Secular Perturba,ons
- Eccentric and Mean anomalies - Kepler’s equa,on - f,g func,ons - Universal variables for hyperbolic and eccentric orbits - Disturbing func,on - Low eccentricity expansions for Disturbing func,on - Secular terms at low eccentricity - Precession of angle of perihelion - Apsidal resonance - Pericenter glow models for eccentric holes in circumstellar disks
(Created by: Zsolt Sandor & Peter Klagyivik,
Eötvös Lorand University)
r
a
f = true anomaly
f
x = r cos f
y = r sin f
Ellipse
b
center of ellipse
Sun is focal point b=semi-‐minor axis a=semi-‐major axis
r =
a(1� e2)
1 + e cos f
E
r a
E = Eccentric anomaly f = true anomaly
f
x = r cos f
y = r sin f
x = a cos E
Orbit from center of ellipse
b = a�
1� e2
x = a(cos E � e)
Ellipse
y = b sinE
= a�
1� e2 sinE
y = a sinE
b
ae
⇣x
a
⌘2
+⇣y
b
⌘2
= 1
Rela,onship between Eccentric, True and Mean anomalies
• Using expressions for x,y in terms of true and Eccentric anomalies we find that
tanf
2=
�1 + e
1� etan
E
2
r = a(1� e cos E)
So if you know E you know f and can find posi,on in orbit
Energy = �GM
2a=
v2
2� GM
rv2 = r2 + r2f2
n =�
GM
a3
Write dr/dt in terms of n, r, a, e Then replace dr/dt with func,on depending on E, dE/dt
E =n
1� e cos E
r2f2 =h2
r2
r = ae sinEE
(r)2 = �GM
a
✓1� 2
(1� e cosE)
+
(1� e2)
(1 + e cosE)
2
◆
h2 = GMa(e2 � 1)
Mean Anomaly and Kepler’s equa,on
• New angle M defined such that or
• Integrate dE/dt finding
E =n
1� e cos E
M = n M = n(t� t0)
M = E � e sin E
Kepler’s equa,on Must be solved to integrate orbit in ,me.
The mean anomaly is not an angle defined on the orbital plane It is an angle that advances steadily in ,me It is related to the azimuthal angle in the orbital plane, for a circular orbit, the two are the same and f=M
– Change t, increase M.
– Compute E numerically using Kepler’s equa,on – Compute f using rela,on between E and f
– Rotate to take into account argument of perihelion
– Calculate x,y in plane of orbit – Rotate two more ,mes for inclina,on and longitude of ascending node
to final Cartesian posi,on
M = E � e sin E Kepler’s equa,on Must be solved to integrate orbit in ,me.
Procedure for integra,ng orbit or for conver,ng orbital elements to a
Cartesian posi,on:
Inclina,on and longitude of ascending node
• Sign of terms depends on sign of hz.
• I inclina,on. – retrograde orbits have π/2<I<π – prograde have 0<I<π/2
• Ω longitude of ascending node, where orbit crosses eclip,c
• Argument of pericenter ω is with respect to the
line of nodes (where orbit crosses eclip,c)
h angular momentum vector hz = h cos I
hx = ± sin I cos �hy = � sin I sin �
Orbit in space • line of nodes: intersec,on of
orbital plane and reference plane (eclip,c)
• Longitude of ascending node Ω: angle between line of nodes (ascending side) and reference line (vernal equinox)
• ω “argument of pericenter” is not exactly the same thing as we discussed before ϖ – longitude of pericenter. ω is not measured in the eclip,c. ϖ=Ω+ω but these angles not in a plane unless I=0
Anomaly : in orbital plane and w.r.t. pericenter Longitude: in eclip,c w.r.t. vernal equinox Argument: some other angle
Orbit in space
Rota,ons
• In plane of orbit by argument of pericenter ω in (x,y) plane
• In (y,z) plane by inclina,on I • In (x,y) plane by longitude of ascending node Ω • 3 rota,ons required to compute Cartesian coordinates given orbital elements
Cartesian to orbital elements
• To convert from Cartesian coordinates to Orbital elements: – Compute a,e using energy and angular momentum – Compute inclina,on and longitude of ascending node from components of angular momentum vector
– From current radius and velocity compute f – Calculate E from f – Calculate M from numerical solu,on of Kepler’s equa,on
– Calculate longitude of perihelion from angle of line of nodes in plane of orbit
The angular momentum vector
h = r⇥ v
h cos i = hz
h sin i sin⌦ = hx
h sin i cos⌦ = �hy
rela,on between inclina,on, longitude of ascending node and angular momentum. Conven,on is to flip signs of hx,hy if hz is nega,ve
F = �r�
F = Fr r+ F✓✓ + Fz z
Force
• component perpendicular to orbital plane
• component in orbital plane perpendicular to r
• component along radius orthogonal coordinate system
h = na2p
1� e2z
Torque
F = Fr r+ F✓✓ + Fz z h = na2p
1� e2z
⌧ = r⇥ F Fr contributes no torque To vary |h| a torque in z direc,on is need, only depends on Fθ -‐ only forces in the plane vary eccentricity To vary direc,on of h a force in direc,on of z is needed
di
dt=
rFz cos(! + f)
hd⌦
dt=
rFz sin(! + f)
h sin i
instantaneous varia,ons Oien integrated over orbit to es,mate precession rate
f and g func,ons
• Posi,on and velocity at a later ,me can be wrijen in terms of posi,on and velocity at an earlier ,me. Numerically more efficient as full orbital solu,on not required.
r(t) = f(t, t0, r0,v0)r0 + g(t, t0, r0,v0)v0
v(t) = f(t, t0, r0,v0)r0 + g(t, t0, r0,v0)v0
f = 1� a
r0[1� cos(E � E0)]
g = (t� t0)�1n
[(E � E0)� sin(E � E0)]
f = �na2
rr0sin(E � E0)
g = 1� a
r[1� cos(E � E0)]
Differen,al form of Kepler’s equa,on Procedure for compu,ng f,g func,ons • Compute a, e from energy
and angular momentum. • Compute E0 from posi,on • Compute ΔE by solving
numerically the differen,al form of Kepler’s equa,on.
• Compute f,g, find new r. • Compute • One of these could be
computed from the other 3 using conserva,on of angular momentum
�M = n�t
= �E � e cos E0 sin �E +e sinE0(1� cos �E
f = 1� a
r0[1� cos(E � E0)]
g = (t� t0)�1n
[(E � E0)� sin(E � E0)]
f = �na2
rr0sin(E � E0)
g = 1� a
r[1� cos(E � E0)]
Subtract Kepler’s equa,on at two different ,mes to find:
f , g
)
Universal variables
• Desirable to have integra,on rou,nes that don’t require tes,ng to see if orbit is bound.
• Conver,ng from ellip,c to hyperbolic orbits is oien of majer of subs,tu,ng sin, cos for sinh, cosh
• Described by Prussing and Conway in their book “Orbital Mechanics”, referring to a formula,on due to Baon.
f = 1� x2
r0C(�r2)
g = (t� t0)�x3
⇥µ
S(�x2)
f =x⇥
µ
rr0
��x2S(�x2)� 1
⇥
g = 1� x2
rC(�x2)
x is determined by Solving a differen,al form of Kepler’s equa,on in universal variables
µ �⇥
GM � � 1a
x
Analogy
f = 1� x2
r0C(�r2)
g = (t� t0)�x3
⇥µ
S(�x2)
f =x⇥
µ
rr0
��x2S(�x2)� 1
⇥
g = 1� x2
rC(�x2)
f = 1� a
r0[1� cos(E � E0)]
g = (t� t0)�1n
[(E � E0)� sin(E � E0)]
f = �na2
rr0sin(E � E0)
g = 1� a
r[1� cos(E � E0)]
x
Differen,al Kepler’s equa,on in universal variables
• x solves (universal variable differen,al Kepler equa,on
• Special func,ons needed:
C(y) =12!� y
4!+
y2
6!� . . .
=1� cos⇥y
y, y > 0
=cosh
⇥�y � 1�y
, y < 0
S(y) =13!� y
5!+
y2
7!� . . .
=⇥
y � sin⇥y�
y3, y > 0
=sinh
⇥�y �
⇥�y�
�y3, y < 0
⇥µ(t� t0) =
(r0 · v0)x2
⇥µ
C(�x2) + (1� r0�)x3S(�x2) + r0x
Solving Kepler’s equa,on
• Itera,ve solu,ons un,l convergence • Rapid convergence (Laguerre method is cubic)
• Only 7 or so itera,ons needed for double precision (though this could be tested more rigorously and I have not wrijen my rou,nes with necessarily good star,ng values).
Orbital elements
• a,e,I M,ω,Ω (associated angles)
• As we will see later on ac,on variables related to the first three will be associated with ac,on angles associated with the second 3.
• For the purely Keplerian system all orbital elements are constants of mo,on except M which increases with
• Problem: If M is an ac,on angle, what is the associated momentum and Hamiltonian?
M = n n =�
µ
a3
Keplerian Hamiltonian
• Problem: If M is an ac,on angle, what is the associated ac,on momentum and Hamiltonian?
• Assume that • From Hamilton’s equa,ons
• Energy
�H
��= M = n
H � ��
H = � µ
2a
�H
��� ���1 � n � a�3/2
�� � a�3�
2(��1)
�� � a�1
� = �2
� � a1/2
H � ��2
µ ⇠ GM⇤
Keplerian Hamiltonian
• Solving for constants
• Unperturbed with only 1 central mass • We have not done canonical transforma,ons to do this so not obvious we will arrive exactly with these conjugate variables when we do so.
H = � µ2
2�2
� =⇥
µa
Hamiltonian formula,on
• Poincare coordinates
⇥ = M + ⇤ + �� = �⇤ � �z = ��
⇥ =⇥
µa
� =⇥
µa(1��
1� e2)
Z =�
µ(1� e2)(1� cos I)
Working in Heliocentric coordinates • Consider a central stellar mass M*, a planet mp and a third low
mass body. “Restricted 3 body problem” if all in the same plane
• We start in iner,al frame (R*, Rp, R) and then transform to heliocentric coordinates (rp, r)
M�R� =GmpM�rp
r3p
mpRp = �GmpM�rp
r3p
R = �GM�rr3
� Gmp(r� rp)(r � rp)3
Rp = rp + R�
R = r + R�
Replace accelera,ons in iner,al frame with expressions involving accelera,on of star. Then replace accelera,on of star with this so we gain a term r = �GM�r
r3� Gmp(r� rp)
(r � rp)3� Gmprp
r3p
= ⇥�
GM�r
⇥+⇥R
Disturbing func,on
r = �GM�rr3
� Gmp(r� rp)(r � rp)3
� Gmprp
r3p
= ⇥�
GM�r
⇥+⇥R
R
New poten,al, known as a disturbing func,on – due to planet Gradient w.r.t to r not rp
Direct term Indirect term -‐-‐ because planet has perturbed posi,on of Sun and we are not working an iner,al frame but a heliocentric one—
-‐ Reduces 2:1 resonance strength. -‐ Contributes to slow m=1 eccentric modes of self-‐gravita,ng disks
R = Gmp
�1
|r� rp|� r · rp
r3p
⇥Force from Sun
Direct and Indirect terms
• For a body exterior to a planet it is customary to write
• For a body interior to a planet:
• In both cases the direct term • Conven5on ra,o of semi-‐major axes
R� =µ
a�RD +µ
a��2RI � = ap/a�
� = a/apR =µ�
apRD +
µ��
apRE
RD =a�
|r� � r|
� < 1
Lagrange’s Planetary equa,ons
• One can use Hamilton’s equa,ons to find the equa,ons of mo,on
• If wrijen in terms of orbital elements these are called Lagrange’s equa,ons
• These are ,me deriva,ves of the orbital elements in terms of deriva,ves of the disturbing func,on
• To relate Hamilton’s equa,ons to Lagrange’s equa,ons you can use the Jacobian of deriva,ves of orbital elements in terms of Poincare coordinates
H = Hkep �R
Lagrange’s equa,ons
da
dt=
2na
⇤R⇤�
de
dt= �
⇥1� e2
na2e(1�
p1� e2)
⇤R⇤��⇥
1� e2
na2e
⇤R⇤⇥
d�
dt= � 2
na
⇤R⇤a
+⇥
1� e2
na2e(1�
p1� e2)
⇤R⇤e
+tan(I/2)
na2⇥
1� e2
⇤R⇤I
where ε is mean longitude at t=0 or at epoch
d�dt
=1
na2⇥
1� e2 sin I
⇤R⇤I
d⇥
dt=
⇥1� e2
na2e
⇤R⇤e
+tan(I/2)
na2⇥
1� e2
⇤R⇤I
dI
dt= � tan(I/2)
na2⇥
1� e2
✓⇤R⇤�
+⇤R⇤⇥
◆� 1
na2⇥
1� e2 sin I
⇤R⇤�
� + nt = ⇥ = M + ⌅ = n(t� ⇤) + ⌅
Secular terms
• Expansion to second order in eccentricity • Neglec,ng all terms that contain mean longitudes • Should be equivalent to averaging over mean anomaly
• Indirect terms all involve a mean longitude so average to zero
�12 = a1/a2 a1 < a2
RD,1 = n21a
21
m2
mc + m1
�18�2
12b(1)3/2e
21 �
�14�2
12b(2)3/2e1e2 cos(⇥1 �⇥2)
⇥
Laplace coefficients which are a func,on of α
I have dropped terms with inclina,on here – there are similar ones with inclina,on
Similar term for other body
Evolu,on in e
• Lagrange’s equa,ons (ignoring inclina,on)
• Convenient to make a variable change
• Wri,ng out the deriva,ves
ej = � 1nja2
jej
⇥Rj
⇥�j�j =
1nja2
jej
⇥Rj
⇥ej
hj = ej sin �j kj = ej cos �j
dhj
dt=
⇥hj
⇥ejej +
⇥hj
⇥�j�j
Equa,ons of mo,on • For two bodies
• With solu,ons depending on eigenvectors es,ef and eigenvalues gs,gf of matrix A (s,f: slow and fast)
– Slow: both components of eigenvector with same sign, – Fast: components of eigenvector have opposite sign
h = (h1, h2)
A =
�n1µ2�2
124 b(1)
1/2(�12) �n1µ2�212
4 b(2)3/2(�12)
�n2µ1�124 b(2)
3/2(�12) n2µ1�124 b(1)
1/2(�12)
⇥
h = Akk = �Ah k = (k1, k2)
h = es sin(gst + �s) + ef sin(gf t + �f )
h = �A2h
Solu,ons
One eigenvector
es
e cos �
e sin �
es
e cos �
e sin �ef
Both eigenvectors
e21 = h2
1 + k21 =
[es,1 sin�s + ef,1 sin �f ]2
+ [es,1 cos �s + ef,1 cos �f ]2
e21 = e2
s,1 + e2f,1 + 2es,1ef,1 cos(�s � �f )
Both objects
ef,1
e sin �
ef,2
es,1
e cos �
e sin �
es,2
An, aligned, fast, Δϖ=π Aligned and slow, Δϖ=0
apsidal alignment
How can angular momentum be conserved with this?
Predic,ng evolu,on from orbital elements
• Unknowns – magnitudes of the 2 eigenvectors
– 2 phases
• From current orbital elements
h = es sin(gst + �s) + ef sin(gf t + �f )
e1, e2, �1, �2
Anima,on of the eccentricity evolu,on of HD 128311
(Created by: Zsolt Sandor & Peter Klagyivik, Eötvös Lorand University)
Both together in a differen,al coordinate system
x = e1e2 cos ��
y = e1e2 sin ��
radius = e1e2
Slow only
Apsidal aligned, non circula,ng, can have one object with nearly zero eccentricity. “Libra,on”
Circula,ng
fast only
Note orbits on this plot should not be ellipses
x = h1h2 + k1k2
y = h1k2 � k1h2
Examples of near separatrix mo,on For exatrasolar planets
Libra,on Circula,on
In both cases one planet drops to near zero eccentricity
e e
Δϖ Δϖ
Time Time
one planet
different lines consistent with data
From Barnes and Greenberg 08
Δϖ
e
Simple Hamiltonian systems Terminology
Harmonic oscillator
Pendulum
Stable fixed point
Libration
Oscillation p
Separatrix
p q
I
Mul,ple planet systems RV systems
• No obvious correla,on mass ordering vs semi-‐major axis • Mostly 2 planets but some with 3, 5 • Eccentric orbits, but lower eccentricity than single planet
systems • Rasio, Ford, Barnes, Greenberg, Juric have argued that
planet scajering explains orbital configura,ons • Subsequent evolu,on of inner most object by ,dal forces • Many systems near instability line • Lower eccentricity for mul,ple planet systems • Lower mass systems have lower eccentrici,es
Mass and Eccentricity distribu,on of mul,ple planet systems
High eccentricity planets tend to reside in single planetary systems
Hamiltonian view
RD,1 = n21a
21
m2
mc + m1
�18�2
12b(1)3/2e
21 �
�14�2
12b(2)3/2e1e2 cos(⇥1 �⇥2)
⇥
R1 ⇥ µ�A�1 �B
⇤�1 cos ⇥�
⇥
� � ⇥R⇥�
� � ⇥R⇥�
massless object near a single planet
� = � =0Fixed point at
�R��
= µ
�A� 1
2B��1/2
⇥�f =
�B
2A
⇥2��f = 0
Aligned with planet Eccentricity does not depend on planet mass but does on planet eccentricity
� =pGMa(1�
p1� e2) ⇡
pGMa
e2
2Poincaré momentum
μ = mp/M*
-‐
Expand around fixed point
• First transfer to canonical coordinates using h,k • Then transfer to coordinate system with a shii h’=h-‐hf, k’=k-‐kf
• Harmonic mo,on about fixed point: that’s the free eccentricity mo,on
H = A�(h�2 + k�2)
Free and Forced eccentricity
• Massless body in proximity to a planet
eforced
efree
e sin �
e cos �
�p
Force eccentricity depends on planet’s eccentricity and distance to planet but not on planet’s mass. Mass of planet does affect precession rate. Free eccentricity size can be chosen.
Secular problem with free and forced eccentrici,es
H(�;�⇥) = � + �⇥
� cos ⇥
Pericenter Glow
• Mark Wyaj, developed for HR4796A system, later also applied to Fomalhaut system
Mul,ple Planet system Hamiltonian view
• Single interac,on term involving two planets • All semi-‐major axes and eccentrici,es are converted to mometa
• Three low order secular terms, involving Γ1,Γ2,(Γ1Γ2)1/2
• Hamilton’s equa,on give evolu,on consistent with two eigenvectors previously found.
Epicyclic approxima,on
⇥L⇥�
= Ld
dt
⇥L⇥�
=⇥L⇥�
= 0
L(x,x) = x
2
2+
y
2
2� �(r)
H(pr, r;L, �) =p2r2
+L2
2r2+ �(r)
u =r � rcrc
H(pr, r;L, �) =p2r2
+L2
2r2c(1 + u)�2 + �(rc(1 + u))
These cancel to be consistent with a circular orbit
L = r2c� ⇥2 = �0(rc)/rc
H =p2u2
+L2
2r2c
�1� 2u+ 3u2
�+ �(rc) + �0(rc)u+ �00(rc)
u2
2
Epicyclic frequency
�2 =3L2
r4c+ �00(rc)
H =p2u2
+L2
2r2c
�1� 2u+ 3u2
�+ �(rc) + �0(rc)u+ �00(rc)
u2
2
H =1
2(p2u + �2u2) + ...
H = �L+ �J + aL2 + bJ2 + cLJ
to higher order in epicyclic amplitude (Contopoulos)
More generally on epicyclic mo,on
H = f(L) + �(L)J + g(L)J2 + ...
f 0(L) = �
Low epicyclic amplitude expansion
For a good high epicyclic amplitude approxima,on see a nice paper by Walter Dehnen using a second order expansion but of the Hamiltonian ,mes a carefully chosen radial func,on
As long as there are no commensurabili,es between radial oscilla,on periods and orbital period this expansion can be carried out
Low eccentricity Expansions
• Func,ons of radius and angle can be wrijen in terms of the Eccentric anomaly
M = E � e sinE
e sin E = E �M
e sin E =��
s=1
bs(e) sin sM
is an odd func,on
bs(e) =2sJs(se) Bessel func,on of the first kind
E = M + e sin M + e2 12
sin 2M + e3
�38
sin 3M � 18
sin M
⇥.....
E = M +��
s=1
2sJs(se) sin sM
This can be shown by integra,ng and using Kepler’s equa,on (see page 38 M+D)
Low eccentricity expansions con,nued
cos nE = �e
2�n,1 +
⇥�
k=1
n
k[Jk�n(ke)� Jk+n(ke)] cos kM
sin nE =⇥�
k=1
n
k[Jk�n(ke) + Jk+n(ke)] sin kM
r = a(1� e cos E)r
a= 1� e cos E
r
a= 1 + e2 �
⇥�
k=1
e
k[Jk�1(ke)� Jk+1(ke)] cos kE
found by integra,ng Fourier coefficients by parts and using integral forms for the Bessel func,on
Con,nued
a
r=
11� e cos E
=dE
dM
E = M +��
s=1
2sJs(se) sin sM
using Kepler’s equa,on
a
r= 1 +
��
s=1
2Js(se) cos sM
� r
a
⇥2= (1 + e cos E)
= 1� 2e cos E + e2 cos2 E
= 1� 2e cos E +e2
2(1 + cos 2E)
now can expand cos func,ons
Expansion of the Disturbing func,on
• in plane
RD =a�
|r� r�| |r� r�| =�
r2 + r�2 � rr� cos �
xx�
rr� = cos(� + f) cos(�� + f �)
yy�
rr� = sin(� + f) sin(�� + f �)
cos � = cos(⇥ + f � ⇥� � f �)
I = � = 0
� = ⇤ + f = ⇥ + f I = � = 0if
cos ⇥ = cos(� � ��)
wri,ng the dot product in terms of orbital angles
ψ angle between defined here
Angular factors
• When inclina,on is not zero we define
• Ψ is small if inclina,ons are small and can be expanded in powers of the sin of the inclina,ons
� ⌘ cos � cos(✓ � ✓0)
Expansion of the disturbing func,on -‐Con,nued
• Expand the disturbing func,on as a series of
,mes inclina,ons, angular factors and some radii • Use low eccentricity and inclina,on expansions for these factors
• Expand powers of Δ0 in terms of powers of ρ0 assuming that
��10 =
⇥r2 + r02 + 2rr0 cos(� � �0)
⇤�1/2
⇥0 =
⇥a2 + a02 + 2aa0 cos(� � �0)
⇤1/2����r0
a0� 1
���� ⌧ 1
���r
a� 1
��� ⌧ 1
these are sa,sfied at low eccentricity
Laplace coefficients
• This is a Fourier expansion
• These are called Laplace coefficients, closely related to ellip,c func,ons
• Can be evaluated by series expansion in α or in 1-‐α • They diverge as α 1
⇥0 ⇥�a2 + a�2 � 2aa� cos(� � ��)
⇥1/2
⇤�(2i+1)0 = a⇥�(2i+1) 1
2
⇤�
�⇤b(j)i+ 1
2(�) cos j(⇥ � ⇥⇥)
12b(j)s (�) ⇥ 1
2⇥
� 2�
0
cos j⇤ d⇤
(1� 2� cos ⇤ + �2)s
separa,ng the radial informa,on from the angle informa,on
Expansion of Disturbing func,on
• Disturbing func,on is wrijen in terms of an expansion of deriva,ves of Laplace coefficients and cosines of arguments
(r2 + r⇥2 � 2rr⇥ cos(� � �⇥)�1/2 = ⇥�1/20 + (r � a)
⇤
⇤a⇥�(2i+1)0 +
(r⇥ � a⇥)⇤
⇤a⇥⇥�(2i+1)0 + ...
Dm,n = ama�n �m+n
�am�a�n
)
First term!
• Useful rela,ons can be found by manipula,ng the integral defini,on of the Laplace coefficients, e.g.
RD =12b(j)
12
+18(e2 + e�2)[�4j2 + 2�D + �2D2]b(j)
12
cos(j⇥� j�⇥)
RD =12b(0)
12
+18(e2 + e�2)[2�D + �2D2]b(0)
12
for j=0
D � ⇥
⇥�
b(�j)s = b(j)
s
�
Second term
• j=1, j=-‐1
18ee�[2j + 4j2 � 2�D � �2D2]b(j)
12
cos((1 + j)⇥� (1 + j)⇥� �⇤ + ⇤�)
18ee�[�2j + 4j2 � 2�D � �2D2]b(j)
12
cos((1� j)⇥� (1� j)⇥� �⇤ + ⇤�)
14
�2� 2�D � �2D2
⇥b(1)1/2 cos(⇥ �⇥�)
Simplifica,on Using rela,ons between coefficients derived by Brouwer & Clemens
2�db(0)
1/2
d�+ �2
d2b(0)1/2
d�2= �b(1)
32
2b(1)1/2 � 2�
db(1)1/2
d�� �2
d2b(1)1/2
d�2= ��b(2)
32
RD,1 = n21a
21
m2
mc + m1
�18�2
12b(1)3/2e
21 �
�14�2
12b(2)3/2e1e2 cos(⇥1 �⇥2)
⇥
As we used in our discussion of secular perturba,ons
Using them we find that secular low order secular terms are
More generally
• Expanding the disturbing func,on in terms of Poincare coordinates
• D’Alembert rules – flipping signs of all angles preserves series so only cosines needed
– rota,ng coordinate system preserves series
⇥ = M + ⇤ + �� = �⇤ � � = �⌅
z = ��
R =�
j,k,l
cijl cos(j⇥ + k� + lz)
j � k � l = 0
Expansion of the Disturbing func,on In Summary
Expansion of the disturbing func,on assuming • low eccentrici,es, low inclina,ons Radial factors wrijen in terms of Laplace coefficients and their deriva,ves
Each argument and order of e,i gives a func,on
Both direct and indirect terms can be expanded
Expansion func,ons listed in appendix by M+D
Reading:
• Murray and Dermoj Chap 2 • Murray and Dermoj Chap 6,7 • Prussing and Conway Chap 2 on universal variables • Wright et al. 2008, “Ten New and Updated Mul,-‐planet Systems, and a Survey of Exoplanetary Systems” astroph-‐arXiv:0812.1582v2
• Malhotra, R. 2002, ApJ, 575, L33, A Dynamical Mechanism for Establishing Apsidal Resonance
• Barnes and Greenberg, “Extrasolar Planet Interac,ons”, astro-‐ph/0801.3226