In the Hamiltonian Formulation, the generalized coordinate q k & the generalized momentum p k are called Canonically Conjugate quantities. Hamilton’s Equations are: q k = (  H/  p k ); p k = - (  H/  q k ) (1)  In the (often occurring) case where H does not contain a particular q k, then, by (1), the corresponding p k = 0 & p k = constant or p k is conserved (is a constant or an integral of the motion) Coordinates q k not appearing in the Hamiltonian H are called Cyclic or Ignorable Coordinates

Note that, if q k 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 q k, will still appear in L. L = L(q 1,..,q k-1,q k+1,…q s,q 1,.…q s,t)  The number of degrees of freedom s in Lagrangian Mechanics will not be changed. We still must set up & solve s 2 nd 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 q k is a cyclic or ignorable coordinate:  p k = constant  α k H = H(q 1,..,q k-1,q k+1,..q s,p 1,..p k-1,α k,p k+1,. p s,t)  It is only necessary to solve 2s st order differential equations.  This means an (effective) reduction in complexity to s-1 degrees of freedom In this case, the cyclic coordinate q k is completely separated.  q k is ignorable as far as rest of solution for the dynamics of the system is concerned.

If q k is cyclic  p k = constant  α k We can calculate the constant α k from the initial conditions. Then we can compute the cyclic coordinate by solving a simple differential equation: q k = (  H/  p k ) = (  H/  α k )  ω k (t) Given the initial conditions, this integrates to give: q k (t) = ∫ω 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.

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 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(q j,q j,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(q j,q j,t) dt = 0 where L  ∑ j q j (  L/  q j ) – H. Or: L  ∑ j q j p j - H & by letting q j & p j be varied independently. (See Eqtn (7.189), p 273).

Hamiltonian Dynamics: Treats the generalized coordinates q j & the generalized momenta, p j as independent. But they aren’t really so, in the true sense! If the time dependence of each coordinate q j (t) is known, then we have completely solved the problem!  We can calculate the generalized velocities from: q j (t)  [dq j (t)/dt] (1) & We can calculate the generalized momenta from: p j (t)  [  L(q j,q j,t)/  q j ] (2) Bottom Line: q j & q j are related by a simple time derivative (1), independent of the manner in which the system behaves. On the other hand, the relations between q j & p j (Hamilton’s Eqtns) are eqtns of motion themselves! Finding the relations between q j & p j is equivalent to solving the problem!

Phase Space Sect 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 q j & s generalized momenta p j 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 q j - q j configuration space). This is important in statistical mechanics, in which ρ is the many particle distribution function!

Virial Theorem Sect Skim discussion. Read more details on your own! Consider a many particle system. Positions r α & momenta p α. Bounded. Define: S  ∑ α r α  p α Take the time derivative of S: (dS/dt) = ∑ α [(dr α /dt)  p α + r α  (dp α /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 (dp α /dt)  r α ] ave = [∑ α p α  (dr α /dt)] ave (4) Now, use the KE theorem from before:  On the right side of (4) we can write: p α  (dr α /dt) = 2T α so the right side of (4) becomes: = [2 ∑ α T α ] ave = 2[T] ave (5) Here T α = KE of particle α & T = total KE of the system Newton’s 2 nd Law:  (dp α /dt) = F α = force on particle α  the left side of (4) is = -[∑ α (F α  r α )] ave (6)

Combine (5) & (6):  [T] ave = - (½) [∑ α (F α  r α )] ave (7)  The Virial Theorem - (½) [∑ α (F α  r α )] 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 (F α  r α )] ave  The Virial Theorem Application to classical dynamics: For a conservative system in which a PE can be defined: F α  -  U α  [T] ave = - (½) [∑ α (  U α  r α )] ave In the special case of a Central Force, in which (for each particle α): |F|  r n, n any power & r = distance between particles  U = k r n+1   U  r = (dU/dr)r =k(n+1) r n+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 r n, U(r) = k r n+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