Ch. 7: Sect. 7.10 (Part II): Cyclic or Ignorable Coordinates. Sect. 7.11: Comments. Sect. 7.12: Phase Space

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Cyclic or Ignorable Coordinates

• In the Hamiltonian Formulation, the generalized
  coordinate qk the generalized momentum pk are
  called Canonically Conjugate quantities.
• Hamiltons 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

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• 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,qk1,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!

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• In the Hamiltonian (canonical) formulation of
  mechanics If qk is a cyclic or ignorable
  coordinate ? pk constant ? ak
• H H(q1,..,qk-1,qk1,..qs,p1,..pk-1,ak,pk1,.
  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.

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• If qk is cyclic ? pk constant ? ak
• We can calculate the constant ak from the initial
  conditions. Then we can compute the cyclic
  coordinate by solving a simple differential
  equation
• qk (?H/?pk) (?H/?ak) ? ?k(t)
• Given the initial conditions, this integrates to
  give
• qk(t) ??k 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.

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• 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 Goldsteins graduate mechanics text.

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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!

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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 Hamiltons Equations from
  Lagranges Equations. We got Lagrange Equations
  from Hamiltons Principle the calculus of
  variations applied to
• d ? L(qj,qj,t) dt 0 where L T - U
• The authors show that we can get Hamiltons
  Equations directly from Hamiltons Principle
  the calculus of variations applied to d ?
  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).

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• Hamiltonian Dynamics Treats the generalized
  coordinates qj the generalized momenta, pj as
  independent. But they arent 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 (Hamiltons Eqtns) are
  eqtns of motion themselves! Finding the relations
  between qj pj is equivalent to solving the
  problem!

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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.
• ? ? phase space density points per unit (2s
  dimen.) volume
• The authors prove Liouvilles Theorem
• (d?/dt) 0 ? ? constant
• Proved using Hamiltonian dynamics. Cannot use
  Lagrangian dynamics (Liouvilles Theorem is not
  valid in qj - qj configuration space).
• This is important in statistical mechanics, in
  which ? is the many particle distribution
  function!

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Virial Theorem Sect. 7.13

• Skim discussion. Read more details on your own!
• Consider a many particle system. Positions ra
  momenta pa. Bounded. Define S ? ?a ra?pa
• Take the time derivative of S
• (dS/dt) ?a (dra /dt)?pa ra? (dpa /dt)
  (1)
• The time average of (dS/dt) in the time interval
  t
• (dS/dt)ave ? (1/ t) ? (dS/dt)dt (0 lt t lt t)
  (2)
• (dS/dt)ave S(t) - S(0)/t (3)
• If the motion is periodic with period t
• ? S(t) S(0) (3) ? (dS/dt)ave 0

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• 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 t very
  large. ? Either for a periodic system or for a
  non-periodic system with large t we can (in
  principle) make (S)ave 0. When (S)ave 0,
  (understood to be the long time average) (1)
  (2) combine to give
• - ?a(dpa/dt)?raave ?a pa?(dra/dt)ave
  (4)
• Now, use the KE theorem from before ? On the
  right side of (4) we can write pa? (dra/dt)
  2Ta so the right side of (4) becomes
  2 ?a Ta ave 2Tave (5)
• Here Ta KE of particle a T total KE of the
  system
• Newtons 2nd Law ? (dpa/dt) Fa force on
  particle a ? the left side of (4) is
  -?a(Fa?ra)ave (6)

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• Combine (5) (6)
• ? Tave - (½) ?a(Fa?ra)ave (7)
• ? The Virial Theorem
• - (½) ?a(Fa?ra)ave ? The Virial
• The time average kinetic energy of a system is
• equal to its virial
• Application to Statistical Mechanics See Example
  7.14.

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• Tave - (½) ?a(Fa?ra)ave ? The Virial
  Theorem
• Application to classical dynamics For a
  conservative system in which a PE can be defined
  Fa ? - ?Ua
• ? Tave - (½) ?a(?Ua?ra)ave
• In the special case of a Central Force, in which
  (for each particle a) F ? rn, n any power r
  distance between particles ? U k rn1
• ? ?U?r (dU/dr)r k(n1) rn1 or ?U?r
  (n1)U
• ? The Virial Theorem gives
• Tave (½)(n1) Uave
• Conservative Central forces ONLY!

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• Virial Theorem, Conservative Central Forces
• (F(r) k rn , U(r) k rn1)
• Tave (½)(n1) Uave
• Example 1 Gravitational (or Coulomb) Potential
• n - 2 ? Tave - (½) Uave
• Example 2 Isotropic Simple Harmonic Oscillator
  Potential
• n 1 ? Tave Uave
• Example 3 n -1 ? Tave 0 !
• Example 4 n ? integer (say, real power x)
• n x ? Tave (½) (x1) Uave