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.