3Hamiltonian vector fields
III Symplectic Geometry
3.3 Classical mechanics
As mentioned at the beginning, symplectic geometry was first studied by physi-
cists. In this section, we give a brief overview of how symplectic geometry
arises in physics. Historically, there have been three “breakthroughs” in classical
mechanics:
(i)
In
∼
1687, Newton wrote down his three laws of physics, giving rise to
Newtonian mechanics.
(ii) In ∼ 1788, this was reformulated into the Lagrangian formalism.
(iii) In ∼ 1833, these were further developed into the Hamiltonian formalism.
Newtonian mechanics
In Newtonian mechanics, we consider a particle of mass
m
moving in
R
3
under
the potential
V
(
x
). Newton’s second law then says the trajectory of the particle
obeys
m
d
2
x
dt
2
= −∇V (x).
Hamiltonian mechanics.
To do Hamiltonian mechanics, a key concept to introduce is the momentum:
Definition (Momentum). The momentum of a particle is
y = m
dx
dt
.
We also need the energy function
Definition (Energy). The energy of a particle is
H(x, y) =
1
2m
|y|
2
+ V (x).
We call
R
3
the configuration space and
T
∗
R
3
the phase space, parametrized
by (x, y). This has a canonical symplectic form ω = dx
i
∧ dy
i
.
Newton’s second law can be written as
dy
i
dt
= −
∂V
∂x
i
.
Combining with the definition of y, we find that (x, y) evolves under.
dx
i
dt
=
∂H
∂y
i
dy
i
dt
= −
∂H
∂x
i
So physical motion is given by Hamiltonian flow under
H
.
H
is called the
Hamiltonian of the system.
Lagrangian mechanics
Lagrangian mechanics is less relevant to our symplectic picture, but is nice to
know about nevertheless. This is formulated via a variational principle.
In general, consider a system with
N
particles of masses
m
1
, . . . , m
N
moving
in
R
3
under a potential
V ∈ C
∞
(
R
3N
). The Hamiltonian function can be defined
exactly as before:
H(x, y) =
X
k
1
2m
k
|y
k
|
2
+ V (x),
where
x
(
t
) = (
x
1
, . . . , x
n
) and each
x
i
is a 3-vector; and similarly for
y
with
y
k
=
m
k
dx
t
dt
. Then in Hamiltonian mechanics, we say (
x, y
) evolves under
Hamiltonian flow.
Now fix
a, b ∈ R
and
p, q ∈ R
3N
. Write
P
for the space of all piecewise
differentiable paths γ = (γ
1
, . . . , γ
n
) : [a, b] → R
3N
.
Definition (Action). The action of a path γ ∈ P is
A
γ
=
Z
b
a
X
k
m
k
2
dγ
k
dt
(t)
2
− V (γ(t))
!
dt.
The integrand is known as the Lagrangian function. We will see that
γ
(
t
) =
x(t) is (locally) a stationary point of A
γ
iff
m
k
d
2
x
t
dt
= −
∂V
∂x
k
,
i.e. if and only if Newton’s second law is satisfied.
The Lagrangian formulation works more generally if our particles are con-
strained to live in some submanifold
X ⊆ R
3n
. For example, if we have a
pendulum, then the particle is constrained to live in
S
1
(or
S
2
). Then we set
P
to be the maps
γ
: [
a, b
]
→ X
that go from
p
to
q
. The Lagrangian formulation
is then exactly the same, except the minimization problem is now performed
within this P.
More generally, suppose we have an
n
-dimensional manifolds
X
, and
F
:
T X → R is a Lagrangian function. Given a curve γ : [a, b] → X, there is a lift
˜γ : [a, b] → T X
t 7→
γ(t),
dγ
dt
(t)
.
The action is then
A
γ
=
Z
b
a
(˜γ
∗
F )(t) =
Z
b
a
F
γ(t),
dγ
dt
(t)
dt.
To find the critical points of the action, we use calculus of variations. Pick a
chart (
x
1
, . . . , x
n
) for
X
and naturally extend to a chart (
x
1
, . . . , x
n
, v
1
, . . . , v
n
)
for T X. Consider a small perturbation of our path
γ
ε
(t) = (γ
1
(t) + εc
1
(t), . . . , γ
n
(t) + εc
n
(t))
for some functions
c
1
, . . . , c
n
∈ C
∞
([
a, b
]) such that
c
i
(
a
) =
c
i
(
b
) = 0. We think
of this as an infinitesimal variation of γ. We then find that
dA
γ
ε
dε
ε=0
= 0 ⇔
∂F
∂x
k
=
d
dt
∂F
∂v
F
for k = 1, . . . , n.
These are the Euler–Lagrange equations.
Example. In X = R
3N
and
F (x
1
, . . . , x
n
, v
1
, . . . , v
n
) =
X
k
m
k
2
|v
k
|
2
− V (x
1
, . . . , x
n
).
Then the Euler–Lagrange equations are
−
∂V
∂x
k
i
= m
k
d
2
x
k
i
dt
2
.
Example.
On a Riemannian manifold, if we set
F
:
T X → R
be (
x, v
)
7→ |v|
2
,
then we obtain the Christoffel equations for a geodesic.
In general, there need not be solutions to the Euler–Lagrange equation.
However, if we satisfy the Legendre condition
det
∂
2
F
∂v
i
∂v
j
6= 0,
then the Euler–Lagrange equations become second order ODEs, and there is a
unique solution given
γ
(0) and
˙γ
(0). If furthermore this is positive definite, then
the solution is actually a locally minimum.