3Hamilton's principle
IB Variational Principles
3.3 Symmetries and Noether’s theorem
Given
F [x] =
Z
β
α
f(x, ˙x, t) dt,
suppose we change variables by the transformation
t 7→ t
∗
(
t
) and
x 7→ x
∗
(
t
∗
).
Then we have a new independent variable and a new function. This gives
F [x] 7→ F
∗
[x
∗
] =
Z
β
∗
α
∗
f(x
∗
, ˙x
∗
, t
∗
) dt
∗
with α
∗
= t
∗
(α) and β
∗
= t
∗
(β).
There are some transformations that are particularly interesting:
Definition
(Symmetry)
.
If
F
∗
[
x
∗
] =
F
[
x
] for all
x
,
α
and
β
, then the transfor-
mation ∗ is a symmetry.
This transformation could be a translation of time, space, or a rotation, or
even more fancy stuff. The exact symmetries
F
has depends on the form of
f
.
For example, if
f
only depends on the magnitudes of
x
,
˙x
and
t
, then rotation
of space will be a symmetry.
Example.
(i)
Consider the transformation
t 7→ t
and
x 7→ x
+
ε
for some small
ε
. Then
F
∗
[x
∗
] =
Z
β
α
f(x + ε, ˙x, t) dx =
Z
β
α
f(x, ˙x, t) + ε
∂f
∂x
dx
by the chain rule. Hence this transformation is a symmetry if
∂f
∂x
= 0.
However, we also know that if
∂f
∂x
= 0, then we have the first integral
d
dt
∂f
∂ ˙x
= 0.
So
∂f
∂ ˙x
is a conserved quantity.
(ii)
Consider the transformation
t 7→ t − ε
. For the sake of sanity, we will also
transform x 7→ x
∗
such that x
∗
(t
∗
) = x(t). Then
F
∗
[x
∗
] =
Z
β
α
f(x, ˙x, t − ε) dt =
Z
β
α
f(x, ˙x, t) − ε
∂f
∂t
dt.
Hence this is a symmetry if
∂f
∂t
= 0.
We also know that if
∂f
∂t
= 0 is true, then we obtain the first integral
d
dt
f − ˙x
∂f
∂ ˙x
= 0
So we have a conserved quantity f − ˙x
∂f
∂ ˙x
.
We see that for each simple symmetry we have above, we can obtain a first
integral, which then gives a constant of motion. Noether’s theorem is a powerful
generalization of this.
Theorem
(Noether’s theorem)
.
For every continuous symmetry of
F
[
x
], the
solutions (i.e. the stationary points of
F
[
x
]) will have a corresponding conserved
quantity.
What does “continuous symmetry” mean? Intuitively, it is a symmetry we
can do “a bit of”. For example, rotation is a continuous symmetry, since we can
do a bit of rotation. However, reflection is not, since we cannot reflect by “a
bit”. We either reflect or we do not.
Note that continuity is essential. For example, if
f
is quadratic in
x
and
˙x
,
then
x 7→ −x
will be a symmetry. But since it is not continuous, there won’t be
a conserved quantity.
Since the theorem requires a continuous symmetry, we can just consider
infinitesimally small symmetries and ignore second-order terms. Almost every
equation will have some O(ε
2
) that we will not write out.
We will consider symmetries that involve only the
x
variable. Up to first
order, we can write the symmetry as
t 7→ t, x(t) 7→ x(t) + εh(t),
for some
h
(
t
) representing the symmetry transformation (and
ε
a small number).
By saying that this transformation is a symmetry, it means that when we
pick
ε
to be any (small) constant number, the functional
F
[
x
] does not change,
i.e. δF = 0.
On the other hand, since
x
(
t
) is a stationary point of
F
[
x
], we know that if
ε
is non-constant but vanishes at the end-points, then
δF
= 0 as well. We will
combine these two information to find a conserved quantity of the system.
For the moment, we do not assume anything about
ε
and see what happens
to F [x]. Under the transformation, the change in F [x] is given by
δF =
Z
f(x + εh, ˙x + ε
˙
h + ˙εh, t) − f(x, ˙x, t)
dt
=
Z
∂f
∂x
εh +
∂f
∂ ˙x
ε
˙
h +
∂f
∂ ˙x
˙εh
dt
=
Z
ε
∂f
∂x
h +
∂f
∂ ˙x
˙
h
dt +
Z
˙ε
∂f
∂ ˙x
h
dt.
First consider the case where
ε
is a constant. Then the second integral vanishes.
So we obtain
ε
Z
∂f
∂x
h +
∂f
∂ ˙x
˙
h
dt = 0
This requires that
∂f
∂x
h +
∂f
∂ ˙x
˙
h = 0
Hence we know that
δF =
Z
˙ε
∂f
∂ ˙x
h
dt.
Then consider a variable
ε
that is non-constant but vanishes at end-points. Then
we know that since x is a solution, we must have δF = 0. So we get
Z
˙ε
∂f
∂ ˙x
h
dt = 0.
We can integrate by parts to obtain
Z
ε
d
dt
∂f
∂ ˙x
h
dt = 0.
for any ε that vanishes at end-points. Hence we must have
d
dt
∂f
∂ ˙x
h
= 0.
So
∂f
∂ ˙x
h is a conserved quantity.
Obviously, not all symmetries just involve the
x
variable. For example, we
might have a time translation
t 7→ t − ε
. However, we can encode this as a
transformation of the x variable only, as x(t) 7→ x(t − ε).
In general, to find the conserved quantity associated with the symmetry
x
(
t
)
7→ x
(
t
) +
εh
(
t
), we find the change
δF
assuming that
ε
is a function of time
as opposed to a constant. Then the coefficient of ˙ε is the conserved quantity.
Example.
We can apply this to Hamiltonian mechanics. The motion of the
particle is the stationary point of
S[x, p] =
Z
(p ·
˙
x − H(x, p)) dt,
where
H =
1
2m
|p|
2
+ V (x).
(i)
First consider the case where there is no potential. Since the action depends
only on
˙
x (or p) and not x itself, it is invariant under the translation
x 7→ x + ε, p 7→ p.
For general ε that can vary with time, we have
δS =
Z
h
p · (
˙
x +
˙
ε) − H(p)
−
p ·
˙
x − H(p)
i
dt
=
Z
p ·
˙
ε dt.
Hence p (the momentum) is a constant of motion.
(ii)
If the potential has no time-dependence, then the system is invariant under
time translation. We’ll skip the tedious computations and just state that
time translation invariance implies conservation of
H
itself, which is the
energy.
(iii)
The above two results can also be obtained directly from first integrals
of the Euler-Lagrange equation. However, we can do something cooler.
Suppose that we have a potential
V
(
|x|
) that only depends on radius. Then
this has a rotational symmetry.
Choose any favorite axis of rotational symmetry
ω
, and make the rotation
x 7→ x + εω × x
p 7→ p + εω × p,
Then our rotation does not affect the radius
|x|
and momentum
|p|
. So
the Hamiltonian
H
(
x, p
) is unaffected. Noting that
ω ×p ·
˙
x
= 0, we have
δS =
Z
p ·
d
dt
(x + εω × x) − p ·
˙
x
dt
=
Z
p ·
d
dt
(εω × x)
dt
=
Z
p ·
ω ×
d
dt
(εx)
dt
=
Z
(p · [ω × ( ˙εx + ε
˙
x)]) dt
=
Z
( ˙εp · (ω × x) + εp · (ω ×
˙
x)) dt
Since p is parallel to
˙
x, we are left with
=
Z
( ˙εp · (ω × x)) dt
=
Z
˙εω · (x × p) dt.
So
ω ·
(
x ×p
) is a constant of motion. Since this is true for all
ω
,
L
=
x ×p
must be a constant of motion, and this is the angular momentum.