Transcript 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 = pδr + pδr = 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)