Separability
Prinicipal Function

In some cases Hamilton’s principal function can be
separated.
• Each W depends on only one coordinate.
• This is totally separable.
N
S (q ,  , t )   Wk (q k ,  j )  Et
j
j
k 1
m
S (q ,  , t )   Wk (q k ,  j )  Wm (q j  m ,  j , t )
j
j
k 1
Function can be partially separable.
Hamiltonian Separation


Simpler separability occurs
when H is a sum of
independent parts.
The Hamilton-Jacobi
equation separates into N
equations.
N
H ( q , p j , t )   H k ( q k , pk )
j
k 1
 k W
H k  q , k
q

N
E  k
k 1

   k

Staeckel Conditions

Specific conditions exist for separability.
• H is conserved.
• L is no more than quadratic in dqj/dt, so that in matrix form:
H=1/2(p  a)T-1(p  a)+V(qj)
• The coordinates are orthogonal, so T is diagonal.
• The vector a has aj = aj (qj)
• The potential is separable.
• There exists a matrix F with Fij = Fij(qi)
T 
1
jj
1

T jj
V
V j (q j )
T jj
F 
1
1j

1
T jj
Combined Potentials

k
V (r , z )    gz
r
Particle under two forces
• Attractive central force
• Uniform field along z
r  x2  y2  z 2

Z
Y
X
Eg: charged particle with
another fixed point charge in
a uniform electric field.
Parabolic Coordinates

Select coordinates
• Constant value x h describe
paraboloids of revolution
• Other coordinate is f
• Equate to cartesian system

Find differentials to get
velocity.
1
dx  
2
1
dy  
2
x rz
h rz
x  xh cos f
y  xh sin f
z  (x  h ) / 2
h
x 
dx 
dh  cos f  xh sin fdf
x
h 
h
x 
dx 
dh  sin f  xh cos fdf
x
h 
dz  (dx  dh ) / 2
Energy and Momentum

 x
1  h 
ds 2  dx 2  dy 2  dz 2   1   dx 2  1   dh 2   xhdf 2
4  x 
 h

2
m  ds 
T  
2  dt 
dT m  h  
 1  x

dx 4  x 
dT m  x 
ph 
 1  h
dh 4  h 
px 
pf 
dT


m
xh
f
df
Substituting for the new variables:


2
pf
2
2
2
T
xpx  hph 
m(x  h )
2mxh
V 
2k
g
 (x  h )
(x  h ) 2
Separation of Variables

Hamiltonian is not directly separable.
• Set E = T + V
• Multiply by x  h/2
xpx2
pf2
hph2 g 2
pf2
g 2
E
E
 x 
 x
 h 
 h k
m 4
4mx 2
m 4
4mh 2


There are parts depending just on x, h.
There is a cyclic coordinate f.
• Constant of motion pf
• Reduce to two degrees of freedom
Generator Separation

Set Hamilton’s function.
• Use momentum definition
• Expect two constants , b

Find one variable
W  Wx  Wh  pff
xpx2
pf2
g 2
E

 x 
 x
m 4
4mx 2
Wx
px 
x


Do the same for the other
variable.
And get the last constant.
xph2
pf2
g 2
E
k  
 h 
 h
m 4
4mh 2
pf2
1
k  g
Wh 
dx E 
 h

mh
2
2mh 2
2
pf2
1
 g
Wx 
dx E 
 x

mx 2
2mx 2
2
b
W

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