now we introduce: L⃗ (spherical potentials)

• Integrals constrain geometry of orbits. • Constant of motion: a function C(⃗x,⃗v, t) which is constant along any orbit: examples:

C(⃗x(t1),⃗v(t1),t1)=C(⃗x(t2),⃗v(t2),t2) • 1. Spherical potentials: E,L ,L ,L are integrals of motion, but also E,|L| C is a function of ⃗x, ⃗v, and time t. x y z and the direction of L⃗ (given by the unit vector ⃗n, Integral of motion which is defined by two independent numbers). ⃗n de- • : a function I(x, v) which is fines the plane in which ⃗x and ⃗v must lie. Define coor- constant along any orbit: dinate system with z axis along ⃗n I[⃗x(t1),⃗v(t1)] = I[⃗x(t2),⃗v(t2)] ⃗x =(x1,x2, 0) I is not a function of time ! Thus: integrals of motion are constants of motion, ⃗v =(v1,v2, 0) but constants of motion are not always integrals of motion! → ⃗x and ⃗v constrained to 4D region of the 6D phase space. In this 4 dimensional space, |L| and E are con- E.g.: for a circular orbit ψ = Ω t + ψo, so that C = served. This constrains the orbit to a 2 dimensional t − ψ/Ω. space. Hence the velocity is uniquely defined for a C is constant of motion, but not an integral as it de- given ⃗x pends on t. 2 2 vr = ± 2(E − Φ) − L /r Constants of motion ! 6 for any arbitrary orbit: vψ = ±L/r Initial position (⃗x0,⃗v0) at time t = t0.Canalways be calculated back from ⃗x,⃗v, t. 12-10-07 see http://www.strw.leidenuniv.nl/˜ franx/college/ mf-sts-07-c4-7 12-10-07 see http://www.strw.leidenuniv.nl/˜ franx/college/ mf-sts-07-c4-8

• 2. Integrals in 2 dimensional flattened potentials 4 4 2 2 Examples: 0 0 -2 -2 Circular potential V (x, y)=V (⃗r) -4 -4 -4 -2 0 2 4 -4 -2 0 2 4 Two integrals: E,Lz. x0=0.5 y0=0 vx0=0 vy0=1 x0=1 y0=0 vx0=0 vy0=1 2 2 y 4 4 Flattened potential V (x, y)=ln(x + a + 1) Only “classic” integral of motion: E 2 2 0 0

-2 -2 Figures on the next page show the orbits that one gets by integrating the equations of motion. -4 -4 -4 -2 0 2 4 -4 -2 0 2 4 Clearly the orbits are regular and do not fill equipoten- x0=1.1 y0=0 vx0=0 vy0=1 x0=1.2 y0=0 vx0=0 vy0=1 tial surface the first 3 are box orbits (no net angular momen- tum) Furthermore, they do not traverse each point in a ran- dom direction, but generally only in 2 directions avoid outer x-axis Conclusion: the orbits do not occupy a 3 dimensional space in the 4-dimensional phase-space, but they 4 4 occopy only a 2-dimensional space ! 2 2 This indicates that there is an additional integral of 0 0 motion: ’a non-classical integral’ -2 -2 The non-classical integral, plus the regular ’Energy’, -4 -4 -4 -2 0 2 4 -4 -2 0 2 4 constrain the orbit to lie on a 2 dimensional surface x0=2 y0=0 vx0=0 vy0=1 x0=4 y0=0 vx0=0 vy0=1 in the 4 dimensional phase-space. loop orbits (with net angular momentum) avoid inner x-axis can circulate in two directions. 12-10-07 see http://www.strw.leidenuniv.nl/˜ franx/college/ mf-sts-07-c4-9 12-10-07see http://www.strw.leidenuniv.nl/˜ franx/college/ mf-sts-07-c4-10

A homogeneous ellipsoid Orbits for general homogeneous ellipsoid All orbits are box orbits The homogeneous ellipsoid helps us to understand how additional integrals of motions, and box orbits, exist. 4 4 Consider a density distribution: 2 2

0 0 2 ρ = ρ0H(1 − m ), with -2 -2 2 x2 y2 z2 -4 -4 m = a2 + b2 + c2 -4 -2 0 2 4 -4 -2 0 2 4 x0=0.5 y0=0 vx0=0 vy0=1 x0=1 y0=0 vx0=0 vy0=1 H(x)=1for x ≥ 0, H(x)=0for x<0 4 4

2 2

Potential inside the ellipsoid: 0 0

2 2 2 -2 -2 V = Axx + Ayy + Azz + C0 -4 -4

-4 -2 0 2 4 -4 -2 0 2 4 x0=1.1 y0=0 vx0=0 vy0=1 x0=1.2 y0=0 vx0=0 vy0=1

Forces are of the form Fi = −Aixi,i.e.3inde- pendent harmonic oscillators: 4 4 2 2

0 0 xi = ai cos(ωit + ψ0,i) -2 -2

-4 -4

-4 -2 0 2 4 -4 -2 0 2 4 3 integrals of motion, Ei x0=2 y0=0 vx0=0 vy0=1 x0=4 y0=0 vx0=0 vy0=1

Hence the orbits are as shown on the next page 12-10-07see http://www.strw.leidenuniv.nl/˜ franx/college/ mf-sts-07-c4-11 12-10-07see http://www.strw.leidenuniv.nl/˜ franx/college/ mf-sts-07-c4-12

4 • 3. Axisymmetric potentials (BT 3.2) Φ = Φ(R, |z|), with R2 = x2 + y2.

2 For z=0: orbits as if potential were circular For (R,z): 3D orbit can be described as 2D orbit in 0 meridional plane:

d2⃗r -2 = −∇Φ(R, z) dt2

-4 with ⃗er,⃗eψ,⃗ez unit vectors in r, ψ and z direction:

-4 -2 0 2 4 x0=0.5x0=1.1x0=1.2x0=1x0=2x0=4 y0=0y0=0 vx0=0vx0=0 vy0=1vy0=1 ⃗r = R ⃗e + z ⃗e Special case: a=b r z all orbits are loop orbits δΦ δΦ ∇Φ = ⃗e + ⃗e δR r δz z The equations of motion reduce to

d2R −δΦ − Rψ˙ 2 = dt2 δR

d δΦ R2ψ˙ = − =0 dt δψ d2z δΦ = − dt2 δz

From the second equation we see that Lz is constant. ˙ 2 2 Hence ψ = Lz/R . If we use this in the upper equa- tion we obtain for the motion in two dimensions R, z: 12-10-07see http://www.strw.leidenuniv.nl/˜ franx/college/ mf-sts-07-c4-13 12-10-07see http://www.strw.leidenuniv.nl/˜ franx/college/ mf-sts-07-c4-14

d2R δΦ d2z δΦ = − eff = − eff dt2 δR dt2 δz

2 Lz with Φeff = Φ(R, z)+ 2R2 . Hence 3D motion can be reduced to motion in (R,z) plane or meridional plane, under influence of the effective potential Φeff . Total energy:

2 2 E = Φeff +1/2R˙ +1/2˙z

. Allowed region in meridional plane For energy E motion only allowed in area where Φeff

2 2 1 2 2 z Lz Φ = v0 ln(R + )+ eff 2 q2 2R2

, i.e. only motion if E>Φeff . 12-10-07see http://www.strw.leidenuniv.nl/˜ franx/college/ mf-sts-07-c4-15 12-10-07see http://www.strw.leidenuniv.nl/˜ franx/college/ mf-sts-07-c4-16

Lines of constant Φeff are shown in Figure 3.2. Stars integral of motion must play a role → dimensionality with energy E have zero velocity at curves of Φeff = reduced to 3 (e.g. R, z, ψ). E. This integral is a non-classical integral of motion

Homework Assignments: 1) Box orbits are characterized by the fact they they go through the center, and have no net angular momen- tum. 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. 2) 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 ? 4) Why are there no loop orbits in a homogeneous el- lipsoid ? 5) Calculate at least 3 orbits in the 2-dimentional po- tential Φ =ln(1+x2 + y2/2). Do this as follows: Start the star at a given location (x0, 0) with velocity (0,v0). Calculate the shift in position at time dt by dt*⃗v.Cal- culate the shift in velocity at time dt by dt*F⃗ .And keep integrating !

Again: not all orbits fill the space Φeff

Two integrals (E,Lz) reduce the dimensionality of the orbit from 6 to 4 (e.g. R, z, ψ, vz). Therefore another 12-10-07see http://www.strw.leidenuniv.nl/˜ franx/college/ mf-sts-07-c4-17 12-10-07see http://www.strw.leidenuniv.nl/˜ franx/college/ mf-sts-07-c4-18

A general 3-dimensional potential A Simple recipe to build galaxies Schwarzschild’s method:

2 2 • Define density ρ St¨ackel potential( ρ =1/(1 + m ) ) • Calculate potential, forces • Integrate orbits, find orbital densities ρi • Calculate weights wi > 0 such that

ρ = ρiwi !

Examples: build a 2D galaxy in a logarithmic potential Φ =ln(1+x2 + y2/a).

• 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, nega- tive, or “neutral”.

Hence: a maximum rotation is defined if all loop orbits rotate the same way. The rotation can vary between zero, and this maximum rotation 12-10-07 see http://www.strw.leidenuniv.nl/˜ franx/college/ mf-sts-07-c4-1 12-10-07 see http://www.strw.leidenuniv.nl/˜ franx/college/ mf-sts-07-c4-2

rewrite equations of motion in polar coordinates 2 4 Orbits in stationary Potentials (BT 3 to r¨ − rψ˙ = F (r) page 107) 2˙rψ˙ + rψ¨ = Fψ Because of the circular symmetry, we have Fψ =0. Now we have seen how to calculate forces and po- Hence: tentials from the smoothed density ρ.Wecannow 2 ˙ 1 dr ψ 2 st analyse how stars move in this potential. Because two 2˙rψ˙ + rψ¨ = =0 ⇒ r ψ˙ = rv⊥ = L =c body interactions can be ignored, we can analyse each r dt star by itself. We therefore speak of “orbits” L2 dΦ r¨ − rψ˙ 2 =¨r − = − r3 dr 4.1 Orbits in spherical potentials where Φ is the potential. Multiply the last equation by r˙, and integrate w.r.t. t:

Potential function of r = |⃗r|: Φ = Φ(r) 2 1 2 L equation of motion for star with unit mass r˙ = E − Φ − = E − Φeff (r) 2 2r2 d2⃗r 2 = F (r)⃗er with E the energy. dt This recall that ⃗r × ⃗r =0for any ⃗r equation governs radial motion effective poten- 2 in d d⃗r d⃗r d⃗r d ⃗r tial Φeff (r) r × = × + ⃗r × = F (r)⃗r × ⃗er =0 dt ! dt " dt dt dt2 Motion possible 2 Hence L⃗ = ⃗r × ⃗r˙ is constant with time. L⃗ = angu- only when r˙ ≥ 0 lar momentum/unit mass . L⃗ is always perpendicular rmin ≤ r ≤ rmax to the plane in which ⃗r and ⃗v lie. Since it is constant pericenter apocenter with time, these vectors always lie in the same plane. Hence the orbit is constrained to this plane. Use polar coordinates (r, ψ) in orbital plane: 12-10-07 see http://www.strw.leidenuniv.nl/˜ franx/college/ mf-sts-07-c4-3 12-10-07 see http://www.strw.leidenuniv.nl/˜ franx/college/ mf-sts-07-c4-4

Typical orbit in a spherical potential is a planar rosette ⃗¨r = −Ω2⃗r or in cartesian coordinates x, y

x¨ = −Ω2x y¨ = −Ω2y Hence solutions are

x = X cos(Ωt + cx) y = Y cos(Ωt + cy)

where X, Y, cx and cy are arbitracy constants. Hence, Angle ∆ψ between successive apocenter passages de- even though energy and angular momentum restrict pends on mass distribution: 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 ! π < ∆ψ < 2π The same holds for Kepler potential. But beware, for the homogeneous sphere the particle does two radial homogeneous sphere point mass excursions per cycle around the center, for the Kepler potential, it does one radial excursion per angular cy- cle. Special cases We now wish to “classify” orbits and their density dis- 2 tribution in a systematic way. For that we use Integrals v⊥ dΦ GM(r) rmin = rmax circular orbit = = of motion. r dr r2

1 2 L =0 ⇒ radial orbit 2 r˙ = E − Φ(R)

Homogeneous sphere

1 2 2 Φ(r)= 2 Ω r + Constant

In radial coordinates Key Point #16 from Bachelor Course: Solving the Collisionless

15-10-07 see http://www.strw.leidenuniv.nl/˜ franx/college/Boltzmann mf-sts-07-c5-7 Equation15-10-07 see http://www.strw.leidenuniv.nl/˜ franx/college/ mf-sts-07-c5-8

This process is called phase-mixing.Itisrelevantin Spherical systems BT 4.4 2 (p 221+) non-equilibrium situations.

Equilibrium models BT 4.4 1 use classical integrals of motion • ∂f In equilibrium models, we have by definition ∂t =0. classical integrals are Energy (E) and angular momen- Furthermore, we have seen from the CBE df/dt =0 tum L⃗ compare to integrals of motion I[⃗x(t),⃗v(t)] for exact spherical symmetry, DF can only depend on d L2, not on L⃗ I[⃗x(t),⃗v(t)] = 0 dt implication: not individual orbits are used to build hence integrals of motion satisfy CBE ! models, but sets of orbits ! the distribution function for an equilibrium model is an Isotropic models integral of motion !

as a consequence: Jeans theorem distribution function f(E) Any steady state solution of CBE depends on w • only through integrals of motion, and any function Define of integrals yields steady state solution of CBE ψ = Φ + Φ first part obvious: f itself is an integral of motion − 0 1 2 second part also obvious: since dI/dt =0,wehave ε = E + Φ0 = ψ 2 v df (I)/dt =0. − − choose Φ0 such that A consequence of Jeans theorem: f>0 for ε > 0 f =0for ε 0 A galaxy can be constructed by adding up orbits. ≤ • Along each orbit, the DF is constant.

this is another justification of Schwarzschild’s method. 15-10-07 see http://www.strw.leidenuniv.nl/˜ franx/college/ mf-sts-07-c5-9 15-10-07see http://www.strw.leidenuniv.nl/˜ franx/college/ mf-sts-07-c5-10

The density is integral of distribution function over all What is “basic set of building blocks” for models of velocities: type f(ε) ?

ρ(r)= f(ε)d⃗v fill in delta function f(ε)=δ(ε ε0) ! • − vmax ρ =4√2π ψ ε0 2 = f(ε)4πv dv " − !0 1 2 maximum in the center, decreasing outwards, to zero substitute ε = ψ 2 v , dε = vdv.Sinceε will run − − at ψ(r)=ε0. from 0 (for v = vmax) to ψ (for v =0): These are NOT individual orbits, but combinations of ψ orbits. ρ(r)=4π f(ε) 2(ψ ε) dε Homework assignments ! − 0 " 1. In the case described above, what kind of orbits manage to reach the maximum distance to the cen- We can make models by specifying f(ε).Wefinda ter given by ψ(r)=ε0. What is the velocity of stars relation of the form ρ(r)=F (ψ). We then have to at this radius ? 2. Calculate the exact relation between ρ and ψ for find solutions of this equation, which also satisfy the γ Poisson equation: polytropes, assuming f(ε)=ε 3. Calculate the density related to the potential ψ = 2 ⃗ 2 1/√1+r . Show that this model is a polytrope 4πGρ(⃗x)= Φ(⃗x) [i.e., that we can write ρ = cnst ψg]. ∇ × Solutions can be constructed. For example, assume f(ϵ)=ϵγ . These give densities of the form ρ = cnst ψ(γ+1.5). These models are analogous to gas polytropes.× 15-10-07see http://www.strw.leidenuniv.nl/˜ franx/college/ mf-sts-07-c5-11 15-10-07see http://www.strw.leidenuniv.nl/˜ franx/college/ mf-sts-07-c5-12

Distribution function f(ε) from density BT p Intermezzo: Abel integral equation 236-237 Let As we saw on the previous pages, we can calculate the x g(t)dt density of a model from the distribution by the integral f(x)= equation: ! (x t)1/2 0 − then ψ 1 d t f(x)dx ρ(r)=4π f(ε) 2(ψ ε) dε g(t)= ! " − π dt ! (t x)1/2 0 0 − Proof: We now show how to derive f from ρ. Write ρ as substitute the first equation into the right hand side of ρ(ψ). This is always possible when we know ρ.Then: the second:

ψ 1 d t x g(s) 1 1 dρ f(ε) dε F (t)= dsdx = π dt ! ! (x s)1/2 (t x)1/2 2π√2 dψ ! √ψ ε 0 0 − − 0 − interchange order of integration This is of Abel form, with solution (Eddington): 1 d t t g(s) 1 F (t)= dxds ε π dt ! ! (x s)1/2 (t x)1/2 1 d dρ dψ 0 s − − f(ε)= Use 2π2√2 dε#! dψ √ε ψ $ 0 − t 1 1/2 1/2 dx = π Thus, f(ε) 0 if and only if ... increases mono- !s (x s) (t x) tonically with≥ ε. This is true for{ models} that are more − − Hence centrally concentrated than ρ(r)=√1 r2 − 1 d t F (t)= g(s)πds π dt !0 It is easy to see that this results into F (t)=g(t) 15-10-07see http://www.strw.leidenuniv.nl/˜ franx/college/ mf-sts-07-c5-13 15-10-07see http://www.strw.leidenuniv.nl/˜ franx/college/ mf-sts-07-c5-14

Velocity moments Isotropic models: Examples: Plummer model BT p 223-225

2 2 2 define velocity dispersions vr , vθ , vφ : 1 GM v2 = f(⃗v)v2d⃗v Potential: ψ = i ρ ! i √r2 + a2 If f = f(ε), f is isotropic: at a given location ⃗x, the Mr3 Mass: M(r)= distribution in velocities is the same in all directions: (r2 + a2)3/2 f = f(ε)=f(ψ(r) 1 v2) 3Ma2 − 2 Density: ρ(r)= 4π(r2 + a2)5/2 Hence v2 = v2 = v2 σ2 r θ φ ≡ We can write for such systems Plummer model = polytrope with index n =5: 1 1 ∼ σ2 = f(⃗v)4π v2dv 3 ρ ! 3a2 ρ = Cψ5 with C = 4π 4πG5M 4 = f(⃗v)(2(ψ ε))3/2dε 3ρ ! − Eddington formula:

64C f(ε)= ϵ7/2 0 7√2π2 ≥

GM velocity dispersion: v22 = 1 ψ = r 6 6√r2 + a2 GM(r) GMr2 circular velocity: v2 = = c r (r2 + a2)3/2 Key Point #17 from Bachelor Course: Solving the Collisionless

12-10-07seeBoltzmann http://www.strw.leidenuniv.nl/˜ Equation franx/college/ mf-sts-07-c4-17 using Schwarzschild’s12-10-07see http://www.strw.leidenuniv.nl/˜ Method franx/college/ mf-sts-07-c4-18

A general 3-dimensional potential A Simple recipe to build galaxies Schwarzschild’s method:

2 2 • Define density ρ St¨ackel potential( ρ =1/(1 + m ) ) • Calculate potential, forces • Integrate orbits, find orbital densities ρi • Calculate weights wi > 0 such that

ρ = ρiwi !

Examples: build a 2D galaxy in a logarithmic potential Φ =ln(1+x2 + y2/a).

• 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, nega- tive, or “neutral”.

Hence: a maximum rotation is defined if all loop orbits rotate the same way. The rotation can vary between zero, and this maximum rotation 12-10-07see http://www.strw.leidenuniv.nl/˜ franx/college/ mf-sts-07-c4-19

• Is the solution unique ? the density is two-dimensional function box orbits are defined by 2 integrals of motion, say the coordinates of the corner loop orbits have two integrals of motion

Hence, we have two construct a 2 dimensional function from the superposition of two 2-2dimentional functions

ρ(⃗x, ⃗y)=wbox(I1,I2)ρbox(I1,I2)+

wloop(I1,I2)ρloop(I1,I2)

The unknown functions are wbox(I1,I2) and wloop(I1,I2). The system is underdetermined. Hence, many solutions are possible.

Homework Assignment: 6) It is easy to think of two integrals of motion for a box orbit: the x,y coordinates of the “corner” of the box. Can you define two integrals of motion for the loop orbits in a similar way ? Key Point #5 from Bachelor Course: Mass Determinations

23-10-07 see http://www.strw.leidenuniv.nl/˜Using Stellar franx/college/ Velocities mf-sts-07-c6-1 and23-10-07 Jeans see http://www.strw.leidenuniv.nl/˜ Equations franx/college/ mf-sts-07-c6-2

Velocity Moments and the Jeans equations Integrate CBE over velocities. This results in: BT 4.2 p 195-198 3 3 ∂f ∂f ∂Φ ∂f d⃗v + vi d⃗v d⃗v =0 ∂t ∂x − ∂x ∂v We usually don’t observe the motions of individual i=1 i i=1 i i ! " ! " ! stars, but we can observe the average motions, and The last term is zero by straight integration over dvi the spread in velocities (the velocity dispersion). Here The second term can be simplified by moving the deriva- we derive equations for the densities, average velocities, tive outside the integral: and dispersions. We can derive these WITHOUT taking into account 3 ∂ν ∂ + ν v =0 the full distribution function. Assume a population of ∂t ∂x i i=1 i objects with density ν and distribution function f in a " potential Φ. Notice that ν is not necessarily the same This is a continuity equation for the mean streaming as ρ, which is the total matter density. Integrate distri- motion ⃗v of the stars in configuration space bution function f(⃗x,⃗v) over velocities. This gives the Integrate CBE times v over velocities velocity moments: j 3 3 ν(⃗x)= f(⃗x,⃗v) d3⃗v ∂ ∂f ∂Φ ∂f fvjd⃗v+ vivj d⃗v vj d⃗v =0 ∂t ∂x − ∂x ∂v ! i=1 i i=1 i i 3 ! " ! " ! ν vi (⃗x)= vif(⃗x,⃗v) d ⃗v ! The last term can be simplified. Do the partial integra- 3 ν vivj (⃗x)= vivjf(⃗x,⃗v) d ⃗v tion over dvi and use the fact that f vanishes for large ! v: where vi is the mean velocity, etc. 2 Define the velocity dispersions σij by: ∂f ∂vj 2 vj dvi =[vjf] fdvi =0 δijfdvi, σij = vivj vi vj ∂v − ∂v − − ! i ! i ! 23-10-07 see http://www.strw.leidenuniv.nl/˜ franx/college/ mf-sts-07-c6-3 23-10-07 see http://www.strw.leidenuniv.nl/˜ franx/college/ mf-sts-07-c6-4 where δij =1for i = j and 0 for i = j.Hence (BT eq 4-27) ̸ These are the Jeans equations. ∂f vj d⃗v = δijfd⃗v = δijν ∂v − − ! i ! Hence we obtain Almost the same as Euler equations for fluid, but • instead of ∂p/∂xj we have the summation over 2 3 3 the stress tensor ∂νσ ∂x . For a stationary model ∂ ∂f ∂Φ ij i fvjd⃗v + vivj d⃗v + δijν =0 the left terms disappear completely, and the veloc- ∂t ∂x ∂x i=1 i i=1 i ! " ! " ity dispersion tensor counter-acts gravity, just like or for a star made of gas. Note that the pressure in a galaxy is anisotropic ! But notice: no equation of 3 ∂(ν vj ) ∂ ∂Φ state for our “gas” in a galaxy, in contrast to stars + (ν vivj )+ν =0 (j =1, 2, 3) ∂t ∂x ∂x ! i=1 i j " (BT eq 4-24a) Generally 3 equations for 6 unknowns: many solu- Multiply continuity equation by vj , and subtract from • tions! the last equation:

Caveat: solutions are not guaranteed to be physi- 3 3 • ∂ vj ∂ν vi ∂ν vivj ∂Φ cal, since no check that f 0 ν vj + = ν ≥ ∂t − ∂x ∂x − ∂x i=1 i i=1 i j " " This can be rewritten as

3 3 ∂ vj ∂ vj ∂Φ ∂ 2 ν + ν vi = ν νσ ∂t ∂x − ∂x − ∂x ij i=1 i j i=1 i " " 23-10-07 see http://www.strw.leidenuniv.nl/˜ franx/college/ mf-sts-07-c6-5 23-10-07 see http://www.strw.leidenuniv.nl/˜ franx/college/ mf-sts-07-c6-6

Velocity ellipsoid Jeans equations for spherical models 2 The tensor σij is symmetric it is diagonal in locally ⇒ BT 4.2 d) page 203-209 orthogonal coordinates (˜x1, x˜2, x˜3):

σ˜11 00 Assume a coordinate system (r, θ, φ). We assume the 0˜σ22 0 system is invariant under rotations about the center. ⎛ ⎞ 00˜σ33 Hence we have ⎝ ⎠ The ellipsoid with semi-axes σ˜11, σ˜22,andσ˜33,ori- vr = vθ = vφ =0 ented along the local axes x˜1, x˜2,andx˜3, is called the vrvθ = vrvφ = vθvφ =0 velocity ellipsoid. It is sometimes used to describe the 2 2 vθ = vφ local velocity distribution so that velocity ellipsoid is everywhere aligned with (r, θ, φ) coordinates. Now the Jeans equations reduce to

d(ν v2 ) ν dΦ r + 2 v2 2 v2 = ν dr r r − θ − dr % & Define the anisotropy function

β(r)=1 v2 / v2 − θ r . Clearly β 1. We obtain one non-trivial Jeans equa- ≤ tion 1 d β dΦ ν v2 +2 v2 = ν dr r r r − dr 23-10-07 see http://www.strw.leidenuniv.nl/˜ franx/college/ mf-sts-07-c6-7 23-10-07 see http://www.strw.leidenuniv.nl/˜ franx/college/ mf-sts-07-c6-8

2 Given β(r), vr and ν(r) we can derive the potential Total enclosed mass and rotation curve and mass distribution. Full knowledge of the full distri- bution function is not necessary to interpret observable For a circular orbit with velocity vc(r) we have: parameters such as the velocity dispersion. dΦ GM(

So the Jeans equation can be written as

2 2 GM(

Measure: ν(r), v2 and β determine enclosed r ⇒ mass Key Point #5 from Bachelor Course: Alternate Derivation of Jeans Equations (Huub’s Version)

11. Velocity Moments and the Jeans Integrate distribution function f(~x, ~v) over equations BT 4.2 p 195-198 velocities. This gives three velocity moments: We usually don’t observe the motions of individual stars, but we can observe the 0. Spatial density of stars / 0th moment: average motions, and the spread in ⌫(~x)= f(~x, ~v)d3~v velocities (the velocity dispersion). Here we Z derive equations for the densities, average velocities, and dispersions. 1. Mean stellar velocity / first moment:

1 3 vi (~x) vif(~x, ~v)d ~v, i =1, 2, 3 We can derive these without taking into ⌘ ⌫ Z account the full distribution function. 2. Second moments: Assume a population of objects with density 1 3 vivj (~x) vivjf(~x, ~v)d ~v, j =1, 2, 3 ⌫ and distribution function f in a potential ⌘ ⌫ Z . Plus: Notice that ⌫ is not necessarily the same as Velocity dispersion tensor: ⇢, which is the total matter density. 2 (v v )(v v ) = v v v v ij ⌘ i i j j i j i j

1 2 Jeans equation 1 Much like fluid dynamics, the three (the Continuity equation) moments and the velocity dispersion tensor This equation is obtained by taking the 0th are constrained by 3 equations: the Jeans moment of the Collisionless Boltzmann equations. These three equations are: Equation (CBE) in v. Recall the CBE: Jeans equation 1 (the Continuity equation): @f 3 @f @ @f + vi =0 (CBE). @t i=1 @xi @xi @vi @⌫ + 3 @ ⌫ v =0 X @t i=1 @xi i P The 0th moment in v of the CBE:

Jeans equation 2 (the Force equation): CBE d~v or: Z @(⌫ vj ) 3 @ @ + (⌫ v v )+⌫ =0 3 3 @t i=1 @xi i j @xj @f @f @ @f d~v + vi d~v d~v =0. P @t @xi @xi @vi Z iX=1 Z iX=1 Z

Jeans equation 3 (a common rewrite of Using the divergence theorem, we can Jeans-2): rewrite the last term as a surface integrale: @f 3 @f 3 @ @ vj 3 @ vj @ 3 @ 2 d~v+ v d~v [f] dS~ =0. ⌫ + i=1 ⌫ vi = ⌫ i=1 ⌫ij i 1 @t @xi @xj @xi @t @xi @xi 1 Z iX=1 Z iX=1 Z P P Since f(~x, ~v) = 0 for “infinite” velocities, the last term is zero. 4 3 Jeans equation 2 + 3 The second term can be simplified by (the Force equation) moving the derivative outside the integral:

The first moment in v of the CBE:

CBE v d~v @⌫ 3 @ + ⌫ vi =0 Jeans 1 Z @t i=1 @xi or: or: P @ 3 @f 3 @ @f @⌫ fvjd~v+ vivj d~v vj d~v =0 + (⌫~v )=0 @t @xi @xi @vi @t r · Z iX=1 Z iX=1 Z

This is a continuity equation for the mean The last term can be simplified. Do the partial integration over dv and use the fact streaming motion ~v of the stars in i that f vanishes for large v: configuration space @f vj d~v = @v Note the similarity with the the continuity Z i equations for fluid mechanics: 2 @vj vj(f(vi = ) f(vi = ))d v=i fd~v = ZZ 1 1 6 Z @vi

0 fd~v = ⌫ @⇢ ij ij + ~ (⇢~v)=0 Z @t r · where =1fori = j and 0 for i = j. ij 6 6 5 Hence: We obtain the more frequently used variant of Jeans-2, Jeans equations 3: @ 3 @f 3 @ fv d~v + v v d~v + ⌫ =0 j i j ij @ vj 3 @ vj @ 3 @ 2 @t Z i=1 Z @xi i=1 @xi ⌫ + ⌫ vi = ⌫ ⌫ X X @t i=1 @xi @xj i=1 @xi ij or P P(Jeans 3) @(⌫ vj ) 3 @ @ + (⌫ vivj )+⌫ =0 @t i=1 @xi @xj Hence we obtain the analogue of the Euler P (Jeans 2) equation: ⇢@~v Multiply continuity equation (Jeans-1) by + ⇢(~v ~ )~v = ⇢~ ~ p =0 @t · r r r vj , and subtract from the last equation (Jeans-2): Almost the same as Euler equations for • fluid, but instead of ~ p we have the r 2 summation over the stress tensor @⌫ij@xi. 3 3 @ vj @⌫ vi @⌫ vivj @ For a stationary model the left terms ⌫ vj + + ⌫ =0 disappear completely, and the velocity @t i=1 @xi i=1 @xi @xj X X dispersion tensor counter-acts gravity, just like for a star made of gas. Note that the Using pressure in a galaxy is anisotropic ! But 2 @⌫ij @ notice: no equation of state for our “gas” = ⌫( vivj vi vj )= in a galaxy, in contrast to stars ! @xi @xi

@(⌫ vivj ) @(⌫ vi ) @ vj Generally 3 equations for 6 unknowns: vj ⌫ vi • @xi @xi @xi many solutions! 8 7 Jeans equations for spherical models BT 4.2d, page 203-209 Caveat: solutions are not guaranteed to • be physical, since no check that f 0 Assume a coordinate system (r, ✓, ). We assume the system is invariant under Velocity ellipsoid rotations about the center. Hence we have

vr = v✓ = v =0 2 The tensor ij is symmetric it is diagonal ) v v = v v = v v =0 in locally orthogonal coordinates (˜x1, x˜2, x˜3): r ✓ r ✓ ˜ 00 2 2 11 v✓ = v 0˜ 0 0 22 1 so that velocity ellipsoid is everywhere 00˜ B 33 C aligned with (r, ✓, ) coordinates. @ A The ellipsoid with semi-axes ˜11,˜22, and ˜33, oriented along the local axesx ˜1,˜x2, Now the Jeans equation(-2/3) in the andx ˜3, is called the velocity ellipsoid. It is stationary case reduces to: sometimes used to describe the local 2 d(⌫ vr ) ⌫ 2 2 d velocity distribution + 2 vr 2 v✓ = ⌫ dr r  dr

Define the anisotropy function:

(r)=1 v2 /v2 . ✓ r 9 10 Total enclosed mass and rotation curve

For a circular orbit with velocity vc(r)we Clearly 1. We obtain one non-trivial have:  Jeans equation:

d GM(

Measure: ⌫(r), v2 and determine r ) enclosed mass

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