Download Sect. 7.9

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
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).
• 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)
• 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.
• 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!
• 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!
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).
• 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)