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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.

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Presentation on theme: "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."— Presentation transcript:

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2 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  p i  (∂L/∂q i ) = constant. The momentum conjugate to a cyclic coordinate q i is conserved. Hamilton’s Eqtns get the same result. q i is cyclic  From (1) q i doesn’t appear in H either. Hamilton’s Eqtn p i = - (∂H/∂q i ) = 0  p i = constant. The momentum conjugate to a cyclic coord q i is conserved.

3  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.

4 What about energy conservation? For H = H(q,p,t) (dH/dt)  (  H/  q i )q i + (  H/  p i )p i + (  H/  t) Hamilton’s Eqtns: q i = (  H/  p i ), p i = - (  H/  q i )  (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)

5 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!

6 Our discussion in Ch. 2 about the energy function h, also holds for the Hamiltonian H = q j p j - L Special cases : a.) The transformation equations from Cartesian to Generalized Coordinates are time independent b.) The potential V is velocity independent. Only if a.) & 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)

7 Summary: Different Conditions: The Hamiltonian: H = q j p j - L ALWAYS: (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 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 & = 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.

9 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 & 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 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!).

10 Note: L is independent of the choice of generalized coordinates. However, H depends on this choice! That is, for one choice of the q i, 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 v 0.

11 Choose the origin so the cart is there at t = 0. Lagrangian: L(x,x) = T – V = (½)mx 2 - (½)k(x – v 0 t) 2 Lagrange’s Equation: mx = -k(x – v 0 t) 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 = p 2 (2m) -1 + (½)k(x – v 0 t) 2 However, H = H(t)  (dH/dt)  0  H  constant H is NOT conserved, but it IS the total energy!

12 Another way to solve this: Change coordinates from x to x´ = x – v 0 t  The Lagrangian: L(x´,x´) = (½)m(x´+ v 0 ) 2 - (½)k(x´) 2 = (½)m(x´) 2 + mx´v 0 + (½)m(v 0 ) 2 - (½)k(x´) 2  mx´ = -kx´ The Hamiltonian? There is a term in L (mx´v 0 ) which is linear in x´. Corresponds to the qa term in the general formalism we’ve discussed. Here, a = mv 0. Using that formalism (or otherwise) the Hamiltonian: H = H(x´,p´) = (p´- mv 0 ) 2 (2m) -1 + (½)k(x´) 2 – (½)m(v 0 ) 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)

13 SUMMARY: 1 Problem, 2 Hamiltonians! Different in magnitude, t dependence, & functional behavior! 1. H = H(x,p,t) = E = T + V = p 2 (2m) -1 + (½)k(x – v 0 t) 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´- mv 0 ) 2 (2m) -1 + (½)k(x´) 2 - (½)mx´(v 0 ) 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!

14 Similar problem: A “dumbbell” of 2 masses, m 1 & m 2 connected by a spring of constant k (see figure): The center of mass (CM) moves to the right at constant velocity v 0. Oscillations are only along this same direction. Solve the oscillation problem in the usual way. The oscillations aren’t affected by uniform motion at v 0.  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.

15 A slightly different problem: The same “dumbbell” (see figure): m 2 moves to the right at constant velocity v 0. The CM & m 1 oscillate relative to m 2. This requires an external periodic force! Oscillations are affected by uniform motion at v 0.  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!


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