Conservation Theorems Section 7.9 CONSERVATION OF ENERGY Consider the general, total time derivative of the Lagrangian: L = L(qj,qj,t) = T - U (dL/dt) = (L/t) + ∑j qj (L/qj) + ∑j qj(L/qj) (1) Recall from Ch. 2: In an inertial reference frame, time is homogeneous (the same for all space!)  In a closed system the Lagrangian L = T - U cannot depend EXPLICITLY on the time!  (L/t) = 0 (2) Also, Lagrange’s Equations are: (L/qj) = (d/dt)[(L/qj)] (3)

Conservation of Energy Use (2) & (3) in (1): (dL/dt) = ∑j qj(d/dt)[(L/qj)] + ∑j qj(L/qj) (4) The right side of (4) is (from the chain rule!): = ∑j(d/dt)[qj (L/qj)] = (d/dt)[∑j qj (L/qj)]  (4) becomes: (d/dt)[L - ∑j qj (L/qj)] = 0 (5) Or: [L - ∑ j qj (L/qj)] = constant in time Define H  ∑j qj (L/qj) - L (6) (5)  (dH/dt) = 0 Or: H = constant in time

Define H  ∑j qj(L/qj) - L (6)  (dH/dt) = 0, H = constant in time H  The Hamiltonian of the system. Defined formally like this. See the next section for more details. Of course L = T – U. In the usual case, the PE is independent of generalized the velocities U = U(qj)  (L/qj) = ([T-U]/qj) = (T/qj) Put this into (6): H = ∑j qj(T/qj) - (T - U) In the previous section, we proved: ∑j qj(T/qj)  2T  H = 2T - (T - U) or H = T + U = E = TOTAL MECHANICAL ENERGY!!

Summary: For a closed system in which KE is a homogeneous, quadratic function of the generalized velocities: 1. The Lagrangian L = T - U = constant in time. 2. The Hamiltonian H is defined: H  ∑j qj (L/qj) - L 3. H = T + U = E = Total Mechanical Energy 4. H = E = constant in time (“A constant of the motion”)  Conservation of Total Mechanical Energy!

The Hamiltonian H (its main use discussed soon!): H  ∑j qj (L/qj) – L Example: Particle, mass m in a plane, subject to gravitational force. Use plane polar coordinates. L = T – U = (½)m(r2 + r2θ2) – mgr cosθ  H = r(L/r) + θ(L/θ) – L H = m(r2 + r2θ2) - L  H = (½)m(r2 + r2θ2) + mgr cosθ = T + U

The definition of the Hamiltonian H  ∑j qj (L/qj) – L is general! Discussion The definition of the Hamiltonian H  ∑j qj (L/qj) – L is general! The relation H = T + U = E is valid ONLY under the conditions of the derivation (!): The eqtns of transformation connecting rectangular & generalized coords must be independent of time  T is a homogeneous, quadratic function of qj. Then potential energy U must be independent of the generalized velocities qj.

Two questions pertaining to any system: 1. Does the Hamiltonian H = E for the system? 2. Is energy E conserved for the system? These are two DIFFERENT aspects of the problem! Could have H  E, but also have energy E conserved. For example: In a conservative system, using generalized coordinates which are in motion with respect to fixed rectangular axes: the Transformation eqtns will contain the time  T will NOT be a homogeneous, quadratic function of the generalized velocities!  H  E, However, because the system is conservative the total energy E is conserved! (This is a physical fact about the system, independent of coordinate choices).

Momentum Conservation Recall from Ch. 2: In an inertial reference frame, space is homogeneous:  In a closed system the Lagrangian L = T - U cannot be affected by a uniform translation of the system! i.e., Changing every coordinate vector rα by an infinitesimal translation: rα  rα + δr leaves L unchanged. The displacement δr is a displacement in the virtual (variational) sense (as opposed to a real, physical displacement dr).

δL= ∑i(L/xi)δxi+ ∑i(L/xi) δxi (1) For simplicity, consider a closed system (so that L has no explicit time dependence: (L/t) = 0) with a single particle & use rectangular coordinates: L = L(xi,xi) Consider the change in L caused by an infinitesimal displacement δr  ∑i ei δxi : δL= ∑i(L/xi)δxi+ ∑i(L/xi) δxi (1) This leaves L unchanged  δL = 0 (2) Consider varied displacements only  the δxi are time independent!  δxi = δ(dxi/dt) = (d[δxi]/dt)  0 (3)

(L/xi) = (d/dt)[(L/xi)] (6) Combine (1), (2), (3): δL= ∑i (L/xi)δxi = 0 (4) Each δxi is an independent displacement  (4) is valid only if (L/xi) = 0 (i =1,2,3) (5) Lagrange’s Equations are: (L/xi) = (d/dt)[(L/xi)] (6) (5) & (6) together  (d/dt)[(L/xi)] = 0 Or: (L/xi) = constant in time (7)

(L/xi) = constant in time (A) Physically, what is (L/xi)? L = T - U, T = T(xi), U = U(xi),  (L/xi) = ([T - U]/xi) = (T/xi) Note: (T/xi) = ([(½)m∑j (xj)2]/xi) = mxi = pi  (L/xi) = pi = LINEAR MOMENTUM! Summary: (A)  The homogeneity of space  The linear momentum p of a closed system (no external forces) is constant in time! (Momentum is conserved!) Or, if the Lagrangian of a system is invariant with respect to uniform translation in a certain direction, the component of linear momentum of the system in that direction is conserved (constant in time).

Angular Momentum Conservation Recall from Ch. 2: In an inertial reference frame, space is isotropic (the same in every direction!)  In a closed system the Lagrangian L = T - U cannot be affected by a uniform, infinitesimal rotation of the system! Rotation through an infinitesimal angle δθ is shown. Angle δθ has vector direction δθ  to plane, as shown.

The change in the Lagrangian due to δθ is: For simplicity, again consider a closed system (so that L has no explicit time dependence: (L/t) = 0) with a single particle & use rectangular coordinates: L = L(xi,xi) Consider the change in L caused by a rotation through an infinitesimal angle δθ as in the figure.  Each radius vector (vector arrows left off!) r changes to r + δr where (from Ch. 1) δr  δθ  r (1) Velocity vectors also change on rotation: Velocities r change to r + δr, where (from Ch. 1) δr  δθ  r (2) The change in the Lagrangian due to δθ is: δL= ∑i(L/xi)δxi+ ∑i(L/xi) δxi (3)

From previous discussion: (L/xi) = pi (linear momentum) Lagrange’s Equations are: (L/xi) = (d/dt)[(L/xi)]  (L/xi) = (dpi/dt) = pi (3) on the previous page becomes: δL= ∑ i pi δxi+ ∑ i pi δxi = 0 Or, in vector notation (arrows left off!): δL = pδr + pδr = 0 (4) Using (1) & (2) in (4): p(δθ  r) + p(δθ  r) = 0 (5) Using vector identity (triple scalar product properties), (5) is: δθ [(r  p) + (r  p)] = 0 Or: δθ [d(r  p)/dt] = 0 (6) δθ is arbitrary:  [d(r  p)/dt] = 0 Or: (r  p) = constant in time!

 The angular momentum about that symmetry axis is conserved! For arbitrary δθ  (r  p) = constant in time! Using the definition of angular momentum: L  (r  p)  L = constant in time Distinguish angular momentum L from the Lagrangian L! We have shown that, for a closed system, the isotropy of space  the angular momentum is a constant in time: Angular Momentum is conserved! Corollary: If system is in external force field with an axis of symmetry, the Lagrangian is invariant with respect to rotations about the symmetry axis.  The angular momentum about that symmetry axis is conserved!

In general, in physical systems: A Symmetry Property of the System Symmetry Properties; Conservation Laws Repeat of a Discussion from Ch. 2! In general, in physical systems: A Symmetry Property of the System  Conservation of Some Physical Quantity Also: The conservation of Some Physical Quantity  A Symmetry Property of the System This isn’t just valid in classical mechanics! Also in quantum mechanics! This forms the foundation of modern theories (Quantum Field Theory, Elementary Particles,…)

Summary: Conservation Laws In a closed system, in an inertial reference frame, we’ve shown that symmetry properties & conservation laws are directly related. This is often called Noether’s Theorem (after Emmy Noether, see footnote, p. 264).

The Total Mechanical Energy (E) For a closed system (no external forces) in an inertial reference frame, there are 7 “Constants of the Motion”  “Integrals of the Motion”  Quantities which are Conserved (constant in time): The Total Mechanical Energy (E) The 3 vector components of the Linear Momentum (p) The 3 vector components of the Angular Momentum (L)