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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. 103-107)
9.2 Constants and Integrals of motion
(BT p. 110-117)
9.2.1 Spherical potentials
9.2.2 Integrals in 2 dimensional flattened
potentials
9.2.3 Axisymmetric potentials
9.3 A general 3-dimensional potential
1
9.4 Schwarzschild’s method
9.1 Orbits in spherical potentials
Appropriate for for example globular
clusters.
Potential function
Φ = Φ(r),
r = |~
r|
with
Equation of motion for a star with unit mass
d2~
r
~
=
F
(r)~
e
=
−
∇Φ
r
dt2
Conservation of angular momentum
Define angular momentum per unit mass:
r
~ ≡~
L
r × d~
dt
using that ~
r×~
r = 0 for any ~
r:
d~
d
d~
r
L=
~
r×
dt
dt
dt
!
d~
r d~
r
d2~
r
=
×
+~
r× 2
dt
dt
dt
= F (r)~
r × ~er = 0
2
~ =~
~ is
Hence L
r×~
r˙ is constant with time. L
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
2~
d
Using BT (App. B.2), the acceleration dt2r
in cylindrical coordinates can be written as:
d2~
r
˙ 2)~er + (2r˙ψ˙ + rψ)~
¨ eψ
=
(¨
r
−
r
ψ
2
dt
2
With the equation of motion ddt2~r = F (r)~er ,
this implies:
r¨ − rψ˙ 2 = F (r)
(∗)
¨=0
2r˙ψ˙ + rψ
(∗∗)
3
Hence:
1 dr2ψ˙
¨=
2r˙ψ˙ + rψ
= 0 ⇒ r2ψ˙ = rv⊥ = L = cst
r dt
Using ψ˙ = rL2 and equation (**):
L2
dΦ
2
r¨ − rψ˙ = r¨ − 3 = −
r
dr
3 where Φ is the potential.
Integrate last equation to obtain:
2
1 r˙2 = E − Φ − L = E − Φ (r)
eff
2
2r2
with E the energy. E is the integration
constant obtained for r → ∞
This equation governs radial motion
through the effective potential Φeff :
L2
Φeff = Φ + 2
2r
4
Veff(r)
rmin
rmax
r
E
Motion possible only when r˙2 ≥ 0
rmin ≤ r ≤ rmax
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 < 2π (pointmass)
6
Special cases
Circular orbit: rmin = rmax
2
v⊥
dΦ
GM (r)
=
=
r
dr
r2
Radial orbit: L = 0
1 r˙2 = E − Φ(R)
2
Homogeneous sphere
1 Ω2 r 2 + Constant
Φ(r) = 2
Equation of motion in in radial coordinates:
¨
~
r = −Ω2~
r
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.
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
9.2 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(t1), ~v (t1), t1) = C(~
x(t2), ~v (t2), t2)
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(t1), ~v (t1)] = I[~
x(t2), ~v (t2)]
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 + ψo, 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 (~
x0, ~v0) at time t = t0.
Can always be calculated back from ~
x, ~v , t.
Hence (~
x0, ~v0) can be regarded as six
constants of motion.
10
Integrals of motion
are often hard or impossible to define.
Simple exceptions include”
− For all static potentials: Energy
1 v2 + Φ
E(~
x, ~v ) = 2
− For axisymmetric potentials: Lz
− 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:
4.2.1 Spherical potentials
4.2.2 Integrals in 2 dimensional attened
potentials
11
4.2.3 Axisymmetric potentials
9.2.1. Spherical potentials
E, Lx, Ly , Lz are integrals of motion, but also
~ (given by the
E, |L| and the direction of L
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 = (x1, x2, 0)
~v = (v1, v2, 0)
→ ~
x and ~v constrained to 4D region of the
6D phase space. In this 4 dimensional
space, |L| and E are conserved. This
constrains the orbit to a 2 dimensional
space. Hence the velocity is uniquely
defined for a given ~
x (see page 4).
q
vr = ± 2(E − Φ) − L2/r2
vψ = ±L/r
12
9.2.2. Integrals in 2 dimensional
flattened potentials
Examples:
•
•
Circular potential V (x, y) = V (~
r)
Two integrals: E, Lz .
Flattened potential
y2
2
V (x, y) = ln(x + 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,
(Lz = x ∗ vy − y ∗ vx)
avoid outer x-axis
4
4
4
4
2
2
0
0
4
4
2
2
0
0
2
2
0
0
-2
-2 -2
-2 -2
-2
-4
-4 -4
-4
-4 -4
-4 -4
-2
-4
-2-20
00 2
22 4
44
-4 -2 -2-2 0 0 0 2 2 2 4 4 4
-4 -4
x0=1
y0=0
vx0=0
vy0=1
x0=1
y0=0
vx0=0
vy0=1
x0=1
y0=0
vx0=0
vy0=1
x0=0.5
y0=0vx0=0
vx0=0
vy0=1
x0=0.5
vy0=1
x0=0.5
y0=0y0=0
vx0=0
vy0=1
4
4 4
4
44
2
2 2
2
22
0
0
-2
-2
-4
-4
0
0
-2
0
-2
-2 -2
-4
-4
-4 -4
-4
-4
0
-2
0
2
4
-4 x0=1.1
0 2 vy0=1
2 4 4
-2 -2 y0=0
0 vx0=0
x0=1.1
vx0=0
vy0=1
x0=1.1
y0=0y0=0
vx0=0
vy0=1
-4
-2
0
2
-4 -2 y0=0
-2 0vx0=0
0 2vy0=1
2 4
-4 x0=1.2
x0=1.2
vx0=0
vy0=1
x0=1.2
y0=0y0=0
vx0=0
vy0=1
4
4
14
4
2
0
-2
Loop orbits with
net angular momentum
-4
4
-2
0 avoid
2
4inner x-axis
-4
-2
0
2
4
0=0.5 y0=0 vx0=0 vy0=1
y0=0 vx0=0 vy0=1
can circulate in x0=1
two
directions.
4
4
2
0
-2
-4
-2
0
2
4
-4
0=1.1 y0=0 vx0=0 vy0=1
-2
0
2
4
x0=1.2 y0=0 vx0=0 vy0=1
4
4
2
2
0
0
-2
-2
-4
-4
-4
-2
0
2
x0=2 y0=0 vx0=0 vy0=1
4
-4
-2
0
2
4
x0=4 y0=0 vx0=0 vy0=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 2 directions
Conclusion: the orbits do not occupy a 3
dimensional space in the 4-dimensional
phase-space, but they occopy only a
2-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 2
dimensional surface in the 4 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 = ρ0H(1 − m2),
2
2
y2
x
z
2
with m = a2 + b2 + c2
and H(x) = 1 for x ≥ 0, H(x) = 0 for
x<0
Potential inside the ellipsoid:
Φ = Axx2 + Ay y 2 + Az z 2 + C0
Forces are of the form Fi = −Aixi, i.e. 3
independent harmonic oscillators:
xi = ai cos(ωit + ψ0,i)
3 integrals of motion, Ei
17
Orbits for general homogeneous ellipsoid
All orbits are box orbits
4
4
2
2
0
0
-2
-2
-4
-4
-4
-2
0
2
4
-4
x0=0.5 y0=0 vx0=0 vy0=1
4
2
2
0
0
-2
-2
-4
-4
-2
0
2
x0=1.1 y0=0 vx0=0 vy0=1
0
2
4
x0=1 y0=0 vx0=0 vy0=1
4
-4
-2
4
-4
-2
0
2
x0=1.2 y0=0 vx0=0 vy0=1
18
4
4
4
2
2
0
0
-2
-2
-4
-4
-4
-2
0
2
x0=2 y0=0 vx0=0 vy0=1
4
-4
-2
0
2
x0=4 y0=0 vx0=0 vy0=1
19
4
Special case: a=b All orbits are loop orbits
4
2
0
-2
-4
-4
-2
0
2
4
x0=0.5
x0=1.1
x0=1.2
x0=1 y0=0
x0=2
x0=4
y0=0vx0=0
vx0=0vy0=1
vy0=1
20
9.2.3. Axisymmetric potentials
Φ = Φ(R, |z|), with R2 = x2 + y 2.
For z=0: orbits as if potential were circular.
For the general case, the equation of
motions of a star is:
d2~
r
~
=
−
∇Φ(R,
z).
dt2
21
With ~er , ~eφ, ~ez unit vectors in r, φ and z
direction, we can write:
~
r = R ~eR + z ~ez ,
~ = ∂Φ~eR + ∂Φ~eφ + ∂Φ~ez
∇Φ
∂R
∂φ
∂z
with
∂Φ
= 0.
∂φ
2
Using BT (App. B.2), the acceleration ddt2~r
in cylindrical coordinates can be written as:
d2~
r
¨)~eφ + z¨~ez .
¨ − Rφ˙ 2)~eR + (2R˙ φ˙ + Rφ
=
(
R
2
dt
Note that:
d
d
¨ = 0.
Lz ≡ R2φ˙ = 2R˙ φ˙ + Rφ
dt
dt
and conclude that the angular momentum
about the z-axis is conserved.
˙ we obtain for
If we use this to eliminate φ,
the equation of motions in the ~er and ~ez
directions:
22
d2 R
∂Φeff
=
−
dt2
∂R
with
;
∂Φeff
d2z
=
−
dt2
∂z
L2
Φef f = Φ(R, z) + z2 .
2R
Hence 3D motion can be reduced to motion
in (R,z) plane or meridional plane, under
influence of the effective potential Φeff .
Note that this meridonial plane can rotate.
Application to a logarithmic potential
Recall: the axisymmetric logarithmic
potential (Rc taken to be zero):
z2
2
2
1
Φ(R, z) = 2 v0 ln(R + 2 ).
q
Then:
1 2
z2
L2
2
Φeff = v0 ln(R + 2 ) + z2 .
2
q
2R
23
24
Total energy:
1 2
˙ 2 + z˙ 2] + Φ
E = [R˙ + (rφ)
2
1 2
= (R˙ + z˙ 2) +
2
=
L2
Φ + z2
2R
!
1 2
(R˙ + z˙ 2) + Φeff .
2
Allowed region in meridional plane
Since the kinetic energy is non-negative,
orbits are only allowed in areas where:
Φeff < E. E.g.
Lines of constant Φef f are shown in Figure
3.2. Stars with energy E have zero velocity
at curves of Φeff = E
25
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 Φef f < E fully!
26
Two integrals (E, Lz ) reduce the
dimensionality of the orbit from 6 to 4 (e.g.
R, z, ψ, vz ). 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.
27
9.3 A general 3-dimensional potential
St¨
ackel potential: ρ = 1/(1 + m2)2
2
2
2
with m2 = xa2 + yb2 + zc2 .
28
29
9.4 Schwarzschild’s method
A simple recipe to build galaxies
• Define density ρ.
• Calculate potential, forces.
• Integrate orbits, find orbital densities ρi.
• Calculate weights wi > 0 such that
ρ=
X
ρiwi.
Examples: build a 2D galaxy in a
logarithmic potential Φ = ln(1 + x2 + y 2/a).
30
• 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 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-dimentional
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.
32
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.
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 do we expect virtually no loop
orbits in a homogeneous ellipsoid ?
5) Figure 3-2 (page 24) shows contours
plot for effective potentials related to an
axisymmetric logarithmic potential given on
33
page 23. When does the minimum occur?
What kind of orbit does this minimum
represent?
6) Calculate at least 3 orbits in the
2-dimentional potential
Φ = ln(1 + x2 + y 2/2). 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 . Calculate the
~ . And
shift in velocity at time dt by dt*F
keep integrating !
Also plot the equipotential curve where
Energy E = Φ.