Transcript Lagrangian and Hamiltonian Dynamics

Lagrangian and
Hamiltonian Dynamics
Chapter 7
Claude Pruneau
Physics and Astronomy
Minimal Principles in Physics
• Hero of Alexandria 2nd century BC.
– Law governing light reflection minimizes
the path length.
• Fermat’s Principle
– Refraction can be understood as the path
that minimizes the time - and Snell’s law.
• Maupertui’s (1747)
– Principle of least action.
• Hamilton (1834, 1835)
Hamilton’s Principle
Of all possible paths along which a
dynamical system may move from one
point to another within a specified time
interval (consistent with any constraints),
the actual path followed is that which
minimizes the time integral of the
difference between the kinetic and
potential energy.
Hamilton’s Principle
• In terms of calculus of variations:
t2
  T  U dt  0
t1
• The  is a shorthand notation which represents a variation as
discussed in Chap 6.
• The kinetic energy of a particle in fixed, rectangular coordinates
is of function of 1st order time derivatives of the position
T  T ( x&i )
• The potential energy may in general be a function of both
positions and velocities. However if the particle moves in a
conservative force field, it is a function of the xi only.
U  U(xi )
Hamilton’s Principle (cont’d)
• Define the difference of T and U as the Lagrange
function or Lagrangian of the particle.
L  T  U  L(xi , x&i )
• The minimization principle (Hamilton’s) may thus be
written:
t2
  L(xi , x&i )dt  0
t1
Derivation of Euler-Lagrange Equations
• Establish by transformation…
x2
  f {y, y'; x}dx  0
x1
xt
t2
  L(xi , x&i )dt  0
t1
yi (x)  xi (t)
yi '(x)  x&i (t)
f {yi (x), yi '(x); x}  L(xi (t), x&i (t))
f
d f

0
yi dx yi '
L d L

 0,
xi dt x&i
i  1,2, 3
Lagrange Equations of Motion
L d L

 0,
xi dt x&i
i  1,2, 3
• L is called Lagrange function or Lagrangian for the
particle.
• L is a function of xi and dxi/dt but not t explicitly (at
this point…)
Example 1: Harmonic Oscillator
Problem: Obtain the Lagrange Equation of motion for
the one-dimensional harmonic oscillator.
Solution:
• Write the usual expression for T and U to determine L.
L  T U  12 m&
x2  12 kx2
• Calculate derivatives.
L
 kx
x
L
 mx&
x&
d L
 m&
x&
dt &
x
Example 1: Harmonic Oscillator (cont’d)
• Combine in Lagrange Eq.
L d L

 0,
xi dt x&i
m&
x& kx  0
i  1,2, 3
Example 2: Plane Pendulum
Problem: Obtain the Lagrange Equation of motion for
the plane pendulum of mass “m”.
l

Solution:
• Write the expressions for T and U to determine L.
T  1 m(x&2  y&2 )  1 I 2  1 ml 2&2
2
2
2
U  mgl(1  cos )
L  T U  12 ml 2&2  mgl(1 cos )
Example 2: Plane Pendulum (cont’d)
• Calculate derivatives of L by treating  as if it were a
rectangular coordinate.
L  1 2 &2
  2 ml   mgl(1 cos )  mgl sin 
 
L  1 2 &2
  2 ml   mgl(1 cos )  ml 2&
& &
d L
 ml 2&&
dt &
• Combine...
mglsin  ml 2&& 0
g
  sin   0
l
Remarks
• Example 2 was solved by assuming that 
could be treated as a rectangular coordinate
and we obtain the same result as one obtains
through Newton’s equations.
• The problem was solved by involving kinetic
energy, and potential energy. We did not use
the concept of force explicitly.
Generalized Coordinates
• Seek generalization of coordinates.
• Consider mechanical systems consisting of a
collection of n discrete point particles.
• Rigid bodies will be discussed later…
• We need n position vectors, I.e. 3n quantities.
• If there are m constraint equations that limit the
motion of particle by for instance relating some of
coordinates, then the number of independent
coordinates is limited to 3n-m.
• One then describes the system as having 3n-m
degrees of freedom.
Generalized Coordinates (cont’d)
• Important note: if s=3n-m coordinates are required to
describe a system, it is NOT necessary these s
coordinates be rectangular or curvilinear coordinates.
• One can choose any combination of independent
parameters as long as they completely specify the
system.
• Note further that these coordinates (parameters)
need not even have the dimension of length (e.g.  in
our previous example).
• We use the term generalized coordinates to describe
any set of coordinates that completely specify the
state of a system.
• Generalized coordinates will be noted: q1, q2, …, qn.
Generalized Coordinates (cont’d)
• A set of generalized coordinates whose number equals the
number s of degrees of freedom of the system, and not
restricted by the constraints is called a proper set of generalized
coordinates.
• In some cases, it may be useful/convenient to use generalized
coordinates whose number exceeds the number of degrees of
freedom, and to explicitly use constraints through Lagrange
multipliers.
– Useful e.g. if one wishes to calculate forces due to constraints.
• The choice of a set of generalized coordinates is obviously not
unique.
– They are in general (infinitely) many possibilities.
• In addition to generalized coordinates, we shall also consider
time derivatives of the generalized coordinates called
generalized velocities.
Generalized Coordinates (cont’d)
Notation:
q1,q2 ,L ,qs
q&1, q&2 ,L , q&s
or
or
{qi }
{q&i }
i  1,..., s
i  1,..., s
Transformation
• Transformation: The “normal” coordinates can be
expressed as functions of the generalized coordinates - and
vice-versa.
x ,i  x ,i (q1 , q2 ,L , qs ,t),
 x ,i (q j ,t),
 =1,2,...,n

i  1, 2, 3
j  1, 2,..., s
Transformation (cont’d)
• Rectangular components of the velocties depend on
the generalized coordinates, the generalized
velocities, and the time.
x ,i  x& ,i (q j ,q&j ,t)
• Inverse transformations are noted:
q j  q j (x ,i ,t)
q&j  q&j (x ,i , x& ,i ,t)
• There are m=3n-s equations of constraint…
fk (x ,i ,t)  0,
k  1,2,..., m
Example: Generalized coordinates
• Question: Find a suitable set of generalized
coordinates for a point particle moving on the surface
of a hemisphere of radius R whose center is at the
origin.
• Solution: Motion on a spherical surface implies:
x 2  y2  z2  R2  0,
z0
• Choose cosines as generalized coordinates.
x
q1  ,
R
y
q2  ,
R
q12  q22  q23  1
z
q3 
R
Example: Generalized coordinates (cont’d)
• q1, q2, q3 do not constitute a proper set of generalized
of coordinates because they are not independent.
• One may however choose e.g. q1, q2, and the
constraint equation
x 2  y2  z2  R2
Lagrange Eqs in Gen’d Coordinates
• Of all possible paths along which a dynamical system
may move from one point to another in configuration
space within a specified time interval, the actual path
followed is that which minimizes the time integral of
the Lagrangian for the system.
Remarks
• Lagrangian defined as the difference between kinetic and
potential energies.
• Energy is a scalar quantity (at least in Galilean relativity).
• Lagrangian is a scalar function.
• Implies the lagrangian must be invariant with respect to
coordinate transformations.
• Certain transformations that change the Lagrangian but leave
the Eqs of motion unchanged are allowed.
• E.G. if L is replaced by L+d/dt f(qi,t), for a function with
continuous 2nd partial derivatives. (Fixed end points)
• The choice of reference for U is also irrelevant, one can add a
constant to L.
Lagrange’s Eqs
• The choice of specific coordinates is therefore immaterial
L  T ( x& ,i )  U(x ,i )
 T (q j , q&j ,t)  U(q j ,t)
 L(q j , q&j ,t)
• Hamilton’s principle becomes
2
  L(q j , q&j ,t)  0
1
Lagrange’s Eqs
xt
yi (x)  qi (t)
yi '(x)  q&i (t)
f {yi (x), yi '(x); x}  L(qi (t), q&i (t))
L d L

 0,
qi dt q&i
i  1,2,..., s
“s” equations
“m” constraint equations
Applicability:
1. Force derivable from one/many potential
2. Constraint Eqs connect coordinates, may be fct(t)
Lagrange Eqs (cont’d)
• Holonomic constraints
fk (x ,i ,t)  0,
k  1,2,..., m
• Scleronomic constraints
– Independent of time
• Rheonomic
– Dependent on time
Example: Projectile in 2D
• Question: Consider the motion of a projectile under
gravity in two dimensions. Find equations of motion in
Cartesian and polar coordinates.
• Solution in Cartesian coordinates:
1 2
mv
2
U  mgy
with U  0 at y  0.
T
L  T U 
1 2
mv  mgy
2
L d L

0
x dt &
x
d
0  m&
x0
dt
&
x& 0
L d L

0
y dt &
y
d
mg  m&
y0
dt
&
y& g
Example: Projectile in 2D (cont’d)
• In polar coordinates…


1
m r&2  r 2&2
2
U  mgr sin 
with U  0 at   0.
T
L d L

0
r dt &
r
d
mr&2  mg sin   m&
r 0
dt
r&2  g sin   &
r& 0
L  T U 


1
m r&2  r 2&2  mgr sin 
2
L d L

0
 dt &
d
mgr cos 
mr 2&  0
dt
gr cos  2r&
r& r 2&& 0


Example: Motion in a cone
• Question: A particle of mass “m” is constrained to move on the
inside surface of a smooth cone of hal-angle a. The particle is
subject to a gravitational force. Determine a set of generalized
coordinates and determine the constraints. Find Lagrange’s Eqs
of motion.
z
Solution:
Constraint:
z  r cot   0
r
2 degrees of freedom only!
2 generalized coordinates.

y
x

Example: Motion in a cone (cont’d)
• Choose to eliminate “z”.
v 2  r&2  r 2&2  z&2
 r&2  r 2&2  r&2 cot 2 
z  rcot 
U  mgz
 mgr cot 
 r&2 csc 2   r 2&2
L is


1 2 1
mv  m r&2 csc 2   r 2&2  mgr cot 
2
2
L
d L
independent of .
0

dt &
L
 constant=mr 2&
&
L  T U 
mr 2& mr 2
is the angular momentum relative to the
axis of the cone.
Example: Motion in a cone (cont’d)
• For r:
L d L

0
r dt &
r
r  r&2 sin2   gsin cos  0
Lagrange’s Eqs with underdetermined multipliers
• Constraints that can be expressed as
algebraic equations among the coordinates
are holonomic constraints.
• If a system is subject to such equations, one
can always find a set of generalized
coordinates in terms of which Eqs of motion
are independent of these constraints.
• Constraints which depend on the velocities
have the form f x ,i , x& ,i ,t  0
Non holonomic constraints unless eqs can be
integrated to yield constrains among the coordinates.
• Consider
 A x&  B  0
i i
i  1,2, 3
i
• Generally non-integrable, unless
Ai 
f
,
xi
Bi 
f
 0,
t
f  f (xi ,t)
• One thus has:
f
f
&
x

 x i t  0
i
i
• Or…
df
0
dt
• Which yields…
f (xi ,t)  constant  0
• So the constraints are actually holonomic…
Constraints…
• We therefore conclude that if constraints can be
expressed
fk
f
 q dqi  t dt  0
i
i
• Constraints Eqs given in differential form can be
integrated in Lagrange Eqs using undetermined
multipliers.
• For:
fk
 q dqi  0
i
i
• One gets:
L d L
fk

  k (t)
0
q j dt q&j
q j
k
Forces of Constraint
• The underdetermined multipliers are the forces of
constraint:
fk
Q j   k
q j
k
Example: Disk rolling incline plane
Example: Motion on a sphere
7.6 Equivalence of Lagrange’s and Newton’s Equations
• Lagrange and Newton formulations of mechanic are
equivalent
• Different view point, same eqs of motion.
• Explicit demonstration…
L d L

 0,
xi dt &
xi
i=1,2,3
 T  U  d  T  U 

 0,
xi
dt
&
xi
T
0
xi
and

U d T

,
xi dt &
xi

U
 Fi ,
xi
U
 0,
&
xi
i=1,2,3
i=1,2,3
i=1,2,3
i=1,2,3
d T
d   3 1 2 d

x j   m&
xi   p&i
 m&
dt &
xi dt &
xi  j 1 2
 dt
Fi  p&i ,
i=1,2,3
xi  xi (qi ,t)
xi  
j
xi
x
q&j  i
q j
t
&
xi xi

q&j q j
Generalized momentum
pj 
T
q&j
Generalized force defined through virtual work W
W   Fi xi
i
xi
W   Fi
qj
q j
i, j
W  Qj q j
j
xi
Q j   Fi
q j
i
For a conservative system:
U
Qj  
q j
pj 
T
  1 2

xi 

2 m&


q&j q&j  i
p j   m&
xi
i
remember
p j   m&
xi
i
&
xi
q&j
&
xi xi

q&j q j
xi
q j
 xi
d xi 
p j    m&x&i  mx&i

q
dt
q
i 
j
j
d xi
2 xi
2 xi

q&k 
dt q j k qk q j
q j t
xi
2 xi
2 xi
p j   m&x&i   mx&i
q&k   mx&i
q j i,k qk q j
q j t
i
i
Qj
T
p j  Qj 
q j
T
pj 
q&j
T
&
xi
  m&
xi
q j i
q j
d  T  T
U
  Qj  


dt  q&j  q j
q j
d  T  T
U
  Qj  


dt  q&j  q j
q j
Because U does not depend on q&j , one has
d   T U   T U 

0


dt  q&j 
q j
And with L  T -U,
d  L  L
 0


dt  q&j  q j
7.10 Canonical Equations of Motion – Hamilton Dynamics
Whenever the potential energy is velocity independent:
pj 
L
x j
Result extended to define the Generalized Momenta:
pj 
L
q j
Given Euler-Lagrange Eqs:
One also finds:
L d L

0

q j dt q j
L
p j 
q j
p j
The Hamiltonian may then be considered a function of the
generalized coordinates, qj, and momenta pj:
H   p j q j  L
j
… whereas the Lagrangian is considered a function of the
generalized coordinates, qj, and their time derivative.
H (qk , pk , t )   p j q j  L(qk , q k , t )
j
To “convert” from the Lagrange formulation to the
Hamiltonian formulation, we consider:
 H
 H
H
dH   
dqk 
dpk  
dt
pk
j  qk
 t
But given: H   p j q j  L
j
One can also write:
dH    pk dq k  q k dpk   dL
j

 L
L
L




   pk dqk  qk dpk 
dqk 
dqk  
dt

qk
qk
j 
 t
p k
 pk
L
dH   q&k dpk  p&k dqk 
dt
t
j
That must also equal:
 H
 H
H
dH   
dqk 
dpk  
dt
pk
j  qk
 t
We then conclude:
H
 q k
pk
H
  p k
qk
H
L

t
t
Hamilton Equations
Let’s now rewrite:
 H
 H
H
dH   
dqk 
dpk  
dt
pk
j  qk
 t
 p k
q k
And calculate:
dH
H
   p k q k  q k p k  
dt
t
j
0
Finally conclude:
If :
H
0
t
dH H

dt
t
H is a constant of motion
If, additionally, H=U+T=E, then E is a conserved quantity.:
Some remarks
• The Hamiltonian formulation requires, in general, more
work than the Lagrange formulation to derive the equations
of motion.
• The Hamiltonian formulation simplifies the solution of
problems whenever cyclic variables are encountered.
•Cyclic variables are generalized coordinates that do not
appear explicitly in the Hamiltonian.
• The Hamiltonian formulation forms the basis to powerful
extensions of classical mechanics to other fields e.g. Beam
physics, statistical mechanics, etc.
• The generalized coordinates and momenta are said to be
canonically conjugates – because of the symmetric nature
of Hamilton’s equations.
More remarks
• If qk is cyclic, I.e. does not appear in the Hamiltonian, then
p k 
L
H

0
q k q k
• And pk is then a constant of motion. p  
k
k
• A coordinate cyclic in H is also cyclic in L.
• Note: if qk is cyclic, its time derivative “q-dot” appears explicitly
in L.
•No reduction of the number of degrees of freedom in the
Lagrange formulation: still “s” 2nd order equations of motion.
•Reduction by 2 of the number of equations to be solved in the
Hamiltonian formulation – since 2 become trivial…
q k 
H
k
 k
where k is possibly a function of t.
One thus get the simple (trivial) solution:
qk (t )   k dt
The solution for a cyclic variable is thus reduced to a simple
integral as above.
The simplest solution to a system would occur if one could
choose the generalized coordinates in a way they are ALL
cyclic. One would then have “s” equations of the form :
qk (t )   k dt
Such a choice is possible by applying appropriate
transformations – this is known as Hamilton-Jacobi Theory.
Some remarks on the calculus of variation
t2
Hamilton’s Principle:
  L(qk , q k , t )dt  0
t1
Evaluated:
 L

L


  q qk  q qk dt  0
k
k

t1 
t2
Where qk and q k are not independent!
 dqk  d
  q k
 dt  dt
q k   
The above integral becomes after integration by parts:
 L d L
  q  dt q
k
k
t1 
t2

qk dt  0


Which gives rise to Euler-Lagrange equations:
L d L

0
q k dt q k
Alternatively, Hamilton’s Principle can be written:




    pk qk  H dt  0

t1  k
t2
Which evaluates to:


H
H
   pk q k  q k pk  q qk  p pk dt  0
k
k

1 j 
2
2
2
1 j
1 j
  pk q k dt    pk
Consider:
Integrate by parts:
d
q k
dt
2
2
1 j
1 j
  pk q k dt    p k qk dt
The variation may then be written:

H


q

   k p
k
1 k 
2


H
pk   p k 


qk


 
qk 
dt  0
 
 


H


q

   k p
k
1 k 
2


H
pk   p k 


qk


 
qk 
dt  0
 
 

H
 q k
pk
H
  p k
qk
Hamilton Equations