Transcript Cyclic or Ignorable Coordinates • In the Hamiltonian Formulation, the generalized coordinate qk & the generalized momentum pk are called Canonically Conjugate quantities. •

Cyclic or Ignorable Coordinates
• In the Hamiltonian Formulation, the generalized
coordinate qk & the generalized momentum pk are
called Canonically Conjugate quantities.
• Hamilton’s Equations are:
qk = (H/pk); pk = - (H/qk)
(1)
 In the (often occurring) case where H does not
contain a particular qk, then, by (1), the
corresponding pk = 0 & pk = constant or pk is
conserved (is a constant or an integral of the motion)
• Coordinates qk not appearing in the Hamiltonian H are called
Cyclic or Ignorable Coordinates
• Note that, if qk is a cyclic coordinate (not appearing in
the Hamiltonian H) it also will not appear in the
Lagrangian L! However, in general, the corresponding
generalized velocity qk, will still appear in L.
L = L(q1,..,qk-1,qk+1,…qs,q1,.…qs,t)
 The number of degrees of freedom s in
Lagrangian Mechanics will not be changed.
We still must set up & solve s 2nd order differential equations!
• However, as we discuss now, in Hamiltonian
Mechanics, a cyclic coordinate reduces the complexity of the
math by reducing the number of differential equations we have
to deal with. This, in fact is one of (the only) advantages of
Hamiltonian Mechanics over Lagrangian Mechanics!
• In the Hamiltonian (canonical) formulation
of mechanics: If qk is a cyclic or ignorable
coordinate:  pk = constant  αk
H = H(q1,..,qk-1,qk+1,..qs,p1,..pk-1,αk,pk+1,. ps,t)
 It is only necessary to solve 2s - 2 1st order
differential equations.  This means an (effective)
reduction in complexity to s-1 degrees of freedom
• In this case, the cyclic coordinate qk is completely
separated.  qk is ignorable as far as rest of
solution for the dynamics of the system is concerned.
• If qk is cyclic  pk = constant  αk
• We can calculate the constant αk from the initial
conditions. Then we can compute the cyclic
coordinate by solving a simple differential equation:
qk = (H/pk) = (H/αk)  ωk(t)
Given the initial conditions, this integrates to give:
qk(t) = ∫ωk dt
 The Hamiltonian Formulation is well suited
(much better than the Lagrangian formulation!) to solve
problems with one or more cyclic coordinates.
My Opinion: This is one of the FEW CASES where the
Hamiltonian method is superior to the Lagrangian method.
• Its worth noting: the fact that the Hamiltonian
Formulation is well suited to solve problems with
cyclic coordinates has led to the development of still
other formulations of mechanics!
– For example, it can be shown that it is always
possible to find transformations of the coordinates
such that in the new coordinate system, ALL
coordinates are cyclic! (None are in the Hamiltonian!)
– This is another formulation of mechanics called
the Hamilton-Jaocbi formulation.
– It forms the foundation of some modern theories of matter
& is beyond the scope of the course. See Goldstein’s
graduate mechanics text.
Example 7.12
• Use the Hamiltonian
method to find the
equations of motion for
the spherical pendulum of
mass m & length b.
(Figure). Worked on the
board!
Brief Comments on Dynamical Variables &
Variational Calculations in Physics Sect. 7.11
• Skip most discussion in class. Read the details on your own!
• Recall: We got Hamilton’s Equations from Lagrange’s
Equations. We got Lagrange Equations from Hamilton’s
Principle & the calculus of variations applied to
δ ∫ L(qj,qj,t) dt = 0 where L = T - U
• The authors show that we can get Hamilton’s Equations
directly from Hamilton’s Principle & the calculus of
variations applied to: δ ∫ L(qj,qj,t) dt = 0 where
L  ∑j qj (L/qj) – H. Or: L  ∑j qj pj - H & by letting
qj & pj be varied independently. (See Eqtn (7.189), p 273).
• Hamiltonian Dynamics: Treats the generalized
coordinates qj & the generalized momenta, pj as
independent. But they aren’t really so, in the true sense!
• If the time dependence of each coordinate qj(t) is known,
then we have completely solved the problem!
 We can calculate the generalized velocities from:
qj(t)  [dqj(t)/dt]
(1)
& We can calculate the generalized momenta from:
pj(t)  [L(qj,qj,t)/qj] (2)
• Bottom Line: qj & qj are related by a simple time derivative
(1), independent of the manner in which the system behaves.
On the other hand, the relations between qj & pj (Hamilton’s
Eqtns) are eqtns of motion themselves! Finding the relations
between qj & pj is equivalent to solving the problem!
Phase Space Sect. 7.12
• Skip most discussion in class. Read details on your own!
• For a system with s degrees of freedom (many particles).
Consider an abstract 2 s dimensional Hamiltonian phase space
in which s generalized coordinates qj & s generalized
momenta pj are represented by a single point.
ρ  phase space density = # points per unit (2s dimen.) volume
• The authors prove Liouville’s Theorem
(dρ/dt) = 0 
ρ = constant
• Proved using Hamiltonian dynamics. Cannot use Lagrangian
dynamics (Liouville’s Theorem is not valid in qj - qj
configuration space).
• This is important in statistical mechanics, in which ρ is the
many particle distribution function!
Virial Theorem Sect. 7.13
• Skim discussion. Read more details on your own!
• Consider a many particle system. Positions rα &
momenta pα. Bounded. Define: S  ∑α rαpα
• Take the time derivative of S:
(dS/dt) = ∑α [(drα /dt)pα+ rα (dpα /dt)]
(1)
• The time average of (dS/dt) in the time interval τ:
(dS/dt)ave  (1/ τ) ∫ (dS/dt)dt (0 < t < τ)
(2)
(dS/dt)ave = [S(τ) - S(0)]/τ
(3)
• If the motion is periodic with period τ:
 S(τ) = S(0); (3)  (dS/dt)ave = 0
• Now some manipulation! If the system motion is not periodic,
we can still make (dS/dt)ave = (S)ave as small as we want by
taking τ very large.  Either for a periodic system or for a
non-periodic system with large τ we can (in principle) make
(S)ave = 0. When (S)ave = 0, (understood to be the long time
average) (1) & (2) combine to give:
- [∑α(dpα/dt)rα]ave = [∑α pα(drα/dt)]ave (4)
• Now, use the KE theorem from before:  On the right side
of (4) we can write: pα (drα/dt) = 2Tα so the right side of (4)
becomes:
= [2 ∑α Tα ]ave = 2[T]ave
(5)
Here Tα = KE of particle α & T = total KE of the system
• Newton’s 2nd Law:  (dpα/dt) = Fα = force on particle α 
the left side of (4) is =
-[∑α(Fαrα)]ave (6)
• Combine (5) & (6):

[T]ave = - (½) [∑α(Fαrα)]ave (7)
 The Virial Theorem
- (½) [∑α(Fαrα)]ave  The Virial
The time average kinetic energy of a system is
equal to its virial
• Application to Statistical Mechanics: See Example 7.14.
[T]ave = - (½) [∑α(Fαrα)]ave  The Virial Theorem
• Application to classical dynamics: For a conservative
system in which a PE can be defined: Fα  - Uα
 [T]ave = - (½) [∑α(Uαrα)]ave
• In the special case of a Central Force, in which (for
each particle α): |F|  rn, n any power & r = distance
between particles  U = k rn+1
 Ur = (dU/dr)r =k(n+1) rn+1 or: Ur = (n+1)U
 The Virial Theorem gives:
[T]ave = (½)(n+1) [U]ave
Conservative Central forces ONLY!
• Virial Theorem, Conservative Central Forces:
(F(r) = k rn , U(r) = k rn+1)
•
•
•
•
[T]ave = (½)(n+1) [U]ave
Example 1: Gravitational (or Coulomb) Potential:
n=-2 
[T]ave = - (½) [U]ave
Example 2: Isotropic Simple Harmonic Oscillator Potential:
n=+1 
[T]ave = [U]ave
Example 3: n = -1 
[T]ave = 0 !
Example 4: n  integer (say, real power x):
n=x

[T]ave = (½) (x+1) [U]ave