The Two-Body Problem. The two-body problem The two-body problem: two point objects in 3D interacting...
63
The Two-Body Problem
-
Upload
morgan-charles -
Category
Documents
-
view
250 -
download
1
Transcript of The Two-Body Problem. The two-body problem The two-body problem: two point objects in 3D interacting...
The Two-Body Problem
The two-body problem
• The two-body problem: two point objects in 3D interacting with each other (closed system)
• Interaction between the objects depends only on the distance between them
• The number of degrees of freedom: 6
• Phase space dimensions: 12
3.1
The two-body problem
• The Lagrangian of the system in Cartesian coordinates:
• It is a very non-trivial problem if we try to deal with the Lagrangian in this format: all the 6 independent coordinates are entangled in the potential function
• Let us look for a different configuration space
3
1
221
3
1
2
1
2
)(2
)(
iii
i j
ijj rrVrm
L
3.1
221
222
211 )(
2
)(
2
)(rrV
rmrm
New generalized coordinates
• Let us introduce new system of coordinates:
• R – center of mass vector
• Then
• And
21
221121 ;
mm
rmrmRrrr
3.1
21
12
21
21
mm
rmRr
mm
rmRr
2
21
122
2
21
221 )(;)(
mm
rmRr
mm
rmRr
New generalized coordinates
• The Lagrangian in the new coordinates:
3.1
221
222
211 )(
2
)(
2
)(rrV
rmrmL
2
2
21
12
2
21
21
)(22
rVmmrm
Rmmmrm
Rm
rVmm
rmm
mm
rRmmRm
mm
rmm
mm
rRmmRm
221
2
12
21
212
22
21
2
21
21
212
1
)(2
2
)(
)(22
)(
2
))(( 221 Rmm
rVmm
rmm
)(2
)(
21
221
New generalized coordinates
• The Lagrangian in the new coordinates:
• The center of mass coordinates are cyclic!
• Three Euler-Lagrange equations for them can be solved immediately
• Total momentum of the system is conserved: three integrals of motion
3.1
rVmm
rmmRmmL
)(2
)(
2
))((
21
221
221
ii R
L
dt
d
R
L const
R
L
i
iRmm )( 21 iP
New generalized coordinates
• The Lagrangian in the new coordinates:
• Let’s re-gauge the Lagrangian
• Constant term
3.1
rVmm
rmmRmmL
)(2
)(
2
))((
21
221
221
rVmm
rmm
mm
P
)(2
)(
)(2
)(
21
221
21
2
rVmm
rmm
mm
PLL
)(2
)(
)(2
)('
21
221
21
2
New generalized coordinates
• The re-gauged Lagrangian:
• We reduced the two-body problem to a one-body problem in a central potential (potential that depends only on the distance from the origin)
• m: reduced mass
• The number of degrees of freedom: 3
• Phase space dimensions: 6
3.1
rVmm
rmmL
)(2
)('
21
221 rV
rm
2
)( 2
21
21
mm
mmm
Spherical coordinates
• Central potential is spherically symmetric
• It is convenient to work in spherical coordinates
• Then
cos ;sinsin ;cossin rrrrrr zyx
)(2
) sin('
222222
rVrrrm
L
222
2222
2
)(
2
)(' zyx
zyx rrrVrrrm
rVrm
L
r
8.1
Spherical coordinates
• The Euler-Lagrange equation for φ
• The φ coordinate is cyclic
• Since the system is spherically symmetric, we have a freedom of choosing the reference frame
• We chose it as follows: the initial velocity vector belongs to a plane φ = const
• Then
)(2/) sin(' 222222 rVrrrmL
3.2
'' LL
dt
d constpmr
L
sin' 22
22 sin
mr
p
0 0 0 p 0
Spherical coordinates
• The Euler-Lagrange equation for θ
• The θ coordinate is also cyclic
• Momentum conjugate to the θ coordinate
• Angular momentum in the plane of motion relative to the origin is conserved
)(2/) sin(' 222222 rVrrrmL
3.2
constmr 2
'' LL
dt
d
constrprmvrmrmr 2
L
p
Spherical coordinates
• The Euler-Lagrange equation for r
• Momentum conjugate to the r coordinate
• Now we can write a Hamiltonian
)(2/) sin(' 222222 rVrrrmL
3.28.1
'LpprpH r )(
2
12
22 rV
r
pp
m r
r
L
r
L
dt
d
'' r
rVrmrm
)(
2
rmr
Lpr
'
The effective potential
• The effective potential
• The Hamiltonian effectively depends only on 1 coordinate now
• We reduced the two-body problem to a 1D problem of a particle with a reduced mass m in the effective potential
• The number of degrees of freedom: 1
• Phase space dimensions: 2
3.2
)(2
2 2
22
rVmr
p
m
pH r constp
)(2
)(2
2
rVmr
prVeff
The effective potential
• Hamilton equations of motion:
3.2
)(2
2 2
22
rVmr
p
m
pH r
dr
dV
mr
p
r
Hpr
3
2
m
p
p
Hr r
r
2
mr
p
p
H
0
H
p
0
t
H
dt
dH EconstH
E
)(
22
2
2
rVmr
pEmpr
mr
The orbit equation
• On the other hand
• Orbit equation
3.23.5
)(
2
2
2
2
rVmr
pE
mr
)(2
2
2
2
rVmrp
Em
drdt
2
mr
p
p
dmrdt
2
)(2
2
2
2
2
rVmrp
Em
dr
p
dmr
r
rrV
mrp
Emr
drp
0 )(2
2
2
22
0
The orbit equation
• The orbit equation can be integrated for potentials with the power dependence on the distance
If n = 2, - 1, - 2, the integral can be expressed in trigonometric functions
• If n = 6, 4, 1, - 3, - 4, - 6, the integral can be expressed in elliptic functions
3.5
r
rrV
mrp
Emr
drp
0 )(2
2
2
22
0
narrV )(
The orbit equation
• From Hamilton’s equations of motion:
• If the orbit is known, the potential can be calculated
3.5
dr
dV
mr
ppr
3
2
m
pr r mrpr mr2
mr
p
dr
dV
mrm
pr
132
2
dmr
p
dt
1
12
dt
dr
dt
d
dr
dV
mrm
p
d
dr
mr
p
d
d
mr
p 132
2
22
dr
dV
p
mr
rd
dr
rd
d2
2
2
11
Example
• Restore a potential for a spiral orbit:
err 0
dr
dV
p
mr
rd
dr
rd
d2
2
2
11
dr
dV
p
erm
ere
r 2
2
0
00
11
dr
dV
p
mr
rr 2
211
3
22
mr
p
dr
dV 03
22V
mr
drpV
2
2
0 mr
pV
Stable circular orbits
• For a circular orbit:
• On the other hand
• For the extremum of the effective potential
• Extremum of the of the effective potential corresponds to a circular orbit
3.6
0 3
2
dr
dV
mr
p
r
Hpr
m
p
p
Hr r
r
0r
0 rp
dr
dV
mr
p
3
2
constr
0)(2 2
2
rV
mr
p
dr
d 0)(
3
2
dr
rdV
mr
p
Stable circular orbits
• For a stable circular orbit, the second derivative of the of the effective potential should be positive:
• For potentials with the power dependence on the distance
3.6
0)(2 2
2
2
2
rV
mr
p
dr
d 0
)(32
2
4
2
dr
rVd
mr
p
4
2
2
2 3)(
mr
p
dr
rVd
narrV )(
dr
dV
mr
p
3
2
dr
dV
rdr
rVd 3)(2
2
12 3)1( nn ar
r
narnn 31 n 2n
Application of the Hamilton-Jacobi theory
• The problem of spherically symmetric potential can be neatly treated employing Hamilton-Jacobi theory
• Then equation for Hamilton’s characteristic function
10.5
)(2/) sin(' 222222 rVrrrmL
'LpprpH r
)(sin
2
122
2
2
22 rV
r
p
r
pp
m r
1
2
22
2
2
2
sin
11
2
1
VW
r
W
rr
W
m
Application of the Hamilton-Jacobi theory
• Let us assume that the variables can be separated
• Then
•The φ coordinate is cyclic, therefore
10.5
1
2
22
2
2
2
sin
11
2
1
VW
r
W
rr
W
m
)()()(),,( WWrWrW r
12
22
2
2
2 'sin
1'
1'
2
1
VW
rW
rW
m r
'W W
Application of the Hamilton-Jacobi theory
• The circled part should be constant, since it contains only the θ dependence
• Then, finally
•The variables are completely separated!
• The resulting equation can be integrated in quadratures and is equivalent to the orbit equation
10.5
12
22
2
2 22sin
'1
'
mmVW
rWr
2
2
22
sin'
W
12
22 22' mmV
rWr
The Kepler problem
• Kepler potential:
• Mediating gravitational and electrostatic interactions
• Attraction:
• Repulsion:
• Integral orbit equation:
3.7
rk
mrp
Emr
drp
2
22
0
22
Johannes Kepler(1571-1630)
1)( rrV
0;)( kr
krV
0;)( kr
krV
The Kepler problem
• Let us consider an attractive potential:
• Table integral:
3.7
0
)(
kr
krV
2
22
0
22
mrp
rk
Emr
drp
u
r
1
220
/)(2
upkuEm
du
4
2arccos
22
u
uu
du
2
2
p
mE
2
2
p
mk
The Kepler problem3.7
ru
1
2
2
2
2
0
24
2
22
arccos
p
mE
p
mk
p
mku
2
2
2
02
1
1arccos
mkEp
km
up
km
up
mk
Ep 2
02
2
)cos(2
11
rmk
Ep
p
km 1 )cos(
211 02
2
2
The Kepler problem
• We obtained an explicit expression for the orbit
• Depending on the values of C and e, the orbits can assume qualitatively different shapes
• For a positive C (attraction), the shapes of the orbits represent all possible conic sections
3.7
)cos(
211
102
2
2
mk
Ep
p
km
r
)cos(1 1
0 eCr
Cp
km2
emk
Ep
21
2
2
Classification of Kepler’s orbits
• Effective potential for the attractive Kepler case:
3.33.7
r
k
mr
prVeff
2
2
2)(
0)( rVeffr
k
mr
p
2
2
2
km
pr
2
2
002
)0(2
2 k
m
pVeff
r 0
2
2
effeff V
km
pV
)0(effV
Classification of Kepler’s orbits
• Effective potential for the attractive Kepler case:
• Minimum point of the effective potential
3.33.7
r
k
mr
prVeff
2
2
2)(
0)(
dr
rdVeff 023
2
r
k
mr
pkm
pr
2
min
min2
min
2
min2
)(r
k
mr
prVeff
2
2
2
2
2 p
mk
p
mk 2
2
2 p
mk
2
2
min2
)(p
mkrVeff
Classification of Kepler’s orbits
• Effective potential for the attractive Kepler case:
• The simplest case
• Circular orbit
3.33.7
0e
r
k
mr
prVeff
2
2
2)(
)cos(1 1
0 eCr
Cr
1
Cr
1
02
12
2
mk
Epe
2
2
2 p
mkE
2
2
min2
)(p
mkrVeff
)( minrVeff
Classification of Kepler’s orbits
• Circular orbit
• Circle is one of conic sections
3.33.7
km
p
Cr
21 2
2
2 p
mkE
0e
Classification of Kepler’s orbits
• If
3.33.7
)cos(1 1
0 eCr
0e
r
rer
Cx1
1xerr
r
rx )cos( 0
xerC
r 1
2222 2 xx reherhr 22yx rr
xerh
2222 2)1( hrherre yxx
1 1
1
1 2
222
2
22
yx r
h
e
e
ehr
h
e
ae
h
21
be
h
21
021 xre
eh
1 22
0
b
r
a
rr yxx
Classification of Kepler’s orbits
• If
• Then is real and is positive
• The orbit is an ellipse with its center shifted from the origin by and two semi-major axes and
• Ellipse is also a conic section
3.33.7
10 e
21 e
1 22
0
b
r
a
rr yxxb
e
h
212b
ba0xr
Classification of Kepler’s orbits
• Elliptic motion is limited by two values of r
3.33.7
10 e 12
102
2
mk
Ep 02 2
2
Ep
mk
2
2
min2
)(p
mkrVeff
0)( min ErVeff
1 22
0
b
r
a
rr yxx
PerihelionAphelion
Classification of Kepler’s orbits
• This parameter is known as an eccentricity of an ellipse
• For a constant energy, perihelion is decreasing with increasing eccentricity
3.33.7
1 22
0
b
r
a
rr yxx
ae
h
21b
e
h
21e
a
ba
22
021 xre
eh
e
hra x
10
3.33.7
1e21 e
1 22
0
b
r
a
rr yxxb
e
h
212b
1 '
22
0
b
r
a
rr yxx
'ibb
Classification of Kepler’s orbits
• If
• Then is imaginary and is negative
• The orbit is a hyperbola
• Hyperbola is a conic section as well
Classification of Kepler’s orbits
• Hyperbolic motion is limited by one value of r - perihelion
3.33.7
1e 12
12
2
mk
Ep 0E
1 '
22
0
b
r
a
rr yxx
Classification of Kepler’s orbits
• Finally, if
• Then
• The orbit is a parabola with its center shifted from the origin by
• Parabola is a conic section
3.33.7
1e
2/h
2222 2)1( hrherre yxx
h
rhr yx 22
2
22 2 hrhr yx
Classification of Kepler’s orbits
• Parabolic motion is also limited by one value of r - perihelion
3.33.7
1e 12
12
2
mk
Ep 0E
h
rhr yx 22
2
Synopsys for orbit classification
Motion in time
• In Kepler’s case:
• Substitution:
• For ψ = 2π (one period):
3.8
)cos2
11(2
2
2
mk
Ep
E
kr
)(2
2
2
2
rVmrp
Em
drdt
r
r
rk
mrp
Em
drt
0
2
2
22
)cos1( ea
0
3
)cos1( dek
mat )sin(
3
ek
ma
2 3
k
mat
A bit of history: Kepler’s laws
• First law: “The planets move in elliptical orbits with the sun at one focus”
3.8
Johannes Kepler(1571-1630)
Tycho Brahe/Tyge Ottesen
Brahe de Knudstrup(1546-1601)
10 e02/ 22 Epmk
A bit of history: Kepler’s laws
• Second law: “The radius vector to a planet sweeps out area at a rate that is independent of its position in the orbit”
• Third law: “The square of the period of an orbit is proportional to the cube of the semi-major axis length”
3.8
2 3
k
ma
k
ma322 4
2/)( rdrdA 2/
dt
drr
dt
dA
2
2 rA
constmrp 2
constm
p
The Kepler problem in action-angle variables
• We consider periodic motion in the case of attraction
• By definition, the action variables are
10.8
Er
kW
r
W
rr
W
m
2
22
2
2
2
sin
11
2
1
dW
dpJ
dW
dpJ
drr
WdrpJ rr
The Kepler problem in action-angle variables
• We found earlier for the two-body problem
• Therefore
10.8
ddW
J
ddW
J2
22
sin
2
22
2
sin
W
2
22
2rr
kEm
r
W
W
2
)(2
The Kepler problem in action-angle variables
• Frequencies
10.8
drrr
kEmdr
r
WJ r 2
2
2
dr
r
JJ
r
kEm
22
2
42
E
mkJJ
2
2
22
)(
2
JJJ
mkE
r H
3
22
)(
4
JJJ
mk
J
Hv
J
Hv
J
Hv
rrr
The Kepler problem in action-angle variables
• Degenerate case: the frequencies for all three variables coincide
• Hence it takes the same time for all three variables to return to the same value – the same point on the orbit
• Therefore, a completely degenerate solution corresponds to a closed orbit
• We did not have to obtain an explicit expression for the orbit to realize that it is closed
10.8
3
22
)(
4
JJJ
mk
J
Hv
J
Hv
J
Hv
rrr
The Kepler problem in action-angle variables
• To lift the degeneracy, we can introduce another canonical transformation employing the following generating function:
• Then
• And the Hamiltonian
10.8
321 )()( JwJwwJwwF rr
rr wwwwwwww 321 ;;
JJJJJJJJJ r 321 ;;
2
22
)(
2
JJJ
mkH
r 2
3
222
J
mk
The Kepler problem in action-angle variables
• This Hamiltonian is cyclic in 5 variables, therefore their 5 corresponding conjugates are conserved:
• We obtained 5 constants of motion for a system with 6 degrees of freedom (the last two can be shown to be related to certain orbit parameters)
10.8
constw 1
constpJJ 21
23
222
J
mkH
constppppJJJ 2)(222
constE
mkJ
23 constw 2
Repulsive Kepler potential
• Let us consider a repulsive potential:
• Total mechanical energy:
• Orbit equation
3.10
0;)( kr
krV
2
22
0
22
mrp
rk
Emr
drp
1)cos( 1
0 eCr
2p
kmC
21
2
2
mk
Epe
VTE 0T 0V 0E
1
Repulsive Kepler potential
• This is a hyperbola
• Therefore, hyperbolic orbits correspond to the case of a positive total energy for both the attractive and the repulsive interactions
• Scattering – the orbit is never closed for E > 0
3.10
1)cos( 1
0 eCr
1 22
0
b
r
a
rr yxx
)1( 20 eC
erx
1
12
eC
b
1 22
0
b
r
a
rr yxx
E
ka
2
Scattering
• Scattering angle:
• Impact parameter:
• On the other hand:
3.10
s
b
1 22
0
b
r
a
rr yxx
2s 2
b
as
1tan22
E
ka
2
Eb
ks 2
tan2 1
Scattering
• For a beam of (noninteracting) particles incident on the scattering center, intensity (flux density) is the number of particles crossing unit area normal to the beam in unit time
• Scattering cross-section in a given direction is the ratio of the number of particles scattered into a solid angle per unit time to the incident intensity
• Differential scattering cross-section:
3.10
I
dNd
Scattering cross-section
• Conservation of the number of particles:
3.10
ssdd sin2
) 2( bdbI dN dI ssdI sin2
I
dNd
ss d
dbb
sin
Eb
ks 2
tan2 1
2tan2 sE
kb
Scattering cross-section
• Rutherford scattering cross-section
• It is independent of the sign of k!
3.10
ss d
dbb
sin
2tan2sin
2tan2 ss
ss E
k
d
d
E
k
2sin4 42
2
sE
k
2tan2 sE
kb
Ernest Rutherford(1871 – 1937)
Total scattering cross-section
• Total scattering cross-section
• It diverges because of the long-range nature of Kepler’s potential
• All the particles in an incident beam of infinite lateral extent will be scattered to some extent and must be included in the total scattering cross-section
3.10
0 42
2
0
2sin4
sin2
s
ssT
E
dkd
Laboratory coordinates
• Let us recall the initial transformation of coordinates:
• Let us recall the re-gauged Lagrangian:
• All the results obtained so far are in the re-gauged center-of-mass system, in which the center of mass is at rest:
3.11
21
12
21
21 ;
mm
rmRr
mm
rmRr
rVmm
rmm
mm
PLL
)(2
)(
)(2
)('
21
221
21
2
21
12
21
21 ;
mm
rmr
mm
rmr
Laboratory coordinates
• In the center-of-mass system, the scattering process of two particles will look like this:
• Often, while the incident particle is moving, the second one is initially at rest
• We introduce the laboratory system of coordinates, in which the center of mass is moving with a constant velocity
3.11
21
12
21
21 ;
mm
rmr
mm
rmr
21
12
21
21 ;
mm
rmRr
mm
rmRr
Laboratory coordinates
• In the laboratory system, the scattering process of two particles will look like this:
• Let us introduce notations:
• Then
• Initially
3.11
21
12
21
21 ;
mm
rmRr
mm
rmRr
21
12
21
21 ';'
mm
rmr
mm
rmr
';' 2211 rRrrRr
constmm
rmrmR
21
202101
Laboratory coordinates
• Taking the ratio of these two equations:
3.11
'121
101 rmm
rm
21
101
mm
rmR
'1
2
10 rm
rm
'12
101 v
m
vmv
svv sin'sin 11 2
1011 cos'cos
m
mvvv s
'cos
sintan
12
10
vmmv
s
s
'11 rRr
Laboratory coordinates
• Now we can write the differential scattering cross section expressed in laboratory system
• Conservation of the number of particles:
3.11
sss dI sin2)( dI L sin2)(
d
d sssL sin
sin)()(
s
ss
cos1
cos21)(
2/32
'12
10
vm
mv
'cos
sintan
12
10
vmmv
s
s
The three-body problem
• The Lagrangian of the system in Cartesian coordinates:
• This problem has 9 independent coordinates entangled by the 3 potential functions
• This Lagrangian cannot be re-gauged to a one-particle Lagrangian
• No general explicit solution is known
3.12
232
231
221
233
222
211
)()()(
2
)(
2
)(
2
)(
rrVrrVrrV
rmrmrmL
The three-body problem
• Constants of motion (not independent): total energy, three components of the center of mass linear and angular momenta
• If the three objects are allowed to move freely in 3D the orbits become very complex and sensitive to initial conditions
• Even after fixing positions of two particles and letting the third particle move in a plane, the orbit still can not be found explicitly
3.12
