Download Lecture 14: Noether`s Theorem

Lecture 14: Noether’s Theorem
• In our review of Newtonian Mechanics, we were reminded
that some quantites (energy, linear momentum, and angular
momentum) are conserved
– That is, they are constant if no external influence acts on a
system
• Since we can think of the universe as one (big) system, this
means that energy and momentum can never be created or
destroyed
• And we found that physicists cling to this idea of
conserved quantities, even when the experimental evidence
seems to indicate otherwise
• Now we are ready to learn why we feel these quantities
must be conserved
Symmetries
• Assume that the Lagrangian is symmetric under some
transformation of variables
– That is, all of the q’s change according to some rule:
q → q (s)
but the Lagrangian doesn’t change, no matter what value of
s is used:
d
L {q ( s ) , q ( s ) ; t} = 0
ds
• Noether claimed that for any such symmetry, the quantity
dq ( s )
C = pq
ds
must be conserved
Proof
• Let’s see what happens when we take the derivative of C
with respect to time (here we assume that s has no time
dependence):
dC
dq ( s )
d dq ( s )
= pq
+ pq
dt
ds
dt ds
• Using the definition of pq, this becomes:
dC
d ∂L dq ( s ) ∂L d dq ( s )
=
+
dt
dt ∂qi
ds
∂qi dt ds
• We now use Lagrange’s Equation of Motion:
dC
∂L dq ( s ) ∂L dq ( s )
=
+
dt
ds
∂qi
∂qi ds
• The right-hand side of the previous equation is equivalent
to
d
L {q ( s ) , q ( s ) ; t}
ds
• But we required this derivative to be 0!
• So we’ve shown that:
dC
=0
dt
• In other words, C is conserved
This result is one of the most important theorems in physics.
It holds not only for classical mechanics, but also for quantum
mechanics, relativity, and quantum field theories!
Examples of Conserved Quantities
• If we express the Lagrangian in rectangular coordinates,
and find that it is invariant under the transformation
xi → xi + si
what quantity is conserved?
• We know that the generalized momentum associated with
xi is just the linear momentum
• So the conserved quantity is:
dxi
C = pi
= pi (1) = pi
dsi
For any Lagrangian symmetric under spatial translations,
linear momentum is conserved
• Now suppose we write the Lagrangian in terms of angular
variables:
1
L = Iθ i2 − U (θ i )
2
• Now let θ i → θ i + si
• In this case, the generalized momentum associated with θ
is:
∂L
pθi =
= Iθ i
∂θ i
which we recognize as Iω, the angular momentum
For any Lagrangian symmetric under rotations,
angular momentum is conserved
What About Symmetries in Time?
• This is outside the scope of our proof of Noether’s Theorem
(since we assumed s was time-independent)
– though a more general and thorough proof does include it!
• Consider a Lagrangian that has no explicit time dependence:
∂L
=0
∂t
then the total time derivative is:
dL {qi , qi }
=
dt
i
∂L
qi +
∂qi
i
∂L
qi
∂qi
• But Lagrange’s Equation of Motion requires that:
∂L d ∂L
=
∂qi dt ∂qi
• So we have:
dL
=
dt
i
d ∂L
qi +
dt ∂qi
∂L
d ∂L
qi +
qi
∂qi
dt ∂qi
=
i
=
i
i
∂L
qi
∂qi
d
∂L
d
qi
=
dt
∂qi
dt
i
∂L
qi
∂qi
• Which can be rearranged to show:
d
dt
j
∂L
qi
−L =0
∂qi
We call this
quantity the
Hamiltonian of
the system (H)
Physical Interpretation of H
• We’ve proven that if the Lagrangian doesn’t depend
explicitly on the time, the quantity H is conserved
– But the definition of H doesn’t give us much intuition into
what it represents
– And since you’ll be seeing Hamiltonians for the rest of your
lives as physicists (especially in quantum mechanics…)
we’d better get an idea of what it is
• Let’s start with the usual case where the potential energy
doesn’t depend on velocity, and doesn’t depend explicitly
on time
• Also assume that the relationship between rectangular and
generalized coordinates has no explicit time dependence
– Systems with this property are called scleronomic
• That means that:
U = U ( qi )
∂U
and
=0
∂qi
• In such cases, we have
∂L ∂ ( T − U ) ∂T
=
=
+0
∂qi
∂qi
∂qi
so we can rewrite the Hamiltonian as:
H≡
qj
j
∂L
−L=
∂q j
qj
j
∂T
−L
∂q j
• To see what the first term represents, consider the total
kinetic energy of a system of N particles:
1
T=
2
α ,i
mα xi2
• That’s fine if we’re using rectangular coordinates, but what
about generalized coordinates? Let’s substitute:
xα ,i = xα ,i ( q j , t )
∂xα ,i
xα ,i =
j
1
T=
2
α
∂q j
qj +
∂xα ,i
mα
i
j
∂q j
∂xα ,i
∂t
qj +
∂xα ,i
∂t
2
• Expanding the squared term, we find:
1
T=
2
1
=
2
α
∂xα ,i
mα
i
j
∂q j
∂xα ,i ∂xα ,i
α ,i
j ,k
∂q j ∂qk
∂xα ,i
qj
k
∂qk
1
T=
2
=
j ,k
α ,i
qk + 2
j
∂xα ,i ∂xα ,i
q j qk + 2
∂q j
j
• For a scleronomic system,
j ,k
1
2
∂xα ,i
∂t
∂q j ∂qk
∂xα ,i ∂xα ,i
α ,i
∂q j ∂qk
∂t
∂q j
qj
qj +
∂xα ,i
∂t
∂xα ,i
+
2
∂t
= 0, so T reduces to:
∂xα ,i ∂xα ,i
mα
∂xα ,i
q j qk
q j qk =
a j ,k q j qk
j ,k
∂xα ,i
∂t
2
• We now recall our expression for the Hamiltonian:
H≡
qj
j
d
∂T
=
∂ql
a j ,k q j qk
=
j ,k
dql
=
j ,k
al ,k qk +
k
∂T
−L
∂q j
(δ jl + δ kl ) a j ,k q j qk
a j ,l q j
j
ql
l
l
∂T
= al ,k qk ql +
∂ql l ,k
a j ,l q j ql
l, j
∂T
ql
= 2 al ,k qk ql = 2T
∂ql
l ,k
These two terms are
identical – the indices
are different, but we
sum over all of them