2Sect. 8.2: Cyclic Coordinates & Conservation Theorems Definition of the Hamiltonian: (sum on i)H(q,p,t) qipi – L(q,q,t) (1)A cyclic coordinate qi one which doesn’t appear in the Lagrangian L. Lagrange Eqtns pi (∂L/∂qi) = 0 pi (∂L/∂qi) = constant. The momentum conjugate to a cyclic coordinate qi is conserved.Hamilton’s Eqtns get the same result. qi is cyclic From (1) qi doesn’t appear in H either. Hamilton’s Eqtn pi = - (∂H/∂qi) = 0 pi = constant. The momentum conjugate to a cyclic coord qi is conserved.
3All momentum conservation theorems from Ch All momentum conservation theorems from Ch. 2 hold also in the Hamiltonian formalism. In Sect. 2.6, merely replace L with H & all else carries over directly!This statement includes the connections between the invariance or symmetry properties of the system & the conserved generalized momenta.If the system is invariant (symmetrical) under uniform translation A Linear momentum is conserved.System is invariant (symmetrical) under uniform rotation An Angular momentum is conserved.
4What about energy conservation? For H = H(q,p,t) (dH/dt) (H/qi)qi + (H/pi)pi + (H/t)Hamilton’s Eqtns: qi = (H/pi), pi = - (H/qi) (dH/dt) = (H/t) (a)In the derivation of Hamilton’s Eqtns, we also showed:(H/t) = - (L/t) (b)(a) & (b) together: (dH/dt) = - (L/t)If L doesn’t depend explicitly on the time(L/t) = 0 neither does H: (dH/dt) = 0 H = constant (conserved)
5Does H = constant (conserved) mean that energy is conserved Does H = constant (conserved) mean that energy is conserved? Not necessarily!As in our discussion of the energy function h in Ch. 2, (Sect. 2.7) the questions1) Whether H = the total energy E&2) Whether H is conservedare 2 separate questions!
6a.) The transformation equations from Cartesian to Our discussion in Ch. 2 about the energy function h, also holds for the Hamiltonian H = qjpj - LSpecial cases:a.) The transformation equations from Cartesian toGeneralized Coordinates are time independentb.) The potential V is velocity independent.Only if a.) & b.) are BOTH satisfied is it true that: H = T + V = E Total Mechanical EnergyUnder these conditions, if V is time independent, (∂L/∂t) = (dH/dt) = 0 H = E = constant (conserved)
7Summary: Different Conditions: The Hamiltonian: H = qjpj - LALWAYS: (dH/dt) = - (∂L/∂t)SOMETIMES: V does not depend on t (dH/dt) = 0 & H = constant (conserved)USUALLY:H = T + V = E Total Mechanical Energy Conservation Theorem for Mechanical Energy:If H = E AND it does not depend on t, E is conserved!
8 We can have conditions in which: 1. H is conserved & = E Clearly, the conditions for conservation of the Hamiltonian H are DISTINCT from those which make it the total mechanical energy E. We can have conditions in which:1. H is conserved & = E2. H is not conserved & = E3. H is conserved & E4. H is not conserved & EThe most common case in classical (& quantum) mechanics is case 1.
9Stated another way, there are two questions: 1. Does the Hamiltonian H = E for the system?2. Is the mechanical energy E conserved for the system?These are 2 aspects of the problem & are DIFFERENT questions!We may have cases where H E, but E is conserved.For example: A conservative system, using generalized coordinates which are in motion with respect to fixed rectangular axes: The transformation equations will contain the time T will NOT be a homogeneous, quadratic functionof the generalized velocities! H E, However, because the system is conservative, E is conserved! (This is a physical fact about the system, independent of coordinate choices!).
10A point mass m is attached to an ideal spring, constant k. Note: L is independent of the choice of generalized coordinates. However, H depends on this choice! That is, for one choice of the qi, H may be conserved & for another it may not!Example: Consider the 1d system in the figure:A point mass m is attached to an ideal spring, constant k.Other end of the spring is fixed on a massless cart moving(by external means) to the right at constant speed v0.
11H is NOT conserved, but it IS the total energy! Choose the origin so the cart is there at t = 0.Lagrangian: L(x,x) = T – V= (½)mx2 - (½)k(x – v0t)2Lagrange’s Equation:mx = -k(x – v0t)Hamiltonian?x is the Cartesian coordinate of the mass, V doesn’t involve velocity x H = total energy of the system:H = H(x,p,t) = E = T + V = p2(2m)-1 + (½)k(x – v0t)2However, H = H(t) (dH/dt) 0 H constantH is NOT conserved, but it IS the total energy!
12H IS conserved, but it IS NOT the total energy! Another way to solve this:Change coordinates from x tox´ = x – v0tThe Lagrangian:L(x´,x´) = (½)m(x´+ v0)2 - (½)k(x´)2= (½)m(x´)2 + mx´v0 + (½)m(v0)2 - (½)k(x´) mx´ = -kx´The Hamiltonian?There is a term in L (mx´v0) which is linear in x´. Corresponds to the qa term in the general formalism we’ve discussed. Here, a = mv0. Using that formalism (or otherwise) the Hamiltonian:H = H(x´,p´) = (p´- mv0)2(2m)-1 + (½)k(x´)2 – (½)m(v0)2Now, H H(t) (dH/dt) = 0 H = constantH IS conserved, but it IS NOT the total energy!(It is the energy of MOTION of m relative to the moving cart)
13SUMMARY: 1 Problem, 2 Hamiltonians! Different in magnitude, t dependence, & functional behavior!1. H = H(x,p,t) = E = T + V = p2(2m)-1 + (½)k(x – v0t)2H = H(t) (dH/dt) 0 H constantH is NOT conserved, but it IS the total energy!2. H = H(x´,p´) = (p´- mv0)2(2m)-1 + (½)k(x´)2 - (½)mx´(v0)2H H(t) (dH/dt) = 0 H = constantH IS conserved, but it IS NOT the total energy!(It is the relative energy of MOTION of m with respect to the cart)Both Hamiltonians obviously lead to the SAME equation of motion for the particle!
14Similar problem: A “dumbbell” of 2 masses, m1 & m2 connected by a spring of constant k (see figure): The center of mass (CM) movesto the right at constant velocityv0. Oscillations are only alongthis same direction. Solve theoscillation problem in the usualway. The oscillations aren’taffected by uniform motion at v0. The motion separates into the CM motion & therelative oscillatory motion about the CM. Once themotion is started, the energy E is conserved & theHamiltonian H = E & is conserved.
15A slightly different problem: The same “dumbbell” (see figure): m2 moves to the right at constant velocity v0. TheCM & m1 oscillate relative tom2. This requires an externalperiodic force! Oscillationsare affected by uniformmotion at v0. The motion doesn’t separate into the CM motion & the relative oscillatory motion. Because of the external oscillation force, the energy E is NOT conserved, the Hamiltonian H is NOT conserved & H E!