2. 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) = (2)
Also, Lagrange’s Equations are:
(L/qj) = (d/dt)[(L/qj)] (3)

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)] = (5)
Or: [L - ∑ j qj (L/qj)] = constant in time
Define H  ∑j qj (L/qj) - L (6)
(5)  (dH/dt) = Or: H = constant in time

4. 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!!

5. 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!

6. 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

7. 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.

8. 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).

9. 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).

10. 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)  (3)

11. (L/xi) = (d/dt)[(L/xi)] (6)
Combine (1), (2), (3):
δL= ∑i (L/xi)δxi = (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)

12. (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).

13. 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.

14. 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)

15. 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 = (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] = (6)
δθ is arbitrary:  [d(r  p)/dt] = 0
Or: (r  p) = constant in time!

16.  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!

17. 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,…)

18. 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).

19. 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)