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Sect. 8.2: Cyclic Coordinates & Conservation Theorems

Sect. 8.2: Cyclic Coordinates & Conservation Theorems. Definition of the Hamiltonian: (sum on i ) H(q,p,t)  q i p i – L (q,q,t) (1) A cyclic coordinate q i  one which doesn’t appear in the Lagrangian L . Lagrange Eqtns  p i  ( ∂ L/ ∂ q i ) = 0

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Sect. 8.2: Cyclic Coordinates & Conservation Theorems

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  1. Sect. 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 qiis conserved. • Hamilton’s Eqtns get the same result. qiis cyclic  From (1)qidoesn’t appear in H either.Hamilton’s Eqtnpi = - (∂H/∂qi) = 0 pi =constant.The momentum conjugate to a cyclic coord qiis conserved.

  2. 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.

  3. What 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)

  4. 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 questions 1) Whether H = the total energy E & 2) Whether H is conserved are 2 separate questions!

  5. Our discussion in Ch. 2 about the energy function h, also holds for the HamiltonianH= qjpj - L • Special cases: a.) The transformation equations from Cartesian to Generalized Coordinates are time independent b.) The potential V is velocity independent. Only ifa.) & b.) are BOTH satisfied is it true that:  H= T + V = E Total Mechanical Energy • Under these conditions, if V is time independent,(∂L/∂t) = (dH/dt) = 0 H = E =constant(conserved)

  6. Summary:Different Conditions: The Hamiltonian:H= qjpj - L ALWAYS:(dH/dt) = - (∂L/∂t) SOMETIMES:Vdoes 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 = EAND it does not depend on t, E is conserved!

  7. Clearly, the conditions for conservation of the HamiltonianH are DISTINCTfrom those which make it the total mechanical energyE.  We can have conditions in which: 1.H is conserved & = E 2.H is not conserved & = E 3.H is conserved &  E 4.H is not conserved &  E The most common case in classical (& quantum) mechanics is case 1.

  8. Stated 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 & areDIFFERENTquestions! • 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 function of 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!).

  9. 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.

  10. Choose the origin so the cart is there at t = 0. Lagrangian:L(x,x) = T – V = (½)mx2 - (½)k(x – v0t)2 Lagrange’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)2 However, H = H(t)  (dH/dt)  0  H  constant H is NOT conserved, but it IS the total energy!

  11. Another way to solve this: Change coordinates from x to x´ = x – v0t • The 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´)2  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)2 Now, H  H(t)  (dH/dt) = 0  H = constant H IS conserved, but it IS NOT the total energy! (It is the energy of MOTION of m relative to the moving cart)

  12. SUMMARY: 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)2 H = H(t)  (dH/dt)  0  H  constant H 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)2 H  H(t)  (dH/dt) = 0  H = constant H 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!

  13. Similar problem: A “dumbbell” of 2 masses, m1 & m2 connected by a spring of constant k (see figure): • The center of mass (CM) moves to the right at constant velocity v0. Oscillations are only along this same direction. Solve the oscillation problem in the usual way. The oscillations aren’t affected by uniform motion at v0.  The motion separates into the CM motion & the relative oscillatory motion about the CM. Once the motion is started, the energy E is conserved & the Hamiltonian H = E & is conserved.

  14. A slightly different problem:The same “dumbbell” (see figure): m2 moves to the right at constant velocity v0. The CM & m1 oscillate relative to m2. This requires an external periodic force! Oscillations are affected by uniform motion at v0.  The motion doesn’t separate into the CM motion & the relative oscillatory motion.Because of the external oscillation force, theenergy E is NOTconserved, the Hamiltonian H is NOT conserved & H  E!

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